Instruction:
q63.
Value at time 2 of annuity benefits= 200(0.64)(0.9)p63.

Response: 1 Introduction and motivation 1.1 Risk and insurance In this book we deal with certain mathematical models. This opening chapter, however, is a nontechnical introduction, designed to provide background and motivation. In particular, we are concerned with models used by actuaries, so we might first try to describe exactly what it is that actuaries do. This can be difficult, because a typical actuary is concerned with many issues, but we can identify two major themes dealt with by this profession. Thefirstis risk,awordthatitselfcanbedefinedindifferentways.Acommonlyaccepted definition in our context is that risk is the possibility thatsomethingbad happens. Of course, many bad things can happen, but in particular we are interested in occurrences that result in financial loss. A person dies, depriving family of earned income or business partners of expertise. Someone becomes ill, necessitating large medical expenses. A home is destroyed by fire or an automobile is damaged in an accident. No matter what precautions you take, youcannotridyourselfcompletelyofthepossibilityofsuchunfortunateevents,butwhatyou can do is take steps to mitigate the financial loss involved. One of the most commonly used measures is to purchase insurance. Insuranceinvolvesasharingorpoolingofrisksamongalargegroupofpeople.Theorigins go back many years and can be traced to members of a community helping out others who sufferedlossinsomeformorother.Forexample,peoplewouldhelpoutneighbourswhohad suffered a death or illness in the family. While such aid was in many cases no doubt due to altruisticfeelings,therewasalsoamotivationofself-interest.Youshouldbepreparedtohelp out a neighbour who suffered some calamity, since you or your family could similarly be aidedbyotherswhenyourequiredsuchassistance.Thiseventuallybecamemoreformalized, giving rise to the insurance companies we know today. Withthe institutionof insurance companies, sharing is no longer confined to the scope of neighbours or community members one knows, but it could be among all those who chose to purchase insurance from a particular company. Although there are many different types FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 4 INTRODUCTION AND MOTIVATION of insurance, the basic principle is similar. A company known as theinsureragrees to pay out money, which we will refer to asbenefits, at specified times, upon the occurrence of specified events causing financial loss. In return, the person purchasing insurance, known as theinsured,agreestomakepaymentsofprescribedamountstothecompany.Thesepayments are typically known aspremiums. The contract between the insurer and the insured is often referred to as theinsurancepolicy. The risk is thereby transferred from the individuals facing the loss to the insurer. The insurer in turn reduces its risk by insuring a sufficiently large number of individuals, so that the losses can be accurately predicted. Consider the following example, which is admittedly vastly oversimplified but designed to illustrate the basic idea. Suppose that a certain type of event is unlikely to occur but if so, causes a financial loss of 100 000. The insurer estimates that about 1 out of every 100 individuals who face the possibilityofsuchlosswillactuallyexperienceit.Ifitinsures1000people,itcanthenexpect 10 losses. Based on this model, the insurer would charge each person a premium of 1000. (We are ignoring certain factors such as expenses and profits.) It would collect a total of 1 000 000 and have precisely enough to cover the 100 000 loss for each of the 10 individuals whoexperiencethis.Eachindividualhaseliminatedhisorherrisk,andinsofarastheestimate of 10 losses is correct, the insurer has likewise eliminated its own risk. (We comment further on this statement in the next section.) We conclude this section with a few words on the connection between insurance and gambling. Many people believe that insurance is really a form of the latter, but in fact it is exactly the opposite. Gambling trades certainty for uncertainty. The amount of money you haveinyourpocketistherewithcertaintyifyoudonotgamble,butitissubjecttouncertainty ifyoudecidetoplaceabet.Ontheotherhand,insurancetradesuncertaintyforcertainty.The uncertain drain on your wealth, due to the possibility of a financial loss, is converted to the certainty of the much smaller drain of the premium payments if you insure against the loss. 1.2 Deterministic versus stochastic models The example in Section 1.1 illustrates what is known as adeterministicmodel. The insurer in effect pretends it will know exactly how much it will pay out in benefits and then charges premiums to match this amount. Of course, the insurer knows that it cannot really predict these amounts precisely. By selling a large number of policies they hope to benefit from the diversification effect. They are really relying on the statistical concept known as the ‘law of large numbers’, which in this context intuitively says that if a sufficiently large number of individuals are insured, then the total number of losses will likely be close to the predicted figure. Tolookatthisideainmoredetail,itmayhelptogiveananalogywithflippingcoins.Ifwe flip100faircoins,wecannotpredictexactlythenumberofthemthatwillcomeupheads,but weexpectthatmostofthetimethisnumbershouldbecloseto50.But‘mostofthetime’does notmeanalways.Itispossibleforexample,thatwemaygetonly37heads,orasmanyas63, orevenmoreextremeoutcomes.Intheexamplegiveninthelastsection,thenumberoflosses maywellturnouttobemorethantheexpectednumberof10.Wewouldliketoknowjusthow unlikely these rare events are. In other words, we would like to quantify more precisely just what the words ‘most of the time’ mean. To achieve this greater sophistication a stochastic model for insurance claims is needed, which will assign probabilities to the occurrence of ADEQUACY AND EQUITY 5 various numbers of losses. This will allow adjustment of premiums in order to allow for the risk that the actual number of losses will deviate from that expected. We will however begin thestudyofactuarialmathematicsbyfirstdevelopingadeterministicapproach,asthisseems tobethebestwayoflearningthebasicconcepts.Aftermasteringthis,itisnotdifficulttoturn to the more realistic stochastic setting. We will not get into all the complications that can arise. In actual coin flipping it seems clear that the results of each toss are independent of the others. The fact that one coin comes up heads, is not going to affect the outcomes of the others. It is this independence which is behindthelawoflargenumbers,andwhichresultsinoutcomesthatareusuallyclosetowhat is expected. There are some risks, often referred to assystematicor non-diversifiable,where the independence assumption fails, and which can adversely affect all or a large number of members of a group at the same time. For example, a spreading epidemic could cause life or health insurers to pay more in claims than they expected. Selling more policies in order to diversifywouldnothelptheirfinancialsituation.Itcouldinfactmakeitworse,ifthepremiums werenotsufficienttocovertheextralosses.Severeclimaticdisturbancescausingstormscould impact property insurance in the same way. In 2008, falling real estate prices in the United States affected mortgage lenders and those who insured mortgage lenders against bad debts, to the extent that this helped trigger a global financial crisis. A detailed discussion of these matters is not within the scope of this work, and for the most part, the stochastic model we present will confine attention to the usual insurance model where the risks are considered as independent.Itshouldbekeptinmindhoweverthatthedetectionandavoidanceofsystematic risk are matters that the actuary must always be aware of. 1.3 Finance and investments The second theme involved in an actuary’s work is finance and investments. In most of the types of insurance that we focus on in this book, an additional complicating factor is the long-termnatureofthecontracts.Benefitsmaynotbepaiduntilseveralyearsafterpremiums arecollected.Thisiscertainlytrueinlifeinsurance,wherethelossisoccasionedbythedeath ofanindividual.Premiumsreceivedareinvestedandtheresultingearningscanbeusedtohelp provide the benefits. Consider the simple example given above, and suppose further that the benefitsdonothavetobepaiduntil1yearafterthepremiumsarecollected.Iftheinsurercan investthemoneyat,say,5%interestfortheyear,thenitdoesnotneedtochargethefull1000 in premium, but can collect only 1000/1.05 from each person. When invested, this amount willprovidethenecessary1000tocoverthelosses.Again,thisexampleisoversimplifiedand there are many more complications. We will, in the next chapter, consider a mathematical model that deals with the consequences of the payments of money at various times. A much more elaborate treatment of financial matters, incorporating randomness, is presented in Chapter 20. 1.4 Adequacy and equity We can now give a general description of the responsibilities of an actuary. The overriding taskistoensurethatthepremiums,togetherwithinvestmentearnings,are adequatetoprovide forthepaymentofthebenefits.Ifthisisnottrue,thenitwillnotbepossiblefortheinsurerto 6 INTRODUCTION AND MOTIVATION meet its obligations and some of the insureds will necessarily not receive compensation for their losses. The challenge in meeting this goal arises from the several areas of uncertainty. The amount and timing of the benefits that will have to be paid, as well as the investment earnings,areunknownandsubjecttorandomfluctuations.Theactuarymakessubstantialuse of probabilistic methods to handle this uncertainty. Anothergoalistoachieve equityinsettingpremiums.Ifaninsureristoattractpurchasers, itmustchargeratesthatareperceivedasbeingfair.Herealso,therandomnessmeansthatitis not obvious how to define equity in this context. It cannot mean that two individuals who are charged thesameamount inpremiumswillreceiveexactly thesamebackinbenefits, forthat wouldnegatethesharingarrangementinherentintheinsuranceidea.Whiletherearedifferent possible viewpoints, equity in insurance is generally expected to mean that the mathematical expectation of these two individuals should be the same. 1.5 Reassessment Actuariesdesigninsurancecontractsandmustinitiallycalculatepremiumsthatwillfulfillthe goals of adequacy and equity, but this is not the end of the story. No matter how carefully one makes an initial assessment of risks, there are too many variables to be able to achieve complete accuracy. Such assessments must be continually re-evaluated, and herein lies the real expertise of the actuary. This work may be compared to sailing a ship in a stormy sea. It is impossible to avoid being blown off course occasionally. The skill is to detect when this occurs and to take the necessary steps to continue in the right direction. This continual monitoring and reassessing is an important part of the actuary’s work. A large part of this involvescalculatingquantitiesknownas reserves.WeintroducethisconceptinChapter2and then develop it more fully in Chapter 6. 1.6 Conclusion We can now summarize the material found in the subsequent chapters of the book. We will describe the mathematical models used by the actuary to ensure that an insurer will be able to meet its promised benefits payments and that the respective purchasers of its contracts are treated equitably. In Part I, we deal with a strictly deterministic model. This enables us to focus on the main principles while keeping the required mathematics reasonably simple. In Part II, we look at the stochastic model for an individual insurance contract. In Part III, we lookatmoreadvancedstochasticmodelsandintroducethemathematicsoffinancialmarkets. In Part IV, we consider models that encompass an entire portfolio of insurance contracts. 2 The basic deterministic model 2.1 Cash flows As indicated in the previous chapter, a basic application of actuarial mathematics is to model the transfer of money. Insurance companies, banks and other financial institutions engage in transactions that involve accepting sums of money at certain times, and paying out sums of money at other times. To construct a model for describing this situation, we will first fix a time unit. This can be arbitrary, but in most applications it will be taken as some familiar interval of time. For conveniencewewillassumethattimeismeasuredinyears,unlessweindicateotherwise.We will let time 0 refer to the present time, and timetwill then denotettime units in the future. Wealsoselectanarbitraryunitofcapital.Inthischapter,weassumethatallfundsarepaidout or received at integer time points, that is, at time 0,1,2,…. The amount of money received or paid out at timekwill be called thenetcashflow at timekand denoted byc k. A positive value ofck denotesthatasumistobereceived,whereasanegativevalueindicatesthatasum is paid out. The entire transaction is then described by listing the sequence of cash flows. We w i l lr e f e rt ot h i sa sacashflowvector , c = (c 0,c1,…,cN), whereN is the final duration for which a payment is made. For example, suppose I lend you 10 units of capital now and a further 5 units a year from now. You repay the loan by making three yearly payments of 7 units each, beginning 3 years from now. The resulting cash flow vector from my point of view is c = (−10,−5,0,7,7,7). From your point of view, the transaction is represented by−c = (10,5,0, −7,−7,−7). FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 8 THE BASIC DETERMINISTIC MODEL One of our main goals in this chapter is to provide methods for analyzing transactions in terms of their cash flow vectors. There are several basic questions that could be asked: r When is a transaction worthwhile undertaking? r How much should one pay in order to receive a certain sequence of cash flows? r How much should one charge in order to provide a certain sequence of cash flows? r How does one compare two transactions to decide which one is preferable? All of these questions are related, and we could answer all of them if we could find a method to put a value on a sequence of future cash flows. If all cash flows were paid at the same time, or if the value of money did not depend on the time that a payment was made, the problem would reduce to one of simple addition. We could simply value a cash flow sequence by adding up all the payments. We cannot proceed in this naive way, however and must consider thetime value of money. It is a basic economic fact that we prefer present to future consumption. We want to eat the chocolate bar now, rather than tomorrow. We want to enjoy the new car today, rather than next month. This means of course that money paid to us today is worth more than money paid in the future. We are no doubt all very familiar with this fact. We pay interest for the privilege of borrowing money today, which lets us consume now, or we advance money to others, giving up our present consumption and expecting to be compensated with interest earnings. In addition, there is the effect of risk. If we are given a unit of money today, we have it. If we forego it now in return for future payments, there could be a chance that the party who is supposed to make remittance to us may be unable or unwilling to do, and we expect to be compensated for the possible loss. A major step in answering the above questions is to quantify this dependence of value on time. Readers who have taken courses on the theory of compound interest will be familiar with many of the ideas. However, our treatment will be somewhat different than that usually given. One reason for this is that we want to develop the concepts in such a way that they are applicable to more general situations, as given in Chapters 3–5. A second reason is that our approach is designed to be compatible with modern-day computing methods such as spreadsheets. To conclude this section, we remark that many complications arise when the cash flows are not exactly known in advance. They may depend on several factors, including random elements.Theremaybecomplicatedinterrelationshipsbetweenthevariouscashflows.These matters involve advanced topics in finance and actuarial mathematics and for the most part willnotbedealtwithinthisbook.InPartIwedealmainlywithasimplifiedmodel,whereall cashflowsarefixedandknowninadvance. Inlaterpartsofthebookwewillconsidercertain aspects of randomness, but will not get into the full extent of complications that can arise. 2.2 An analogy with currencies To motivate the basic ideas, we will consider first a completely different problem, which is nonetheless related to that introduced above. Suppose that I give you 300 Canadian dollars, 200 US dollars and 100 Australian dollars. How much money did I give you? It would be naive indeed to claim that you received 600 dollars, for clearly the currencies are of different value. To answer the question we will need conversion factors that allow us to deduce the DISCOUNT FUNCTIONS 9 valueofeachtypeofdollarintermsofothers.Let v(c,u)denotethevalueinCanadiandollars of one US dollar. Assume thatv(c,u) = 1.05, which means that a US dollar is worth 1.05 Canadiandollars.(Ournumbershereareforpurposesofillustrationonly.Theyareclosetothe conversion rates at the time of writing, but they may well have changed considerably by the timeyouarereadingthis.)Similarly,letting astandforAustralian,wewillassumethat v(c,a) equals 0.95, which means 95 cents Canadian will buy one Australian dollar. The convention we are using here, which should be noted for later use, is that thevfunction returns the value of oneunit of thesecondcoordinate currency in terms of thefirstcoordinate currency. Therearefourmoreconversionfactorsofinterest,butanimportantfactisthattheycanall bededucedfromjustthesetwo(orindeedfromanytwothathaveacommonfirstorcommon second coordinate). We note first that if it takes 1.05 Canadian dollars to buy 1 US dollar, then a single Canadian dollar is worth 1∕1.05= 0.9524 US dollars. That is, v(u,c) = v(c,u) −1 = 0.9524, v(a,c) = v(c,a)−1 = 1.0526, where we use similar reasoning for the Australian dollar. Next considerv(u,a). We want the amount of US dollars needed to buy one Australian dollar. We could conceivably effect this purchase in two stages, first using US money to buy Canadian, and then using Canadian to buy Australian. Working backwards, it will take 0.95 Canadiantobuy1Australian,anditwilltake v(u,c)0.95USdollarstobuythe0.95Canadian. To summarize, v(u,a) = v(u,c)v(c,a) = 0.9048. Our calculations are completed with v(a,u) = v(u,a) −1 = 1.1052. The reader may notice, given a typical real-life listing of currency prices, that the rela- tionships we state here do not hold exactly, but that is due to commissions and other charges. In the absence of these, they must necessarily hold. Let us now return to the original problem of determining of how much I paid you. We must first select a currency to express the answer in. For example, we could say that the total was equivalent to 300+200v(c,u)+100v(c,a) = 605 Canadian dollars. We could also say thatthetotalwasequivalentto300v(u, c)+200+100(u,a) = 576.20USdollars.Noticeasa shortcut,thatwedidnotneedtocomputethelattersum(whichcouldbeasignificantsavingin calculationifwehadseveralratherthanjustthreecurrencies).Ifthetotalamountisequivalent to605Canadian,thenitmustalsobeequivalentto605 v(u,c) = 576.20USdollars.Similarly, the total in Australian dollars can be computed as 605v(a,c) or alternatively as 576.20v(a,u), both of which are equal to 637 (approximately as there are some rounding differences). 2.3 Discount functions Wenowgobacktotheoriginalsituation.Wewanttovalueasequenceofcashflows,whichare all in the same currency, but which are paid at different times. Conversion factors are needed toconvertthevalueofmoneypaidatonetimetothatpaidatanother.Theprinciplesinvolved 10 THE BASIC DETERMINISTIC MODEL are exactly the same as in Section 2.2. Letv(s,t) denote the value at times, of 1 unit paid at timet. (Note again that our convention is that the 1 unit goes with the second coordinate. In other words, 1 unit paid at timetis equivalent tov(s,t) paid at times.) In the case where s< t, you can interpretv(s,t) as the amount you must invest at timesin order to accumulate 1 at timet. In the case wheres> tyou can interpretv(s,t) as the amount that you will have accumulated at times from an investment of 1 at timet. The fundamental relationship that must be satisfied is the same as we noted with currencies, namely v(s,t)v(t,u) = v(s,u), for all s,t,u. (2.1) Due to its importance, we will repeat the reasoning for this fundamental fact in the current context.Itissimplythat1unitattime uisequivalentto v(t,u)attime t,andthis v(t,u)attime t is equivalent tov(s,t)v(t,u) at times , showing that 1 unit at timeuis indeed equivalent to v(s,t)v(t,u) at times. We now make a formal definition. Definition 2.1 A discount functionis a positive valued functionv, of two nonnegative variables, satisfying (2.1) for all values ofs,t,u. Otherdesirablefeaturesfollowimmediatelyfrom(2.1).Taking s= t= u,wededucethat v(s,s)v(s,s) = v(s,s) and, sincev(s,s) is nonzero, we verify the obvious relationship v(s,s) = 1, for all s. (2.2) From this we deduce thatv(s,t)v(t,s) = v(s,s) = 1 so that we recover the relationship, noted in the currency case, that v(s,t) = v(t,s)−1. (2.3) Although we have calledva ‘discount’ function, the common English usage of the word really applies to the case wheres< t. In that case,v(s,t) will be normally less than 1, and the function is returning thediscounted amountof 1 unit paid at a later date. Some authors wouldprefertodefinethediscountfunctiontoapplyonlytothiscaseandthendefineanother function, called anaccumulationfunction, to cover the case wheresis greater thant. In that case, the function returns the accumulated amount from an investment of 1 unit at an earlier date,whichwillnormallybegreaterthanorequalto1.Wefinditmoreconvenienttouseonly the one function as given above, since the ideas, and the key relationship (2.1), are the same regardless of the ordering ons,tand u. The concept of a discount function will be a key ingredient in what follows. We will supposethat,givenanyfinancialtransaction,thereisasuitablediscountfunctionthatgoverns theinvestment ofallfunds.Wewilldeal onlybrieflywiththeimportantproblem ofchoosing the discount function. There are many factors governing this choice and it will depend on the nature of the transaction. It may be chosen to simply reflect the preferences of the parties for present as opposed to future consumption. It may reflect the desired return that an investor wishes to achieve. In many cases it is based on a prediction of market conditions that will determine what returns can be expected on invested capital. CALCULATING THE DISCOUNT FUNCTION 11 Apossiblecomplication that wewill not deal withto any great extent isthat which arises when two parties to a transaction have different choices of a suitable discount function. We will assume, unless indicated otherwise, that the same function applies to both parties. 2.4 Calculating the discount function Wenowprovideaprocedureforcalculatingthevaluesof vinasystematicway.Thecurrency example indicated that we can calculate all the values of a discount function just by knowing thoseatpointswithafixedfirstcoordinate,thatis,withacommoncomparisonpoint.Inmost applications it is convenient to take this as time 0. To simplify notation, we drop the first coordinate in this case and define v(t) = v(0,t). It follows from (2.1) thatv(s,t) = v(s,0)v(0,t) and then from (2.3) that v(s,t) = v(t) v(s). (2.4) In the first few chapters of this book, we will need to know the value ofv(s,t) only for integral values ofs and t. From (2.4), it will be sufficient to knowv(n) wheren is any nonnegative integer. To calculatev(n), note first that the key relationship (2.1) can be extended from one involving three terms to an arbitrary number. That is, given timest1,t2,…,tn, v(t1,t2)v(t2,t3)⋯ v(tn−1,tn) = v(t1,tn). (2.5) To see this, take for example, n= 4. The quantity v(t1,t2)v(t2,t3)v(t3,t4) is equal to v(t1,t3)v(t3,t4) by applying (2.1) to the first two terms. By another application of (2.1) it is equal tov(t1,t4). We have extended (2.1) to a formula involving four terms. A similar step extends from four to five, and so on. Formally, we are using mathematical induction. It follows from (2.5) that v(n) = v(0,1)v(1,2) ⋯ v(n−1,n), (2.6) so we need only knowv(n−1,n) for all positive integersn. Given such values, we can use the recursion formula v(n) = v(n−1)v(n−1,n), v(0) = 1, (2.7) to calculate all values ofv(n). The information we need is then summarized by the vector v = (v(0),v(1),v(2),…,v(N)), whereN is the final duration at which a nonzero cash flow occurs. 12 THE BASIC DETERMINISTIC MODEL 2.5 Interest and discount rates In practice, rather than specifyingv(k−1,k) directly, it is more common to deduce this quantity from the corresponding rates of interest or discount. Given any discount functionv and a nonnegative integerk, these are defined as follows. Definition 2.2 The rateofinterest for the time intervalktok+1 is the quantity ik = v(k+1,k)−1. Definition 2.3 The rateofdiscount for the time intervalktok+1 is the quantity dk = 1−v(k,k+1). Note that an investment of 1 unit at timekwill producev(k+1,k) = 1+ik units at time k+1.Similarly,aninvestmentof1 −dk = v(k,k+1)unitsattime kwillaccumulateto1unit at timek+1. Sodk is the amount you must take off from each unit paid at the end of the period, to get the equivalent amount at the beginning of the period. Given any of the three quantitiesv(k,k+1),ik or dk we can easily obtain the other two. For example, using the definitions and (2.3), it is straightforward to deduce that dk = ikv(k,k+1) = ik 1+ik , ik = dkv(k+1,k) = dk 1−dk . (2.8) Remark In most cases we expectik will be a nonnegative number, although from the definition it could in theory take on any value greater than−1. Indeed, people who make investments yielding less than the rate of inflation are in effect receiving a negative interest rate.Wewillnotbeconcernedwithinflationinthistexthowever,andthereadercangenerally assume that interest and discount rates are nonnegative, unless otherwise specified. Remark The reader is cautioned that some other authors use a different convention for the subscriptk in interest and discount rates. They would refer to ouri 0 as i1, since it is the interestrateforthe firsttimeinterval,andingeneralwoulduse ik+1 forour ik.Wefinditmore convenient to start all indexing at 0. 2.6 Constant interest Readers who have previously studied compound interest will be familiar with one particular familyofdiscountfunctions.Supposewebelievethataccumulationofinvestedfundsdepends only on the length of time for which the capital is invested, and not on the particular starting time. That is, we postulate that for all nonnegatives,t,h, v(s,s+h) = v(t,t+h). (2.9) VALUES AND ACTUARIAL EQUIVALENCE 13 If this holds, then v(s+t) = v(0,s+t) = v(0,s)v(s,s+t) = v(0,s)v(0,t) = v(s)v(t). Wehaveafamiliarfunctionalequation,anditiswellknownthatifweassumesomeregularity condition, for example, thatvbe continuous, we must havev(t) = v t and therefore that v(s,t) = vt−s for some constantv. For such a discount function, the rate of interestik is a constanti= v−1 −1, and the rate of discountdk is a constantd= 1−v. The discount function is therefore conveniently given by simply stating a single parameter, which is usually taken asi, the constant rate of interest. So, for example, if we want to know how much we will accumulate at timenfrom an investment of 1 at time 0, this is justv(n,0) = (1+i)n, which is the usual starting point for the subject of compound interest in elementary texts. In the pre-calculator, pre-computer age, this constant interest family of discount functions was almost always used, mainly to facilitate the computation. Many current textbooks on this subject still deal largely with this constant interest case. However, with modern computing methods, such as spreadsheets, the extra effort involved in using a general discount function is negligible, and there is no reason to restrict the flexibility that one can achieve. Throughout this book we will use arbitrary discount functions, although occasionally we will restrict discussion to the constant interest caseinordertosimplifythenotation.Oneofthemainadvantagesofnotrestrictingourselves to constant interest will be apparent in Chapters 4 and 5 where we will be able to incorporate the contingencies of life and death into the discount function. A key point in these chapters is that the calculation of premiums for life insurance and life annuities can be expressed as a special case of interest theory, using a general discount function. 2.7 Values and actuarial equivalence Wenowputtogetherthetwokeyconceptsof cashflowvector anddiscountfunction.Suppose we are given a cash flow vectorc = (c0,c1,…,cN) and a discount functionv. We want to calculate the single payment at time zero that is equivalent to all the cash flows, assuming that the time value of money is modeled by the given discount functionv. This amount is commonly referred to as thepresent valueof the sequence of cash flows, and sometimes abbreviated as P.V. We can think of it as the amount we would pay at time zero in order to receive all of the cash flows, or equivalently as the single payment that we would be willing toacceptnowinlieuofallthesefuturecashflows.Thecashflowattime khasapresentvalue ofc kv(k)bydefinitionofthediscountfunction.When ck > 0thisiswhatwehavetopaynow in order to receiveck at time k. Whenck < 0, receiving−ckv(k) now will let us pay out−ck at timek. We then add up the individual present values to get Present value of all cash flows= N∑ k=0 ckv(k). (2.10) 14 THE BASIC DETERMINISTIC MODEL Example 2.1 To make sure this is understood, take a very simple example. Letv(k) = 2−k for allk. This is a constant interest rate per period of 100 %. In other words, money doubles itself every period. This is of course not very realistic for a period of a year, but it could hold for a sufficiently long interval. Suppose we are to receive 12 units at time 2, but will be required to pay out 8 units at time 3. Find the present value, and verify that it makes sense. Solution. From(2.10)thepresentvalueis12 v(2)−8v(3) = 12(1∕4)−8(1∕8) = 2.To verify this, we note that the 12 units received at time 2 will accumulate to 24 at time 3. We then have topayout8,leaving anaccumulation of16unitsbytime3.Contrastthiswithreceiving instead asinglepayment of2at time0. Thiswillaccumulate to4attime1,8at time2and at 16attime3.Thereforeassuming(aswedothroughout)thatallmoneyaccumulatesaccording to the given discount function, we are in exactly the same position in both cases. Remark Unrealistic interest rates will be frequently used in examples and exercises throughout the book, in order to simplify the numerical computation, and allow the reader to concentrate on the underlying concepts. So for example, we will often takei k = 20%, 25%,50%,100%,whichcorrespondrespectivelyto v(k,k+1)equalto5∕6, 4∕5, 2∕3, 1∕2. Calculationiseveneasierwhenwehaveaninterestrateof0,inwhichcase v(k) = 1forall k. Notethatitwasconvenient intheaboveexampletocomparetheamountsaccumulatedat thetimeofthelastpayment.Thisisknownasthe accumulatedvalue andingeneralisgivenby N∑ k=0 ckv(N,k). It represents the amount we will have at timeN, resulting from all the cash flows. More generally, we can calculate a value atanytime between 0 andN. In this chapter we concentrate on integer times. We take the present value of the future cash flows at that time, plus the accumulated values of the past cash flows. The following definition formulates this precisely. Definition2.4 Foranytime n= 0,1, …,N,the valueattimen ofthecashflowvector cwith respect to the discount functionvis given by Val n(c;v) = N∑ k=0 ckv(n,k). It represents that single amount that we would accept at timenin place of all the other cash flows, assuming that everything accumulates according to the discount functionv. The values at various times are related in a simple way. Sincev(m,k) = v(m,n)v(n,k), it follows immediately that Valm(c;v) = Valn(c;v)v(m,n). (2.11)‡ (Formulas marked with‡ denote key facts, of particular importance.) VALUES AND ACTUARIAL EQUIVALENCE 15 Formula (2.11) should be intuitively clear. The single amount we would accept at time m in place of all the cash flows must be the value at timem of the single payment that we would accept at timen, in place of all the cash flows. We can then easily deduce all values from the value at a particular time. Normally it will be convenient to take this as time 0. The same point was illustrated in the currency example of Section 2.2, where we pointed out that thetotalamountinanyonecurrencywaseasilyconvertedtothetotalinanyotherbyasingle multiplication. Notation For the particular case of values at time 0 we will use a special symbol. Let ̈a(c;v) = Val 0(c;v). The letterais a standard actuarial symbol that is used to stand forannuity, another name for a sequence of periodic payments. See Section 2.14 for an explanation of the two dots. Whenthereisonlyonediscountfunctionunderconsiderationweoftensuppressthe vand just write Valn(c)o r̈a(c). We can express and calculatëaconveniently by expressing it in vector form: ̈a(c) = v⋅ c = vcT. (2.12) The second term is the (scalar) inner product of the two vectors. The third views the vectors v andc as 1×N matrices, with the superscript T denoting a matrix transpose. Formulas (2.11) and (2.12) make it clear that calculating values is a linear operation, a fact we will often exploit. That is, Val k(c+d) = Valk(c)+Valk(d), Val k(𝛼c) = 𝛼Valk(c), (2.13) for any cash flow vectorsc and d, scalar𝛼 and durationk. We have been comparing the values of a sequence of cash flows with a single payment at a particular time. We often wish to compare the values of two sequences of cash flows. For this we have the following definition. Definition 2.5 Two cash flow vectorsc and e are said to beactuarially equivalentwith respect to the discount functionvif, for some nonnegative integern, Val n(c;v) = Valn(e;v). From (2.11), if the above holds for somen, it holds for alln. Take,forexample, n= N.Weseethatapersontakingthepaymentsgivenby candletting them accumulate according to the given discount functionvwill eventually be in exactly the samefinancialpositionasonetakingthepaymentsgivenby e.Thisisthemeaningofactuarial equivalence. Manyproblemsinactuarialmathematicsreducetothefollowing.Wearegivenacashflow vectorc,andanothercashflowvector ethatdependsonsomeunknownparameters.Wehave to solve for the unknown parameters in order to make the two vectors actuarially equivalent. 16 THE BASIC DETERMINISTIC MODEL We have already done asimpleexample of thisincalculating Valn(c).Inthat case wewanted a single payment at timenthat is actuarially equivalent toc. In another common application, a lender advances payments to a borrower, and the borrower must return the money by loan repayments. The lender is therefore trading one sequence of payments (the advances) for another (the repayments) and wants the two to be actuarially equivalent. We look at a simple example. Example 2.2 A lends B 20 units now and another 10 units at time 1. B promises to repay theloanbytwopayments,madeattimes2andtime3.Therepaymentattime3istobetwice as much as that at time 2. If A wishes to earn interest of 25% per period, what should these repayments be? Solution. Let K be the unknown payment at time 2. We want to findK so that the vectors c = (20, 10, 0, 0) ande = (0,0,2 K,3K) are actuarially equivalent. There are many possible calculation methods. We could determine the vectorv and use formula (2.12), which is essentially the best approach for a spreadsheet method, as we describe later in Section 2.14. For small problems to be done by hand calculation, it is convenient to make use of a time diagram, where we indicate the payments and the 1 year discount factorsv(k,k+1) which areallequalto(1.25) −1 = 0.8.SeeFigure2.1.Readersmayfindtousefultowritedowntheir own time diagrams for the examples in the book, if they are not given. Advances 0 0 10 20 Time 3 2 1 0 Repayments 0 0 K 2K vs 0.80 .80 .8 Figure 2.1 Example 2.2 Making use of formula (2.6), we calculate values at time 0 by multiplying each payment ck by v(k), which we calculate as the product of all the preceding discount factorsv(i,i+1). So, P.V. of advances=20 + 10 (0.8)=28 P.V. of repayments=K[(0.82 +2(0.83)] = 1.664K. Equatingvaluestomaketheadvancesandrepaymentsactuariallyequivalent, K = 28∕1.664 = 16.83. The borrower pays 16.83 at time 2 and 33.66 at time 3. Weconcludethissectionbydescribingausefultechniquethatwewillcallthe replacement principle.Supposewearegivenacashflowvectorandsomesubsetoftheentries(0,1, …,N). Takethevalueattime kofjustthosecashflowsinthesubsetandthenreplaceallentriesinthe subsetbyasinglepaymentattime kequaltothatvalue.Thisleavesavectorthatisactuarially equivalent to the original. A formal derivation can be given by writing the vector as the sum of two vectors and using linearity. We will leave the details to the interested reader. VECTOR NOTATION 17 Benefits 42− 3 − 5 Time 3 2 1 0 1 Benefits 02− 3 0 Time 3 2 1 0 Figure 2.2 Example 2.3 Thefollowingisasimple,althoughartificialexample.Wewillgivemorerelevantusesof this principle later. See Figure 2.2. Example 2.3 Let c = (2,4, −3,−5). Assumeaconstantinterestrateof0.25.Findtheactuariallyequivalentvectorbyapplyingthe replacement principle with timek= 2 and the subset{1,3}. Solution. Thevalueof c 1 andc3 attime2is4(1.25) −5(1.25)−1 = 1.Makingthereplacement, weobtainthevector(2,0, −2,0)thatisactuariallyequivalentto c,ascanbeverifiedbydirect calculation. 2.8 Vector notation We now introduce some convenient notation for vectors. In all examples, we will have a maximum durationN, and all of our vectors will be (N+1)-dimensional, with entries indexed from 0 toN. However, we will often write a vector of lower dimension with the understanding that all subsequent entries will be zero. For example ifN = 8, then (1, 3, 2) will denote the vector (1, 3, 2, 0, 0, 0, 0, 0, 0). For vectorsc and d, written as above, we will write (c, d) to denote the vector consisting of the entries ofc f o l l o w e db yt h o s eo f d. For example ifc = (2,2), and d = (3,7,4,1)then( c,d) = (2,2,3,7,4,1,0,… ,0).(Note that the juxtaposition must comebeforefilling in the ending zero entries.) For any number r and integer k we will write (r k) to denote the vector consisting ofk entries of r.F o r example,(13,25) = (1,1,1,2,2,2,2,2,0,… ,0).W eletei denotethestandard ithbasicvector, i= 0,1,2, …,N, that is, a vector with an entry of 1 in positioniand zeros elsewhere. For a given vectorb = (b0,b1,…,bn), let Δb = (b0,b1 −b0,b2 −b1,…,bn −bn−1,−bn). Thatis, Δbisobtainedfrom bbysubtractingfromeachentrytheimmediateprecedingentry, except for the entry in position 0, which remains the same. (Think of the entry inposition−1 18 THE BASIC DETERMINISTIC MODEL as 0). Note also that by our notational convention,bj = 0f o rj> n, which accounts for the final entry of−bn. As a check, the sum of all the entries inΔbmust be 0. (The reader should also note that our definition differs from some authors’ usage, wherebyΔb(k) would equal bk+1 −bk, rather thanbk −bk−1 as we have defined it.) We use the symbol * to denote pointwise multiplication of vectors, that is, (a1,a2,…,an) ∗ (b1,b2,…,bn) = (a1b1,a2b2,…,anbn). The remainder of this section can be omitted on first reading. We develop an identity which is somewhat technical but which will be very useful in later chapters. For any vectorb = (b0,b1,…,bn) and a discount functionv, define a new vector∇b whose entry of indexkis given by ∇bk = bk −v(k,k+1)bk+1. (2.14) So∇is something likeΔexcept we subtract in the reverse order and discount the second term. Note that∇ depends onvand it could be denoted as∇v if there is any confusion. Our main result is that for any other vectorc = (c0,c1,…) ̈a(∇b ∗ c) = ̈a(b ∗Δ c). (2.15) To see this, we just note that the left hand side is (b0 −v(1)b1 )c0 +(b1 −v(1,2)b2 )c1v(1)+(b2 −v(2,3)b(3))c2v(2)+⋯ . Expanding and using the fact thatv(k)v(k,k+1) = v(k+1), this equals b0c0 +b1(c1 −c0)v(1)+b2(c2 −c1)v(2)+⋯ which is the right hand side. We can remember this formula as saying that when computing the present value of a pointwise product of one vector multiplied byΔ applied to a second vector, theΔcan slide off the second vector and become a∇ on the first. 2.9 Regular pattern cash flows Before the introduction of computers and calculators, finding values of cash flows could involve a lengthy computation. Accordingly, the emphasis was not only on constant interest rates, but also on vectors where all nonzero cash flows were of the same amount, usually referred to as the case oflevel cash flows, or where the cash flows followed some regular pattern, such as entries increasing in arithmetic progression. Values in this case were fairly easy to compute by algebraic means. One would then try to handle more general cash flow vectors by expressing them in terms of those with regular patterns. Readers who have had previousexperiencewithcompoundinterestcourseshavenodoubtseensuchtechniques.With modern computing methods there is little need for these methods for computing purposes, althoughtheyaresometimesusefulfortheoreticalmatters.Wegiveabriefillustrationofsome REGULAR PATTERN CASH FLOWS 19 of the more useful formulas of this type. We assume throughout the section that the discount function is given byv(n) = v n for some constantv. Consider the vectors (1n) andjn = (1,2, …,n−1,n). Then ̈a(1n) = 1+v+v2 +⋯ +vn−1. Multiplying byv, v̈a(1n) = v+v2 +⋯ +vn. Subtracting the second equation from the first and dividing by 1−vgives ̈a(1n) = 1−vn 1−v . (2.16) Many readers will have seen this technique for summing a geometric progression. A similar trick handles the vectorj by reducing to the level payment case. ̈a(j n) = 1+2v+3v2 +⋯ +nvn−1. Multiplying byv, v̈a(jn) = v+2v2 +⋯ +nvn. Subtracting the second equation from the first gives (1−v)̈a(jn) = (1+v+v2 +⋯ +vn−1)−nvn, and dividing by (1−v), ̈a(jn) = ̈a(1n)−nvn 1−v . (2.17) There is an alternate way to derive (2.16) and (2.17), as well as many other similar formulas,whichdoesnotinvolveanyseriessummation.Itisbasedonthefactthataloanmay be paid off by paying interest each period on the prior amounts advanced, and then paying off the total principal at the end. This is intuitively clear. A formal derivation will be given in Section 2.11. Suppose you receive a loan of 1 unit. You could repayi at the end of each period forn periods and then eventually repay the principal at timen. The repayment vector is therefore (0,i,i,…,i+1), which by the replacement principle is actuarially equivalent to (d,d,…, d,1) =d(1 n)+en,sinceacashflowof datanytime kisactuariallyequivalentto d(1+i) = iat timek+1.(Inotherwords,ifyoupayinterestatthebeginningoftheyear,theappropriaterate isdratherthan i.)Equatingthepresentvalueoftheadvancesandrepayments,1 = d̈a(1n)+vn, which leads to (2.16). Forthesecondformula,supposeyouaretoreceiveloanadvancesof1unitatthebeginning of each year fornyears. According to our scheme you will pay 1 unit of interest at time 1, 20 THE BASIC DETERMINISTIC MODEL 2 units of interest at time 2, etc. You then will repay the total principal ofn at timen.T h e repayment vector is (0,i,2i,…(n−1)i,ni+n). By replacing eachiby adone period earlier, this is actuarially equivalent todjn +nen. Equating present values of this with the vector of loan advances gives̈a(1n) = d̈a(jn)+nvn, which leads to (2.17). 2.10 Balances and reserves 2.10.1 Basic concepts This section will introduce one of the most fundamental actuarial concepts, that of a reserve. Supposeweenterintoafinancialtransaction,representedbythecashflowvector c.Atany future timekthere are two fundamental quantities to compute. First, we would like to know the total amount accumulated from all payments of the transaction up to this point. Second, we would like to know how much money we willneedat that time in order to discharge our future obligations under the transaction. We illustrate with a simple example. Example 2.4 Let c = (3,6,1,2,−20), v(0,1) = 0.6, v(1,2) = 0.5, v(2,3) = 0.4, v(3,4) = 0.5. How much do we have, and how much will we need at time 2, justbeforethe 1-unit payment due at that time? SeeFigure2.3,wherewehaveinsertedanarrowtoindicatethetimethatvaluesaretaken. Payments 2 1 6 3−20 Time 4 3 2 1 0 v s 0.6 0.5 0.4 0.5 Figure 2.3 Example 2.4 Solution. Theamountwehaveisclearlythe3unitspaidtousattime0,accumulatedfortwo periods, and the 6 units paid at time 1 accumulated for one period, for a total amount of 3v(2,0) +6v(2,1) = 3v(2,1)v(1,0) +6v(2,1) = 3 0.6×0.5 + 6 0.5 = 22. For the second question, note first that we have an obligation to pay out 20 units at time 4, and the amount we need at time 2 to provide for this is 20v(2,4) = 20v(2,3)v(3,4) = 20×0.4×0.5 = 4. We can offset this with the positive cash flows that we will acquire after time 2. The value at time 2 of the 1 unit due immediately at time 2 is just 1, and the value at time 2 of the 2 units payable at time 3 is just 2v(2,3)= 0.8. The total needed to ensure we can meet our obligations is 4−1−0.8 = 2.2. Let us verify this directly. Suppose we have 2.2 units at time 2, just before the payment due at that time. The payment of 1 will come in to give us 3.2. This will accumulate to BALANCES AND RESERVES 21 (3.2/0.4) = 8 units at time 3. We will receive another 2 units at time 2 for a total of 10, and this will accumulate to the 20 units that we need at time 4 in order to meet our obligation at that time. We now introduce some notation and terminology to express these concepts in a gen- eral case. Notation Given any cash flow vectorc and nonnegative integerk,l e t kc = (c0,c1,…,ck−1,0,… ,0), kc = (0,0, …,0,c k,ck+1,…,cN) so that c = kc+kc. (2.18) For example, for the vectorc in Example 2.4,2c = (3,6,0,0,0)while 2c = (0,0,1,2,−20). The idea is thatkc represents the past cash flows andkc the future cash flows when measured from timek. It is important to note that the payment at exact timek is by our convention taken as future. (We will see in Chapter 6 that this fits in with the usual treatment for life insurance contracts.) Note also that 0cis the zero vector while0cis justc. Definition 2.6 For k= 0,1, …,N,t h ebalanceat timekwith respect toc and v, is defined by Bk(c;v) = Valk(kc;v) = k−1∑ j=0 cjv(k,j). (Wewill,asbefore,suppressthe vwhenthereisnoconfusion.)Thebalanceattime kissimply the accumulated amount at timek resulting from all the payments received up to that time, and answers the question of how much we will have. Note again that by our conventional treatment, balances are computed just before the payment at exact timekis made, so that the timekpayment isnotincluded inB k. Definition2.7 Fork= 0,1, …,N,the reserveattime kwithrespectto candvisdefinedby kV(c;v) =− Valk(kc;v) =− N∑ j=k cjv(k,j). Thereserveattime kisthe negativeofthevalueattime kofthefuturepayments,andisequal to the amount we will need in order to meet future obligations. It is important to remember thenegativesign,toreflectthefactthatwearemeasuringthevalueofnetobligations,thatis, amounts to be paid out of the fund. Referring again to Example 2.5, we calculated thatB 2(c;v) = 22 and2V(c;v) = 2.2. 22 THE BASIC DETERMINISTIC MODEL Theterminologyandnotationforreserveswereborrowedfromaparticularlifeinsurance application (which we will discuss in detail in Chapter 6). In keeping with this standard reserve notation, we have placed duration as a left rather than right subscript. We introduce the reserve concept now, in order to stress that its use is more general than just this particular life insurance application and can apply to any sequence of cash flows. One can think of this quantity as representing capital that must be set aside or ‘reserved’ for future use. For those familiar with accounting terminology, reserves, if positive, representliabilities,which are amounts that we owe to other parties. It is possible of course for the reserve to be negative, which indicates that at timek the presentvalueofthefutureamountscominginforthistransactionwillexceedthepresentvalue ofthefutureamountstobepaidout.Insuchacase,theabsolutevalueofthereserveisa‘receiv- able’, that is,anamount that isowed tous and that willbe received through futurepayments. 2.10.2 Relation between balances and reserves From (2.16) and the linearity of Valk, Valk(c) = Bk(c)−kV(c), (2.19)‡ which says that the value of the transaction at any point is equal to the difference between what you actually have accumulated and what you need to set aside from that accumulation to meet future obligations. An important consequence of this is that for azero-valuecash flow vectorc, B k(c) = kV(c), (2.20)‡ where a zero-value vector is one whose value is zero at all durations. Such vectors arise frequently. Ifc andd are actuarially equivalent, then for any durationk, Val k(c−d) = Valk(c)−Valk(d) = 0, so thatc−d is a zero-value vector. Foratypicalexample,supposethatanindividualborrowsmoney,andagreestorepaywith a sequence of repayments. Consider the net cash flows from theborrower’spoint of view, that is, advances less repayments. If the advances and repayments are actuarially equivalent, these net cash flows form a zero-value vector. What is the outstanding balance on the loan at timek?Thisisthevalueofallthepaymentsyettobemade,whichbydefinitionisjust kV(c). In view of (2.18) it is also equal toBk(c). This is perfectly logical. It says that the amount still owing at timek is the value of all the money that been received, offset by the value of the repayments that have been made. Calculating the outstanding balance by means of the reserveisoftencalledthe prospectivemethod sinceitlookstothefuture,whilecalculatingby means of the balance is often called theretrospective method, since it looks to the past. For this reason, many authors refer to what we have called a balance as aretrospectivereserve. Example 2.5 (Figure 2.4). An individual borrows 1000 now and another 2000 at the end of 1 year. This loan will be repaid by yearly payments for 10 years, beginning 5 years from the present. The yearly payment doubles after 5 years. The interest rate is 0.06 for the first BALANCES AND RESERVES 23 Advances 20001000 0 0 0 0 0 0 0 0 0 0 0 0 0 Time 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Repayments 0 0 0 0 0 K K K K K2K 2K 2K 2K 2K Figure 2.4 Example 2.5 5 years and 0.07 thereafter. How much is still owing 8 years from now, prior to the payment at that time? How much is still owing after the payment at time 8? Solution. Wemust firstdetermine therepayment amounts. LetKbe theinitial repayment. If c denotes the advances andd denotes repayments, then c = (1000,2000), d = (0,0,0,0,0,K ,K,K,K,K,2K,2K,2K,2K,2K) = K(0 5,15,25), ̈a(c) = 2886.79, ̈a(d) = 7.95326K, and equating, K = 362.97 (see Section 2.15 for the calculation details). So the first five instalments are each 362.97, and the second five are each 725.94. The amount owing at time 8, just prior to the payment due on that date, can be calculated either as B 8(c−d) = Val8(e), where e = (1000,2000,0 3,−362.973) or as 8V(c−d) = Val8(f), where f = (08,362.972,725.945). In either case the amount is 3483.97. The outstanding balance at time 8,after the payment due at that time, is just 3483.97−362.97 = 3121.00. It is of interest to note that the lower payments at the beginning have the effect that the borrower is not repaying enough to handle the interest on the loan, so the amount owing is greater than the total amount borrowed. 2.10.3 Prospective versus retrospective methods For the model that we have described so far, either the prospective or retrospective approach can be used to determine the amount owing on a loan. In pre-computer days, one often chose the one that allowed easier calculation, but with modern methods it makes little difference. Thereare,however,situationswhereoneapproachortheotherdominates.Theretrospec- tive approach must be used in cases where the time and amounts of loan repayments are not scheduled in advance, but can be made at the option of the borrower. ConsideravariationonExample2.5.Supposetheborrowerhadmadepaymentsof300at time5,400attime6and500attime7,andthepaymentsafterthatwerenotyetdetermined.To 24 THE BASIC DETERMINISTIC MODEL obtaintheoutstandingbalanceattime8wewouldnecessarilyusetheretrospectivemethodand calculate the balance. This is Val 8(b), whereb = (1000,2000,0,0,0, −300,−400,−500,0). Ontheotherhand,theprospectiveapproachfiguresprominentlyincaseswherethechoice of an appropriate discount function may have changed since the transaction was originally entered into. In that case the reserve calculated with respect to anewdiscount function that is realistic at that time, could well give an amount that differs from that given by the balance as calculated retrospectively. After all, theequality of these twodepended on the fact that the same discount function was used for both. We will not dwell too much on this complication, but would like to describe two very familiar situations. Suppose the borrower wishes to discharge a loan at some point prior to the scheduled completion. The natural amount to pay would be the outstanding balance on the loan at that date, as calculated by either method. Suppose, however, that the loan contract requires that payments be made as scheduled and does not permit you to repay early. This is often the casewithhomemortgages.Suppose,further,thatinterestrateshavegonedownsincetheloan was originallytaken out. The lender willnot welcome earlyrepayment asthese amounts will have to be reinvested at a lower rate. Borrowers may however be permitted to repay, if they include an extra amount as a penalty. What is really happening is that the lender is revaluing the reserve, by applying the new discount function to the future repayments. If interest rates decrease, the value of the functionv increases. Assuming the normal case where entries in the vector kc are nonpositive as they represent repayments, the negative of these values are nonnegative, and multiplying by the higher values ofv and summing will lead to a higher reserve. The penalty represents the excess value of the reserve as calculated by an up-to-date discount function, over the original value. (It should be noted that, in practice, the penalty amounts are often determined by approximate formulas rather than an exact recalculation of the reserve.) As a second example, suppose you purchase a bond, which means that you are in the position of the lender. The nature of a bond is that this debt is assignable to another party by tradinginthebondmarket.Yousellthebondtosomeonewhowillthencollecttherepayments by the issuer. What is a fair price? The answer is obviously the outstanding balance, not as calculated originally, but rather with respect to the rates of interest applicable at the time of sale.Asinthecaseabove,lowerinterestrateswillcausethisoutstandingbalancetoincrease. Traders who buy bonds always hope that interest rates will fall, so the outstanding balances increase, and they can sell the bonds for a profit in the market. 2.10.4 Recursion formulas If f is a function defined on the nonnegative integers, it is often useful to derive a recursion formula that expressesf(k+1) in terms off(k). Given an initial valuef(0), one can then successivelycalculatevaluesof f(k)forall k.Thisisknownasa recursionformula.Arecursion formula often leads to adifferenceformula that gives an expression forf(k+1)−f(k). There are many recursion formulas in actuarial mathematics. For the most part, they can be derived from the basic recursions for balances and reserves that we present in this section. We will first expressBk+1 in terms ofBk. For any cash flow vectorc and durationk,i ti s clear from the definitions that k+1c = kc+ckek. BALANCES AND RESERVES 25 The value at timek of the second term on the right isck and the value at timek of the first term on the right isBk(c).Invoking linearity and multiplying byv(k+1,k) to get the value at timek+1, gives the recursion Bk+1(c) = (Bk(c)+ck)v(k+1,k), (2.21) where we start the recursion with the initial value,Bk = 0. Subtracting Bk from each side and expressingv(k+1,k)a s1 +ik yields the difference formula Bk+1(c)−Bk(c) = ik(Bk(c)+ck)+ck. (2.22) Difference equations are not normally efficient for calculating numbers. Rather, they are useful for analyzing how quantities change from one period to another. For example, the above formula expresses the increase (or decrease if negative) in the balance over a period as a sum of two quantities, the interest earned plus the additional cash flow. Recall that our convention was to treat cash flows at timek as future with respect tok. This ties in well withinsurance contracts, as we will see later. However, the other convention is normally used when dealing with loans. Accordingly, we define ̃B k(c) = Bk(c)+ck, theaccumulatedamountattime kafter thecashflowattime kispaid.Weleaveittothereader to verify the corresponding recursion and difference equations as ̃Bk+1(c) = ̃Bk(c)v(k+1,k)+ck+1, (2.23) ̃Bk+1(c)− ̃Bk(c) = ik̃Bk(c)+ck+1. (2.24) Rewriting (2.22), we get −ck+1 = ik̃Bk(c)+[̃Bk(c)− ̃Bk+1(c)], (2.25) which forms the basis for the usual loan amortization schedules. (These are schedules that show the balances at each duration and how they change.) The left hand side above is the amountrepaidontheloanattime k+1(thenegativeofthenegativecashflow)andtheformula gives the split of this repayment into two parts. The first is the interest on the outstanding balance, and the second is the amount of principal reduction. Let us now turn to reserves. They satisfy exactly the same recursion relation as balances. That is, k+1V(c) = (kV(c)+ck)v(k+1,k), (2.26) ‡ wherewestarttherecursionwiththeinitialvalue 0V(c) =− ̈a(c).Thisisobviousinthecaseof zero-valuevectorswherereservesequalbalances,butitholdsingeneral,asseenforexample from the relation, k+1c= kc−ckek, or alternatively derived from (2.19), substituting from (2.17) and using (2.11). 26 THE BASIC DETERMINISTIC MODEL 2.11 Time shifting and the splitting identity In this section we provide a background for following some classical formulas that appear in the actuarial literature. From (2.16), taking values at time 0 and applying (2.11), we derive ̈a(c) = ̈a( kc)+v(k)Valk(kc). (2.27) We call this thesplitting identity, since it splits the calculation of a present value into two parts, considering first those cash flows before timek, and then those after timek. It can also beobtainedasaspecialcaseofthereplacementprinciple.Replaceallthecashflowsaftertime kby the single payment of Val k(kc) at timek. In classical actuarial mathematics, the spitting identity was often used as a calculation tool when there was distinct change in the cash flow sequence or discount function occurring at timek. ( It can still be useful in this regard in the continuous case discussed in Chapter 8.) The remainder of this section deals mainly with notational issues, and could be deferred until we apply it later in Section 4.3.2. The splitting identity often appears in a somewhat different form, since the symbol Val k is not standard. In traditional actuarial notation, it is commontoexpressallformulasintermsof ̈a(orsimilarsymbolswhichweintroducelater).So thequestionisthen,howdowewriteVal k intermsof ̈a?Thisisquitesimplewhenpayments and interest rates are constant. In general case we must introduce some new notation for timeshifting. Suppose we are at timekand we wish to consider this as the ‘new’ time 0. Given a cash flow vectorc, define the cash flow vectorc◦ kby (c◦ k) n = ck+n. For example, ifc = (2,2,3,4,5),then c◦ 2 = (3,4,5). In other words,c◦ ksimply gives the cashflowsinorder,butstartingwiththeoneattime k.Anotherwayoflookingatitisthat c◦ k is justkcwith the firstkzeros removed. This notation was not needed with constant payment vectors, since whenc is (1n), thenc◦ k= (1n−k). Similarly, given a discount functionv, define a new discount functionv◦ kby v◦ k(n,m) = v(n+k,m+k) so that v◦ k(n) = v(k,k+n). Once again, the idea is simply that we are treating timekas time 0 and measuring time from that point. It is quite simple to calculatev◦ k from the factors for periods of length 1 as we did in (2.6). We multiply as before, but start withv(k,k+1) rather thanv(0,1). An important point to notice is that, from (2.9) v◦ k= v if interest is constant, CHANGE OF DISCOUNT FUNCTION 27 which explains why this concept does not appear in classical works where interest is almost always taken as constant. We can now write Val k(kc) = N∑ i=k v(k,i)ci = N∑ i=k v◦ k(i−k)(c◦ k)i−k = N−k∑ j=0 v◦ k(j)(c◦ k)j = ̈a(c◦ k,v◦ k), (2.28) whichisintuitivelyclearsincewearestartingattime k,treatingitasanewtime0andvaluing the future payments. We can then write the splitting identity in its usual form ̈a(c;v) = ̈a(kc;v)+v(k)̈a(c◦ k;v◦ k). (2.29) Under constant interest, we can write this in an easier fashion as ̈a(c) = ̈a(kc)+vk̈a(c◦ k). (2.30) The splitting identity can sometimes be useful in determining the effect of changes in certain quantities. Example 2.6 Suppose thaẗa(c) = 19.6 and̈a( 10c) = 10. If v(9,10) increases from 0.8 to 0.81 while all other values ofv(k,k+1) remain the same what is the new value of̈a(c)? Solution. From (2.29), withk = 10, we know thatv(0,9)(0.8) ̈a(c◦ 10,v◦ 10) = 9.6, so v(0,9)(0.81) ̈a(c◦ 10,v◦ 10) = 9.72 and the new value of̈a(c) is 19.72. ∗2.11 Change of discount function This section is somewhat technical. Its purpose is to derive a useful relationship that allows ustoreplaceonediscountfunctionwithanother,whichcouldbemoreconvenientforthetask at hand. Suppose we are given a cash flow vectorc and a discount functionv.L e tv ′ be another discount function. Define a new cash flow vectorc′ whosekth entry is given by c′ k = ck +[v′(k,k+1)−v(k,k+1)]Bk+1(c;v). (2.31) It follows that Bk(c;v) = Bk(c′;v′), for all k. (2.32) 28 THE BASIC DETERMINISTIC MODEL We prove (2.30) inductively. To simplify the notation, denoteBk(c;v)b yBk,B′ k (c′;v′)b yB′ k , v(k,k+1) byv and v′(k,k+1) byv′.A tk = 0, the balances are both equal to 0. We will assume thatBk =B′ k and show thatBk+1 = B′ k+1 : Bk+1 = Bk +ck v = B′ k +c′ k v + ck −c′ k v = B′ k+1 v′ v + v−v′ v Bk+1. Formula (2.30) fork+1 follows after some minor algebraic manipulation. A corollary of this is that ifchas zero value with respect tov, thenc′ has zero value with respect tov′. This follows since a cash flow vector has zero value if and only if its balances eventually equal to 0. We will later apply this result to a key idea in life insurance. In this chapter we will give a formal proof of the assertion used in Section 2.9. Consideraloanfor Nperiodsmadeaccordingtoadiscountfunction,whichwewilldenote byv′ totieinwiththenotationabove.Welet d′ andi′ refertothecorrespondingdiscountand interest rates. Suppose the loan advances are given by vectorc = (c0,c1,…,cN−1). Let s= N−1∑ k=0 ck, r = (d′ 0(c0),d′ 1(c0 +c1),d′ 2(c0 +c1 +c2),…,d′ N−1s). We want to show thatc is actuarially equivalent to (r+seN) with respect tov′. This will mean that we can repay any loan by paying interest each year on the prior advances, and then repaying the principal at the end. While this may be intuitively clear, it is important to verify it formally to ensure that our model captures the idea of interest the way one normally perceives it. Letv(k) = 1 for allk, the discount function for a constant interest rate of zero. It follows that B k(c,v) = k−1∑ i=0 ck so that the vectorc′ as given by (2.31) is justc−r. Invoking (2.32), this gives ValN(c−r−seN;v′) = BN(c′;v′)−s= BN(c;v)−s= 0, showing the desired actuarial equivalence. 2.12 Internal rates of return In this section we take the point of view of an investor or lender, who is undergoing a transactionthatinvolvesinvestingfunds,hopingtogetbackamountsthataregreaterinvalue thanthoseputin.Let c = (c 0,c1,…,cN)denotethenetcashflowvectorofthetransaction.We assume throughout thatcN ≠ 0. We are not given a discount function but rather want to find the constant interest rate that will makec a zero-value vector. In this section it is convenient to deal with balances that include the payment due at the time the balance is computed. That is, we want to consider̃Bas defined above. INTERNAL RATES OF RETURN 29 Definition 2.8 An internal rate of return(often abbreviated asi.r.r.) of the transaction is a number iin the interval (−1,∞) such that, for the discount functionv(n) = (1+i)−n, (i) ̃BN(c;v) = 0, and (ii) ̃Bk(c;v) ≤ 0f o rk= 0,1, …,N−1. An i.r.r. is sometimes referred to as ayieldrate. Remark The definition is unchanged if we useBin place of̃B, sinceBk+1 = ̃Bk(1+ik). Theorem 2.1 Ifani.r.r.ofatransactionexists,itisunique. Proof. Suppose, to the contrary, that we have two such rates,iandi′, withi< i′.L e t̃Bk and ̃B′ k denotetheresultingbalances. Suppose that cj isthefirstnonzero entryin c.W emustha v e j< N since j= N would imply that̃BN = cN ≠ 0. We will now show by induction that, for k≥ j+1, we have ̃Bk > ̃B′ k . (2.33) If (2.33) holds for some indexk then it will hold fork+1 since the nonpositivity of balances implies that ̃Bk+1 = ̃Bk(1+i)+ck+1 > ̃B′ k (1+i′)+ck+1 = ̃B′ k+1 . Applyingtheabovefor k= j,andnotingthat cj = ̃Bj = ̃B′ j < 0,verifiestheinitialstep.So we have shown that̃BN > ̃B′ N , contradicting the fact that both rates satisfy (i). The reader familiar with previous literature on this subject may well be surprised at this theorem, since a great deal has been written on the non-uniqueness of the i.r.r. In our opinion this occurs from a faulty definition. The standard way of defining the i.r.r is to require only point (i) of Definition 2.8 and not (ii). The problem then reduces to finding the roots of the polynomial ∑ N k=0 ck(1+i)N−k, and there may indeed be several. However, when balances become positive, it means that the status of the individual has changed from that of a lender to that of a borrower. The transaction therefore involves both borrowing and lending, and it is difficult to give any interpretation to the roots. After all, a lender is seeking a high i.r.r. whereas a borrower wants it to be low. We feel that the best approach is to give the definition above and concede that other transactions just do not possess an i.r.r. under this definition. Other means are needed for their analysis. Note however that no problem is presented by the usual type of transaction, whereby the loan advances, represented by negative entries, all precede the loan repayments, as given by positive entries. The balances will begin as negative. If they ever become positive, it could only be when the entries change to positive. But as the entries then remain positive, the balance could then never become 0. Therefore, in this case, if an interest rateisatisfies (i) it will necessarily be an i.r.r. The remaining material in this section can be omitted as it is not used in the remainder of the book. To handle cases where the i.r.r. does not exist by the definition given above, a 30 THE BASIC DETERMINISTIC MODEL generalization of the definition was given by Teichroewetal. (1965a, 1965b), often referred to as the TRM method. Their approach was to recognize that when the status of a lender or an investor reverts to that of a borrower, as indicated by a positive balance, then that balance shouldaccumulateatsomerate rfixedinadvanceandnotattheunknown,presumablyhigher rate that one is trying to solve for. The rater could be, for example, the rate at which the individualcouldobtainfinancing.Theythendefineageneralizationofbalanceswithrespectto thisrate r,buildingontherecursiveformula(2.21).Let ̃B 0(c;v;r) = c0 anddefineinductively ̃Bk+1(c;v;r) = { ̃Bk(c;v;r)(1+ik)+ck+1,i f ̃Bk(c;v;r) ≤ 0, ̃Bk(c;v,r)(1+r)+ck+1,i f ̃Bk(c;v;r) > 0. Definethei.r.r.of crelativeto rasthevalueof iforwhich ̃BN(c;v;r) = 0when v(k) = (1+i)−k. We will denote this quantity asir. Note that ifi is an i.r.r. by Definition 2.8, then it will be an i.r.r. with respect torfor anyr. In this case balances are never positive and̃Bk(c;v;r) will simply equal̃Bk(c;v). The same type of inductive calculation as in the proof of Theorem 2.1 shows that asi increases, whilek,r and a nonzero vectorc are held fixed, the value of̃Bk(c;r;v) is strictly decreasing. There is therefore at most one i.r.r. relative tor. Moreover, there will always be oneexcept when either̃BN(c;v;r) < 0or ̃BN(c;v;r) > 0forall iin(−1, ∞).Defining ir tobe −1 in the first case and∞in the second case, we can state the following theorem. Theorem 2.2 Foranynonzerovector candanyr > −1,thereisauniquevalueofi r inthe interval[0,∞]. This shows that we do obtain a unique i.r.r. for any transaction after we first postulate the deposit rater. This gives rise to a simple criterion to decide if it is worthwhile to enter into a transaction.Thegeneralruleisthatthetransactionisworthwhileif ir > r,andnotworthwhile if ir < r. ∗2.13 Forward prices and term structure Aninterestingexampleofadiscountfunctionisfurnishedbythe forwardprices on risk-free zero-coupon bonds.A zero-coupon bond is a financial instrument that promises to pay a fixed sum at some future date, known as the maturity date, and which makes no intervening payments (coupons) before that time. By risk-free we mean something like a government bond, where we can safely assume that the bond is sure to be redeemed with no chance of default. Wefirstdiscussthegeneralideaofa forwardcontract.Thisisanagreementbetweentwo parties, whereby one party, the seller, agrees to deliver a certain specified asset at a specified future time to the other party, the buyer. At the time of delivery, the buyer, will remit a sum of money that is agreed upon at the time when the contract is entered into and is independent of the prevailing price at the time of exchange. The price agreed upon at the contract date is known as theforward priceof the contract. The buyer is often referred to as taking along positionand the seller is referred to as takingashortposition. Therearevariousreasonsforsuchanarrangement.Thebuyermaybesomeonewhoneeds theparticularitematsomefuturedate,andwishestolockinapricenow,toprotectthemselves FORWARD PRICES AND TERM STRUCTURE 31 from a future increase in the price. Alternatively, the buyer may be a speculator with no need oftheitem,butwhopredictsthatpriceswillrise,therebygivingthemaprofitastheycanbuy the asset at the delivery date and immediately sell it for a higher price. Similarly, the seller may be one who owns the asset, wishes to dispose of it at a future date, and wants to lock in the amount they will receive as protection against falling prices. Alternatively, the seller may beaspeculatorwhopredictsthatpriceswillfall,andhopestoprofitbybuyingtheitematthe delivery date for less than they have agreed to sell it for. Let ̃v(0,t) = ̃v(t) be the price at time 0 of a 1-unit, zero-coupon, risk-free bond maturing attime t.Forany s≤ t,let ̃v(s,t)betheforwardpricefora1-unitzero-couponbond,maturing at timet, where the delivery date is times. Now what will these forward prices be? It turns out that under a certain natural assumption they are determined from the time zero prices by the rule ̃v(s,t) = ̃v(t)∕̃v(s) (2.34) which implies immediately that̃v is a discount function. (Note that in the notation of Sec- tion 2.11 forward prices as determined from times are given by the time shifted discount function ̃v◦ s.) Theassumptionmadeisthe noarbitrage hypothesis,whichisamajorconceptinmodern day financial economics. Anarbitrage opportunity is one where a party can make a sure profit by buying and selling certain financial assets with no risk of a loss. The hypothesis in question says that in an environment when all parties have perfect information, arbitrage opportunitiescannotpersist.Thisisasimpleresultofsupplyanddemandlawsineconomics. Thearbitrageopportunitiesoccurwhensomeassetsareovervaluedandothersareundervalued. The argument is that if such an opportunity should ever rise, individuals will, in an attempt to take advantage of this, rush to buy the undervalued assets, thereby raising their prices, and rush to sell the overvalued assets, thereby lowering their prices, which restores equilibrium and eliminates the arbitrage. We also make some other idealized assumptions. One is that an arbitrary number of units of a bond can be bought or sold at any time. This includes the possibility ofshort selling whereby an individual can sell an asset that they do not own by borrowing the asset from another party. They plan to acquire the asset at a later date (hopefully at a lower price than theysolditfor)forreturntothelender.Anotherassumption,isthatbuyingorsellingdoesnot involve any transaction costs such as commissions. To verify our claim above we will show by an example, that in the absence of Equation (2.34) an arbitrage opportunity would arise. Suppose, for example, that̃v(10) = 0.75 and ̃v(5) = 0.90 but ̃v (5,10) = 0.8 < ̃v(10)∕̃v(5). An individual could sell a 1-unit bond maturing attime10,receiving0.75andthen(i)usetheproceedstobuy5/6ofa1-unitbondmaturingat time5,and(ii)takealongpositiononaforwardcontractfora1-unitbondmaturingattime10 withadeliverydateoftime5.Attime5,theindividualreceives5/6fromthematuring5-year bond, uses 4/5 of that to settle the forward contract, and now owns a 1-unit bond maturing at time 10, which they use to settle the short sale. A sure profit of 5/6−4/5 is made at time 5. The reader is invited to find a corresponding example of an arbitrage opportunity in the case that̃v(s,t) > ̃v(t)∕̃v(s) and then to provide a general proof. Theanalysisaboveindicatesthatforanindividualorcorporationwhoseinvestmentenvi- ronment consists of risk-free zero-coupon bonds, the forward prices constitute a reasonable 32 THE BASIC DETERMINISTIC MODEL choiceofdiscountfunction.Forexample,theindividualcouldensure(underourassumptions) thatapaymentreceivedattime swouldaccumulateto ̃v(s,t)atasubsequenttime tbyentering into a suitable forward contract. We conclude this section by introducing some conventional terminology. We have given theaboveanalysisintermsofprices,whichseemsthemostconvenientway,buttypicallythe economic and finance literature refers to rates instead of prices. Assume we are given prices of risk-free zero-coupon bonds for all maturities. Definition 2.9 The spot rate of interest y t is the yield rate for the bond maturing at timet. That is ̃v(t) = (1+yt)−t so that yt = ̃v(t)−1∕t −1. (2.35) Definition 2.10 For s≤ t,t h eforwardrateofinterestf (s,t) is the yield rate earned for the bond maturing at timetand acquired on a forward contract with delivery at times. That is ̃v(s,t) = [1+f(s,t)]s−t so that from (2.34), f(s,t) = ( ̃v(t) ̃v(s) )1∕(s−t) −1. (2.36) The relationship between the various maturity dates and the spot rates is known as the term structure of interest rates. A graph which shows values ofyt for various values oft is known as ayieldcurve. Examples of yield curves can be found in the financial and business sections of many daily newspapers, as well as being freely available online. The study of the various possible shapes of yield curves is an important topic for economists which is beyond the scope of this text. Asafinalword,notethatalthoughwehavediscussedspotandforwardratesinthecontext ofrisk-freebonds,theycanbedefined,usingformulas(2.35)and(2.36)forageneraldiscount function. Example 2.7 Given the spot ratesy 2 = 0.05 andy5 = 0.07, find the forward ratef(2,5). Solution. f(2,5) = ( 1.07−5 1.05−2 )−1∕3 −1 = .084. STANDARD NOTATION AND TERMINOLOGY 33 2.14 Standard notation and terminology There is a system of standard international actuarial notation that can seem quite complex at first. One goal in this book is to simplify notation as much as possible. In order to read this book,itwillnotbenecessarytolearnanyofthissystemotherthanthatwhichisintroducedin the text. However, the student who wishes to read other actuarial literature, or write actuarial examinations,orpursueanactuarialcareer,willbeexpectedtobefamiliarwiththefullextent of the system. Accordingly, we will, at the end of each relevant chapter, review the standard notation, indicating how it ties in with the notation in the main text. 2.14.1 Standard notation for cash flows discounted with interest The vector notation is peculiar to this book. In the standard notation the same basic symbola is used for the present value of a cash flow stream, but the particular vector of cash flows is indicated by embellishments of this symbol. A prime example is a n = ̈a(0,1n). A few comments are in order. The ‘angle’ around then is intended to signify a duration of time, as opposed to an age (which will be encountered in the chapters on life annuities and insurances). The vector above arises frequently. It is usual for a loan contract to stipulate that the first repayment is madeoneperiodafter receiving the loan. One does not usually make a repaymentasoftheloandate,sincethatwouldineffectjustmeanyouweregettingasmaller loan. Therefore, the simplest unadorned symbol was reserved for the vector with first entry equal to 0. When there is a payment at time zero, standard notation denotes this by placing two dots above thea. That is, ̈a n = ̈a(1n). Theformercaseistermedan immediateannuity,andthelattera dueannuity.Theterminology is a bit unusual since the immediate annuity does not start immediately but after one period. The name arose as a contrast to the generaldeferred annuitywhere there are several zero entries at the beginning. Once we move away from level payments, the distinction between due and immediate annuities no longer applies. From our point of view, there isalways a payment at time zero although it may be of zero amount. To avoid conflicts with standard notation we use the two dots as a general symbol. Another traditional way of looking at the difference between due and immediate was that theformerappliedwhenpaymentsweremadeatthebeginningoftheyear,andthelatterwhen payments were made at the end of the year. Once again, these distinctions no longer apply when payments are not level, since the beginning of 1 year is the end of the preceding one. (We will, however, have need to refer to this point again, when we discuss fractional period payments in Chapter 7.) Some other symbols are Ia n = ̈a(0,1,2,… ,n,0,… ,0), Ïan = ̈a(1,2, …,n,0,… ,0), 34 THE BASIC DETERMINISTIC MODEL using the vector that we denoted byjn in Section 2.9, Dan = ̈a(0,n,(n−1),(n −2),…,1,0,… ,0), D̈an = ̈a(n,(n−1),(n −2),…1,0, …,0). These are simple enough to remember from the fact thatI indicates anincreasingsequence of payments andDindicates adecreasingsequence. The symbolsis used to indicate an accumulated value. So, for example, sn = Valn(0,1n) = v(n,0)an, ̈sn = Valn(1n) = v(n,0) ̈an. Notethatinbothcasesthevaluesareattime n.Fortheunadornedsymbolthisisatthedateof the last payment. For the double-dotted symbol, timenis oneperiod after the last payment, since the first payment was at time 0. 2.14.2 New notation A reader who has already spend a great deal of effort in mastering the standard notation may havefeltsomedismayinhavingtolearnyetmorenotationintroducedinthischapter.Wefeel, however, that the devices introduced here are useful. There are three main innovations: the vectornotation;thesymbolVal;andandthetime-shiftingnotation.Weconsidereachinturn. The vector notation allows us to conveniently refer to an arbitrary sequence of cash flows in a systematic way. The traditional notation uses special embellishments for each particular sequence, and even so, is restricted to a small number of cases, such as payments constant over some period, or payments in arithmetic progression. The system of writing vectors is useful since it tells you exactly how to enter a cash flow vector into a column of a spreadsheet. For example, faced with the vector (1 20,210), one simply enters a 1 in the first column, copies it down for 20 rows, then enters a 2, and copies it down for 10 rows. Thestandardnotationallowsyoutospecifyvalues onlyatthebeginningorendofatrans- action,anditisoftenusefultodenotevaluesatanintermediatepoint,throughtheuseofVal k. Having introduced the vector notation, it is convenient to introduce the lower and upper subscriptstodenotepastandfuture,respectively.Suchadevicewasnotneededintheclassical literaturethatdealtwithlevelcashflowsandinterestrates.Forexample,thevector k1n isjust 1n−k.Similarly,asymbollikeour ◦ fortimeshiftingcouldeasilybeavoidedsince,asnoted, v◦ k= v.Moreover,1 k ◦ hisjustequalto1 k−h.Thereare,however,examplesintheliterature whereitisnecessarytorefertogeneral time-shiftedcashflows,and adhoc symbolsareused (see Bowersetal., 1997, p. 519). We feel it is much preferable to have a consistent notation to handle these. 2.15 Spreadsheet calculations Here and at the end of Chapters 4, 5, 6, 9 and 10 we will describe spreadsheets for doing the basiccalculationspertinenttothematerialoftherespectivechapter.Ourdescriptionswillrefer in particular to Microsoft Excel ®, and we assume the reader is reasonably acquainted with EXERCISES 35 this application. The basic ideas should be easily adaptable to other formats as well. Readers can reproduce our spreadsheets as given, or they may wish to come up with their own. Our general idea is to enter various functions of the durationk in columns starting with row 10 fork= 0. The cells above can then be used for headings or information. We put the durationk in column A. That is, we insert 0 in cell A10, 1 in cell A11, etc. This is done by putting the formula=A10+1 in cell A11 and copying down to cell A10+N. Theusualwayofspecifyingthediscountfunctionisthroughtheone-periodinterestrates i k. We enter these in column B, withik being inserted into cell B10+k. We then calculate the vector ofv(k) values in column C. Insert 1 in cell C10 and the recursive formula (2.7) is entered into cell C11 as=C10/(1+B10). This is copied down column C to cell C10+N. We enter the cash flows in column D. The cash flowck is put into cell D10+k.T h ev a l u e ̈a(c,v) is then calculated in cell D8 through the formula = SUMPRODUCT($C10:$C10 +N∗D10:D10 +N) where we substitute for the particular value ofN. As a check, take a constant interest rate of 0.06,andthevector(1 10,25).Theansweris12.7883.BycopyingcellD8totheright,wecan do calculations for several different cash flow vectors. Values at timekare easily obtained by dividing̈a(c)b yv(k), which is in cell C(10+k). Notes and references Thischapterisnotintendedtoprovideanexhaustivetreatmentofthemathematicaltheoryof interest.Thegoalwastoprovidethatportionofthesubjectthatisneededfortheremainderof thebook.SomeadditionalmaterialswillalsobegiveninChapters7and8.Readersinterested in a more detailed account can consult Broverman (2010) or Daniel & Vaaler (2009). Additional work on internal rates of return can be found in Promislow (1980, 1997) and Teichroewetal. (1965a, 1965b). The concept of the generalized i.r.r. is due to the latter. Exercises Type A exercises 2.1 You are given a discount functionvwherev(1,3) = 0.9, v(3,6) = 0.8, v(8,6) = 1.2 (a) How much must you invest at time 1, in order to accumulate 10 at time 8? (b) If you invest 100 at time 3, how much will have accumulated by time 8? 2.2 If v(t) = 2−t, and you are given cash flow vectorsc = (1,2,3)and e = (2,K,1), find Kso thatc and e are actuarially equivalent with respect tov. 2.3 Youaregiveninterestrates i0 = i1 = 0.25, i2 = i3 = 1.Youhaveenteredintobusiness transaction where you will receive 2 at time 0, 5 at time 3 and 10 at time 4, in return for a payment by you of 3 at time 2. In place of all these cash flows you are offered a single payment made to you at time 1. What is the smallest payment you would accept? 36 THE BASIC DETERMINISTIC MODEL 2.4 You are given rates of discount as follows:dk = 1∕3, fork= 0,1,2, and dk = 1∕4, for k= 3,4. The vectorc = (62,93,12). Find (a) the discount vectorv,( b )̈a(c) and (c) Val2(c). 2.5 Let c = (1,2,4,−3,8, −12). Suppose v(0,1) = v(1,2) = 0.8, v(2,3) = v(3,4) = 0.75,v(4,5) = 0.5.Find (a)3V,( b )B3. 2.6 A discount function satisfies v(k) = 2−k [ 1− k 6 ] , k= 0,1,2, …,5. For the vectorc = (1,−2,4,3, −3,−5) find (a)3V(c), (b)B3(c). 2.7 Given that 8V(c) = 100,c8 = 60,c9 = 70,v(8,9) = 0.8,v(9,10) = 0.75,v(0,10) = 0.5, and thaẗa(c) = 40, find (a)10V(c), (b)B10(c). 2.8 Given the vector c = (1,22,73) and a discount function v satisfying v(k) = 1− k∕10,k= 0,1, …,5, find v◦ 3(k)f o rk= 0,1,2; the vectors 3c and c◦ 3; and the present values̈a(c;v), ̈a(3c;v), ̈a(c◦ 3;v◦ 3). Verify that formula (2.29) holds. Type B exercises 2.9 Define a two variable function fors,t≥ 0b y v(s,t) = { (1+(t−s))−1,i f s≤ t, 1+(s−t), if s≥ t. Show thatvisnota discount function. 2.10 Show that, given any positive-valued functiongof one variable, v(s,t) = g(t) g(s) defines a discount function. 2.11 Suppose that v1 andv2 are two discount functions. Are either of the functionsv1v2 or v1 +v2 discount functions? These are defined by v1v2(s,t) = v1(s,t)v2(s,t), [v1 +v2](s,t) = v1(s,t)+v2(s,t). 2.12 Assume constant interest. (a) Show thaẗa(0,1n) = ̈a(1n)−1+vn. (b) Showthat ̈a(0,1n) = (1−vn)∕i.Trytodothisfrompart(a)andalsodirectlyfrom (2.14). EXERCISES 37 2.13 Let kn denote the vector (n,n−1,n−2,…,2,1) Show that ̈a(kn) = n− ̈a(0,1n) d in three different ways: (a) From first principles, as in the first derivation of (2.17); (b) Bymakinguseofalinearrelationshipbetweenthevectors(1 n),kn andthevector jn of Section 2.9; (c) By using the loan-interest argument given at the end of Section 2.9. 2.14 Suppose that v(2,7) = 0.5,v(2,8) = 0.4,v(2,9) = 0.3. Find a vector d actuarially equivalent to c = (1,3,5,2,9,10,6,4,8,3) such that di = ci for i< 7 and di = 0 for i> 7. 2.15 A loan of 2300 is to be repaid bynyearly payments of 230 beginning at time 5. The borrower is given the option of repaying only 115 at time 5, but must then pay 240 in each of the subsequent payments. Findv(5). (An exact numerical answer is required. It should not be a function ofn.) 2.16 A loan of 20 000, made at an interest rate of 6%, is to be repaid by level yearly payments for 10 years, beginning 1 year after the loan is advanced. Just before making the seventh repayment, the borrower wishes to repay the entire loan. (a) If interest rates remain unchanged, what is the outstanding balance? (b) Suppose interest rates have dropped to 5%. How much will the borrower have to pay if the lender uses the lower interest rate to calculate the outstanding balance? 2.17 Apersonhas1000nowandplanstoinvestitfor5years.Heistryingtodecidebetween two alternatives. The first is to buy a bond that matures in 5 years. The second is to buy a bond that matures in 10 years and sell it at the end of 5 years. Assume that in both cases the bonds have no payments before maturity and can be purchased today at an interest rate of 6%. How much better or worse off is the individual at the end of 5 years if he chooses the second alternative instead of the first, assuming that at time 5 the interest rate for this class of bonds is (a) 4%, (b) 7%? 2.18 For a certain cash flow vectorc, c 0 = 1, c1 = 5, ̈a(c) = 15. If the discount function is changed so thatv(1,2) is decreased by 0.1, while all other of values ofv(n,n+1) remain unchanged, then ̈a(c) decreases by 2.4. One the other hand, ifv(0,1) is decreased by 0.1 while all other values ofv(n,n+1) remain unchanged, then̈a(c) decreases by 2. Findv(0,1) andv(1,2). 2.19 canddareactuariallyequivalentvectorssuchthat cisconstantand disnondecreasing. Show thatkV(c−d) ≥ 0, fork= 0,1, …,N. *2.20 For the cash flow vector (−1,3, −2), (a) show that the i.r.r. does not exist; (b) findir as a function ofr. 38 THE BASIC DETERMINISTIC MODEL *2.21 Suppose that for a cash flow vectorc, the numberr satisfies ̃BN(c;v) = 0, where v(n) = (1+r)−n. Show thatir = r. *2.22 Show that when interest is constant at rater, then̈a(c) ≥ 0 if and only ifir ≥ r.( T h i s shows that the criteria for determining if it is worthwhile to enter into a transaction given after Theorem 2.2 are in accordance with the remarks in Section 2.1.) *2.23 Suppose that current interest yields on risk-free bonds are 4% for a 5-year bond and 5%fora10-yearbond.Calculatetheforwardpricetobepaidattheendof5yearsfor a zero-coupon bond that pays 1000 units, 10 years from today. Suppose that instead ofthenumberyoujustcalculated,thisforwardpriceis800.Illustratehowyouwould make a sure profit (under the idealized conditions given in Section 2.12). *2.24 The current price of a 1-unit zero-coupon bond maturing in 10 years is 0.610. The forward ratef(6,10) = 0.06.Find the spot ratey 6. Spreadsheet exercises 2.25 A loan contract involves advances of 10 000 initially, 20 000 one year later, and 30 000 one year after that. This is to repaid by 20 yearly instalments beginning at time 3. The payments reduce by 5% each year (so, for example, if the first payment was 1000, the second would be 950, the third would be 902.50). Interest rates are 6% for the first 5 years, 7% for the next 5 years, and 8% after that. Find all payments and the outstanding balances at the end of each year, until the loan is fully discharged. Balances should be calculated after the payment due at the particular time is made. 3 The life table 3.1 Basic definitions Fortheactuaryworkinginthelifeinsurancefield,amajorobjectiveistoestimatethemortality pattern which will be exhibited by a group of individuals. A basic device for accomplishing this is known as alife table. (It is also known as amortality table– an interesting example of a word and its opposite being used interchangeably.) Let𝓁 0 beanarbitrarynumber,usuallytakentobearoundfiguresuchas100000.Suppose we start with a group of𝓁0 newly born lives. We would like to predict how many of these individuals will still be alive at any given time in the future. Of course, we cannot expect to computethisexactly,butwecanhopetoarriveatacloseestimateifwehavesufficientlygood statistics. In the first part of this book we will make the assumption that we can indeed arrive at exact figures. This is in keeping with the concept of adeterministic modelintroduced in Chapter 1. In Part II, we introduce thestochastic modelfor mortality, where we investigate the more realistic assumption that the quantities we want are random variables. Let𝓁 x be the numberofthoseoriginallivesaged0whowillstillbealiveatage x,andlet dx bethenumber of those original lives aged 0 who die between the ages ofx andx +1. The basic relationship between these quantities is 𝓁x+1 = 𝓁x −dx. (3.1) A life tableis a tabulation of𝓁x and dx where x is a nonnegative integer. The following is an example of a portion of a life table (this is an illustration only, and the figures are not intended to be realistic): Fundamentals of Actuarial Mathematics, Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 40 THE LIFE TABLE x 𝓁x dx 0 100 000 2000 1 98 000 1500 2 96 500 1000 3 95 500 900 ⋮⋮ ⋮ 𝜔 0 The table will end at some age, traditionally denoted by𝜔, such that𝓁𝜔 = 0. This is the limiting age of the table, and denotes the first age at which all of the original group will have died. The actual value of𝜔 will vary with the particular life table, but it is typically taken to be around 110 or higher. 3.2 Probabilities Although we assume we can predict𝓁x exactly, there is still randomness in our model, since it is not known whether or not any given individual will be among the survivors at a particular point of time. It is convenient to introduce some elementary probabilistic notions. For nonnegative integersn andx,l e t npx = 𝓁x+n 𝓁x . (3.2) Whatisthemeaningofthisterm?Considerthe 𝓁x survivorsage x.Outofthisgroup, 𝓁x+n will survivetoage x +n.Thequotientthengivesustheprobabilitythatapersonaged x,hereafter denoted just by the symbol (x), will be alive at agex +n.L e t nqx = 𝓁x −𝓁x+n 𝓁x . (3.3) Thisgivesustheprobabilitythat( x)willdiebetweentheagesof x andx +n.Itisclearthat nqx = 1−n px. (3.4) As an example, in the table given above we would have2p0 = 965∕1000, 2q1 = 25∕980. Since a left subscript of ‘1’ occurs frequently it is omitted for notational convenience. Thatis, px denotes1px,and qx denotes1qx.Thequantity qx isoftenreferredtoasthe mortality rate at agex. Whatistheprobabilitythat( x)willdiebetweentheagesof x +n andx +n +k?Thisisa quantity which we will use frequently. There are three main ways of expressing it: 𝓁x+n −𝓁x+n+k 𝓁x , (3.5a) CONSTRUCTING THE LIFE TABLE FROM THE VALUES OFqx 41 or npx − n+kpx, (3.5b) or npxk qx+n. (3.5c) The reader should verify, by substituting values of𝓁, that all three expressions are equal. They can each be explained intuitively. Consider the first. The numerator is the number of people living at agex +n, less the number living at agex +n +k. This difference must be the number of people who died between the two ages. Dividing by the number of people that we start with will give us the required probability. In the second expression we express this quantity as the probability that (x) will liven years, but willnot liven +k years. In the third expressionweconsidertwostages.Todiebetweenthespecifiedages,( x)mustfirstlivetoage x +n. The individual, then being agex +n, must die within the nextk years. We will have occasion to use all three of these expressions, choosing the one which is most convenient for the purpose at hand. Another useful identity, which we will refer to as themultiplication rule,i s n+kpx = npxk px+n, (3.6) forallnonnegativeintegers n,k andx.Itcanbeverifieddirectlyfrom(3.2).Intuitively,itsays that in order for (x)t ol i v en+k years, the individual must first liven years, and then, being agex +n, must live anotherk years. 3.3 Constructing the life table from the values ofqx Thelifetableisconstructedinpracticebyfirstobtainingthevaluesof qx forx = 0,1, …,𝜔−1. Obtainingthesevaluesisastatisticalproblemwhichwewillnotdiscussindetail.Itisbasically donebycarryingoutastudyinwhichweobservehowlongpeopleofdifferentageswilllive. Forexample, ifweobserveagroupof1000peopleofexact age50and10ofthemdiewithin 1 year, then we could estimateq 50 as 0.01. This of course is an extreme simplification and the process is much more complicated. It is not practical to gather a group of people exactly age 50 at one point of time, and then to observe them for an entire year. In practice, people will enter the study at various times, and leave for reasons other than death. In addition, we must achieve consistency between values at different ages. The statistical subject known as survival analysisdeals with these problems. In this book we will take the values ofq x as given. The life table can then be constructed inductively, starting with𝓁0, from the formulas dx = 𝓁xqx, 𝓁x+1 = 𝓁x −dx, (3.7) which follow immediately from (3.1) and (3.3). We will see, however, that it is not usually necessary to actually calculate𝓁 x and dx. In practice, life tables are specified by just giving the values ofqx, which is sufficient for the necessary computations. The advantage of the traditional form lies mainly in its intuitive appeal, rather than its use as a calculating tool. 42 THE LIFE TABLE 3.4 Life expectancy Life expectancy is one of the most frequently quoted actuarial concepts. Probably for this reasonitisoftenusedincorrectly,aswewillexplainbelow.Abasicquestionisthefollowing. How long can a person agex expect to live? Of course there is much variety in the future lifetime of various individuals of the same age. Some will live for several years, and some will die immediately, but we can attempt to arrive at some sort of average figure. One approach would be to take a large number of people agex and observe them until all have died. We could then compute the total future lifetime of all these individuals. Dividing by the number of people in the original group would give an estimate of the desired average. For a drastically oversimplified example, take three people exactly age 60. Suppose one dies at age 62, another at age 72 1 2 and the third at age 911 4. The total future lifetime would be 2+121 2 +311 4 = 453 4. Dividing by 3 we could estimate that a person age 60 could expect to live on average another 151 4 years. Of course to be statistically accurate we would need many more than three people. Moreover, the length of time needed for such a study makes it completely impractical. Remarkably however, once we have the life table, we can obtain the figure directly, without carrying out the observations. To see this, we look at another approach to obtaining total future lifetime. Suppose we start with𝓁 x people agex.A f t e r1 year, there will𝓁x+1 survivors who would have each contributed 1 year of lifetime to this total. At the end of the 2 years there will𝓁x+2 survivors who would have each contributed anotheryeartothetotal.Continuinginthisway,wecanestimatethetotalfuturelifetimeofall lives as 𝓁 x+1 +𝓁x+2 +𝓁x+3 +⋯ +𝓁𝜔−1, and, dividing through by𝓁x, we obtain the quantity ex = 𝜔−x−1∑ k=1 𝓁x+k 𝓁x = 𝜔−x−1∑ k=1 kpx. (3.8) Thequantity ex isknownasthe curtate life expectancyorcurtateexpectationoflifeatage x.Thew ordcurtate meansreduced ortruncated,reflectingthefactthatthisisnotexactlythe quantity that we want. We have cheated a little in our alternate measurement scheme, for it measures only whole future years of lifetime and ignores the fraction of the year lived in the yearofdeath.Inourillustrationabove,forexample,underthealternatecountingmethod,the 60-year-old who died at age 72 1 2 would be credited withonly 12years of total lifetimerather than the actual 121 2. The person who died at 911 4 would be credited with only 31 years rather than the actual 311 4. We are undercounting between 0 and 1 years for each individual, and it seems reasonable to take this to be one-half on average. The true life expectancy, usually referredtoasthe complete life expectancyatage x anddenotedby e ◦ x isgivenapproximatelyby e ◦ x = ex + 1 2. We will give a more formal treatment ofe ◦ x in Chapter 8 and again in Chapter 15. LIFE EXPECTANCY 43 There is a simple recursion formula to computeex for all values ofx. From the second formula in (3.8), ex = px +2px +3px +⋯ +𝜔−x−1px = px(1+px+1 +2px+1 +⋯ +𝜔−x−2px+1) (3.9) = px(1+ex+1). The second line is obtained by using the multiplication rule (3.6). Note that this is a backward recursion formula as it gives the value of the function in terms of the next higher argument. The recursion is then started from the initial value ofe𝜔 = 0. Itisinstructivetogiveanintuitiveexplanationof(3.9).Toliveanywholenumberofyears in the future, (x) must first live to agex +1, as reflected by the factorpx. The individual will then have completed 1 year of lifetime, and in addition, being now agex +1, will complete, on average, the expected number of future whole years for a person of that age. The reader should note carefully that life expectancy is a function of age. For each agex, the life expectancy at that age gives the average number of future years that (x) will live. A majorsourceofmisquotingistoexpressthisasasinglefigureratherthanafunction.Onecan oftenfindstatementsinnewspapersorsimilarsources,statingsomethinglike‘lifeexpectancy hasincreasedfrom75.3to75.8years’.Thewriterisinvariablyreferringtothelifeexpectancy at age 0 only. This is certainly of interest, but it conveys somewhat limited information. A person already aged 80 who wishes to estimate how much longer he/she can expect to live is not helped by a statement which claims that newborn lives live on average to age 75.8. Thereadershouldalsonotethatlifeexpectancyistheaverage duration andnottheaverage age a person can expect to live to. We say, for example, that the life expectancy at age 50 is 31.2, meaning that on average a person age 50 can expect to live to age 81.2. There is often confusion on this point, which is again a result of the tendency to report only the age 0 figure where duration and age are the same. There are many other quantities obtainable from the life table which are of interest, but onlyoneinparticularthatwewilldiscusshere.Sometimeswemaybeinterestedintheaverage duration lived by (x) over the nextn years, wheren is some fixed duration. The quantity n∑ k=1 𝓁x+k 𝓁x = n∑ k=1 kpx (3.10) isknownasthe curtate n-year temporary life expectancyatage x.Thewordtemporary comes from an analogy with life annuities which we discuss in the next chapter. It gives us the expected complete number of years livedover the next n yearsby people now agex.T h i si s what we would compute if we repeated the alternate measurement system described above, but ended the observations aftern years. To adjust this for the undercounting in the year of death requires some care. Those who lived to agex +n will have contributed the correct total ofn years, and therefore it is only the (𝓁 x −𝓁x+n) people who died during then-year period who must be considered in the adjustment. To get a more accuraten-year temporary life expectancy at agex we add to the quantity in (3.10) not1 2, but rather1 2(𝓁x −𝓁x+n)∕𝓁x,t o get an approximation to thecomplete n-year temporary life expectancyat agex of n∑ k=1 kpx + 1 2nqx. 44 THE LIFE TABLE 3.5 Choice of life tables Thelifetablereflectsthefactthatageisthemajordeterminantoffuturemortality.Thereareof course several other factors that affect future lifetime, such as gender, health status, lifestyle, and geographical location. In practice, the effect of these factors is handled by producing several different life tables. One confines the mortality study that produces the table to a particulargroupofpeople,namelythosewiththecharacteristicthatyouwanttoseparateout. The following are a few of the more important distinctions made in practice. It is observed, for reasons that nobody has fully explained, that females live longer than males.Forthemiddlerangeofages,itistypicalforthelifeexpectancyofafemaletobefrom 5 to 7 years more than that of a male of the same age. To reflect this, it is usual to produce separate male and female life tables. In recent years, there has been overwhelming statistical evidence to show the dangers of smoking. This has led insurance companies to construct different life tables for smokers and nonsmokers. The choice of life table will also depend heavily on the type of contract that is being sold.Thelifetablesproducedforthegeneralpopulationfromcensusdataarenotsuitablefor insurance purposes. People accepted for life insurance policies are usually screened by the insurancecompanytomakesuretheyareinreasonablehealth.Theycanexpecttolivelonger than a person of the same age taken from the population as a whole. Life tables for insurance purposes are constructed by looking at insurance company data only. Purchasers of life annuities (discussed in Chapter 4) will on the average live even longer, since a person in poor health would be unlikely to buy such a product. Separate tables are needed for annuity purposes. Still another distinction to be made is the difference between individual contracts and group contracts. In the former case the buyers makes a definite decision to enter into the insurance or annuity contract and are presumably aware of their health conditions and acting in their best interests. In the latter cases, an employer purchases the contract to cover a large group of employees. Different mortality patterns in the two cases can be expected. There are many other examples which we will not discuss here, although some will be pointed out briefly in succeeding chapters. The reader should be aware that selecting an appropriate table for a particular use is an important actuarial task. Sometimes a very simple method, known asmultiples of standard mortality, is employed toconstructmanydifferenttablesfromagivenone,knownasthestandardtable.Forexample, itmightbedecidedthatforrisksofacertaintype,themortalityis150%ofstandardmortality. Thelifetableforsuchrisksisconstructedbymultiplyingeach q x inthestandardtableby1.5. This method is justified more by simplicity of calculation rather than any scientific rationale. 3.6 Standard notation and terminology We have already introduced the standard symbolsnpx,nqx,ex,e ◦ x and𝜔. The symbol n|kqx denotes the probability that (x) will die between the ages ofx +n and x +n +k, the quantity that we already have three ways of writing as shown in (3.5). A subscript of 1 is omitted, so thatn|qx denotes n|1qx. This use of a vertical bar is a typical actuarial device to denote a ‘waiting period’. In this case, the symbol is intended to indicate that the person will waitn years and then die in the followingk years. EXERCISES 45 The quantity in (3.10) is denoted byex:n and the complete temporary life expectancy is denoted bye ◦ x:n. 3.7 A sample table Todothespreadsheetproblemsinthefollowingchaptersthatrequirealifetable,weintroduce a sample table. It is given by q x = { 1−e−0.00005(1.09)x , x = 0,1, …,118, 1, x = 119. (3.11) We have𝜔 = 120.Theformulaiseasilyprogrammedintoaspreadsheet,andthatisthereason forgivingthetableinthisform.Givinganexistingtablewouldnecessitateenteringthefigures individually. Another advantage of this parametric form is that the two constants of 0.00005 and 1.09 can be changed toprovide a variety of different lifetables for comparison purposes. Our table is, as stated, a sample table, and it should not be taken as being a realistic picture of modern-day mortality. This is true especially at the younger or very old ages, as will be discussed later in the text. Chapter 14 will provide some motivation for the formula; see, in particular, Exercise 14.11. Notes and references London (1997) provides an introduction to survival analysis, and gives more details on the construction of life tables. Exercises Type A exercises 3.1 You are given thatq60 = 0.20,q61 = 0.25,q62 = 0.25,q63 = 0.30,q64 = 0.40. (a) Find𝓁x for ages 60–65, beginning with𝓁60 = 1000. (b) Find the probabilities of the following: (i) (61) will die between the ages of 62 and 64. (ii) (62) will live to age 65. (c) Given thate65 = 0.8, findex forx = 60–64. 3.2 You are given that5p40 = 0.8,10p45 = 0.6,10p55 = 0.4. Find the probability that (40) will die between the ages of 55 and 65. 3.3 Supposethatoutofatypicalgroupof100peopleage70,10willdieinthefirstyear,15 will die in the second year, and 20 will die in the third year. Calculateq70,q71,q72 and 3p70. 46 THE LIFE TABLE Type B exercises 3.4 Suppose that 𝓁x = 100−x, x = 0,1, …,100. Findexpressionsfor(a) npx,(b) nqx,(c)theprobabilitythat(x )willdiebetweentheages of x +n andx +n +k. 3.5 You are given that e60 = 17,10p50 = 0.8, and that the 10-year curtate temporary life expectancy at age 50 is 9.2. Finde50. 3.6 Prove the following statement, assuming the approximation given in the text, and give an intuitive explanation: e ◦ x = 1 2qx +px(1+e ◦ x+1). 3.7 Suppose that qx is equal to a constantq for allx. (Note that in this case𝜔 =∞ ). Find expressionsintermsof q andn for(a) npx,(b)e x.Doyouthinkthatthisgivesarealistic life table? Why or why not? 3.8 In constructing a life table for heavy smokers, Actuary A decides to take a standard table, and double each value ofqx, using a value of 1 ifqx > 0.5. Actuary B takes the sametableandsquareseachvalueof px.ShowthattheresultingtableofActuaryB,has lower mortality rates at all ages than that of Actuary A. Spreadsheet exercise 3.9 Taking the sample life table as given by (3.11), use recursion to find ex for x = 0,1, …,𝜔−1.Focuson e0.Howmuchisthisreducediftheconstant0.00005ischanged to 0.00006. What happens if 0.00005 is kept the same, but 1.09 is changed to 1.092? 4 Life annuities 4.1 Introduction The financial losses that we listed at the beginning of Chapter 1 are no doubt familiar to all readers.Anothertypeofrisk,whichmaynotbesoobvious,isthatof livingtoolong.Givenour definitionofriskasthepossibilityofsomethingbadhappening,thereadermaywonderabout thisstatement.Afterall,isnotlivingalonglifeagoodratherthanbadoccurrence?Itcertainly can be, but it does carry with it the possibility of financial hardship if one does not have adequate income. Imagine the decision faced by a retired individual who has accumulated savings of 1 000 000, which is invested at an annual rate of 5%, providing an annual income of 50 000. Imagine also that the individual decides that this is an insufficient return, so it is necessary to consume a portion of the capital each year, as well as the interest. Doing so, however, runs the risk that the entire capital may be depleted before death, leaving the individual with no source of income for his/her remaining lifetime. Life annuities are a means of insuring against such a risk. A life annuity is a contract between an insurance company, known as theinsurer, and another party, known as the annuitant,which provides the following. In return for the payment by the annuitant of pre- scribedpremiums,theinsurerwillprovideasequenceofpayments,knownas annuitybenefits of prescribed amounts and at prescribed times. The unique provision is that the annuitant must be alive to receive each benefit payment. These will terminate upon the death of the annuitant. People who purchase such a contract are investing capital that they have decided will not be needed for any dependents after they die. They agree to give up these funds upon their death in return for the greater yield they can achieve on their investment by sharing the amounts forfeited by those who predecease them. A good way to picture the workings of a life annuity is to imagine a room with a number of small boxes, belonging to one annuitant each.Eachannuitantpaysthesamepremiumintohis/herbox,andinterestearningsareadded FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 48 LIFE ANNUITIES totheamountsinhis/herbox.Eachreceivesthesameannuitybenefits,whichareprovidedby takingmoneyoutofthebox.Whenoneparticipantdies,his/herboxisopened,andtheamount is spread evenly among all the other boxes in the room. Participants then receive not only interest earnings, but also these forfeited amounts of those who die before them. We refer to thelatteras survivorshipearnings.Thisisafairlyaccurateaccountofwhatactuallyhappens, except that amounts are calculated on paper. There is obviously no need for the physical rooms or boxes. More precisely, it is an account of what would happen if mortality followed the life table exactly. For most annuities sold in practice, the accumulation is guaranteed in advance, so it does not depend on the actual deaths. If fewer people die than expected, the insurer would still have to add the extra survivorship amounts to people’s ‘boxes’, and would experience a loss, while more deaths than expected would result in a gain to the insurer. Tobeequitable,allparticipantsinanyoneroommusthaveroughlythesameriskofdying, andinparticularmustbeofthesameage.A70-year-old,putinthesameroomas20-year-olds, would be much more likely to die early on and would clearly be at a disadvantage. We will thereforecalculateannuitypremiumsprimarilyasafunctionofage.Thereareofcoursemany other factors to consider, such as gender, which we discuss briefly in Section 4.6. Insurance companies are not the only source of life annuities. Many pension plans pay retirement benefits in the form of a life annuity. 4.2 Calculating annuity premiums We now turn to the mathematical aspects of annuities. The reader should review the vector notation given in Section 2.8, which we will make frequent use of. Weassumethatwehavefixedthroughoutanappropriatelifetableandadiscountfunction vto model the effect of investment earnings. Consideracontractsoldto(x )–or,moreaccurately, onthelifeof (x),sincethepurchaser could be a different party than the annuitant, such as an employer. It is assumed throughout that (x) is an integer, which allows us to conveniently use a life table if desired. Indeed, the common practice is to classify an individual by their age on their last birthday, which is just the normal way that one thinks of age. We suppose that benefit payments are made yearly. Suppose that the amount of benefit paid at timekis c k. This could of course be zero, indicating that no payment is to be made. The contract is then described by the cash flow vector c = (c 0,c1,…,c𝜔−x−1). In this case we will refer toc as theannuity benefitvector. (Notethatthefinalcashflowismadeatage 𝜔−1,whichthelastageatwhichanyoneisliving, according to our model.) Suppose that the benefits are to be purchased by a single premium paid at the beginning of the contract at agex. Our first problem is to calculate the premium that the annuitant should pay. The criterion is that the total premiums from all annuitants, together with the interest earned, should be sufficient to provide all of the required benefits, assuming that invested capital accumulates according to the given discount function and that mortality follows that of the given life table. Consider the simplest possible case, namelyc = e k. This is an annuity consisting of a single benefit payment of 1 to be made at timekif (x) is then alive. Such a contract is usually referred to as apure endowment.L e tE denote the single premium that should be paid to provide for this. Suppose that𝓁x people, each agex, buy such a contract. The total amount collected in premiums at time zero would be𝓁xE, and this will accumulate to𝓁xEv(k,0)at CALCULATING ANNUITY PREMIUMS 49 time k. There will be𝓁x+k survivors at timek, and each of these must receive 1 unit. This means that 𝓁xEv(k,0) = 𝓁x+k. Solving, we obtain E= v(k)𝓁x+k 𝓁x = v(k)kpx. This can be verified intuitively. If the benefit of 1 unit at timek were guaranteed, we know from Chapter 2 that we would need to investv(k). In this case, where the benefit is not guaranteed, we multiply by the probability of receiving it to arrive at the premium. Thegeneral caseeasilyfollows.Let ̈ax(c)denote thesinglepremium foralifeannuity on (x)withbenefitvector c.Thiscontractcanbeviewedasasequenceofpureendowments,one foreach value ofk,wherethe kthpureendowment pays ck attime k.Thepremiumforsucha pureendowmentisjust ckv(k)kpx,andthetotalpremiumisobtainedbysummation.Wehave ̈ax(c) = 𝜔−x−1∑ k=0 ckv(k)kpx. (4.1)‡ Life annuities have been classified into various types. An annuity for which benefit payments are made for a fixed number of years and then cease is known as atemporarylife annuity(an example for those who collect oxymorons). The most common type of annuity is one for which benefits continue for as long as the annuitant lives. This is known as awhole lifeannuity. Purchasers of whole life annuities protectthemselvescompletelyagainstoutlivingtheiravailablecapital,fornomatterhowlong theylive,theincomefromtheannuitywillcontinue.Wehavenoneedtodistinguishbetween temporary and whole life annuities in our mathematical model. As indicated, we consider all annuities on (x) as running from time 0 to time𝜔−1−x. In the case of a temporary life annuitywesimplytakebenefitsequalto0afterthelastpositivepayment.Inreallife,however, there is a difference. The insurer can obviously not inform a person who has reached the age of𝜔,thataccordingtotheinsurer’smodeltheyarenolongeraliveandannuitypaymentswill stop. This means that as a practical matter the benefit payments on a whole life annuity must eventually be constant (or at least follow some regular pattern) so it is clear what amount to continue to pay should an annuitant live beyond𝜔. In most tables,𝜔 is set sufficiently high that such an occurrence is extremely rare. Deferredannuities arecontractswheretheinitialpaymentdoesnotcommenceforseveral years. For example, a person wishes to provide for an income beginning at retirement. Math- ematically, this simply means that the initial entries in the benefit vector will be zero, so no special treatment is required. The annuity benefit vector will be of the formc = (0 k,d), and the durationkis known as thedeferredperiod. Example4.1 Atemporarylifeannuityon(50)providesfor1payableatage50,2payableat age51,3payableatage52,and4payableatage53.Suppose q50 = 0.1, q51 = 0.2, q52 = 0.25. The interest rate is 50% for the first year and 100% thereafter. Find the single premium. 50 LIFE ANNUITIES Solution. Forasmallproblemsuchasthis,itisconvenient tomakeuseofatimediagramas introduced in Chapter 2. See Figure 4.1. Benefits 4 3 2 1 Time 3 2 1 0 vs 2/31 /21 /2 qs 0.10 .20 .25 Figure 4.1 Example 4.1 Here we have inserted values ofqy, for relevant agesy, as well as the 1-year interest discountfactors.Wenowmustmultiplyeachpayment ck bytheproductofallpreviousvalues of v(i,i+1) as we did in Chapter 2, andalsoby kpx which we calculate as the product of all the previous values ofpy. The single premium equals 1+ ( 2× 2 3 ×0.9 ) + ( 3× 2 3 × 1 2 ×0.9×0.8 ) + ( 4× 2 3 × 1 2 × 1 2 ×0.9×0.8×0.75 ) = 3.28. An alternate way of calculating the survival probabilities, which some may prefer, is to use (3.2). We can construct part of a life table, utilizing (3.7). In this example, starting with say 𝓁 50 = 1000 we would get in turnd50 = 100, 𝓁51 = 900, d51 = 180,𝓁52 = 720, d52 = 180, 𝓁53 = 540.Thisautomaticallyperformsallthemultiplicationinvolvingthevaluesof py, and gives us the factors of 0.9, 0.72, 0.54, used above. 4.3 The interest and survivorship discount function 4.3.1 The basic definition Itisimportanttonotethattheformulafor ̈ax(c)isthesameasthatgivenfor ̈a(c)inChapter2, but withv(k) replaced byv(k)kpx for each value of k. This suggests that we are in the general situation of Chapter 2, but with a different discount function. This is indeed the case. Define yx(n) = v(n)npx and extend to a two-variable function yx(k,n) = yx(n) yx(k) forallnonnegativeintegers k,and n.Itisclearthat yx satisfies(2.1)andthereforeisadiscount function. This fact is borne out by the discussion of the pure endowment in the previous section, which showed that the value at agexof 1 paid at agex+k, when accumulation is by interest and survivorship, is preciselyy x(k). THE INTEREST AND SURVIVORSHIP DISCOUNT FUNCTION 51 Using the multiplication rule we can verify that fork≤ n yx(k,n) = yx(n) yx(k) = v(k,n)n−kpx+k, (4.2) and fork> n yx(k,n) = v(k,n) k−npx+n . (4.3) So for example, assuming constant interest yx(2,5) = v3 3px+2, yx(5,2) = (1+i)3 3px+2 . Note that the last quantity, giving the amount at time 5 resulting from an investment of 1 at time 2, is greater than (1+i) 3, the amount resulting from interest only. This reflects the extra earnings due to survivorship. Wewillreferto yx asan interestandsurvivorship discountfunction.From(4.1)itfollows that ̈ax(c) = ̈a(c;yx), (4.4) andisthereforejustthepresentvalueofthepayments,withresecttothediscountfunction yx. The significance of the above is that it shows that all the results of Chapter 2 can be carried over directly to life annuities. As a matter of terminology we will interchangeably refer töa x(c) and similar quantities introduced later, asnetsinglepremiums orpresentvalues. When calculating ̈ax(c) by spreadsheets it is convenient to make use of the recursion formula (2.7) adapted to the interest and survivorship function. We have yx(k+1) = yx(k)v(k,k+1)px+k. (4.5) Notation Considerthevector1 𝜔−x whichreferstoa1-unitcashflowsequencecontinuingfor thelifeof(x ).Duetotheprevalenceofthissymbol,itissuppressedfornotationalconvenience. This is a standard convention of actuarial notation that will be followed throughout the book. Ingeneral,wheneveravectorisomitted,itisunderstoodtobe1 𝜔−x.Forexample, ̈ax denotes thenetsinglepremiumforawholelifeannuityon( x),paying1peryearforlife,beginningat time 0. In cases where payments run for life, but are not necessarily constant, it is sometimes convenient to avoid specifying𝜔 by using the symbol 1∞ to refer to a vector with entries of 1, running to age𝜔−1 For example if𝜔 = 120 we could writëa40(215,1∞) in place of ̈a40(215,165). Inwhatfollowswewillkeep vasthenotationfortheinvestmentdiscountfunction,which reflects the discounting resulting from earnings on invested capital. We will use the lettery to denote a general discount function, which could bev or yx or any other such function. 52 LIFE ANNUITIES Life annuity symbols will be written as given with the age as a subscript. For present values involving discounting at interest only, we will then incorporate thevand writëa(c;v). 4.3.2 Relations between yx for various values ofx The reader should note that we have several interest and survivorship functions, one for each value of the initial agex. They are, however, related in the following way. From (4.2) y x◦ k(n) = yx(k,k+n) = v(k,k+n)npx+k = v◦ k(n)npx+k. This tells us that yx◦ k(calculated withv) = yx+k(calculated withv◦ k). (4.6) Therefore, if interest is constant, yx+k = yx◦ k. (4.7) In particular, with constant interestyx = y0 ◦ x, so that knowing the interest and survivorship function for initial age 0 determines it for all initial ages. In the remainder to this section, we continue with the notational ideas first introduced in Section 2.11, with the intention of providing a background for following classical actuarial formulas. It can be safely omitted at first, as everything can be conveniently expressed by using the new notation that we introduced in Chapter 2 and discussed in Section 2.14.2. We do however occasionally refer back to this section as well as Section 2.14.2 In particular the symbols v◦ k, andc◦ k prove to be convenient and will be used in various places. In the first place, since we do not assume constant interest as was traditionally done, it is necessary to introduce a new notational device into the classical life annuity symbols in order to handle time shifting. The idea is to separate age from duration in the quantity x+k.W el e ẗa [x]+k(c) denote the present value of a life annuity on (x+k), paying ci at time i, and calculated with respect to the investment discount functionv◦ k. Of course with constantinterest, v◦ k= v,sothat ̈a[x]+k(c) = ̈ax+k(c)forall k,andthisnotationisnotneeded. (However even with constant interest we will have need of this notation with a more refined mortality model which we discuss in Chapter 9.) From (2.28) we can write ̈a [x]+k(c◦ k) = Valk(kc;yx), (4.8) so the symbol on the left hand side is conveniently interpreted as the value at timek,o ft h e benefits at andaftertimek, paid on a life annuity issued to (x) with benefit vectorc. The above equation leads to the traditional form of the life annuity splitting identity. ̈ax(c) = ̈ax(kc)+yx(k)̈a[x]+k(c◦ k). (4.9) GUARANTEED PAYMENTS 53 For a particular case of (4.9), assume constant interest and letc = 1∞. Then kc = 1k and c◦ k= 1∞. Using (3.2) we can write ̈ax = ̈ax(1k)+vk𝓁x+k 𝓁x ̈ax+k. (4.10) Thiswasapopularformulainpre-computerdays.Itwasusedtocalculatevaluesoftemporary annuities ̈a x(1k) froma table of values of whole lifeannuities and the lifetable. Withmodern spreadsheets, we do not need this or similar formulas for calculation purposes. They are however sometimes useful to explain relationships between various quantities. Formula (4.9) is often applied to express the present value of a deferred annuitÿa x(c) where c = 0k◦ d, directly in terms of the vectord. Since kc = 0 and c◦ k= d,w eh a v e immediately that ̈ax(c) = yx(k)̈a[x]+k(d). (4.11) Example4.2 Thisexampleisdesignedtofurtherillustratethesquarebracketnotation.Sup- pose thatq60 = 0.05, q61 = 0.10, q62 = 0.15, i0 = i1 = 0.04 andik = 0.08 fork≥ 2. Calcu- late ̈a60(14), ̈a[59]+1(14), ̈a[58]+2(14). Solution. ̈a60(14) = 1+(1.04)−10.95+(1.04)−2(0.95)(0.90) +(1.04)−2(1.08)−1(0.95)(90)(0.85) = 3.33 ̈a[59]+1(14) = 1+(1.04)−10.95+(1.04)−1(1.08)−1(0.95)(0.90) +(1.04)−1(1.08)−2(0.95)(90)(0.85) = 3.27. This could be interpreted as the amount that a person now age 59 would have to pay for a 1-unit 4-year annuity if the purchase was made at the end of 1 year (assuming no change in the life table). It is lower than the first amount, since there is only 1 year of low interest rates rather than two. ̈a [58]+2(14) = 1+(1.08)−10.95+(1.08)−2(0.95)(0.90) +(1.08)−3(0.95)(90)(0.85) = 3.19. This could be interpreted as the amount that a person now age 58 would have to pay for a 1-unit 4-year annuity if the purchase was made at the end of 2 years. It is lower yet, since the buyer would avoid both years of low interest. 4.4 Guaranteed payments Supposeanindividualpurchasesalifeannuityanddiesimmediatelyafterpayingthepremium. The person will get nothing back, except the benefit payment at time zero if that is positive. While we have seen that this is fair, the situation is not always well understood by the dependentsoftheannuitant,whomaywellcomplainthattheinsurerhasconfiscatedthefunds. 54 LIFE ANNUITIES Asidefromthis,manyprospectivepurchasersareuneasywiththepossibilitythatalloralarge portion of their money could be lost. To provide additional flexibility, insurers frequently offer life annuities with aguaranteedperiod. A typical contract of this type would stipulate that for a certain duration (commonly 10 or 15 years) the benefits will be paid regardless of whethertheannuitantisaliveornot.Afterthisguaranteedperiodthecontractrevertstoalife annuity and benefits are conditional on survival. The guaranteed period of course means that a higher premium must be paid for the same level of benefits. To calculate the premium, it is best to consider such an annuity as two separate contracts, one for the guaranteed payments, the other for the contingent payments, and add the respective premiums. Examples follow. Example 4.3 A person age 40 purchases a life annuity that provides 10 000 each year for life,withthefirstpayment startingatage41.Thefirst10payments willbepaidregardless of whether the annuitant is alive or not. Find a formula for the single premium. Solution. The premium for the guaranteed annuity is 10000 ̈a(c;v), c = (0,1 10). The premium for the non-guaranteed annuity is 10000 ̈a40(f), f = (011,1∞). Thus total premium is just the sum 10000 [̈a(c;v)+ ̈a40(f)]. Letusverifythedurationinthevector f.Thefirstguaranteedpaymentisattime1,sothelast guaranteed payment is at time 10, and the first non-guaranteed payment is at time 11. As we start the indexing with 0, there will be 11 zeros in the non-guaranteed vector. Example4.4 Apersonage40purchasesalifeannuitythatprovides1peryearforlifewith the first payment at age 65. If (40) lives to age 65 he/she will receive at least 10 payments. Nothing is paid if death occurs before age 65. Find a formula for the premium. Solution. The premium for the non-guaranteed annuity is̈a 40(035,1 ∞). One must be careful with the ‘guaranteed’ portion here, which is deferred and therefore not completely guaranteed. The value of this at time 25 is Val25(025,110;v). To get the value at time 0, we multiply this byy40(25). It would be wrong to multiply byv(25), since these payments are not made if (40) dies before age 65, so that accumulation is at interest and survivorship for the first 25 years. The total premium is therefore ̈a 40(035,1∞)+y40(25)Val25(025,110;v). The more traditional way of writing this, incorporating our time shifting notation to handle non-constant interest is v(35) 35p40̈a[40]+35 +v(25)25p40̈a(110;v◦ 25). DEFERRED ANNUITIES WITH ANNUAL PREMIUMS 55 To handle the general case of guaranteed payments, letu denote a vector of benefit paymentsthatareguaranteedprovidedtheannuitantlivestotime g.SoinExample4.3, g= 0 and u = (110), whereas in Example 4.4,g= 25 andu = (025,110). Clearly, we can assume thatthefirst gentriesareequalto0sothat uisoftheform(0 g,w).Then,reasoningasabove, the present value of the guaranteed payments is gpẍa(u;v) = v(g)gpẍa(w; v◦ g), which equals the present value with respect to the discount functionv, multiplied by the probability that the payments will be made. SeeExercise4.15foronevariationontheguaranteed payment concept. Wepresentmore modifications in the next chapter. See Example 5.11 and Exercise 5.19. 4.5 Deferred annuities with annual premiums Deferred annuities are often purchased by a series of annual premiums rather than a single premium. The premium payment period can be any length that does not exceed the deferred period, so that premiums stop when the benefit payments commence. Deferred annuities with annual premiums are frequently used to provide pensions. An individual, together with his/heremployer,willpaypremiumsduringhis/herworkingyearsinordertoprovideincome beginning at retirement. The sequence of annual-premium payments in the contract can be described by means of a premium pattern vector. This is a vector𝜌 with 𝜌 0 = 1. Then,𝜌k denotes the ratio of the premiumpayableattime ktothepremiumpayableattime0.Ifweknowthepremiumpayable at time 0, often referred to as theinitialpremium, the premium pattern vector will determine all premiums. Namely, if𝜋k denotes the premium payable at timek, 𝜋k = 𝜋0𝜌k. (4.12) The vector𝜋 = (𝜋0,𝜋1,…) will be called thepremiumvector. Supposethat( x)purchasesalifeannuitytobeginatage x+n.Themostcommonpatternin practicewouldbe 𝜌 = (1n),signifyingalevelpremiumpayableuntiltheincomecommences. However, many other patterns are encountered. Some annuitants may prefer to pay a higher premium but for a shorter period, adopting a pattern of (1 m)f o rs o m em< n. Others may prefer to pay a lower premium for an initial period, and then more at the end. For example, the pattern𝜌 = (1 k,2n−k) would call for the premium to double afterkyears. Thepremiumpaymentsalsoconstitutealifeannuityon(x )sincetheywillceaseupon( x)’s death. (In this case they are paidby(x) rather than to (x).) To achieve the goal that premiums togetherwithinvestmentearningsaresufficienttoprovidetherequiredannuitypayments,the actuary will set premiums to be actuarially equivalent to benefits, with respect to the interest and survivorship discount function. Example 4.5 (Figure 4.2). An annuity on (40) provides for 1 annually for life, beginning at age 65. Nothing is paid for death before 65. Annual premiums are payable for 25 years 56 LIFE ANNUITIES beginning at age 40. The premium reduces by one-half after 15 years. Find a formula for the initial premium, assuming𝜔 = 110. Benefits 0 00. . . 0 0. . . 1 10. . . 1 Time 2 10 ... ... 151 4 ... ... 26 252 4 ...69 Premiums π 0 π0 π0 π0 π0/ 2 ... π0/2 Figure 4.2 Example 4.6 Solution. Equating present values, we have ̈a40(c) = ̈a40(𝜋) = 𝜋0̈a40(𝜌), where c = (025,145), 𝜌 = (115,0.510). Solving, we obtain 𝜋0 = ̈a40(c) ̈a40(𝜌). 4.6 Some practical considerations 4.6.1 Gross premiums Theannuitypremiums,bothsingleandannual,thatwecalculatedintheprevioussectionsare referred to asnetpremiums orbenefitpremiums. They are the premiums that are required to provide the benefits. The premiums that are actually charged in practice are known asgross premiums, and involve other factors intheircalculation. The insurer must make provision for amounts to cover expenses and profits. In addition, a contingency charge is added to provide for adverse experience. After all, interest earnings may fall short of those predicted by the discount function or people may live longer than predicted by the given life table. Moreover, marketingconsiderationsinevitablyplayarole.Ifaninsurerwantstostayinoperationitmust ensure that the premiums it actually charges are competitive. The many details that go into the calculation of the actual gross premiums are beyond the scope of this book, although we willtouchonthistopicinChapter12.Inanyevent,theinitialbasicstepistocalculatethenet premiums as done in the examples of this chapter. 4.6.2 Gender aspects The role of gender in life annuities is a controversial issue. As noted above, the participants in a ‘female agexroom’ can expect to live longer than those in the ‘male agexroom’. The females will receive less money in forfeiture due to death, and therefore must invest more to STANDARD NOTATION AND TERMINOLOGY 57 receive back the same amounts. If the insurer uses separate life tables for males and females, theresultisthatfemalesmustpayhigherpremiumstoreceivethesameannuitybenefits.This has in particular caused controversy in connection withdefined contributionpension plans. In such a plan, the employee and employer make specified contributions into a fund and the accumulatedamountsarethenpaidoutassomeformoflifeannuitywhentheemployeeretires. (This is as opposed to adefinedbenefit plan, where the annuity income is fixed as a function oflengthofserviceandsalary.)Ifseparatelifetablesareused,thesamelevelofcontributions will purchase smaller pension benefits for a female than for a male of the same age. This has resultedinchargesofdiscriminationbygender.Therehavebeenlengthydebatesonthisissue and many points of view put forward. Some question the fairness of considering only gender while ignoring other factors. For example, a female smoker may actually expect to live a shorterperiodthanamalenonsmokerofthesameage,butwillstillreceiveasmallerpension ifonlygenderistakenintoaccount.Theissueiscomplicatedandwewillnotgointoitfurther here.Thetrendlatelyhasbeentoagreethatthegenderdiscriminationshouldbeavoided,and manypensionplansnowuse‘unisexlifetables’thatareablendofthecorrespondingmaleand femaletables.Theresultisthat,fordefinedcontributionplans,malesreceivesomewhatlower pensionbenefits,andfemalesreceivesomewhathigherpensionbenefits,thantheywouldhave received had separate male and female tables been used. 4.7 Standard notation and terminology Standard notation for life annuities generally follows that for general annuities as introduced in Section 2.14, except that the agexis inserted as a right subscript, as we have done. So, for example, ̈a x: n = ̈ax(1n), ax: n = ̈ax(01,1n); ̈sx: n = yx(n)−1̈ax(1n), sx: n = yx(n)−1̈ax(0,1n); Iax: n = ̈ax(0,1,2,… ,n), Ïax: n = ̈ax(1,2, …,n); Dax: n = ̈ax(0,n,(n−1),(n −2),…,1), D̈ax: n = ̈ax(n,(n−1),(n −2),…,1); k|n̈ax = k−1|nax = ̈ax(0k,1n). The last item is the present value of a deferred annuity where the first payment is at timek, and there arenpayments. In some cases, as here, a duration appears as a left subscript rather than under an ‘angle’. To understand the second symbol, note that the unadornedaindicates a first payment at time 1, and thek−1 before the deferred symbol indicates that payments are k−1 years later than this, that is at timek, which is the same as the first symbol. The present value of ak-year pure endowment of 1 is denoted bykEx, another example where duration is given by a left subscript. Wehavealreadyintroducedthesymbol ̈ax,inwhichtheomitteddurationsymbolindicates a whole life annuity with payments of 1 for life. In the standard notation this carries over to the other symbols as well. That is, ̈a x, ax, Ïax, Iax, k|̈ax = k−1|ax are all defined as the corresponding symbols above, with benefits continuing for as long as (x) is alive. In other words,nis taken to be𝜔−x−j, wherejis the time of the first nonzero payment. 58 LIFE ANNUITIES Traditional actuarial texts include many identities relating these quantities. For the most part,theyallfollowfromthesplittingidentity(4.9).Forexample(assumingconstantinterest), we get the following variation of (4.10), written in standard notation as ̈a x:n+k = ̈ax:k +vk kpẍax+k: n. The termannuitycertain or fixedtermannuity is often used to describe annuities where the payments are certain to be made, as distinguished from life annuities. 4.8 Spreadsheet calculations In order to handle problems involving a life table, we proceed as follows. As in the Chapter 2 spreadsheet we put duration in column A, and interest rates in Column B, starting at row 10.WeinsertthelifetableincolumnN(columnsinbetweenarereservedforotherpurposes, described in later chapters ), withq y entered in cellN(10+y). Our sample table can be input as follows. Enter the two parameters 0.0005 in cell N3, 1.09 in cell N4. This allows us to change parameters when desired. Then enter the formula = 1− EXP(−N$3 ∗ N$4̂$A10) in cell N10 and copy down to cell N128. Enter 1 in cell N129. It is important to also ensure that the remaining cells in Column N are filled with zeros up to at least N250. In cell C1 we insert the agex. We then use the index function in Excel to select the pertinent mortality rates, puttingq x+t in cell C (10+t) This is done by inserting the formula =INDEX(N$10 :N$250,C$1+$A11) in cell C10 and copying down. Entries of zero will appear after duration𝜔−xcoming from the extra zeros in Column N. Wecalculate v(k)incolumnDasinSpreadsheet2,entering1incellD10, =D10/(1+B10) in column D11 and copying down. We calculate the vectoryx in column E, using the recursion (4.5). The entry in E10 is 1, the entry in E11 is = E10 ∗ (1+B10)̂−1 ∗ (1−C10) and we copy down. We insert the cash flow vectorcin column F, and calculate the valuëax(c) in cell F8 with the formula. =SUMPRODUCT ($E10 : $E129,F10 :F129). For example at a constant interest rate of 6%,̈a50(120) = 11.5957. For annual-premium deferred annuities, we enter the premium pattern vector in Column H. We copy the formula in F8–H8 (reserving column G for a later purpose). In I6, we enter =F8/H8, the initial premium. In I10, we enter=$I$6*H10 and copy down, which inserts the premium vector in column I. Copying the formula in H8–I8 provides a check. We should get the same value as in F8. EXERCISES 59 Asacheck,at6%interest,foranannuityon(50)providing1perlifebeginningatage60, the level annual premium payable for 10 years is 0.855. We will leave it to the interested reader to modify this spreadsheet to handle guaranteed periods. Notethatthisspreadsheetcanbeusedtohandleshortdurationproblemswiththemortality rates given individually, such as we have in the examples and exercises. We can ignore the given age, and simply insert the values ofq x+t that we want directly in column C. Remember howevernotto save changes when closing, or alternatively, copy up in column C to restore the formulas when you next use the spreadsheet. Exercises Type A exercises 4.1 Redo Example 4.1, only suppose now that interest rates are 50% for the first2years and 100% after that. 4.2 Agroupofindividualsage40,eachinvest1000inafundearninginterestat5%.Atthe end of 20 years the fund is divided equally among the survivors. If20p40 = 0.8, how much does each get? 4.3 You are givenq60 = 0.20, q61 = 0.25, q62 = 0.40, q63 = 0.50. The interest rate is a constant 5% for the first 2 years, and 7% after that. A 5-year life annuity on (60) provides for payments of 100(1+k) at timek, wherek= 0,1,2,3,4. (a) Find the present value. (b) Supposethatinsteadofbeingastraightlifeannuity,thefirstthreeannuitypayments are guaranteed regardless of whether (60) is alive or not. Find the present value now. 4.4 You are given that interest is constant and that ̈a 40(130) = 21, ̈a60(110) = 6, v20 20p40 = 0.8. Calculate ̈a40(120). 4.5 The present values of a 10-year life annuity on (60), and a 10-year compound inter- est annuity, with annual payments of 1, are given, respectively bÿa60(110) = 5 and ̈a(110;v) = 6. A person age 60 has 100 000 that he plans to pay as a single premium for a life annuity, beginning at age 60. This will provide a level income of 12 500 per year for life. What will the yearly income be if, instead of a straight life annuity, he purchases a life annuity with a 10-year guaranteed period? Type B exercises 4.6 Mortalityisgivenby q52 = 0.1, q53 = 0.2.Theinvestmentdiscountfunction visgiven by i0 = 0.2, i1 = 0.25, i2 = 0.25, i3 = 0.5. Calculatey50 ◦ 2(2) andy52(2), usingvin both cases, to show thatyx◦ kis not equal toyx+k when interest is not constant. 60 LIFE ANNUITIES 4.7 If interest is constant andqx is a constantqfor allx, find an expression for̈ax in terms of vandq. 4.8 Showthat,ataconstantzerorateofinterest, ̈ax(0,1∞) = ex.Giveanintuitiveexplana- tion of this fact. 4.9 A life annuity on (x) provides for annual benefit payments for life beginning at age x. The initial benefit payment is 1000 and each subsequent payment increases by 6% (i.e.,thebenefitpaymentatage x+1is1060,thepaymentatage x+2is1000(1.06) 2, and so on). The first 10 benefit payments areguaranteedand will be made regardless of whether (x) is alive or not. The interest rate is a constant 6%. You are given that 10px = 0.9 andex+10 = 10. Find the present value of this contract. 4.10 Alifeannuitycontracton(80)hasapresentvalueof3.14.Theannuitybenefitsatboth age 80 and 81 are 1. The interest rate in the first year is 25% andp80 = 0.8. Suppose thatthevalueof p81 increasesby10%whileallothermortalityratesremainunchanged. What is the new present value of the contract? 4.11 Given that ̈a40 = 15, ̈a[40]+25 = 10, v(25) = 0.5, 25p40 = 0.4, find the net annual pre- mium, payable for 25 years beginning at age 40, for a deferred annuity, paying 1000 yearly for life, with the first benefit payment at age 65. 4.12 Suppose that for all x,q x+1 ≥ qx. Show thaẗax+1 ≤ ̈ax. Does this remain true if we remove the monotone condition onqx? Spreadsheet exercises The following exercises are to be done using the sample life table of Section 3.7. 4.13 A deferred life annuity on (40) provides for income of 10 000 per year beginning at age 65. Nothing is paid if death occurs before age 65. This is purchased by annual premiums payable for 25 years beginning at age 40. Premiums increase each year by 10%oftheinitialpremium,thatis, 𝜋 k = (1+0.1k)𝜋0.Interestratesare5%forthefirst 10 years and 6% thereafter. Find the initial premium𝜋0. 4.14 Agroupofindividualsage30eachagreetoinvest1000peryearfor40yearsbeginning at age 30. At time 40, the fund is divided up among all the survivors. If interest rates are 5% for the next 10 years, 6% for the following 10 years and 8% after that, how much does each survivor receive? 4.15 A single-premium life annuity on (x) provides 1 unit per year for life, beginning at time 1, with an-year guaranteed period, wheren is the smallest integer greater than or equal to the premium. (This is known as aninstalmentrefund annuity. It guarantees thatatleastthefullamountofthepremium,withoutinterest,willbereturned.)Assume a constant interest rate of 4%. Findn and the single premium if (a)x = 40, (b)x = 70. (Note that there is no direct method of calculation. A trial and error procedure is called for). 4.16 Interest rates are 8% for the first 25 years and 6% thereafter. Comparëa [40]+20 and ̈a[50]+10. Which one of these is smaller? Answer this before any calculation, and then do the actual calculation to verify your answer. 5 Life insurance 5.1 Introduction A life insurance policy is a contract between theinsurer and another party known as the policyholder. In return for a payment of premiums, the insurer will pay a predetermined amount of money, known as adeathbenefit, upon the death of the policyholder. The amount ofthebenefitcanvarywiththetimeofdeath.Inpractice,thismoneywillbepaidimmediately upon death (or more realistically, a short time after, to allow for processing the claim) but for mathematicalconvenienceweassumeinthischapterthatitwillbepaidat theendoftheyear of death. For example, if the policyholder purchases a policy on January 1 and dies a week later,ourassumptionmeansthatthedeathbenefitwillnotbepaiduntilDecember31.Wewill consider the more realistic situation of payment at the moment of death in Chapter 8. The reader should distinguish carefully between life annuity and life insurance contracts. The life annuity provides a sequence of periodic benefit payments. The typical life insurance contract provides only a single benefit payment, paid on the occasion of death. 5.2 Calculating life insurance premiums Considerapolicyon(x ).Let bk betheamountthatwillbepaidattime k+1fordeathbetween timekand timek+1. We will refer to the vector b = (b0,b1,…,b𝜔−x−1) as thedeathbenefitvector. Notation The reader is cautioned that many authors use the subscript onbto refer to the timeofpayment. Whatwecall bk,theywouldcall bk+1,sinceitispaidattheendoftheyear, which is timek+1. We prefer the convention above. All our vectors are then indexed from 0 FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 62 LIFE INSURANCE to 𝜔−x−1. In particular, this facilitates matters when dealing with contracts that combine annuity and insurance benefits. Suppose we have fixed an investment discount functionv and a life table. We want to calculate thenetsinglepremium fortheabove policy, whichwewilldenoteby Ax(b).(The A is a standard symbol that probably came from the word ‘assurance’, an older version of the word ‘insurance’.) The principle to determine this is the same as that used for annuities. The total premiums, together with all investment earnings, must be sufficient to provide all the death benefits. To derive the formulas, we will follow the annuity procedure and begin with the case where b = e k. This is a policy in which 1 is paid at timek+1 providing (x) dies between the ages ofx+k and x+k+1. All of the other death benefits are of zero amount. Suppose that we have𝓁x people agexwho each purchase this same contract. Out of these, there will be dx+k individuals who die between the ages ofx+k and x+k+1, and each of them will receive1attime k+1.Thetotalpresentvalueofallthesedeathbenefitswillbe v(k+1)dx+k. This must be equal to the total amount collected in premiums, which will be𝓁xAx(ek). Therefore Ax(ek) = v(k+1)dx+k 𝓁x . The general policy can be viewed as a sequence of 1-year policies as above, one for each value ofk, where thekth policy paysbk at timek+1 for death in the previous year. The premiumforsucha1-yearpolicyisjust bkv(k+1)dx+k∕𝓁x andthetotalpremiumisobtained by summation. We have Ax(b) = 𝜔−x−1∑ k=0 bkv(k+1)dx+k 𝓁x = 𝜔−x−1∑ k=0 bkv(k+1)(kpx − k+1px) (5.1)‡ = 𝜔−x−1∑ k=0 bkv(k+1)kpxqx+k. The reader should note that the expression above follows the same pattern as the annuity singlepremium.Itisapatternthatwewillencountermanymoretimesinsubsequentmaterial. Namely, we sum up a number of terms, each of which consists ofthreefactors, amount × interest discount factor× probability that payment is made. (5.2) In the case of insurance, formulas [3.5(a)–(c)] give three different expressions for the probabilitythatthepaymentwillbemade,givingusthethreedifferentwaysofwriting A x(b). Each will be useful in certain cases. Aswithannuities,thenotationwillsuppressthedeathbenefitvector1 ∞ = 1𝜔−x.Accord- ingly,Ax will be the net single premium for a policy paying 1 at the end of the year of death, whenever it occurs. CALCULATING LIFE INSURANCE PREMIUMS 63 Example 5.1 Suppose thatq60 = 0.2, q61 = 0.4.q62 = 0.5 andi= 100%. A policy sold to (60) provides for benefits at the end of the year of death of 80 for death in the first year, 75 for death in the second year, and 100 for death in the third year. If the insured lives to age 63, the policy terminates and nothing is paid. Find the net single premium. Benefits 80 75 100 Time 3210 Pattern 0.7521 vs 1/21 /21 /2 qs 0.20 .40 .5 Figure 5.1 Example 5.1 Solution. WeagaincanmakeuseofatimediagramasshowninFigure5.1.Thedeathbenefit amount is put at thebeginningof the year to which it is applicable. That is,bk is inserted above timek. Our convention here is to enclose these amounts in a box, to distinguish them from annuity benefits. The amountbk must be multiplied by the previous interest discount factors,aswellasbytheinterestdiscountfactorfortheyearstartingattime k,sinceitispaid at the end of the year. Moreover it is multiplied by the previouspvalues and as well byqx+k, asindicatedinthethirdformulain(5.1),whichtogether givetheprobabilityoflivingtotime kand then dying in the following year. In this case, the net single premium is ( 80× 1 2 ×0.2 ) + ( 75× 1 4 ×0.8×0.4 ) + ( 100× 1 8 ×0.8×0.6×0.5 ) = 17 As with life annuities, one can as an alternative construct a partial life table, which automatically gives you the multiplication of the probabilities, and then use the first formula in (5.1). So for example in this case, starting with𝓁 60 = 1000, we have in turn,d60 = 200, 𝓁61 = 800, d61 = 320, 𝓁62 = 480, d62 = 240. Normally, the policyholder does not pay for insurance by a single premium but rather by asequenceofperiodicpremiums.Weassumeinthischapterthatpremiumsarepaidannually. The premiums will be given by a premium vector𝜋 = (𝜋 0,𝜋1,…,𝜋𝜔−x−1) as defined for deferred annuities in Section 4.5. Following the principle used there, we want the net single premium to be equal to the present value of the premiums, with respect to the interest and survivorship accumulation function. This means that A x(b) = ̈ax(𝜋) = 𝜋0̈ax(𝜌), so that 𝜋0 = Ax(b) ̈ax(𝜌), (5.3) ‡ where𝜌 is the premium pattern vector. 64 LIFE INSURANCE We follow the terminology used for single premiums and call these premiumsnetannual premiumsor annual benefit premiums. We will later encounter examples of premiums that are different from net premiums, but our convention is that unless otherwise mentioned, all premiums will be net. Example5.2 SupposethattheinsuranceinExample5.1istobepurchasedbythreeannual premiums, beginning at age 60, where the second premium is double the first, and the third premium is three quarters of the first. Find the premiums. Solution. The premium pattern vector is given by𝜌 = (1,2,0 .75). We have ̈a 60(𝜌) = 1+ ( 2× 1 2 ×0.8 ) + ( 0.75× 1 4 ×0.8×0.6 ) = 1.89. so that the initial premium will be 17/1.89= 8.99. The insured will then pay 8.99 in the first year, 17.99 in the second year and 6.75 in the third year. 5.3 Types of life insurance Lifeinsurancepoliciesaretraditionallyclassifiedintodifferenttypes. Terminsuranceprovides death benefits for a fixed number of years (similar to the temporary annuity). After the expiration of the term, coverage ceases and there are no more benefits. Example 5.1 involved such a policy, with a term of 3 years.Wholelife insurance provides death benefits for life, so some payout on the policy is certain to occur. In our mathematical model we will not need to distinguish between the two types. We will assume that all policies will continue to age𝜔. Forterminsurancerunningfor nyears(whichwerefertoas n-yearterm)wesimplywillhave b k = 0f o rk≥ n. Nonetheless, in the next chapter we will see that there are differences inthe natureofthesetwotypesandthatwholelifeinsurancehasa savingscomponentaswellasan insurancecomponent. Another common type of contract isendowmentinsurance, which we will discuss in detail in the next section. A more modern development, known asuniversal lifewill be described in Chapter 13. 5.4 Combined insurance–annuity benefits It is possible to combine both life insurance benefits and life annuity benefits in the same contract. One of the most popular types of such a policy is known asendowmentinsurance. This provides for a payment at some future timenif (x) is then alive, and in addition a death benefit if (x) dies before timen. Such a policy is known asn-year endowment insurance or endowment insuranceat agex+n, since it combines apure endowment(as defined in Section4.2withinsurance.Itisusuallymarketedasasavingsplan,wherebythepolicyholder, by paying premiums each year, will accumulate a certain sum at timen. In addition, the policyholder is protected with life insurance if he/she dies before accumulating the desired amount. It should be noted that a whole life policy is similar to endowment insurance since the policyholderisguaranteedsomedeathbenefit.Mathematicallyitcanbeviewedasendowment COMBINED INSURANCE–ANNUITY BENEFITS 65 insurance at age𝜔, and in practice it actually is interpreted in this way. Since it is to the advantage of the policyholder, the insurer will assume that everybody dies by age𝜔, unlike the case of a life annuity, and pay the death benefit at age𝜔 to all survivors. To calculate the single premium for endowment insurances, we simply view it as two separate contracts and add the premiums. Example 5.3 An insurance policy on (x) provides 1 unit payable at timen if (x) is then alive,plus1unitpayableattheendoftheyearofdeathif( x)diesbeforetime n.Levelannual premiums ofPare payable fornyears. Find a formula forP. Solution. The single premium for the death benefit isA x(1n). The single premium for the pure endowment isyx(n). The premium pattern vector𝜌is (1n). The total single premium for the contract is thenAx(1n)+yx(n) and P= Ax(1n)+yx(n) ̈ax(1n) . Example 5.3 is a very common type of endowment insurance with a level death benefit equal to the pure endowment amount, and level premiums payable for the full term. This is not essential, however, and many other combinations are possible. Example 5.4 Consider a 20-year endowment insurance with the death benefit equal to 1 unit for the first 10 years, and 2 units for the second 10 years. The amount of the pure endowment is 3 units. Level annual premiums ofPare payable for 15 years. Find a formula for P. Solution. Calculating as in the previous example, P= A x(110,2 10)+3yx(20) ̈ax(115) . Example5.5 (Figure5.2).Supposethat q60 = 0.2,q61 = 0.4and i= 100%.A2-yearpolicy provides for benefits at the end of the year of death of 80 for death in the first year, and 75 for death in the second year. In addition, there is a pure endowment of 70 paid at age 62 if thepolicyholderisthenalive.Thisispurchasedbytwo-levelannualpremiums.Calculatethe premium. Benefits 80 70 75 Time 210 Pattern 11 vs 1/21 /2 qs 0.20 .4 Figure 5.2 Example 5.5 66 LIFE INSURANCE Solution. The present value of death benefits is A60(80,75) = ( 80× 1 2 ×0.2 ) + ( 75× 1 4 ×0.8×0.4 ) = 14. The present value of the pure endowment is ̈a60(0,0,70) = 70× 1 4 ×0.8×0.6 = 8.4. The present value for the premium pattern vector is ̈a60(𝜌) = ̈a60(1,1) = 1+ 1 2 ×0.8 = 1.4 By (5.3) 𝜋0 = 𝜋1 = 14+8.4 1.4 = 16. Another common type of combined policy is a deferred annuity that provides for death benefits during the deferred period. This is somewhat similar in nature to endowment insur- ance. The difference is that the accumulated savings are paid out as an annuity rather than as a single payment. Example 5.6 (Figure 5.3). A contract on (40) provides for annuity benefits of 1 per year for life, beginning at age 65. If (40) dies before age 65, a death benefit of 10 will be paid at the end of the year of death. Level annual premiums ofP are payable for 25 years. Find a formula forP. Benefits 10 10 10 ... 1 1 10 ... Time 210 ... 26 25 24 ... Pattern 111 ... 1 Figure 5.3 Example 5.6 Solution. Wecalculatethis,asintheexamplesabove,byaddingthesinglepremiumsforthe death benefits and the annuity benefits and dividing by the annuity for the premium pattern vector. The result is P= 10A 40(125)+ ̈a40(025,1∞) ̈a40(125) . The second term in the numerator can be written alternatively asv(25)25p40̈a[40]+25. An interesting variation on the above involves a popular marketing device, which is to issue policies with a return-of-premium feature. COMBINED INSURANCE–ANNUITY BENEFITS 67 Example 5.7 (Figure 5.4). A contract on (40) provides for annuity payments of 1 per year for life beginning at age 65. Should (40) die before age 65, there will be a return of all premiums (without interest) paid prior to death. Level annual premiums ofPare paid for 25 years. Find a formula forP. Benefits P 2P 3P ... 25P 1 1 ... Time 210 ... 26 25 24 ... Pattern 111 ... 1 Figure 5.4 Example 5.7 Solution. Thisisbasicallythesametypeofproblemasabove,exceptthatthebenefitsdepend on the unknownP. We cannot obtainP directly, but we can set up an equation to solve for it. Suppose (x) dies between timek and k+1, wherek is between 0 and 24 inclusive. The insuredwillhavepaid k+1premiumsofP andthesewillbereturnedattime k+1.Thedeath benefit vector is then b = (P,2P,3P,…,25P) = Pj, wherej = (1,2, …,25). Equating present values of the premiums and benefits, P̈a40(125) = PA40(j)+ ̈a40(025,1∞), from which P= ̈a40(025,1∞) ̈a40(125)−A40(j). (5.4) Example 5.8 Consider the same policy as above, except that the premiums are returned withinterest as determined by the functionv. Find a formula forP. Solution. One method is to simply use (5.4), where the vectorj is changed fromjk = k+1 to jk = Valk+1(1k+1;v). Another approach, which produces the solution in a much easier form for calculation is found through the following reasoning. In the first 25 years the accumulation is according to the discount functionvrather thany 25. To see this clearly, consider the room-box description of Chapter 4. Upon death of a participant before age 65, the person’s box is opened up, but the amount paid back as a death benefit is exactly the same as the amount that was there in the first place. The other participants will receive no survivorship earnings. Therefore, each 68 LIFE INSURANCE survivor will have accumulated at age 65 the amountPVal25(125;v). This amount must be sufficient to provide an annuity of 1 unit per life beginning at age 65. Thus P= ̈a[40]+25 Val25(125;v). We will present an alternative solution to this problem in Section 6.7. A generalization of the above is concerned with partial refunds. Example 5.9 Now consider Example 5.8, except that the payment upon death isk times the accumulated premiums with interest, where 0≤ k≤ 1. For each premiumP, the portionkP accumulates at interest and (1−k)P accumulates with interest and survivorship. Equating values at time 25, kP Val25(125,v)+(1−k)PVal25(125;y40) = ̈a[40]+25 and we can solve to get P= ̈a[40]+25 kVal25(125;v)+(1−k)V a l25(125;y40). For k= 1 this reduces to the solution to Example 5.8 and fork= 0, it is a special case of Example 4.5. Remark For those preferring an alternative to the square bracket notation, there are var- ious other ways in which to write the numerator of the last two examples. For example, Val 25(251∞;y40)or ̈a40(025,1∞)∕y40(25).Ofcoursewithconstantinterest,thesimplestchoice is jusẗa65. The following variation is suggested by provisions in some pension plans where an employee must remain in the plan for a minimum period in order to get credit for premiums paid by the employer. A more realistic version will be given in Section 13.3. Example 5.10 Redo Example 5.8 only with the provision that premiums are returned with interest upon death only if (40) lives to age 45. Nothing is returned if death occurs in the first 5 years. Solution. Theredoesnotappeartobeanyconvenientwaytoapplythesimplifyingapproach of the last two examples. We can however use (5.4) with the vectorj given by j k = {0i f 0 ≤ k< 5, Valk+1(1k+1;v)i f 5 ≤ k< 25. Indeed, (5.4) is a flexible formula which can be adapted to a variety of premium return provisions,includingcaseswheretheinterestrateonthepremiumsmaydifferfromthatused in the discount function. INSURANCES VIEWED AS ANNUITIES 69 We now present an example involving a more complex calculation. Example 5.11 An annuity on (x), purchased by a single premiumS, provides for 1 per year for life, beginning at time 1. If (x) dies before a total income ofShas been paid out, the difference betweenSand the total income received will be refunded at the end of the year of death, so that the policyholder will at least receive total income equal to the single premium. (e.g., ifSis 20 and (x) dies between time 5 and time 6, a death benefit of 15 would be paid at time 6.) Describe a procedure for calculatingS. Solution. The premiumSmust satisfy S= ̈a x(0,1∞)+Ax(b), where bk =m a x{S−k,0}. Sincebdependson S,thereisnodirectwaytosolvethisandaniterativenumericalprocedure must be employed. One guesses at an initial value ofS (it will be close to but greater than ̈ax(0,1∞)), and then continues to adjust the value ofS until the right hand side above also equals this value. This can be carried out automatically in Excel with thegoal-seekfunction. Exercise 5.20 gives some particular examples. ThecontractinExample5.9isusuallytermeda cashrefundannuity anditisavariationon theinstalmentrefundideaintroducedinExercise4.15.Inpractice,bothofthesearesomewhat more complicated than described since the refund is based on the gross premium rather than the net. 5.5 Insurances viewed as annuities A comparison of (4.1) and the third expression in (5.1) shows that Ax(b) = ̈ax(c), where ck = v(k,k+1)bkqx+k. (5.5)‡ Let wx denote the vector with entries (wx)k = v(k,k+1)qx+k. We can then write (5.5) in compact form as Ax(b) = ̈ax(wx ∗ b). (5.6) We can verify this intuitively as follows. Suppose (x) wishes to pay a single premium for a life insurance policy running over several years, based on the death benefit vectorb. The insurer refuses, claiming that it only sells life insurance policies for 1 year at a time. It does, however, sell life annuities. The person therefore purchases a life annuity with benefits of v(k,k+1)b kqx+k at timek if she is alive. She does not keep this annuity payment, but immediately returns it tothe insurer topurchase a 1-year lifeinsurance policy paying adeath benefit at timek+1 if death occurs in the next year. This single premium will purchase a death benefit of exactlyb k. This follows from our previous discussion, but to re-emphasize the point, note that this premium will accumulate tobkdx+k∕𝓁x+k at the end of the year, and assuminglx people engage in this scheme, this amount collected from each of𝓁x+k survivors at timekwill be enough to providebk to each of thedx+k people who die during the year. 70 LIFE INSURANCE Thisillustrationisofcoursefanciful,asallinsurerssellpoliciesforperiodsofmorethan1 year.(Moreover,asapracticalmatteritoverlooksthefactthat1-yearpremiumscouldchange over time, as well as the fact that annuity and insurance premiums are based on different tables.) It does, however, provide a useful point of view. Each policyholder can look upon a life insurance policy as a life annuity, providing what is essentially that person’s share of the deathbenefits.Onecouldsupposethattheseannuitypaymentsarethencollectedinaseparate fundandusedtopayallthebenefitstothosewhodieduringtheyear.Thisisoftenavaluable way of looking at the situation, since all policies can be thought of as life annuities if we wish. The calculation of premiums and other quantities can then be reduced to the general principles outlined in Chapter 2. Thisviewpointallowsustoimmediatelyadaptallresultsobtainedforlifeannuitiestothe insurancesetting.Forexample,the splittingidentity forlifeannuities(4.8)takesthefollowing form for insurances: A x(c) = Ax(kc)+yx(k)A[x]+k(c◦ k), (5.7) where the bracket onxindicates, as with annuities, that the discount function used isv◦ k. We can also maintain the room-box visualization for life insurance policies, which we introduced for annuities in the previous chapter, and which will provide a useful guide in the next chapter. The reader is cautioned that while (5.6) is useful conceptually and also for spreadsheet calculation (as illustrated in Section 5.9), it is not always the best for hand calculation. For suchpurposes,theprocedureusedinExample5.1isusuallylesssubjecttoarithmeticalerrors. 5.6 Summary of formulas Inthissectionwesummarizetheprocedure,developedinthelasttwochapters,forcalculating anannualpremiumonagenerallifeinsurance–annuitycontract.Wefirstidentifyfourpertinent vectors: b, the death benefit vector;c, the life annuity benefit vector;u = (0 g,w), the vector of payments that are guaranteed provided (x) lives to agex+g; and𝜌, the premium pattern vector. In many cases only one or two of the first three vectors will be applicable, and the others will be set equal to the zero vector. We then calculate the initial premium from the equation 𝜋 0̈ax(𝜌) = Ax(b)+ ̈ax(c)+v(g)gpẍa(w;v). This allows for cases where the vectors on the right hand side can themselves depend on𝜋0. 5.7 A general insurance–annuity identity 5.7.1 The general identity There is another useful relationship between insurances and annuities. For the vectorwx introduced in Section 5.5, we can write (wx)k = v(k,k+1)−v(k,k+1)px+k = 1−v(k,k+1)px+k −dk, A GENERAL INSURANCE–ANNUITY IDENTITY 71 and it follows that wx =∇ 1∞ −d in the notation of (2.14) with respect to the discount functionyx. From (5.6) Ax(b) = ̈ax(wx ∗ b) = ̈ax(∇1∞∗ b)− ̈ax(d ∗ b) and from (2.15) we obtain our main identity Ax(b) = ̈ax(Δb)− ̈ax(d ∗ b). (5.8) 5.7.2 The endowment identity We will use (5.8) to derive a well-known actuarial formula, which we call theendowment identity. Assume constant interest. The constant discount ratedthen factors out as a constant multiple and ̈ax(d ∗ b) = d̈ax(b). Supposeb = (1n). ThenΔb = (1,0,0,… ,−1) where the −1 is in position indexed withn(i.e., in the (n+1)th entry since we start with 0). Equation (5.8) says that Ax(1n) = 1−yx(n)−d̈ax(1n). Let Ax:n be the net single premium for ann-year, 1-unit endowment insurance. That is, 1 is paid either at timen, or at the end of the year of death if that occurs before timen. (This is the standard symbol for such a premium.) Addingyx(n) to both sides of this equation, we get the endowment identity, Ax:n = 1−d̈ax(1n). (5.9) This identity is analogous to (2.16) and its derivation as given at the end of Section 2.9. For an interpretation, suppose I lend you 1 unit now, to be repaid in full at the end ofnyears, orattheendoftheyearofyourdeathifthisoccursbefore nyears.Youmustalsopayinterest at the beginning of each year until the loan is paid. The present value of the loan must be equal to the present value of the principal repayments, plus the present value of the interest. The latter is just a temporary life annuity payingdunits yearly fornyears, beginning at time zero. The present value of the loan is just 1 and the present value of the principal repayments is justA x:n, the net single premium for the endowment insurance. We obtain the equation 1 = Ax:n +d̈ax(1n), which gives (5.9). If Pdenotes the level annual premium payable fornyears for the 1-unit,n-year endow- ment insurance, we can also expressP in term of annuities by dividing bÿa(1n) in (5.9) to obtain P= 1 ̈ax(1n) −d. (5.10) This identity can be interpreted as follows. Suppose you invest 1 unit. This will provide you with interest earnings ofd at the beginning of each year, until such time as you wish to 72 LIFE INSURANCE terminate the investment and take back your principal of 1. An alternate scheme is to use the 1 unit to purchase ann-year life annuity, paying 1∕̈a x(1n) at the beginning of each year for nyears, and to use part of these proceeds to purchase a 1-unit,n-year endowment insurance, carrying level annual premiums ofP for n years. The insurance will pay you back your principalattheendof nyears,oratdeathifearlier.Beforerecoveringyourprincipalyouwill havenetannualearningsof1 ∕̈ax(1n)−P,andthismustequaltheincomeof dthatyouwould get from the first alternative. This yields the given identity. Remark This back-to-back annuity–insurance combination has recently become popular as an investment vehicle. At first glance, it appears that in practice it will produce a lower return than the straight investment, since one must pay expenses on both policies and, in addition, the different lifetables used for annuity and insurances will work to the purchaser’s disadvantage.(Oursimplifiedmodelassumesnoexpensesandthatthelifetableisthesamein all cases.) However, the fact that the proceeds of the insurance and annuity contracts receive favourable tax treatment in many jurisdictions often means that the net after-tax return can actually be higher than that of the straight investment. The above formulas are of course true for a 1-unit whole life insurance, which is just endowment insurance at age𝜔.W eh a v e A x = 1−d̈ax, Px = 1 ̈ax −d, (5.11) where Px is the net level annual premium payable for life for a 1-unit whole life contract on (x). 5.8 Standard notation and terminology 5.8.1 Single-premium notation The standard symbol for an insurance single premium isA, as we have given it. As with annuities, thisisembellished withsuperscriptsand subscriptstohandle thecommon types of death benefit vectors. For example: r Ax denotes Ax(1∞) as we have already indicated r A1 x: n denotes Ax(1n), the net single premium for a 1-unit,n-year term insurance. The superscript 1 above thexsignifies that (x) must die before the expiration of then-year period in order to collect. r Ax:n denotes Ax(1n)+yx(n), the net single premium for a 1-unitn-year endowment insurance, as we have indicated above. The subscriptx: n signifies that the death benefit is paid upon the first ‘failure’ of (x)o rt h en-year period. The life (x) fails upon death and then-year period fails at the end ofnyears. Recall that the same subscript is used in the temporary annuity standard symbol to signify that benefits are paid as long as both the life (x) and then-year period are ‘surviving’. STANDARD NOTATION AND TERMINOLOGY 73 r Ax: 1 n isanothersymbolfor nEx (denotedby yx(n)inournotation).Thisisthenetsingle premium for a 1-unit,n-year pure endowment. The superscript 1 is now is above then to signify that then-year period must fail before (x) does in order for the contract to pay. r k|nAx denotes Ax(0k,1n). This isdeferred insurance. A level death benefit of 1 unit begins afterk years and continues forn years. Such a policy would not normally be sold by itself but may be combined with other policies. r k|Ax stands for k|𝜔−xAx in accordance with the usual practice of omitting duration symbols when the contract continues for life. r DA1 x: n denotes Ax(n,n−1,…,1).The Dstands for decreasing. r IA1 x: n denotes Ax(1,2, …,n). TheIstands for increasing. r DAx and IAx are respectively the above two symbols withn= 𝜔−x, in keeping with the general notational principle discussed in Section 4.7. 5.8.2 Annual-premium notation The standard symbol for a level net annual premium isP. This is followed by the single- premiumsymbol.Forthosesinglepremiumsthatbeginwithacapital A,the Aisomitted.The premium payment durationtappears before thePon the lower left. If omitted, it means that premiums are paid for the natural duration of the contract. Examples follow: r Px is the annual premium, payable for life, for a 1-unit whole life policy on (x). r tPx is the annual premium payable fortyears for a 1-unit whole life policy on (x). r P1 x: n is the annual premium payable fornyears for a 1-unit,n-year term policy on (x). r tPx:n istheannualpremiumpayablefor tyearsfora1-unit n-yearendowmentinsurance on (x). r P(IA1 x: n ) is the annual premium payable forn years for ann-year increasing term policy on (x). (In this case we insert the full single-premium symbol since it does not begin with anA.) r P(n|̈ax)isthelevelannualpremiumpayablefor nyears,foradeferredannuityproviding income of 1 unit per year beginning at agex+n. Note here that the missing premium paymentdurationsymbolistakenas n,thenaturalpremiumpaymentduration.Although the contract continues for life, it is not natural to continue paying premiums when the annuity payments begin. Alloftheseannualpremiumsymbolsareevaluatedbytakingthecorrespondingnetsingle premium and dividing bÿa x:t, wheretis the premium paying duration. For example, 10P50:20 = A50:20 ̈a50:10 . 74 LIFE INSURANCE 5.8.3 Identities Thesametypes ofidentitythatwediscussedinSection4.7arisewithinsurances aswell.For example, A x:n = A1 x: k +v(k)kpxA{x}+k: n−k. The derivation will be left to the reader. 5.9 Spreadsheet applications We modify the Chapter 4 Spreadsheet to handle death benefits. The death benefit vector is entered into Column G. The vectorw x is calculated in column L by putting the following formula in L10 and copying down = G10 ∗ (1+B10)̂(−1) ∗ C10 In G8 we then put the same formula as in F8 except with L replacing F, and this will return A x(b). The entry in I6 is changed to=(F8+G8)/H8. The premium vector appears in Column I. Foracheck,theformulainH8canbecopiedtoI8andthiscellshouldreturnatotalofF8 and G8. For a sample problem, compute the initial premium for a 1000-unit, 30-year endowment policyon(40)withpremiumspayablefor20years,andthepremiumtodoubleafter10years. The interest rate is a constant 6%. The answer is 12.68. Exercises Type A exercises 5.1 Given that 𝓁70 = 1000, 𝓁71 = 960, 𝓁72 = 912, and that interest rates are a constant 10%, calculateA70(12). 5.2 A 3-year endowment insurance policy on (60) provides for benefits paid at the end of the year of death of: 500 if death occurs in the first year (i.e., between time 0 and time 1); 800 if death occurs in the second year; and 1000 if death occurs in the third year. In addition, there is a pure endowment of 1000 payable at age 63 if (60) is then alive. Thisispurchasedbythreeannualpremiumsbeginningatage60.Thesecondpremium is double the initial premium and the third premium is three times the initial premium. You are given that thatq 60 = 0.1,q61 = 0.2 andq62 = 0.25. The interest rate is 25% for the first 2 years and 20% after that. Find the initial premium. 5.3 A 2-year term insurance policy on (60) provides for a death benefit of 100 payable at theendoftheyearofdeath.Thisispurchasedbyasinglepremium.If(60)livestoage EXERCISES 75 62, the single premium is returned without interest. Given thatq60 = 0.1,q61 = 0.15, and interest is a constant 10%, find the single premium. 5.4 Youaregiven qx = 0.053, qx+1 = 0.054, qx+2 = 0.055, i0 = 0.06,i1 = 0.08,i2 = 0.10. Ifb = (1,2,3), find the vectorc = b ∗ wx. Type B exercises 5.5 A deferred life annuity on (40) provides for a yearly income of 1000 beginning at age 65 and continuing for life. It is to be purchased by a single premium ofS payable at age 40. If death occurs during the deferred period (i.e., during the first 25 years), then the single premium is refunded without interest at the end of the year of death. (a) Give a formula forSusing the symbols̈aandA. (b) You are now given the following information. The same annuity contract without thepremium-refundfeature(i.e.,nothingispaidduringthedeferredperiod)canbe purchased for a single premium of 2000. In addition, a contract that provides the same annuity benefits, plus a level death benefit of 2000, payable at the end of the year of death, for death during the deferred period, can be purchased for a single premium of 2200. Calculate an exact numerical value forS. 5.6 Acertainelectricalapplianceissoldwitha5-yearguarantee.Thisprovidesthatthefull purchase price is refunded if the product fails within 2 years, and half of the purchase price is refunded if the product fails in the following 3 years. A study shows that out ofatypicalbatchof100items,therewillbe2failuresinthefirstyear,3failuresinthe second year and 4 failures per year after that. Assuming that interest is a constant 5% andthatreimbursementismadeattheendoftheyearoffailure,whatisthecostofthis guarantee to the manufacturer, as a percentage of the purchase price? 5.7 What is A x(1n) if the constant interest ratei= 0? Give both a formal derivation, and a proof by general reasoning. 5.8 Suppose that interest is constant andqy is a constantqfor ally. Find an expression for Ax in terms ofqand y. 5.9 An actuary calculates a single premium for a certain life insurance policy on (40), and then discovers there were two errors made. In the first place, the life table used showedavalueof q 40 thatwasonlyone-halfofwhatthecorrectfigurewas.Second,the first-year death benefit was taken as 20 when it should have been 10. Will the correct premium be the same as, lower, or higher than the one calculated? 5.10 For a certain insurance contract, on (50), the death benefit for the first year of the contract is 1100, payable at the end of the year of death. The single premium for the whole contract is 600. This is based on an interest rate of 10% for the first year and a mortality table withq 50 = 0.20. If the value ofq50 is changed to 0.25, while all other value ofqx are unchanged, what is the new single premium? 5.11 There is a constant interest rate of 20%, and A50 = 0.300, v10 10p50 = 0.10, A61 = 0.400, q60 = 0.20. 76 LIFE INSURANCE Suppose thatq60 is changed to 0.23, while all other values ofqx remain unchanged. What is the new value ofA50? 5.12 The cash flow vectorj = (1,2,3,… ,10). You are given that interest is constant and that ̈ax(j) = 30, Ax(110) = 0.10, ̈ax(110) = 7, v10 10px = 0.48. FindAx(j). 5.13 Consider a whole life policy on (x) with a level death benefit of 1. Suppose that, for someage y> x,thevalueof qy isincreasedwhileallothervaluesof qremainthesame. (a) Show thatAx is increased. (b) Show thatPx is increased, wherePx is the level annual premium payable for life, for this policy. (c) Showbyexamplethattheabovestatementsarenotnecessarilytrueforawholelife policy with a non-constant death benefit. 5.14 (a) Suppose that, for allx, qx+1 ≥ qx. Show thatAx+1 ≥ Ax. (b) Does the above remain true if we remove the monotone condition onqx? 5.15 Show that for the vectorcgiven by formula (5.5), andk= 0,1,2 …,w eh a v e (a) ̈ax(kc) = Ax(kb); (b) ̈a[x]+k(c◦k) = A[x]+k(b◦k). 5.16 A deferred annuity on (40) provides for an income of 1000 per year for life beginning atage60.If(40)livestoage60,thefirst10annuitypaymentsareguaranteedregardless of whether (40) is alive or not. Level annual premiums ofPare payable for 10 years, beginning at age 40. If (40) dies before the annuity begins, all premiums paid prior to death are returned at the end of the year of death, and no annuity payments are made. Find a formula forP, assuming: (a) premiums are returned without interest; (b) premiums are returned with interest. 5.17 One could generalize our definition of an insurance contract by stipulating that for death at timet, the death benefit is paid at time𝜏(t) which is some function oft. (e.g., whenbenefitsarepayableattheendoftheyearofdeath, 𝜏(t) = [t]+1,where[ ⋅]isthe greatest integer function.) Show that a pure endowment contract can be considered as an insurance, in this sense. 5.18 An annuity on (x) provides 1 per year for life beginning at time 0, with the further provision that, n additional payments will be made after (x) dies, beginning at the end of the year of death. Interest is constant. Show that the net single premium is ̈a(1 n;v)+vn̈ax.Derivethisformulaintwoways.(a)Byusing(5.11).(b)Byusing(5.2). 5.19 Redo Example 5.8 only now assuming that for death during the first 5 years, one-half of all premiums paid prior to death are returned at the end of the year of death. EXERCISES 77 5.20 A contract on (70) provides for a payment of 1000 at the end of 3 years if (70) is then alive, and is to be paid for by three-level annual premiums. For death in the first year, nothing is paid. For death in the 2nd or 3rd years, all premiums paid are returned with interestattheendoftheyearofdeath.Youaregiventhat q 70 = 0.1,q71 = 0.2,q73 = 0.3 andi=25%. Find the annual premium. 5.21 Provide an algebraic proof that the two methods given for the solution of Example 5.8 produce the same answer. (Hint: Change the order of summation in a double sum.) 5.22 Provide an algebraic proof and give an intuitive explanation of the following identity. For any vectorc = (c0,c1,…,cn) ̈ax(c) = v(n)npxValn(c;v)+Ax(j) wherejk = Valk+1(kc;v), k= 0,1, …,n−1 andkcis as defined before Equation 2.18. Spreadsheet exercises The following exercises are to be done using the sample life table of Section 3.7. 5.23 A contract on (40) provides for death benefits if death occurs in the next 25 years. The amount of the death benefit is 50 000 for the first 10 years and 100 000 for the next 15 years. If (40) lives to age 65, he/she will receive a life annuity of 10 000 per year for life beginning at age 65. Premiums are payable for 15 years beginning at age 40. The premium doubles after 5 years. Interest rates are 5% for the first 20 years and 6% thereafter. Find the initial premium. 5.24 Interestratesareaconstant6%.Acontracton(50)providesforapaymentof10000at age 70 if then alive. Level premiums ofPare paid only at even-numbered times, that is, at age 50, 52, 54,…. If (50) dies before age 70 there is a return at the end of the year of death of all premiums paid prior to death. FindP. 5.25 Assume a constant interest rate of 4%. Find the premium for the annuity in Example 5.9 if: (a)x= 40 ;(b)x= 70. 6 Insurance and annuity reserves 6.1 Introduction to reserves Givenaninsuranceorannuitycontractandaduration k,thereserve attime kisdefinedexactly as in Definition 2.7. It is the amount that the insurer needs at timek in order to ensure that obligationsunderthecontractcanbemet.Calculatingreservesforeachpolicyisanimportant responsibilityoftheactuary,knownas valuation.Theinsurerwantstobeconfidentthatfunds on hand, together with future premiums and investment earnings, are sufficient to pay the promised future benefits. It is important to thoroughly master the concept of insurance and annuity reserves in order to properly understand and analyze the nature of these contracts. Throughout this chapter we will deal with the following model. As usual we start with a fixed investment discount functionvand a life table. We have a contract issued on (x) with death benefit vectorb, annuity benefit vectorc and premium vector𝜋. (For simplicity we will omit the possibility of guaranteed payments in our discussion, but this feature can easily be incorporated if desired.) Recall from Section 5.5 that we can view the death benefits as a vector of annuity benefitsb ∗ w x, where (wx)k = v(k,k+1)qx+k. We can then form thenet cashflow vector f = 𝝅 −b ∗ wx −c, (6.1) which indeed represents the net cash flow on the contract from the insurer’s viewpoint. The insurerwillcollectpremiumsof 𝜋andpayoutdeathbenefitsintheformof b ∗ w x andannuity benefits ofc. The reserve at timekon the contract is the reserve for the vectorf with respect to the interest and survivorship discount function, as given in Definition 2.6. Denoting the reserve by kV,w eh a v e kV = kV(f;yx) =− Valk (kf;yx ). (6.2)‡ FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial INTRODUCTION TO RESERVES 79 Itisoftenusefultowritethisinanalternatewaythatkeepsbenefitsandpremiumsseparate. Thatis,wenotethatthereserveattime kisthe valueattimekoffuturebenefitslessthevalue attimekoffuturepremiums .Thiscanbewrittenintermsofthestandard Aand ̈asymbolsas kV = 1 yx(k) [Ax(kb)+ ̈ax(kc)− ̈ax(k𝝅)] (6.3) or, equivalently (and the way in which it usually appears in the literature) as kV = A[x]+k(b◦ k)+ ̈a[x]+k(c◦ k)− ̈a[x]+k(𝝅 ◦ k). (6.4) Here we are using (4.8) and the corresponding statement forA. Under the assumption that premiums are actuarially equivalent to benefits, we can also calculate the reserve retrospectively as Bk(f;yx) = 1 yx(k) [̈ax(k𝝅)− ̈ax(kc)−Ax(kb)]. (6.5) Underthisformulationthereserveisthevalueattime kofthepastpremiumslessthevalueat time kof past benefits. The reader is cautioned that the reserve cannot be so calculated when premiums and benefits are not actuarially equivalent. Example 6.1 For the policy of Example 5.5, find the reserves at time 1 and time 2. See Figure 6.1. As in Chapter 2 we use an arrow to mark the point at which values are computed. Benefits 80 70 75 Time 210 Premiums 1616 v s 1/21 /2 qs 0.20 .4 Figure 6.1 Example 6.1 Solution. Value at time 1 of future death benefits=75(1/2)(0.4) =15. Value at time 1 of future annuity benefits=70(1/2)(0.6) =21. Value at time 1 of future premiums=16. 1V = 15+21−16 = 20. Theaboveapproachisrecommendedforhandcalculation.Allreservecalculationscanbe handledinexactlythesameway,althoughtherewillgenerallybemorethanonesummandin each of the three items. 80 INSURANCE AND ANNUITY RESERVES Additional information can be obtained by first computing the net cash flow vector, which also provides the best procedure for spreadsheet calculation. To illustrate, the vector w 60 = (0.1,0.2, −). (Wedon’t have informationtocompute the thirdentrybut itisirrelevant, since it is multiplied by 0.) Then,b ∗ w60 = (8,15,0), c = (0,0,70), 𝝅 = (16,16,0), so that the net cash flow vectorf is (8,1, −70). For present purposes, we can forget about the particular death benefits and premiums. From the insurer’s viewpoint the contract can be viewedassimplyoneofcollecting8attime0,1attime1,andpayingback70attime2.From (6.2), 1V is just the negative of the value at time 1, with respect to interest and survivorship, of the payments at times 1 and 2. This equals−1 +70(0.5)(0.6)=20, as above. Similarly,2V is just the value at time 2 of the payment at time 2 which is 70. In general, foracontractrunningfor nyears,thereserveattime nisjustthepaymentdueattime ntothe survivors. (Recall that from the convention introduced in Chapter 1, reserves are calculated before this payment.) Forn-year term insurance, when there is nothing payable to survivors at the end, thenth year reserve will equal to 0. Since we have used net premiums, we can calculate balances as a check. B 1(f) = 8 y60(1) = 8 0.4 = 20, B2(f) = 1 y60(2)̈a60(2f) = 1 0.12[8+0.4] = 70, which agree with our previous calculations. The following example exhibits a point of interest. Example 6.2 For a 4-year endowment insurance on (60),b2 = 100,b3 = 200 and there is a pure endowment of 200 paid at age 64 if the insured is then alive. You are given that 𝜋 2 = 10,𝜋3 = 20, q62 = 0.1. The interest rate after 2 years is 25%. Find 2V and 3V. See Figure 6.2. Benefits 100 200 200 Time 43210 Premiums 2010 v s 0.80 .8 qs 0.1? Figure 6.2 Example 6.2 Solution. Value at time 2 of death benefits= 100(0.8)(0.1)+200(0.64)(0.9)Value at time 2 of premiums= 10+20(0.8)(0.9) = 24.40. THE GENERAL PATTERN OF RESERVES 81 We don’t knowq63, but it is not needed. Sinceq63 +p63 = 1, the two unspecified terms sum to 200(0.64)(0.9)=115.20 and 2V = 80+115.20−24.40 = 170.80. Similarly,3V = 200(0.8)−20 = 140. Thisexampleshowsthatforanendowmentinsurance(withbenefitspaidattheendofthe yearofdeath)wherethepureendowmentisthesameamountasthefinaldeathbenefit,wedo not need toknow themortalityrate forthe final year. This isin fact evident, sincethe insured gets that amount whether they live or not. Remark There is a subtle point involved with Equation (6.4), which is important to note fortheactualcalulationofreserves.WehavealreadyintroducedtheideainSection2.10.3.In ourmodelwechooseaninvestmentdiscountfunctionandamortalitytable,andtheseremain fixedforthedurationofacontract.Inpractice,whenonereachestime kandactuallywantsto compute kVtheseassumptionsmaywellhavechanged.Thereservecomputedattime kwillbe A′ x+k (b◦ k)+ ̈a′ x+k (c◦ k)− ̈a′ x+k (𝜋◦ k) wheretheprimesindicatequantitiesthatarecalculatedwiththenewinterestandsurvivorship functiony ′ x+k ascomputedunderwhatcouldbechangedconditionsattime k.Ofcoursewhen there is no change in assumptions,y′ x+k will just equalyx◦ k and formula (4.6) shows that bothformulasarethesame.Thepointisthen,thatinpractice,onedoesnotreallyknowwhat kV will be before timek. In this more realistic setting, we can view the original formula (6.4) as the best estimate one could make ofkV, if asked to compute it at time 0. 6.2 The general pattern of reserves Are insurance reserves generally positive or negative? Paradoxically, we will motivate the answer to this question by providing an example where they are neither. Example 6.3 An insurance policy provides a death benefit of 1 paid at the end of the year of death. Level annual premiums are payable for life. The interest rate is constant and the value ofq x is a constantqfor allx. Find the reserves and give an explanation for the answer. Solution. Let p= 1−qbe the constant value ofpx. Thenkpx = pk, which is never 0. There is a positive probability of living to any age, so we have an example in which𝜔 =∞ .O u r vectors will be of infinite length. Let vdenote the constant value ofv(k,k+1). Thenwx is a vector with a constant entry of vqand the premium pattern vector𝜌 is a vector with a constant entry of 1. The net level premium 𝜋 is clearlyvq, since that will make the vectorswx ∗ b and 𝜋 not only actuarially equivalent but actually equal to each other. The net cash flow vectorf is then not just a zero-value vector butactuallyequalto the zero vector. All reserves will be 0. Whatishappeninghereisthatthepremiumof vq,collectedeachyear,willaccumulateto qat the end of the year, and this will be exactly sufficient to pay the death benefits due at that time. There will be nothing left over, so balances, and therefore reserves, equal zero. 82 INSURANCE AND ANNUITY RESERVES Thesituationaboveistypicalformanyformsofinsurance,suchasautomobileorproperty coverage, and insurers of such risks have little in the way of reserves. The given scenario is, however, not realistic for life insurance. The values ofq x are not constant but increase with age. If one paid for the insurance 1 year at a time, the yearly premium per unit ofvqx would rapidly increase and eventually become prohibitively high. As we noted in Chapter 5, the typical life insurance policies will level the premiums out. In most cases, policyholders are paying more in premiums than they need to in the early years, but not enough in the later years. At any point of time after time zero, future premiums will not be sufficient to cover the remaining benefits. The excess collected in the early years is used to cover this deficit. It is expected therefore that reserves are usually positive. This has important implications for the life insurance industry. It means that investing becomes a major activity, as life insurance companiestendtoaccumulatelargeamountsofassets.Somecritics,withlittleunderstanding of insurance, look at these holdings of real estate, stocks and bonds, and claim that they represent unfair profits made at the expense of the policyholder. The truth is, however, that a large portion of these assets represent reserves, which in effect belong to the policyholders, as they will be used to pay the future benefits. Negativereservescanariseonpolicieswherethecostoftheinsurancebenefitsisdecreas- ingeachyear.Anexampleisapolicywitharapidlydecreasingbenefitamount,wheredespite theincreasein q x,thequantity bkv(k,k+1)qx+k decreases.Someexamplesappearintheexer- cises. Similarly, negative reserves can arise in the case where the premiums increase rapidly rather than remaining level. Insurers try to avoid such a situation if possible. A negative reservemeans,viewingthingsprospectively,thatthepolicyholderowesmoneytotheinsurer, whichwillbeprovidedbyfuturepremiums,orequivalently,lookingatthingsretrospectively, that the policyholder has received coverage but not yet paid for it. The problem is that the policyholder may stop paying premiums on the policy, leaving an unpaid debt. 6.3 Recursion In this section, we develop some important recursion formulas. It is convenient to make a slightalterationinnotation.Wewillincorporatetheannuitypaymentswiththepremiumsand let 𝝅 denote 𝝅 −c. In other words, we think of annuity benefits as just negative premiums, which is really what they are, since the policyholder is receiving rather than paying these amounts. Our net cash flow vectorf has entriesf k = 𝜋k −bkv(k,k+1)qx+k, and from our basic recursion formula (2.26), k+1V = ( kV+𝜋k −bkv(k,k+1)qx+k )yx(k+1,k). (6.6) Since yx(k+1,k) = (1+ik)∕px+k this is sometime written as k+1V = (kV+𝜋k)1+ik px+k −bk qx+k px+k . (6.7) Therecursionisstartedwiththeinitialvalue 0V =− ̈ax(f)whichwillbe0underourstandard assumption of net premiums. DETAILED ANALYSIS OF AN INSURANCE OR ANNUITY CONTRACT 83 It is instructive to note that the last term ofqx+k∕px+k is equal todx+k∕𝓁x+k+1. From this, weseeeasilythatitistheamount,perunitofdeathbenefit,thateachsurvivormustpayatthe endoftheyear to provide the benefits paid to those who died during the year. Formula (6.7) takes a retrospective viewpoint, and says that the reserve at timek+1i s obtainedfromthatattime k,byaddingthepremium,accumulatingatinterestandsurvivorship for 1 year, and then subtracting enough to pay the death benefits. It is known as theFackler reserveaccumulationformula,namedafteroneoftheearlyNorthAmericanactuaries,David Parks Fackler. In pre-computer days it was a popular method for calculating reserves. It is used infrequently for calculating purposes now, but is still useful for illustrating how the life insurance reserve changes from one period to the next. Remark Thequantity kV+𝜋k intheaboveformulaisoftencalledthe initialreserve attime kas it represents the reserve at the beginning of the year, after the payment of the premium. In contrast,kV is sometimes referred to as theterminal reserveat timek, reflecting the fact thatitthereserveattheendoftheyear,priortothepaymentofthepremiumforthefollowing year. Alternateversionsofthisformulaprovideinstructiveinformation.Wefirstgiveanimpor- tant definition. Definition 6.1 The quantityb k − k+1V is known asthenetamountatrisk for the (k+1)th yearandwillbedenotedby 𝜂k (thesubscriptischosentocorrespondto bk).(Therearevarious other names in the literature, such asdeathstrainatrisk.) Now, multiplying (6.7) bypx+k = 1−qx+k and rearranging, we get k+1V = (kV+𝜋k)(1+ik)−qx+k𝜂k. (6.8) Formula (6.8) reflects the fact that we can also view the accumulation of funds on an insurance policy as an interest-only investment, rather than as an interest and survivorship investment.Fromthispointofview,thepolicyholder keepsthereservewhenhe/shedies(the amount accumulated in her box), but then the insurer only needs to make up the difference as a death benefit. The insurer is therefore at risk only for the difference between the death benefit and reserve, which is the source of the name. Readers who looked at Section 2.11 will note that this is a special case of the change of discount function that we investigated there. In this casey x and bk are replaced byvand 𝜂k, respectively. 6.4 Detailed analysis of an insurance or annuity contract In this section we use (6.8) to provide a detailed discussion of the workings of an insurance policy. 6.4.1 Gains and losses In practice, interest and mortality rates will not conform exactly to those provided by our model. In a particular year, the insurer may earn more interest than predicted by the given 84 INSURANCE AND ANNUITY RESERVES discount function, which will result in gains. For an insurance contract, there may be fewer deaths than predicted by the given life table, also causing gains. If the insurer earns less interestthanexpected,and/ortherearemoredeathsthanexpected,therewillbelosses.Inany year,theactuarywantstoanalyzethesegainsorlossesandseehowmuchisduetoinvestment earnings and how much is due to mortality. Suppose we wish to measure the gain for a particular policyholder over the period run- ning from timek to timek+1. At the beginning of the year, before premium payment the policyholder’s box will contain the amount kV. At the end of the year the insurer must make certain that the box hask+1V in order that future obligations can be met. Anything in excess of that amount can be considered as a gain, taken out and added to general surplus funds. On the other hand, if there is less than k+1V, the insurer will have to make up the deficit from generalsurplusfundsandtherewillbealoss.Wewillderivesomegeneralformulas.Suppose the actual interest rate earned during this year wasi ∗ k rather thanik, and the actual rate of mortality wasq∗ x+k rather thanqx+k. Then, the actual amount accumulated at timek+1 will be the right hand side of (6.8) with starrediand q. If we subtract the reserve, we obtain the total gainGk from that policy for that year as Gk = (kV+𝜋k)(1+i∗ k ) −q∗ x+k 𝜂k − k+1V. (6.9) If we substitute fork+1V with the left hand side of (6.8) we can write this as Gk = (qx+k −q∗ x+k )𝜂k +(i∗ k −ik )(kV+𝜋k). (6.10) which gives us a decomposition of the gain by source. The first term gives the gain due to mortality, and the second term gives the gain due to interest. That is, the mortality gain is the difference between the expected and actual mortality rates times thenetamountatrisk.T h e interest gain is the difference between actual and expected interest rates times the amount of funds at the beginning of the year, after payment of the premium. Example 6.4 Refer back to the policy of Example 6.1. Suppose that in the first year of the policy, the interest earned was 50% instead of the predicted 100%, and the actual rate of mortality was 0.1 instead of the predicted 0.2. Find the total gain for the year, split into the portion due to interest and the portion due to mortality. Solution. Substituting directly from (6.9), the mortality gain is (0.2−0.1)(80−20) = 6, while the interest gain is (0.5−1)(0+16) =− 8. There is a total gain of−2, or in other words a loss of 2, for each policy. We will verify this by working out an example in the aggregate. Suppose that 10 people age60buythispolicyatacertaintime.Theinsurercollects160,andthisaccumulatesat50% DETAILED ANALYSIS OF AN INSURANCE OR ANNUITY CONTRACT 85 interest to 240 at the end of the year. Out of this, the insurer will pay a death benefit of 80 for the one death that occurred, leaving a total of 160. They have to put aside a total of 180, which is the reserve of 20 for each of the 9 survivors. There is a aggregate shortfall of 20, or 2 from each policy, which has to be drawn from surplus. (Note that the gain given by (6.8) is for each policy in force at thebeginningof the year, not at the end.) Formula (6.9) assumes that the premium paid is the net premium, calculated as in Chap- ter 5. In practice, the premiums actually charged on a policy will normally be different from the valuation premiumswhich are net premiums determined from the interest and mortality assumptions used to compute reserves. (See Section 6.5 for more detail on this.) This neces- sitatesanadjustmentinouranalysis.Supposethepremiumpaidattime kisactually 𝜋 ∗ k rather than valuation premium𝜋k. The first term on the right of (6.8) is then ( kV+𝜋∗ k )(1+i∗ k ) = (kV+𝜋k)(1+i∗ k ) +(𝜋∗ k −𝜋k )(1+i∗ k ), which leads to an extra source of gain or loss. We now have Gk = (qx+k −q∗ x+h )𝜂k +(i∗ k −ik )( kV+𝜋k ) +(𝜋∗ k −𝜋k )(1+i∗ k ). (6.11) Thethirdtermrepresentsthegainorlossarisingfrompremiumsthatdifferfromthevaluation premium. (We will elaborate on this point in Section 12.4.) One application of the formulas in this section is todividend calculation. Insurers fre- quentlyissuewhataretermed participatingpolicies,inwhichgainsresultingfromfavourable investmentandmortalityexperiencearereturnedtothepolicyholderintheformofdividends. We will not go into further detail on this topic, but note that formula (6.10) is a basic tool in computing these amounts. Anotheruseofgainandlossanalysisistodeterminehowchangesinthebasisassumptions will affect the reserves. As an example, suppose that the reserve interest rate decreases. It is clear that valuation premiums will increase. In order to provide the same benefits with decreased earnings from investments, more must be collected from the policyholder. On the otherhand,itisnotclearhowthiswillaffectreserves,sinceatanytime,boththepresentvalue of the future benefits and the present value of the future premiums will increase. Indeed, the effectwilldependonthenatureofthepolicy.Intheusualcasehoweverwithalevelpremium 𝜋 andreservesthatincreasewithtime,theloweringoftheinterestratewillincreasereserves. To see this, suppose that interest rates are constant and there is a change of rate fromi to i ∗ < i. This will cause an interest loss of (kV+𝜋)(i−i∗) for the year running from timekto k+1.Sincethereservesareincreasingwithtime,thelosseswillalsobeincreasingwithtime. Therewillbeofcourseanewlevel valuation premiumof 𝜋 +Δ tocover theselosses.Dueto theincreasinglosses,itmustbethat Δisgreaterthanthelossesinearlyyearsandlessinlater years.Intheearlyyears,thatportionof Δnotusedtocovertheloss,willbesetasideasanextra reserve,tocoverthegreaterlossestocomeinthefuture,andthiswillcausereservestoincrease. 6.4.2 The risk–savings decomposition Wenowlookatausefuldecompositionofthepolicyintoariskportionandasavingsportion. Multiply Equation (6.8) byv(k,k+1) and rearrange to obtain 𝜋 k = v(k,k+1)qx+k𝜂k +[v(k,k+1)k+1V−k V]. 86 INSURANCE AND ANNUITY RESERVES This formula decomposes each premium into two parts. The first term is known as therisk portion of the premium, as it is that amount needed to buy insurance for 1 year for the net amount at risk. The remainder provides for the difference in reserves and is known as the savingsportion of the premium. Example 6.5 Find the decomposition for the premiums of Example 6.1. Solution. In the first year the net amount at risk is 80−20 = 60. The risk portion of the premium is then 1 2 ×0.2×60 = 6. The savings portion is1 2 ×20−0 = 10. (As a check, the two portions add up to the total premium.) In the second year the net amount at risk is 75−70 = 5. The risk portion of the premium is 1 2 ×0.4×5 = 1 and the savings portion is 1 2 ×70−20 = 15. This decomposition shows that any policy can be viewed as being composed of two separate policies, the pure insurance part and the savings part. For the pure insurance part, the premium is the risk premium and the death benefit paid is the net amount at risk. This part of the policy has zero reserves, as the risk premium is just sufficient to purchase coverage for the net amount at risk for 1 year. In the above example, thepolicyholderpays6inthefirstyear,whichisexactlyenoughtopurchasethecoveragefor the net amount at risk of 60. He/she then pays 1 in the second year, which is exactly enough to purchase coverage for the net amount at risk of 5. Thesavingspartofthepolicyoperatesjustlikeabankaccount,withamountsaccumulating at interest only. In the example above, the savings portion of 10 from the first premium accumulatesto40attime2andthesavingportionof15fromthesecondpremiumaccumulates to30attime2.Thetotalsavingsof70arethenpaidoutasthepureendowmenttoallsurvivors at time 2. This is typical of endowment policies, including whole life, which operates as an endowmentatage 𝜔,aswehaveindicated.Theaccumulatedamountsfromthesavingsportion of the premium increase steadily to the pure endowment amount. It is instructive to compare this with term insurance. Example 6.6 Redo Example 6.1 for the corresponding 2-year term policy without the endowment. That is,b still equals (80, 75, 0), butc = 0. Solution. In this case the premium𝜋 will equal (8+15(0.4))∕1.4 = 10, and we havef = (2,−5,0).Then 1V = 5andweknowfromthediscussionfollowingExample6.1that 2V = 0. Inthefirstyear,thenetamountatriskis75,sotheriskportionofthepremiumis 1 2 ×0.2×75 = 7.5, and the savings portion is 2.5. In the second year the net amount at risk is 75, so the risk portion of the premium is 1 2 ×0.4×75 = 15 and the savings portion is−5. This shows the typical savings pattern on term policies. Modest savings are built up in early years, but these must be drawn on in later years when the premium is insufficient to pay for the insurance, resulting in a negative savings portion. As a check, the 2.5 deposited into the savings fund at time 0 will increase to 5 at time 1, which will then be withdrawn to make up for the deficit in the premium payable at that time. Exactlythesameanalysisasaboveholdsforlifeannuityaswellaslifeinsurancecontracts. In this case, the death benefits are all of zero amount, so that the net amount at risk will be negative.Thisisnaturalenoughandreflectsthefactsthatextradeathsinthecaseofannuities BASES FOR RESERVES 87 result in gains. On single-premium annuities, we have𝜋k =− ck < 0, fork> 0. Examples appear in the exercises. 6.5 Bases for reserves Choosingtheappropriatediscountfunctionandlifetableforthepurposeofreservecalculation is another complex subject that we will only discuss briefly. As mentioned, the insurer must ensure that it has sufficient assets to cover their reserves, for if not, it will be in danger of being unable to meet future obligations. It is usually thought that reserve calculations should use conservative assumptions, so there is a built-in safety margin should experience prove to be adverse. In other words, the discount function would use somewhat lower interest rates thanactuallyexpectedandthemortalitytablewouldshowmoredeathsthanactuallyexpected (orfewerinthecaseofannuitycontracts).Inseveraljurisdictions,thebasesusedforreserves are specified by insurance regulatory bodies, whose main goal is to ensure protection for the policyholders. The following example illustrates the resulting effect on profitability. Example 6.7 For the policy in Example 6.1, the company is required by legislation to compute reserves using a 50% interest rate rather than 100%. It actually does achieve the estimated100%returnonitsinvestments,andmortalityfollowsthepredictedratesexactly.It stillchargesthepremiumof16basedontherealisticinterestrateof100%.Analyzetheeffect of the conservative interest rate assumption on the company’s gains and losses. Solution. Redoing the calculations with an interest rate of 50% in place of 100% leads to a valuation premium of 544/23 and a first year reserve of 560/23, as the reader can verify. We willdoanaggregateanalysis.Supposetheinsurersells2300policiesatage60.Itwillcollect totalpremiumsof36800,whichwillaccumulatewithinterestto73600attheendoftheyear. Outofthisitwillpay460peopleadeathbenefitof80units,foratotaldeathbenefitpayment of 36 800. This leaves 36 800. It now must set up a total reserve of 44 800, consisting of 560/23foreachofthe1840survivors.Thereforethelossshownforthefirstyearofthepolicy is 8000. Looking at formula 6.11, the large loss in the third term, more than offsets the gain from the second term. In the second year, it starts with a reserve of 44 800. It collects another 16-unit premium from each of the 1840 survivors for a total of 29 440. This leaves a total of 74 240, which accumulates with interest to 148 480. Out of this it must pay a death benefit of 75 to each of the736 = 1840×0.4peopleremaining1104peoplewhosurvived.Thetotalbenefitpayments are 132 480, which leaves a gain for the year of 16 000. For this group of policies, the insurer will show a loss of 8000 for the first year, and a gain of 16 000 for the second year. If the insurer had used the realistic interest rate of 100% interest, there would have been no gains or losses. In effect the more conservative reserve requires the insurer to borrow 8000 from surplus at the end of the first year, and then repay it with the 16 000 at the end of the second year, which is what the amount should be in view of the 100% interest rate. This example shows that for our present model, with a single discount function, reserve assumptions do not affect the ultimate profitability of the insurer, since this depends solely on what actually happens. It can, however, change the incidence of this profit from year to 88 INSURANCE AND ANNUITY RESERVES year. This means that reserve assumptions can have an effect on profitabiltly when there are different interest rates involved, and we discuss this further in Section 12.4.3. 6.6 Nonforfeiture values Whatshouldhappentopolicyholderswhostoppayingpremiumsbeforethetermstatedinthe contract?Thisisknowntechnicallyas withdrawalorlapse,orsurrender.Inconsiderationfor the premiums they have already paid, they should be entitled to some reduced benefits under the policy. These are known asnonforfeiture benefitssince they are benefits that were not forfeitedbythecessationofpremiums.Infact,inoursimplifiedmodel,theyshouldbeentitled to take the reserve on their policy at any time they wish. Looking at this retrospectively, the reserveistheexcessoftheaccumulatedamountoftheirpremiumsovertheaccumulatedcost oftheinsuranceprotectionthattheyhavereceived.Itwillbetheamountthattheyhaveintheir ‘box’. In practice, insurers pay an amount that is somewhat less than the reserve for several reasons. This is a complex topic that we will only comment on briefly here. One reason is the high incidence of expenses in the early years of the policy (we discuss this more fully in Chapter 12). While these are accounted for by adding an amount to the premiums, the total amount of the initial expenses may not be recovered at the time of withdrawal. Another factor is the phenomenon known asanti-selection. This is a well-established concept in insurance which is simply a recognition of the fact that policyholders will make choices according to their own self-interest, acting on knowledge that they have, but that the insurer may not have. (The prefix ‘anti’ refers to the fact that it is the policyholder doing the selectingagainsttheinsurer.)Onalifeinsurancecontract,theoptiontowithdrawislesslikely to be exercised by an unhealthy policyholder than a healthy one. After all, if someone is told theywilldiewithinafewmonthsfromaterminaldisease,theywouldbefoolishtogiveupthe policy. Consequently, the group that doesnotwithdraw can be expected to experience higher mortalitythannormal.Thereisthereforeananti-selectionexpensetowithdrawal,intheform of these higher mortality rates of the remaining policyholders. The principle followed here is that this expense should be borne equitably by all the policyholders, not just by those who remain. This is done by adjusting the amounts paid out in the case of withdrawal. The cash amount that will be paid to the withdrawing policyholder on a life insurance contract is known as thecashsurrendervalue and is usually guaranteed at the time of issue for all durations. Normally, the policyholder is given the option of taking the nonforfeiture benefitsintheformofareducedlevelofinsuranceratherthanincash.Thereductioncantake the form of either a reduced amount of benefits, or a reduced term for the same benefits. Lifeannuities alsopresent anobvious possibilityforanti-selection. Unhealthy annuitants would find it worthwhile to end the contract, take their reserve, leaving a group who could be expected to live longer than expected, and mortality losses would result. For this reason, nonforfeiturebenefitswouldnotbeofferedonsingle-premiumlifeannuitycontracts,oncethe benefitscommence.Dependingonthecontract,theymightbepresentduringadeferredperiod. 6.7 Policies involving a return of the reserve At the time of death, the policyholder (or, more accurately, the estate of the policyholder) receives the death benefit. Had he/she decided to lapse the policy an instant before death, he/she would have received an amount close to the reserve. Some people, who do not have POLICIES INVOLVING A RETURN OF THE RESERVE 89 a complete understanding of life insurance, have raised the complaint that the company is confiscating the person’s reserve, since it is only returning the death benefit, and not the reserve, at death. The answer to this is that one must adopt one of two points of view. One can view the insurance policy as an interest and survivorship investment. In that case, it is true that the reserve is taken and spread among the survivors, but this is completely fair, as discussed in Chapter 4. Alternatively, one can view the policy as an interest-only investment. In this case the reserve is indeed available at death. But now, one must view the death benefit as the net amount as risk, rather than the originally stated amount. Itispossibletodesignapolicywhereafixedamount,plusthereserve,ispaidatdeath.The policyholder might then be told that the reserve is being paid in addition to the death benefit. This is of course just playing with words. What you really have is a policy with the pattern of death benefits worked out so that when you subtract the reserve you get some prescribed amount. Consider the following example. Example 6.8 A policy on (x) provides for a payment, at the end of the year of death, of 1plusthereserve,shoulddeathoccurwithin nyears.Levelannualpremiumsof Parepayable for nyears. Find a formula forP. Solution. A direct solution of this problem by the method outlined in Chapter 5 will cause difficulties, sinceP depends on the reserves, but the reserves in turn depend onP. While it maybepossibletosolvetheresultingequationsforsmallvaluesof n,itismuchbettertotake the interest-only view for the accumulation of money. That is, we use the discount function vin place ofy x and the net amount at risk𝜂k in place ofbk. Expressing the insurance as an annuity, we wish to solve P̈a(1n;v) = ̈a(𝜂 ∗ wx;v) where 𝜂 is the vector (𝜂0, 𝜂1,…,𝜂n−1). While this could be done on any policy, it would normally be completely impractical, since we would not know the net amounts at risk in advance. In this case, however, it works perfectly. We are given thatb k = 1+k+1V, so that 𝜂k = 1 for allk. This leads to P̈a(1n;v) = n−1∑ k=0 v(k+1)qx+k, and we easily solve forP. For another application of this idea, we can give a more formal solution to Example 5.8. We calculate the balance at time 25 and equate it to the reserve at time 25, which we know is ̈a {40}+25. To calculate the balance, we use the discount functionvand death benefits of𝜂k. In this case, the actual death benefit is just equal to the reserve, so that𝜂k = 0 for allk.T h e balance isjusttheaccumulated value of thepremiums atinterest, whichisPv(25,0) ̈a(125;v), as we had before. Notethatreservecalculationbyrecursionisquitesimpleinthistypeofpolicy.Onesimply uses (6.8) where now𝜂k is just the face amount of the policy. 90 INSURANCE AND ANNUITY RESERVES 6.8 Premium difference and paid-up formulas Consider any policy on (x) with death benefit vectorb and with level annual premiums of P payable forh years, so that the premium vector𝜋 = P(1h). In this case, there are some other formulas that are useful for providing additional insight into the nature of balances and reserves. Throughout this sectionPis arbitrary and not necessarily the net premium. 6.8.1 Premium difference formulas Fixaduration k< h.LetP s bethelevelpremiumthat shouldbechargedforapolicywiththe same remaining benefits, if issued at timek to a person agex+k. That is, the value at time k of these new premiums should equal the value at timekof the death benefits after timek. Equating values at timek, A[x]+k(b◦ k) = Ps̈a[x]+k(1h−k). (6.12) SincekV = A[x]+k(b◦ k)−P̈a[x]+k(1h−k), we substitute in (6.10) to get kV = (Ps −P)̈a[x]+k(1h−k). (6.13) Formula(6.11)isknownasthe premiumdifferenceformula forreserves.Thequantity Ps −P is the difference between what should be charged and what is actually charged after timek. The value at timek of this yearly deficit over the remaining premium payment period gives the reserve. We can also obtain a retrospective premium difference formula. LetPc be the premium thatcouldhave been charged to provide the benefits that were provided up to timek. That is, Ax(kb) = Pc̈ax(1k). (6.14) Since Bk = [P̈ax(1k)−Ax(kb)]∕yx(k), we substitute in (6.14) to get Bk = (P−Pc)̈ax(1k) yx(k) . (6.15) The quantityP−Pc is the difference between what was actually charged and what could have been charged up to timek. The accumulated amount of this yearly excess gives the balance. WhenPis a net premiumBk = kV and (6.15) gives another formula for the reserve. 6.8.2 Paid-up formulas Substituting for the annuity rather the insurance in (6.12) gives kV = ( 1− P Ps ) A{x}+k(b◦ k). (6.16) STANDARD NOTATION AND TERMINOLOGY 91 Similarly, from (6.15), Bk = ( P Pc −1 ) Ax(kb) yx(k) . (6.17) To interpret formula (6.16), note that the fraction of the future benefit that is purchased by the actual premiums isP∕Ps. For example, ifPs = 36 andP= 24 then the policyholder is only paying two-thirds of what they should be for the future benefits. The difference of (1−P∕P s) must be that portion of the future benefits that has already been provided by the excess past premiums, so that multiplying this ratio by the present value of future benefits gives the reserve. In formula (6.17) the ratioP∕P c −1 is that portion of the past benefits that were purchased but not needed. For example, ifP= 24 andPc = 16, then the policyholder has paid 1.5 times what they could have paid to provide those benefits. Multiplying this ratio by the value of these past benefits gives the balance. Formula (6.16) is known as thepaid-upformula. Suppose a policyholder lapses at time k and is given nonforfeiture benefits equal in value to the reserve. If the individual elects to takepaid-upinsuranceforareducedamount,thisformulashowsthattheappropriatefraction is (1−P∕P s). 6.8.3 Level endowment reserves Wecanusethepremiumdifferenceformulatoderiveaverysimpleexpressionforreserveson level endowment insurance where a net level premium is paid for the full period and interest is constant. At durationkof ann-year contract, we know from (5.10) that P= 1 ̈ax(1n) −d, Ps = 1 ̈ax+k(1n−k) −d, and, substituting in (6.13), kV = 1− ̈ax+k(1n−k) ̈ax(1n) , (6.18) reducing the calculation of reserves on such a policy to the calculation of annuity values. 6.9 Standard notation and terminology The standard notation for reserves closely follows the notation for annual premiums as describedinSection5.8.Thebasicsymbolforthereserveattime kis kVaswehaveadopted. This is embellished in exactly the same way as the symbolPwas for annual premiums, with one exception. The premium payment period is moved to the upper left, from the lower left, since the latter is now used for the duration. If the upper left is empty, it signifies again that level premiums are paid for the natural duration of the contract. The following are some examples. 92 INSURANCE AND ANNUITY RESERVES kVx isthereserveattime kfora1-unitwholelifepolicyon( x),withlevelannualpremiums paid for life. The basic prospective formula (6.4) for this policy in standard symbols reads kVx = Ax+k −Pẍax+k. The corresponding retrospective formula is kVx = Pẍsx: k −A1 x: k (1+i)k kpx . To simplify retrospective reserve formulas, a symbol for the last term was introduced. Let tkx = A1 x: t (1+i)t tpx . This is often called theaccumulated cost of insuranceand denotes the single premium that each survivor would pay per unit of death benefit fort years, if this single premium were collected attheend of thek-year period, rather than at the beginning. (This would never be done in practice as it is not feasible to charge people at a time when they have no chance of collecting.) tV1 x: n is the reserve at timet for a 1-unit,n-year term policy on (x) with level annual premiums payable fornyears. Fort≤ n, the prospective formula for this quantity is tV1 x: n = A1 x+t: n−t −P1 x: n ̈ax+t: n−t , while the retrospective formula is tV1 x: n = P1 x: n̈sx: t −t kx. h t Vx: n is the reserve at timetfor a 1-unit,n-year endowment insurance policy on (x), with level annual premiums payable forhyears. Fort≤ n, this is given prospectively as h t Vx: n = { Ax+t: n−t −hPx: n ̈ax+t: h−t if t< h, Ax+t: n−t if t≥ h, or retrospectively as h t Vx: n = {hPx: n ̈sx: t −t kx if t< h, hPx: n ̈sx: h (1+i)t−h t−hpx −t kx if t≥ h. kV(n|̈ax) denotes the reserve at timek for a deferred annuity onx providing 1 unit for life beginning at agex+n and with annual premiums payable forn years. Fork< n,t h e retrospective formula is the easiest and is given by tV(n|̈ax) = P(n|̈ax)̈sx: t . SPREADSHEET APPLICATIONS 93 Fork≥ n, the prospective formula is the easiest and is given by kV(n|̈ax) = ̈ax+k. 6.10 Spreadsheet applications For reserve calculations we add two columns J and K to Chapter 5 spreadsheet. In cell J10, we enter = I10−F10−L10 and copy down to get the net cash flow vector. Then in cell K10 we enter the formula =− SUMPRODUCT(E10 :E$129,J10 :J$129)∕E10, and copy down. This calculates kV in cell K 10+k, as the negative of the value of future net cash flows, divided byyx(k). (Division by 0 error terms will appear for large enough durations, but these occur after age𝜔 and can be ignored If one prefers, the formula in K10 can be suitably modified with an IF statement to replace them with a blank.) As a test calculate15V for the test problem given in Section 5.9. The answer is 333.16. Wenowhaveafinalspreadsheetthatwillcalculatepremiumsandreservesonallinsurance and annuity contracts without guaranteed payments. Here is a complete summary. INPUT FORMULAS Column D: 1 in cell D10, Section 2.14 formula in cell D11. Copy down. Column E: 1 in cell E10, Section 4.8 formula in cell E11. Copy down. Column F: Section 4.8 formula in cell F8. Column G: Section 5.9 formula in cell G8. Column H: Copy cell F8 to cell H8. Column I: Insert the formula=(F8 + G8)/H8 in I6. Copy Cell H8 to I8. Insert the formula =$I$6*H10 in I10 and copy down. Column J: Copy down the formula for cell J10. Column K: Copy down the formula for cell K10. INPUT DATA FOR EACH PARTICULAR PROBLEM Column B: Interest rates. Cell C1: Age at issue. Column F: Annuity benefit vectorc. Column G: Death benefit vectorb. Column H: Premium pattern vector𝜌. Column N: Life table. (For sample table, insert parameters in N3 and N4, Section 4.8 formula in N10. Copy down). 94 INSURANCE AND ANNUITY RESERVES OUTPUT ̈ax(c) in F8. Ax(b)i nG 8 . Present value of all premiums in I8. The premium vector𝜋 in column I. Reserves in column K. Exercises Type A exercises 6.1 A3-yearendowmentinsuranceon(60)providesfordeathbenefitspayableattheendof theyearofdeath.Thedeathbenefitis1000fordeathinthefirstyear,2000fordeathinthe secondyearand3000fordeathinthethirdyear.Inaddition,thereisapureendowment of 4000 paid at time 3 if the insured is then alive. This is purchased by three annual premiums,beginningatage60.Thefirsttwopremiumsareequalandthethirdisdouble the amount of the initial premium. You are givenq 60 = 0.10, q61 = 0.20, q62 = 0.25. The interest rates are 25% for the first 2 years and 100% in the third year. Find the initial premium. Find kV for k= 1,2,3. 6.2 Youaregiven q60 = 0.20,q61 = 0.25.Interestratesare20%inthefirstyear,25%inthe secondyearand50%inthethirdyear.A3-yearendowmentinsurancepolicyissuedto (60)providesfordeathbenefits,attheendoftheyearofdeath,of1000ifdeathoccurs inthefirstyearand2000ifdeathoccurs inthethirdorsecondyears.Inaddition,there is a pure endowment of 2000 paid at age 63 if the insured is then alive. Level annual premiums are payable for 3 years. Find the premium. Find kV,f o rk= 1,2,3. Do this first by the basic prospective reserve formula, and check your answers by using the recursion formula (6.7). 6.3 For a 3-year term insurance policy on (x), level premiums are payable for 3 years. The following data are given. kq x+k ik bk 00 . 1 00 . 2 05 0 0 0 0 10 . 1 50 . 2 52 0 0 0 0 20 . 2 00 . 3 01 5 0 0 0 (a) Find the vectoryx. (b) Find the net cash flow vector. (c) Using your answers to (a) and (b), compute1V and2V. (d) Explain briefly why your answers to (c) are negative. 6.4 A 10-year endowment policy on (40) has death benefits of 900, payable at the end of the year of death, should this occur within 10 years, plus a pure endowment of 900 at age 50 if the insured is then alive. Level annual premiums of 20 are payable for EXERCISES 95 10 years. The interest rate is a constant 50% andq48 = 0.25. You are not givenq49. Find tV for t= 8,9,10. 6.5 An insurance policy has level premiums of 20 payable for the duration of the contract. If the reserve is calculated with a premium of 15, then0V = 100. What is0V if the reserve is calculated with a premium of 16? 6.6 Refer to Exercise 6.2. (a) Decompose each premium into the risk portion and savings portion. (b) Suppose that during the second year of the policy the actual interest rate earned was20%ratherthan25%andtheactualrateofmortalitywas0.20ratherthan0.25. Find the per policy gain during this year from both interest and mortality. 6.7 RefertoExercise6.1.Decomposeeachofthethreepremiumsintotheriskportionand savings portion. 6.8 Foracertaincontractissuedatage60, 5V = 200,thepremiumpayableatage65is40, q65 = 0.20,thedeathbenefitpayableatage66fordeathbetweenage65and66is800, the interest rate for the sixth year of the contract (between age 65 and 66) is 20%. (a) Find 6V. (b) Decompose the premium payable at age 65 into the risk portion and the savings portion. (c) Suppose that during the sixth year, the actual interest rate earned was 25% instead of20%andtheactualrateofmortalityatage65was0.15insteadof0.20.Findthe gain or loss during this year, from interest and from mortality, for each policy in existence at the beginning of the year. Type B exercises 6.9 Refer again to Exercise 6.7. Explain briefly in words why the third premium has a negative risk portion. 6.10 A deferred life annuity on (60) provides for annuity benefits payable for 3 years, beginning at age 63, provided that (60) is alive. The first annuity payment is 1000, the second is 2000 and the third is 3000. Premiums are payable for 3 years, beginning at age 60. The second and third premiums are equal in amount and eachdouble the amount of the initial premium. If (60) dies before age 63, there will be, at the end of the year of death, a return of all premiums paid prior to death, without interest. You are given thatq 60 = 0.1, q61 = 0.2,q62 = 0.25,q63 = 0.3,q64 = 0.4. The interest rates are 25% per year for the first 4 years, and 50% for the fifth year. Find (a) the initial premium, (b) 1V and (c)4V. 6.11 A life insurance policy on (x) has level death benefits of 1000. A life insurance on (y) has exactly the same premiums and reserves as the policy on (x) for the first 10 years, but different death benefits. Suppose thatqy+k = 2qx+k for k= 0,1, …,9.Ifthe commonreserve 8V = 300,whatisthedeathbenefiton(y)’spolicyfortheyearrunning from time 7 to time 8? 96 INSURANCE AND ANNUITY RESERVES 6.12 Use formula (6.8) to derive formula (4.9) withk= 1. (Note that for annuity contracts the death benefits are 0, and the premiums are the negative of the annuity benefits.) 6.13 You are given thaẗa81 = 3.2,q80 = 0.10,i= 0.20. (a) Find ̈a80. (b) A single-premium whole life annuity on (80) provides for a constant benefit of 1 per year. Due to improvements in mortality, the actual mortality rate experienced during the first year of the contract is 0.07 rather than 0.1. On the other hand, the interest earned during that year was 22% rather than 20%. What is the total per contract gain for this year? 6.14 A 2-year term insurance policy on (50), with a constant death benefit of 1000, is purchasedbytwo-levelannualpremiums.Policyholderswhochoosetolapsethepolicy attime1willreceive 1V asacash value. Assumethat mortalityand interestfollowthe projectedpattern,andthatthepremiumchargedisthenetpremium.Suppose,however, that a typical group of people age 51 who have purchased insurance at age 50 can be divided into two groups. Half of them are healthy and half are not. The non-healthy group can expect to have twice as many deaths over the following year as the healthy group. Suppose thatq 51 = 0.03. What is the loss on each remaining policy for the second year, assuming: (a) Allofthehealthypolicyholdersatage51lapsethepolicyatthattime,andnoneof the unhealthy ones do; (b) One-half of the healthy policyholders, and one-quarter of the unhealthy policy- holders, lapse the policy at age 51. 6.15 For a certain whole life policy on (x), given by a death benefit vectorb and premium pattern vector𝜌, reserves are calculated according to two mortality tables, with the 1-yearprobabilitiesofdyingdenotedby q x+k andq∗ x+k ,respectively.Thesamepositive interest rates are used in both calculations, and in each case the premiums are the net premiums as determined by the particular table being used. Suppose the two ‘curves’ cross at one point. That is q x+k ≥ q∗ x+k , k= 0,1, …,n, qx+k ≤ q∗ x+k , k= n,n+1,…,𝜔−x−1. Assumethatdeathbenefitsarenonincreasing.Showthatthereserveattime nishigher for the starred rates (the steeper curve). 6.16 (a) Use (6.8) to derive the formula Ax+1(1n−1)−Ax(1n) = iAx(1n)−qx (1−Ax+1(1n−1)). (b) As an actuary for an insurance company, you receive an angry letter from a pol- icyholder. The person, age 50, has just purchased a single-premium 3-year term insurance policy with a constant death benefit of 1. The complaint is that the per- son’s friend, age 49, purchased a single-premium 4-year term insurance policy, EXERCISES 97 with the same death benefits, for a lower single premium. The letter writer argues that both policies provide coverage for the ages 50–53, but that the friend gets an extra year of coverage, for the year running from age 49 to 50. Therefore, it is claimed,thefriend’spremiumshouldbehigher,notlower.Whatisyourresponse? 6.17 Aspecial2-yearterminsurancepolicyon(70)istobepurchasedbyasinglepremium. Shoulddeathoccurinthefirstyear,theinsuredwillreceive,attheendoftheyear,1000 plusthereserveatthattime.Ifdeathoccursinthesecondyear,theinsuredwillreceive at time 2 only a return of the single premium paid without interest. Supposeq 70 = 0.36, q71 = 0.40, and the interest rate is 100%. Find (a) the single premium, (b)1V. 6.18 A whole life insurance policy on (x) provides for death benefits, paid at the end of the year of death, of 1000 plus the reserve at that time. Level annual premiums ofP are payable for life. Suppose thatqx = 0.2, qx+1 = 0.2 andqy = 0.4f o rally ≥ x+2. The interest rate is a constant 100%. Find (a)P(b)kV, k= 1,2, …. 6.19 Two people, A and B, are both agex. A buys a 20-year endowment policy, with a constant death benefit of 1, and a pure endowment of 1 at time 20, paying level annual premiums for 20 years. B buys a whole life policy with a constant death benefit 1, payinglevelannualpremiumsforlife,andinaddition,eachyear,investsthedifference between his/her premium and A’s premium, in a savings account. At time 20, the reserve on B’s policy, plus the amount in his savings account total 1. All premiums were calculated at a constant interest rate ofi. B earned a constant rate ofj on his investments. Isjgreater than, less than or equal toi?W h y ? 6.20 A whole life policy with a level death benefit of 10 000 carries net annual premium of 100payableforlife.Therateofdiscountisaconstant0.04.Attime n,thepolicyholder wishes to reduce their annual premium payment to 25, and is told they can do so, but the death benefit will be reduced to 7000. Assuming that the entire reserve is available as a nonforfeiture benefit, find nV. 6.21 Supposethatamortalitytableusedtocalculatereservesisalteredbyaddingapositive constant to each value ofqx. Explain how the reserves will be affected, for whole life policies with a constant death benefit and level premiums. Spreadsheet exercise 6.22 SupposethatmortalityfollowsthesamplelifetableofSection3.7andinterestratesare 5% in the first 15 years, 6% for the next 15 years, and 7% thereafter. A contract issued at 40 provides for a life annuity beginning at age 65. The annual annuity payment is 1000 for the first 10 years and 2000 thereafter for life. If death occurs before annuity payments begin, there is a death benefit of 10 000 paid a the end of the year of death. Level annual premiums are payable for 15 years. (a) Find the premium and all reserves. (b) Suppose the interest rate in the first 15 years increases from 5% to 5.5%. Do reserves increase or decrease? (c) If the interest rate in the first 15 years decreases to 4%, what happens to reserves? 7 Fractional durations 7.1 Introduction Up to now, we have considered cash flows where the payments were at integer times. We nowdealwiththecasewherecashflowscanoccuratfractionaldurations.Forexample,ifthe basic time unit was a year, and payments were made monthly, these would be at times that are multiples of 1/12. In practice this is a common occurrence. Purchasers of life annuities oftenwanttheincometobepaidmonthly.Manypurchasersofinsurancepolicieswishtopay premiums monthly, or possibly quarterly or semiannually. One obvious method of handling thisfeatureistochangethetimeunit.Ifwearedealingwithannuitieswithmonthlypayments, we could just take our unit of time as 1 month, and payments would be at integer times. This option was not always feasible in pre-computer days. For cash flows discounted at interest only, calculations were done from tables showing yearly rates of interest, and elaborate formulas were developed to handle the fractional durations. This is no longer necessary, and the change-of-period approach is the modern way to handle the valuation of cash flows at compound interest. For life annuities however it is still common to keep the time unit as a year. There are several reasons for this. In the first place, the year has been so ingrained as a measure of age that it seems hard to break away from this tradition. Another more important reason is that insurers often do not calculate premiums exactly for the different periods, but rely on approximate conversion formulas. For this purpose it is convenient to have some simple relationships established between the yearly annuity values and those with more frequent payments.Stillanotherreasonisthatwealsowanttoconsiderannuitieswithpaymentsmade continuously,whichwillbediscussedinChapter8.Todothis,itisconvenienttofirstdiscuss the case in which payments are madem times per year, and then view continuous annuities as a limiting case asm approaches infinity. We will follow the standard usage of referring to annuities with payments made at times which are multiples of 1∕mas mthly annuities or annuities payablemthly. Fundamentals of Actuarial Mathematics, Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial CASH FLOWS DISCOUNTED WITH INTEREST ONLY 99 We will restrict attention to sequences of cash flows where theperiodic payments during each year are constant,but can vary from year to year. If payments change during the year (e.g., they increase each month), then usually the only feasible approach is to change the period.Thebasicmodelisasfollows.Wearegivenapositiveinteger m,whichisthenumber of payments made each year. We are given a cash flow vector,c = (c 0,c1,…,cN) where, as before,ck isthe total amountpaidinyear k +1,thatis,betweentime k andk +1.Ratherthan being paid as a single cash flow however, there are payments of ck m at timek + j m, j = 0,1, …,m −1. (7.1) Let y denote an arbitrary discount function. We denote the present value of the sequence of cash flows as described bÿa(m)(c;y). The task at hand is to relate this quantity to the case where m = 1, namelÿa(c;y). Our first observation is that it is sufficient to find a formula for ̈a(m)(c;y) in the case thatc = e0, a single payment of 1 at time zero. This follows from the replacementprinciple.Let uk bethevalueattime k ofasequenceofpaymentsof1∕m attime k +(j∕m),j = 0,1, …,m −1. We can write uk = ̈a(m)(e0;y ◦k). The replacement principle tells us that we can replace all the payments between timek and k +1 by a single payment ofckuk at timek. It follows that ̈a(m)(c;y) = ̈a(c ∗ u;y). (7.2) whereu = (u0,u1,…,uk,…). Thekeytoevaluatingfractionalannuitiesistocalculatethequantities uk.Wewilldiscuss particular cases in the following two sections. 7.2 Cash flows discounted with interest only Supposethatwearegivenyearlyinterestrates,andwewanttoconsiderachangeofperiodto 1∕m of a year. We use primed symbols to denote the quantities applicable to the new period, that is,v ′,i′ and d′ in place ofv,i and d, respectively. We will start as usual by supposing that we are given the interest ratesik for each nonnegative integerk. To determine the rates applicabletothenewperiods,thestandardassumptionmadeisthatinterestratesareconstant over each year. To be precise, one postulates is that for all nonnegative integersk, (2.9) holds whenevers,t,s +h,t +h all lie in the interval [k,k +1). Leti ′ k denotetheconstantvalueof v(s +1∕m,s)−1andlet d′ k denotetheconstantvalueof 1−v(s,s +1∕m)f o rk ≤ s < s +1∕m < k +1. An investment of 1 at time 0 will accumulate to 1+i′ k at time 1∕mand then to (1+i′ k )2 at time 2∕m,( 1+i′ k )3 at time 3∕mand so on. Arguing in the same way for the discount rate gives (1+i′ k )m = 1+ik, (7.3)‡ (1−d′ k )m = 1−dk, (7.4)‡ 100 FRACTIONAL DURATIONS since in (7.3) each side of the equation is the value at timek +1 of 1 unit invested at timek, and in (7.4) each side of the equation is the value at timek of 1 unit paid at timek +1. In standard actuarial notation it is common to ‘annualize’ these rates by multiplying by m. That is, one defines i(m) k = mi′ k , d(m) k = md′ k. (7.5) The quantitiesi(m) k and d(m) k are often known asnominal rates, to reflect the fact that they are rates in ‘name only’, and not really applicable to any period. They are sometimes referred to as the yearly interest (respectively discount) ratesconvertible or compounded mthly. The actual rate applicable to eachmthly period in the yeark to k +1 is found by dividing these nominal rates bym. Inthecaseofconstantinterestthesubscript k isnotneededandwesimplyusethesymbols i (m) and d(m) for the nominal rates. Example 7.1 Suppose thati(m) = 0.12. Findi, the yearly rate of interest, if (a)m = 2, (b) m = 12. Solution. (a) We must divide by 2, to deduce that the rate per half-year period is 0.06. From (7.3), (1+i) = 1.062 = 1.1236 so that the corresponding yearly interest rate is 12.36%. (b) Similarly, the corresponding yearly interest rate= (1+0.01)12 −1 = 12.68%. Itisimportanttodistinguishbetweentheactualyearlyratesandnominalratesfordifferent periods, when comparing quoted figures. For example, by custom, the rates on Canadian mortgages are quoted as nominal rates withm = 2, while for US mortgages,m = 12. These provisions would be evident in the actual written contract, but are not always made clear to a prospective borrower. As the above example indicates, the actual annual rates paid on Canadian or US mortgages would be higher than the quoted nominal rates. We now deduce the present value of fractional annuities for the investment discount function. To simplify the notation, we first assume constant interest. A 1-year, 1-unit annuity payable mthly is simply an annuity paying 1∕m units for m periods, discounted with the primed rates. From formulas (2.16) and (7.4) we can write u k = ̈a(m)(e0) = 1 m (1−(1−d′)m d′ ) = d d(m) (7.6) for allk, where the last quantity is taken as 1 ifd = 0. From (7.2), ̈a(m)(c;v) = d d(m) ̈a(c;v). For the general case of non-constant interest we haveuk = dk∕d(m) k , which is taken equal t o1i fd(k) = 0. LIFE ANNUITIES PAIDmTHLY 101 Notethatinthecaseofpositivecashflows(andassumingtheusualcaseofpositiveinterest rates)thepresentvaluewithpaymentspaid mthlyislessthanthecorrespondingyearlypresent value. This can be shown mathematically by noting thatdk is less thand(m) k . For example, if dk = 0.36 then, from(7.4),d′ k = 1−0.641∕2 = 0.20 andd(2) k = 0.40.(The reader is invited to provide a general proof.) This conclusion is also evident from a comparison of payments. In theyearlycase,afullpaymentof c 0 wouldbemadeattime0.Inthe mthlycase,theannuitant receivesc0∕m each period of length 1∕m, and will not have collected the fullc0 amount until one mth of a year before the year end. This is true for each year of the annuity. There is therefore a loss of interest, which is reflected in the lower present value for themthly case. 7.3 Life annuities paidmthly We now consider the case where we have the interest and survivorship discount functionyx. We will usually writëa(m) x (c)f o r̈a(m)(c;yx) as we did whenm = 1. As we are normally given mortality data only for integer ages, we are faced with the necessity of making some assumption in order to extrapolate for the data that we do not have.Manyapproachesarepossible.Inthenextsectionwedescribethemostcommonlyused method. Other possibilities that have been proposed are described in Exercises 7.6, 7.13 and 7.14, as well as later in Chapter 8. 7.3.1 Uniform distribution of deaths This preferred method is known as the assumption of auniform distribution of deaths over each year of age, abbreviated as UDD. It can be viewed as just linear interpolation for𝓁y. Supposeweassumethat 𝓁y,ratherthanbeingdefinedjustforintegervalues,isdefinedforall nonnegativey. The UDD assumption is that for any integerx and 0< t < 1, 𝓁x+t = (1−t)𝓁x +t𝓁x+1. The name comes from the fact that given 0< t < t +h < 1, 𝓁x+t −𝓁x+t+h = h(𝓁x −𝓁x+1) = hdx. Inotherwords,duringanyintervalof h yearswhichliesbetweentwointegralages,thenumber of deaths occurring in that period is justh times the total number of deaths for the yearly interval. We can therefore say that the deaths are spread uniformly over the year. Example7.2 Suppose𝓁 60 = 1000,𝓁61 = 940.AssumingUDD,whatistheprobabilitythat a person age 601 3 will die before reaching age 601 2? Solution. The number of deaths in the1 6-year period between 601 3 and 601 2 equals 10, one-sixth of the total deaths in the year. Out of 1000 lives at age 60, there will 20 who die before reaching age 60 1 3 (one-third of the yearly deaths), leaving 980 alive at age 601 3.T h e probability is therefore 10/980. 102 FRACTIONAL DURATIONS An equivalent form of the UDD assumption is tqx = tqx, for any integer x and 0< t < 1, (7.7) which follows immediately by writingtqx as (𝓁x −𝓁x+t)∕𝓁x. AssumingUDD,wecaneasilycalculate tpy forall t andy (notnecessarilyanintegerage) by the formula tpx+h = t+hpx hpx , which follows from the multiplication rule. This points out an inconsistency in the UDD assumption. Suppose that the life table follows the expected pattern of increasing mortality byage.Thatis,forallintegers x ≤ y wehave q x ≤ qy.ThenitcanbeshownthatunderUDD, tqx ≤t qy for integersx ≤ y and allt ≥ 0, as we would expect. However, this inequality need not hold ifx andy are not integers. See Exercise 7.15. 7.3.2 Present value formulas WenowderiveformulasforfractionaldurationlifeannuitiesbymakingtheUDDassumption for mortality and the normal assumption for interest that we used in Section 7.2. This is the standardprocedure,butthedifferenttreatmentofthetwofactorsleadstosomecomplications in the formulas. We first assume constant interest. For our purpose we require some new interest quantities. Let 𝛽(m) = 1 vm2[v1∕m +2v2∕m +⋯ +(m −1)v(m−1)∕m], 𝛼(m) = d d(m) +d𝛽(m). We will later derive a simplified expression for𝛽(m). We have ̈a(m) x (e0) = 1 m[1+v1∕m 1∕mpx +v2∕m 2∕mpx +⋯ +v(m−1)∕m m−1∕mpx]. From (7.7),j∕mpx = 1−(j∕m)qx, so we can write u0 = ̈a(m) x (e0) = 1 m[1+v1∕m +v2∕m +⋯ +v(m−1)∕m] − qx m2[v1∕m +2v2∕m +⋯ +(m −1)v(m−1)∕m] = ̈a(m)(e0)−𝛽(m)vqx = d d(m) −𝛽(m)vqx, where we refer back to (7.6). Similarly, for any positive integerk, uk = d d(m) −𝛽(m)vqx+k. LIFE ANNUITIES PAIDmTHLY 103 From (7.2) and (5.6) can now write ̈a(m) x (c) = d d(m) ̈ax(c)−𝛽(m)Ax(c). (7.8) Somewhat surprisingly, an insurance term has come into an annuity value formula, but it is easilyexplained.Apartfrominterestloss,areasonforthe mthlyannuitytobeworthlessthan the yearly annuity is the loss of annuity income in the year of death. The annuitant receiving benefits mthly will only receive payments up to the time of death in their final year. For example, if they die shortly after the beginning of the year they will only get 1∕m of the total yearly income. Of course, they could receive the total income by living until the lastmth of the year. On average, annuitants should lose slightly less than half of the income in the year of death, which is reflected by the second term in (7.8). Under UDD the actual proportion is 𝛽(m). This is close to (m−1)∕2m for small interest rates, as shown in Section 7.5 below. To derive an alternate expression to (7.8) that uses only annuity present values we apply the general insurance-annuity identity (5.8) to the second term in (7.8) and obtain ̈a (m) x (c) = 𝛼(m)̈ax(c)−𝛽(m)̈ax(Δc). (7.9)‡ In the general case where the interest rate is constant over each year, but can vary from year to year,𝛼(m) and𝛽(m) will be vectors with the entry corresponding to indexk, equal to the corresponding value for the interest ratei k. The general form of (7.9) will be ̈a(m) x (c) = ̈ax(c ∗ 𝛼(m))− ̈ax(Δ(c ∗ 𝛽(m))). For the particular case of the level payment annuity with constant interest we have ̈a(m) x (1n) = 𝛼(m)̈ax(1n)−𝛽(m)(1−v(n)npx) Example 7.3 A woman age 60, earns a salary of 60 000 per year, payable monthly, and would normally expect to receive a raise of 2000 per year until retirement at age 65. She is injured and unable to work again. She plans to sue the parties responsible for her injury, for anamountequaltothepresentvalueofherlostsalary.Findaformulatocalculatethispresent value. Assume a constant interest rate. Solution. We will assume the accident occurred just before her 60th birthday, and that the first raise is due 1 year later. The present value is 1000 [ ̈a (12) 60 (c) ] = 1000[𝛼(12)̈a60(c)−𝛽(12)̈a60(Δc)], wherec = (60,62,64,66,68), so that Δc = (60,2,2,2,2, −68). 104 FRACTIONAL DURATIONS k… … … … … …c k /4 ck /4 ck /4 ck /4 ck /4 ck /4 ck /4 ck /4 ck+1 /4 ck+1 /4 ck +1 /4 ck +1 /4 ck +1 /4 ck +1 /4ck –1 /4 ck+1 /4ck+1 /4 ck+2 /4 Due payments Time Immediate pa yments k + 1/4 k + 2/4 k + 3/4 k + 5/4 k +6 / 4 k + 7/4 k +2k +1 Figure 7.1 Due and immediate quarterly annuity payments 7.4 Immediate annuities We now discuss an important variation. Refer back to (7.1). Suppose that instead of this scheme we assume payments of c k m at timek + j m, j = 1,2, …,m. (7.10) Wecanviewthisashavingpaymentsmadeatthe end ratherthanthe beginning ofeachperiod of length 1∕m. Annuities satisfying (7.10) are often referred to asimmediate, while those satisfying (7.1) are referred to asdue. The present value of the annuity with benefit vector given by (7.10) and discount functiony is denoted bya(m)(c;y). That is, the two dots are removed for immediate annuities. ThereadermaywishtoreferbacktoSection2.14.1wherewecommentedonthesomewhat unusual terminology. We also pointed out that the distinction between the two types is not needed in the annual case. However, when we postulate a constant rate of payment over the year, we do need to make the distinction in themthly case. See Figure 7.1 for a comparison of the payments on a due and immediate quarterly annuity. To go from the due annuity to the immediate annuity, we must do the following. At time zero, subtract a payment ofc 0∕m. At time 1, subtract a payment ofc1∕m and add a payment of c0∕m. At time 2, subtract a payment ofc2∕m and add a payment ofc1∕m, etc. We then have, forany discount functiony, a(m)(c;y) = ̈a(m)(c;y)− 1 m ̈a(Δc;y). (7.11) Example 7.4 Find a formula for the present value of a life annuity on (40) consisting of payments of $500 at the end of each month for 10 years. Assume a constant interest rate. Solution. In this case, c = 6000(1 10), Δc = 6000(1,09,−1), ̈a40(Δc) = 6000[1−y40(10)]. The present value is 6000 [ a(12) 40 (110) ] = 6000 [ ̈a(12) 40 (110)− 1 12(1−y40(10)) ] = 6000 [ 𝛼(12)̈a40(110)− ( 𝛽(12)+ 1 12 ) (1−y40(10))] . APPROXIMATION AND COMPUTATION 105 7.5 Approximation and computation Intheprevioussectionwedevelopedsomebasicformulas.Wenowconcentrateontheproblem of obtaining numerical values. For this purpose it is convenient to develop a closed-form formula for𝛽(m) in terms of standard interest symbolsi (m) and d(m). Assume for the moment constant interest. From the definition of𝛽(m) we can write, using the discount functionv(k) = (1+i′)−k, 𝛽(m) = 1 m2Valm−1(j), wherej = (1,2, …,m −1). From formulas (2.16) and (2.17), Valm−1(j) = (1+i′)(m−1)̈a(j) = [(1+i′)m−1 −1]∕d′ −(m −1) d′ . We can simplify the numerator since (1+i′)m−1 −1 d′ = (1+i′)m −(1+i′) i′ = i i′ −1. Substituting and noting thatmi′ = i(m),md′ = d(m), we obtain the formula 𝛽(m) = i −i(m) i(m)d(m). (7.12) It then follows that 𝛼(m) = d d(m) +d𝛽(m) = id i(m)d(m). For varying interest we get the same relationships holding on a year-by-year basis. We need merely insert subscripts ofk on all quantities. Exact calculations of𝛼(m) and𝛽(m) are not always done in practice. It is common to use an approximation that assumes a constant interest rate of 0. This implies that 𝛽(m) = 1+2+⋯ +m −1 m2 = m −1 2m . Moreoverd∕d(m) = 1 and𝛼(m) = 1, so we can write the approximate formula ̈a(m) x (c) ≐ ̈ax(c)− m −1 2m ̈ax(Δc). (7.13) In much of the actuarial literature this approximation is not found by setting the interest rate equal to 0 as we have done, but rather by using linear interpolation for the discount function y x. That is, it is assumed that for all nonnegative integersk and 0< s < 1, yx+k(s) = (1−s)+syx+k(1). 106 FRACTIONAL DURATIONS Under this assumption ̈a(m) x (e0) = 1 m [ 1+yx ( 1 m ) +yx ( 2 m ) +⋯ +yx (m −1 m )] = m +1 2m + m −1 2m vpx = 1− m −1 2m [1−vpx], which is the same as̈a(m) x (e0) calculated at 0 interest. By (7.2) both assumptions lead to the same approximation. This reason why this occurs is that the linearity ofyx, in addition to the UDD assumption, necessarily implies that the interest rate must be 0. The advantage of stating the latter condition from the outset is that it makes it clear that the approximation shouldbereasonablewithrelativelylowinterestrates.However,ifratesaretoolargerelative to mortality rates, this approximation can lead to inconsistencies. Example 7.5 Suppose thati =0.21, andq x =0.004. Calculatëa2(e0;v) and̈a(2) x (e0), using (7.13) for the latter. Solution. ̈a (2)(e0;v) = 1 2[1+1.10−1] = 0.955. Using (7.13), ̈a(2) x (e0) = 1− 1 4 [ 1− 0.996 1.21 ] = 0.956. It is, however, inconsistent for an annuity in which payments are contingent upon survival to cost more than an annuity with identical payments that are certain to be made. Equation 7.13 is part of a series of other possible approximation methods known as Woolhouse’s formulas. This one is the two-term formula. The next in the series, the three- term formula, is not used often but appears to give very good results in certain cases. After developing the appropriate tools in the next chapter, we will investigate further the two-term formula inconsistency and derive the three-term formula. See Section 8.11. ∗7.6 Fractional period premiums and reserves Annual premium and reserve formulas remain basically unchanged when premiums are paid mthly. The only difference is that we replacëa x with ̈a(m) x throughout. The purpose of this section is to derive a few useful identities. Consider all policies with a particular issue age (x) and with level premiums payable for h years. The simplest type of such a policy will be a 1-unit,h-year term policy with annual premiums. LetP denote the net annual premium andkV denote the reserve at timek for this term policy. Given any other such policy, let̃P denote the net annual premium if paid yearly, RESERVES AT FRACTIONAL DURATIONS 107 and let̃P(m) denote the net annual premium for this policy if paidmthly (i.e.,̃P(m)∕m is paid at the beginning of eachmth of the year). Letk ̃V denote the reserve at timek on this policy if premiums are paid annually and letk ̃V(m) denote the reserve at timek if premiums are paid mthly. For simplicity, we will assume constant interest. We have ̃P(m)̈a(m) x (1h) = ̃P̈ax(1h), as both sides are equal to the present value of the benefits, so that ̃P ̃P(m) = ̈a(m) x (1h) ̈ax(1h) = d d(m) −𝛽(m)Ax(1h) ̈ax(1h), using (7.8) for the last equality, and we can write ̃P ̃P(m) = d d(m) −𝛽(m)P. (7.14) Since the value of insurance benefits is independent of the premium frequency, we have, for k < h, k ̃V(m) −k ̃V = ̃P̈ax+k(1h−k)− ̃P(m)̈a(m) x+k(1h−k) = ̃P(m) [ ̃P ̃P(m) ̈ax+k(1h−k)− ̈a(m) x+k(1h−k) ] . Now substitute from (7.8) for ̈a(m) x+k(1h) and from (7.14) for ̃P∕̃P(m).T h et e r m(d∕d(m)) ̈ax+k(1h−k) cancels out and we are left with k ̃V(m) −k ̃V = 𝛽(m)̃P(m) kV. (7.15) The term on the right represents the additional reserve that must be held at timek in order to provide for the premiums that will not be collected in the year of death. The significance of (7.14) and (7.15) is as follows. We consider a collection of policies with level premiums, common issue age and premium payment period. The benefits can be anything at all. We know the premium and reserve for the particular case of the level term policy. Then, using only those two quantities, we can easily adjust premiums and reserves from annual mode tomthly mode, for all other policies in the collection. 7.7 Reserves at fractional durations InsurancecompaniesdoacompletecalculationofreservesonDecember31(orwhenevertheir particular fiscal year ends). However, policyholders are not always accommodating enough to purchase their policy on January 1. This means that most of the policy durations at the calendaryearendwillbefractional.Inthissectionwederiveformulasforcalculatingreserves attime k +s wherek isanintegerand0 < s < 1.Considerthegeneralannualpremiumpolicy 108 FRACTIONAL DURATIONS with death benefit vectorb and premium vector𝝅. Suppose first that the death benefit for those dying between timek and k +s is paid at timek +s. The reserve at timek +s would then be, by a calculation analogous to that used in deriving (6.7), (kV +𝜋k)yx(k +s,k)−bk sqx+k spx+k . Thatis,thinkingretrospectively,thebalanceattime k +s isobtainedbytakingthebalanceat time k, adding the premium, accumulating up to timek +s at interest and survivorship, and thensubtractingfromeachsurvivorenoughtopaythedeathbenefitsforthosedyingbetween timek and timek +s. Wemust,however,adjustthistoreflectthefactthatthedeathbenefitwillnotbepaiduntil the end of the year, and to do so we multiply the second term byv(k +s,k +1). Note now that v(k +s,k +1) spx+k = yx(k +s,k +1) (1−spx+k+s)(spx+k) = yx(k +s,k +1) px+k . FromtheUDDassumption, sqx+k = s(qx+k)andsubstitutingtheselasttwoidentitiesgives the desired reserve formula as k+sV = (kV +𝜋k)yx(k +s,k)−sbk qx+k px+k yx(k +s,k +1) = (1−s)(kV +𝜋k)yx(k +s,k)+s [ (kV +𝜋k)yx(k +1,k)−bk qx+k px+k ] yx(k +s,k +1). Finally, substituting from (6.7) yields k+sV = (1−s)[(kV +𝜋k)yx(k +s,k)]+s[k+1Vy x(k +s,k +1)]. In other words, under UDD, the reserve at some intermediate point in the year is obtained by linearly interpolating between the initial reserve, accumulated to that point with interest and survivorship, and the end-of-year terminal reserve, discounted to that point with interest and survivorship. Care must be taken not to use the terminal instead of the initial reserve at time k, as that would involve interpolating over a discontinuity. A common simplification is to assume that interest and mortality rates are sufficiently smallsothatwecantake y x asidenticallyequalto1.Thisresultsintheapproximateformula k+sV = (1−s)(kV +𝜋k)+s(k+1V) = (1−s)kV +s(k+1V)+(1−s)𝜋k. The third term above is known as theunearned premium.It is simply the total premium paid at the beginning of the year multiplied by the fraction of the year remaining. When premiums are paidm times a year rather than annually, the same type of approxi- mation as above is normally used. That is, we calculate the reserve at timek +s by linearly interpolatingbetweentheterminalreservesattime k andk +1andaddtheunearnedpremium. EXERCISES 109 To be precise, suppose thats = h∕m +r where h is an integer and 0< r < 1∕m and that the annual premium𝜋k is paidm times per year. Then a commonly used approximation is k+sV = (1−s)kV +s(k+1V)+ ( 1 m −r ) 𝜋k. Example7.6 Foracertainpolicy 3V = 108, 4V = 180andforthefourthyearofthepolicy the premium is 60, paid quarterly (i.e.,15 is paid at time 3, 31 4,3 1 2 and 33 4. Find the reserve at time 3 years and 7 months. Solution. We haves =7/12=2/4 + 1/12. The unearned premium is ( 1 4 − 1 12 ) 60 = 10. The above formula gives 3+7∕12V = 5 12 ×108+ 7 12 ×180+10 = 160. An alternate way of calculating the unearned premium is to note that the reserve is taken at a time one-third through the quarter-year period, so the unearned premium is two-thirds of the premium payable at the beginning of the quarter, which is 2 3 ×15 = 10. 7.8 Standard notation and terminology Most of the standard notation for fractional payments has been discussed already. The one remaining symbol isA (m) which is used to denote net single premiums for an insurance in which the death payments are paid at the end of themth of the year in which death occurs. For example ifm =12, the death benefit would be made at the end of the month of death. The assumption of Exercise 7.14 below is known as theBalducci hypothesis, named for G. Balducci, an Italian actuary. It arose in conjunction with mortality studies designed to produce life tables. Exercises Assume UDD unless otherwise indicated. Type A exercises 7.1 You are given thaẗa60 = 12,10p50 = 0.8, and interest is constant at 6%. (a) Findthenetsinglepremiumforanannuityon(50)withlevelbenefitpaymentsof 1000 per month, beginning at age 50 and continuing for life, with the provision that the first 120 monthly payments are guaranteed, regardless of whether (50) is alive or not. (b) Redo (a), assuming now that the first payment is made at the end of 1 month. 110 FRACTIONAL DURATIONS 7.2 You are given thatq70 = 0.2,q71 = 0.25,q72 = 0.30. Interest rates are 20% for the first2years and 30% thereafter. A 3-year lifeannuity on(70) provides forbenefits of 1000inthefirstyear,2000inthesecondyearand3000inthethirdyear,providedthat (70) is then alive. Suppose that payments are to be made quarterly. Find the present value if: (a) the first payment is to be made at age 70; (b) the first payment is to be made at age 70 1 4. 7.3 An annuity on (50) provides for yearly payments for 30 years. The amount of the income is 10 000 for the first 10 years, 8000 for the next 10 years and 5000 for the last 10 years. The net single premium for this annuity is 100 000. You are given that theinterestrateisaconstant5%andthat 10p50 = 0.9, 20p50 = 0.8, 30p50 = 0.7.Find the premium if the same income is to be paid monthly instead of annually, assuming payments are made (a) at the beginning of each month, (b) at the end of each month. 7.4 You are given the following figures from a life table.𝓁 60 = 1000,𝓁61 = 700,𝓁62 = 500.Findtheprobabilitythatapersonnow age60 1 3 willdiebetween theagesof60 1 2 and 613 4. 7.5 Given q70 = 0.2, q71 = 0.3, q72 = 0.4,findtheprobabilitythatapersonage70 1 2 will live to age 721 4. 7.6 Instead of assuming UDD, suppose that, for an integerx and 0< t < 1, we assume that tpx = pt x .G i v e nl60 = 100000, l61 = 81000, l62 = 41472, find the probability that (601 2) will die between the ages of 611 3 and 612 3. 7.7 A policy is issued on March 20, 1990. Given that10V = 2000, 11V = 3000 and that the premium is a level 105 per month, find the reserve on December 31, 2000, using the standard approximation. Assume that each month is 30 days. 7.8 For a policy on (x ), 5V = 220, 6V = 80, b5 = 1000, i5 = 0.10, qx+5 = 0.2. Find 5+1∕4V. Type B exercises 7.9 Let r bethepresentvalueofalifeannuityon(x )providing1atthebeginningofeach periodoflength1∕m,withpaymentsguaranteedfor n years.Let s bethepresentvalue ofthesameannuitybutwiththepaymentsatthe end ofeachperiod.If v(n) = 0.9and npx = 0.6, findr −s. You are not given the value ofm. 7.10 Show that under a constant interest rate ofi,a(m)(1n) = (i∕i(m))a(1n). 7.11 You want to evaluate ̈a(4) x (k) at a constant interest rate of 20%, where k = (10,9,8, …,1). Assuming that UDD holds, what error is made if you use the zero- interest approximation? You are given thaẗax(k) = 35 and̈ax(111) = 6. *7.12 A 20-year term insurance policy on (50) has a constant death benefit of 1000, and net level annual premiums of 10, payable for 20 years. The reserve at time 15 is 60. Anotherpolicyon(50)hasthesamepremiumpatternbutdifferentdeathbenefits.On this latter policy, the annual premium is 100, and 15V = 500. The interest rate is a EXERCISES 111 constant 6%. What is the value of the annual premium and the reserve at time 15 on this latter policy, if the premiums are payable monthly instead of annually? 7.13 Suppose we assume that for all integersx and 0< t < 1, 𝓁 x+t = 𝓁1−t x 𝓁t x+1. (a) Show that this is equivalent to the assumption of Exercise 7.6. (b) Show that if (2.9) holds forv, then it also holds for the discount functionyx whenever s,t,s +h,t +h are all in the interval [k,k +1) for some nonnegative integerk. (See Exercise 8.21 for more on this assumption.) 7.14 Suppose we assume that for all integersx and 0< t < 1, 𝓁−1 x+t = (1−t)𝓁−1 x +t𝓁−1 x+1. Show that for all integersx and 0< t < 1,1−tqx+t = (1−t)qx. 7.15 Assume UDD. Suppose that, a life table is such thatqx < qx+1 for all nonnegative integersx. (a) Show that for anyt > 0 and integersx < y we havetqx ≤ tqy. (b) Suppose thatq80 = 0.4, q81 = 0.45.Show that0.5q80.5 >0.5 q81. (c) Show that under the assumption of Exercise 7.6, it is true that for allt > 0 and positive numbersx,y,w eh a v etqx ≤ tqy. Spreadsheet exercise 7.16 ModifytheChapter4spreadsheetsothatbyenteringthefrequency m,thespreadsheet will calculate both due and immediate annuities payablem times per year. 8 Continuous payments 8.1 Introduction to continuous annuities In this section we consider annuities where payments are made continuously. Naturally, this is not physically possible, but we can picture these as a limiting case ofmthly annuities as mgoes to infinity. Suppose, for example, that you are to receive a total of 36 500 units each year. This could be done by paying 100 units per day, or 4 1∕6 units per hour, or 0.001157 units per second and so on. If you can imagine payments coming in every nanosecond, you may get some feeling for what a continuous annuity would be like. This may seem as a somewhat artificial concept, but there are many uses for continu- ous annuities. They can be used to approximatemthly annuity values for large values ofm. Moreover, we will show that insurance contracts with the realistic provision of benefit pay- mentsatthemomentofdeath,canbeviewedascontinuousannuities.Forinsurancecontracts purchased by continuous premium payments we can derive some interesting mathematical relationships that are analogues of those that appeared in Chapters 5 and 6. Inthecontinuouscase,wecannotspecifyanactualpaymentatanypointoftime.Asshown by the figures above, this approaches zero as the frequency of payment increases. Instead we must speak of theperiodicrate of payment. Consider a monthly annuity. If the payment in 1 monthis100,wecoulddescribethisbysayingthattheannual rateofpaymentforthatmonth is1200.Thiswouldmeanthatifthepaymentsremainedatthesamemonthlylevelforayear, then the total payment for that year would be 1200. Each of the annuities described above wouldbepaidattheannual rateof36500.Ofcourseinanannuitypaid mtimesperyear,the actual payment and therefore the periodic rate of payment could change everymth period. In our continuous annuity, it could change from moment to moment. (Picture a person standing under a chute, receiving money that is flowing down continuously. The speed at which it comes down could change at each instant.) Inplaceofacashflowvectorwenowhaveacashflowfunction cdefinedon[0, N],where c(t) is the periodic rate of payment at timet. FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial THE FORCE OF DISCOUNT 113 Wewillassumethatallourcashflowfunctionsare piecewisecontinuous.Thatis,thereis a partition, 0= a1 < a2 < ⋯ < an = N, such thatcis continuous on the intervals (ai−1,ai), i= 1,2, …,n−1. Supposewehaveanarbitrarydiscountfunction yandacontinuousannuitywithcashflow functionc.Thepresentvalueoftheannuitywillbedenotedby a(c;y)orjust a(c).(The‘bar’is astandardactuarialnotationaldevicetodenotecontinuousasopposedtodiscretequantities.) Consider an approximating annuity with payments mademthly. That is, payments are made at timej∕mwhere jis an integer between 0 andmN−1. We assume thatN is an integer and theannual rateofpaymentattime j∕misc(j∕m).Theactualpaymentattime j∕mistherefore (1∕m)c(j∕m), so the present value of thismthly annuity is mN−1∑ j=0 c(j∕m) m y(j∕m). (8.1) Taking the limit asmgoes to∞gives one of the key formulas of this chapter, a(c;y) = ∫ N 0 c(t)y(t)dt, (8.2) ‡ since themthly present values are Riemann sums for this integral. 8.2 The force of discount Throughoutthebookweuse‘log’torefertothenaturallogarithm(oftendenotedby lnorloge.) The evaluation of continuous streams of cash flows leads naturally to a new function, which we will spend some time investigating before returning to the main topic. It can be written in several equivalent forms. Definition 8.1 Given an arbitrary discount functiony, define another function,𝛿 y(t), which we will call theforceofdiscountassociatedwithy,b y 𝛿y(t) = lim h→0 y(t+h,t)−1 h = lim h→0 y(t+h,0) −y(t,0) hy(t,0) = d dty(t,0) y(t,0) = d dtlogy(t,0) =− d dtlogy(t) =− d dty(t) y(t) . (8.3) (To derive the second last equality, recall thaty(t) = y(0,t) = y(t,0)−1.) From the first equality in (8.3) we see that𝛿y represents arelativerate of growth. Each unit invested at timetwill accumulate toy(t+h,t) units over the nexthtime periods, so the numerator gives us the relative growth rate over this time interval. We divide byhto get the relative rate of growth perperiod, and take limits to get the instantaneous periodic relative rate of growth at timet. We are often given𝛿y and wish to recovery. 114 CONTINUOUS PAYMENTS Proposition 8.1 y(t) = e−∫ t 0 𝛿y(r)dr. Proof. From the second last expression in (8.3), we integrate to get ∫ t 0 𝛿y(r)dr=− ∫ t 0 d dr[logy(r)]dr=−l o gy(t)+log y(0). We use the fundamental theorem of calculus for the second equality. Note thatlogy(0) = log1 = 0, multiply by−1, and take exponentials to complete the proof. An immediate consequence of Proposition 8.1 is that fors≤ t, y(s,t) = y(t)∕y(s) = e−∫ t s 𝛿y(r)dr = e−∫ t−s 0 𝛿y(s+r)dr. (8.4) Remark Thistraditionaluseoftheforceofdiscountisfortheinvestmentdiscountfunction v. In this case𝛿v(t) will be denoted by just𝛿(t). The standard actuarial notation is𝛿t,b u tw e find it more convenient to use the bracket in place of a subscript to parallel our treatment for thesymbol v.Thisquantityiscommonlyreferredtoasthe forceofinterest. Theterminologyis justifiedsinceforaninstantaneousperiodthereisnodifferencebetweeninterestanddiscount. (This is shown precisely by the equality of the fourth and last expressions in (8.3)). 8.3 The constant interest case Consider the constant interest discount function given byv(t) = vt. Then𝛿(t) is also constant anddenotedjustby 𝛿.Since log(vt) = tlog(v),thesecondlastexpressionin(8.3)showsthat 𝛿 =−l o g(v) =l o g(1+i), (8.5) and Proposition 8.1 gives v(t) = e−𝛿t,( 1 +i)t = e𝛿t. (8.6) The second expression reflects the fact that under constant interest, invested capital grows exponentially and𝛿 is just the usualrateofexponentialgrowth. Additionalinsightmaybegainedbycomparing 𝛿 withthefractionalinterestquantitiesof Chapter 7. We have that 𝛿 = lim m→∞ i(m) = lim m→∞ d(m). This can be seen by writing i(m) = (1+i)1∕m −1 1∕m , d(m) = 1−v1∕m 1∕m , CONTINUOUS LIFE ANNUITIES 115 so that lim m→∞ i(m) = d dt(1+i)t|t=0 = (1+i)t log(1+i)|t=0 = 𝛿. The calculation ford(m) is similar. The quantity𝛿 is often thought of as an instantaneous rate of interest, but it is important to note that just like the ratesi(m) and d(m), its value depends on the underlying period. The transformationforchangeofperiodthoughismucheasierthanfortheinterestrates,asitjust involves a proportional change. For example, if the annual force of interest is 0.06, then the force of interest for a half year period will be simply 0.03. We now return to continuous annuities, using a constant interest rate, and show that in certain cases we get expressions analogous to discrete annuities, but with𝛿 replacing d.W e will derive the continuous versions of (2.16) and (2.17). Define a function 1 n by 1n(t) = { 1, if t≤ n, 0, if t> n. This is the same notation as for the analogous vector, but the meaning should be clear from the context. We also define the functionI n(t)b y In(t) = { t,i f t≤ n, 0, if t> n. From (8.2) and (8.6), a(1n) = ∫ n 0 e−𝛿tdt= 1−e−𝛿n 𝛿 . (8.7) Integrating by parts, a(In) = ∫ n 0 te−𝛿tdt=− te−𝛿t 𝛿 | | | | n 0 + 1 𝛿 ∫ n 0 e−𝛿tdt= a(1n)−ne−𝛿n 𝛿 . (8.8) 8.4 Continuous life annuities 8.4.1 Basic definition The present value of an annuity issued to (x) that provides benefits paid continuously at the annualrateof c(t)attime t,provided(x )isalive,willbedenotedby ax(c).Formula(8.2)gives the exact formula ax(c) = ∫ 𝜔−x 0 c(t)v(t)tpxdt. (8.9)‡ Note that this formula follows essentially the same pattern that we observed after formula (5.1). Instead of summing we integrate, and for eacht the integrand is the product of three 116 CONTINUOUS PAYMENTS factors: the interest discount factor; the ‘amount paid ’ at timet, (which we can intuitively think of asc(t)dt); and the probability that the payment is made. 8.4.2 Evaluation We will generally only have available the values of𝓁x for integer values of (x), and must interpolatefortheremainingvaluesinordertoevaluatetheintegral.AsinChapter7,wewill assumeUDD,whichwillallowustoexpresscontinuousannuitiesintermsofannualannuities. The procedure closely parallels that formthly annuities and we could simply write down the formulas as limiting cases of those in Chapter 7. It is instructive however to proceed a new. For simplicity of notation, we will at first assume constant interest. The initial step is to define the continuous analogues of the functions𝛽 and𝛼 introduced in Chapter 7. Let 𝛽(∞) = e 𝛿a(I1) = e𝛿 −1−𝛿 𝛿2 = i−𝛿 𝛿2 , invoking (8.7) and (8.8). Let 𝛼(∞) = a(11)+d𝛽(∞) = di 𝛿2. It is clear that𝛽(∞) and𝛼(∞) are the respective limits of𝛽(m) and𝛼(m)a smapproaches∞. WenowusethesameprocedureasweusedinSection7.1andreplaceallthepaymentsin any 1 year by a single payment, equal to the value of these payments at the beginning of the year.Thisreplacementprincipleholdsinthecontinuouscaseaswell.Foraformalverification use the standard integration result ∫ b 0 = ∫ a 0 +∫ b a for a< b. Analogously to (7.2) this leads to ax(c) = ̈ax(z), (8.10) where zk = ∫ 1 0 c(k+s)vs spx+kds= ∫ 1 0 c(k+s)vs(1−s⋅ qx+k)ds. (8.11) This is easily evaluated when the rate of payment is constant over each year. For a nonnegativeinteger k,let ck denotetheconstantvalueof c(k+s),0 ≤ s< 1.Wecanfactor ck outside of the integral sign in (8.11), which gives zk = ck[a(11)−qx+ka(I1)] = ck [d 𝛿 −𝛽(∞)vqx+k ] . Then, from (8.10), ax(c) = d 𝛿 ̈ax(c)−𝛽(∞)Ax(c), CONTINUOUS LIFE ANNUITIES 117 where c = (c0,c1,…). Finally, from Theorem 5.1, ax(c) = 𝛼(∞)̈ax(c)−𝛽(∞)̈ax(Δc). (8.12) As promised, this reduces the calculation to finding the present value of yearly annuities. Asinthe mthlycase,itisusualinpracticetoapproximatethisbycalculatingtheadjusting factorsatzerointerest.Wethenhave 𝛼(∞) = 1,𝛽(∞) = 1 2,andtheapproximatingformulais ax(c) ≐ ̈ax(c)− 1 2̈ax (Δc). In the case wherec = 1∞ so thatΔc = (1,0,0,…), we get the familiar equation ax ≐ ̈ax − 1 2. (8.13) This concludes the development of formulas for the constant interest case. When rates are only constant over each year, but can vary from year to year, we must replace the constants d∕𝛿,𝛼(∞) and𝛽(∞) by vectors, analogously to the treatment in the discrete case. 8.4.3 Life expectancy revisited Comparing the formula for life expectancy (3.8) with formula (4.1), we see thatex is simply thepresentvalueofalifeannuityof1peryear,beginningattime1,ataconstantinterestrate of0.Thismakesenseintuitively,sinceatzerorateofinterest,thepresentvalueoftheannuity will simply be the total amount paid, which in turn is just the whole number of years lived, whichonaverageisjust e x.Wecannowwritedownamorerigorousformulaforthe complete life expectancy, as the present value of a life annuity paying 1 per year for life continuously, at zero interest. This gives ◦ex = ∫ ∞ 0 tpxdt. Assuming UDD, we apply the formulas given above. Withc = 1𝜔−x, and an interest rate of zero, ̈ax = 1+ex, ̈ax(Δc) = 1, and from (8.13), ◦ex = 1+ex − 1 2 = ex + 1 2, recovering the approximation we obtained intuitively in Chapter 3. Similarly,thecomplete n-yeartemporarylifeexpectancyisgivenby ∫ n 0 tpxdt,whichunder UDD, is equal to∑n k=1 k px + 1 2nqx as we obtained previously. Hereisainterestinguseforlifeexpectancy.Supposeyouaregiven ◦ex andtheinvestment discountfunction v,butnottheactuallifetable,andyouwanttoapproximate ax.Onenatural waytodoitissimplytake a(1◦ex ;v)forthisapproximation.Indeed,somemighteventhinkthat 118 CONTINUOUS PAYMENTS this (or an appropriate modification in the discrete case) should be exact, reasoning that the cost of a life annuity on a person should be the cost of providing the individual with income uptotheirpreciselifeexpectancy.Theapproximationshouldbeclose,butinthenormalcase of positive interest rates, this method always overstates the true value. A mathematical proof isoutlinedinExercise22.15butwecaneasilyexplainthisfactbygeneralreasoning.Suppose an insurer sells 1-unit continuous life annuities to a group of people agexand charges them each the price of a fixed period annuity with the period equal to their life expectancy. The insurercanexpecttogainmoneyonthosewhodiebeforelivingtheirlifeexpectancy,butlose moneyonthosewholivelonger,andthesegainsandlossesshouldcancelout.However,since thegainscomeearlier,theywillproduceextrareturnsduetointerestearningsandtheinsurer will end up with more than they need to provide the benefits. The break even premium will then be somewhat less than the premium for the fixed period annuity. 8.5 The force of mortality Beforeproceedingwithinsurances,weintroduceanewquantitythatisanalogoustotheforce of discount. Wefirstmotivatethisintuitively.Whatistheprobabilitythat( x)willdieinthenextinstant of time? We could approximate this by looking at hqx, but if we assume continuity and take thelimitas happroaches0,wewouldget0.Thequestionisanswerablebuttheanswerof0is uninteresting and gives us no information about the mortality of (x) at that point. Instead, we can compute an annualrateof mortality at age (x), by dividing by h before taking the limit. This leads to the following definition. Definition 8.2 The forceofmortality at age (x) is the quantity 𝜇(x) = lim h→0 hqx h = lim h→0 𝓁x −𝓁x+h h𝓁x =− d dx𝓁x 𝓁x =− d dxlog𝓁x. (8.14) The expression after the third equality is useful for obtaining additional insight. We have a group of lives declining over time due to mortality. The quantity𝜇(x) gives us therelative rate of decline in this group at agex. In many cases we are looking at a fixed agexand we want the variable to be the timet. We will accordingly define 𝜇x(t) = 𝜇(x+t). We can view𝜇x(t) as the force of mortality at timetfor an individual agexat time 0. From the fourth expression in (8.14), 𝜇x(t) =− d dttpx tpx , (8.15) and we imitate the proof of Proposition 8.1 to derive tpx = e−∫ t 0 𝜇x(r)dr. (8.16)‡ INSURANCES PAYABLE AT THE MOMENT OF DEATH 119 Example 8.1 Suppose that the force of mortality is given by 𝜇(x) = 1 𝜔−x, x<𝜔 . Find an expression fortpx. Solution. Fort<𝜔 −x, −∫ t 0 1 𝜔−x−rdr=l o g(𝜔−x−t)−log(𝜔−x) =l o g (𝜔−x−t 𝜔−x ) . We then substitute in (8.16) to get tpx = (𝜔−x−t 𝜔−x ) = 1− t 𝜔−x. The following quantity is often useful. Definition8.3 Let𝜆x(t)denotetheforceofdiscountfortheinterestandsurvivorshipfunction at agex Note that 𝜆x(t) =− d dtlogyx(t) =− d dtlogv(t)− d dtlog tpx = 𝛿(t)+𝜇x(t). (8.17) Inotherwords,theforceofdiscountforinterestandsurvivorshipisjustthesumoftheforces forthesetwocomponents.Wecanthenviewthefunction 𝜇 x astheparticularcaseof 𝛿yx when interest rates are zero. 8.6 Insurances payable at the moment of death 8.6.1 Basic definitions Unlikeannuities,thecontinuousversionistherealisticmodelforinsurancepolicies.Itcaptures the idea that, in practice, claims are paid at the moment of death rather than the end of the yearofdeath.Wecanalsoallowforthedeathbenefittovarycontinuouslywithtime.Inplace of our death benefit vector, we have adeathbenefitfunction. This is a piecewise continuous function bdefined on [0,𝜔−x) whereb(t) is the payment made at timetshould death occur at that time. The net single premium for such a policy on (x) will be denoted by Ax(b). To calculate this, we will consider an approximating policy. Letmbe a positive integer. If death occursbetweentime j∕mand( j+1)∕m,where j= 0,1, …,m(𝜔−x)−1,ourapproximating policy will payb(j∕m) at time (j+1)∕m. Up to now we have looked at the case ofm= 1. If m= 365, for example, it would mean that death benefits were paid at midnight on the day of 120 CONTINUOUS PAYMENTS death, clearly getting close to what we want. If we letA(m) x (b) denote the net single premium for this approximating policy, we will have Ax(b) = lim m→∞ A(m)(b). Reasoning as in the middle expression of formula (5.1), we have A(m) x (b) = b(0)v(1∕m)(1− 1∕mpx ) +b(1∕m)v(2∕m)( 1∕mpx − 2∕mpx ) +⋯ . Taking limits asmgoes to∞, Ax(b) =− ∫ 𝜔−x 0 b(t)v(t)dtpx =− ∫ 𝜔−x 0 b(t)v(t) d dttpxdt, and from (8.15) we obtain our final formula Ax(b) = ∫ 𝜔−x 0 b(t)v(t)tpx𝜇x(t)dt. (8.18)‡ Onceagainwehavethefamiliarthree-factorproduct;theamountpaid;theinterestfactor; andtheprobabilitythatpaymentismade.Wecanwritethelatterasbeingitselftheproductof two terms: tpx, the probability of living to timet; and𝜇x(t)dt, which we can intuitively think of as the probability of dying at exact momentt. Aswell,thisformulaillustratesourstatementabovethatinsurancespayableatthemoment of death can be viewed as continuous annuities. Analogously to (5.5), Ax(b) = ax(c), where c(t) = b(t)𝜇x(t). (8.19)‡ 8.6.2 Evaluation Aswithcontinuousannuities,wenormallywillhaveonlythelifetableavailableandwilluse UDD in order to evaluate the integral in (8.18). We again will simplify the notation by first using constant interest. The most useful consequence of UDD for this purpose is tpy𝜇y(t) =− d dttpy =− d dt(1−tqy) = qy. (8.20) whereyis an integer and 0< t< 1. As in the annuity case, we have Ax(b) = ̈ax(z), (8.21) where zk is the value at timek of the benefits paid in the yeark to k+1. Substituting from (8.20) zk = ∫ 1 0 vtb(k+t)tpx+k𝜇x+k(t)dt= vqx+k∫ 1 0 (1+i)1−tb(k+t)dt. INSURANCES PAYABLE AT THE MOMENT OF DEATH 121 This leads to Ax(b) = Ax(̃b), where ̃bk = ∫ 1 0 (1+i)1−tb(k+t)dt, (8.22) thereby reducing the evaluation of moment of death insurances to insurances paid at the end of the year of death. This is further simplified in the case where the benefit functionb is constant over each year.If b k istheconstantvalueof b(k+t)where kisanintegerand0 < t< 1,then bk factors out of the integral in (8.22) and ̃bk = bk∫ 1 0 (1+i)(1−t)dt= bk i 𝛿, leading to the very simple formula Ax(b) = i 𝛿Ax(b), (8.23)‡ whereb = (b0,b1,…). Additional insight can be obtained by observing that i 𝛿 = e𝛿 −1 𝛿 ≐ 1+ 𝛿 2 ≐ 1+ i 2. (Here, ≐ stands for ‘approximately equal to’.) This reflects the fact that in the case where benefits are constant over the year, the difference between paying benefits at the end of the year of death and at the moment of death is that in the latter case the insurer will not earn interest from the time of death to the end of the year, and the premium must be higher to account for this. This time interval will vary with time of death, ranging from 0 for those dying at the end of the year, to 1 year for those dying at the beginning. On average it should be half a year. In the case where interest is only constant over each year, we replace the right hand side above in (8.23) byA x(i∕𝛿 ∗ b) where nowi∕𝛿 is a vector with entries ofik∕𝛿k. Example 8.2 An insurance contract on (x) provides for 2, paid at the moment of death if this occurs withinnyears, plus a pure endowment of 3 ifxis alive at the end ofnyears. Find the present value given that interest rates are a constant 6%,vnnpx = 0.4 andAx(1n) = 0.2. Solution. Care must be taken to apply the adjustment factor only to the term insurance, and not to the pure endowment, which is always paid at timen.A t6 % ,w eh a v ei∕𝛿 = 1.0297 so the present value is equal to 2(0.2)(1.0297)+3(0.4) = 1.612. Whenbenefitpaymentsarenotconstantovereachyear,wehavetoresortdirectlyto(8.22) and try to evaluate the integral. Example8.3 Aninsurancepolicyon(x )providesforadeathbenefitof tpaidatthemoment of death, should this occur at timet. Assuming UDD, find a formula for the present value in terms of end-of-the-year-of-death insurances. 122 CONTINUOUS PAYMENTS Solution. For an integerk and 0< t< 1, we have the death benefitb(k+t) = k+t= (k+ 1)−(1−t).Wecanthinkofthisastwopolicies,onepaying k+1fordeathbetweentime kand timek+1,andtheotherpaying −(1−t)fordeathattime k+t,where k= 0,1, ….(Negati ve deathbenefitsmaybeunrealisticinpractice,buttheyarefinefromamathematicalviewpoint.) From (8.23), the present value of the first policy is (i∕𝛿)A x(j), wherej = (1,2,3 …). From (8.22) the present value of the second policy is the same as that with a constant end of the year death benefit of −∫ 1 0 (1+i)1−t(1−t)dt=− (1+i 𝛿 − i 𝛿2 ) =− i 𝛿 (1 d − 1 𝛿 ) after integration by parts. The total premium is therefore i 𝛿 [ Ax(j)− (1 d − 1 𝛿 ) Ax ] . It may appear more natural to have taken the split of the death benefit ask plus t.T h e split above was done to get an expression involving the death benefit vectorj, representing a linearly increasing benefit starting at 1 unit. This is often encountered in practice. The alternativemethodwouldresultinapolicybasedonavectorwithinitialentry0,whichwould not normally be found in a real-life policy. In a more complicated case the integration could be difficult to carry out, necessitating the use of an approximate integration formula. This can arise in the case of death benefits in pension plans that may depend on salary and length of service and could vary continuously over each year. A frequently used, very simple rule, is simply to assume that the death benefit for the year from timek to k+1 is constant at the value which it takes at time k+1∕2. 8.7 Premiums and reserves The realistic model for life insurance calls for death benefits to be paid at the moment of deathandpremiumstobepaidperiodically(yearly,monthly,quarterly,etc.).Thismeansthat we have to treat death benefits separately when computing premiums and reserves, so we cannot conveniently calculate the net cash flow vectorf as we did in Section 6.1. However, thisusuallycauseslittledifficultyaswecansimplychange A y(b)to Ay(b)inallpremiumand reserve formulas. We also must define the shifted functionsb◦ kanalogous to that for vectors. That is (b◦ k)(t) = b(k+t), 0 ≤ t≤ 𝜔−x−k−t. Example 8.4 Assume a constant interest rate. Consider a policy on (40) with death benefit function b(constant over each year). The total premium for the yearktok+1i s𝜋k, payable monthly. The interest rate is a constant 0.06. Write a formula for the reserve at time 10. Assume UDD. THE GENERAL INSURANCE–ANNUITY IDENTITY IN THE CONTINUOUS CASE 123 Solution. We have 10V = [A50(b◦ 10)− ̈a(12) 50 (𝜋◦ 10)], where𝜋 = (𝜋0,𝜋1,…). Fori= 0.06, i 𝛿 = 1.0297, 𝛼 = 1.00028, 𝛽 = 0.46810, and we can write 10V = 1.0297A50(b◦ 10)−1.00028 ̈a50(𝜋◦ 10)+0.46810 ̈a50 (Δ(𝜋◦ 10)), withb = (b0,b1 …), wherebk is the constant value ofb(t)f o rk≤ t< k+1. Inordertoavoidthisdualtreatment,itissometimesconvenienttoconsiderapolicywhere premiumsarepayablecontinuously.Thisisunrealistic,butareasonableapproximationinthe case of monthly premiums. It often results in simplifying the mathematics. We will look at cases where this is useful in the next two sections. 8.8 The general insurance–annuity identity in the continuous case Inthissectionwedevelopaconnectionbetweenmoment-of-deathinsurancesandcontinuous annuities, which is analogous to that obtained in Section 5.7. We now assume that the associated benefit functionbis differentiable except possibly at the endpoints of an interval. That is, we suppose there exist pointsr and s such thatb(t) is equal to 0 for 0< t< r and s< t<𝜔 −x, and that the derivativeb ′(t)e x i s t sf o rr< t< s. Then Ax(b) = ∫ s r b(t)v(t)tpx𝜇x(t)dt=− ∫ s r b(t)v(t) d dttpxdt. Integration by parts gives Ax(b) = b(t)v(t)tpx| | r s +∫ s r [b′(t)v(t)−b(t)v(t)𝛿(t)]tpxdt. Simplifying, Ax(b) = b(r)v(r)rpx −b(s)v(s)spx +ax(b′)−ax(𝛿b). (8.24) Here 𝛿bis just the function defined by𝛿b(t) = 𝛿(t)b(t). This is in fact a continuous analogue of (5.8). We need the two additional terms at the beginning to handle the possible jumps in the functionb. 124 CONTINUOUS PAYMENTS Wecanusethistoderiveacontinuousversionofthe endowmentidentity ofSection5.7.2. Assume a constant force of interest of𝛿. For a 1-unit,n-year endowment insurance with benefits paid at the moment of death,Ax:n denotes the net single premium, andPx:n denotes the rate of annual premium, payable continuously fornyears. In this case the functionbis constant on the interval (0,n), sob′ = 0. Takingb= (1n), the first two terms in (8.24) are 1−vn npx, which leads to Ax:n = Ax(1n)+v(n)npx = 1−𝛿ax(1n), and Px:n = Ax:n ax(1n) = 1 ax(1n) −𝛿. 8.9 Differential equations for reserves The goal of this section is to develop a differential equation for reserves, which is analogous to the recursion formulas of Sections 2.10.4 and 6.3. Suppose we have a transaction with continuous cash flows defined by the cash flow function cdefined on [0,N], and a general discount functiony.T h ereserveat timet, which wewilldenoteby tV(c;y),isdefinedexactlyasinthediscretecase,namelyasthenegativeof the value at timetof all future cash flows. So tV(c;y) =− ∫ N−t 0 c(t+r)y(t,t+r)dr. It is convenient to make a change of variable fromrtos= r+tand write tV(c;y) =− ∫ N t c(s)y(t,s)ds=− y(t,0) ∫ N t c(s)y(s)ds. Differentiating by the product rule, using (8.3) to differentiate the first factor in the last expression,andthefundamentaltheoremofcalculustodifferentiatethesecondfactor,gives d dttV(c;y) = 𝛿y(t)tV(c;y)+c(t), (8.25) a continuous analogue of (2.26). Considernowaninsurancepolicyon( x)withdeathbenefitsof b(t)paidatthemomentof death,andpremiumspaidcontinuouslyattherateof 𝜋(t)attime t.Let c(t) = 𝜋(t)−b(t)𝜇x(t), the annual rate of continuous cash flow. The reserve at timeton this policy is just−tV(c;yx), which we will denote by justtV. Substituting in (8.25), and using (8.17), d dttV = tV(𝜇x(t)+𝛿(t))+𝜋(t)−b(t)𝜇x(t), SOME EXAMPLES OF EXACT CALCULATION 125 or d dttV = 𝛿(t)(tV)+𝜋(t)−𝜇x(t)(b(t)−tV). These last two formulas are the continuous versions of (6.7) and (6.8), respectively. In the first one, we view accumulation at interest and survivorship. In the second, we view the accumulation at interest only, and take the death benefit as the net amount at risk,b(t)− tV. Thisresultisoftenreferredtoas Thiele’sdifferentialequation namedaftertheDanishactuary T. N. Thiele, who discovered it in the late 19th century. In certain simple cases, for example, constant𝜇 and 𝛿, this equation can be solved analytically.See,forexample,Exercise8.21(b).Ingeneralhowever,theonlyfeasibleapproach is a numerical one. It is of interest to see what the solution will look like when we employ Euler’smethod, one of the main procedures for numerical solutions of differential equations. We describe this first for a general first order equation of the form f ′(t) = g(t,f(t)) given an initial valuef(t0). We pick a step sizeh> 0 and then simply replace the derivative with the difference quotient f(t+h)−f(t)∕hleading to a recursion f(t+h) ≐ f(t) +hg(t,f(t)) (8.26) Wethenworkbackwardsandforwardsfrom t0,startingwithgivenvalueof f(t0).Forexample, from(8.26)wegetanapproximatevalueof f(t0 +h)andwethenagainapply(8.26)withthat value to get an approximate value off(t0 +2h), and so on. Iterating the procedure we can compute approximate values off(t), wheretdiffers fromt0 by an integral multiple ofh.F o r sufficientlysmall h,thisshouldgiveareasonableapproximationstothetruevalues.Applying this to Thiele’s equation we get t+hV = tV(1+h𝛿t)+h𝜋t −h𝜇x(t)(bt −tV). (8.27) Note that what we get is very close to Equation (6.8) applied to time units of lengthh, wherefortheperiodrunningfromtime ttotime t+h,wetaketheinterestrateforthisperiod ash𝛿t,themortalityrateforthisperiodas h𝜇x(t),andthepremiumpaidforthisperiodas h𝜋t. The only difference is that (8.27) does not apply the interest factor to the premium, assuming in effect that it is paid at the end of the period. For very short periods this will make little difference. 8.10 Some examples of exact calculation Asweindicatedabove,thenormalapproachtocalculatingvaluesinthecontinuouscaseisto invoke UDD and reduce to the yearly case. Conceivably, one could have some nice analytic expressions for mortality functions and therefore be able to calculate values exactly. This is rarely done in practice. In the first place, it seems to be extremely difficult to find functions inclosedformthatprovideanaccuratepictureofobservedmortality.Second,evenwithsuch 126 CONTINUOUS PAYMENTS functions, it may be difficult to actually carry out the integration. For illustrative purposes, however, we will look at a few particularly simple cases. These are not meant to be typical of modern human mortality. In Chapter 14, we will introduce some more realistic mortality functions. 8.10.1 Constant force of mortality Supposethat 𝜇(x)isaconstant 𝜇forall x.Thisofcourseisunrealistic,sincewewouldexpect, as verified by observed data, that in general𝜇(x) increases withx. This is precisely what we mean by theaging process. There are exceptions. In the very early years there appears to be a decrease in𝜇(x), reflecting the effect of early childhood diseases. There is also a noted decrease in the early adult ages, around 25 or so. Many attribute this to the effect of a large number of automobile accident deaths for people in their early twenties. From age 30 onwards there is a rapid, exponential growth in𝜇(x), until around age 70, where𝜇(x) continues to increase but the exponential growth dampens out. There are some who predict that with medical and genetic advances, we may be approaching a situation where we can freely transplant or replace any genes or body parts that go awry. They postulate in essence that we will then have eliminated aging and achieved a constant force of mortality. This does not mean that people will no longer die, but only that the 90-year-old will be no more likely to die at any time than the 20-year-old. At the present time, however, we are not at all close to this state and it is still in the realm of science fiction. In any event, we look at some of the mathematical consequences of this assumption. In the first place, note that𝜔 is now∞, since tpx = e−𝜇t > 0, which means that there is at any time, a positive probability of continuing to survive. In our examples we will also assume a constant force of interest𝛿. Our first observation is that with these assumptions, ax = ∫ ∞ 0 e−𝛿te−𝜇tdt= ∫ ∞ 0 e−(𝜇+𝛿)tdt= 1 𝜇 +𝛿. This is easily verified intuitively. As we have seen, the force of discount for the interest and survivorship discount function under our assumptions is the constant𝜇 +𝛿. Therefore, an investment of 1∕(𝜇+𝛿) will provide a continuous return at the annual rate of [1∕(𝜇 +𝛿)]⋅ [𝜇 +𝛿] = 1. Taking𝛿 =0, we see as well that ◦ex = 1 𝜇, which again makes sense, since as𝜇 increases, life expectancy should decrease. Turning to insurance, the constant forces of interest and mortality lead to Ax = ∫ ∞ 0 e−𝛿te−𝜇t𝜇dt= 𝜇 𝜇 +𝛿. (8.28) Note, as a check, that when𝛿 = 0 we getAx = 1. This is obvious, since the insured is sure to die sometime and receive 1 unit, which has a present value of 1 under a zero interest rate. SOME EXAMPLES OF EXACT CALCULATION 127 Thisformulahasaninterestingalternatederivationwhichavoidsanyintegration.Suppose you have wealthAinvested at a constant force of interest𝛿 and each instant of time, you use theinterestearningsof 𝛿Atobuyaninstantofterminsurance,whichwillcostyou 𝜇 perunit. So the amount of insurance purchased at each unit of time will be (𝛿∕𝜇)A. At death you will have the principle ofAplus the insurance benefit, so to make the total wealth at death equal to 1, we want (𝛿∕𝜇)A= 1−A. Solving forAgives precisely the above quantity. For a 1-unit whole-life policy, with level premiums payable continuously for life, the annual rate of premium payment is given by P= Ax ax = 𝜇, and the reserve on such a policy at any timetis given by Ax+t −Pax+t = 0, asweseefromsubstitutingfromabove.Infact,thisisjustthecontinuousanalogueofExample 6.2.Sincethecostofprovidingtheinsurancedoesnotincreasewithtime,theamountcollected atanyinstantisexactlywhatisneededatthatinstant,andthereisnoaccumulationofareserve. 8.10.2 Demoivre’s law Another extremely simple mortality assumption states that for all agesx, tpx = 1− t 𝜔−x,0 ≤ t≤ 𝜔−x. We have already encountered this law in Example 8.1. It is known asDemoivre’s law, named after Abraham Demoivre, an eighteenth-century mathematician, perhaps best known forhistrigonometricidentities.Heisreputedtohaveusedtheaboveformulainmakingsome life table calculations. We can look upon his reasoning as follows. We know that for fixedx, the function tpx decreases from a value of 1 att= 0t oav a l u eo f0a tt= 𝜔. Equipped with nootherinformationatall,Demoivremadethesimplestpossibleassumption,namelythatthe graph of the function was a straight line. It is obviously far too simplisticto be representative of actual mortality, but we will nonetheless investigate its mathematical consequences. We get a very simple expression for life expectancy: ◦ex = ∫ ∞ 0 ( 1− t 𝜔−x ) dt= 1 2(𝜔−x). That is, at any age, one can expect to live one-half of the maximum remaining lifetime. Similarly,takinganupperlimitof nintheaboveintegral,wecalculatethe n-yeartemporary life expectancy asn−n2∕2(𝜔−x) = n( n∕2px ). That is, we multiply the maximum possible future lifetime over the nextnyears by the probability of living to the middle of this period. The force of mortality takes a particularly simple form. 𝜇x(t) =− d dt tpx tpx = 1 𝜔−x−t,0 ≤ t<𝜔 −x 128 CONTINUOUS PAYMENTS whichverifiesExample8.1,wherewedidtheinversecalculation.Thisatleasthasthedesirable feature of being increasing. This leads to tpx𝜇x(t) = 1 𝜔−x = qx,0 ≤ t<𝜔 −x. WenoticedabovethatthisformulaheldunderUDDforaninteger xandtintheinterval(0,1). Under Demoivre’s law, it holds for allxandt. As a consequence, Ax = 1 𝜔−x∫ 𝜔−x 0 v(t)dt= 1 𝜔−xa(1𝜔−x;v). (8.29) Once again, we can check that under zero interest, we getAx = 1. Similarly, Ax(1n) = 1 𝜔−xa(1n;v). A variation of Demoivre’s law, which we will call themodifiedDemoivre’slaw ,i sg i v e n by 𝜇x(t) = 𝛼 𝜔−x−t,0 ≤ t<𝜔 −x−t, for some𝛼> 0. Our original form was with𝛼 = 1. Using (8.16), this means that tpx = ( 1− t 𝜔−x )𝛼 ,0 ≤ t<𝜔 −x, which leads to a simple life expectancy formula, ◦ex = ∫ 𝜔−x 0 ( 1− t 𝜔−x )𝛼 dt= 𝜔−x 𝛼 +1. 8.10.3 An example of the splitting identity The following is an example showing the use of the splitting identity in the continuous case. Example 8.5 Suppose that the force of mortality is given by 𝜇(x) = {0.02, 30 ≤ x≤ 50, 1 100−x,5 0 ≤ x< 100. The force of interest𝛿 is a constant 5 %. CalculateA30. Solution. By the splitting identity, A30 = A30(120)+v(20)20p30A50. FURTHER APPROXIMATIONS FROM THE LIFE TABLE 129 Proceeding as in the previous examples of this section, A30(120) = 𝜇 𝜇 +𝛿(1−e−20(𝜇+𝛿)), A50 = 1 50 ( 1−e−50𝛿 𝛿 ) , v(20)20p30 = e−20(𝜇+𝛿). Substituting for𝜇 and𝛿, we getA30 = 0.3058. 8.11 Further approximations from the life table Suppose you want to approximate the force of mortality given only a life table. For ages that are not integers, the UDD assumption gives an easy answer. For an integerxand 0< t < 1, we obtain from (8.20) that 𝜇(x+t) ≐ q x 1−tqx . Suppose now we want to approximate𝜇(x) wherexis an integer. We could taket= 0i nt h e above, to get an estimated value of𝜇(x)a sj u s tqx, but we could also replacexby x−1 and take t= 1 to get an estimated value of𝜇(x)a sqx−1∕(1−qx−1). These are unlikely to be the same.Itiseasytoseewheretheproblemlies.FromUDD,thefunction tpx ispiecewiselinear asafunctionof tandthederivativewillnotexistatintegervaluesof t.Weareineffectgetting the different right and left hand forces of mortality at the integer points. To get a unique answer we need to consider an interval containingxas a interior point, such as (x−1,x+1). One such approach is to note that p x−1px = 2px−1 = e−∫ 2 0 𝜇x−1(t)dt. We now approximate the integral by one of the simplest methods for doing so, namely the mid-point rule. This simply says that to approximate an integral of a nonnegative function overanintervalyoutakethevalueofthefunctionatthemidpointoftheintervalandmultiply by the length of the interval. (This approximation is exact when the function is linear.) With this method, the right hand side above is approximated bye −2𝜇(x), and solving gives the formula 𝜇(x) ≐ −logpx−1 +log px 2 . (8.30) (See Exercise 8.24 for a check on the accuracy of this approximation.) This can be furthered simplified by using the fact that for small values ofr, −log(1−r) is very close tor, which leads to the approximation 𝜇(x) ≐ q x−1 +qx+1 2 , (8.31) a average of two values ofqat surrounding ages, which has a certain intuitive appeal. We now return to developing the three-term Woolhouse formula introduced in 7.5, which offers an alternative method to the𝛼(m)−𝛽(m) method for fractional annuity computations. 130 CONTINUOUS PAYMENTS FormotivationwerevisitFormula(7.13)whicharosefromtheassumptionthatforallintegers x,and0 ≤ s≤ 1,thequantity y x(s)isoftheform1 −aswherea= 1−vpx.Thisassumption, together with constant interest, implies that spx = e𝛿s(1−as), so that d dsspx = e𝛿s[𝛿(1−as)−a], andat s= 0thisderivativeequals 𝛿 −a.Thiswillbepositivewhen 𝛿> a= 1−vpx whichis equivalent after some manipulation to px > e𝛿(1−𝛿). This explains the inconsistency observed in Example 7.5. If the above condition holds, and we assume the piecewise linearity ofyx(s) as given above, the positivity of the derivative at s= 0 means thatspx actually takes values greater than 1 for sufficiently small values ofs! Writing this in terms of𝓁x we can interpret it fancifully as meaning that people who have died are coming back to life in even greater numbers than were there before, causing the life annuity to be worth more than the one with fixed payments. Suppose we instead make the next simplest assumption. Assume thaty x is a quadratic on [0,1] so that for some constantsa,b, yx(s) = 1−as−bs2,0 ≤ s≤ 1. (8.32) This is the only assumption we make now. We do not need constant interest. Chooseanyinteger mandinvokewell-knownformulasforthesumofconsecutiveintegers (as we did in Chapter 7) as well as the formula for the sum of their squares. These result in m−1∑ i=0 i m = m−1 2 , m−1∑ i=0 ( i m )2 = m−1 2 − m2 −1 6m . We now follow the method introduced in Section 7.1 to compute the relevant vectoru.I t follows from (8.32) that u0 = 1 m m−1∑ j=0 yx(j∕m) = 1−(a+b) (m−1 2m ) +2b ( m2 −1 12m2 ) . Next, calculate the coefficientsa+band 2b. From (8.32) a+b= 1−yx(1) = (∇ 1∞)0. (8.33) using the notation of (2.14) with respect to the discount functionyx. STANDARD ACTUARIAL NOTATION AND TERMINOLOGY 131 Wenowconsidertheinterestandsurvivorshipforceofdiscount 𝜆x(s) =− y′ x (s)∕yx(s),and let 𝝀x denote the vector whose entry at indexkis 𝜆x(k). From (8.32) y′ x (s) =− a−2bs, so that assuming differentiability at integer points, 2b= y′ x (0)−y′ x (1) =− 𝜆x(0)+yx(1)(𝜆x(1) =− (∇𝜆x)0. (8.34) From (8.33) and (8.34) u0 = 1− (m−1 2m ) (∇1∞)0 − ( m2 −1 12m2 ) (∇𝜆x)0. Similarly we can verify that uk = 1− (m−1 2m ) (∇1∞)k − ( m2 −1 12m2 ) (∇𝜆x)k. Then by invoking (7.2) and (2.15) we arrive at the three-term Woolhouse formula. ̈a(m) x (c) ≐ ̈ax(c)− m−1 2m ̈ax(Δc)− m2 −1 12m2 ̈ax(𝝀x ∗Δ c). (8.35) The particular form in the case of level payments is ̈a(m) x (1n) ≐ ̈ax(1n)− m−1 2m (1−v(n)npx) − m2 −1 12m2 (𝜆x(0)−𝜆x(1)v(n)npx ). A direct calculation from (8.35) is not practical, as normally we will be computing from the life table and will not have available the exact values of𝜇(x) needed to compute𝝀x. However we can simply use the approximations given by (8.30) or (8.31) in order to derive numerical results. Formula(8.35)seemstoavoidforthemostparttheinconsistencyofalifeannuityhaving a higher present value than the corresponding fixed interest period annuity, but this can still arise in extreme cases. See Exercise 8.27. 8.12 Standard actuarial notation and terminology We first review some terminology. Policies where benefits are paid at the moment of death, and premiums are paid continuously are sometimes referred to asfully continuous.Policies where benefits are paid at the end of the year of death, and premiums are paid yearly, are sometimesrefereedtoas fullydiscrete.Neitherofthesearerealistic.Thenormallifeinsurance contract involves benefits at the moment of death, and premium payable yearly (or possibly monthly, quarterly, etc). These are referred to assemi-continuous. 132 CONTINUOUS PAYMENTS The single premium notation for standard insurance and annuity contracts is the same as in the yearly case, except, as we have already introduced,A is replaced byA and ̈a is replaced bya. For annual premium contracts, the rules given in 5.8.2 apply. When premiums are payable continuously at a level rate, the same notation is used withPreplacingP. So, for example, r hP(Ax:n)denotesthenetannuallevelpremium,payablefor hyears,fora,1-unit n-year endowmentinsuranceon( x)withdeathbenefitspayableatthemomentofdeath.(Note that we do not omit the single premium symbol, as it starts withArather thanA.) r P(A 1 x:n ) denotes the annual rate of premium payment, when premiums are paid contin- uouslyatalevelratefor nyears,fora1-unit n-yearterminsuranceon( x)withbenefits payable at the moment of death. SymbolsforreservesfollowtherulesinSection6.9,whichsayessentiallythatyousimply takethepremiumsymbolandreplace PbyV,or PbyV,andmovethedurationsymboltothe top rather than bottom left. So, for example, thetth year reserve for the two above policies would be denoted byh t V(Ax:n) andV(A1 x:n ), respectively. TheassumptionofExercise7.13issometimesreferredtoasthe constantforceassumption since it implies that the force of mortality is constant overeach year, as shown by Exercise 8.21 below. It should not be confused with the assumption of a constant force over the entire span of life, as we discussed in Section 8.10. Notes and references The derivation of (8.18) uses a Riemann–Stieljes integral (see Rudin 1976, Theorem 6.17). For an alternate derivation of the general Woolhouse formula (which follows the original derivationofW.Woolhousein1859)seeDickson etal. (2013),AppendixB.2.InSection5.13 of that work, the author’s take a known function for𝜇x and show that the three-term formula gives remarkably close approximations to the true value. See also Exercise 8.26 for more confirmation of this. Exercises Type A exercises 8.1 Alifeannuityon( x)providesforbenefitsmadecontinuouslyfor2years,providedthat (x) is alive. The annual rate of payment at timetis c(t), defined by c(t) = {t,0 < t< 1, 1, 1 < t< 2. Theinterestrateis0, qx = 0.3and qx+1 = 0.4.Findthepresentvalue,assumingUDD. 8.2 Given qx = 0.2, qx+1 = 0.25, i= 0.25 and a vectorc = (100,300), calculate ax(c), assuming UDD. EXERCISES 133 8.3 You are given thatq40 = 0.1,q41 = 0.2, the rate of interest is a constant 25% and that UDD holds. Find the present value of a 2-year endowment insurance on (40), paying 1000 at the moment of death if this occurs before age 42, and 1000 at age 42 if (40) is alive at that time. 8.4 Suppose that q 50 = 0.2. Assume UDD. (a) Calculate𝜇50(0.2). (b) Given that the interest rate is 25%, find the single premium for a 1-year term insurance policy on (50) that pays 1 unit at the moment of death should this occur within 1 year. 8.5 Suppose that the force of mortality is given by 𝜇 x = B(1.09)x and (a) Find a value forB such the resulting mortality table is the illustrative table of Chapter 3. (b) Using this value ofBcompare the true value of𝜇 50 with the estimated value using formula (8.30). 8.6 Alifeannuityon(80)providesforbenefitsmadecontinuouslyfor2years,attheannual rate ofc(t) at timet, provided that (x) is alive. Suppose thatc(t) = t,0 < t< 2, and the interest rate is 0. Find the present value under each of the following assumptions. (a) q 80 = 0.09,q81 = 0.12, and UDD holds. (b) The force of mortality is given by 𝜇80(t) = 1 10−t,0 < t< 10. 8.7 For a certain mortality basis,q50 = 0.2. Suppose that the force of mortality𝜇50(t)i s changed to a new force of mortalitŷ 𝜇given by ̂ 𝜇50(t) = 2𝜇50(t)−0.1, 0 < t<𝜔 −50. Find the new value ofq50. 8.8 The force of mortality is a constant 0.04 and the force of interest is a constant 0.06. A term insurance policy has a level death benefit of 1 payable at the moment of death if this occurs within 40 years. Level annual premiums are payable continuously for 20 years at the annual rate of of𝜋.F i n d( a )𝜋,( b ) 10V. 8.9 Suppose that the force of mortality for ages over 60 is given by 𝜇60(t) = { t,0 < t< 1, 1, 1 < t. Find the probability that (60) will die within 2 years. 134 CONTINUOUS PAYMENTS 8.10 Suppose mortality follows Demoivre’s law with𝜔 = 110 and that the interest rate is 0. A term insurance policy on (60) has a level death benefit of 1000 payable at the momentofdeathprovidedthisoccurswithin40years.Netlevelpremiumsarepayable continuously for 20 years at the annual rate of𝜋. (a) Find𝜋. (b) Find 10V by both the prospective and retrospective methods. 8.11 Two actuaries, A and B, agree that the probability that a female age 60 will die within 10 years is 0.36. (a) ActuaryAdecidesthattheforceofmortalityformalelivesis1.5timestheforceof mortality for females lives, at all ages over 60. What is the probability that a male age 60 will die within 10 years, under A’s assumption? (b) Actuary B disagrees and decides that the force of mortality for males lives is obtained by adding a constant of 0.01 to the force of mortality for female lives, at allagesover60.Whatistheprobabilitythatamaleage60willdiewithin10years under B’s assumption? 8.12 An insurance policy provides for death benefits payable at the moment of death. The amount payable for death at timet is e 0.08t, for allt> 0. The force of mortality is a constant 0.06 and the force of interest is a constant 0.04. Net level premiums are payable at a constant rate for life. Find the rate of premium payment. Type B exercises 8.13 Show that under UDD, for an integerxand 0< t< 1,𝜇x(t) = qx∕(1−tqx). 8.14 The central rate of mortality at agexis defined by mx = dx ∫ 1 0 𝓁x+tdt . (a) Show thatmx can be expressed as a continuous weighted average of values of the force of mortality. (b) Show that under UDD,mx = 𝜇x ( 1 2 ) . 8.15 A whole-life insurance, with net level premiums for life payable continuously at the annual rate of𝜋, provides for death benefits of e𝛾t at the moment of death, for death at timet. Given that the force of mortality is a constant𝜇 and the force of interest is a constant 𝛿, such that𝜇 +𝛿>𝛾 , find𝜋 in terms of𝛾,𝜇 and𝛿. 8.16 A 1-year deferred, 1-year life annuity on (x) provides for continuous payments from time1totime2provided( x)isalive.Therateofpaymentattime1 +tise0.1t,0 < t< 1. This is purchased by a single premium. For death during the first year, the single premium is returned without interest at the moment of death. You are given that q x = 0.1, qx+1 = 0.2, and that the force of interest𝛿 is a constant 0.10. Assuming UDD, find the single premium. EXERCISES 135 8.17 An n-year term insurance policy on (30) has a constant death benefit of 1, payable at themomentofdeath.Premiumsarepayablecontinuouslyataconstantratefor nyears. MortalityfollowsDemoivre’slawwith 𝜔 = 100,andtheinterestrateis i= 0.Findthe rate of premium payment as a function ofn. 8.18 Theforceofinterestisaconstant 𝛿.Theforceofmortalityisaconstant 𝜇1 forthefirst nyears, and a constant𝜇2 after nyears. Derive expressions forax and Ax in terms of 𝜇1,𝜇2,𝛿 andn. 8.19 Suppose Demoivre’s law holds and the force of interest is a constant𝛿.F i n da n expression forax(1n) in two ways. First use formula (8.9) directly, and second apply (8.23) to the expression given forAx(1n) under the same assumptions. Verify that as𝛿 approaches 0, you get the formula for then-year temporary life expectancy. 8.20 A ‘sawtooth’ life insurance policy on (x) provides for a death benefit oftpaid at death if this occurs at timek+t, wherek is an integer and 0< t< 1. Assume UDD and constant interest. Show that the net single premium for this policy is equal to𝛽(∞)Ax. 8.21 Supposethattheforceofinterestisaconstant 𝛿 andtheforceofmortalityisaconstant 𝜇. Consider ann-year endowment insurance with net premiums payable continuously at the constant annual rate of𝜋. Death benefits ofbt are payable at the moment of death for death at timet< n. There is a payment of 1 at timenif the insured is then alive. (a) Supposethat b t = 1forall t< n.Usethereservedifferentialequationtoshowthat tV = (𝜋 −𝜇)et(𝜇+𝛿) −1 𝛿 +𝜇 , 𝜋 = 𝜇 + 𝜇 +𝛿 en(𝜇+𝛿) −1. Give an intuitive explanation of these results. Verify that these agree with the formulas given at the end of Section 8.8. Show that the reserve formula is a continuous version of formula (6.15). (b) Suppose now thatb t = 1+tV for allt< n. Use Thiele’s differential equation to derive formulas fortV and𝜋. 8.22 Show that under the assumption of Exercise 7.13, the force of mortality is constant over each year. That is,𝜇x(t) = 𝜇x(s) wherexis a nonnegative integer andsand tare between 0 and 1. 8.23 Give an alternate derivation of (8.12) by using the continuous endowment identity, as given at the end of Section 8.8, and (8.23). First prove it forc = e0. 8.24 Suppose that modified Demoivre’s law holds with𝛼 = 2. If𝜇x = 30 what is◦ex? 8.25 Suppose that the force of mortality is given by 𝜇(x) = B(1.09)x. (a) Find a value forB such the resulting mortality table is the illustrative table of Chapter 3. 136 CONTINUOUS PAYMENTS (b) Using this value ofB compare the true value of𝜇(50) with the estimated value using formula (8.30). 8.26 Suppose that the force of mortality is a constant 0.2 and the force of interest is a constant 0.1. Computëa(4) x (11) by four methods. (a) formula (7.9), (b) formula (7.13), (c) formula (8.35) and (d) exactly. Compare the results. 8.27 Find an example of some life table values and an interest rate, such that using (8.35) together with ( 8.30), yields an inconsistent result. Hint: Choose life table values so that the estimated value𝜇 x+1 is much higher than the estimated values of𝜇x. 8.28 A life is subject to a constant force of mortality of 0.15. The force of interest is a constant0.10.Aninsurancecontractonthislifeprovidesforadeathbenefitof e.05t for death at timet, with level premiums payable continuously for life. (a) FindtV. (b) Write down Thiele’s differential equation for the reserve, and verify that your answer to (a) satisfies this equation. Spreadsheet exercise 8.29 Modifypreviousspreadsheetstohandlethecaseofdeathbenefitspayableatthemoment of death. 9 Select mortality 9.1 Introduction In this chapter we will discuss a refinement to our basic model. In previous chapters we assumed that for a sufficiently homogeneous group – North American male nonsmokers, for example – mortality rates are a function of attained age only. This may be a reasonably accurate assumption for the totality of people in this group, but it need not apply for certain subsets. An insurer, after all, is not interested in the mortality of the general population, but only the subset of those who buy policies. Observed data show that for purchasers of life insurance, mortality depends not only on age, but onduration since the policy was purchased. To see why this should be true, let us compare two people both age 60, alike in all respects except that A just recently purchased an insurance policy, and B purchased a policy 1 year ago at age 59. During the next year, we would expect that A will have a higher probability of living to age 61 than B will. This arises from the fact that the insurer does not have to accept everyone who applies for a policy. Applicants can be requested to undergo medical exams, or to furnish information regarding their health, lifestyle and other factors, in order to verify that they are reasonable risks. Individual A has just been so certified. Individual B, on the other hand, was certified as a good risk 1 year ago, but this condition could have changed. B may have contracted a fatal disease, or taken up life- threatening habits. Moreover, if we consider a third individual, C, also age 60, similar to A and B, except that he/she was sold a policy at age 58, his/her chance of living over the next year can be expected to be even less that that of B. This person has had 2 years in which to deteriorate. Thestandardactuarialmethodtohandlethissituationistousethesamenotationaldevice that we introduced in Section 4.3.2. We keep the convention that subscripts denote attained age, but put square brackets around the age that the policy was issued, in order to separate age and duration. Therefore, the symbol sq[x]+t denotes the probability that a person now age x +t, who was sold insurance at agex, will die within the nexts years. This is known Fundamentals of Actuarial Mathematics, Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 138 SELECT MORTALITY as aselect mortality rate, as it incorporates the effects of the insurer’s ability to select. We define sp[x]+t = 1−sq[x]+t. As usual we omit left subscripts of 1 forq and p. Our discussion above showed that q[60] ≤ q[59]+1 ≤ q[58]+2 …, and, in general, q[x]+t ≤ q[x−s]+t+s, (9.1) for all nonnegativex andt and 0≤ s ≤ x. 9.2 Select and ultimate tables To fully model this phenomenon we would need a separate mortality table for each agex to cover policies issued at that age. Of course the tables would get shorter with increasing issue age. The agex table would consist of the𝜔−x entries {q [x]+t,t = 0,1, …,𝜔−x −1}. Observationsshow,however,thattheselectioneffectsdecreasewithtimeandcanbeassumed to wear off after a certain duration, known asthe select period. If the select period isr,w e can expect that two individuals of the same age, who have been issued insurancer or more years ago, will exhibit the same mortality rates, even if the issue age is different. This allows for some simplification. We can revert to the symbolq y to mean the probability that a person now agey who was issued insurancer or more years ago will die within a year. In other words, we postulate that equality will hold in (9.1) whent ≥ r, so we can remove the square bracket and denote the common value byqx+t.T h er a t e sqy are known asultimate rates, and the resulting life table is known as aselect and ultimate table. A common choice for select periodis15years,althoughsomerecenttableshaveusedasmuchas25.Thesimpleexample below has a select period of 2, but that is enough to illustrate the basic idea. xq [x] q[x]+1 qx+2 x +2 60 0.10 0.12 0.15 62 61 0.11 0.14 0.17 63 62 0.12 0.15 0.18 64 Tofindthemortalityratesapplicabletoapolicyissuedatage x,westartattheleftcolumn, readacrossuntilwegettotheultimatecolumn,andthenreaddown.So,forexample,therates usedforapolicyissuedatage60wouldbeinorder q [60], q[60]+1, q[60]+2 = q62, q63, q64,…, whichinthisexampleare0.10,0.12,0.15,0.17,0.18,….Wecanpicturetheflowofmortality rates as several streams, all running into the same river, which is the ultimate column. Therefore, we do not need a complete separate table for each age, but onlyr +1 columns. CHANGES IN FORMULAS 139 Tableswithaselectperiodof0years,whichispreciselywhatwehavebeendiscussingin previous chapters, are known asaggregate tables. Some people prefer to construct a select and ultimate table of𝓁x, although this is not really necessary, for as we have seen, one can always work directly with the values ofqx. The process is straightforward, although it requires a little care. One completes the ultimate column first, proceeding as in Chapter 3 to calculate values of𝓁 x starting with an arbitrary value for𝓁r which is the first entry in this column. One then simply works backwards to complete the other entries, using the recursion 𝓁[x]+k = 𝓁[x]+k+1 1−q[x]+k . Forexample,intheabovetable,if 𝓁63 = 5000,then 𝓁[61]+1 = 5000∕0.86 = 5814and 𝓁[61] = 5814∕0.89 = 6353. 9.3 Changes in formulas To change our previous formulas to incorporate select mortality is usually just a matter of inserting a square bracket around the issue age. We will look at some of these in detail. Chapter 3 The multiplication rule (3.6) becomes s+tp[x] = sp[x] tp[x]+s. (9.2) Chapter 4 We will denote the interest survivorship discount function for a policy issued at agex byy[x]. From (9.2) we derive y[x](k,k +n) = y[x](k +n) y[x](k) = v(k,k +n)np[x]+k. (9.3) Takingn = 1, the recursion formula (4.5) now becomes y[x](k +1) = y[x](k)v(k,k +1)p[x]+k. It is important to note that with select mortality, (4.6) and (4.7) no longer hold. As mentioned, we need a separate table for each issue age, and there is no longer any necessary relation between the ages. Note, for example, that y [x+k](n) = v(n)np[x+k], and even ifv(k,k +n) = v(n) this is not the same as the right hand side of (9.3) unlessk is greater than or equal to the select period. 140 SELECT MORTALITY The subscript of [x]+k on annuity symbols is used in the same way as introduced in Section 4.3.2, except that now, modification to the mortality as well as the interest is needed. That is ̈a [x]+k(c) = ̈a(c;y[x] ◦ k) = 𝜔−x−k−1∑ j=0 cjv(k,k +j)jp[x]+k. Notethat ̈a[x]+k(c) = ̈ax+k(c),if k isgreaterthantheselectperiod,andinterestisconstant. With non-constant interest however, these are not necessarily equal even ifk is greater than the select period, as we observed in Chapter 4. Chapter 5 In the third formula in (5.1),qx+k becomes q[x]+k. The subscript [x]+k is used for insurances in the same way to that noted above for annuities. That is A[x]+k(b) = 𝜔−x−k−1∑ j=0 bjv(k,k +j +1)jp[x]+k q[x]+k+j. Chapter 6 Nochangesarenecessaryinthischapterexcepttonotethatthesubscriptsof[ x]+k inreserve formulas are modified as we noted for Chapters 4 and 5 above. Chapter 8 We define 𝜇x(t) = lim h→0 hq[x]+t h = lim h→0 𝓁[x]+t −𝓁[x]+t+h h𝓁[x]+t . (We are following standard notation here. The square bracket is not necessary since thex is already separated from thet.) The basic identity (8.16) can then be written tp[x] = e−∫ t 0 𝜇x(r)dr. Of course, if the select period is 0 then𝜇x(t) is just equal to𝜇(x +t) and agrees with the notation that we have already introduced in Chapter 8. Spreadsheets The interested reader may want to adapt the previous spreadsheets that have been introduced to incorporate a select and ultimate table. However, without doing this fully, one can still do calculations directly on the existing spreadsheet for each particular problem. For a policy issued at age (x) on a table with a select period ofr years, one simply replaces the entries of q x+t withq[x]+t for t = 0,1, …,r −1. PROJECTIONS IN ANNUITY TABLES 141 9.4 Projections in annuity tables There are other situations where a two-dimensional approach to mortality rates is called for. One such example is annuity tables. We need however a different formulation than that presented above. The select tables we have described would not normally be used for annuities. Purchasers of annuities are generally people who expect to live longer than the average or they would be unlikely to enter into such a contract. In any event the insurer is not worried about deteriorating mortality with duration. The problem here is exactly the opposite, namely, people may live longer than the tables predict, requiring that more must be paid out in benefits than the premiums allow for. This has proved to the case in recent years, as advances in medical and public health knowledge, better lifestyles and other factors have led to continually improving mortality, causing issuers of annuities as well as pension plans, to be confronted with what has been termedlongevity risk. In addition to age,year of birth becomes an important variable when pricing annuities. A 60 year old born say in 1950 could beexpectedtohavelesschanceofdyinginthenextyear,thana60yearoldbornin1930,due to improvements in mortality that have come about over the 20 year period. In place of the rates q x, we would like a two variable functionqx,n where x denotes age andn denotes year of birth. We will alter this notation somewhat in order to point out the parallel nature of this situationwithouroriginaldiscussionofselectmortality.Supposewehavepickedabaseyear, say the year a mortality study was completed. We will now letq [x]+t denote the probability of dying within 1 year, for a person now agex +t who was agex in the base year. In other words we have just extended the select notation by allowingt to denote time elapsed since some particular event, which could be anything at all. In our original application, the event wasobservationforinsurancepurposes.Forannuitytables,itistheyearofthemortalitystudy that produced the table. In this case, we have in place of (9.1) q [x]+t ≥ q[x−s]+t+s, since,letting b denotethebaseyear,thetermontheleftreferstosomeoneborninyear(b −x) while that on the right refers to some one born in a later year (b −x +s). There is a however a major problem in constructing a table to show such rates. Suppose for example that a mortality study of annuitants was completed in the year 2000. The result would be a table showingq [x], the one year mortality rates for people agex in the year 2000. Now suppose we want to know whatq[60]+1 is. This would be the rate for a person age 61 who was born in 1940, but in the year of the study no such person would ever have existed. The rateq [60]+1 (as well as all rates witht > 0) could not be based purely by a statistical analysis of past mortality data, but would involve the prediction of future trends in mortality improvement. This is rather difficult task, and present actuarial usage involves rather simple tools. The usual method is to assign a yearly projection factorr y for each age, which represents the reduction in mortality expected at that age for each year. That is, it is assumed that, q [x]+t = q[x+t](1−rx+t)t. Following is an example taken from an actual table known as the Uninsured Pensioner (UP) 1994 Table with projection scale AA. 142 SELECT MORTALITY Example 9.1 The UP 1994 table shows male mortality rates for age 60–62 of 0.008576, 0.009633, 0.010911, respectively and projection factors of 0.016, 0.015, 0.015, respectively. Amaleisage60intheyear 2010. Usingthistablefindtheprobabilitythathewilllivetoage 63.Comparethiswiththeprobabilitythatamalewhowasage60in1994wouldlivetoage63. Solution. The required probability is [(1−(0.008576)(0.984) 16][(1−0.009633)(0.985)17][(1−0.010911)(0.985)18] = 0.9778. For the male age 60 in 1994, the probability would be [(1−(0.008576)][(1−0.009633)(0.985)][(1−0.010911)(0.985)2] = 0.9716. There is no reason to believe that mortality improvement will stop, so the concept of a select and ultimate table does not appear in this context. Therefore it is not usual in the actuarial literature to view the table that we have described as a select table. Rather it is viewed as a collection of separate tables, one for each agex in the base year. Each of these separate tables is often referred to as agenerational table, as it traces the mortality of people all born in the yearb −x. In the construction above each generational table would appear as the row of the select table as described. Each column of this select table can be considered as a version of the original table projected to a future year, and is referred to as thestatic table for that year. It practice, insurers often simplify calculation by making use of the appropriate static table for setting annuity rates in a particular year, rather than using the more accurate generational tables, which would involve a separate table for each age. 9.5 Further remarks There are many other situations in which the concept of select mortality applies. For another example involving rates decreasing with duration, suppose we are studying the effects of a certain treatment for cancer patients. This could depend both on age and the duration since treatment. In this case, if a person was treated a year ago, and is still alive, there is some indication that the treatment is working, and that such person is more likely to live than someone of the same age who has just received the treatment. Exercises 9.1 A person now age 62 was sold insurance 1 year ago at the age of 61. Mortality is given by the table at the beginning of the chapter. Find the probability that this individual will die between the ages of 63 and 65. 9.2 A4-yearterminsurancepolicysoldtoapersonage61providesfor1000paidattheend oftheyearofdeathshouldthisoccurbeforeage65.Levelannualpremiumsarepayable for 4 years. The interest rate is a constant 5% and mortality is given by the table at the beginning of the chapter. (a) Find the premium. (b) Find 2V. EXERCISES 143 9.3 Explain in words what the following indicates: (6p[30]+10)(8q[30]+16). 9.4 For thetablegiven inSection 9.2, construct the corresponding table of𝓁[x]+t, given that 𝓁62 = 5000. 9.5 An annuity mortality table, with projection factors for improvement, constructed in the year 2000, shows q70 = 0.02, q71 = 0.03 q72 = 0.04, r70 = 0.02, r71 = 0.018, r72 = 0.016. Find the probability that a person age 70 in the year 2010 will die between the ages of 72 and 73. Spreadsheet exercises 9.6 Consider a select and ultimate version of the sample table, with a 15-year select period, as follows. Forx ≥ 50, andt = 0,1, …,14wetake q[x]+t = 1−(1.00001)15−te−0.00005(1.09)x+t . Interestratesareaconstant6%.Fora20-yearendowmentinsuranceon(50)withalevel deathbenefitof100000andapureendowmentof100000atage70,purchasedbylevel net premium for 20 years, calculate (a) the premium, (b) 10V. 9.7 Using the same table as in Exercise 9.6, calculate (a)e[60],( b )e[50]+10. 10 Multiple-life contracts 10.1 Introduction Inpreviouschapterswehavestudiedinsuranceandannuitycontractssoldtoasingleindivid- ual. There are many situations when such contracts are sold to a group of several people. We will concentrate on the case of two lives, the most common arrangement, which is sufficient to exhibit most of the complications. The following are a few of the more usual situations. A married couple may wish to purchase an annuity that pays income while either of them is alive. They may also want an insurance policy that pays benefits on the second death, which in some jurisdictions enables themtopassonassetswithoutpayingestatetaxes.Twobusinesspartnersmaydesireapolicy thatpaysadeathbenefituponthefirstdeathofthetwo.Apersonmayarrangeforanannuityto bepaidtoadependent,wherepaymentsbeginonlyonthedeathofthesupportingindividual. Wewilldiscusstheseandmoreinthefollowingsections.Forsimplicityweassumeaggregate mortalityinthischapter.Itisnottoodifficulttomodifytheresultsfortheeffectsofselection. 10.2 The joint-life status Consider a pair of lives, (x) and (y). Suppose that we consider the pair to be in a state of survival when both of them are alive. In other words, the pairfails on the first death. Such a pair is commonly known as ajoint-lifestatus. An annuity sold on a joint-life status would provide income as long as the status survived, that is, as long as both individuals were living. Annuity benefits would stop upon the first death. An insurance policy sold on a joint-life status would pay a death benefit upon the first death. Whileitispossibletoconstructlifetablesinthiscase,itisawkward,anditisbesttowork directly with probabilities. The basic quantity we need is the probability thatbothof (x) and (y) will be alive intyears. We will denote this probability by tpxy. FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial THE JOINT-LIFE STATUS 145 We will now derive an expression fortpxy that can be calculated from the life table for single lives. The method discussed will be well known to the reader who is familiar with the concept of independence in probability theory (see Section A.2). Suppose, for example, that tpx = 0.9, tpy = 0.8. Imagine that we start by observing 100 pairs of lives, where one is agex and the other is agey. Of these 100 pairs, we can expect that in 90 cases, the life agex will be alive int years. Out of these 90 cases we can expect that 80% of the time the life ageywill survivetyears. This leaves on average 72 pairs of the 100 where both lives survive, so we should havetpxy =0.72. It is instructive to derive this in another way by looking at deaths rather than survivors. We expect 20 deaths altogether from the agey individuals. On average, two of these agey deaths should occur among the 10 cases where the agexindividual dies. This leaves 18 age ydeaths among the 90 cases of agexsurvivors, and therefore 90−18 = 72 cases where both survive. The particular numbers 0.9 and 0.8 do not matter and the same arguments show that in general we should have tpxy = tpxt py. (10.1) We define tqxy = 1−tpxy, the probability that the joint-life status will fail withintyears, which is the probability that at least one of (x) and (y) will die withintyears. Thereadershouldcarefullynotethat tqxy isnotequalto tqxt qy.Thelatterexpressionisthe probability thatbothlives will die withintyears, which is certainly less than the probability that at least one will die in this time interval. To expresstqxy in terms of single-life rates, we havetqxy = 1−tpxt py = 1−(1−tqx)(1−tqy), which gives tqxy = tqx +tqy −tqxt qy. (10.2) Wenowconfessthatalthough(10.1)and(10.2)arealmostuniversallyused,theargument wegaveaboveisnotquiteaccurate.Itwouldbeperfectlyvalidifthetwoliveswerecompletely independent of each other. This is unlikely to be true, however, for two people who would choose to buy an insurance or annuity contract together, such as a married couple or two business partners. One can expect that their lives are intertwined to the extent that life- threatening occurrences for one would also affect the other. Forexample,supposethatthetwolivesarebusinesspartnerswhofrequentlyflytogether. If one of them were to die in a plane crash, it is likely that the other one would as well. To illustrate the point, we take an extreme case. Return to the example given above, but suppose that the 100 pairs we start with consist in every case of two people who always fly together. We claimed above that there would be on average 2 agey deaths from the 10 cases where the individual agedx died. But if we expect that the lives are likely to die together, we can certainly expect more than this. For example, if all the agexdeaths were from plane crashes, we would necessarily have 10 agey deaths as well. This is extreme, but suppose that we have even 3 instead of 2 ageydeaths. This leaves 17 instead of 18 deaths from the 90 agex 146 MULTIPLE-LIFE CONTRACTS survivorsandwewouldhave tpxy =0.73.Thisshowsthatratherthan(10.1)wewouldexpect for the typical buyers of an insurance or annuity contract that tpxy > tpx tpy. (10.3) Formarriedcouplesthereisanadditionalfactorthatsupports(10.3).Statisticsbearoutthe fact that upon the death of one partner, the survivor, deprived of the care and companionship of the deceased, often tends to die earlier than otherwise expected. This has been referred to as the ‘broken heart syndrome’. For convenience, we will assume (10.1), or equivalently (10.2), unless otherwise stated. The reader should note, however, that most of the key formulas remain true without this assumption. The main use of (10.1) is to conveniently calculate numerical quantities, given the single-life data. Generalizationsofthejoint-lifestatuscanbefoundinChapters17and19.Onthequestion of dependence, see in particular Sections 17.5 and 17.6. 10.3 Joint-life annuities and insurances Consider a joint-life annuity contract payingck each year provided that both (x) and (y)a r e alive. In order for a payment to be made at timekwe need both lives to have survived to that time, and the probability of this is justkpxy. The present value of the benefits, denoted by ̈axy(c), is given by a formula exactly the same as (4.1), exceptkpxy replaces kpx: ̈axy(c) = N−1∑ k=0 ckv(k)kpxy, (10.4)‡ whereN =m i n{𝜔−x,𝜔−y}. We useN in this way throughout the chapter. To verify that it is the appropriate upper bound, note for example that ifx= 40,y= 70 and𝜔 = 110, the last possible payment would be at time 39. At time 40, (70) would not be living according to our model. As with single lives, we will use 1 ∞ to denote a vector with entries of 1, running to durationN−1, and by convention the omission of a benefit vector or function implies that it is 1∞. In the continuous case we have the formula ̄axy(c) = ∫ N 0 c(t)v(t)tpxydt, (10.5) ‡ for the present value of an annuity, with payments made continuously at the annual rate of c(t) at timet, provided both (x) and (y) are alive. We can also consider ajoint-life insurancepolicy that paysb k at the end of the year of the first death of (x) and (y) when this occurs between timekand timek+1. In other words, paymentwillbemadeattime k+1ifbothlivessurvive kyears,butbothdonotsurvive k+1 years. The probability of this is just kpxy −k+1 pxy. LAST-SURVIVOR ANNUITIES AND INSURANCES 147 This probability can also be expressed as kpxyqx+k:y+k, since for such an event to happen, both lives must survive fork years and then one must die in the subsequent year. (A colon is used to separate the two ages in a joint-life status if neededforclarity.)Thislatterexpressioncanalsobederivedfromthejoint-lifeversionofthe multiplication formula, k+1pxy = kpxypx+k:y+k, sinceforboth(x )and(y)tosurvive k+1years,theymustfirstbothsurvive kyears,andthen, being agex+kandy+krespectively, they must both survive one more year. Thepresentvalueofthejoint-lifeinsurance,denotedby Axy(b),canbewritten,analogously to the second and third expressions in (5.1), as Axy(b) = N−1∑ k=0 bkv(k+1)(kpxy − k+1pxy) = N−1∑ k=0 bkv(k+1)kpxyqx+k:y+k. (10.6)‡ Suppose that such a joint-life insurance were sold with annual premiums, based on the premiumpatternvector 𝜌.Premiumswillceaseuponthefirstdeath,sothepremiumpayments form a joint-life annuity. We determine the initial premium𝜋0 in exactly the same way as with a single-life insurance, namely, 𝜋0 = Axy(b) ̈axy(𝜌). (10.7)‡ 10.4 Last-survivor annuities and insurances 10.4.1 Basic results Consider a pair of lives (x) and (y), which we consider to be in state of survival ifeither of them is alive. In other words, the pair fails upon thesecond death. This is known as a last-survivorstatus. Thestandardsymbolforthisis xy,whichdistinguishesitfromajoint-life status.Theprobabilitythatthisstatuswillbeinastateofsurvivalattime tisdenotedby tpxy. To calculate this, we use (A.3), a basic fact from probability theory, already illustrated above in (10.2), which says that the probability that at least one of two events occurs is given by the sum of the probabilities of each occurring, less the probability that they both occur. This gives tpxy = tpx +tpy −tpxy. (10.8) 148 MULTIPLE-LIFE CONTRACTS We lettqxy denotetheprobabilitythat xywillfailbeforetime t.Clearly tqxy = 1−t pxy,so that if our independence assumption (10.1) holds, we can substitute in (10.8) to obtain tqxy = tqxt qy. (10.9) Alternatively,thiscanbederiveddirectlybynotingthatinorderforthisstatustofailwithin t years, we need to have both lives fail withintyears. The probability that (x) fails between timetand times+tis given either bytpxy −s+t pxy or bys+tqxy −t qxy. Assuming independence, the latter expression simplifies to (s+tqx)(s+tqy)−(tqx)(tqy). Formulasforannuitiesandinsurancesbasedonthelastsurvivorstatusareeasilyobtained. Asinthejoint-lifecasewesimplyreplacethesubscript( x)by( xy).So,foranannuitypaying ck at timek, provided that both (x) and (y) are alive, the present value is just ̈axy(c) = M−1∑ k=0 v(k)ckk pxy = ̈ax(c)+ ̈ay(c)− ̈axy(c), (10.10) where the upper limitMnow denotes the maximum of𝜔−xand𝜔−y.The final expression results directly from (10.8). The analogous formula holds for continuous annuities with̄areplacing ̈a. Similarly, the present value of an insurance contract payingbk at the end of the year of the second death of (x) and (y), if this occurs between timekandk+1 is given by Axy(b) = M−1∑ k=0 v(k+1)bk(kpxy −k+1 pxy) = Ax(b)+Ay(b)−Axy(b), (10.11) whereweagainuse(10.8),aswellasthesecondformulain(5.1)toderivethefinalexpression. Formula(10.11)canbeverifieddirectly.Suppose,forexamplethat( x)diesfirst,atatime betweenjandj+1. Then the first term on the right hand side paysbj and the third term pays −bj, so a net amount of 0 is paid on the first death. When (y) dies later at a time betweenk andk+1 only the second term pays andywill get the required death benefit ofbk.T h es a m e typeofreasoningcanbeappliedtoverify(10.10)aswellasmanyothertypesofmultiple-life contracts. These last survivor annuities and insurances are a special case of more general contracts that we will investigate in Sections 10.6 and 10.7. 10.4.2 Reserves on second-death insurances Reserves on last survivor insurances present a particular problem. For a second-death insur- ance, the true reserve at timek depends on the state of the pair. There are three possible cases. (For simplicity in notation we assume constant interest. In the general case one uses the investment discount functionv◦ kin computingAand ̈a.) kV(0) = Ax+k:y+k(b◦ k)− ̈ax+k:y+k(𝜋◦ k) if both lives survivekyears. kV(1) = Ax+k(b◦ k)− ̈ax+k(𝜋◦ k) if (x) only surviveskyears. kV(2) = Ay+k(b◦ k)− ̈ay+k(𝜋◦ k) if (y) only surviveskyears. MOMENT OF DEATH INSURANCES 149 Theabove formulasalsoassumethatthepremium patternremains unchanged on thefirst death.Ifthecontractcalledforreducedpremiumsinsuchanevent,thensuitableadjustments would be required in the second term of the formulas for kV(1) and kV(2). Itmayseempuzzlingatfirstthatthereserveonthiscontractshouldmakeasuddenchange on the first death, but a moment’s reflection will verify that thisis perfectly logical. The pairs for which the first death has already occurred present a greater liability and a higher reserve is needed. Taking a retrospective, point of view, what happens is that there is a transfer each periodfromthefundsofthepairswithbothindividualsstillliving,tothefundsofthosewhere one has already died. The problem is however that there is no reason for a pair of lives to report the first death to the insurer, unless so stipulated in the contract. This means that the insurer might well not beawareofthestateofthevariouspairs.Inthiscaseanappropriatesolutionwouldbetohold a weighted average of the different reserves. That is, for each contract the reserve at timek would be kpxy kV(0) +(kpx −k pxy) kV(1) +(kpy −k pxy) kV(2) kpxy . If the actual mortality experience is close to that of the weights used above, then this method should produce aggregate reserves which are close to what they should be by the more exact method. Note however that it would not be reasonable to base cash values on such a blended reserve figure, which will clearly be higher than kV(0). This would create a definite anti- selection opportunity as those pairs for which both are still living, could cash in the policy and receive an amount which is more than that which they are entitled to. 10.5 Moment of death insurances To handle moment of death insurances for two-life statuses, we need the joint-life analogue of the force of mortality, Definition 10.1 The forceoffailure for the joint-life status (xy) at timetis the quantity 𝜇 xy(t) =− d∕dttpxy tpxy =− d dtlog(tpxy). Theindependenceassumption(10.1)letsuscalculatethisfromthesinglelifefunctions. 𝜇xy(t) =− d dt(log tpx +log tpy) = 𝜇x(t)+𝜇y(t). (10.12) Consider an insurance policy that paysb(t) at timetif the first death of (x) and (y) occurs atthattime.Thepresentvaluewillbedenotedby ̄Axy(b)andgiven,analogouslyto(8.18),as ̄Axy(b) = ∫ N 0 b(t)v(t)tpxy𝜇xy(t)dt. (10.13)‡ 150 MULTIPLE-LIFE CONTRACTS The present value of an insurance payingb(t) at timetif theseconddeath of (x) and (y) occurs at that time is denoted bȳAxy(b). The same reasoning as used after formula (10.11) establishes the analogous formula ̄Axy(b) = ̄Ax(b)+ ̄Ay(b)− ̄Axy(b). (10.14) Aswiththesingle-lifecase,wearefacedwiththeproblemofevaluating ̄Axy fromthelife table. Assume thatbtakes the constant value ofbk over the year running from timekto time k+1. It is natural to ask if the samei∕𝛿 correction that we used for single-life insurances is also applicable to the joint-life case. Recall that this correction came from (8.20), so the question is whether the corresponding statement is true for joint-life statuses. That is, is it truethat tpxy (𝜇x(t)+𝜇y(t)) = qxy for0 < t< 1?AssumeUDDandtherefore(8.20)forsingle lives. Then, for 0< t< 1, tpxy (𝜇x(t)+𝜇y(t)) = tpyqx +tpxqy = (1−tqy)qx +(1−tqx)qy = qxy +(1−2t)qxqy, where we use (10.2) for the last equality. We see that the required condition does not hold exactly, due to the extra term of (1−2t)q xqy. Arguing as in Section 8.6.2, we deduce that ̄Axy(b) = Axy ( i 𝛿 ∗ b ) +R, (10.15) whereb = (b0,b1,…) and R= N−1∑ k=0 r(k)b(k)v(k+1)kpxyqx+kqy+k, (10.16) withr(k) = ∫ 1 0 v(k+1,k+t)(1−2t)dt.Analysisshowsthat r(k) > 0sothat Rispositive,but likely to be quite small and safely ignored in practice. Using thei∕𝛿 correction will slightly understate the value of the insurance, but it is a reasonable approximation.R is the present valueofacontractthatpaysoffonlyif( x)and( y)dieinthesameyear,aneventthatwillhave smallprobabilityformostages.Moreover, since1 −2tispositivefor0 ≤ t< 1 2 andnegative for 1 2 < t≤ 1, the cancellation in the integral will tend to maker(k) small. 10.6 The general two-life annuity contract There is a demand for two-life annuities that are more flexible than ones we described in previous sections. One popular arrangement is for benefits to continue while either person is alive, but for the amount to reduce when only one of the parties is alive. For example, a married couple may wish to provide for an annuity at retirement that will pay 60 000 yearly whilebotharealive,reducingto40000yearlywhenonlyoneisalive.Thisone-thirdreduction is a common one, reflecting the fact that while one person can live more cheaply than two, he/she will need more than half of the amount, since many expenses, such as housing, will THE GENERAL TWO-LIFE ANNUITY CONTRACT 151 not necessarily reduce. Many variations are possible. Benefits need not be symmetric and could vary according to the particular survivor. A common provision in pension plans is that the income stays level as long as the employee is alive, but will continue to the spouse at a reduced level upon the death of the employee. We present a single formula that covers all possibilities. A general life annuity on the pair (x) and (y) can be described by three annuity benefit vectors:f, wheref k is the amount paid at timekifonly(x)i sa l i v e ;g, wheregk is the amount paid at timekif only(y)i sa l i v e ;a n dh, wherehk is the amount paid at timekif both(x) and (y) are alive. Letj = h−f −g. We can think of the contract as three separate annuities. One isalifeannuityon( x)withbenefitvector f.Anotherisalifeannuityon(y)withbenefitvector g. These two will provide for the required payments when only one of the pair is living. The thirdcontractmustadjustthepaymentwhenbotharealive.Ifbotharelivingattime k,thefirst two annuities will providef k +gk, so the third must provide the difference,jk = hk −fk −gk. Thepresentvalueofthecompletecontractisthenthesumofthethreeseparatepresentvalues, which is just ̈a x(f)+ ̈ay(g)+ ̈axy(j). (10.17)‡ This formula reduces the calculation of present values for the general two-life annuity to calculatingthoseforsingleorjoint-lifeannuities.(SeeSection10.12forspreadsheetmethods to calculate the latter.) Example 10.1 Verify that (10.17) reduces to (10.10), when the yearly benefit is a constant 1 unit. Solution. We havef k = gk = hk = 1 for allk,s ojk =− 1 for allkand (10.17) gives ̈axy = ̈ax + ̈ay − ̈axy. Example 10.2 An annuity on (x) and (y) provides yearly payments as long as either (x)o r (y) are alive. Payments begin at 12, but reduce to 8 if (x)o n l yi sa l i v eo rt o( 6 )i f(y) only is alive. Find the present value Solution. f k = 8, gk = 6, andhk = 12, sojk =− 2 for allk. This gives a present value of 8̈ax +6̈ay −2̈axy. Example10.3 Anannuitypays1uniteachyear,providedthatatleastoneof(50)and(65) is living and is over age 70, but not if (50) is alive and under age 65. Find the present value. Solution. Lettingxrefer to (50) andyto (65), f = (0 20,1∞), g = (05,1∞), h = (015,1∞), 152 MULTIPLE-LIFE CONTRACTS so j =− (05,110,0 5,1∞). The present value is then ̈a50(020,1 ∞)+ ̈a65(05,1 ∞)− ̈a50:65(05,110,05,1∞). Example 10.4 Find a formula for the present value of an annuity that paysck at timek, provided that (y)i sa l i v e ,b u t(x) is not alive. Solution. fk = 0, gk = ck, hk = 0, so thatjk =− ck, for allk. The present value is ̈ay(c)− ̈axy(c). A contract of the type described in Example 10.4 is known as areversionaryannuity.I t provides a life annuity on one life, which does not begin until another life has died. It can be used for a similar purpose as life insurance, except that the proceeds are paid out as a life annuitytoanotherperson,ratherthanasalumpsum.Aswell,reversionaryannuitiesareoften used to put a value on certain inheritances that stipulate that the income from a certain asset will first go to a certain individual and then, upon the death of that party, revert to another person. Remark We comment briefly here on the effect of using assumption (10.1) in which tpxy was approximated by a quantity that is normally too low, as we saw in (10.3). It follows that this approximation will give values for̈a xy(b) that are slightly too low (assuming the entries of b are nonnegative). Since the coefficient of this term is negative in the most common examplesoftwo-lifeannuities,asillustratedabove,thestandardassumptionofindependence meansthatmosttwo-lifeannuitypremiumsaresomewhathigherthantheywouldbeifamore realistic model was used. 10.7 The general two-life insurance contract Themostcommontypeofpolicysoldtotwolivesisthejoint-lifeinsurancedescribedabove, where payment is made on the first death. In some cases, however, people want policies that will pay on the second death. As we indicated in the introduction to this chapter, this can be usedasameansofminimizingestatetaxes.Wewillinfactconsiderageneraltypeofcontract where death benefits can be paid on both deaths. Assuming payment at the moment of death, such a policy will be described by two death benefit functions:b(t), the amount paid at the timeofthefirstdeath;and d(t),theamountpaidatthetimeoftheseconddeath.Wecanview this as two separate insurances, one on the statusxywith benefit functionb, and one on the status xywith benefit vectord. Substituting from (10.14), the present value of benefits is ̄Axy(b)+ ̄Axy(d) = ̄Ax(d)+ ̄Ay(d)+ ̄Axy(b−d). (10.18) Thiscanbeverifieddirectlybythesametypeofreasoningasthatusedafterformula(10.11). CONTINGENT INSURANCES 153 10.8 Contingent insurances In some insurance policies written on two lives, the benefits depend on theorderof death. In the most common cases, a designated individual must die first or second in order to receive benefits. If this does not occur, then no benefits are paid. For this reason these are known as contingentinsurances. We will first consider such contracts with benefits payable at the end of the year of death. It is assumed throughout this section that there is zero probability that two lives will die at exactly the same instant of time. 10.8.1 First-death contingent insurances We first need to compute some new probabilities. Letq1 xy denote the probability that in the next year (x) will die and that (y) will be alive at the time of the death of (x). The symbol of 1 abovexindicatesthat( x)isthefirst todie.Notethat(y)doesnothavetodiewithinthe1-year period in order for the event in question to occur. The first problem is to estimate this probability from the life table. We will derive the answer here intuitively. A more formal derivation appears in Chapter 17. We note thatq1 xy can be approximated by another probability. Consider the following events: eventAconsists of (x) dying within the year, before the death of (y); eventBconsists of (x) dying within the year, and (y) surviving to the middle of the year. Thenq1 xy is the probability ofA. We claim that this is close to the probability of eventB. These events are of coursenotthesame.If( x)and( y)bothdieinthefirsthalfoftheyear,with( x)dyingfirst,then Aoccurs but notB. On the other hand, if (x) and (y) both die in the second half of the year with(y)dyingfirst,then Boccursbutnot A.However,boththeselattersituationsarerelatively rare, involving both lives dying in the same year. In any event, as long as the probabilities of these two death situations are roughly the same, they will tend to cancel each other out, and we can assume that the probability ofAis approximately the same as the probability ofB. The probability ofB can be readily computed from the life table. The probability that (x) dies during the year is justq x. Assuming UDD, the probability that (y) will be alive in the middle of the year is1∕2py = 1−qy∕2. Making the usual independence assumption that we did in deriving (10.1), we deduce that the probability that both of these occurrences will happen is the product of probabilities. In other words, assuming UDD q 1 xy ≐ qx − 1 2qxqy. (10.19)‡ As a check on this, let us compute another expression forqxy, the probability that at least oneofthetwoliveswilldiewithintheyear.Thiscanhappenintwomutuallyexclusiveways. Either (x) dies during the year, with (y) alive at that time, or (y) dies during the year, with (x) aliveatthattime.Bybasicrulesofprobabilitytheorywecansumthetwoprobabilitiestoget the probability that either will happen. We must have that q xy = q1 xy +q1 yx . (10.20) Adding the right hand side of (10.19) to that with (x) and (y) interchanged does indeed give qxy, as shown by (10.2). 154 MULTIPLE-LIFE CONTRACTS Consider now an insurance contract where a designated life must die before another in order to collect. This arises in the following situation. In any insurance policy, a particular individual, known as thebeneficiary, is designated to receive the death benefit upon death of the insured. If the beneficiary dies before the insured, a new beneficiary is normally chosen. Suppose, however, that an insured has no other beneficiary in mind. To handle this, insurers are willing to offer a contract that only pays if a stated beneficiary is alive at the death of the insured. Suppose that a policy sold on the lives (x) and (y) provides for benefits to be paid at the endoftheyearofthedeathof(x )providedthat(y)isaliveatthetimeofsuchdeath.If(y)dies before (x), nothing is paid. For a death benefit vectorb, the present value of the benefits for this contract will be designated byA 1 xy (b). We calculate this by the standard formula for an insurancepresentvalue,asgivenin(5.1)andagainin(10.6).Weneedonlyinsert,foreach k, the probability that the conditions for payment will occur in the yearktok+1. Focus on the secondexpressionintherighthandsideof(10.6)wherewewrotetheprobabilitythatthefirst death of (x) and (y) will occur between timekand timek+1a s kpxyqx+k:y+k. In the present case, in order that the death benefit is paid at timek+1, we still require that both (x) and (y) are alive at time (k) but then we need that (x+k) dies in the next year, and to be the first of the two lives to die. In place ofqx+k:y+k we wantq1 x+k:y+k . This gives A1 xy (b) = N−1∑ k=0 bkv(k+1)kpxyq1 x+k:y+k . (10.21)‡ As a check, we can use (10.20) to see that Axy(b) = A1 xy (b)+A1 yx (b), (10.22) which must be true, since a joint-life insurance can be considered as two separate contingent insurances, each paying benefits if a particular life dies first. 10.8.2 Second-death contingent insurances Suppose we have a policy where death benefits are paid at the end of the year of death only if (x)i st h esecondof the two lives to die. Denote the present value of this byA 2 xy (b). We calculatethisasfollows.Thinkofaregularsingle-lifepolicyon( x)astwoseparatecontracts. The first pays if (x) dies before some other life (y), and the second pays if (x)d i e sa f t e r(y). We must have that Ax(b) = A1 xy (b)+A2 xy (b), so that A2 xy (b) = Ax(b)−A1 xy (b). (10.23)‡ CONTINGENT INSURANCES 155 10.8.3 Moment-of-death contingent insurances Moment-of-deathcontingentinsurancescanbecalculatedbyaformulasimilarto(10.21)but with integrals replacing sums, and𝜇 replacing q.I f̄A 1 xy (b) is the present value of a contract paying b(t) at timetprovided that (x) dies at timetand (y) is alive that time, then ̄A1 xy (b) = ∫ N 0 b(t)v(t)tpxy𝜇x(t)dt. (10.24)‡ Wecanintuitivelyverifythisbynotingthat tpxy𝜇x(t)dtrepresentstheprobabilitythatboth(x ) and (y) survive up to timetand then (x) dies at that time. Note also, from (10.12), that ̄Axy(b) = ̄A1 xy (b)+ ̄A1 yx (b), which is obviously required. For second-death insurance, the same reasoning as in (10.23) shows that ̄A2 xy (b) = ̄Ax(b)− ̄A1 xy (b). (10.25) Example 10.5 Suppose (x) is subject to a constant force of mortality𝜇,( y) is subject to a constant force of mortality𝜈, and the force of interest is a constant𝛿.F i n d̄A1 x:y . Solution. From (10.24), ̄A1 xy = ∫ ∞ 0 e−𝛿te−(𝜇+𝜈)t𝜇dt= 𝜇 𝜇 +𝜈 +𝛿. Normally, we must evaluate contingent insurances directly from the life table. In the case thatbenefitsareconstantovereachyear,theprocedureissimilarto(10.15)exceptthatweget only one-half of the remainder term. This necessarily follows from (10.22) and the fact that Ris symmetric in (x) and (y). That is, assuming UDD for each of the lives (x) and (y), ̄A 1 xy (b) = A1 xy ( i 𝛿 ∗ b ) + R 2, (10.26) whereRis as defined in (10.16). 10.8.4 General contingent probabilities Letsq1 xy denotetheprobabilitythatwithin syears(x)willdieand(y)willbealiveatthetimeof thedeathof(x ).(So q1 xy ,aswehavedefineditaboveisjustthissymbolwith s=1.)Similarly, letsq2 xy denote the probability that withinsyears (x) will die having been predeceased by (y). 156 MULTIPLE-LIFE CONTRACTS Considerfirstthecasewhere sisapositiveinteger n.Wecanevaluatethiseasilybyadding up the probabilities of the required event occurring in each year to obtain nq1 xy = n−1∑ k=0 kpxyq1 x+k:y+k . (10.27)‡ From (10.21) this is justA1 xy (1n) at zero interest. For the general durationswe evaluate the probabilitybytakingthecorrespondingcontinuousinsuranceformula(10.24)atzerointerest: sq1 xy = ∫ s 0 tpxy𝜇x(t)dt. (10.28)‡ Notethatwecantake s=∞ intheabovetogettheprobabilitythat(x )diesbefore(y).(There was no point in doing this in the single life case since∞qx is just equal to 1). Some basic identities are ∞q1 xy +∞ q1 yx = 1, ∞q1 xy +∞ q2 xy = 1. 10.9 Duration problems Certain multiple-life problems involve a duration that runs from the time of death of an individual rather than from time zero. Normally, the best approach to finding present values is to refer to our basic formula where we sum or integrate the product of three factors. Complicationscanarisewhencomputingtheprobabilitythatpaymentwillbemade.Inmany cases a formula for this probability will vary with time, and we will need to break up our total time interval into appropriate subintervals, each with its own formula. Evaluation of the resultingintegralsofteninvolvesthetechniquesofchangingvariablesandreversingtheorder of integration or summation. We will illustrate with several examples involving continuous annuities or moment of death insurances. For simplicity, we assume throughout constant interest and independence of the two lives involved. In all cases we wish to find a formula for either a present value or a probability. Moreover we want the resulting formula to involve either single-life or joint-life statuses, or contingent probabilities and insurances. Example 10.6 An insurance contract provides for 1 to be paid at the moment of the death of(y)providedthat( y)diesatleast nyearsafterthedeathof( x).Findaformulaforthepresent value. Solution. We must consider two separate time periods. For a timetbetween 0 andnthere is no chance of payment being made. For a timet≥ n, payment ismade providedydies at time tandxwas not alive at timet−n. The present value is ∫ ∞ n vt tpy𝜇y(t)(1−t−n px)dt. DURATION PROBLEMS 157 To write this in terms of our standard symbols, we change the variable fromt to s= t−n, which gives us an integral with lower limit 0. Then the integral becomes ∫ ∞ 0 vs+n s+npy𝜇y(s+n)(1−s px)ds= vn npy∫ ∞ 0 vs spy+n 𝜇y+n(s)(1−s px)ds. Inthesecondequalityabove,weusethemultiplicationruleaswellasthefactthatbydefinition, 𝜇 z(r)dependsonlyonthesum z+r.Fromtheaboveformulawecanwritethesolutioninthe required form as vn npy ( ̄Ay+n − ̄A1 y+n:x ) = vn npȳA2 y+n:x . Wecandeducethisintuitivelyifwedosomefancifulreasoning.Wetell( y)toage nyears while we keep (x) at the same age. We then let the two lives, now agey+nand x, ‘race’ to see who dies second. We must multiply the result by an interest factor ofvn to account for the delay in the race, and also bynpy which is the probability thatyw i l ls u r v i v et ob ea b l et o participate in the race in the first place. Example 10.7 What is the probability that (x) and (y) will die at leastnyears apart from each other? Solution. The probability that (y) will dienor more years after the death of (x) is given by the solution to Example 10.6 with 0 interest. This is npy ( 1−∞ q1 y+n:x ) =n py ∞q1 x:y+n . Adding to this the probability that (x)d i e snor more years after the death of (y)g i v e st h e solution of npy ∞q1 x:y+n +n px ∞q1 y:x+n . Example 10.8 A temporary life annuity on (y) provides for payments made continuously at the annual rate of 1 fornyears, but with the payments beginning at the death of (x), rather than at time 0. Payments stop at the death of (y). Find a formula for the present value. Solution. For 0≤ t< n,payment willbemadeprovided(y)isaliveand( x)isnot.For t≥ n, the payment will be made provided (y)i sa l i v e ,(x) is not alive but was alivenyears before. The present value is ∫ n 0 vt tpy(1−t px)dt+∫ ∞ n vt tpy(t−npx −t px)dt= ̄ay(1n)− ̄axy +∫ ∞ n vt tpyt −npxdt, 158 MULTIPLE-LIFE CONTRACTS where we rearrange and combine terms to get the expression on the right. Using the change of variable technique of Example 10.6 on the final term yields the answer of ̄a y(1n)− ̄axy +vn npx̄ax+n:y. Note that this is similar to the usual reversionary annuity formula except (using the same fanciful reasoning as in Example 10.6), afternyears we only award payment to (y)i fl i v i n g jointlywithanindividualwhowasage xandwasthenallowedtoage nyearswhile(y)stayed the same. Another contract of this type, which does not involve a duration directly, but is based on a similar theme, is a reversionary annuity where the benefits can depend on the time of death of the first person as well as the time they are made. Toillustratethetechnique,wefirstrevisitacontinuousversionofthestandardreversionary annuityasgiveninExample10.4.Consideracontractwithcontinuouspaymentsattheannual rate ofc(s) at times, made if (y)i sa l i v eb u t(x) is not. As an alternate approach to that used before, we will view this as a contingent insurance, with a death benefit of̄a y+t(c◦ t) paid at timet,themomentofdeathof(x )ifthisoccursbeforethedeathof(y).Notethetimeshifting necessaryhere.Forexample,if( x)diesattime2,therateofpaymentatatime3periodsafter the annuity begins isc(5) = (c◦ 2)(3). We can write the present value as ∫ ∞ 0 vt tpxt py𝜇x(t)̄ay+t(c◦ t)dt. (10.29) Note now that vt tpȳay+t(c◦ t) = ∫ ∞ t c(s)vs spyds, since each side is the present value of at-year deferred life annuity on (y) where the payment at timesfor s≥ tisc(s). Substituting this into the integral gives us the present value as ∫ ∞ 0 ∫ ∞ t c(s)vs spyt px𝜇x(t)d sdt= ∫ ∞ 0 ∫ s 0 c(s)vs spyt px𝜇x(t)dtds, (10.30) by reversing the order of integration. Now, ∫ s 0 tpx𝜇x(t)ddt= ∫ s 0 − d dttpxdt= 1−s px, the probability thatxdies before times. From this we conclude that the present value of the contract is ∫ ∞ 0 c(s)vs spy(1−s px)ds= ̄ay(c)− ̄axy(c). This appears as a very complicated way of arriving at the same form of answer as that obtained in a much simpler fashion in Section 10.6. However, one benefit of this insurance APPLICATIONS TO ANNUITY CREDIT RISK 159 approach is that it works just as well with the alternative type of reversionary annuities we introducedabove,andprovidesaconvenientmethodforhandlingthose.Suppose,forexample that upon the death of (x), the periodic rate of payment to a surviving (y)i sc(r) at a timeras measuredfromthe startoftheannuity.Theannuitywouldthenbeginwiththeperiodicrateof c(0).In(10.29)wethenhave cinplaceof c◦ t.Fordeathattime ttherateofpaymentatatime s> twould bec(s−t), and that would appear in 10.30 in place ofc(s). This is will normally make for a rather complicated integration, but there are cases where it can be worked out, as in the following example. Example10.9 Supposethattheforceofmortalityfor xisaconstant 𝜇,theforceofmortality for yis a constant𝜈 and the force of interest is a constant𝛿. Upon the death of (x) an annuity will be paid to (y), if surviving, where the rate of payment at a timer after the annuity payments begin is e 𝛾r, with 0≤ 𝛾<𝜈 +𝛿. Find the present value. Solution. Fromtherightsideof10.30,substituting c(s−t) = e𝛾se−𝛾t forc(s)asoutlined,the integral involving the variabletwill be ∫ s 0 e−𝛾te−𝜇t𝜇dt= 𝜇(1−e−(𝜇+𝛾)s) 𝜇 +𝛾 . Substituting and integrating with respect tosgives the present value as 𝜇 𝜇 +𝛾 ∫ ∞ 0 e−𝛿se−𝜈se𝛾s(1−e−(𝜇+𝛾)s)d s= 𝜇 𝜇 +𝛾 [ 1 𝜈 +𝛿 −𝛾 − 1 𝜈 +𝛿 +𝜇 ] . ∗10.10 Applications to annuity credit risk Suppose that you are a prospective annuity purchaser (or an advisor to such), and have a list of various companies and the premiums they charge. The prices will vary, due to different interest and mortality assumptions, as well as a different treatment of many other factors that go into gross premium calculation. A natural tendency would be to select the one with the lowest premium, but that is not necessarily the best choice. This ignorescredit riskwhich is the risk that the company may be unable to meet the promised payments due to financial difficulties. One may want to adjust the prices to reflect the different degrees of this risk among the various issuers. This could be an extensive task, and we will not go into details here. Our goal in this section is to indicate briefly how the theory of multiple-life contracts can be applied towards this endeavour. A first step might be to compare for a fixed interest and mortality basis, the cost of a life annuity calculated in the usual way with one that provides for payments only if a chosen company is viable. For this purpose we just treat the company as another individual, which we’ll denote by𝛾, which is a survival state if it is in position to meet its financial obligations. We need to model tp𝛾, which is the probability that the company is in a survival state at time t. This will produce the associated force of failure,𝜇𝛾(t) =− d∕dtlog tp𝛾. We can then consider two-life contracts based on (x) and𝛾. For simplicity we confine our discussion to a life annuity on (x) that provides continuous payments at a level annual rate ofK. 160 MULTIPLE-LIFE CONTRACTS Takingintoaccountthecreditrisk,theworthofthecontracttothepurchasershouldbethe joint-life annuityK̄ax𝛾 = K∫ ∞ 0 tpxt p𝛾 dtrather thanK̄ax. This assumes that no benefits are paidafterfailureofthecompany.Inpracticehowever,thereisoftenarecoveryrate rwhereby the company, upon financial failure, will continue to pay an income ofr for every 1 unit of promised annuity income. In this case the purchaser will getKrif (x) only is surviving, and K if bothxand𝛾 are surviving so that our general two-life annuity formula gives the value to the purchaser as K[r̄ax +(1−r)̄ax𝛾], which of course reduces toK̄ax when r= 1. Another complication arises in cases where government regulatory bodies provide guar- anteesforannuities.Therearemanypossibleoptionsconcerningtheamountoftheguarantees andhowtheyrelatetotherecoveryratesofthefailingcompany.Wewillconsiderthefollow- ing particular example, which is encountered in some jurisdictions. Upon failure, the insurer continues to pay at their recovery rate. The guarantor makes up the difference, but the value, at interest and survivorship, at the time of failure, of all payments made by the guarantor is limitedtoanamount G.Suppose,forexamplethat K = 30,000, G= 100,000, r= 0.4.Inthe event of default, the issuing company would pay 12,000 each year, and the government body wouldprovidetheannualshortfallof18,000butwouldonlypayitonlyuntilthepresentvalue of100,000wasexhausted,whichshouldbesomewhatover6yearsundernormalinterestand mortality bases. We will arrive at a formula to compute the present value of aK-unit, whole life annuity subject to these default arrangements. We assume that the mortality and interest bases are the same for the issuer and the guaranteeing party. Lets 0 =i n f{s: ̄ax+s ≤ G∕(1−r)K}.Note thats0 would equal 0 if̄ax < G∕(1−r)K.For failure of the issuer at a timet before times0 the guarantee will not cover the full annuity. In that case the annuitant, who, in the absence of default, would have received total future payments with a value at timetofK̄a x+t, will instead receive payments with a reduced value at timet of G+rK̄ax+t. The deficit is (1−r)K̄ax+t −G. We must subtract from the usual present value, the present value of a a contingent insurance that provides this deficit on the failure of𝛾 before times 0, provided that (x) is then alive. Referring to equation (10.24) the credit risk adjusted present value of the annuity is K̄ax −∫ s0 0 vt tpxt p𝛾𝜇𝛾(t)[(1−r)K̄ax+t −G]dt. 10.11 Standard notation and terminology Particularexamplesofthemultiple-lifenotationhavealreadybeenencounteredinthestandard notationfortermandendowment insurancesandtemporaryannuities.Inplaceofthelife( y), we dealt with a period ofn years, denoted by n, which ‘survives’ for exactlyn years and then ‘fails’ at timen. So the standard symbol for the present value of a 1-unit,n-year term insurance premium,A1 x: n, does indeed signify that the policy provides for 1 unit at the end of the year of death provided (x) dies before the ‘failure’ ofn. Similarly, the standard symbol for the present value of a 1-unit,n-year endowment insuranceAx:n indicates that the contract provides for 1 unit to be paid at the death of (x)o ra tt i m en if earlier, that is, on the first failure ofxand n. The 1-unit,n-year, temporary annuity present value symbol̈ax:n indicates EXERCISES 161 thatthecontractprovidesforpaymentstocontinueaslongasboth( x)isaliveand nis‘alive’, meaning that then-year period has not run out. 10.12 Spreadsheet applications WecanadapttheChapter6spreadsheettobeapplicabletojoint-lifeannuitiesandinsurances. In the first place we will make provision to insert two life tables, one for (x), and one for (y). This is common in practice as often one life will be male, and the other female, so different tables apply. Start by inserting a new column D, which we intend show the values ofq y+t. This will move all the existing columns over one to the right. In cell D1 we put the agey. Copy the formula from C10 to D10. It should be the same except with C1 replaced by D1 and a reference to column P. (the reference is column C will be automatically changed from column N to column O). WethencaninsertanewtableinColumnPbyputtingtheparametersinP3andP4,coping the formula from 010 to P10 and copying down. In column F, which now containsy xy, we change the formula in F11 by appending ∗(1−D10) and copy down. In Column M, which now containswxy ∗ b, we change the formula in M10 by replacing C 1 0b y( C 1 0+D 1 0−C10*D10) in order to multiply byqxy rather thatqx. As a test exercise suppose that our example table applies to females and that for the male table the mortality at each age up to 119 is 1.25 times the female rate. Interest rates are a constant 6%. An insurance on a male age 50 and a female age 45 provides 10,000 at the end of the year of the first death. Find the net level premium payable yearly while both lives are living. The answer is 208.75. Notes and references An extensive study of dependence in two-life annuity contracts can be found in Freeset al. (1996). Moreadvancedexamplesofmultiple-lifecontracts,involvinganarbitrarynumberoflives, can be found in Jordan (1967, Part II), or in Bowersetal. (1997, Chapter 18). The result from real analysis that justifies the reversal of integration in formula (10.30) can be found in Royden (1988) Theorem 12.19 (iii). Exercises In all the exercises for this chapter, we assume that (10.1) holds. Type A exercises 10.1 Youaregiventhefollowinginformationfromalifetable: 𝓁80 = 100, 𝓁81 = 80, 𝓁82 = 40, 𝓁83 = 20. You also have the vectorc = (1000,1200,1500) and the vectorb = (1000,2000). Theinterestrateisaconstant25%.Find(a) p80:81 and2p80:81,(b)p 80:81 and 2p80:81,( c )̈a80:81(c), (d)A80:81(b), (e)̈a80:81(c), (f)A80:81(b). 162 MULTIPLE-LIFE CONTRACTS 10.2 Suppose that 5px = 0.8, 6px = 0.7, 5py = 0.6, 6py = 0.4. Find the probability that both (at least one) (exactly one) (neither) of the two lives live 5 years (die within the next 5 years) (die between time 5 and time 6). There are 12 answers required here. 10.3 You are given thatp 70 = 0.9, 2p70 = 0.8, 3p70 = 0.7. The interest rate is 20% for the first year and 25% for the second year. Find the present value of the benefits on a 2-year term insurance policy sold to two lives (70) and (71) that provides benefits upon the first death, payable at the end of the year of death. The amount of the death benefit is 1000 if the first death occurs in the first year, and 2000 if it occurs in the second year. 10.4 An insurance policy pays 1 unit at the moment of the second death of (x) and (y). Level premiums are payable continuously as long as either of the two is alive. Given that (x) is subject to a constant force of mortality of 0.1, (y) is subject to a constant forceofmortalityof(0.3)andtheforceofinterestisaconstant0.1,whatistheannual rate of premium payment? 10.5 For two lives (x) and (y), the forces of mortality are given by𝜇 x(t) = 0.04 for allt, and𝜇y(t) = 1∕(20−t)f o r0≤ t< 20. (a) Find the probability that the joint-life status (xy) will survive to time 10. (b) Find the probability that the last-survivor statusxywill fail before time 10. (c) An insurance contract provides for a death benefit paid at the moment of the first deathof(x )and(y).Theamountofthebenefitise 0.1t fordeathattime t.Theforce of interest is a constant 0.06. Find the present value of the benefits. 10.6 Suppose that Demoivre’s law holds with𝜔 = 100. Consider two lives (80) and (60). (a) Whatistheprobabilitythatthefirstdeathwilloccurbetweentime5andtime10? (b) What is the probability that the second death will occur between time 5 and time 10? 10.7 Thefollowingisaportionofaselectandultimatetablewithaselectperiodof2years. xq [x] q[x]+1 qx+2 x+2 60 0.08 0.14 0.22 62 61 0.09 0.20 0.25 63 62 0.10 0.20 0.30 64 A and B are both age 62. A is a newly selected life, while B was first selected at age 61.WhatistheprobabilitythattheseconddeathofAandBwilloccurbetween2and 3 years from now? EXERCISES 163 10.8 An insurance contract sold to (x) and (y) provides for a death benefit of 1 unit at the moment of the first death and 3 units at the moment of the second death. Annual premiums are payable while at least one of the two is living. The annual rate of premiumpaymentreducestothreequartersoftheinitialrateuponthefirstdeath.You are given that (x) is subject to a constant force of mortality of 0.2, and (y) is subject toaconstant forceofmortalityof0.3.Theforceofinterestisaconstant0.1.Findthe initial annual rate of premium payment. 10.9 A contract on two lives (x) and (y) provides that if (y) dies first, (x) will receive a life annuity of 1 per year, starting at the end of the year of (y)’s death, while if (x)d i e s first,(y)willreceivealifeannuityof2peryear,startingattheendoftheyearof( x)’s death.Levelannualpremiumsof 𝜋 arepayablewhileboth( x)and( y)arealive.Given that ̈a x = 16, ̈ay = 13 and̈axy = 10, find𝜋. 10.10 You are given two independent lives (x) and (y)f o rw h i c h(x) is subject to a constant force of mortality of log(4/3), while (y) is subject to a force of mortality at timetof 1∕(10−t)f o r0≤ t< 10. (a) What is the probability that the first death among these two will occur between time 2 and time 3? (b) Whatistheprobabilitythattheseconddeathamongthesetwowilloccurbetween time 2 and time 3? 10.11 You are given thatq70 = 0.2, q71 = 0.25, q72 = 0.4,q73 = 0.6. The interest rate is a constant25%.Thevector bis(1000,2000,3000).Compute(a) A1 70:71 (b),(b) A1 71:70 (b), (c)A2 70:71 (b), (d)A2 71:70 (b). Type B exercises 10.12 Anannuitycontractissuedtotwolives,(40)and(50),providesforpaymentsof1per year, to be made provided that either or both of the following conditions hold. (i) (40) is alive and under age 60; (ii) (50) is alive and over age 60. Find a formula for the present value of the benefits using terms of the form̈ax(c) or ̈axy(c). 10.13 Repeat Exercise 10.12, but now with (ii) stating that (50) is alive and between age 60 and age 80. 10.14 An insurance policy sold to (x) and (y) provides for a payment of 2 units on the first deathand3unitsontheseconddeath.Levelannualpremiumsarepayableforasalong as either individual is living, with the premiums reducing by one-third upon the first death. Find an expression for the initial premium in terms of̄A x, ̄Ay, ̄Axy, ̈ax, ̈ay, and ̈axy. 10.15 An annuity sold to (x) and (y) provides for annual payments for 20 years, provided at least one of the two is alive. The annual payment begins at 12 units. During the first 164 MULTIPLE-LIFE CONTRACTS 10 years, the payments remain at 12. However, during the second 10-year period, the payment reduces to 6 if (x)o n l yi sa l i v e ,o rt o8i f(y) only is alive. Write the present value in terms of single- and joint-life annuities. 10.16 An annuity contract sold to (40) and (50) provides for yearly payments if either life is living and under age 70. The yearly payment is 6 if both are alive, 4 if (40) only is alive, and 5 if (50) only is alive. Find a formula for the present value of the benefits using terms of the form̈a x(c) and̈axy(c). 10.17 For the policy of Exercise 10.8, calculatetV. 10.18 SupposethatDemoivre’slawholds.Showthat(10.19)holdsexactlyforageneralleft subscriptnin place of 1. That is, nq1 xy = nqx − 1 2nqxn qy. 10.19 For the two lives of Exercise 10.10, find the probability that (x) will die before (y). 10.20 Given two lives (x) and (y) and timess< t, consider the following four events:A,t h e joint-status (xy) fails between timesand timet; B, the last-survivor status (xy) fails between timesand timet;C, at least one of (x) and (y) fails between timesand time t; andD, both (x) and (y) fail between times and timet. Rank the probabilities of these from largest to smallest to the extent that you are able, showing clearly those events that you cannot compare without further information. 10.21 Suppose that interest is a constant 25% and thatq x = 0.2. Assume UDD. Find the ratio of̄Axx(11)t o(i∕𝛿)Axx(11). 10.22 The probability nq1 xy can be generalized tonq1 x 1:x2:…xk , which is the probability that outofagroupof klives,(x1),(x2),…,(xk),(x1)willdiewithin nyearsandbethefirst to die. Express the following in terms of first-death probabilities: (a) theprobabilitythat(x )willdiewithin nyearsandwillbethe secondtodieamong a group of three lives (x),(y), and (z); (b) theprobabilitythat(x )willdiewithin nyearsandbethe thirdtodieamong(x ),(y), and (z). 10.23 Find a formula to calculatēA 2 yx (b) in terms of (a) ̄A1 xy (b), ̄Ay(b), ̄Axy(b), (b) ̄A2 xy (b), ̄Axy(b). 10.24 Suppose that nqx = 0.2, andnqy = 0.3. What isnq1 xy − nq2 yx ? 10.25 An insurance pays 1 at the death of (y) provided that (x) was alivenyears before this death. Assume constant interest. Find a formula for the present value in two ways. Firstbysettingupandevaluatingthenecessaryintegrals.Secondly,byusingExample 10.6 to avoid any integration. EXERCISES 165 10.26 An annuity provides for continuous payments at the annual rate of 1, while both (x) and (y) are alive, and in addition continues the payments forn more years after the firstdeath,aslongasthesurvivorisliving.Paymentsstopcompletelyuponthesecond death. Find a formula for the present value. Verify that your answer gives the correct result in the case thatn= 0. 10.27 In Example 10.9 take𝜈 = 𝛿 = 0 and𝛾< 0. (a) Evaluate the present value and give an intuitive explanation of the answer. (b) Taketheparametersasinpart(a),butnowconsiderthesimplertypeofreversionary annuitywheretherateofpaymentattime r(measuredasusualfrom0)is e 𝛾r.Show that the present value is an increasing function of𝜇 and explain why this is so. 10.28 Inthisproblemalldeathbenefitsarepayableatthemomentofdeath,andallpremiums arepayablecontinuouslyatalevelrate.Alifeinsurancecontracton( x)and( y)provides for death benefits as follows. If (x) dies first at timet, the amount paid is the reserve at timet on a 1-unit whole life insurance on (y) with premiums payable for life. If (y) dies first at timet, the amount paid is the reserve at timetfor a 1-unit whole life insurance on (x) with premiums payable for life. Premiums are payable until the first death. Show that the rate of premium payment is̄Px + ̄Py − ̄Pxy, where ̄Px denotes the rate of premium payment for a 1-unit whole life insurance onx with premiums payable for life, and̄Py and ̄Pxy denote similar quantities foryand xy, respectively. You should demonstrate this mathematically and also give a verbal explanation. 10.29 A contract on ( x) and (y) provides the following death benefits, payable at the moment of death. If (x) dies first, 10 is paid upon the death of (x) and 30 is paid upon the death of (y). If (y) dies first, 20 is paid upon the death of (x) and 40 is paid upon the death of (y). Find the present value given that ̄Ax = 0.30, ̄Ay = 0.45, ̄Axy = 050, ̄A1 xy = 0.20. 10.30 An annuity provides continuous payments at the annual rate of 1 to (y) while living, beginning n years after the death of (x). Assuming constant interest, find a formula for the present value. Spreadsheet exercise 10.31 Suppose that our sample table applies to females and the mortality rate for males is 1.25 times that for females up to age 119. A is a male age 50 and B is a female age 40. The interest rate is a constant 5%. (a) Find the level annual premium for an insurance paying 10 000 at the end of the year of first death of A and B with premiums payable while both are alive. (b) Find the level annual premium for an insurance paying 10 000 at the end of the year of the death of A provided this occurs before the death of B. Premiums are payable while both are alive. (c) Assuming UDD, find the probability that A will die before B. 11 Multiple-decrement theory 11.1 Introduction Our discussion of insurance contracts up to now has been concerned with benefits payable upon the occurrence of death. We now investigate situations when an insured is at the same time subject to several different events that can have financial impact. One example, which we mentioned earlier, is the event ofwithdrawalor lapse, whereby an insured life terminates the contract and receives a cash value. To properly model a life insurance contract, one must consider two causes of termination, death and withdrawal, operating simultaneously. Sometimes the insurer must distinguish between different causes of death. For example, some policies have a feature that provides additional death benefits for accidental death as opposed to death from natural causes. Somepoliciesincludedisabilitybenefits,providingincomeforpeoplewhocannolonger work. For such, the insurer must consider the possibility of disability, as well as death and withdrawal. For employees covered under a pension plan, there are at least four events of interest: disability; death; termination of employment; and retirement. These are just a few of the possibilities, and many more can arise. In this chapter we analyze the general situation. 11.2 The basic model We suppose that we havemdifferent causes offailureoperating simultaneously on a group of lives. These causes are often referred to asdecrements, since they bring about a decrease in the number of lives under observation. The insurance policies introduced in Chapter 5 involved only one decrement, namely that of death. When there are several decrements we are concerned with what is called in the actuarial literature asmultiple-decrementtheory,o r , in biostatistical contexts, the theory ofcompetingrisks. FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial THE BASIC MODEL 167 There is a generalization of these ideas to multi-state insurances and annuities, where individualscantransferfreelybetweenoneofseveralstates.WedealwiththisinChapter19, which requires a basic knowledge of Markov chains, as given in Chapter 18. 11.2.1 The multiple-decrement table It is convenient to make use of a generalized life table, known as amultiple-decrementtable. We will number our causes from 1 tomand use a superscript (j) to refer to causej. We will also use a superscript (𝜏), which the reader can interpret as meaningtotal. We begin with an arbitrary number𝓁(𝜏) 0 of lives age 0. We let𝓁(𝜏) x denote the number of individuals from this group who are still surviving at agex. That is, they have not succumbed to any of them causes. We letd(j) x denote the number of lives who will failfirstfrom cause (j) between the ages ofxandx+1. Let d(𝜏) x = m∑ j=1 d(j) x , (11.1) whichisthenumberofpeoplewhowillfailfrom somecausebetweentheagesof xandx+1. It follows that 𝓁(𝜏) x+1 = 𝓁(𝜏) x −d(𝜏) x . (11.2) That is, the number of survivors at agex+1 is equal to the number of survivors at agex,l e s s thosewhofailedfromsomecausebetweentheagesof xandx+1.Thefollowingisaportion of a sample table with two decrements: x 𝓁(𝜏) x d(1) x d(2) x 0 1000 50 100 1 850 60 105 2 685 70 120 Itisimportanttonotetheword‘first’usedinthedefinitionof d(j) x .Themodelassumesthat any cause of failure results in the individual leaving the group, so they are no longer under observation. For example, if cause 1 denotes death and cause 2 withdrawal, a policyholder whowithdrawsatage60 1 2 andthendiesatage60 3 4 wouldbeincludedin d(2) 60 ,butnotin d(1) 60 . Thisiswhatwewantinthecommonapplications.Inthecaseofwithdrawal,theinsurerwould paythepolicyholdertheircashvalue,thepolicywouldterminate,andtheinsurerwouldhave no further interest in the time of death of this person. Therefore, whenever we refer to failing from a certain causejin the multiple-decrement model, it is understood that this means that this occurs before failure from any other cause. Wedefinedamultiple-decrementtablestartingatage0,butthiscouldbesomeotherage, depending upon the particular application. For example, the multiple-decrement table for an employee pension plan would begin at the first age the employees become eligible for the plan, perhaps age 25 or so. 168 MULTIPLE-DECREMENT THEORY Our model assumes that the multiple-decrement table, as given, can be used for all ages x. That is, it is an aggregate table that does not show selection effects. We could define for eachx, as we did for the single-decrement case, aselectmultiple-decrement table to apply to people first observed at agex. This is important in practice since many causes of decrement, such as withdrawal, will certainly depend more on duration since the policy began than on attained age. However, we will not get into these details in this chapter and will work with the aggregate model. The more general case can be deduced from the material of Chapter 17. 11.2.2 Quantities calculated from the multiple-decrement table We first define probabilities of failure. Let q(j) x = d(j) x 𝓁(𝜏) x be the probability that (x) will fail first from causej within 1 year. In practice, one would start with these probabilities and construct the multiple-decrement table inductively, by calculating d (j) x = 𝓁(𝜏) x q(j) x and then using (11.1) and (11.2) to complete the table. Let kp(𝜏) x = 𝓁(𝜏) x+k 𝓁(𝜏) x (11.3) be the probability that (x) will survive to agex+kwithout succumbing to any cause. Probabilitiesarecalculatedfromthetablewithsimilarformulasasinthesingle-causelife table. One must only remember that the denominators are𝓁(𝜏) x . For example, nq(j) x = ∑n−1 k=0 d(j) x+k 𝓁(𝜏) x is the probability that (x) will fail as a result of causejwithin nyears. Similarly, d(j) x+k 𝓁(𝜏) x =kp(𝜏) x q(j) x+k (11.4) is the probability that (x) will fail from causejin the time intervalktok+1. Another quantity of interest is 𝓁(j) x = 𝜔−x−1∑ k=0 d(j) x+k. INSURANCES 169 This represents the number of people in our group who will fail from causejsometime after age x. We assume in our model that everyone will eventually fail from some cause (which is obvious if death is one of the causes) so we can write 𝓁(𝜏) x = m∑ j=1 𝓁(j) x . Knowing the value of𝓁(j) x for all integral values ofxand alljallows us to complete the table since d(j) x = 𝓁(j) x −𝓁(j) x+1. Note that we can write nq(j) x = 𝓁(j) x −𝓁(j) x+n 𝓁(𝜏) x . (11.5) Another symbol we can define is np(j) x = 𝓁(j) x+n 𝓁(𝜏) x (11.6) which is the probability that (x) will fail first from causej after time n. Here, instead of thinking of the basic symbolnp as denoting survival to timen, we think of the equivalent formulation as failing after timen. 11.3 Insurances Consideraninsurancepolicyon( x)thatprovides,forallrelevantvalues kandj,apaymentof b(j) k attime k+1,providedthatfailureoccursfromcause jintheyear ktok+1.Wecanview this asm separate policies, where thejth policy pays benefits only for failure from causej. The formulas are exactly the same as (5.1), but with the probabilities taken from (11.4). That is, the present value of thejth policy will be ∞∑ k=0 b(j) k v(k+1) d(j) x+k 𝓁(𝜏) x = ∞∑ k=0 b(j) k v(k+1)kp(𝜏) x q(j) x+k, (11.7)‡ and the total present value would be obtained by summing over allj. To handle insurances payable at the moment of failure we will need the following: Definition 11.1 For each causej,l e t 𝜇(j)(x) = lim h→0 hq(j) x h = lim h→0 𝓁(j) x −𝓁(j) x+h h𝓁(𝜏) x (11.8) 170 MULTIPLE-DECREMENT THEORY The forceofdecrementattimet , from cause (j)f o r(x) is the quantity given by 𝜇(j) x (t) = 𝜇(j)(x+t). (11.9) (Inthecaseofselectedmultiple-decrementtableswewouldneedtodefine 𝜇(j) x (t)separately for each agex.) Analogouslyto(8.15)andnotingthatthequantitiesin(11.5)and(11.6)differbyaconstant we can write 𝜇(j) x (t) = −d dt tp(j) x tp(𝜏) x = d dttq(j) x tp(𝜏) x . (11.10) Considerthecontinuousanalogueoftheinsurancewhosepresentvalueisgivenby(11.7). Thisisacontractthatpaysadeathbenefitof b(j)(t)attimet,shouldfailurefromcause joccur at that time. We denote the present value bȳA(j) x (b(j)). It is given by ̄A(j) x (b(j)) = ∫ ∞ 0 b(j)(t)v(t)tp(𝜏) x 𝜇(j) x (t)dt. (11.11)‡ This reflects the fact that in order to collect at timet,(x) must surviveallcauses up to timet, andthensuccumbtocause jattime t.WediscusstheevaluationofthisintegralinSection11.5. For policies which are purchased by annual premiums, payable until the first failure, we determine theinitialpremium as inthe singledecrement case, bysimplydividing thepresent value by ̈a (𝜏) x (𝜌) = ∞∑ k=0 v(k)𝜌kk p(𝜏) x , where𝜌 is the premium pattern vector. 11.4 Determining the model from the forces of decrement The multiple-decrement model is often given from the outset by specifying the forces of decrement, and we must use these to calculate probabilities. Note that in the last expression in(11.8)thenumeratorinvolves 𝓁 (j) whilethedenominatorinvolves 𝓁(𝜏).Thismeansthatwe cannot apply a version of (8.16) directly. We can, however, define the total force 𝜇(𝜏) x (t) = m∑ j=1 𝜇(j) x (t) = lim h→0 𝓁(𝜏) x+t −𝓁(𝜏) x+t+h h𝓁(𝜏) x+t . So𝜇(𝜏) x isthesametypeofquantityasthesingle-lifeforceofmortalityintroducedinChapter8, except based on the total decrement rather than just on failure by death. We can now apply (8.16) to deduce that tp(𝜏) x = e−∫ t 0 𝜇(𝜏) x (r)dr. (11.12)‡ A MACHINE ANALOGY 171 Arguing as we did in deriving (10.28), apply (11.11) with a constant interest rate of 0, and b (j)(t) = 1f o r0≤ t< s, and 0 elsewhere, to conclude that sq(j) x = ∫ s 0 tp(𝜏) x 𝜇(j) x (t)dt. (11.13)‡ Example 11.1 In a model with two decrements, you are given that𝜇(1) x (t) = 0.02 and 𝜇(2) x (t) = 0.03 for allt≥ 0. Find3q(1) x . Solution. 𝜇(𝜏) x = 0.05, so thattp(𝜏) x = e−0.05t and, from (11.13), 3q(1) x = 2 5(1−e−0.15). 11.5 The analogy with joint-life statuses We have already encountered a multiple-decrement model in Chapter 10. The difference is that, rather than single lives, the basic objects were joint-life statuses consisting of a pair of lives (x) and (y). These pairs were subject to two distinct causes of failure, namely the death of (x) and the death of (y). The notation differs somewhat, but the reader should notice that, with this point of view, the contingent insurance present value given in (10.24) is a special caseofthatgivenin(11.11),andtheprobabilityin(10.28)isaspecialcaseofthatin(11.13). It follows that if we assume that the causes of decrement are acting independently, we can evaluate the integral (11.11) from the multiple-decrement table as in Chapter 10. That is, if the benefits are constant over each year, and the particular decrement can be assumed to be uniformly distributed over each year, thei∕𝛿 correction will result in a reasonable approximation. 11.6 A machine analogy Thedifficult partofmultiple-decrement theorydeals withrelationships between thedifferent decrements.Inordertobetterunderstandandmotivatetheideas,wefirstlookatwhatappears tobeadifferentsituation,butwhichwewilleventuallyrelatetoourmodelabove.Tofacilitate this, we will use notation for this new example that parallels that which we have already introduced. Suppose we have a machine with two components, part 1 and part 2, which work com- pletely independently of each other so that the condition of one part does not affect the operation of the other. In order for the machine to work, both parts must be working, so if either part fails, the machine will fail, even though the other part may be in perfect order. We assume also that both parts cannot fail simultaneously, so we can always identify which part caused the machine to fail. Suppose we want to compute probabilities of failure over some time period, say a year. We have four quantities of interest. Forj= 1,2, letq ′(j) be the probability that partj will fail, and letq(j) be the probability that the machine will fail due to the failure of partj, meaning that partjfailed during the year and was the first of the two partstofail.Whataretherelationshipsbetweenthese?Sincethefailureofthemachinedueto 172 MULTIPLE-DECREMENT THEORY the failure of partjobviously implies that partjfailed, it is clear from elementary probability theory that q′(j) ≥ q(j). (11.14) Itisisalsoclearthat,ingeneral,thetwoquantitiesarenotequalandwewouldexpecttheleft handsidetobestrictlygreaterthantheright.Suppose,forexample,thatsometimeduringthe year part 1 fails, causing the machine to fail, and sometime after that but before the end of the year, part 2 fails. Then the event of the failure of part 2 would have occurred, butnotthe event of part 2 causing the failure of the machine. After introducing some additional notation, we will state some further relationships. Let p ′(j) = 1−q′(j) denote the probability that partjwill be working at the end of the year,q(𝜏) denote the probability that the machine will fail within the year, andp(𝜏) = 1−q(𝜏) denote theprobabilitythatthemachineisworkingattheendoftheyear.Themachinecanfailinone of two mutually exclusive ways, namely, failure of part 1 or failure of part 2. By elementary probability theory we have q (𝜏) = q(1) +q(2). Ontheotherhand,inorderthatthemachinebeworkingattheendoftheperiod,bothpartsmust be working. Since the parts work independently, we can multiply probabilities to calculate this (as we illustrated in Chapter 10), which gives p (𝜏) = p′(1)p′(2). From the above two equations we get the fundamental identity q(1) +q(2) = 1−p′(1)p′(2), (11.15) which is often written in the form q(1) +q(2) = q′(1) +q′(2) −q′(1)q′(2). (11.16) Thebasicproblemofinterestisasfollows.Ifwearegiventheunprimedsymbols,canwe calculatetheprimedones,orifwearegiventheprimedsymbolscanwecalculatetheunprimed ones?Ingeneral,wecannotdothisuniquelyandwewillneedadditionalinformationinorder to definitely deduce one set of probabilities from the other. We will, however, present some possible solutions to this problem, and then discuss some conditions that will lead to these solutions. 11.6.1 Method 1 Givenq(j), j= 1,2, we sum to getq(𝜏), then calculatep(𝜏), and then let p′(j) = p(𝜏) ( q(j) q(𝜏) ) , (11.17) A MACHINE ANALOGY 173 which will clearly satisfy (11.15). It is not too difficult to remember this formula. We have to factor the quantityp (𝜏) into two factors, and it is natural to take the factors asp(𝜏) to some exponent where the exponents sum up to one. Two numbers that obviously sum to 1 are q (j)∕q(𝜏), j= 1,2. For the other direction, givenq′(j), j= 1,2, we calculatep′(j), multiply to getp(𝜏), then calculateq(𝜏), and finally take q(j) = logp′(j) logp(𝜏) q(𝜏). (11.18) For this formula we have to split upq(𝜏) as the sum of two terms. We obtain the terms by multiplying by weights that add up to 1. Since we have a multiplicative relationship, we take logs to accomplish this. A straightforward calculation verifies that this approach is consistent. That is, if we are givenq (j), j= 1,2, and use (11.17) to calculateq′(j), j= 1,2, and then apply (11.18) to these numbers, we will end up with the same values ofq(j) as we started with. To see that (11.14) is satisfied we use the inequality (1−x)𝛼 ≤ (1−𝛼x), for 0 ≤ x≤ 1,0 ≤ 𝛼 ≤ 1, which can be demonstrated by looking at the the Taylor polynomial of degree 2 for the left hand side. From (11.17), taking𝛼 = q (j)∕q(𝜏), p′(j) = (1−q𝜏)𝛼 ≤ 1−𝛼q(𝜏) = 1−q(j), and (11.14) follows. The entire discussion above can be generalized to the case ofmindependent parts, where the failure of any one part will cause the machine to fail. Use notation as above, except that jnow takes values 1,2,…,m. Everything is the same with the obvious changes to handlem probabilities instead of two. For example, (11.15) will now read m∑ j=1 q(j) = 1− m∏ j=1 p′(j), (11.19) and the solutions given by (11.17) and (11.18) will satisfy this relationship. 11.6.2 Method 2 We present another method to deduce the unprimed probabilities from the primed, which in certaincasesmaybemorerealisticthanformula(11.18).However,for m> 2itisnoteasyto invert, in order to deduce unprimed from primed. To motivate this method, we look again at the case of the joint-life statuses of Chapter 10. In fact we can directly visualize the machine pointofviewinthissituation.The‘machine’consistsofthejoint-lifestatus( xy)withthetwo distinct components (x) and (y). For our particular problem, suppose we are givenq ′(1) and q′(2),andwanttocomputetheprobabilitythatthefailureofpart1causedthemachinetofail. 174 MULTIPLE-DECREMENT THEORY Reasoning as we did in deriving (10.19), we approximate this by computing the probability that part 1 failed during the year and that at the middle of the year part 2 was still working. Making an assumption of auniform distribution of failures for each part, over each year, which is analogous to our UDD assumption, we have, as in (10.19), q (1) = q′(1) − 1 2q′(1)q′(2), q(2) = q′(2) − 1 2q′(1)q′(2), (11.20) which clearly satisfy (11.14) and (11.16). To calculate the inverse formula, letΔ= q(1) −q(2). From (11.20) q(1) = q′(1) − 1 2 (q′(1))2 + 1 2q′(1)Δ Wehaveaquadraticequationin q′(1),whichcanbesolvedbytheusualformulatoyield q′(1) = (2+Δ)− √ (2+Δ)2 −8q(1) 2 , (11.21) and directly from (11.20) q′(2) = q′(1) −Δ. Example 11.2 Givenq(1) = 0.27, q(2) = 0.17, findq′(1) andq′(2) by Method 2. Solution. q′(1) = 2.1− √ (2.1)2 −2.16 2 = 0.3, q′(2) = 0.3−0.1 = 0.2. As a check we can take these answers and apply (11.20) to get back to where we started. q((1)) = 0.3−(1∕2).06 = 0.27, q((2)) = 0.17. Example 11.3 Givenq′(1) = 0.200,q′(2) = 0.488, estimateq(1) and q(2) by both methods. Solution. ByMethod1: Sincep′(1) = 0.8and p′(2) = 0.512,wehave q(𝜏) = 0.5904and,from (11.18), q(1) = 1 4 ×0.5904 = 0.1476, q(2) = 3 4 ×0.5904 = 0.4428. SolutionbyMethod2: From (11.20), q(1) = 0.2000−0.0488 = 0.1512, q(2) = 0.4880−0.0488 = 0.4392. ASSOCIATED SINGLE-DECREMENT TABLES 175 AswithMethod1,Method2canalsobeextendedtothecaseof mcomponentstodeduce the unprimed from the primed, as we show later in Section 11.7.2. For example, withm= 3 we get q(1) = q′(1) − 1 2 [q′(1)q′(2) +q′(1)q′(3)] + 1 3q′(1)q′(2)q′(3). (11.22) Note that the coefficients of 1/2 and 1/3 are what is needed in order to satisfy (11.19). With m> 2 there is no convenient direct method for the inverse process, and it would have to be done by numerical approximation. 11.7 Associated single-decrement tables 11.7.1 The main methods We now return to the original setting dealing with a group of lives, to which we will apply our ‘machine’ model. For a definite example start with a multiple-decrement table with two decrements; cause 1 is death, and cause 2 is disability. Suppose we wish to use this table to construct a regular single-life table relating only to death, as introduced in Chapter 3. Would webejustifiedinsimplytaking q (1) x asthemortalityrateforage( x),inthesingle-lifetable?The answerisadefinite‘no’.Westressagainthat q(1) x isnottheprobabilitythat(x )willdiewithin the year, but rather the probability that (x) will die within the yearbeforebecoming disabled. Theactualvalueof qxshouldbelargerthan q(1) x .Thesituationisthesameasthemachinemodel intheprevioussection.Apersoncoulddieduringtheyear,afterhavingalreadyleftthegroup by reason of disability. In other words, the mortality rate for age (x) in the single decrement table is analogous to the primed rate in the machine model, and will be denoted byq ′(1) x . We may also want to computeq′(2) x , the probability that (x) will become disabled during the year, assuming that no other causes of failure are operating. This admittedly requires a stretch of the imagination, as the possibility of death would always appear to be present. We have to imagine, however, that we are computing what these probabilities would be if we couldsomehoweliminatethepossibilityofdeath.Thebestapproachissimplytoviewthings in terms of the machine model. That is, we have to think of a person as having a ‘disability component’ that exists independently and could in fact continue to operate even after death. For each causej, the collection of values {q ′(j) x } for various values ofx is known as the associated single-decrement table for cause j. As we have stressed above, these rates give probabilities of failure for the particular causej, assuming thatnoother causes of decrement are operating. Theproblemoftenarisesofgoingfromonesetofratestoanother.Wemight,forexample, construct the multiple-decrement table in the first place by using our knowledge of the associated single-decrement tables. Alternatively, we might have first constructed a multiple tablebyactuallyobservingtheeffectofthevariouscausesactingtogetherandwishtousethat to construct the associated single-decrement tables. In order to do so we can directly apply our machine model and use the twogiven methods, provided we make the key assumption of that model, namely that the various causes are actingindependently.One can certainly argue in many applications that this does not hold, but the standard actuarial model incorporates this independence assumption. The more general theory is discussed in Chapter 17. 176 MULTIPLE-DECREMENT THEORY Consider a numerical example. Suppose that in our death–disability table 𝓁(𝜏) x = 1000, d(1) x = 300, d(2) x = 100. Wewillcomputetheassociatedsingle-decrementratesusingmethod1.Fromformula(11.17), we have q ′(1) x = 1−0.63∕4 = 0.318, q′(2) x = 1−0.61∕4 = 0.120. Supposewearegiventheabovesingle-decrementratesandwanttocomputethemultiple- decrement table. If we use the Method 1 formula (11.18) we necessarily will get back to unprimed rates of 0.300 and 0.100 that we started with, as the reader should verify. If we use the Method 2 formula (11.20) instead, we will get q (1) x = 0.318(1−0.06) = 0.299, q(2) x = 0.120(1−0.159) = 0.101, verifying as we saw in Example 11.3 that the two methods are different, but will generally give answers that are close. 11.7.2 Forces of decrement in the associated single-decrement tables Refer back to the statement made in Section 11.4 that the multiple-decrement model is often constructed from the forces of decrement. A question that naturally arises is, how are these forcesofdecrementobtained?Theansweristhattheycanbetakenastheforcesofdecrement in the associated single-decrement tables, namely 𝜇 ′(j) x (t) =− d dttp′(j) x tp′(j) x . Note that 𝜇′(j) x (t) is defined differently than the quantity𝜇(j) x (t) as given in (11.9) but there is reason to believe that they might be the same. We observed above that we expect that q ′(j) x > q(j) x sincethepersonmayfailwithintheyearfromcause jafterfirstfailingfromanother cause,butthisargumentdoesnotholdwhenwearespeakingof instantaneousratesoffailure rather than failure over a period of positive length. This by itself is not sufficient for equality – see Exercise 11.15 – and we need the independence hypothesis. In Section 17.4.5, we give a formal proof and provide an explanation of the fact that by virtue of independence, 𝜇 ′(j) x (t) = 𝜇(j) x (t). (11.23) We can now summarize the main conclusions which follow from our independence hypothesis. r Given the forces of decrement in the associated single-decrement tables, we can uniquely determine the multiple-decrement model, using (11.23) together with the procedure of Section 11.4. ASSOCIATED SINGLE-DECREMENT TABLES 177 r Givenonlytheratesofdecrementintheassociatedsingle-decrementtables,wecannot uniquely determine the multiple-decrement model, and must choose from different methods, the most notable being the two described in Section 11.7.1. 11.7.3 Conditions justifying the two methods In this section we give conditions to justify the Method 1 and Method 2 introduced above, andalsosomeadditionalproceduresthatapplyinspecialcases.Westartwiththeeasiercase. Method 2 This will hold provided we have auniformdistributionofdecrementsineachof theassociatedsingle-decrementtables. Thatis,foraninteger x,0 < t< 1,and j= 1,2, …,m, tq′(j) x = tq′(j) x . (11.24) To show this, we first note that as in (8.16) tp′(j) x = e−∫ t 0 𝜇′(j) x (r)dr. (11.25) and then, invoking (11.23) and applying (8.20) tp′(j) x 𝜇(j) x (t) =t p′(j) x 𝜇′(j) x (t) = q′(j) x , (11.26) From (11.13) sq(1) x = ∫ 1 0 tp′(1) x (1−t q′(2) x )𝜇(1) x (t)dt. Now substitute from (11.24) and (11.26) and integrate to obtain the Method 2 formula. The sameprocedureworksforthegeneralcaseof mdecrements.Wejustneedtoincludeadditional factors of (1−tq ′(j) x ) in the integrand. Method 1 We need a preliminary definition. We say that theUniform Ratio Hypothesis (UR)holdsataninteger(x )ifthereareconstants Kj forj= 1,2, …msuchthatfor0 < t< 1, 𝜇(j) x (t) 𝜇(𝜏) x (t) = Kj. Asanexampleinthetwodecrementcases,let 𝜇(1) x (t) = 2𝜇(2) x (t)for0 < t< 1.ThenURholds atxwithK1 = 2∕3,K2 = 1∕3. Suppose now that UR holds atx. From (11.13) withs=1, q(j) x = Kj∫ 1 0 tp(𝜏) x 𝜇(𝜏) x dt= Kjq(𝜏) x . (11.27) 178 MULTIPLE-DECREMENT THEORY Finally, (11.23), and (11.27) show that p′(j) x = e−∫ 1 0 𝜇′(j) x (r)dr = e−∫ 1 0 𝜇(j) x (r)dr = e−Kj∫ 1 0 𝜇(𝜏) x (r)dr = p(𝜏) x Kj = p(𝜏) x q(j) x ∕q(𝜏) x , which is the Method 1 formula for going fromq(j) x toq′(j) x . TherearecertainassumptionsthatwillimplyUR.Oneobviousconditionisthattheforces of failure for each decrement are constant over the year from agexto agex+1. That is, for 0 < t< 1 andj= 1,2, …,m, 𝜇x (j)(t) is independent oft. UR also holds at all ages under the assumption of a uniform distribution of deaths in the multiple-decrementtable. That is, for an integer (x), 0< t< 1 andj= 1,2, …,m, tq(j) x = tq(j) x . (11.28) If (11.28) holds, we differentiate (11.13) with respect tosto conclude that for 0< s< 1, q(j) x = sp(𝜏) x 𝜇(j) x (s). Adding this equality for allj q(𝜏) x = sp(𝜏) x 𝜇(𝜏) x (s). Dividing the first equation above by the second shows that UR holds atx. Itshouldbenotedthatcondition(11.28)isasomewhatunnaturalassumptionwhichdiffers from(11.24),andisthereforeinconsistentwiththeusualassumptionsofUDDforeachsingle decrement. To see this, suppose that we assume (11.28) in a situation with two decrements where q ′(j) x = 0.1, for j= 1,2. Then Method 1 applies, so from (11.18),q(j) x = 0.095. Our assumption then gives1∕2q(j) x = 0.0475. Now we can also apply Method 1 to rates over half- yearperiods.From(11.17)wehave 1∕2q′(j) x = 1−0.951∕2 = 0.0487.However,theassumption of UDD for each decrement would give1∕2q′(j) x = 0.05. Special Cases There are decrements that by their nature will not occur uniformly over the year, but rather at definite discrete points. A prime example is withdrawal from an insurance policy.Nobodyislikelytowithdrawinamiddleofapremiumpayingperiodsothatthesecan beassumedtooccuronlyonpremiumduedates.Ifweassumethatalldecrementseitherfollow such a discrete pattern or are uniformly distributed over each year, there is a fairly simple proceduretomovefromtheprimedtounprimedrates.Thisisillustratedbythefollowingtwo examples. Example 11.4 In a double-decrement model with decrements of (d) for death and (w)f o r withdrawal, we are given that deaths are uniformly distributed over each year of age in the single-decrementtable,whileone-thirdofthewithdrawalsinanyyeartakeplaceinthemiddle of the year and two-thirds occur at the end of the year. Given thatq ′(1) x = 0.20,q′(2) x = 0.36 calculateq(1) x andq(2) x . ASSOCIATED SINGLE-DECREMENT TABLES 179 Solution. To calculateq(2) x we add the probability of withdrawal before death in the middle oftheyeartotheprobabilityofwithdrawalbeforedeathattheendofyear.Theprobabilityof a withdrawal in the middle of the year is (1/3)(0.36)= 0.12. In order that this occurs before death, we require that death has not yet occurred by the middle of the year, which by the uniformdistributionassumptionhasaprobabilityof1 −(1∕2)(0.20) = 0.9.Arguingsimilarly for the end of the year we have that q (2) = 0.12(0.9)+0.24(0.8) = 0.300. Since q(𝜏) x = 1−(0.8)(0.64) = 0.488, we know thatq(1) x = 0.188. As a check we can calculate the latter figure directly. The probability of death before withdrawal in the first half of the year is just the probability of death in the first half of the year, since there are no withdrawals during that time. This is equal to 0.10. The probability of death before withdrawal in the second half of the year is the probability of death in the second half, multiplied by the probability that withdrawal didnot take place in the middle of the year. This equals (0.1)(1−0.12) = 0.088. The sum of the two probabilities is indeed equal to 0.188. Note that in both calculations above, we used the independence assumption in order to multiply relevant probabilities. Example 11.5 In a multiple-decrement model with three decrements, failures from causes 1 and 2 both occur uniformly over each year in the single-decrement table. For cause 3, 60% of the failures in any year occur 1/4 of the way through the year and the other 40% occur 3/4 of the way through the year. Find formulas that give the unprimed rates in terms of the primed. Solution. To simply the notation, fix any agex and leta ′,b′,c′ denote q′(j) x for j= 1,2,3, respectively,andlet a,b,cdenoteq(j) x fori= 1,2,3,respectively.ProceedasinExample11.4 only now we need that survival from both decrements 1 and 2 occurred at the two points in question. This gives c= 0.6c ′ ( 1− 1 4a′ )( 1− 1 4b′ ) +0.4c′ ( 1− 3 4a′ )( 1− 3 4b′ ) = c′ − 9 20a′c′ − 9 20b′c′ + 21 80a′b′c′. Formula (11.30) says that a+b+c= 1−(1−a′)(1−b′)(1−c′). Therefore we know by symmetry (since there is nothing to distinguish decrements 1 and 2) that a= a ′ − 1 2a′b′ − 11 20a′c′ + 59 160a′b′c′, b= b′ − 1 2b′a′ − 11 20b′c′ + 59 160a′b′c′. 180 MULTIPLE-DECREMENT THEORY There is a much longer procedure for the first two decrements,which might be done as a check. To illustrate with decrement 1, we use the formula a= ∫ 1 0 a′(1−tb′)tp′(3) x dt, whichfollowsfrom(11.16)aftersubstitutingfrom(11.26).Calculationofthiswillnecessitate integrating over several intervals, since the formula for tp′(3) x changes we get a= a′ [ ∫ 1∕4 0 (1−tb′)dt+∫ 3∕4 1∕4 (1−tb′)(1−0.6c′)dt+∫ 1 3∕4 (1−tb′)(1−c′)dt ] , We leave it to the reader to verify that this gives the same answer as above. 11.7.4 Other approaches We will describe two additional methods, which were popular in pre-computer days for calculating primed quantities from the unprimed. Neither of them can be easily inverted to give the unprimed in terms of the primed. They do satisfy (11.14) but their main drawback is that they donotsatisfy the consistency relation (11.15) which must hold if we assume that the causes operate independently. On the other hand, they give approximate answers quickly and can give additional insights into the difference between the primed and unprimed rates. Looking at the particular data given above in Section 11.7.1, we see that in order to compute q ′(1) x we want to estimate how many of those 100 people who became disabled during the year will die during the year after becoming disabled. Suppose we postulate that, onaverage,peopleleavefromdisabilityinthemiddleoftheyear.Sincethechanceofleaving by death during the year is 0.30, there should be approximately a chance of 0.15 of dying in the second half of the year. In other words, approximately 15 of the 100 disabilities would die during the year after becoming disabled and we could estimateq ′(1) = 0.315, not too far from the result obtained above by method 1. Our final method gives a more plausible argument for decrements other than death, for as we already noted, it is difficult to imagine someone becoming disabled after they have died.Theargumentisasfollows.Theprobabilityofdeathduringtheyearshouldbeobtained by taking the number of people who died during the year and dividing by the number of people who were in the observation period for that year. In the above example we observed 300 deaths. However, we did not really have a full 1000 people under observation for the entire year. Some of the group dropped out due to disability, and were no longer observable as potential deaths. Consider extreme cases. If all the people becoming disabled did so at the beginning of the year we would have only 900 people left. If all of them dropped out at the end of the year, then we would indeed have the full 1000 under observation. We can conclude that on average we would have the equivalent of 950 people to observe and we canestimate q ′(1) x = 300∕950 = 0.316.Similarly,wecouldestimate q′(2) x = 100∕850 = 0.118. Again, we obtain answers close to those of method 1. EXERCISES 181 Both of these alternate methods can be applied to the general case ofmdecrements. To state the formulas it is convenient to define, for each causej, q(−j) x = ∑ i≠j q(i) x , the probability that (x) will leave the group within the next year from some cause other than j. Our first argument above then gives the approximation q′(j) x ≐ q(j) x ( 1+ 1 2q(−j) x ) , (11.29) while the second argument leads to the approximation q′(j) x ≐ q(j) x 1− 1 2q(−j) x . (11.30) Thesegeneralformulasconfirmthatthetwolattermethodsgivedifferentapproximations,but the answers are usually close, since for a small value ofa,1 +ais close to (1−a) −1. The reader may find it instructive to take the case of two decrements and substitute algebraically into the right hand side of (11.16) from both (11.29) and (11.30). In both cases one obtains a result that is somewhat less than the left hand side of (11.16), verifying the inconsistency. However, we see that the difference is likely to be small. It is a sum of terms each of which is close to the product of three or moreq (j)s. Notes and references Promislow (1991) discusses select multiple-decrement tables. Exercises Type A exercises 11.1 You are given the following portion of a double-decrement table. (Blanks indicate data that you must calculate from the given figures if you need them.) x 𝓁(𝜏) x d(1) x d(2) x 50 – 100 300 51 700 50 – 52 470 40 – 53 320 Find the probability that (50) will fail first from cause 2 between the ages of 51 and 53. 182 MULTIPLE-DECREMENT THEORY 11.2 You are given the following portion of a double-decrement table. (Blanks are as in Exercise 11.1.) x 𝓁(𝜏) x d(1) x d(2) x 50 1200 100 300 51 – 200 – 52 300 (a) A 2-year insurance contract on (50) provides for benefits paid at the end of the yearoffailureifthisoccurswithin2years.Thebenefitpayableis1unitiffailure is from cause 1, or 2 units if it is from cause 2. If the interest rate is a constant 50% per year, find the present value of the benefits. (b) Find the associated single-decrement rates,q ′(1) 51 and q′(2) 51 , using the Method 1 approximation. 11.3 In a table with three decrements you are given that, for allx∈ [0,100), 𝓁(1) x = 10(100−x), 𝓁(2) x = 20(100−x), 𝓁(3) x = 30(100−x). Find the probability that (50) will fail first from cause 1 between the ages of 60 and 70. 11.4 Suppose that in a double-decrement model, you are given the associated single- decrement rates q′(1) x = 0.3, q′(2) x = 0.51. Compute q(1) x andq(2) x by Method 1 and Method 2. 11.5 Adisabilityinsurancepolicyprovidesforpaymentsatthemomentofdisabilityshould this occur within 10 years. The amount of the benefit at timetis e0.10t. The policy is purchased by level annual premiums payable continuously for 10 years until either death or disability occurs. Nothing is paid on this policy if the insured dies before becoming disabled. If the force of disability is a constant 0.03, the force of mortality is a constant 0.06, and the force of interest is a constant 0.05, find the annual rate of premium payment. 11.6 Given q (1) x = 0.05,q(2) x = 0.08,q(3) x = 0.10, use Method 1 to findq′(j) x , j= 1,2,3. 11.7 Given q′(1) x = 0.1,q′(2) x = 0.2,q′(3) x = 0.25, use both Method 1 and Method 2 to find q(j) x , j= 1,2,3. Type B exercises 11.8 Suppose that in a double-decrement model,q′(1) x = q′(2) x . Show that Method 1 and Method 2 lead to the same answer for the unprimed rates. EXERCISES 183 11.9 In a double-decrement table, with decrementsdfor death andwwithdrawal, we have that q(d) x is a constant 0.10 for allx and q(w) x is a constant 0.02 for allx. The rate of interest is a constant 0.06. An insurance policy pays 1 at the end of the year of death. Nothing is paid to a withdrawing policyholder. Find the present value of the benefits on this policy. 11.10 Inamultiple-decrementmodelwithtwocausesofdecrement,youaregivenassociated single-decrement rates ofq ′(1) x = 0.20,q′(2) x = 0.36. Calculate the double-decrement rates q(1) x and q(2) x in each of the following three cases: (a) by Method 1; (b) by Method 2; (c) assuming that failures from cause 1 are uniformly distributed over the year, but that the failures from cause 2 all take place at a point three quarters of the way through the year. 11.11 In the case of two decrements, compute the difference between the right hand side and the left hand side of (11.16) in terms ofq (1) x and q(2) x , under each of (11.29) and (11.30). 11.12 Is the following statement true or false? np(j) x +nq(j) x = 1. If false, give a correct version. 11.13 In a model with two decrements, you are given that, for allx, 𝜇(1) x (t) = 0.02 for allt and𝜇(2) x (t) = (10−t)−1 for 0≤ t< 10. Find4q(2) x . 11.14 In a model with two decrements you are given thatq(1) x = 0.1274, q(2) x = 0.1674 and that there is a uniform distribution of decrements in each year for both of the single- decrement tables. Findq ′(1) x andq′(2) x . 11.15 Consider a double-decrement model, where we relax the hypothesis that the causes operate independently. Cause 1 is death, and failure from cause 2 occurs at timetif death occurs at timet−1. Show that 11.23 does not hold. 11.16 In a multiple-decrement model with 4 decrements,q′(i) x = i∕10,i= 1,2,3,4. Each decrement is uniformly distributed over each year in the single-decrement table. Calculate q (j) x , j= 1,2,3,4. 11.17 For an insurance contract with quarterly premiums, withdrawals will occur at times 1/4,1/2,3/4and1.Studiesshowthatataparticularage x,one-thirdofthewithdrawals occur at time 1/4, one-third occur at time 1, one sixth occur at time 1/2 and one sixth occur at time 3/4. Assume that deaths are distributed uniformly over the year. If q ′(d) x = 0.12 andq′(w) x = 0.20, findq(d) x andq(w) x .( H e r e ,ddenotes death andwdenotes withdrawal.) 11.18 Amultiple-decrementmodelhas3decrements.Fordecrement1,60%ofthefailures occurattime1/3,and40%occurattime2/3.Fordecrement2,failuresoccureitherat times1/4or3/4inequalnumbers.Fordecrement3,failuresareuniformlydistributed over each year. Find formulas giving the unprimed rates in terms of the primed rates. 12 Expenses and profits 12.1 Introduction Provision for expenses and profits are two important features that we have ignored in our models up to now. Premiums paid on insurance and annuity contracts must not only provide benefits,butmustalsocontainanextraamounttocovertheexpensesofoperatingthebusiness. Inadditionthepremiumsmustbesufficienttogenerateprofitsasareturntotheinvestorswho provide the capital necessary to start an insurance operation. We will begin with a discussion of expenses, and the first step is to distinguish between three basic categories of regular periodic expenses. These can depend on the premiums, on the amount of the benefits, or be constant per policy. The major expense of the first type is commissions paid to the agents selling the policies. It is traditional for their compensation to take the form of certain percentages of the premiums. There are various expenses that will depend on the amount of the benefits. For example, larger policies will require additional efforts and expenses in the selection procedure to verify that the individual is a sound risk. Finally, there are expenses, such as setting up records for a new policy, that are largely independent of the size of premiums or benefits, and are fixed for each policy. Apeculiarfeatureofexpensesforlifeinsuranceisthatalargeportionoftheseareincurred inthefirstyearofthepolicy.Commissionspaidontheinitialpremiumaretraditionallymuch higher than those paid in subsequent years (known as renewal premiums). The expenses of selectionandofthesettingupofpolicyrecordsalsooccuratthebeginning.Oneconsequence ofthisisthattheamountstobepaiduponwithdrawalbecomeanimportantconsideration.The potential loss on withdrawal must be taken into account, since if a policyholder withdraws early, there may be no chance for the high initial expenses to be recovered through the premiums. Theexpensesmentionedabovewillbepaidatfixedtimes.Buttherearealsoextraexpenses incurredwhenalossoccursandaclaimisreportedtotheinsurer.Theinsurermustinvestigate to ensure that the claim is legitimate and then arrange to disburse the benefits. Some of these FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial INTRODUCTION 185 expenseswoulddependonthepolicyamount,asmorecarewouldbetakenwithalargepolicy. Others would be fixed. These are often referred to assettlementexpenses. For a typical life insurance or annuity policy, the mathematical treatment of expenses is quite straightforward. Suppose the insurer determines that an expensee k will be incurred at each timek= 0,1, …. Then the insurer is simply providing an additional life annuity with benefit vectore = (e0,e1,…). Similarly the expenses of claim settlement would be handled byanincreasetothedeathbenefit.Ofcoursethepolicyholderdoesnotreceivetheseamounts. They are paid to those providing the goods or services involved in the expense, but they still must be provided for from the premiums. Premiums which make provisions for expenses as well as benefits will be referred to as expense-augmentedpremiums. Some authors use the termgrosspremiums. We will however reserve this term for the premiums actually charged, which as pointed out in Section 4.6.1, can take into account factors other than the benefits and expenses. such as profits, which we discuss later on in this chapter. Example 12.1 A whole-life policy issued at agex, with level annual premiums paid for life, provides for a death benefit paid at the moment of death. The expenses are as follows: in the first year, 70% of the initial premium, 1% of the face amount, and 30 per policy; in years 2 to 10, 10% of each premium, 0.5% of the face amount, and 10 per policy; after 10 years, 5% of each premium, 0.2% of the face amount and 5 per policy; the settlement expense is 100 per policy plus 0.5% of the face amount. Assume that the expenses in any year are paid at the beginning of the year. Find a formula to compute the annual expense-augmented premium for a policy with a constant death benefit of 200 000. Solution. Let G be the expense-augmented premium. We simply proceed as we did in Chapter5,equatingthepresentvalueofpremiumswiththepresentvalueofthedeathbenefits plus the present value of the additional expense benefits. Of course the latter depend partly onG, which just means we have to solve a simple equation at the end. It is best to consider two expense vectors. These are e = (2030,1010 9,405∞), which covers the expenses depending on face amount as well as the fixed expenses, and r = (0.07, 0.19,0 .05∞), which gives the proportion of the premium for premium-based expenses. The settlement chargesmeanthattheamountpaidondeathiseffectively201100.Wethenhavetheequation G̈a x = 201100 ̄Ax + ̈ax(e)+G̈ax(r), from which we get G= 201100 ̄Ax + ̈ax(e) ̈ax(0.3, 0.99,0 .95∞). (12.1) Note that the vector in the denominator is just (1∞ −r). 186 EXPENSES AND PROFITS A more realistic model would also involve withdrawal. We would start with a double- decrementtablewithdecrements dfordeathand wforwithdrawal.Wewouldalsoneedacash valuevector cv,where cvk wouldequalthecashvaluegiventoapolicyholderswhowithdraws at timek+1. The subscript is chosen to correspond to the death benefit vector. (Note that we can safely assume that for a policy with premiums paid annually, withdrawal will only occur at an integer time, when a new premium is due.) We need as well a vectore w(standing for theexpensesofwithdrawal ) whereew k would be the expenses incurred for withdrawal at time k, analogous to the settlement expenses added to the death benefit. Formula (12.1) would be modified to G= 201100 ̄A(d) x +A(w) x (cv+ew)+ ̈a(𝜏) x (e) ̈a(𝜏) x (1∞ −r) . 12.2 Effect on reserves Suppose that we compute reserves by adding expenses to the benefits and using expense- augmentedpremiumsinplaceofnet.Theresultingquantitiesareknownas expense-augmented reserves, while the reserves calculated in Chapter 6 are known asnet premium reserves.Is it natural to ask as to how these two quantities compare. In the usual level premium policy, theexpenses constituteadecreasing sequence of benefit payments that arepaid forbyalevel addition to premiums. The extra expense charge included in the premium is less than needed to cover the expenses in early years, and more in later years. This means that typically the expense-augmented reserves will be less than the net premium reserves. To put it another way,thehighinitialexpenses areareceivable totheinsurerthatwillbecollected fromfuture premiums, and this causes a reduction in liabilities. Insurance regulators have usually taken the viewpoint that one should not count on these receivables since the policyholder might withdraw and never pay them. It is traditional that reserves be calculated without taking into accountexpenses,andusingnetinplaceofexpense-augmentedpremiums.Soevenifrealistic mortalityandinterestassumptionswereusedincalculatingreserves,ignoringexpensescauses higher reserves, and will cause losses to be shown in the early years of a policy. Consider a simple example. Suppose that for a certain policy issued to (x) with benefits paid at the end of the year of death,b 0 = 1000,qx = 0.2, i= 0.10, the net premium payable at time 0 is 250, the expense-augmented premium is 300, and the total expenses payable at time 0 are 100. The expense-augmented premium reserve at t time 1 would be (300−100)1.10 0.8 −10000.2 0.8 = 25. while the net premium reserve at time 1 is 2501.10 0.8 −10000.2 0.8 = 93.75. Even if the mortality and interest matched exactly those of the reserve assumptions, the use of net premium reserves will result in a loss of 68.75 per policy in the first year. This arises from the fact that the provision for expenses is made by an extra charge of only 50 to the REALISTIC RESERVE AND BALANCE CALCULATIONS 187 level premium, but this does not cover the actual expense of 100. (Of course, as we showed in Chapter 6, there will be corresponding gains in future years, when the actual expenses are less than the 50 allowed for in the premium.) To alleviate this distortion of the profitability in the first year, many regulatory bodies permit reserves to be calculated by methods that are generally known in North America as modified reserve systems. (Similar procedures are popular in some European countries and are known asZilmerizedreserves.) Recall the termvaluationpremium to refer to premiums used in calculating reserves, which we introduced in Chapter 6. We stress again that these are not necessarily (and in fact almost never) the same as the insurer actually charges. The modifiedsystemsstillcallforexpensestobeignored,buttheyallowforlowerinitialvaluation premiums, followed by higher valuation premiums for the later durations, calculated so that valuationpremiumsremainactuariallyequivalenttobenefits.Thiseffectivelyrecognizesthat the insurer has a smaller premium in the first few years to provide the benefits since a large amount of these premiums must go to pay the high initial expenses. Therearemanysuchsystems.Themostbasicisthe fullpreliminaryterm method.Suppose wehavealifeinsurancepolicywithlevelpremiumspayablefor nyears,andbenefitspayable at the end of the year of death. The initial valuation premium is taken asb 0v(1)qx, the exact amountneededtoprovidethebenefitsforthefirstyear,followedbyalevelvaluationpremium for all subsequent years. This will automatically make 1V = 0. This level valuation premium for years after the first will be A[x]+1(b◦ 1) ̈a[x]+1(1n−1), which will provide for the benefits after the first year. This effectively means that the entire first-yearvaluationpremium,lesstheamountneededforthebenefitsaccordingtothereserve basis, is available to use for expenses. There are several variations to full preliminary term. In some cases there is a recognition that for policies with high initial premiums, such as endowment insurance, or premiums payableoveralimitedperiod,thisprovidesmoreexpenseallowancethanisneededinthefirst year,andthefirst-yearvaluationpremiumistakentobehigherthan b 0vqx.Inothercasesthere may be more than one step. There will be a low premium used in the first year, a somewhat higher one for some duration, for example the next 10 years, and a still higher one after that. We will not discuss the details of the various methods here. Some of these are dealt with in the exercises. 12.3 Realistic reserve and balance calculations As we have indicated, insurers are required to calculate reserves in accordance with the particular regulations that apply in their jurisdiction. These methods are intended mainly to ensure that policyholder will receive their promised benefits. These are often referred to as statutory reserves. There are however many reasons for the insurer, to calculate reserves and balances on a more realistic basis which use expense-augmented or gross premiums and expenses, rather than net premiums with expenses ignored, and which takes into account withdrawals. An acronym that is sometimes used here is that of a ‘GAAP’ reserve, the acronym standing forgenerally accepted accounting principles.Such a reserve would then 188 EXPENSES AND PROFITS be calculated as the present value of all future benefits, including death benefits, expenses, surrender benefits, less the present value of future premiums. It is useful to replace our basic recursion formula (6.8) with the following more general version. For simplicity we assume aggregate mortality, and benefits payable at the end of the year of death. Consider a policy withannualpremiumsof 𝜋 k,deathbenefitsof bk,cashvaluesof cvk,anexpenseof rk perunit of premium paid at timek, fixed expenses at timekof ek, claim settlement expenses ofed k in thecaseofdeathbetweentime kandtime k+1,andsurrenderexpensesof ew k forsurrenderat timek+1.Ourbasicrecursion(6.8),whenmodifiedforexpensesandwithdrawal,becomes k+1V = [kV+𝜋k(1−rk)−ek](1+ik)−q(d) x+k (bk +ed k −k+1 V) −q(w) x+k (cvk +ew k −k+1 V) , (12.2) and solving fork+1V, we obtain the Fackler type equation k+1V = [kV+𝜋k(1−rk)−ek](1+ik)−q(d) x+k (bk +ed k ) −q(w) x+k (cvk +ew k ) 1−q(d) x+k −q(w) x+k . (12.3) We want to discuss three main quantities which arise from the above recursion. They differ in the choice of premiums and initial values. Expense-augmented reserves. These reserves, as defined above, can be calculated from (12.3) with𝜋k equal to the expense-augmented premium. Gross-premium reserves.These are simply the present value of future benefits and expenses less the present value of futuregrosspremiums. They can be calculated from (12.3) with𝜋k equal to the gross premium. AssetShares. Thesearerealisticbalancecalculations.Theygivetheaccumulatedvalueof thepremiumslesstheaccumulatedvalueofthebenefitsandexpenses,Theycanbecalculated from (12.3) using gross premiums, but now with an initial value of 0 at time 0. (Different notation is often used here to distinguish them from reserves. We will write AS k in place of kV.)Thenamecomesfromthefactthattheyrepresentthatshareoftheinsurer’sassetswhich are attributable to the particular contract. We will use the following simplified example to illustrate the difference between these quantities. Example 12.2 Atwo-yearpolicy on( x)provides fordeathbenefits of1000 paidattheend oftheyearofdeath.Premiumsof160arepaidfor2years.Thereareexpenses,incurredatthe beginningofeachyear,of20%ofthepremiumplus18inyear1,and5inyear2.Youaregiven that i= 100%,q x = 0.2, qx+1 = 0.3. There are no withdrawals. Find the expense-augmented reserves, gross-premium reserves, and asset shares. Solution. Let Gdenote the expense-augmented premium. The present value of benefits and expenses is 1000[(0.5)(0.2)+(0.25)(0.8)(0.3)]+18+0.2G+5(0.5)(0.8)] = 180+0.2G. Then G[1+(.5)(.8)] = 180+0.2Gshowing thatG=150. PROFIT MEASUREMENT 189 For the expense-augmented reserves, 0V = 0, 1V = [0.8(150)−18)×2−200]∕0.8 = 5, 2V = 0. For gross-premiums reserves, 0V = 180+0.2(160)−160(1.4) =− 12, 1V = [(−12+0.8(160)−18)×2−200]∕0.8 =− 5, 2V = 0 For the asset shares AS0 = 0, AS1 = [(128−18)×2−200]∕0.8 = 25, 2AS2 = [25+160−5]×2−300]∕0.7 = 85.71. In the following section we will revisit this example, and interpret the figures in terms of profit calculations. Remark For simplicity, some examples involving expenses will ignore withdrawals. This doesnecessarilymeanthatwithdrawalsarenotallowedaswestatedinthepreviousexample. The implicit assumption is that at each duration, the cash value together with any expense of withdrawal is precisely equal to the reserve on the policy. In such a case the withdrawal rates and cash values have no effect on premiums or reserves, and need not be taken into account. 12.4 Profit measurement 12.4.1 Advanced gain and loss analysis WewanttocontinuetheprofitanalysiswebeganinChapter6,butnowincorporatingexpenses and withdrawals, and also by considering a more extensive context. We start with a very general setting. We are given a particular insurance or annuity policy, with a description of the benefits and cash values, and a set of reserves. The latter couldbenetpremiumreserves,expense-augmentedreserves,gross-premiumreserves,oreven somethingelse.Inadditionwearegivenabasisforallrelevantfactors,knownasthe profittest basis. These are interest rates, rates of mortality and withdrawal, expenses and the premiums which will be charged. We want to determine the profit we will have in a specified period, if our actual experience is according to the given profit test basis and we use the specified reserves to measure profit. There are different contexts possible. The profit test basis could be a hypothetical test basis, to see what profits will emerge under our assumptions, and the major goal is to see if the proposed premiums will achieve a desired level of such profits. Alternativelythiscalculationcouldbecarriedoutafterthefactwiththeprofitbasisreflecting the actual observed experience, and the result will be the actual profit that is achieved. 190 EXPENSES AND PROFITS Inanyevent,fortheperiodrunningfromtime ktok+1wehaveaprofitofPr k+1 consisting of the right hand side of (12.2) less the reservek+1V. That is Prk+1 = (kV+𝜋k(1−rk)−ek)(1+ik)−q(d) x+k (bk +ed k −k+1V) −q(w) x+k ( cvk +ew k −k+1V ) −k+1 V, (12.4a) = (kV+𝜋k(1−rk)−ek)(1+ik)−q(d) x+k (bk +ed k ) −q(w) x+k (cvk +ew k ) −p(𝜏) x+k k+1V. (12.4b) The two versions represent two possible points of view, leading to the same result, as we already indicated in Chapter 6. In (12.4a) the insurer sets up the reserve for everyone and then needs to pay only the amount of the benefits above the reserve in the case of death or surrender, while in (12.4b), the insurer pays the benefits in full but then needs to set up a reserve only for the survivors. Version (12.4b) is normally the most efficient for calculation. Yet another version is to write Pr k+1 = (𝜋k(1−rk)−ek)(1+ik)−q(d) x+k (bk +ed k ) −q(w) x+k (cvk +ew k ) −Δk(V), (12.4c) where Δk(V) = p(𝜏) x+k k+1V−k V(1+ik) (12.5) represents the increase in reserves over the year. This version simply says that profit is what is left over after we accumulate the net amounts collected, and provide for benefit payments, expenses and the increase in reserves. Example 12.3 Refer back to Example 12.2 and calculate Pr k when the the underlying reserves are: (a) gross-premium reserves; (b) expense-augmented reserves. In each case, the profit test basis is the same as the reserve basis. Solution. (a) It is clear that Pr 1 = Pr2 = 0. So what happened to the profits? The answer is that with gross-premium reserves, and observed experience the same as the reserve basis, they all occur at time 0. The reserve of−12 at this time, means that the insurer could take 12 out of the company surplus for each such policy and still have enough left to provide for the reserves. This shows that in addition to the definition of Pr k given above we must add the statement that Pr0 =− 0V. (b) Using (12.4(b)), Pr1 = (128−18)2−0.2(1000)−0.8(5) = 16. Pr2 = (5+160−5)2−0.3(1000) = 20. Notethedifferenceintheincidenceofprofitforthedifferenttypesofreserves.Withgross- premium reserves, all the profit is recognized at the beginning. With the expense-augmented PROFIT MEASUREMENT 191 reserves, the profit emerges gradually as the extra amounts above the expense-augmented premiums come in. In this case there is an extra 10 received in the first premium, but this is reduced to 8 after the percentage of premium expense. This accumulates in 1 year to 16. Similarlytheextra10inthesecondpremiumaccumulatesto20attheendofthesecondyear. It will instructive for the reader to verify that in the long run, that the reserve basis will not affect the actual profits received. This example also indicates that we can view asset shares as the sum of the expense- augmented reserves plus an accumulation of the profits at interest and survivorship. For example, AS 2 = 0+8 ( 4 0.56 ) +10 ( 2 0.7 ) = 85.71. as calculated above. 12.4.2 Gains by source We now elaborate on the decomposition of gains by source, as initiated in Section 6.4.1. We willinfactconsiderasomewhatmoregeneralsituation.Supposewehavetwodifferentbases under consideration, one in which quantities are starred and the other unstarred. In the most commontypeofapplicationtheunstarredsymbolsrefertotheexpectedbasis,andthestarred will refer to what actually happened, but mathematically these could be any two bases. We thenhavetwoprofitcalculations,Pr ∗ k andPrk andweareinterestedinthedifferencePr ∗ k −Prk which in the common situation will be the excess of actual profit over the expected profit. In the case where the reserve basis coincides exactly with the unstarred basis, we have that Pr k = 0 for allkand the difference will just be Pr∗ k . This was precisely the point of view we took in Chapter 6. The goal is to decompose this difference into the various gains by source. As well as the mortality and interest gains of the Chapter 6 model, we now have additional sources. Oneisthegainfromwithdrawal,whichiscalculatedanalogouslytothemortalitygainas ( q (w) x+k −q∗(w) x+k ) (cvk −k+1 V). In addition we have gains from the various types of expenses. These present a compli- cation, since they are invariably tied up with other factors. Consider for example the case of death settlement expenses. As a particular example, suppose we haveq (d) x+k = 0.07, q∗(d) x+k = 0.05, ed k = 200 ande∗d k = 100.Soourmortalityexperienceismorefavourablethanexpected and the costs of settling each claim are less than expected. We gain in both ways and in fact the total gain per policy will be 0.07(200)−0.05(100)=9. We would like to divide this gain up into that attributable to mortality, and that attributable to lower settlement expenses. A somewhat surprising fact is that there seems to be no reasonable way to do this uniquely. To illustrate, consider the general situation, which applies as well to withdrawal expenses. (For simplicity in notation we will omit superscripts and subscripts.) We have a total gain of (qe−q ∗e∗).Now we can write this in two distinct ways, namely (q−q∗)e+(e−e∗)q∗,or( q−q∗)e∗ +(e−e∗)q. 192 EXPENSES AND PROFITS In either case it would be reasonable to take the first term to be the mortality gain and the second term to be the expense gain. However, there is nothing to choose between them, and they give different answers. One way to decide is to determine an order in which we will calculate the gains. If we decide to look at mortality first, it would be natural to ignore the expense differential at this point and take (q−q ∗)e as the mortality gain, as in the first expression. Then necessarily we would take (e−e∗)q∗ as the expense gain. If we decide to look at expense first, we would decompose according to the second expression. In short, for the factor considered second, the gain will be expressed with the other factor starred. So in ournumericalexampleabove,wecoulddividethetotalgainof9into4formortalityand5for expenses, or 2 for mortality and 7 for expenses, deciding on which decomposition we used. Anotherwayofviewingthisistonotethatwecanalsowritethetotalgaininasymmetric fashion as qe−q ∗e∗ = (q−q∗)e+q(e−e∗)−(q−q∗)(e−e∗). The first term on the right is strictly mortality gain, the second term is strictly expense gain and the third term, which is a combined effect, can be allocated to one or the other. This seems rather arbitrary and artificial but it is the best we can do if we insist on a decomposition. Indeedwehavealreadyencounteredthisproblembefore,namelyintheexampleinvolving premium difference gain in Chapter 6, a gain which is tied in with interest. Since we had previously talked about interest gain, it seemed natural to do that first, but we could have insteadtakenthepremiumdifferencegainas(𝜋 ∗ −𝜋)(1+i),leavingthetotalinterestgainas (tV+𝜋∗)(i∗ −i). Thesameproblemcomeswithperiodicexpensegainswhichalsoinvolveinterest.Ifwefor example,focusoninterestfirst,thetotalgainof(e k(1+ik)−e∗ k (1+i∗k)willbedecomposed into a part due to interest ofek(ik −i∗ k ) and a part due to expenses of (ek −e∗ k )(1+i∗). (Note thatearningmoreinterestthanexpectedwillresultinthefirstitembeingnegative.Thereader should ensure that they can explain why this is so.) What would happen if there were a source of gain connected withthreedifferent factors? We’ll leave it to the reader to verify that there would now be six possible ways of dividing up the gain. Theoretically this could happen. For example, a percentage of premium expense would be affected by differences in both premiums and interest. In the common situation we have described, this will not occur since the premiums payable on both the actual profit basis and the expected profit basis will be the same, namely the gross premium, and the premium difference gain will be zero. Example 12.4 You are given the following information about a policy issued at age 50. The gross premium payable at age 65 is 2000, and a death benefit of 100 000 is payable at age 66 for death occurring between age 65 and 56. A cash value of 20 000 is paid at age 66 for policyholders surrendering at that time. Moreover 15V = 20 300, and16V = 22 500. The insurer’s best estimate for mortality, withdrawal and interest is given by q(d) 65 = 0.005, q(w) 65 = 0.008, i= 0.04. Anticipated expenses at the beginning of the year are 5% of the gross premium, plus 100. In addition, the expected settlement expenses are 200 for a death claim and 100 for withdrawal. PROFIT MEASUREMENT 193 Nowhereistheactualexperience.Outof1000policiesatthebeginningofthe16thpolicy year, there were 4 deaths and 10 withdrawals. The actual interest rate earned was 0.048. The actual expenses incurred were 230 per policy at the beginning of the year, including both percentage of premium and fixed expenses, 150 to settle each death claim, and 120 for each withdrawal. For the year running from time 15 to time 16, calculate the actual profit less the expected profit, and classify this difference by source. For the latter, assume that all expense gains are calculated afterthe gains from the other sources. Solution. Taking the actual results for the starred quantities, and using (12.3) Pr ∗ 16 = (20 300+2000−230)(1.048)−.004(10 0150)−0.010(20 120)−0.986(22 500) = 342.56 Pr16 = (20 300+2000(1−0.05)−200)(1.04) = 0.005(10 0200)−0.008(20 100) −0.987(22 500) = 114.7 Actual profit−Expected profit = 227.86. This can be decomposed as follows: Gain from mortality (including the portion due to settlement expenses) = (0.005 − 0.004)(10 0200−22 500)=77.70. Gain from withdrawal (including the portion due to withdrawal expenses)= (0.008 − 0.010)(20 100−22 500)=4.80. Gain from interest=(20 300+ 2000−200)(0.048 −0.040) =176.80 Gainfrompercentageofpremiumandperiodicexpenses =(200−230)(1.048)=−31.44 Death settlement expense gain=0.004(200−150)=0.20 Withdrawal expense gain=0.010(100 −120)=− 0.020. Theexpensegainsabove,comingaftertheothercalculations,allusetheactualexperience for the other factors, as we noted in our remarks about order above. The reader should find it instructive to redo the calculations, now assuming that the expense gains are calculated first. Note that since the insurer benefits from a withdrawal, paying out less than the reserve, the higher than expected number of withdrawals caused a positive gain. 12.4.3 Profit testing We now return to the case where we have a proposed profit test basis, and proposed gross premiums. We want to compute the profits over several periods, and see how we can use the resulting figures. Example 12.5 A five-year, 100 000 endowment insurance policy on (60) has benefits payable at the end of the year of death and carries annual premiums of 21 500, payable for 194 EXPENSES AND PROFITS 5 years. Initial expenses are 60% of the premium plus 200, and renewal expenses are 5% of thepremiumplus100.Deathsettlementexpensesforthefirst4yearsare0.2%ofthetheface amount plus 50. The expense of paying any cash value, or the final payment at time 5 is 50. The profit test basis interest rate is 5%. The profit test basis mortality and withdrawal rates, reserves, and cash values are shown in the following table. Find Pr k fork= 0,1, …,5. kq (d) 60+k−1 q(w) 60+k−1 k Vc v k−1 Prk 00 0 0 1 0.009 0.080 18 500 0 −8935.75 2 0.010 0.050 37 500 17 500 3636.25 3 0.012 0.030 57 500 42 500 3151.75 4 0.014 0.020 78 500 70 000 3080.75 5 – – 100 000 0 3716.25 Solution. The values of of Prk are computed directly from (12.3) and are shown in the final column. As an example, here is the computation of Pr2. Pr2 = (18 500+21 500−0.05(21 500)−100)(1.05)−0.01(100 250)−0.05(17 550) −0.94(37 500) = 3636.25 Thevector(Pr 0,Pr1,…)issometimescalledthe profitvector.Itisnothoweverthevector we want in order to accomplish our goal of analyzing profits. Recall that Prk is the profit in the year running from timek−1t ot i m ekfor a policy in force at timek−1. The figure we want is the expected profit for that year. Accordingly we make the following definition. Definition 12.1 The profit signaturefor a policy issued at agex is the vectorΠ, where Π 0 = Pr0, and fork> 0 Πk =k−1 p(𝜏) x Prk. Inotherwordswemultiplytheyearlyprofitsbytheprobabilitythatthepolicyholderwill survive to the beginning of the particular year. Example 12.6 Find the profit signature for Example 12.5. Solution. We first calculate p (𝜏) x = 0.911, 2p(𝜏) x = 0.85634, 3p(𝜏) x = 0.82037, 4p(𝜏) x = 0.79248. A straightforward calculation then yields Π= (0,−8935.75, 3312.62, 2698.97, 2527.37, 2945.06) . PROFIT MEASUREMENT 195 Our problem of analyzing profits is now reduced to the familiar problem, dealt with from the beginning of this book, of analyzing a sequence of cash flows. Forexamplewecansummarizethiscashflowsequencewithasinglenumber,bycomput- ing the present value of the profit signature according to some discount function. This gives the net present value of profits, often abbreviated as NPV. In practice the discounting is at a constant rate of interest which is called therisk discountrate. We can interpret this rate as one which is the desired return by an investor putting up the capital to fund an insurance enterprise.Itwillthereforereflecttheriskinvolvedinsuchaninvestmentandwillaccordingly benormallyhigherthansaytheprofittestrate.Notethatwedonotdiscountbysurvivorshipin calculating the NPV. This has already been considered when calculating the profit signature. RecallnowthatwehavecalculatedthisNPVaccordingtoasinglepolicy,withaspecified amount of benefits and premiums. To interpret it properly and obtain measures which can be used for comparative purposes, we must relate this to some measure of volume. A natural candidate for this is the present value of premiums. Definition 12.2 We define theprofitmarginforthepolicy as the quantity NPV ̈a(𝜏) x (𝜋) where 𝜋 is the premium vector for the policy. The discounting in the denominator is done withbothinterest,atthesameriskdiscountrateusedinthenumerator,andwithsurvivorship as given by the multiple-decrement table. Example 12.7 If the risk discount rate is 0.07, find the profit margin for the policy of Example 12.5. Solution. The denominator is just 21 500̈a 60(15) at 7% interest=83 282.65. The numerator is (1.07−1)(−8935.75)+(1.07)−2(3312.62)+(1.07)−3(2698.97)+(1.07)−4(2527.36) +(1.07)−52984.09 = 773.27 The profit margin is 773.27/ 83292.65=0.0093. This example shows then that for the policy in question, for every dollar in premium received in sales, 0.93 cents would represent profit. This can be used for various decision problems.Ifitdeemedlowerthanthecompanywouldlike,thensomechangesmustbemade in the policy structure, usually by raising the premium charged, or in some case by lowering the surrender benefits if allowed by applicable legislation. The profit margin can also be compared with other policies to see which types of benefit and premium structures are most profitable. Anaturalquestionnowarises.Howistheprofitmarginaffectedbythechoiceofreserves. Our calculations in Section 6.5 suggest that under certain conditions the reserve basis should 196 EXPENSES AND PROFITS notaffecttheprofitmargin,andthisindeedistruewhenthediscountratesusedincomputing the NPV are the same as those of the profit test basis. To see this, look at formulas (12.3) and (12.5). Suppose that the single reserve figure kV is increased by an amountE. ThenΔk−1V is increased bypx+kEso that Prk is decreased bypx+kE. Similarly,Δk(V) is decreased byE(1+ik) so that Prk+1 is increased by (1+ik)E resulting in an decrease inΠk of kpxE and and increase inΠk+1 of kpx(1+ik)E, which will cancel out when we discount, leaving the NPV unchanged. This is not true however in the more typical case where the risk discounting is at higher rates. Suppose thatE> 0. The calculation above shows that discounting the increase inΠk+1 with a rate higher thanik for the year running from timek to k+1 will decrease the NPV. Thisiseasilyexplained.Thehigherreservemeansthatmoremoneymustbesetasideattime k, reducing profits at that time. This money is earning interest and will add to the profits the following year. However under the present scenario, it is not earning sufficient interest to compensate.Theconclusionisthatintheusualcasewhereriskdiscountratesarehigherthan theprofittestrates,increasesinreservescausereducedprofitability,anddecreasesinreserves cause increased profitability. Notes and references See Bowerset al.(1997, Chapter 16) for a detailed discussion of various modified reserve systems. Netpremiumreservesaresometimesreferredtoas benefitreservesintheliteraturesince, they involve only benefits and not expenses. The difference between the expense-augmented reserve and the net premium reserve is sometimes referred to as theexpense reserve.I t represents the present value of the future premiums available for expenses, less the present value of the future expenses. Some authors treat certain expenses incurred right at or just before a policy begins differ- ently than above. They would appear as a negative contribution to Pr 0. In our treatment they are accumulated to the end of the year, and appear as a negative contribution to Pr1,i nt h e same way as the initial premium received at time 0, makes a positive contribution to Pr1. Exercises Type A exercises 12.1 Aninsurancepolicyissuedatage xprovidesforapaymentof100000atthemoment ofdeathprovidedthisoccurswithin20years.Levelannualpremiumsof Garepayable for10years.Theexpensesareasfollows:inthefirstyear,60%oftheinitialpremium, 1%ofthefaceamountand20perpolicy;afterthefirstyear,5%oftheeachpremium, 0.25% of the face amount and 10 per policy; the death benefit settlement expense is EXERCISES 197 50 per policy plus 0.5% of the face amount. Assume that the expenses for any year are paid at the beginning of the year. Find a formula forG. 12.2 Aninsurancepolicyissuedatage xhasdeathbenefitvector bpaidattheendoftheyear of death, and level annual premiums payable fornyears. Assume a constant interest rate. GivenA x(b) = 330, Ax+1(b◦ 1) = 280, ̈ax(1n) = 11, ̈ax+1(1n−1) = 8 and ̈ax+k(1n−k) = 6, find the excess of the reserve at timekcalculated with net premiums over the reserve at timekcalculated by the full preliminary term method. 12.3 A certain policy on (x) has gross-premium reserve of 1000 at time 10. The gross premium payable at time 10 is 100. The death benefit payable at the end of year of death for death between time 10 and time 11 is 1500. The cash value payable at time 11 is 900. Expenses for the year running from time 10 to time 11 are 20, paid at the beginningoftheyear.Theinterestrate i= 10%, q (d) x+10 = 0.08and q(w) x+10 = 0.05.Find the gross premium reserve at time 11. 12.4 A policy on (x) provides death benefits of 1000 paid at the end of the year of death. There is no cash value for withdrawal in the first year. For withdrawal in the second year, an amount of 30 is paid at the end of the year. You are given that q (d) x = 0.1, q(d) x+1 = 0.2, q(w) x = 0.05, q(w) x+1 = 0.06, i= 0.07. Expenses are 120 in the first year and 40 thereafter, paid at the beginning of the year. Thelevelexpense-augmentedpremiumpayableannuallyis250andthecorresponding gross premium is 300. The gross-premium reserve at time 0 is−130. Find (i) the expense-augmentedreserve,(ii)thegross-premiumreserve,(iii)theassetshare,allat the end of 2 years. Type B exercises 12.5 A modified reserve system has a first-year premium ofb0vqx a second-year premium of b1vqx+1, followed by a level renewal premium of𝛽. Assume constant interest and benefits at the end of the year of death. Find an expression for𝛽, when premiums are payable fornyears. 12.6 Considerapolicyissuedatage xwithbenefitspayableattheendoftheyearofdeath. A modified reserve system, used in Canada for tax purposes (sometimes known as the 1 1 2 preliminary term), takes an initial premium tob0vqx, a premium at time 1 of 𝛽,andalevelpremiumof 𝛾 payableaftertime1,suchthat 𝛽 istheaverageof b1vqx+1 and 𝛾. Assume constant interest and benefits paid at the end of the year of death. If premiums are payable for life, show that 𝛾 = Ax+1(b◦ 1)+vpx+1Ax+2(b◦ 2) ̈ax+1 +vpx+1̈ax+2 . 12.7 SupposethatinExample12.3theinsurersells100ofthegivenpolicies.Showthethe total profits accumulated with interest to time 2 are the same, regardless of whether gross premium or expense-augmented reserves are used. 198 EXPENSES AND PROFITS 12.8 Redo Example 12.4 with the expense gains calculatedbeforethose from mortality, withdrawal and interest. 12.9 Refer to Example 12.5 (a) Suppose that1V is changed to 13 000 while all other data remains the same. Calculate the new profit signature and profit margin. (b) Redo part (a) with the risk discount rate equal to 5%. (c) Find the profit margin, for the original problem with a risk discount rate equal to 5%. 12.10 RefertoExample12.5.Supposewedecidetoincreasethepremiumtoachieveaprofit margin of 3%. What should the new premium be? ∗13 Specialized topics FortheconcludingchapterinPartIofthisbook,wecoverafewspecializedtopicsininsurance and annuities that are of current importance. We will not go into any of these topics in depth, but rather provide an introduction to some of the major features. The material here is not needed for any other chapters of the book. 13.1 Universal life 13.1.1 Description of the contract Universallifeisatypeofcontractthatbeganinthe1970s,andnowaccountsforasubstantial portion of life insurance sales. Prior to that time, the mainstay of life insurance was for the most part whole life or endowment contracts. The origin of the change could well have been the advice given by some financial advisors that people should buy term insurance instead of those products, for a much smaller premium, and then invest the difference elsewhere. Now our analysis in Section 6.4.2 shows that this is exactly what happens within the policy itself when one purchases whole life or endowment insurance. The excess premiums, not needed to provide insurance in a particular year, are invested in the savings portion. Of course, the purchaser of endowment or whole life insurance has less flexibility than the term buyer,bothwithregardstotherelativeamountsgoingintotheinsuranceandsavingsportion, and to the types of investments and rates of return. Universal life plans were developed to allow this flexibility within a single contract. The main features involve variable premium payments, and the ability for the policyholder to participate in higher yielding investment opportunities. In the usual type of universal life plan, premiums are not fixed in advanced and may be varied at the option of the purchaser. Each individual has in effect a separate fund that changes over time in the way we describe it by the recursion formula (6.8). Premiums paid FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 200 SPECIALIZED TOPICS are deposited, the fund is credited with interest or investment earnings, while expenses, and the cost of insurance, based on the net amount at risk are deducted. The policyholder need only pay sufficient premiums to ensure that the fund value will cover the cost of insurance. In this type of plan where there is a more direct relationship between policyholders and their individual accounts, there is a more frequent demand for the type of plan that we discussed in Section 6.7. It is usual to allow two options. The purchaser can fix the amount paid at death, or they can fix the net amount at risk, so at death, the amount in their account (representing the reserve) would be returned in addition to the stipulated face amount, giving an increasing death benefit. In the usual terminology the fixed death benefit is often referred to as a Type A policy, whereas the death benefit plus account value is referred to as type B. Some plans provide the flexibility to alter the face amount as well as the premium, with the stipulation that a request to increase the coverage would normally necessitate the type of information as required by new policyholders as to health and other conditions affecting the risk. There do arise cases however when such changes are mandated. In the United States this occurs due to regulations stipulating that in order to receive the favourable tax treatment accordedtolifeinsurance,thedeathbenefitmustexceedthefundvaluebyacertainminimum percentage. If the fund value gets high enough, the insurer will increase the death benefit in order to maintain this so-calledcorridorrequirement. Policyholderscanlapsethepolicyatanytimeandreceivetheirfundvaluelessaso-called surrender chargethat is intended mainly to account for the initial expenses, as we outlined in Section 6.6. The surrender charge will decrease with time and normally will become zero after a certain number of years, often 20 or so. As well as the flexibility in premium payments a major attraction of universal life for the purchaser is the opportunity to earn a higher yield. The credited interest rate is usually not fixed in advance as it essentially is in traditional plans, but it is allowed to vary. There are differentwaysofaccomplishingthis.Insomecasesthefundsofthepolicyholderareactually invested in certain assets, so that the account is very much like a mutual fund investment togetherwithguaranteedbenefitspayableondeath.Usuallythereisachoiceofvarioustypes of assets or funds to invest in, so the policyholder can control the amount of risk they are willing to take in order to obtain higher returns. This type of contract is often referred to as variableuniversallife. In other cases the account of the policyholder need not be invested in anyparticularassetsbutinterestiscreditedaccordingtoearningsonsomereferenceportfolio. Suchcontractsareknownas equityindexedinsurance. Therearenormallysomequalifications tothecreditingofinterest.Therewillbeastipulatedminimumrateofinterestthatiscredited regardless of what the reference portfolio does. It is also common to have limits on the other end. There may be a ‘participation rate’. So for example if the reference portfolio earned 9% interest over a period and the participation rate was 80%, the the policyholder’s account wouldbecreditedwith7.2%.Inadditionthereisoftenastatedmaximumthatwillbecredited regardless of the actual earnings. Universal life contracts often carry other guarantees that mitigate again unfavourable investment experience. These are referred to assecondary guarantees. A popular option of thistype,isthataslongastheinsuredkeepsupacertainstipulatedminimumrateofpremium payment, the death benefit is guaranteed even though lower than expected rates of return bringtheperson’saccountbelowtheamountneededtomeetthecostsofinsurance.Asimilar provision is commonly used with endowment type contracts that have a maturity date, in which case a minimum guaranteed amount paid at maturity provided that the insured keeps up a minimum schedule of premium payments. UNIVERSAL LIFE 201 13.1.2 Calculating account values For each policyholder, the amount of money in their account is calculated periodically, often monthly. For the most part this is handled easily by the basic recursion formulas that we have seen in previous chapters. There are however some aspects of terminology and special features that will be discussed. Given the account value AV k at timek, the account value AVk+1 is calculated using the following familiar quantities. We assume aggregate mortality and payment of benefits at the end of the period of death. (a) 𝜋k, the premium paid at timek. (b) rk,thepremiumexpenserate.Thisisapercentage,oftenaround5%,takenfromeach premium by the insurer to handle expenses. Only the remaining amount (sometimes referred to as theallocated premium) is credited to the account. In the so-called unit linked policiespopular in the United Kingdom, it is common to incorporate this expense by a device termed abid–offer spread. In these cases the policyholder is deemed to have their own separate fund, distinct from the general assets of the company, and the premium is used to buy a number of units of the fund. These are bought at a certainofferprice, and sold back at a lowerbidprice. So for example if the offer price was 100 per unit and the bid price was 95 per unit, and a premium of 1000 was paid, the purchaser would own 10 units of the fund, which would have a value of 950. So the allocated premium would be effectively 950. (c) e k, an additional flat amount that is charged each period for expenses. (d) ck, the cost of insurance rate. This is just a mortality rate which will apply to the time periodrunningfromtime ktotime k+1.Forayearlycalculationwithanissueageof x, we would haveck = qx+k, for an appropriate life table. For a monthly calculation, assuming UDD, we would haveck = (1∕12)qx+m, where 12m≤ k< 12(m+1). It is oftenexpressedasarateofsomuchper1000.Soforexampleacostofinsurancerate of 15 per 1000 would mean thatc k = 0.015. (e) ik, the credited interest rate. This may be specified directly or tied to the performance of some other investments. (f) bk, the death benefit paid at timek+1 for death between timekandk+1. Then the recursion is just AVk+1 = (AVk +𝜋k(1−rk)−ek)(1+ik)−ck𝜂k. (13.1) FortypeApolicieswehave 𝜂k = bk −AVk+1,ifthisispositive,andwesolvetheequation for AVk+1.For type B policies where the death benefit is paid in addition to the face amount, 𝜂k = bk. Another possible type of arrangement is where the fund itself is returned at death, subject to a guaranteed minimum amountbk. (As mentioned above, this may be ruled out in certain jurisdictions by taxation requirements.) In such case if AVk+1 ≥ bk, we would have 𝜂k =0, while if AVk+1 < bk, we would have𝜂k = bk −AVk+1. It must be kept in mind that the account on a universal life policy should be viewed as belonging strictly to the particular policyholder, so while the account value is similar in 202 SPECIALIZED TOPICS nature to a retrospective reserve, it is not quite the same. For example, withdrawal rates and surrendervaluesarenottakenintoaccountinthiscalculationaswasdoneinsomeChapter12 formulas.Todosowouldmeanthatpolicyholderaccountswouldbeaugmentedwithpositive gainsfromsurrender,butthesebelongstrictlytotheinsurer(asofcoursedosurrenderlosses, should they arise). Example 13.1 A universal life policy has a stated death benefit of 100 000, monthly premiums of 5000, monthly expenses of 100, and a further monthly expense charge of 4% of eachpremium.Theannualcreditedinterestrateis8%,andthemonthlycostofinsurancerate in the 5th year of the policy is 30 per thousand. If the account value at the end of the 52nd monthis50000,findtheaccountvalueattheendofthe53rdmonth,assumingthatthepolicy is (a) type B and (b) type A. Solution. We first must calculate the applicable monthly interest rate which is (1.08) 1∕12 − 1 = .00643. Then (a) AV53 = (50000 +5000(0.96)−100)(1.00643)−0.03(100000) = 52051.72. (b) The100000aboveisreplacedwith100000 −AV53,andsolvingtheequationwejust obtain the Type B amount divided by (1−0.03). AV53 = 52051.72∕(1 −0.03) = 53661.57. Another calculation that will normally be done each period is to compute thecost of insurance(as opposed to the cost of insurancerate). This is sometimes abbreviated as COIk, fortheperiodrunningfromtime ktok+1.(Someauthorspreferadifferentindexingmethod and would refer to this as COIk+1. We prefer to have the index correspond to that of the cost of insurance rateck.) This is normally determined as at thebeginningof each period, so if therearenotsufficientfundstopayit,thepolicywouldlapseintheabsenceofanyguarantees to the contrary. In effect it can be viewed as the net single premium paid each period for the death benefit coverage. That is COI k = (1+ik)−1ck𝜂k. So in the above example we would have COI52 =(1.00643)−1(0.03)100000 = 2980.83, for the type B policy and COI52 =(1.00643)−1(0.03)(100000 −53661.57) =1381.27, for the type A policy. There are some variations which can arise. In some circumstances the insurer may use a discountrate jk incalculatingthecostofinsurancewhichisdifferentfromthecreditedinterest rateik. In this case formula (13.1) would be modified to AVk+1 = [AVk +𝜋k(1−rk)−ek −(1+jk)−1ck𝜂k](1+ik) (13.2) which shows directly the insurance costs being deducted at the start of the period. Clearly when j k = ik, the interest factors cancel and we just get back the same formula as before. VARIABLE ANNUITIES 203 Notice however that in general we still have the same formula as before if we adjust theck. We change this to a new cost of insurance rate given by c′ k = ck 1+ik 1+jk . The following illustrates. Example 13.2 RedoExample13.1andcalculatethecostofinsurance,undertheassumption of a 3% discount rate for the cost of insurance. Solution. We can do all calculations by changingc k from 0.03 to c′ k = 0.03(1.08)1∕12∕ (1.03)1∕12 = 0.03012. In the type B policy, this will change the value of AV 53 to 52039.72 and COI52 to (1.00643)−1(0.03012)100 000= 2992.76. In the type A policy, this will change the value of AV53 to 52039∕(1−0.03012) = 53655.83 and COI52 to (1.00643)−1(0.00301)(100000 −53655.83) = 1386.96. Another modification may be necessary because of the corridor requirement. In a close situation, we must test whether an account value is sufficiently low to meet this restriction. Let cor k denote the minimum allowable ratio ofbk∕AVk+1.I fA Vk+1cork > bk, we must redo the calculations. The death benefit will now be corkFk+1 and𝜂k = (cork −1)AVk+1. Example 13.3 For the policies of Example 13.1, find AV53 and COI52 assuming that cor53 = 1.9. Solution. For the type B policy the corridor requirement is clearly satisfied and the answer will remain as calculated above. For the type A policy, we check the original answer to see that 53 661.57(1.9)=101 957 which is above 100 000, so we must recalculate: AV53 = (50000 +5000(0.96)−100)(1.00643)−0.03(0.9AV53), and solving we get AV53 = 53604.40. The death benefit will increase to 101 848.36 COI52 = 1.00643−10.03(0.9×53604.62) = 1438.07. 13.2 Variable annuities A similar attempt to provide additional flexibility has become popular for certain deferred annuitycontracts,whichhavecometobeknownas variableannuities.Theycorrespondtothe contractdescribedinExample5.8totheextentthatthereisnosurvivorshipaccumulationprior to the annuity payments. In fact they operate in the same basic manner as we have described aboveforUniversalLife,exceptthatnodeductionsaremadeforthecostofinsurance.Infact some purchasers use these as a vehicle to accumulate money in a high yielding investment accountwithoutanyintentionofconvertingthefundstoanannuity.Theannuityaspectenters intothecontracthowever,sinceitcommontoincludeprovisionsprovidingfortheconversion of the funds into annuities at guaranteed rates of interest and mortality. 204 SPECIALIZED TOPICS Asintheinsurancecase, anindividual’s account maybeinvestedinparticular assetsorit may be equity-indexed, with the credited interest tied to some reference portfolio. Similarly to Universal Life, their are several types of possible guarantees that have been designed to reduce the losses under bad investment experience. A common provision is a guaranteed minimum death benefit,whereby the account holder is promised a certain minimumreturnupondeath,regardlessoftheaccountvalue.Thedeathbenefitamountcould be the amount originally invested, or it could be that amount together with a certain fixed accumulation rate. A similar type of provision is theguaranteed minimum accumulation benefit whereby the end of a specified period, the account holder is promised a minimum account value, regardless of theinvestment experience. Stillanother option istheguaranteed minimumwithdrawalbenefit. Under this option the account holder is allowed to withdraw a minimumamount(oraminimumpercentage)fromtheiraccount,eachyear,untiltheoriginal investedvalueisreturned.Evaluationofthesebenefitsrequiresaknowledgeofoptionpricing. An introduction to this topic is found in Chapter 20. 13.3 Pension plans Pensionplansaresetupbycompaniestoprovideretirementincometoagroupofemployees. This is a vast subject and we confine ourself here to providing a survey of principal features and definitions. Pension plans can be classified into two main categories,defined benefit plans,abbreviatedasDB,and definedcontribution plans,abbreviatedasDC,asintroducedin Section 4.6.2. We now provide more detail. 13.3.1 DB plans The usual type of DB plan provides that an employee will receive a life annuity, beginning at a specified normal retirement age, with a periodic paymentK. Rules for calculatingK are specified at the outset. Itwillnormally depend onthe employee’s salaryand years ofservice. The type of annuity is often a whole life annuity, although there may be options to elect a guaranteed period, or a joint-life annuity with another individual, such a spouse. The income of course will be adjusted according to the nature of the annuity selected. A typical formula is thatK will be equal tor times the average of the employee’s last h years of salary times the number of years of service, for some specifiedr and h.A sa n example,supposethatthenormalretirementageis65, r=0.02,h =3andthat Kisanannual payment. Consider an employee who is hired at age 30, retires at age 65, and whose last 3 years of salary were 89 000, 93 000, 100 000. The employee would then receive an annual pension of 0.02×35×94000 = 65800. A measure which is used to compare various plans is known as thereplacement ratio and it is simply the ratio of the pension to the final year’s salary. In the above example the replacement ratio would be 65.8%. The quantityhwill often be between 3 and 5 years. A possible variation involves plans, known ascareer average earnings planwhere h is not fixed, but is equal to the complete number of years of service for each employee. To evaluate benefits on such a plan the actuary will use an investment discount functionv andamultiple-decrementtablewhichwilltypicallyshowdecrementsofdisability,withdrawal PENSION PLANS 205 fromservice,deathandretirement.Anadditionalnecessarytoolisaso-called salaryscaleS x defined as follows. For some minimal agex0,w eh a v eSx0 = 1 and then fory> x0. Sy = the expected ratio of salary at ageyto salary at agex0. The scale is then used to estimate future salaries. For example, if an employee is earning an annual salary ofJ at agex, then an estimate of their annual salary as agex+h will be JS x+h∕Sx. Pension plans must specify what benefits if any will be paid to those who leave the employee group for reasons other than retirement at the normal retirement age. Such reasons include disability, death, withdrawal from company employment and retirement at an age otherthanthenormalone.Wefocusnowonearlyretirement.Plansmayspecifythataperson mayreceivepensionbenefitsiftheyretireearly,withsomeminimumcriteriaspecified,which can depend on age and duration of employment. Early retirement means of course that the pensionispaidforalongerperiod,andalsothatcontributionsintotheplanwillnotbemadefor the remaining time to the normal date, so typically the pension income will be appropriately reduced. Rather than calculate the applicable amount of reduction in each case, it is usual to work out approximate figures, known asactuarial reduction factorsand these are specified as part of the provisions of the plan. In the following simplified example, provision is made for early retirement up to 2 years before the normal date, but not for other forms of decrement. Example 13.4 ConsideraDBplanwhichprovidesayearlypensionof2%ofthefinal3-year average salary, times the number of years of service, for retirement at ages 63–65, but with a reduction of 5% per year should retirement occur before 65. An employee now 45 was hired atage35.His/herpresentsalaryis70000.Thesalaryscaleisgivenby S x = 1.03x−30, x≥ 30. (This simply means that salaries are expected to increase by 3% per year.) An employee may retire at any age from 63 to 65, with a reduction in pension income of 5% per year. Find a formula for the actuarial present value of the pension benefits. Solution. For a person retiring at age 63, the estimated 3-year final average salary is given by 70000[1.03 15 +1.0316 +1.0317]∕3 = 112362 so that the estimated annual pension income, accounting for the actuarial reduction factor of 10%, is given by 0.90×0.02×28×112262 = 56630. For person retiring at age 64 a similar calculation, now using a 5% reduction results in annual pension income of 63 769, and for a person retiring at age 65, the figure is 71 523. The actuarial present value is then given by the following (the subscriptr denoting the retirement decrement). 56630 18p(𝜏) 45q(r) 63 ̈a63 +6376919p(𝜏) 45q(r) 64 ̈a64 +7152320p(𝜏) 45q(r) 65 ̈a65. Our solution here incorporates several simplifying assumptions. We have assumed that thepensionwouldbepaidannually,insteadofthemoreusualmonthlyarrangement.Wehave also assumed that all employees are hired and retired on their birthdays, and that each salary increase occurs on birthdays. It is not too difficult to incorporate more realistic assumptions, 206 SPECIALIZED TOPICS using the fractional duration techniques in Chapter 7. One common provision is to assume hiring, retirements, and salary increases at the middle of each year, that is at agesx+1∕2, wherexis an integer. Fundstoprovidetheincomecouldbeprovidedsolelybytheemployer,ormorecommonly sharedbetweentheemployeesandtheemployeronsomespecifiedbasis.Ausualprovisionis thattheemployeeswillcontributeeachperiodaamountof ctimestheirsalaryforthatperiod.A typicalfiguremightbe c= 0.05.Theemployerwillthencontributeadditionalamountswhich are estimated to be sufficient to provide the promised benefits. There are various methods for doing so. The subject of funding DB pensions is complex, and will not be discussed further. 13.3.2 DC plans AtonetimetheDBarrangementwasthemostcommonone.Howeverperiodsoflowinterest rate earnings and improving mortality meant that many employers needed to put in more money then originally estimated in order to ensure the promised level of benefits. In some casesthisbecameprohibitive,resultinginfirmsswitchingtotheDCmode.Insuchapension, theemployeescontributeacertainpercentageoftheirsalary,theemployeraddsanadditional percentage, and the funds are invested and accumulated as an individual account for each participant until retirement. At the time of retirement the total accumulated contributions made on the employee’s behalf are used to purchase an annuity at the then prevailing interest andmortalityrates.AswiththeDBplanthereisoftenatargetedgoal,butthisisnotguaranteed. Iftheinvestmentexperienceisunfavourable,ormortalityhasimprovedsothatthecostoflife annuities go up, the pension income may be short of the projected amount. Of course, things can go the other way in a DC plan. Very favourable investment returns can result in higher pensions than expected. Funding arrangements are usually much easier for the DC plan than in the DB case. A common practice is to deposit into the account a certain fractionc of a employee’s salary eachyear,whichcouldbesharedinsomewaybetweentheemployeeandemployer.Onedoes not have to worry as much about benefits for withdrawal or early retirement. Withdrawing employees can be given the amount in their account, either in cash or as a deferred annuity. For those retiring early the amount can be used to buy an annuity starting on the normal retirement date. Inthefollowingexamples,wecontinuetomakethesimplifyingassumptionsnotedabove, Example 13.5 For an employee hired at age 35 at a salary of 50 000, it is estimated that an amount of 500000 is needed at at the normal retirement age of 65 to buy an appropriate pension. For exit from the plan before age 65 the accumulated amount of contributions with interest is returned to the employee. Find a formula to calculate the contribution ratec, assumed to be made at the beginning of each year. Solution. TheideaissimilartoExample5.8.Wesolvethefollowingequationtodetermine c. 500000 = 50000 cVal 30(g;v) (13.3) where gk = S35+k∕S35, k= 0,1, …,34. (13.4) EXERCISES 207 There are many variations in practice. Many plans have a so-calledvestingrequirement, whichmeansthatanemployeedoesnotreceivebacktheemployer’scontributionsunlessthey remain in the plan for a minimum period. Example 13.6 Suppose that in the situation described above, the contributions are split, withtheemployeepaying40%andtheemployertheremaining60%.Forparticipantsexiting the plan after 5 years, the total of all contributions with interest is returned at the end of the year of leaving. For those exiting in the first 5 years only the employee’s own contributions with interest are returned. Find an equation to determine the contribution ratec. Solution. We use a variation of formula (5.4). c= 500000v(35) 35p(𝜏) 30 ̈a(𝜏) 35(g)−A(𝜏) 35 (j) . whereg is as in (13.4) and jk = {0.4∑k−1 i=0 v(k,i)S35+i∕S35 if 0≤ k< 5, ∑k−1 i=0 v(k,i)S35+i∕S35 if 5≤ k< 25. Exercises 13.1 A universal life policy provides a death benefit at the end of the period of death equal totheaccountvalue,butsubjecttoaminimumof30000.Monthlypremiumsare2000. There is a monthly expense charge of 50 and a further charge of 3% of each premium. Theannualcreditedinterestrateis6%.Theaccountvalueattheendofthe30thmonth is 27 318. The monthly cost of insurance during the third year of the policy is 15 per 1000. Find AV 31 and COI30. 13.2 A universal life policy provides a death benefit at the end of the period of death of 100 000. The account value at time 20 is 2500. For the time period from time 20 to time 21, the credited interest rate is 0.006, the expense charge is 5% of any premium paid plus 100, and the cost of insurance rate is 30 per thousand. (a) Suppose that at time 20, the policyholder makes the minimum premium payment of 500, which according to the guarantee in the contract ensures that the cost of insurance will be paid regardless of the amount in the account. Find the account value at time 21. (b) If the contract does not carry the secondary guarantee of part (a), what is the minimum premium that the policyholder will have to pay at time 20 to ensure that the cost of insurance can be paid. 13.3 At a certain timek, two universal life policies have exactly the same account value, death benefit, premiums, expenses and cost of insurance rates. In both cases the cost 208 SPECIALIZED TOPICS of insurance discount rate is the same as the credited interest rate. One policy is type A and the other is type B. Show that the difference between the type A and type B account values at the end of the year is the same as the difference between the type B and type A COI for that period, accumulated with interest to the end of the year. 13.4 A universal lifepolicy provides a death benefit equal to the account value plus 50 000. Monthly premiums are 3000. There is a monthly expense charge of 100 and a further charge of 4% of each premium. The annual credited interest rate is 8%. The account value at the end of the 73rd month is 49 200. The monthly cost of insurance for the followingmonthis20per1000andiscomputedatanannualinterestrateof4%.There is a corridor requirement which specifies that the death benefit must be at least twice the account value. Find AV 74 and COI73. 13.5 Redo Example 13.4, only assume salary increases of 4% per year, and an actuarial reduction factor of 3% per year. 13.6 RedoExample13.6,onlyassumingthatanemployeewhodiesduringthefirst5years, receivesattheendoftheyearofdeath,boththeirownandtheemployer’scontribution accumulated with interest. 13.7 An employee starts employment in a firm at age 35 and is offered a choice of either a DC or a DB plan. In the DC plan, contributions of 15% of salary are accumulated withinterestat4%untilage65andthenusedtobuyalifeannuity.UndertheDBplan, the employee is given an annual amount, beginning at age 65 of 1.8% of their 3 years average salary times the number of years of service. If the cost of a 1-unit life annuity atage65is10,andsalariesincreaseby4%peryear,findtheratiooftheannualincome under the DC plan to that of the DB plan, for an employee who remains in the plan until age 65. 13.8 A DB plan provides for an annual pension of a certain percentage of the final 5 years average salary times the number of years of service upon retirement. Retirement is based on the80 factorwhich means that an employee can retire at any time that the sumofageplusyearsofserviceisgreaterthanorequalto80.Anemployeeage55who beganworkatage25,iscomparingtheirincomeforretirementnow,tothatwhichthey would receive if they stay for another year. They can expect a 5% increase in salary for the following year, which is the same that they received in each of the last 4 years. Findtheratioofthepensionincomeforretirementin1yeartothatforretirementnow. Part II THE STOCHASTIC LIFE CONTINGENCIES MODEL

Human: q63.
Value at time 2 of annuity benefits= 200(0.64)(0.9)p63.

Actuarial AI: 1 Introduction and motivation 1.1 Risk and insurance In this book we deal with certain mathematical models. This opening chapter, however, is a nontechnical introduction, designed to provide background and motivation. In particular, we are concerned with models used by actuaries, so we might first try to describe exactly what it is that actuaries do. This can be difficult, because a typical actuary is concerned with many issues, but we can identify two major themes dealt with by this profession. Thefirstis risk,awordthatitselfcanbedefinedindifferentways.Acommonlyaccepted definition in our context is that risk is the possibility thatsomethingbad happens. Of course, many bad things can happen, but in particular we are interested in occurrences that result in financial loss. A person dies, depriving family of earned income or business partners of expertise. Someone becomes ill, necessitating large medical expenses. A home is destroyed by fire or an automobile is damaged in an accident. No matter what precautions you take, youcannotridyourselfcompletelyofthepossibilityofsuchunfortunateevents,butwhatyou can do is take steps to mitigate the financial loss involved. One of the most commonly used measures is to purchase insurance. Insuranceinvolvesasharingorpoolingofrisksamongalargegroupofpeople.Theorigins go back many years and can be traced to members of a community helping out others who sufferedlossinsomeformorother.Forexample,peoplewouldhelpoutneighbourswhohad suffered a death or illness in the family. While such aid was in many cases no doubt due to altruisticfeelings,therewasalsoamotivationofself-interest.Youshouldbepreparedtohelp out a neighbour who suffered some calamity, since you or your family could similarly be aidedbyotherswhenyourequiredsuchassistance.Thiseventuallybecamemoreformalized, giving rise to the insurance companies we know today. Withthe institutionof insurance companies, sharing is no longer confined to the scope of neighbours or community members one knows, but it could be among all those who chose to purchase insurance from a particular company. Although there are many different types FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 4 INTRODUCTION AND MOTIVATION of insurance, the basic principle is similar. A company known as theinsureragrees to pay out money, which we will refer to asbenefits, at specified times, upon the occurrence of specified events causing financial loss. In return, the person purchasing insurance, known as theinsured,agreestomakepaymentsofprescribedamountstothecompany.Thesepayments are typically known aspremiums. The contract between the insurer and the insured is often referred to as theinsurancepolicy. The risk is thereby transferred from the individuals facing the loss to the insurer. The insurer in turn reduces its risk by insuring a sufficiently large number of individuals, so that the losses can be accurately predicted. Consider the following example, which is admittedly vastly oversimplified but designed to illustrate the basic idea. Suppose that a certain type of event is unlikely to occur but if so, causes a financial loss of 100 000. The insurer estimates that about 1 out of every 100 individuals who face the possibilityofsuchlosswillactuallyexperienceit.Ifitinsures1000people,itcanthenexpect 10 losses. Based on this model, the insurer would charge each person a premium of 1000. (We are ignoring certain factors such as expenses and profits.) It would collect a total of 1 000 000 and have precisely enough to cover the 100 000 loss for each of the 10 individuals whoexperiencethis.Eachindividualhaseliminatedhisorherrisk,andinsofarastheestimate of 10 losses is correct, the insurer has likewise eliminated its own risk. (We comment further on this statement in the next section.) We conclude this section with a few words on the connection between insurance and gambling. Many people believe that insurance is really a form of the latter, but in fact it is exactly the opposite. Gambling trades certainty for uncertainty. The amount of money you haveinyourpocketistherewithcertaintyifyoudonotgamble,butitissubjecttouncertainty ifyoudecidetoplaceabet.Ontheotherhand,insurancetradesuncertaintyforcertainty.The uncertain drain on your wealth, due to the possibility of a financial loss, is converted to the certainty of the much smaller drain of the premium payments if you insure against the loss. 1.2 Deterministic versus stochastic models The example in Section 1.1 illustrates what is known as adeterministicmodel. The insurer in effect pretends it will know exactly how much it will pay out in benefits and then charges premiums to match this amount. Of course, the insurer knows that it cannot really predict these amounts precisely. By selling a large number of policies they hope to benefit from the diversification effect. They are really relying on the statistical concept known as the ‘law of large numbers’, which in this context intuitively says that if a sufficiently large number of individuals are insured, then the total number of losses will likely be close to the predicted figure. Tolookatthisideainmoredetail,itmayhelptogiveananalogywithflippingcoins.Ifwe flip100faircoins,wecannotpredictexactlythenumberofthemthatwillcomeupheads,but weexpectthatmostofthetimethisnumbershouldbecloseto50.But‘mostofthetime’does notmeanalways.Itispossibleforexample,thatwemaygetonly37heads,orasmanyas63, orevenmoreextremeoutcomes.Intheexamplegiveninthelastsection,thenumberoflosses maywellturnouttobemorethantheexpectednumberof10.Wewouldliketoknowjusthow unlikely these rare events are. In other words, we would like to quantify more precisely just what the words ‘most of the time’ mean. To achieve this greater sophistication a stochastic model for insurance claims is needed, which will assign probabilities to the occurrence of ADEQUACY AND EQUITY 5 various numbers of losses. This will allow adjustment of premiums in order to allow for the risk that the actual number of losses will deviate from that expected. We will however begin thestudyofactuarialmathematicsbyfirstdevelopingadeterministicapproach,asthisseems tobethebestwayoflearningthebasicconcepts.Aftermasteringthis,itisnotdifficulttoturn to the more realistic stochastic setting. We will not get into all the complications that can arise. In actual coin flipping it seems clear that the results of each toss are independent of the others. The fact that one coin comes up heads, is not going to affect the outcomes of the others. It is this independence which is behindthelawoflargenumbers,andwhichresultsinoutcomesthatareusuallyclosetowhat is expected. There are some risks, often referred to assystematicor non-diversifiable,where the independence assumption fails, and which can adversely affect all or a large number of members of a group at the same time. For example, a spreading epidemic could cause life or health insurers to pay more in claims than they expected. Selling more policies in order to diversifywouldnothelptheirfinancialsituation.Itcouldinfactmakeitworse,ifthepremiums werenotsufficienttocovertheextralosses.Severeclimaticdisturbancescausingstormscould impact property insurance in the same way. In 2008, falling real estate prices in the United States affected mortgage lenders and those who insured mortgage lenders against bad debts, to the extent that this helped trigger a global financial crisis. A detailed discussion of these matters is not within the scope of this work, and for the most part, the stochastic model we present will confine attention to the usual insurance model where the risks are considered as independent.Itshouldbekeptinmindhoweverthatthedetectionandavoidanceofsystematic risk are matters that the actuary must always be aware of. 1.3 Finance and investments The second theme involved in an actuary’s work is finance and investments. In most of the types of insurance that we focus on in this book, an additional complicating factor is the long-termnatureofthecontracts.Benefitsmaynotbepaiduntilseveralyearsafterpremiums arecollected.Thisiscertainlytrueinlifeinsurance,wherethelossisoccasionedbythedeath ofanindividual.Premiumsreceivedareinvestedandtheresultingearningscanbeusedtohelp provide the benefits. Consider the simple example given above, and suppose further that the benefitsdonothavetobepaiduntil1yearafterthepremiumsarecollected.Iftheinsurercan investthemoneyat,say,5%interestfortheyear,thenitdoesnotneedtochargethefull1000 in premium, but can collect only 1000/1.05 from each person. When invested, this amount willprovidethenecessary1000tocoverthelosses.Again,thisexampleisoversimplifiedand there are many more complications. We will, in the next chapter, consider a mathematical model that deals with the consequences of the payments of money at various times. A much more elaborate treatment of financial matters, incorporating randomness, is presented in Chapter 20. 1.4 Adequacy and equity We can now give a general description of the responsibilities of an actuary. The overriding taskistoensurethatthepremiums,togetherwithinvestmentearnings,are adequatetoprovide forthepaymentofthebenefits.Ifthisisnottrue,thenitwillnotbepossiblefortheinsurerto 6 INTRODUCTION AND MOTIVATION meet its obligations and some of the insureds will necessarily not receive compensation for their losses. The challenge in meeting this goal arises from the several areas of uncertainty. The amount and timing of the benefits that will have to be paid, as well as the investment earnings,areunknownandsubjecttorandomfluctuations.Theactuarymakessubstantialuse of probabilistic methods to handle this uncertainty. Anothergoalistoachieve equityinsettingpremiums.Ifaninsureristoattractpurchasers, itmustchargeratesthatareperceivedasbeingfair.Herealso,therandomnessmeansthatitis not obvious how to define equity in this context. It cannot mean that two individuals who are charged thesameamount inpremiumswillreceiveexactly thesamebackinbenefits, forthat wouldnegatethesharingarrangementinherentintheinsuranceidea.Whiletherearedifferent possible viewpoints, equity in insurance is generally expected to mean that the mathematical expectation of these two individuals should be the same. 1.5 Reassessment Actuariesdesigninsurancecontractsandmustinitiallycalculatepremiumsthatwillfulfillthe goals of adequacy and equity, but this is not the end of the story. No matter how carefully one makes an initial assessment of risks, there are too many variables to be able to achieve complete accuracy. Such assessments must be continually re-evaluated, and herein lies the real expertise of the actuary. This work may be compared to sailing a ship in a stormy sea. It is impossible to avoid being blown off course occasionally. The skill is to detect when this occurs and to take the necessary steps to continue in the right direction. This continual monitoring and reassessing is an important part of the actuary’s work. A large part of this involvescalculatingquantitiesknownas reserves.WeintroducethisconceptinChapter2and then develop it more fully in Chapter 6. 1.6 Conclusion We can now summarize the material found in the subsequent chapters of the book. We will describe the mathematical models used by the actuary to ensure that an insurer will be able to meet its promised benefits payments and that the respective purchasers of its contracts are treated equitably. In Part I, we deal with a strictly deterministic model. This enables us to focus on the main principles while keeping the required mathematics reasonably simple. In Part II, we look at the stochastic model for an individual insurance contract. In Part III, we lookatmoreadvancedstochasticmodelsandintroducethemathematicsoffinancialmarkets. In Part IV, we consider models that encompass an entire portfolio of insurance contracts. 2 The basic deterministic model 2.1 Cash flows As indicated in the previous chapter, a basic application of actuarial mathematics is to model the transfer of money. Insurance companies, banks and other financial institutions engage in transactions that involve accepting sums of money at certain times, and paying out sums of money at other times. To construct a model for describing this situation, we will first fix a time unit. This can be arbitrary, but in most applications it will be taken as some familiar interval of time. For conveniencewewillassumethattimeismeasuredinyears,unlessweindicateotherwise.We will let time 0 refer to the present time, and timetwill then denotettime units in the future. Wealsoselectanarbitraryunitofcapital.Inthischapter,weassumethatallfundsarepaidout or received at integer time points, that is, at time 0,1,2,…. The amount of money received or paid out at timekwill be called thenetcashflow at timekand denoted byc k. A positive value ofck denotesthatasumistobereceived,whereasanegativevalueindicatesthatasum is paid out. The entire transaction is then described by listing the sequence of cash flows. We w i l lr e f e rt ot h i sa sacashflowvector , c = (c 0,c1,…,cN), whereN is the final duration for which a payment is made. For example, suppose I lend you 10 units of capital now and a further 5 units a year from now. You repay the loan by making three yearly payments of 7 units each, beginning 3 years from now. The resulting cash flow vector from my point of view is c = (−10,−5,0,7,7,7). From your point of view, the transaction is represented by−c = (10,5,0, −7,−7,−7). FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 8 THE BASIC DETERMINISTIC MODEL One of our main goals in this chapter is to provide methods for analyzing transactions in terms of their cash flow vectors. There are several basic questions that could be asked: r When is a transaction worthwhile undertaking? r How much should one pay in order to receive a certain sequence of cash flows? r How much should one charge in order to provide a certain sequence of cash flows? r How does one compare two transactions to decide which one is preferable? All of these questions are related, and we could answer all of them if we could find a method to put a value on a sequence of future cash flows. If all cash flows were paid at the same time, or if the value of money did not depend on the time that a payment was made, the problem would reduce to one of simple addition. We could simply value a cash flow sequence by adding up all the payments. We cannot proceed in this naive way, however and must consider thetime value of money. It is a basic economic fact that we prefer present to future consumption. We want to eat the chocolate bar now, rather than tomorrow. We want to enjoy the new car today, rather than next month. This means of course that money paid to us today is worth more than money paid in the future. We are no doubt all very familiar with this fact. We pay interest for the privilege of borrowing money today, which lets us consume now, or we advance money to others, giving up our present consumption and expecting to be compensated with interest earnings. In addition, there is the effect of risk. If we are given a unit of money today, we have it. If we forego it now in return for future payments, there could be a chance that the party who is supposed to make remittance to us may be unable or unwilling to do, and we expect to be compensated for the possible loss. A major step in answering the above questions is to quantify this dependence of value on time. Readers who have taken courses on the theory of compound interest will be familiar with many of the ideas. However, our treatment will be somewhat different than that usually given. One reason for this is that we want to develop the concepts in such a way that they are applicable to more general situations, as given in Chapters 3–5. A second reason is that our approach is designed to be compatible with modern-day computing methods such as spreadsheets. To conclude this section, we remark that many complications arise when the cash flows are not exactly known in advance. They may depend on several factors, including random elements.Theremaybecomplicatedinterrelationshipsbetweenthevariouscashflows.These matters involve advanced topics in finance and actuarial mathematics and for the most part willnotbedealtwithinthisbook.InPartIwedealmainlywithasimplifiedmodel,whereall cashflowsarefixedandknowninadvance. Inlaterpartsofthebookwewillconsidercertain aspects of randomness, but will not get into the full extent of complications that can arise. 2.2 An analogy with currencies To motivate the basic ideas, we will consider first a completely different problem, which is nonetheless related to that introduced above. Suppose that I give you 300 Canadian dollars, 200 US dollars and 100 Australian dollars. How much money did I give you? It would be naive indeed to claim that you received 600 dollars, for clearly the currencies are of different value. To answer the question we will need conversion factors that allow us to deduce the DISCOUNT FUNCTIONS 9 valueofeachtypeofdollarintermsofothers.Let v(c,u)denotethevalueinCanadiandollars of one US dollar. Assume thatv(c,u) = 1.05, which means that a US dollar is worth 1.05 Canadiandollars.(Ournumbershereareforpurposesofillustrationonly.Theyareclosetothe conversion rates at the time of writing, but they may well have changed considerably by the timeyouarereadingthis.)Similarly,letting astandforAustralian,wewillassumethat v(c,a) equals 0.95, which means 95 cents Canadian will buy one Australian dollar. The convention we are using here, which should be noted for later use, is that thevfunction returns the value of oneunit of thesecondcoordinate currency in terms of thefirstcoordinate currency. Therearefourmoreconversionfactorsofinterest,butanimportantfactisthattheycanall bededucedfromjustthesetwo(orindeedfromanytwothathaveacommonfirstorcommon second coordinate). We note first that if it takes 1.05 Canadian dollars to buy 1 US dollar, then a single Canadian dollar is worth 1∕1.05= 0.9524 US dollars. That is, v(u,c) = v(c,u) −1 = 0.9524, v(a,c) = v(c,a)−1 = 1.0526, where we use similar reasoning for the Australian dollar. Next considerv(u,a). We want the amount of US dollars needed to buy one Australian dollar. We could conceivably effect this purchase in two stages, first using US money to buy Canadian, and then using Canadian to buy Australian. Working backwards, it will take 0.95 Canadiantobuy1Australian,anditwilltake v(u,c)0.95USdollarstobuythe0.95Canadian. To summarize, v(u,a) = v(u,c)v(c,a) = 0.9048. Our calculations are completed with v(a,u) = v(u,a) −1 = 1.1052. The reader may notice, given a typical real-life listing of currency prices, that the rela- tionships we state here do not hold exactly, but that is due to commissions and other charges. In the absence of these, they must necessarily hold. Let us now return to the original problem of determining of how much I paid you. We must first select a currency to express the answer in. For example, we could say that the total was equivalent to 300+200v(c,u)+100v(c,a) = 605 Canadian dollars. We could also say thatthetotalwasequivalentto300v(u, c)+200+100(u,a) = 576.20USdollars.Noticeasa shortcut,thatwedidnotneedtocomputethelattersum(whichcouldbeasignificantsavingin calculationifwehadseveralratherthanjustthreecurrencies).Ifthetotalamountisequivalent to605Canadian,thenitmustalsobeequivalentto605 v(u,c) = 576.20USdollars.Similarly, the total in Australian dollars can be computed as 605v(a,c) or alternatively as 576.20v(a,u), both of which are equal to 637 (approximately as there are some rounding differences). 2.3 Discount functions Wenowgobacktotheoriginalsituation.Wewanttovalueasequenceofcashflows,whichare all in the same currency, but which are paid at different times. Conversion factors are needed toconvertthevalueofmoneypaidatonetimetothatpaidatanother.Theprinciplesinvolved 10 THE BASIC DETERMINISTIC MODEL are exactly the same as in Section 2.2. Letv(s,t) denote the value at times, of 1 unit paid at timet. (Note again that our convention is that the 1 unit goes with the second coordinate. In other words, 1 unit paid at timetis equivalent tov(s,t) paid at times.) In the case where s< t, you can interpretv(s,t) as the amount you must invest at timesin order to accumulate 1 at timet. In the case wheres> tyou can interpretv(s,t) as the amount that you will have accumulated at times from an investment of 1 at timet. The fundamental relationship that must be satisfied is the same as we noted with currencies, namely v(s,t)v(t,u) = v(s,u), for all s,t,u. (2.1) Due to its importance, we will repeat the reasoning for this fundamental fact in the current context.Itissimplythat1unitattime uisequivalentto v(t,u)attime t,andthis v(t,u)attime t is equivalent tov(s,t)v(t,u) at times , showing that 1 unit at timeuis indeed equivalent to v(s,t)v(t,u) at times. We now make a formal definition. Definition 2.1 A discount functionis a positive valued functionv, of two nonnegative variables, satisfying (2.1) for all values ofs,t,u. Otherdesirablefeaturesfollowimmediatelyfrom(2.1).Taking s= t= u,wededucethat v(s,s)v(s,s) = v(s,s) and, sincev(s,s) is nonzero, we verify the obvious relationship v(s,s) = 1, for all s. (2.2) From this we deduce thatv(s,t)v(t,s) = v(s,s) = 1 so that we recover the relationship, noted in the currency case, that v(s,t) = v(t,s)−1. (2.3) Although we have calledva ‘discount’ function, the common English usage of the word really applies to the case wheres< t. In that case,v(s,t) will be normally less than 1, and the function is returning thediscounted amountof 1 unit paid at a later date. Some authors wouldprefertodefinethediscountfunctiontoapplyonlytothiscaseandthendefineanother function, called anaccumulationfunction, to cover the case wheresis greater thant. In that case, the function returns the accumulated amount from an investment of 1 unit at an earlier date,whichwillnormallybegreaterthanorequalto1.Wefinditmoreconvenienttouseonly the one function as given above, since the ideas, and the key relationship (2.1), are the same regardless of the ordering ons,tand u. The concept of a discount function will be a key ingredient in what follows. We will supposethat,givenanyfinancialtransaction,thereisasuitablediscountfunctionthatgoverns theinvestment ofallfunds.Wewilldeal onlybrieflywiththeimportantproblem ofchoosing the discount function. There are many factors governing this choice and it will depend on the nature of the transaction. It may be chosen to simply reflect the preferences of the parties for present as opposed to future consumption. It may reflect the desired return that an investor wishes to achieve. In many cases it is based on a prediction of market conditions that will determine what returns can be expected on invested capital. CALCULATING THE DISCOUNT FUNCTION 11 Apossiblecomplication that wewill not deal withto any great extent isthat which arises when two parties to a transaction have different choices of a suitable discount function. We will assume, unless indicated otherwise, that the same function applies to both parties. 2.4 Calculating the discount function Wenowprovideaprocedureforcalculatingthevaluesof vinasystematicway.Thecurrency example indicated that we can calculate all the values of a discount function just by knowing thoseatpointswithafixedfirstcoordinate,thatis,withacommoncomparisonpoint.Inmost applications it is convenient to take this as time 0. To simplify notation, we drop the first coordinate in this case and define v(t) = v(0,t). It follows from (2.1) thatv(s,t) = v(s,0)v(0,t) and then from (2.3) that v(s,t) = v(t) v(s). (2.4) In the first few chapters of this book, we will need to know the value ofv(s,t) only for integral values ofs and t. From (2.4), it will be sufficient to knowv(n) wheren is any nonnegative integer. To calculatev(n), note first that the key relationship (2.1) can be extended from one involving three terms to an arbitrary number. That is, given timest1,t2,…,tn, v(t1,t2)v(t2,t3)⋯ v(tn−1,tn) = v(t1,tn). (2.5) To see this, take for example, n= 4. The quantity v(t1,t2)v(t2,t3)v(t3,t4) is equal to v(t1,t3)v(t3,t4) by applying (2.1) to the first two terms. By another application of (2.1) it is equal tov(t1,t4). We have extended (2.1) to a formula involving four terms. A similar step extends from four to five, and so on. Formally, we are using mathematical induction. It follows from (2.5) that v(n) = v(0,1)v(1,2) ⋯ v(n−1,n), (2.6) so we need only knowv(n−1,n) for all positive integersn. Given such values, we can use the recursion formula v(n) = v(n−1)v(n−1,n), v(0) = 1, (2.7) to calculate all values ofv(n). The information we need is then summarized by the vector v = (v(0),v(1),v(2),…,v(N)), whereN is the final duration at which a nonzero cash flow occurs. 12 THE BASIC DETERMINISTIC MODEL 2.5 Interest and discount rates In practice, rather than specifyingv(k−1,k) directly, it is more common to deduce this quantity from the corresponding rates of interest or discount. Given any discount functionv and a nonnegative integerk, these are defined as follows. Definition 2.2 The rateofinterest for the time intervalktok+1 is the quantity ik = v(k+1,k)−1. Definition 2.3 The rateofdiscount for the time intervalktok+1 is the quantity dk = 1−v(k,k+1). Note that an investment of 1 unit at timekwill producev(k+1,k) = 1+ik units at time k+1.Similarly,aninvestmentof1 −dk = v(k,k+1)unitsattime kwillaccumulateto1unit at timek+1. Sodk is the amount you must take off from each unit paid at the end of the period, to get the equivalent amount at the beginning of the period. Given any of the three quantitiesv(k,k+1),ik or dk we can easily obtain the other two. For example, using the definitions and (2.3), it is straightforward to deduce that dk = ikv(k,k+1) = ik 1+ik , ik = dkv(k+1,k) = dk 1−dk . (2.8) Remark In most cases we expectik will be a nonnegative number, although from the definition it could in theory take on any value greater than−1. Indeed, people who make investments yielding less than the rate of inflation are in effect receiving a negative interest rate.Wewillnotbeconcernedwithinflationinthistexthowever,andthereadercangenerally assume that interest and discount rates are nonnegative, unless otherwise specified. Remark The reader is cautioned that some other authors use a different convention for the subscriptk in interest and discount rates. They would refer to ouri 0 as i1, since it is the interestrateforthe firsttimeinterval,andingeneralwoulduse ik+1 forour ik.Wefinditmore convenient to start all indexing at 0. 2.6 Constant interest Readers who have previously studied compound interest will be familiar with one particular familyofdiscountfunctions.Supposewebelievethataccumulationofinvestedfundsdepends only on the length of time for which the capital is invested, and not on the particular starting time. That is, we postulate that for all nonnegatives,t,h, v(s,s+h) = v(t,t+h). (2.9) VALUES AND ACTUARIAL EQUIVALENCE 13 If this holds, then v(s+t) = v(0,s+t) = v(0,s)v(s,s+t) = v(0,s)v(0,t) = v(s)v(t). Wehaveafamiliarfunctionalequation,anditiswellknownthatifweassumesomeregularity condition, for example, thatvbe continuous, we must havev(t) = v t and therefore that v(s,t) = vt−s for some constantv. For such a discount function, the rate of interestik is a constanti= v−1 −1, and the rate of discountdk is a constantd= 1−v. The discount function is therefore conveniently given by simply stating a single parameter, which is usually taken asi, the constant rate of interest. So, for example, if we want to know how much we will accumulate at timenfrom an investment of 1 at time 0, this is justv(n,0) = (1+i)n, which is the usual starting point for the subject of compound interest in elementary texts. In the pre-calculator, pre-computer age, this constant interest family of discount functions was almost always used, mainly to facilitate the computation. Many current textbooks on this subject still deal largely with this constant interest case. However, with modern computing methods, such as spreadsheets, the extra effort involved in using a general discount function is negligible, and there is no reason to restrict the flexibility that one can achieve. Throughout this book we will use arbitrary discount functions, although occasionally we will restrict discussion to the constant interest caseinordertosimplifythenotation.Oneofthemainadvantagesofnotrestrictingourselves to constant interest will be apparent in Chapters 4 and 5 where we will be able to incorporate the contingencies of life and death into the discount function. A key point in these chapters is that the calculation of premiums for life insurance and life annuities can be expressed as a special case of interest theory, using a general discount function. 2.7 Values and actuarial equivalence Wenowputtogetherthetwokeyconceptsof cashflowvector anddiscountfunction.Suppose we are given a cash flow vectorc = (c0,c1,…,cN) and a discount functionv. We want to calculate the single payment at time zero that is equivalent to all the cash flows, assuming that the time value of money is modeled by the given discount functionv. This amount is commonly referred to as thepresent valueof the sequence of cash flows, and sometimes abbreviated as P.V. We can think of it as the amount we would pay at time zero in order to receive all of the cash flows, or equivalently as the single payment that we would be willing toacceptnowinlieuofallthesefuturecashflows.Thecashflowattime khasapresentvalue ofc kv(k)bydefinitionofthediscountfunction.When ck > 0thisiswhatwehavetopaynow in order to receiveck at time k. Whenck < 0, receiving−ckv(k) now will let us pay out−ck at timek. We then add up the individual present values to get Present value of all cash flows= N∑ k=0 ckv(k). (2.10) 14 THE BASIC DETERMINISTIC MODEL Example 2.1 To make sure this is understood, take a very simple example. Letv(k) = 2−k for allk. This is a constant interest rate per period of 100 %. In other words, money doubles itself every period. This is of course not very realistic for a period of a year, but it could hold for a sufficiently long interval. Suppose we are to receive 12 units at time 2, but will be required to pay out 8 units at time 3. Find the present value, and verify that it makes sense. Solution. From(2.10)thepresentvalueis12 v(2)−8v(3) = 12(1∕4)−8(1∕8) = 2.To verify this, we note that the 12 units received at time 2 will accumulate to 24 at time 3. We then have topayout8,leaving anaccumulation of16unitsbytime3.Contrastthiswithreceiving instead asinglepayment of2at time0. Thiswillaccumulate to4attime1,8at time2and at 16attime3.Thereforeassuming(aswedothroughout)thatallmoneyaccumulatesaccording to the given discount function, we are in exactly the same position in both cases. Remark Unrealistic interest rates will be frequently used in examples and exercises throughout the book, in order to simplify the numerical computation, and allow the reader to concentrate on the underlying concepts. So for example, we will often takei k = 20%, 25%,50%,100%,whichcorrespondrespectivelyto v(k,k+1)equalto5∕6, 4∕5, 2∕3, 1∕2. Calculationiseveneasierwhenwehaveaninterestrateof0,inwhichcase v(k) = 1forall k. Notethatitwasconvenient intheaboveexampletocomparetheamountsaccumulatedat thetimeofthelastpayment.Thisisknownasthe accumulatedvalue andingeneralisgivenby N∑ k=0 ckv(N,k). It represents the amount we will have at timeN, resulting from all the cash flows. More generally, we can calculate a value atanytime between 0 andN. In this chapter we concentrate on integer times. We take the present value of the future cash flows at that time, plus the accumulated values of the past cash flows. The following definition formulates this precisely. Definition2.4 Foranytime n= 0,1, …,N,the valueattimen ofthecashflowvector cwith respect to the discount functionvis given by Val n(c;v) = N∑ k=0 ckv(n,k). It represents that single amount that we would accept at timenin place of all the other cash flows, assuming that everything accumulates according to the discount functionv. The values at various times are related in a simple way. Sincev(m,k) = v(m,n)v(n,k), it follows immediately that Valm(c;v) = Valn(c;v)v(m,n). (2.11)‡ (Formulas marked with‡ denote key facts, of particular importance.) VALUES AND ACTUARIAL EQUIVALENCE 15 Formula (2.11) should be intuitively clear. The single amount we would accept at time m in place of all the cash flows must be the value at timem of the single payment that we would accept at timen, in place of all the cash flows. We can then easily deduce all values from the value at a particular time. Normally it will be convenient to take this as time 0. The same point was illustrated in the currency example of Section 2.2, where we pointed out that thetotalamountinanyonecurrencywaseasilyconvertedtothetotalinanyotherbyasingle multiplication. Notation For the particular case of values at time 0 we will use a special symbol. Let ̈a(c;v) = Val 0(c;v). The letterais a standard actuarial symbol that is used to stand forannuity, another name for a sequence of periodic payments. See Section 2.14 for an explanation of the two dots. Whenthereisonlyonediscountfunctionunderconsiderationweoftensuppressthe vand just write Valn(c)o r̈a(c). We can express and calculatëaconveniently by expressing it in vector form: ̈a(c) = v⋅ c = vcT. (2.12) The second term is the (scalar) inner product of the two vectors. The third views the vectors v andc as 1×N matrices, with the superscript T denoting a matrix transpose. Formulas (2.11) and (2.12) make it clear that calculating values is a linear operation, a fact we will often exploit. That is, Val k(c+d) = Valk(c)+Valk(d), Val k(𝛼c) = 𝛼Valk(c), (2.13) for any cash flow vectorsc and d, scalar𝛼 and durationk. We have been comparing the values of a sequence of cash flows with a single payment at a particular time. We often wish to compare the values of two sequences of cash flows. For this we have the following definition. Definition 2.5 Two cash flow vectorsc and e are said to beactuarially equivalentwith respect to the discount functionvif, for some nonnegative integern, Val n(c;v) = Valn(e;v). From (2.11), if the above holds for somen, it holds for alln. Take,forexample, n= N.Weseethatapersontakingthepaymentsgivenby candletting them accumulate according to the given discount functionvwill eventually be in exactly the samefinancialpositionasonetakingthepaymentsgivenby e.Thisisthemeaningofactuarial equivalence. Manyproblemsinactuarialmathematicsreducetothefollowing.Wearegivenacashflow vectorc,andanothercashflowvector ethatdependsonsomeunknownparameters.Wehave to solve for the unknown parameters in order to make the two vectors actuarially equivalent. 16 THE BASIC DETERMINISTIC MODEL We have already done asimpleexample of thisincalculating Valn(c).Inthat case wewanted a single payment at timenthat is actuarially equivalent toc. In another common application, a lender advances payments to a borrower, and the borrower must return the money by loan repayments. The lender is therefore trading one sequence of payments (the advances) for another (the repayments) and wants the two to be actuarially equivalent. We look at a simple example. Example 2.2 A lends B 20 units now and another 10 units at time 1. B promises to repay theloanbytwopayments,madeattimes2andtime3.Therepaymentattime3istobetwice as much as that at time 2. If A wishes to earn interest of 25% per period, what should these repayments be? Solution. Let K be the unknown payment at time 2. We want to findK so that the vectors c = (20, 10, 0, 0) ande = (0,0,2 K,3K) are actuarially equivalent. There are many possible calculation methods. We could determine the vectorv and use formula (2.12), which is essentially the best approach for a spreadsheet method, as we describe later in Section 2.14. For small problems to be done by hand calculation, it is convenient to make use of a time diagram, where we indicate the payments and the 1 year discount factorsv(k,k+1) which areallequalto(1.25) −1 = 0.8.SeeFigure2.1.Readersmayfindtousefultowritedowntheir own time diagrams for the examples in the book, if they are not given. Advances 0 0 10 20 Time 3 2 1 0 Repayments 0 0 K 2K vs 0.80 .80 .8 Figure 2.1 Example 2.2 Making use of formula (2.6), we calculate values at time 0 by multiplying each payment ck by v(k), which we calculate as the product of all the preceding discount factorsv(i,i+1). So, P.V. of advances=20 + 10 (0.8)=28 P.V. of repayments=K[(0.82 +2(0.83)] = 1.664K. Equatingvaluestomaketheadvancesandrepaymentsactuariallyequivalent, K = 28∕1.664 = 16.83. The borrower pays 16.83 at time 2 and 33.66 at time 3. Weconcludethissectionbydescribingausefultechniquethatwewillcallthe replacement principle.Supposewearegivenacashflowvectorandsomesubsetoftheentries(0,1, …,N). Takethevalueattime kofjustthosecashflowsinthesubsetandthenreplaceallentriesinthe subsetbyasinglepaymentattime kequaltothatvalue.Thisleavesavectorthatisactuarially equivalent to the original. A formal derivation can be given by writing the vector as the sum of two vectors and using linearity. We will leave the details to the interested reader. VECTOR NOTATION 17 Benefits 42− 3 − 5 Time 3 2 1 0 1 Benefits 02− 3 0 Time 3 2 1 0 Figure 2.2 Example 2.3 Thefollowingisasimple,althoughartificialexample.Wewillgivemorerelevantusesof this principle later. See Figure 2.2. Example 2.3 Let c = (2,4, −3,−5). Assumeaconstantinterestrateof0.25.Findtheactuariallyequivalentvectorbyapplyingthe replacement principle with timek= 2 and the subset{1,3}. Solution. Thevalueof c 1 andc3 attime2is4(1.25) −5(1.25)−1 = 1.Makingthereplacement, weobtainthevector(2,0, −2,0)thatisactuariallyequivalentto c,ascanbeverifiedbydirect calculation. 2.8 Vector notation We now introduce some convenient notation for vectors. In all examples, we will have a maximum durationN, and all of our vectors will be (N+1)-dimensional, with entries indexed from 0 toN. However, we will often write a vector of lower dimension with the understanding that all subsequent entries will be zero. For example ifN = 8, then (1, 3, 2) will denote the vector (1, 3, 2, 0, 0, 0, 0, 0, 0). For vectorsc and d, written as above, we will write (c, d) to denote the vector consisting of the entries ofc f o l l o w e db yt h o s eo f d. For example ifc = (2,2), and d = (3,7,4,1)then( c,d) = (2,2,3,7,4,1,0,… ,0).(Note that the juxtaposition must comebeforefilling in the ending zero entries.) For any number r and integer k we will write (r k) to denote the vector consisting ofk entries of r.F o r example,(13,25) = (1,1,1,2,2,2,2,2,0,… ,0).W eletei denotethestandard ithbasicvector, i= 0,1,2, …,N, that is, a vector with an entry of 1 in positioniand zeros elsewhere. For a given vectorb = (b0,b1,…,bn), let Δb = (b0,b1 −b0,b2 −b1,…,bn −bn−1,−bn). Thatis, Δbisobtainedfrom bbysubtractingfromeachentrytheimmediateprecedingentry, except for the entry in position 0, which remains the same. (Think of the entry inposition−1 18 THE BASIC DETERMINISTIC MODEL as 0). Note also that by our notational convention,bj = 0f o rj> n, which accounts for the final entry of−bn. As a check, the sum of all the entries inΔbmust be 0. (The reader should also note that our definition differs from some authors’ usage, wherebyΔb(k) would equal bk+1 −bk, rather thanbk −bk−1 as we have defined it.) We use the symbol * to denote pointwise multiplication of vectors, that is, (a1,a2,…,an) ∗ (b1,b2,…,bn) = (a1b1,a2b2,…,anbn). The remainder of this section can be omitted on first reading. We develop an identity which is somewhat technical but which will be very useful in later chapters. For any vectorb = (b0,b1,…,bn) and a discount functionv, define a new vector∇b whose entry of indexkis given by ∇bk = bk −v(k,k+1)bk+1. (2.14) So∇is something likeΔexcept we subtract in the reverse order and discount the second term. Note that∇ depends onvand it could be denoted as∇v if there is any confusion. Our main result is that for any other vectorc = (c0,c1,…) ̈a(∇b ∗ c) = ̈a(b ∗Δ c). (2.15) To see this, we just note that the left hand side is (b0 −v(1)b1 )c0 +(b1 −v(1,2)b2 )c1v(1)+(b2 −v(2,3)b(3))c2v(2)+⋯ . Expanding and using the fact thatv(k)v(k,k+1) = v(k+1), this equals b0c0 +b1(c1 −c0)v(1)+b2(c2 −c1)v(2)+⋯ which is the right hand side. We can remember this formula as saying that when computing the present value of a pointwise product of one vector multiplied byΔ applied to a second vector, theΔcan slide off the second vector and become a∇ on the first. 2.9 Regular pattern cash flows Before the introduction of computers and calculators, finding values of cash flows could involve a lengthy computation. Accordingly, the emphasis was not only on constant interest rates, but also on vectors where all nonzero cash flows were of the same amount, usually referred to as the case oflevel cash flows, or where the cash flows followed some regular pattern, such as entries increasing in arithmetic progression. Values in this case were fairly easy to compute by algebraic means. One would then try to handle more general cash flow vectors by expressing them in terms of those with regular patterns. Readers who have had previousexperiencewithcompoundinterestcourseshavenodoubtseensuchtechniques.With modern computing methods there is little need for these methods for computing purposes, althoughtheyaresometimesusefulfortheoreticalmatters.Wegiveabriefillustrationofsome REGULAR PATTERN CASH FLOWS 19 of the more useful formulas of this type. We assume throughout the section that the discount function is given byv(n) = v n for some constantv. Consider the vectors (1n) andjn = (1,2, …,n−1,n). Then ̈a(1n) = 1+v+v2 +⋯ +vn−1. Multiplying byv, v̈a(1n) = v+v2 +⋯ +vn. Subtracting the second equation from the first and dividing by 1−vgives ̈a(1n) = 1−vn 1−v . (2.16) Many readers will have seen this technique for summing a geometric progression. A similar trick handles the vectorj by reducing to the level payment case. ̈a(j n) = 1+2v+3v2 +⋯ +nvn−1. Multiplying byv, v̈a(jn) = v+2v2 +⋯ +nvn. Subtracting the second equation from the first gives (1−v)̈a(jn) = (1+v+v2 +⋯ +vn−1)−nvn, and dividing by (1−v), ̈a(jn) = ̈a(1n)−nvn 1−v . (2.17) There is an alternate way to derive (2.16) and (2.17), as well as many other similar formulas,whichdoesnotinvolveanyseriessummation.Itisbasedonthefactthataloanmay be paid off by paying interest each period on the prior amounts advanced, and then paying off the total principal at the end. This is intuitively clear. A formal derivation will be given in Section 2.11. Suppose you receive a loan of 1 unit. You could repayi at the end of each period forn periods and then eventually repay the principal at timen. The repayment vector is therefore (0,i,i,…,i+1), which by the replacement principle is actuarially equivalent to (d,d,…, d,1) =d(1 n)+en,sinceacashflowof datanytime kisactuariallyequivalentto d(1+i) = iat timek+1.(Inotherwords,ifyoupayinterestatthebeginningoftheyear,theappropriaterate isdratherthan i.)Equatingthepresentvalueoftheadvancesandrepayments,1 = d̈a(1n)+vn, which leads to (2.16). Forthesecondformula,supposeyouaretoreceiveloanadvancesof1unitatthebeginning of each year fornyears. According to our scheme you will pay 1 unit of interest at time 1, 20 THE BASIC DETERMINISTIC MODEL 2 units of interest at time 2, etc. You then will repay the total principal ofn at timen.T h e repayment vector is (0,i,2i,…(n−1)i,ni+n). By replacing eachiby adone period earlier, this is actuarially equivalent todjn +nen. Equating present values of this with the vector of loan advances gives̈a(1n) = d̈a(jn)+nvn, which leads to (2.17). 2.10 Balances and reserves 2.10.1 Basic concepts This section will introduce one of the most fundamental actuarial concepts, that of a reserve. Supposeweenterintoafinancialtransaction,representedbythecashflowvector c.Atany future timekthere are two fundamental quantities to compute. First, we would like to know the total amount accumulated from all payments of the transaction up to this point. Second, we would like to know how much money we willneedat that time in order to discharge our future obligations under the transaction. We illustrate with a simple example. Example 2.4 Let c = (3,6,1,2,−20), v(0,1) = 0.6, v(1,2) = 0.5, v(2,3) = 0.4, v(3,4) = 0.5. How much do we have, and how much will we need at time 2, justbeforethe 1-unit payment due at that time? SeeFigure2.3,wherewehaveinsertedanarrowtoindicatethetimethatvaluesaretaken. Payments 2 1 6 3−20 Time 4 3 2 1 0 v s 0.6 0.5 0.4 0.5 Figure 2.3 Example 2.4 Solution. Theamountwehaveisclearlythe3unitspaidtousattime0,accumulatedfortwo periods, and the 6 units paid at time 1 accumulated for one period, for a total amount of 3v(2,0) +6v(2,1) = 3v(2,1)v(1,0) +6v(2,1) = 3 0.6×0.5 + 6 0.5 = 22. For the second question, note first that we have an obligation to pay out 20 units at time 4, and the amount we need at time 2 to provide for this is 20v(2,4) = 20v(2,3)v(3,4) = 20×0.4×0.5 = 4. We can offset this with the positive cash flows that we will acquire after time 2. The value at time 2 of the 1 unit due immediately at time 2 is just 1, and the value at time 2 of the 2 units payable at time 3 is just 2v(2,3)= 0.8. The total needed to ensure we can meet our obligations is 4−1−0.8 = 2.2. Let us verify this directly. Suppose we have 2.2 units at time 2, just before the payment due at that time. The payment of 1 will come in to give us 3.2. This will accumulate to BALANCES AND RESERVES 21 (3.2/0.4) = 8 units at time 3. We will receive another 2 units at time 2 for a total of 10, and this will accumulate to the 20 units that we need at time 4 in order to meet our obligation at that time. We now introduce some notation and terminology to express these concepts in a gen- eral case. Notation Given any cash flow vectorc and nonnegative integerk,l e t kc = (c0,c1,…,ck−1,0,… ,0), kc = (0,0, …,0,c k,ck+1,…,cN) so that c = kc+kc. (2.18) For example, for the vectorc in Example 2.4,2c = (3,6,0,0,0)while 2c = (0,0,1,2,−20). The idea is thatkc represents the past cash flows andkc the future cash flows when measured from timek. It is important to note that the payment at exact timek is by our convention taken as future. (We will see in Chapter 6 that this fits in with the usual treatment for life insurance contracts.) Note also that 0cis the zero vector while0cis justc. Definition 2.6 For k= 0,1, …,N,t h ebalanceat timekwith respect toc and v, is defined by Bk(c;v) = Valk(kc;v) = k−1∑ j=0 cjv(k,j). (Wewill,asbefore,suppressthe vwhenthereisnoconfusion.)Thebalanceattime kissimply the accumulated amount at timek resulting from all the payments received up to that time, and answers the question of how much we will have. Note again that by our conventional treatment, balances are computed just before the payment at exact timekis made, so that the timekpayment isnotincluded inB k. Definition2.7 Fork= 0,1, …,N,the reserveattime kwithrespectto candvisdefinedby kV(c;v) =− Valk(kc;v) =− N∑ j=k cjv(k,j). Thereserveattime kisthe negativeofthevalueattime kofthefuturepayments,andisequal to the amount we will need in order to meet future obligations. It is important to remember thenegativesign,toreflectthefactthatwearemeasuringthevalueofnetobligations,thatis, amounts to be paid out of the fund. Referring again to Example 2.5, we calculated thatB 2(c;v) = 22 and2V(c;v) = 2.2. 22 THE BASIC DETERMINISTIC MODEL Theterminologyandnotationforreserveswereborrowedfromaparticularlifeinsurance application (which we will discuss in detail in Chapter 6). In keeping with this standard reserve notation, we have placed duration as a left rather than right subscript. We introduce the reserve concept now, in order to stress that its use is more general than just this particular life insurance application and can apply to any sequence of cash flows. One can think of this quantity as representing capital that must be set aside or ‘reserved’ for future use. For those familiar with accounting terminology, reserves, if positive, representliabilities,which are amounts that we owe to other parties. It is possible of course for the reserve to be negative, which indicates that at timek the presentvalueofthefutureamountscominginforthistransactionwillexceedthepresentvalue ofthefutureamountstobepaidout.Insuchacase,theabsolutevalueofthereserveisa‘receiv- able’, that is,anamount that isowed tous and that willbe received through futurepayments. 2.10.2 Relation between balances and reserves From (2.16) and the linearity of Valk, Valk(c) = Bk(c)−kV(c), (2.19)‡ which says that the value of the transaction at any point is equal to the difference between what you actually have accumulated and what you need to set aside from that accumulation to meet future obligations. An important consequence of this is that for azero-valuecash flow vectorc, B k(c) = kV(c), (2.20)‡ where a zero-value vector is one whose value is zero at all durations. Such vectors arise frequently. Ifc andd are actuarially equivalent, then for any durationk, Val k(c−d) = Valk(c)−Valk(d) = 0, so thatc−d is a zero-value vector. Foratypicalexample,supposethatanindividualborrowsmoney,andagreestorepaywith a sequence of repayments. Consider the net cash flows from theborrower’spoint of view, that is, advances less repayments. If the advances and repayments are actuarially equivalent, these net cash flows form a zero-value vector. What is the outstanding balance on the loan at timek?Thisisthevalueofallthepaymentsyettobemade,whichbydefinitionisjust kV(c). In view of (2.18) it is also equal toBk(c). This is perfectly logical. It says that the amount still owing at timek is the value of all the money that been received, offset by the value of the repayments that have been made. Calculating the outstanding balance by means of the reserveisoftencalledthe prospectivemethod sinceitlookstothefuture,whilecalculatingby means of the balance is often called theretrospective method, since it looks to the past. For this reason, many authors refer to what we have called a balance as aretrospectivereserve. Example 2.5 (Figure 2.4). An individual borrows 1000 now and another 2000 at the end of 1 year. This loan will be repaid by yearly payments for 10 years, beginning 5 years from the present. The yearly payment doubles after 5 years. The interest rate is 0.06 for the first BALANCES AND RESERVES 23 Advances 20001000 0 0 0 0 0 0 0 0 0 0 0 0 0 Time 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 Repayments 0 0 0 0 0 K K K K K2K 2K 2K 2K 2K Figure 2.4 Example 2.5 5 years and 0.07 thereafter. How much is still owing 8 years from now, prior to the payment at that time? How much is still owing after the payment at time 8? Solution. Wemust firstdetermine therepayment amounts. LetKbe theinitial repayment. If c denotes the advances andd denotes repayments, then c = (1000,2000), d = (0,0,0,0,0,K ,K,K,K,K,2K,2K,2K,2K,2K) = K(0 5,15,25), ̈a(c) = 2886.79, ̈a(d) = 7.95326K, and equating, K = 362.97 (see Section 2.15 for the calculation details). So the first five instalments are each 362.97, and the second five are each 725.94. The amount owing at time 8, just prior to the payment due on that date, can be calculated either as B 8(c−d) = Val8(e), where e = (1000,2000,0 3,−362.973) or as 8V(c−d) = Val8(f), where f = (08,362.972,725.945). In either case the amount is 3483.97. The outstanding balance at time 8,after the payment due at that time, is just 3483.97−362.97 = 3121.00. It is of interest to note that the lower payments at the beginning have the effect that the borrower is not repaying enough to handle the interest on the loan, so the amount owing is greater than the total amount borrowed. 2.10.3 Prospective versus retrospective methods For the model that we have described so far, either the prospective or retrospective approach can be used to determine the amount owing on a loan. In pre-computer days, one often chose the one that allowed easier calculation, but with modern methods it makes little difference. Thereare,however,situationswhereoneapproachortheotherdominates.Theretrospec- tive approach must be used in cases where the time and amounts of loan repayments are not scheduled in advance, but can be made at the option of the borrower. ConsideravariationonExample2.5.Supposetheborrowerhadmadepaymentsof300at time5,400attime6and500attime7,andthepaymentsafterthatwerenotyetdetermined.To 24 THE BASIC DETERMINISTIC MODEL obtaintheoutstandingbalanceattime8wewouldnecessarilyusetheretrospectivemethodand calculate the balance. This is Val 8(b), whereb = (1000,2000,0,0,0, −300,−400,−500,0). Ontheotherhand,theprospectiveapproachfiguresprominentlyincaseswherethechoice of an appropriate discount function may have changed since the transaction was originally entered into. In that case the reserve calculated with respect to anewdiscount function that is realistic at that time, could well give an amount that differs from that given by the balance as calculated retrospectively. After all, theequality of these twodepended on the fact that the same discount function was used for both. We will not dwell too much on this complication, but would like to describe two very familiar situations. Suppose the borrower wishes to discharge a loan at some point prior to the scheduled completion. The natural amount to pay would be the outstanding balance on the loan at that date, as calculated by either method. Suppose, however, that the loan contract requires that payments be made as scheduled and does not permit you to repay early. This is often the casewithhomemortgages.Suppose,further,thatinterestrateshavegonedownsincetheloan was originallytaken out. The lender willnot welcome earlyrepayment asthese amounts will have to be reinvested at a lower rate. Borrowers may however be permitted to repay, if they include an extra amount as a penalty. What is really happening is that the lender is revaluing the reserve, by applying the new discount function to the future repayments. If interest rates decrease, the value of the functionv increases. Assuming the normal case where entries in the vector kc are nonpositive as they represent repayments, the negative of these values are nonnegative, and multiplying by the higher values ofv and summing will lead to a higher reserve. The penalty represents the excess value of the reserve as calculated by an up-to-date discount function, over the original value. (It should be noted that, in practice, the penalty amounts are often determined by approximate formulas rather than an exact recalculation of the reserve.) As a second example, suppose you purchase a bond, which means that you are in the position of the lender. The nature of a bond is that this debt is assignable to another party by tradinginthebondmarket.Yousellthebondtosomeonewhowillthencollecttherepayments by the issuer. What is a fair price? The answer is obviously the outstanding balance, not as calculated originally, but rather with respect to the rates of interest applicable at the time of sale.Asinthecaseabove,lowerinterestrateswillcausethisoutstandingbalancetoincrease. Traders who buy bonds always hope that interest rates will fall, so the outstanding balances increase, and they can sell the bonds for a profit in the market. 2.10.4 Recursion formulas If f is a function defined on the nonnegative integers, it is often useful to derive a recursion formula that expressesf(k+1) in terms off(k). Given an initial valuef(0), one can then successivelycalculatevaluesof f(k)forall k.Thisisknownasa recursionformula.Arecursion formula often leads to adifferenceformula that gives an expression forf(k+1)−f(k). There are many recursion formulas in actuarial mathematics. For the most part, they can be derived from the basic recursions for balances and reserves that we present in this section. We will first expressBk+1 in terms ofBk. For any cash flow vectorc and durationk,i ti s clear from the definitions that k+1c = kc+ckek. BALANCES AND RESERVES 25 The value at timek of the second term on the right isck and the value at timek of the first term on the right isBk(c).Invoking linearity and multiplying byv(k+1,k) to get the value at timek+1, gives the recursion Bk+1(c) = (Bk(c)+ck)v(k+1,k), (2.21) where we start the recursion with the initial value,Bk = 0. Subtracting Bk from each side and expressingv(k+1,k)a s1 +ik yields the difference formula Bk+1(c)−Bk(c) = ik(Bk(c)+ck)+ck. (2.22) Difference equations are not normally efficient for calculating numbers. Rather, they are useful for analyzing how quantities change from one period to another. For example, the above formula expresses the increase (or decrease if negative) in the balance over a period as a sum of two quantities, the interest earned plus the additional cash flow. Recall that our convention was to treat cash flows at timek as future with respect tok. This ties in well withinsurance contracts, as we will see later. However, the other convention is normally used when dealing with loans. Accordingly, we define ̃B k(c) = Bk(c)+ck, theaccumulatedamountattime kafter thecashflowattime kispaid.Weleaveittothereader to verify the corresponding recursion and difference equations as ̃Bk+1(c) = ̃Bk(c)v(k+1,k)+ck+1, (2.23) ̃Bk+1(c)− ̃Bk(c) = ik̃Bk(c)+ck+1. (2.24) Rewriting (2.22), we get −ck+1 = ik̃Bk(c)+[̃Bk(c)− ̃Bk+1(c)], (2.25) which forms the basis for the usual loan amortization schedules. (These are schedules that show the balances at each duration and how they change.) The left hand side above is the amountrepaidontheloanattime k+1(thenegativeofthenegativecashflow)andtheformula gives the split of this repayment into two parts. The first is the interest on the outstanding balance, and the second is the amount of principal reduction. Let us now turn to reserves. They satisfy exactly the same recursion relation as balances. That is, k+1V(c) = (kV(c)+ck)v(k+1,k), (2.26) ‡ wherewestarttherecursionwiththeinitialvalue 0V(c) =− ̈a(c).Thisisobviousinthecaseof zero-valuevectorswherereservesequalbalances,butitholdsingeneral,asseenforexample from the relation, k+1c= kc−ckek, or alternatively derived from (2.19), substituting from (2.17) and using (2.11). 26 THE BASIC DETERMINISTIC MODEL 2.11 Time shifting and the splitting identity In this section we provide a background for following some classical formulas that appear in the actuarial literature. From (2.16), taking values at time 0 and applying (2.11), we derive ̈a(c) = ̈a( kc)+v(k)Valk(kc). (2.27) We call this thesplitting identity, since it splits the calculation of a present value into two parts, considering first those cash flows before timek, and then those after timek. It can also beobtainedasaspecialcaseofthereplacementprinciple.Replaceallthecashflowsaftertime kby the single payment of Val k(kc) at timek. In classical actuarial mathematics, the spitting identity was often used as a calculation tool when there was distinct change in the cash flow sequence or discount function occurring at timek. ( It can still be useful in this regard in the continuous case discussed in Chapter 8.) The remainder of this section deals mainly with notational issues, and could be deferred until we apply it later in Section 4.3.2. The splitting identity often appears in a somewhat different form, since the symbol Val k is not standard. In traditional actuarial notation, it is commontoexpressallformulasintermsof ̈a(orsimilarsymbolswhichweintroducelater).So thequestionisthen,howdowewriteVal k intermsof ̈a?Thisisquitesimplewhenpayments and interest rates are constant. In general case we must introduce some new notation for timeshifting. Suppose we are at timekand we wish to consider this as the ‘new’ time 0. Given a cash flow vectorc, define the cash flow vectorc◦ kby (c◦ k) n = ck+n. For example, ifc = (2,2,3,4,5),then c◦ 2 = (3,4,5). In other words,c◦ ksimply gives the cashflowsinorder,butstartingwiththeoneattime k.Anotherwayoflookingatitisthat c◦ k is justkcwith the firstkzeros removed. This notation was not needed with constant payment vectors, since whenc is (1n), thenc◦ k= (1n−k). Similarly, given a discount functionv, define a new discount functionv◦ kby v◦ k(n,m) = v(n+k,m+k) so that v◦ k(n) = v(k,k+n). Once again, the idea is simply that we are treating timekas time 0 and measuring time from that point. It is quite simple to calculatev◦ k from the factors for periods of length 1 as we did in (2.6). We multiply as before, but start withv(k,k+1) rather thanv(0,1). An important point to notice is that, from (2.9) v◦ k= v if interest is constant, CHANGE OF DISCOUNT FUNCTION 27 which explains why this concept does not appear in classical works where interest is almost always taken as constant. We can now write Val k(kc) = N∑ i=k v(k,i)ci = N∑ i=k v◦ k(i−k)(c◦ k)i−k = N−k∑ j=0 v◦ k(j)(c◦ k)j = ̈a(c◦ k,v◦ k), (2.28) whichisintuitivelyclearsincewearestartingattime k,treatingitasanewtime0andvaluing the future payments. We can then write the splitting identity in its usual form ̈a(c;v) = ̈a(kc;v)+v(k)̈a(c◦ k;v◦ k). (2.29) Under constant interest, we can write this in an easier fashion as ̈a(c) = ̈a(kc)+vk̈a(c◦ k). (2.30) The splitting identity can sometimes be useful in determining the effect of changes in certain quantities. Example 2.6 Suppose thaẗa(c) = 19.6 and̈a( 10c) = 10. If v(9,10) increases from 0.8 to 0.81 while all other values ofv(k,k+1) remain the same what is the new value of̈a(c)? Solution. From (2.29), withk = 10, we know thatv(0,9)(0.8) ̈a(c◦ 10,v◦ 10) = 9.6, so v(0,9)(0.81) ̈a(c◦ 10,v◦ 10) = 9.72 and the new value of̈a(c) is 19.72. ∗2.11 Change of discount function This section is somewhat technical. Its purpose is to derive a useful relationship that allows ustoreplaceonediscountfunctionwithanother,whichcouldbemoreconvenientforthetask at hand. Suppose we are given a cash flow vectorc and a discount functionv.L e tv ′ be another discount function. Define a new cash flow vectorc′ whosekth entry is given by c′ k = ck +[v′(k,k+1)−v(k,k+1)]Bk+1(c;v). (2.31) It follows that Bk(c;v) = Bk(c′;v′), for all k. (2.32) 28 THE BASIC DETERMINISTIC MODEL We prove (2.30) inductively. To simplify the notation, denoteBk(c;v)b yBk,B′ k (c′;v′)b yB′ k , v(k,k+1) byv and v′(k,k+1) byv′.A tk = 0, the balances are both equal to 0. We will assume thatBk =B′ k and show thatBk+1 = B′ k+1 : Bk+1 = Bk +ck v = B′ k +c′ k v + ck −c′ k v = B′ k+1 v′ v + v−v′ v Bk+1. Formula (2.30) fork+1 follows after some minor algebraic manipulation. A corollary of this is that ifchas zero value with respect tov, thenc′ has zero value with respect tov′. This follows since a cash flow vector has zero value if and only if its balances eventually equal to 0. We will later apply this result to a key idea in life insurance. In this chapter we will give a formal proof of the assertion used in Section 2.9. Consideraloanfor Nperiodsmadeaccordingtoadiscountfunction,whichwewilldenote byv′ totieinwiththenotationabove.Welet d′ andi′ refertothecorrespondingdiscountand interest rates. Suppose the loan advances are given by vectorc = (c0,c1,…,cN−1). Let s= N−1∑ k=0 ck, r = (d′ 0(c0),d′ 1(c0 +c1),d′ 2(c0 +c1 +c2),…,d′ N−1s). We want to show thatc is actuarially equivalent to (r+seN) with respect tov′. This will mean that we can repay any loan by paying interest each year on the prior advances, and then repaying the principal at the end. While this may be intuitively clear, it is important to verify it formally to ensure that our model captures the idea of interest the way one normally perceives it. Letv(k) = 1 for allk, the discount function for a constant interest rate of zero. It follows that B k(c,v) = k−1∑ i=0 ck so that the vectorc′ as given by (2.31) is justc−r. Invoking (2.32), this gives ValN(c−r−seN;v′) = BN(c′;v′)−s= BN(c;v)−s= 0, showing the desired actuarial equivalence. 2.12 Internal rates of return In this section we take the point of view of an investor or lender, who is undergoing a transactionthatinvolvesinvestingfunds,hopingtogetbackamountsthataregreaterinvalue thanthoseputin.Let c = (c 0,c1,…,cN)denotethenetcashflowvectorofthetransaction.We assume throughout thatcN ≠ 0. We are not given a discount function but rather want to find the constant interest rate that will makec a zero-value vector. In this section it is convenient to deal with balances that include the payment due at the time the balance is computed. That is, we want to consider̃Bas defined above. INTERNAL RATES OF RETURN 29 Definition 2.8 An internal rate of return(often abbreviated asi.r.r.) of the transaction is a number iin the interval (−1,∞) such that, for the discount functionv(n) = (1+i)−n, (i) ̃BN(c;v) = 0, and (ii) ̃Bk(c;v) ≤ 0f o rk= 0,1, …,N−1. An i.r.r. is sometimes referred to as ayieldrate. Remark The definition is unchanged if we useBin place of̃B, sinceBk+1 = ̃Bk(1+ik). Theorem 2.1 Ifani.r.r.ofatransactionexists,itisunique. Proof. Suppose, to the contrary, that we have two such rates,iandi′, withi< i′.L e t̃Bk and ̃B′ k denotetheresultingbalances. Suppose that cj isthefirstnonzero entryin c.W emustha v e j< N since j= N would imply that̃BN = cN ≠ 0. We will now show by induction that, for k≥ j+1, we have ̃Bk > ̃B′ k . (2.33) If (2.33) holds for some indexk then it will hold fork+1 since the nonpositivity of balances implies that ̃Bk+1 = ̃Bk(1+i)+ck+1 > ̃B′ k (1+i′)+ck+1 = ̃B′ k+1 . Applyingtheabovefor k= j,andnotingthat cj = ̃Bj = ̃B′ j < 0,verifiestheinitialstep.So we have shown that̃BN > ̃B′ N , contradicting the fact that both rates satisfy (i). The reader familiar with previous literature on this subject may well be surprised at this theorem, since a great deal has been written on the non-uniqueness of the i.r.r. In our opinion this occurs from a faulty definition. The standard way of defining the i.r.r is to require only point (i) of Definition 2.8 and not (ii). The problem then reduces to finding the roots of the polynomial ∑ N k=0 ck(1+i)N−k, and there may indeed be several. However, when balances become positive, it means that the status of the individual has changed from that of a lender to that of a borrower. The transaction therefore involves both borrowing and lending, and it is difficult to give any interpretation to the roots. After all, a lender is seeking a high i.r.r. whereas a borrower wants it to be low. We feel that the best approach is to give the definition above and concede that other transactions just do not possess an i.r.r. under this definition. Other means are needed for their analysis. Note however that no problem is presented by the usual type of transaction, whereby the loan advances, represented by negative entries, all precede the loan repayments, as given by positive entries. The balances will begin as negative. If they ever become positive, it could only be when the entries change to positive. But as the entries then remain positive, the balance could then never become 0. Therefore, in this case, if an interest rateisatisfies (i) it will necessarily be an i.r.r. The remaining material in this section can be omitted as it is not used in the remainder of the book. To handle cases where the i.r.r. does not exist by the definition given above, a 30 THE BASIC DETERMINISTIC MODEL generalization of the definition was given by Teichroewetal. (1965a, 1965b), often referred to as the TRM method. Their approach was to recognize that when the status of a lender or an investor reverts to that of a borrower, as indicated by a positive balance, then that balance shouldaccumulateatsomerate rfixedinadvanceandnotattheunknown,presumablyhigher rate that one is trying to solve for. The rater could be, for example, the rate at which the individualcouldobtainfinancing.Theythendefineageneralizationofbalanceswithrespectto thisrate r,buildingontherecursiveformula(2.21).Let ̃B 0(c;v;r) = c0 anddefineinductively ̃Bk+1(c;v;r) = { ̃Bk(c;v;r)(1+ik)+ck+1,i f ̃Bk(c;v;r) ≤ 0, ̃Bk(c;v,r)(1+r)+ck+1,i f ̃Bk(c;v;r) > 0. Definethei.r.r.of crelativeto rasthevalueof iforwhich ̃BN(c;v;r) = 0when v(k) = (1+i)−k. We will denote this quantity asir. Note that ifi is an i.r.r. by Definition 2.8, then it will be an i.r.r. with respect torfor anyr. In this case balances are never positive and̃Bk(c;v;r) will simply equal̃Bk(c;v). The same type of inductive calculation as in the proof of Theorem 2.1 shows that asi increases, whilek,r and a nonzero vectorc are held fixed, the value of̃Bk(c;r;v) is strictly decreasing. There is therefore at most one i.r.r. relative tor. Moreover, there will always be oneexcept when either̃BN(c;v;r) < 0or ̃BN(c;v;r) > 0forall iin(−1, ∞).Defining ir tobe −1 in the first case and∞in the second case, we can state the following theorem. Theorem 2.2 Foranynonzerovector candanyr > −1,thereisauniquevalueofi r inthe interval[0,∞]. This shows that we do obtain a unique i.r.r. for any transaction after we first postulate the deposit rater. This gives rise to a simple criterion to decide if it is worthwhile to enter into a transaction.Thegeneralruleisthatthetransactionisworthwhileif ir > r,andnotworthwhile if ir < r. ∗2.13 Forward prices and term structure Aninterestingexampleofadiscountfunctionisfurnishedbythe forwardprices on risk-free zero-coupon bonds.A zero-coupon bond is a financial instrument that promises to pay a fixed sum at some future date, known as the maturity date, and which makes no intervening payments (coupons) before that time. By risk-free we mean something like a government bond, where we can safely assume that the bond is sure to be redeemed with no chance of default. Wefirstdiscussthegeneralideaofa forwardcontract.Thisisanagreementbetweentwo parties, whereby one party, the seller, agrees to deliver a certain specified asset at a specified future time to the other party, the buyer. At the time of delivery, the buyer, will remit a sum of money that is agreed upon at the time when the contract is entered into and is independent of the prevailing price at the time of exchange. The price agreed upon at the contract date is known as theforward priceof the contract. The buyer is often referred to as taking along positionand the seller is referred to as takingashortposition. Therearevariousreasonsforsuchanarrangement.Thebuyermaybesomeonewhoneeds theparticularitematsomefuturedate,andwishestolockinapricenow,toprotectthemselves FORWARD PRICES AND TERM STRUCTURE 31 from a future increase in the price. Alternatively, the buyer may be a speculator with no need oftheitem,butwhopredictsthatpriceswillrise,therebygivingthemaprofitastheycanbuy the asset at the delivery date and immediately sell it for a higher price. Similarly, the seller may be one who owns the asset, wishes to dispose of it at a future date, and wants to lock in the amount they will receive as protection against falling prices. Alternatively, the seller may beaspeculatorwhopredictsthatpriceswillfall,andhopestoprofitbybuyingtheitematthe delivery date for less than they have agreed to sell it for. Let ̃v(0,t) = ̃v(t) be the price at time 0 of a 1-unit, zero-coupon, risk-free bond maturing attime t.Forany s≤ t,let ̃v(s,t)betheforwardpricefora1-unitzero-couponbond,maturing at timet, where the delivery date is times. Now what will these forward prices be? It turns out that under a certain natural assumption they are determined from the time zero prices by the rule ̃v(s,t) = ̃v(t)∕̃v(s) (2.34) which implies immediately that̃v is a discount function. (Note that in the notation of Sec- tion 2.11 forward prices as determined from times are given by the time shifted discount function ̃v◦ s.) Theassumptionmadeisthe noarbitrage hypothesis,whichisamajorconceptinmodern day financial economics. Anarbitrage opportunity is one where a party can make a sure profit by buying and selling certain financial assets with no risk of a loss. The hypothesis in question says that in an environment when all parties have perfect information, arbitrage opportunitiescannotpersist.Thisisasimpleresultofsupplyanddemandlawsineconomics. Thearbitrageopportunitiesoccurwhensomeassetsareovervaluedandothersareundervalued. The argument is that if such an opportunity should ever rise, individuals will, in an attempt to take advantage of this, rush to buy the undervalued assets, thereby raising their prices, and rush to sell the overvalued assets, thereby lowering their prices, which restores equilibrium and eliminates the arbitrage. We also make some other idealized assumptions. One is that an arbitrary number of units of a bond can be bought or sold at any time. This includes the possibility ofshort selling whereby an individual can sell an asset that they do not own by borrowing the asset from another party. They plan to acquire the asset at a later date (hopefully at a lower price than theysolditfor)forreturntothelender.Anotherassumption,isthatbuyingorsellingdoesnot involve any transaction costs such as commissions. To verify our claim above we will show by an example, that in the absence of Equation (2.34) an arbitrage opportunity would arise. Suppose, for example, that̃v(10) = 0.75 and ̃v(5) = 0.90 but ̃v (5,10) = 0.8 < ̃v(10)∕̃v(5). An individual could sell a 1-unit bond maturing attime10,receiving0.75andthen(i)usetheproceedstobuy5/6ofa1-unitbondmaturingat time5,and(ii)takealongpositiononaforwardcontractfora1-unitbondmaturingattime10 withadeliverydateoftime5.Attime5,theindividualreceives5/6fromthematuring5-year bond, uses 4/5 of that to settle the forward contract, and now owns a 1-unit bond maturing at time 10, which they use to settle the short sale. A sure profit of 5/6−4/5 is made at time 5. The reader is invited to find a corresponding example of an arbitrage opportunity in the case that̃v(s,t) > ̃v(t)∕̃v(s) and then to provide a general proof. Theanalysisaboveindicatesthatforanindividualorcorporationwhoseinvestmentenvi- ronment consists of risk-free zero-coupon bonds, the forward prices constitute a reasonable 32 THE BASIC DETERMINISTIC MODEL choiceofdiscountfunction.Forexample,theindividualcouldensure(underourassumptions) thatapaymentreceivedattime swouldaccumulateto ̃v(s,t)atasubsequenttime tbyentering into a suitable forward contract. We conclude this section by introducing some conventional terminology. We have given theaboveanalysisintermsofprices,whichseemsthemostconvenientway,buttypicallythe economic and finance literature refers to rates instead of prices. Assume we are given prices of risk-free zero-coupon bonds for all maturities. Definition 2.9 The spot rate of interest y t is the yield rate for the bond maturing at timet. That is ̃v(t) = (1+yt)−t so that yt = ̃v(t)−1∕t −1. (2.35) Definition 2.10 For s≤ t,t h eforwardrateofinterestf (s,t) is the yield rate earned for the bond maturing at timetand acquired on a forward contract with delivery at times. That is ̃v(s,t) = [1+f(s,t)]s−t so that from (2.34), f(s,t) = ( ̃v(t) ̃v(s) )1∕(s−t) −1. (2.36) The relationship between the various maturity dates and the spot rates is known as the term structure of interest rates. A graph which shows values ofyt for various values oft is known as ayieldcurve. Examples of yield curves can be found in the financial and business sections of many daily newspapers, as well as being freely available online. The study of the various possible shapes of yield curves is an important topic for economists which is beyond the scope of this text. Asafinalword,notethatalthoughwehavediscussedspotandforwardratesinthecontext ofrisk-freebonds,theycanbedefined,usingformulas(2.35)and(2.36)forageneraldiscount function. Example 2.7 Given the spot ratesy 2 = 0.05 andy5 = 0.07, find the forward ratef(2,5). Solution. f(2,5) = ( 1.07−5 1.05−2 )−1∕3 −1 = .084. STANDARD NOTATION AND TERMINOLOGY 33 2.14 Standard notation and terminology There is a system of standard international actuarial notation that can seem quite complex at first. One goal in this book is to simplify notation as much as possible. In order to read this book,itwillnotbenecessarytolearnanyofthissystemotherthanthatwhichisintroducedin the text. However, the student who wishes to read other actuarial literature, or write actuarial examinations,orpursueanactuarialcareer,willbeexpectedtobefamiliarwiththefullextent of the system. Accordingly, we will, at the end of each relevant chapter, review the standard notation, indicating how it ties in with the notation in the main text. 2.14.1 Standard notation for cash flows discounted with interest The vector notation is peculiar to this book. In the standard notation the same basic symbola is used for the present value of a cash flow stream, but the particular vector of cash flows is indicated by embellishments of this symbol. A prime example is a n = ̈a(0,1n). A few comments are in order. The ‘angle’ around then is intended to signify a duration of time, as opposed to an age (which will be encountered in the chapters on life annuities and insurances). The vector above arises frequently. It is usual for a loan contract to stipulate that the first repayment is madeoneperiodafter receiving the loan. One does not usually make a repaymentasoftheloandate,sincethatwouldineffectjustmeanyouweregettingasmaller loan. Therefore, the simplest unadorned symbol was reserved for the vector with first entry equal to 0. When there is a payment at time zero, standard notation denotes this by placing two dots above thea. That is, ̈a n = ̈a(1n). Theformercaseistermedan immediateannuity,andthelattera dueannuity.Theterminology is a bit unusual since the immediate annuity does not start immediately but after one period. The name arose as a contrast to the generaldeferred annuitywhere there are several zero entries at the beginning. Once we move away from level payments, the distinction between due and immediate annuities no longer applies. From our point of view, there isalways a payment at time zero although it may be of zero amount. To avoid conflicts with standard notation we use the two dots as a general symbol. Another traditional way of looking at the difference between due and immediate was that theformerappliedwhenpaymentsweremadeatthebeginningoftheyear,andthelatterwhen payments were made at the end of the year. Once again, these distinctions no longer apply when payments are not level, since the beginning of 1 year is the end of the preceding one. (We will, however, have need to refer to this point again, when we discuss fractional period payments in Chapter 7.) Some other symbols are Ia n = ̈a(0,1,2,… ,n,0,… ,0), Ïan = ̈a(1,2, …,n,0,… ,0), 34 THE BASIC DETERMINISTIC MODEL using the vector that we denoted byjn in Section 2.9, Dan = ̈a(0,n,(n−1),(n −2),…,1,0,… ,0), D̈an = ̈a(n,(n−1),(n −2),…1,0, …,0). These are simple enough to remember from the fact thatI indicates anincreasingsequence of payments andDindicates adecreasingsequence. The symbolsis used to indicate an accumulated value. So, for example, sn = Valn(0,1n) = v(n,0)an, ̈sn = Valn(1n) = v(n,0) ̈an. Notethatinbothcasesthevaluesareattime n.Fortheunadornedsymbolthisisatthedateof the last payment. For the double-dotted symbol, timenis oneperiod after the last payment, since the first payment was at time 0. 2.14.2 New notation A reader who has already spend a great deal of effort in mastering the standard notation may havefeltsomedismayinhavingtolearnyetmorenotationintroducedinthischapter.Wefeel, however, that the devices introduced here are useful. There are three main innovations: the vectornotation;thesymbolVal;andandthetime-shiftingnotation.Weconsidereachinturn. The vector notation allows us to conveniently refer to an arbitrary sequence of cash flows in a systematic way. The traditional notation uses special embellishments for each particular sequence, and even so, is restricted to a small number of cases, such as payments constant over some period, or payments in arithmetic progression. The system of writing vectors is useful since it tells you exactly how to enter a cash flow vector into a column of a spreadsheet. For example, faced with the vector (1 20,210), one simply enters a 1 in the first column, copies it down for 20 rows, then enters a 2, and copies it down for 10 rows. Thestandardnotationallowsyoutospecifyvalues onlyatthebeginningorendofatrans- action,anditisoftenusefultodenotevaluesatanintermediatepoint,throughtheuseofVal k. Having introduced the vector notation, it is convenient to introduce the lower and upper subscriptstodenotepastandfuture,respectively.Suchadevicewasnotneededintheclassical literaturethatdealtwithlevelcashflowsandinterestrates.Forexample,thevector k1n isjust 1n−k.Similarly,asymbollikeour ◦ fortimeshiftingcouldeasilybeavoidedsince,asnoted, v◦ k= v.Moreover,1 k ◦ hisjustequalto1 k−h.Thereare,however,examplesintheliterature whereitisnecessarytorefertogeneral time-shiftedcashflows,and adhoc symbolsareused (see Bowersetal., 1997, p. 519). We feel it is much preferable to have a consistent notation to handle these. 2.15 Spreadsheet calculations Here and at the end of Chapters 4, 5, 6, 9 and 10 we will describe spreadsheets for doing the basiccalculationspertinenttothematerialoftherespectivechapter.Ourdescriptionswillrefer in particular to Microsoft Excel ®, and we assume the reader is reasonably acquainted with EXERCISES 35 this application. The basic ideas should be easily adaptable to other formats as well. Readers can reproduce our spreadsheets as given, or they may wish to come up with their own. Our general idea is to enter various functions of the durationk in columns starting with row 10 fork= 0. The cells above can then be used for headings or information. We put the durationk in column A. That is, we insert 0 in cell A10, 1 in cell A11, etc. This is done by putting the formula=A10+1 in cell A11 and copying down to cell A10+N. Theusualwayofspecifyingthediscountfunctionisthroughtheone-periodinterestrates i k. We enter these in column B, withik being inserted into cell B10+k. We then calculate the vector ofv(k) values in column C. Insert 1 in cell C10 and the recursive formula (2.7) is entered into cell C11 as=C10/(1+B10). This is copied down column C to cell C10+N. We enter the cash flows in column D. The cash flowck is put into cell D10+k.T h ev a l u e ̈a(c,v) is then calculated in cell D8 through the formula = SUMPRODUCT($C10:$C10 +N∗D10:D10 +N) where we substitute for the particular value ofN. As a check, take a constant interest rate of 0.06,andthevector(1 10,25).Theansweris12.7883.BycopyingcellD8totheright,wecan do calculations for several different cash flow vectors. Values at timekare easily obtained by dividing̈a(c)b yv(k), which is in cell C(10+k). Notes and references Thischapterisnotintendedtoprovideanexhaustivetreatmentofthemathematicaltheoryof interest.Thegoalwastoprovidethatportionofthesubjectthatisneededfortheremainderof thebook.SomeadditionalmaterialswillalsobegiveninChapters7and8.Readersinterested in a more detailed account can consult Broverman (2010) or Daniel & Vaaler (2009). Additional work on internal rates of return can be found in Promislow (1980, 1997) and Teichroewetal. (1965a, 1965b). The concept of the generalized i.r.r. is due to the latter. Exercises Type A exercises 2.1 You are given a discount functionvwherev(1,3) = 0.9, v(3,6) = 0.8, v(8,6) = 1.2 (a) How much must you invest at time 1, in order to accumulate 10 at time 8? (b) If you invest 100 at time 3, how much will have accumulated by time 8? 2.2 If v(t) = 2−t, and you are given cash flow vectorsc = (1,2,3)and e = (2,K,1), find Kso thatc and e are actuarially equivalent with respect tov. 2.3 Youaregiveninterestrates i0 = i1 = 0.25, i2 = i3 = 1.Youhaveenteredintobusiness transaction where you will receive 2 at time 0, 5 at time 3 and 10 at time 4, in return for a payment by you of 3 at time 2. In place of all these cash flows you are offered a single payment made to you at time 1. What is the smallest payment you would accept? 36 THE BASIC DETERMINISTIC MODEL 2.4 You are given rates of discount as follows:dk = 1∕3, fork= 0,1,2, and dk = 1∕4, for k= 3,4. The vectorc = (62,93,12). Find (a) the discount vectorv,( b )̈a(c) and (c) Val2(c). 2.5 Let c = (1,2,4,−3,8, −12). Suppose v(0,1) = v(1,2) = 0.8, v(2,3) = v(3,4) = 0.75,v(4,5) = 0.5.Find (a)3V,( b )B3. 2.6 A discount function satisfies v(k) = 2−k [ 1− k 6 ] , k= 0,1,2, …,5. For the vectorc = (1,−2,4,3, −3,−5) find (a)3V(c), (b)B3(c). 2.7 Given that 8V(c) = 100,c8 = 60,c9 = 70,v(8,9) = 0.8,v(9,10) = 0.75,v(0,10) = 0.5, and thaẗa(c) = 40, find (a)10V(c), (b)B10(c). 2.8 Given the vector c = (1,22,73) and a discount function v satisfying v(k) = 1− k∕10,k= 0,1, …,5, find v◦ 3(k)f o rk= 0,1,2; the vectors 3c and c◦ 3; and the present values̈a(c;v), ̈a(3c;v), ̈a(c◦ 3;v◦ 3). Verify that formula (2.29) holds. Type B exercises 2.9 Define a two variable function fors,t≥ 0b y v(s,t) = { (1+(t−s))−1,i f s≤ t, 1+(s−t), if s≥ t. Show thatvisnota discount function. 2.10 Show that, given any positive-valued functiongof one variable, v(s,t) = g(t) g(s) defines a discount function. 2.11 Suppose that v1 andv2 are two discount functions. Are either of the functionsv1v2 or v1 +v2 discount functions? These are defined by v1v2(s,t) = v1(s,t)v2(s,t), [v1 +v2](s,t) = v1(s,t)+v2(s,t). 2.12 Assume constant interest. (a) Show thaẗa(0,1n) = ̈a(1n)−1+vn. (b) Showthat ̈a(0,1n) = (1−vn)∕i.Trytodothisfrompart(a)andalsodirectlyfrom (2.14). EXERCISES 37 2.13 Let kn denote the vector (n,n−1,n−2,…,2,1) Show that ̈a(kn) = n− ̈a(0,1n) d in three different ways: (a) From first principles, as in the first derivation of (2.17); (b) Bymakinguseofalinearrelationshipbetweenthevectors(1 n),kn andthevector jn of Section 2.9; (c) By using the loan-interest argument given at the end of Section 2.9. 2.14 Suppose that v(2,7) = 0.5,v(2,8) = 0.4,v(2,9) = 0.3. Find a vector d actuarially equivalent to c = (1,3,5,2,9,10,6,4,8,3) such that di = ci for i< 7 and di = 0 for i> 7. 2.15 A loan of 2300 is to be repaid bynyearly payments of 230 beginning at time 5. The borrower is given the option of repaying only 115 at time 5, but must then pay 240 in each of the subsequent payments. Findv(5). (An exact numerical answer is required. It should not be a function ofn.) 2.16 A loan of 20 000, made at an interest rate of 6%, is to be repaid by level yearly payments for 10 years, beginning 1 year after the loan is advanced. Just before making the seventh repayment, the borrower wishes to repay the entire loan. (a) If interest rates remain unchanged, what is the outstanding balance? (b) Suppose interest rates have dropped to 5%. How much will the borrower have to pay if the lender uses the lower interest rate to calculate the outstanding balance? 2.17 Apersonhas1000nowandplanstoinvestitfor5years.Heistryingtodecidebetween two alternatives. The first is to buy a bond that matures in 5 years. The second is to buy a bond that matures in 10 years and sell it at the end of 5 years. Assume that in both cases the bonds have no payments before maturity and can be purchased today at an interest rate of 6%. How much better or worse off is the individual at the end of 5 years if he chooses the second alternative instead of the first, assuming that at time 5 the interest rate for this class of bonds is (a) 4%, (b) 7%? 2.18 For a certain cash flow vectorc, c 0 = 1, c1 = 5, ̈a(c) = 15. If the discount function is changed so thatv(1,2) is decreased by 0.1, while all other of values ofv(n,n+1) remain unchanged, then ̈a(c) decreases by 2.4. One the other hand, ifv(0,1) is decreased by 0.1 while all other values ofv(n,n+1) remain unchanged, then̈a(c) decreases by 2. Findv(0,1) andv(1,2). 2.19 canddareactuariallyequivalentvectorssuchthat cisconstantand disnondecreasing. Show thatkV(c−d) ≥ 0, fork= 0,1, …,N. *2.20 For the cash flow vector (−1,3, −2), (a) show that the i.r.r. does not exist; (b) findir as a function ofr. 38 THE BASIC DETERMINISTIC MODEL *2.21 Suppose that for a cash flow vectorc, the numberr satisfies ̃BN(c;v) = 0, where v(n) = (1+r)−n. Show thatir = r. *2.22 Show that when interest is constant at rater, then̈a(c) ≥ 0 if and only ifir ≥ r.( T h i s shows that the criteria for determining if it is worthwhile to enter into a transaction given after Theorem 2.2 are in accordance with the remarks in Section 2.1.) *2.23 Suppose that current interest yields on risk-free bonds are 4% for a 5-year bond and 5%fora10-yearbond.Calculatetheforwardpricetobepaidattheendof5yearsfor a zero-coupon bond that pays 1000 units, 10 years from today. Suppose that instead ofthenumberyoujustcalculated,thisforwardpriceis800.Illustratehowyouwould make a sure profit (under the idealized conditions given in Section 2.12). *2.24 The current price of a 1-unit zero-coupon bond maturing in 10 years is 0.610. The forward ratef(6,10) = 0.06.Find the spot ratey 6. Spreadsheet exercises 2.25 A loan contract involves advances of 10 000 initially, 20 000 one year later, and 30 000 one year after that. This is to repaid by 20 yearly instalments beginning at time 3. The payments reduce by 5% each year (so, for example, if the first payment was 1000, the second would be 950, the third would be 902.50). Interest rates are 6% for the first 5 years, 7% for the next 5 years, and 8% after that. Find all payments and the outstanding balances at the end of each year, until the loan is fully discharged. Balances should be calculated after the payment due at the particular time is made. 3 The life table 3.1 Basic definitions Fortheactuaryworkinginthelifeinsurancefield,amajorobjectiveistoestimatethemortality pattern which will be exhibited by a group of individuals. A basic device for accomplishing this is known as alife table. (It is also known as amortality table– an interesting example of a word and its opposite being used interchangeably.) Let𝓁 0 beanarbitrarynumber,usuallytakentobearoundfiguresuchas100000.Suppose we start with a group of𝓁0 newly born lives. We would like to predict how many of these individuals will still be alive at any given time in the future. Of course, we cannot expect to computethisexactly,butwecanhopetoarriveatacloseestimateifwehavesufficientlygood statistics. In the first part of this book we will make the assumption that we can indeed arrive at exact figures. This is in keeping with the concept of adeterministic modelintroduced in Chapter 1. In Part II, we introduce thestochastic modelfor mortality, where we investigate the more realistic assumption that the quantities we want are random variables. Let𝓁 x be the numberofthoseoriginallivesaged0whowillstillbealiveatage x,andlet dx bethenumber of those original lives aged 0 who die between the ages ofx andx +1. The basic relationship between these quantities is 𝓁x+1 = 𝓁x −dx. (3.1) A life tableis a tabulation of𝓁x and dx where x is a nonnegative integer. The following is an example of a portion of a life table (this is an illustration only, and the figures are not intended to be realistic): Fundamentals of Actuarial Mathematics, Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 40 THE LIFE TABLE x 𝓁x dx 0 100 000 2000 1 98 000 1500 2 96 500 1000 3 95 500 900 ⋮⋮ ⋮ 𝜔 0 The table will end at some age, traditionally denoted by𝜔, such that𝓁𝜔 = 0. This is the limiting age of the table, and denotes the first age at which all of the original group will have died. The actual value of𝜔 will vary with the particular life table, but it is typically taken to be around 110 or higher. 3.2 Probabilities Although we assume we can predict𝓁x exactly, there is still randomness in our model, since it is not known whether or not any given individual will be among the survivors at a particular point of time. It is convenient to introduce some elementary probabilistic notions. For nonnegative integersn andx,l e t npx = 𝓁x+n 𝓁x . (3.2) Whatisthemeaningofthisterm?Considerthe 𝓁x survivorsage x.Outofthisgroup, 𝓁x+n will survivetoage x +n.Thequotientthengivesustheprobabilitythatapersonaged x,hereafter denoted just by the symbol (x), will be alive at agex +n.L e t nqx = 𝓁x −𝓁x+n 𝓁x . (3.3) Thisgivesustheprobabilitythat( x)willdiebetweentheagesof x andx +n.Itisclearthat nqx = 1−n px. (3.4) As an example, in the table given above we would have2p0 = 965∕1000, 2q1 = 25∕980. Since a left subscript of ‘1’ occurs frequently it is omitted for notational convenience. Thatis, px denotes1px,and qx denotes1qx.Thequantity qx isoftenreferredtoasthe mortality rate at agex. Whatistheprobabilitythat( x)willdiebetweentheagesof x +n andx +n +k?Thisisa quantity which we will use frequently. There are three main ways of expressing it: 𝓁x+n −𝓁x+n+k 𝓁x , (3.5a) CONSTRUCTING THE LIFE TABLE FROM THE VALUES OFqx 41 or npx − n+kpx, (3.5b) or npxk qx+n. (3.5c) The reader should verify, by substituting values of𝓁, that all three expressions are equal. They can each be explained intuitively. Consider the first. The numerator is the number of people living at agex +n, less the number living at agex +n +k. This difference must be the number of people who died between the two ages. Dividing by the number of people that we start with will give us the required probability. In the second expression we express this quantity as the probability that (x) will liven years, but willnot liven +k years. In the third expressionweconsidertwostages.Todiebetweenthespecifiedages,( x)mustfirstlivetoage x +n. The individual, then being agex +n, must die within the nextk years. We will have occasion to use all three of these expressions, choosing the one which is most convenient for the purpose at hand. Another useful identity, which we will refer to as themultiplication rule,i s n+kpx = npxk px+n, (3.6) forallnonnegativeintegers n,k andx.Itcanbeverifieddirectlyfrom(3.2).Intuitively,itsays that in order for (x)t ol i v en+k years, the individual must first liven years, and then, being agex +n, must live anotherk years. 3.3 Constructing the life table from the values ofqx Thelifetableisconstructedinpracticebyfirstobtainingthevaluesof qx forx = 0,1, …,𝜔−1. Obtainingthesevaluesisastatisticalproblemwhichwewillnotdiscussindetail.Itisbasically donebycarryingoutastudyinwhichweobservehowlongpeopleofdifferentageswilllive. Forexample, ifweobserveagroupof1000peopleofexact age50and10ofthemdiewithin 1 year, then we could estimateq 50 as 0.01. This of course is an extreme simplification and the process is much more complicated. It is not practical to gather a group of people exactly age 50 at one point of time, and then to observe them for an entire year. In practice, people will enter the study at various times, and leave for reasons other than death. In addition, we must achieve consistency between values at different ages. The statistical subject known as survival analysisdeals with these problems. In this book we will take the values ofq x as given. The life table can then be constructed inductively, starting with𝓁0, from the formulas dx = 𝓁xqx, 𝓁x+1 = 𝓁x −dx, (3.7) which follow immediately from (3.1) and (3.3). We will see, however, that it is not usually necessary to actually calculate𝓁 x and dx. In practice, life tables are specified by just giving the values ofqx, which is sufficient for the necessary computations. The advantage of the traditional form lies mainly in its intuitive appeal, rather than its use as a calculating tool. 42 THE LIFE TABLE 3.4 Life expectancy Life expectancy is one of the most frequently quoted actuarial concepts. Probably for this reasonitisoftenusedincorrectly,aswewillexplainbelow.Abasicquestionisthefollowing. How long can a person agex expect to live? Of course there is much variety in the future lifetime of various individuals of the same age. Some will live for several years, and some will die immediately, but we can attempt to arrive at some sort of average figure. One approach would be to take a large number of people agex and observe them until all have died. We could then compute the total future lifetime of all these individuals. Dividing by the number of people in the original group would give an estimate of the desired average. For a drastically oversimplified example, take three people exactly age 60. Suppose one dies at age 62, another at age 72 1 2 and the third at age 911 4. The total future lifetime would be 2+121 2 +311 4 = 453 4. Dividing by 3 we could estimate that a person age 60 could expect to live on average another 151 4 years. Of course to be statistically accurate we would need many more than three people. Moreover, the length of time needed for such a study makes it completely impractical. Remarkably however, once we have the life table, we can obtain the figure directly, without carrying out the observations. To see this, we look at another approach to obtaining total future lifetime. Suppose we start with𝓁 x people agex.A f t e r1 year, there will𝓁x+1 survivors who would have each contributed 1 year of lifetime to this total. At the end of the 2 years there will𝓁x+2 survivors who would have each contributed anotheryeartothetotal.Continuinginthisway,wecanestimatethetotalfuturelifetimeofall lives as 𝓁 x+1 +𝓁x+2 +𝓁x+3 +⋯ +𝓁𝜔−1, and, dividing through by𝓁x, we obtain the quantity ex = 𝜔−x−1∑ k=1 𝓁x+k 𝓁x = 𝜔−x−1∑ k=1 kpx. (3.8) Thequantity ex isknownasthe curtate life expectancyorcurtateexpectationoflifeatage x.Thew ordcurtate meansreduced ortruncated,reflectingthefactthatthisisnotexactlythe quantity that we want. We have cheated a little in our alternate measurement scheme, for it measures only whole future years of lifetime and ignores the fraction of the year lived in the yearofdeath.Inourillustrationabove,forexample,underthealternatecountingmethod,the 60-year-old who died at age 72 1 2 would be credited withonly 12years of total lifetimerather than the actual 121 2. The person who died at 911 4 would be credited with only 31 years rather than the actual 311 4. We are undercounting between 0 and 1 years for each individual, and it seems reasonable to take this to be one-half on average. The true life expectancy, usually referredtoasthe complete life expectancyatage x anddenotedby e ◦ x isgivenapproximatelyby e ◦ x = ex + 1 2. We will give a more formal treatment ofe ◦ x in Chapter 8 and again in Chapter 15. LIFE EXPECTANCY 43 There is a simple recursion formula to computeex for all values ofx. From the second formula in (3.8), ex = px +2px +3px +⋯ +𝜔−x−1px = px(1+px+1 +2px+1 +⋯ +𝜔−x−2px+1) (3.9) = px(1+ex+1). The second line is obtained by using the multiplication rule (3.6). Note that this is a backward recursion formula as it gives the value of the function in terms of the next higher argument. The recursion is then started from the initial value ofe𝜔 = 0. Itisinstructivetogiveanintuitiveexplanationof(3.9).Toliveanywholenumberofyears in the future, (x) must first live to agex +1, as reflected by the factorpx. The individual will then have completed 1 year of lifetime, and in addition, being now agex +1, will complete, on average, the expected number of future whole years for a person of that age. The reader should note carefully that life expectancy is a function of age. For each agex, the life expectancy at that age gives the average number of future years that (x) will live. A majorsourceofmisquotingistoexpressthisasasinglefigureratherthanafunction.Onecan oftenfindstatementsinnewspapersorsimilarsources,statingsomethinglike‘lifeexpectancy hasincreasedfrom75.3to75.8years’.Thewriterisinvariablyreferringtothelifeexpectancy at age 0 only. This is certainly of interest, but it conveys somewhat limited information. A person already aged 80 who wishes to estimate how much longer he/she can expect to live is not helped by a statement which claims that newborn lives live on average to age 75.8. Thereadershouldalsonotethatlifeexpectancyistheaverage duration andnottheaverage age a person can expect to live to. We say, for example, that the life expectancy at age 50 is 31.2, meaning that on average a person age 50 can expect to live to age 81.2. There is often confusion on this point, which is again a result of the tendency to report only the age 0 figure where duration and age are the same. There are many other quantities obtainable from the life table which are of interest, but onlyoneinparticularthatwewilldiscusshere.Sometimeswemaybeinterestedintheaverage duration lived by (x) over the nextn years, wheren is some fixed duration. The quantity n∑ k=1 𝓁x+k 𝓁x = n∑ k=1 kpx (3.10) isknownasthe curtate n-year temporary life expectancyatage x.Thewordtemporary comes from an analogy with life annuities which we discuss in the next chapter. It gives us the expected complete number of years livedover the next n yearsby people now agex.T h i si s what we would compute if we repeated the alternate measurement system described above, but ended the observations aftern years. To adjust this for the undercounting in the year of death requires some care. Those who lived to agex +n will have contributed the correct total ofn years, and therefore it is only the (𝓁 x −𝓁x+n) people who died during then-year period who must be considered in the adjustment. To get a more accuraten-year temporary life expectancy at agex we add to the quantity in (3.10) not1 2, but rather1 2(𝓁x −𝓁x+n)∕𝓁x,t o get an approximation to thecomplete n-year temporary life expectancyat agex of n∑ k=1 kpx + 1 2nqx. 44 THE LIFE TABLE 3.5 Choice of life tables Thelifetablereflectsthefactthatageisthemajordeterminantoffuturemortality.Thereareof course several other factors that affect future lifetime, such as gender, health status, lifestyle, and geographical location. In practice, the effect of these factors is handled by producing several different life tables. One confines the mortality study that produces the table to a particulargroupofpeople,namelythosewiththecharacteristicthatyouwanttoseparateout. The following are a few of the more important distinctions made in practice. It is observed, for reasons that nobody has fully explained, that females live longer than males.Forthemiddlerangeofages,itistypicalforthelifeexpectancyofafemaletobefrom 5 to 7 years more than that of a male of the same age. To reflect this, it is usual to produce separate male and female life tables. In recent years, there has been overwhelming statistical evidence to show the dangers of smoking. This has led insurance companies to construct different life tables for smokers and nonsmokers. The choice of life table will also depend heavily on the type of contract that is being sold.Thelifetablesproducedforthegeneralpopulationfromcensusdataarenotsuitablefor insurance purposes. People accepted for life insurance policies are usually screened by the insurancecompanytomakesuretheyareinreasonablehealth.Theycanexpecttolivelonger than a person of the same age taken from the population as a whole. Life tables for insurance purposes are constructed by looking at insurance company data only. Purchasers of life annuities (discussed in Chapter 4) will on the average live even longer, since a person in poor health would be unlikely to buy such a product. Separate tables are needed for annuity purposes. Still another distinction to be made is the difference between individual contracts and group contracts. In the former case the buyers makes a definite decision to enter into the insurance or annuity contract and are presumably aware of their health conditions and acting in their best interests. In the latter cases, an employer purchases the contract to cover a large group of employees. Different mortality patterns in the two cases can be expected. There are many other examples which we will not discuss here, although some will be pointed out briefly in succeeding chapters. The reader should be aware that selecting an appropriate table for a particular use is an important actuarial task. Sometimes a very simple method, known asmultiples of standard mortality, is employed toconstructmanydifferenttablesfromagivenone,knownasthestandardtable.Forexample, itmightbedecidedthatforrisksofacertaintype,themortalityis150%ofstandardmortality. Thelifetableforsuchrisksisconstructedbymultiplyingeach q x inthestandardtableby1.5. This method is justified more by simplicity of calculation rather than any scientific rationale. 3.6 Standard notation and terminology We have already introduced the standard symbolsnpx,nqx,ex,e ◦ x and𝜔. The symbol n|kqx denotes the probability that (x) will die between the ages ofx +n and x +n +k, the quantity that we already have three ways of writing as shown in (3.5). A subscript of 1 is omitted, so thatn|qx denotes n|1qx. This use of a vertical bar is a typical actuarial device to denote a ‘waiting period’. In this case, the symbol is intended to indicate that the person will waitn years and then die in the followingk years. EXERCISES 45 The quantity in (3.10) is denoted byex:n and the complete temporary life expectancy is denoted bye ◦ x:n. 3.7 A sample table Todothespreadsheetproblemsinthefollowingchaptersthatrequirealifetable,weintroduce a sample table. It is given by q x = { 1−e−0.00005(1.09)x , x = 0,1, …,118, 1, x = 119. (3.11) We have𝜔 = 120.Theformulaiseasilyprogrammedintoaspreadsheet,andthatisthereason forgivingthetableinthisform.Givinganexistingtablewouldnecessitateenteringthefigures individually. Another advantage of this parametric form is that the two constants of 0.00005 and 1.09 can be changed toprovide a variety of different lifetables for comparison purposes. Our table is, as stated, a sample table, and it should not be taken as being a realistic picture of modern-day mortality. This is true especially at the younger or very old ages, as will be discussed later in the text. Chapter 14 will provide some motivation for the formula; see, in particular, Exercise 14.11. Notes and references London (1997) provides an introduction to survival analysis, and gives more details on the construction of life tables. Exercises Type A exercises 3.1 You are given thatq60 = 0.20,q61 = 0.25,q62 = 0.25,q63 = 0.30,q64 = 0.40. (a) Find𝓁x for ages 60–65, beginning with𝓁60 = 1000. (b) Find the probabilities of the following: (i) (61) will die between the ages of 62 and 64. (ii) (62) will live to age 65. (c) Given thate65 = 0.8, findex forx = 60–64. 3.2 You are given that5p40 = 0.8,10p45 = 0.6,10p55 = 0.4. Find the probability that (40) will die between the ages of 55 and 65. 3.3 Supposethatoutofatypicalgroupof100peopleage70,10willdieinthefirstyear,15 will die in the second year, and 20 will die in the third year. Calculateq70,q71,q72 and 3p70. 46 THE LIFE TABLE Type B exercises 3.4 Suppose that 𝓁x = 100−x, x = 0,1, …,100. Findexpressionsfor(a) npx,(b) nqx,(c)theprobabilitythat(x )willdiebetweentheages of x +n andx +n +k. 3.5 You are given that e60 = 17,10p50 = 0.8, and that the 10-year curtate temporary life expectancy at age 50 is 9.2. Finde50. 3.6 Prove the following statement, assuming the approximation given in the text, and give an intuitive explanation: e ◦ x = 1 2qx +px(1+e ◦ x+1). 3.7 Suppose that qx is equal to a constantq for allx. (Note that in this case𝜔 =∞ ). Find expressionsintermsof q andn for(a) npx,(b)e x.Doyouthinkthatthisgivesarealistic life table? Why or why not? 3.8 In constructing a life table for heavy smokers, Actuary A decides to take a standard table, and double each value ofqx, using a value of 1 ifqx > 0.5. Actuary B takes the sametableandsquareseachvalueof px.ShowthattheresultingtableofActuaryB,has lower mortality rates at all ages than that of Actuary A. Spreadsheet exercise 3.9 Taking the sample life table as given by (3.11), use recursion to find ex for x = 0,1, …,𝜔−1.Focuson e0.Howmuchisthisreducediftheconstant0.00005ischanged to 0.00006. What happens if 0.00005 is kept the same, but 1.09 is changed to 1.092? 4 Life annuities 4.1 Introduction The financial losses that we listed at the beginning of Chapter 1 are no doubt familiar to all readers.Anothertypeofrisk,whichmaynotbesoobvious,isthatof livingtoolong.Givenour definitionofriskasthepossibilityofsomethingbadhappening,thereadermaywonderabout thisstatement.Afterall,isnotlivingalonglifeagoodratherthanbadoccurrence?Itcertainly can be, but it does carry with it the possibility of financial hardship if one does not have adequate income. Imagine the decision faced by a retired individual who has accumulated savings of 1 000 000, which is invested at an annual rate of 5%, providing an annual income of 50 000. Imagine also that the individual decides that this is an insufficient return, so it is necessary to consume a portion of the capital each year, as well as the interest. Doing so, however, runs the risk that the entire capital may be depleted before death, leaving the individual with no source of income for his/her remaining lifetime. Life annuities are a means of insuring against such a risk. A life annuity is a contract between an insurance company, known as theinsurer, and another party, known as the annuitant,which provides the following. In return for the payment by the annuitant of pre- scribedpremiums,theinsurerwillprovideasequenceofpayments,knownas annuitybenefits of prescribed amounts and at prescribed times. The unique provision is that the annuitant must be alive to receive each benefit payment. These will terminate upon the death of the annuitant. People who purchase such a contract are investing capital that they have decided will not be needed for any dependents after they die. They agree to give up these funds upon their death in return for the greater yield they can achieve on their investment by sharing the amounts forfeited by those who predecease them. A good way to picture the workings of a life annuity is to imagine a room with a number of small boxes, belonging to one annuitant each.Eachannuitantpaysthesamepremiumintohis/herbox,andinterestearningsareadded FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 48 LIFE ANNUITIES totheamountsinhis/herbox.Eachreceivesthesameannuitybenefits,whichareprovidedby takingmoneyoutofthebox.Whenoneparticipantdies,his/herboxisopened,andtheamount is spread evenly among all the other boxes in the room. Participants then receive not only interest earnings, but also these forfeited amounts of those who die before them. We refer to thelatteras survivorshipearnings.Thisisafairlyaccurateaccountofwhatactuallyhappens, except that amounts are calculated on paper. There is obviously no need for the physical rooms or boxes. More precisely, it is an account of what would happen if mortality followed the life table exactly. For most annuities sold in practice, the accumulation is guaranteed in advance, so it does not depend on the actual deaths. If fewer people die than expected, the insurer would still have to add the extra survivorship amounts to people’s ‘boxes’, and would experience a loss, while more deaths than expected would result in a gain to the insurer. Tobeequitable,allparticipantsinanyoneroommusthaveroughlythesameriskofdying, andinparticularmustbeofthesameage.A70-year-old,putinthesameroomas20-year-olds, would be much more likely to die early on and would clearly be at a disadvantage. We will thereforecalculateannuitypremiumsprimarilyasafunctionofage.Thereareofcoursemany other factors to consider, such as gender, which we discuss briefly in Section 4.6. Insurance companies are not the only source of life annuities. Many pension plans pay retirement benefits in the form of a life annuity. 4.2 Calculating annuity premiums We now turn to the mathematical aspects of annuities. The reader should review the vector notation given in Section 2.8, which we will make frequent use of. Weassumethatwehavefixedthroughoutanappropriatelifetableandadiscountfunction vto model the effect of investment earnings. Consideracontractsoldto(x )–or,moreaccurately, onthelifeof (x),sincethepurchaser could be a different party than the annuitant, such as an employer. It is assumed throughout that (x) is an integer, which allows us to conveniently use a life table if desired. Indeed, the common practice is to classify an individual by their age on their last birthday, which is just the normal way that one thinks of age. We suppose that benefit payments are made yearly. Suppose that the amount of benefit paid at timekis c k. This could of course be zero, indicating that no payment is to be made. The contract is then described by the cash flow vector c = (c 0,c1,…,c𝜔−x−1). In this case we will refer toc as theannuity benefitvector. (Notethatthefinalcashflowismadeatage 𝜔−1,whichthelastageatwhichanyoneisliving, according to our model.) Suppose that the benefits are to be purchased by a single premium paid at the beginning of the contract at agex. Our first problem is to calculate the premium that the annuitant should pay. The criterion is that the total premiums from all annuitants, together with the interest earned, should be sufficient to provide all of the required benefits, assuming that invested capital accumulates according to the given discount function and that mortality follows that of the given life table. Consider the simplest possible case, namelyc = e k. This is an annuity consisting of a single benefit payment of 1 to be made at timekif (x) is then alive. Such a contract is usually referred to as apure endowment.L e tE denote the single premium that should be paid to provide for this. Suppose that𝓁x people, each agex, buy such a contract. The total amount collected in premiums at time zero would be𝓁xE, and this will accumulate to𝓁xEv(k,0)at CALCULATING ANNUITY PREMIUMS 49 time k. There will be𝓁x+k survivors at timek, and each of these must receive 1 unit. This means that 𝓁xEv(k,0) = 𝓁x+k. Solving, we obtain E= v(k)𝓁x+k 𝓁x = v(k)kpx. This can be verified intuitively. If the benefit of 1 unit at timek were guaranteed, we know from Chapter 2 that we would need to investv(k). In this case, where the benefit is not guaranteed, we multiply by the probability of receiving it to arrive at the premium. Thegeneral caseeasilyfollows.Let ̈ax(c)denote thesinglepremium foralifeannuity on (x)withbenefitvector c.Thiscontractcanbeviewedasasequenceofpureendowments,one foreach value ofk,wherethe kthpureendowment pays ck attime k.Thepremiumforsucha pureendowmentisjust ckv(k)kpx,andthetotalpremiumisobtainedbysummation.Wehave ̈ax(c) = 𝜔−x−1∑ k=0 ckv(k)kpx. (4.1)‡ Life annuities have been classified into various types. An annuity for which benefit payments are made for a fixed number of years and then cease is known as atemporarylife annuity(an example for those who collect oxymorons). The most common type of annuity is one for which benefits continue for as long as the annuitant lives. This is known as awhole lifeannuity. Purchasers of whole life annuities protectthemselvescompletelyagainstoutlivingtheiravailablecapital,fornomatterhowlong theylive,theincomefromtheannuitywillcontinue.Wehavenoneedtodistinguishbetween temporary and whole life annuities in our mathematical model. As indicated, we consider all annuities on (x) as running from time 0 to time𝜔−1−x. In the case of a temporary life annuitywesimplytakebenefitsequalto0afterthelastpositivepayment.Inreallife,however, there is a difference. The insurer can obviously not inform a person who has reached the age of𝜔,thataccordingtotheinsurer’smodeltheyarenolongeraliveandannuitypaymentswill stop. This means that as a practical matter the benefit payments on a whole life annuity must eventually be constant (or at least follow some regular pattern) so it is clear what amount to continue to pay should an annuitant live beyond𝜔. In most tables,𝜔 is set sufficiently high that such an occurrence is extremely rare. Deferredannuities arecontractswheretheinitialpaymentdoesnotcommenceforseveral years. For example, a person wishes to provide for an income beginning at retirement. Math- ematically, this simply means that the initial entries in the benefit vector will be zero, so no special treatment is required. The annuity benefit vector will be of the formc = (0 k,d), and the durationkis known as thedeferredperiod. Example4.1 Atemporarylifeannuityon(50)providesfor1payableatage50,2payableat age51,3payableatage52,and4payableatage53.Suppose q50 = 0.1, q51 = 0.2, q52 = 0.25. The interest rate is 50% for the first year and 100% thereafter. Find the single premium. 50 LIFE ANNUITIES Solution. Forasmallproblemsuchasthis,itisconvenient tomakeuseofatimediagramas introduced in Chapter 2. See Figure 4.1. Benefits 4 3 2 1 Time 3 2 1 0 vs 2/31 /21 /2 qs 0.10 .20 .25 Figure 4.1 Example 4.1 Here we have inserted values ofqy, for relevant agesy, as well as the 1-year interest discountfactors.Wenowmustmultiplyeachpayment ck bytheproductofallpreviousvalues of v(i,i+1) as we did in Chapter 2, andalsoby kpx which we calculate as the product of all the previous values ofpy. The single premium equals 1+ ( 2× 2 3 ×0.9 ) + ( 3× 2 3 × 1 2 ×0.9×0.8 ) + ( 4× 2 3 × 1 2 × 1 2 ×0.9×0.8×0.75 ) = 3.28. An alternate way of calculating the survival probabilities, which some may prefer, is to use (3.2). We can construct part of a life table, utilizing (3.7). In this example, starting with say 𝓁 50 = 1000 we would get in turnd50 = 100, 𝓁51 = 900, d51 = 180,𝓁52 = 720, d52 = 180, 𝓁53 = 540.Thisautomaticallyperformsallthemultiplicationinvolvingthevaluesof py, and gives us the factors of 0.9, 0.72, 0.54, used above. 4.3 The interest and survivorship discount function 4.3.1 The basic definition Itisimportanttonotethattheformulafor ̈ax(c)isthesameasthatgivenfor ̈a(c)inChapter2, but withv(k) replaced byv(k)kpx for each value of k. This suggests that we are in the general situation of Chapter 2, but with a different discount function. This is indeed the case. Define yx(n) = v(n)npx and extend to a two-variable function yx(k,n) = yx(n) yx(k) forallnonnegativeintegers k,and n.Itisclearthat yx satisfies(2.1)andthereforeisadiscount function. This fact is borne out by the discussion of the pure endowment in the previous section, which showed that the value at agexof 1 paid at agex+k, when accumulation is by interest and survivorship, is preciselyy x(k). THE INTEREST AND SURVIVORSHIP DISCOUNT FUNCTION 51 Using the multiplication rule we can verify that fork≤ n yx(k,n) = yx(n) yx(k) = v(k,n)n−kpx+k, (4.2) and fork> n yx(k,n) = v(k,n) k−npx+n . (4.3) So for example, assuming constant interest yx(2,5) = v3 3px+2, yx(5,2) = (1+i)3 3px+2 . Note that the last quantity, giving the amount at time 5 resulting from an investment of 1 at time 2, is greater than (1+i) 3, the amount resulting from interest only. This reflects the extra earnings due to survivorship. Wewillreferto yx asan interestandsurvivorship discountfunction.From(4.1)itfollows that ̈ax(c) = ̈a(c;yx), (4.4) andisthereforejustthepresentvalueofthepayments,withresecttothediscountfunction yx. The significance of the above is that it shows that all the results of Chapter 2 can be carried over directly to life annuities. As a matter of terminology we will interchangeably refer töa x(c) and similar quantities introduced later, asnetsinglepremiums orpresentvalues. When calculating ̈ax(c) by spreadsheets it is convenient to make use of the recursion formula (2.7) adapted to the interest and survivorship function. We have yx(k+1) = yx(k)v(k,k+1)px+k. (4.5) Notation Considerthevector1 𝜔−x whichreferstoa1-unitcashflowsequencecontinuingfor thelifeof(x ).Duetotheprevalenceofthissymbol,itissuppressedfornotationalconvenience. This is a standard convention of actuarial notation that will be followed throughout the book. Ingeneral,wheneveravectorisomitted,itisunderstoodtobe1 𝜔−x.Forexample, ̈ax denotes thenetsinglepremiumforawholelifeannuityon( x),paying1peryearforlife,beginningat time 0. In cases where payments run for life, but are not necessarily constant, it is sometimes convenient to avoid specifying𝜔 by using the symbol 1∞ to refer to a vector with entries of 1, running to age𝜔−1 For example if𝜔 = 120 we could writëa40(215,1∞) in place of ̈a40(215,165). Inwhatfollowswewillkeep vasthenotationfortheinvestmentdiscountfunction,which reflects the discounting resulting from earnings on invested capital. We will use the lettery to denote a general discount function, which could bev or yx or any other such function. 52 LIFE ANNUITIES Life annuity symbols will be written as given with the age as a subscript. For present values involving discounting at interest only, we will then incorporate thevand writëa(c;v). 4.3.2 Relations between yx for various values ofx The reader should note that we have several interest and survivorship functions, one for each value of the initial agex. They are, however, related in the following way. From (4.2) y x◦ k(n) = yx(k,k+n) = v(k,k+n)npx+k = v◦ k(n)npx+k. This tells us that yx◦ k(calculated withv) = yx+k(calculated withv◦ k). (4.6) Therefore, if interest is constant, yx+k = yx◦ k. (4.7) In particular, with constant interestyx = y0 ◦ x, so that knowing the interest and survivorship function for initial age 0 determines it for all initial ages. In the remainder to this section, we continue with the notational ideas first introduced in Section 2.11, with the intention of providing a background for following classical actuarial formulas. It can be safely omitted at first, as everything can be conveniently expressed by using the new notation that we introduced in Chapter 2 and discussed in Section 2.14.2. We do however occasionally refer back to this section as well as Section 2.14.2 In particular the symbols v◦ k, andc◦ k prove to be convenient and will be used in various places. In the first place, since we do not assume constant interest as was traditionally done, it is necessary to introduce a new notational device into the classical life annuity symbols in order to handle time shifting. The idea is to separate age from duration in the quantity x+k.W el e ẗa [x]+k(c) denote the present value of a life annuity on (x+k), paying ci at time i, and calculated with respect to the investment discount functionv◦ k. Of course with constantinterest, v◦ k= v,sothat ̈a[x]+k(c) = ̈ax+k(c)forall k,andthisnotationisnotneeded. (However even with constant interest we will have need of this notation with a more refined mortality model which we discuss in Chapter 9.) From (2.28) we can write ̈a [x]+k(c◦ k) = Valk(kc;yx), (4.8) so the symbol on the left hand side is conveniently interpreted as the value at timek,o ft h e benefits at andaftertimek, paid on a life annuity issued to (x) with benefit vectorc. The above equation leads to the traditional form of the life annuity splitting identity. ̈ax(c) = ̈ax(kc)+yx(k)̈a[x]+k(c◦ k). (4.9) GUARANTEED PAYMENTS 53 For a particular case of (4.9), assume constant interest and letc = 1∞. Then kc = 1k and c◦ k= 1∞. Using (3.2) we can write ̈ax = ̈ax(1k)+vk𝓁x+k 𝓁x ̈ax+k. (4.10) Thiswasapopularformulainpre-computerdays.Itwasusedtocalculatevaluesoftemporary annuities ̈a x(1k) froma table of values of whole lifeannuities and the lifetable. Withmodern spreadsheets, we do not need this or similar formulas for calculation purposes. They are however sometimes useful to explain relationships between various quantities. Formula (4.9) is often applied to express the present value of a deferred annuitÿa x(c) where c = 0k◦ d, directly in terms of the vectord. Since kc = 0 and c◦ k= d,w eh a v e immediately that ̈ax(c) = yx(k)̈a[x]+k(d). (4.11) Example4.2 Thisexampleisdesignedtofurtherillustratethesquarebracketnotation.Sup- pose thatq60 = 0.05, q61 = 0.10, q62 = 0.15, i0 = i1 = 0.04 andik = 0.08 fork≥ 2. Calcu- late ̈a60(14), ̈a[59]+1(14), ̈a[58]+2(14). Solution. ̈a60(14) = 1+(1.04)−10.95+(1.04)−2(0.95)(0.90) +(1.04)−2(1.08)−1(0.95)(90)(0.85) = 3.33 ̈a[59]+1(14) = 1+(1.04)−10.95+(1.04)−1(1.08)−1(0.95)(0.90) +(1.04)−1(1.08)−2(0.95)(90)(0.85) = 3.27. This could be interpreted as the amount that a person now age 59 would have to pay for a 1-unit 4-year annuity if the purchase was made at the end of 1 year (assuming no change in the life table). It is lower than the first amount, since there is only 1 year of low interest rates rather than two. ̈a [58]+2(14) = 1+(1.08)−10.95+(1.08)−2(0.95)(0.90) +(1.08)−3(0.95)(90)(0.85) = 3.19. This could be interpreted as the amount that a person now age 58 would have to pay for a 1-unit 4-year annuity if the purchase was made at the end of 2 years. It is lower yet, since the buyer would avoid both years of low interest. 4.4 Guaranteed payments Supposeanindividualpurchasesalifeannuityanddiesimmediatelyafterpayingthepremium. The person will get nothing back, except the benefit payment at time zero if that is positive. While we have seen that this is fair, the situation is not always well understood by the dependentsoftheannuitant,whomaywellcomplainthattheinsurerhasconfiscatedthefunds. 54 LIFE ANNUITIES Asidefromthis,manyprospectivepurchasersareuneasywiththepossibilitythatalloralarge portion of their money could be lost. To provide additional flexibility, insurers frequently offer life annuities with aguaranteedperiod. A typical contract of this type would stipulate that for a certain duration (commonly 10 or 15 years) the benefits will be paid regardless of whethertheannuitantisaliveornot.Afterthisguaranteedperiodthecontractrevertstoalife annuity and benefits are conditional on survival. The guaranteed period of course means that a higher premium must be paid for the same level of benefits. To calculate the premium, it is best to consider such an annuity as two separate contracts, one for the guaranteed payments, the other for the contingent payments, and add the respective premiums. Examples follow. Example 4.3 A person age 40 purchases a life annuity that provides 10 000 each year for life,withthefirstpayment startingatage41.Thefirst10payments willbepaidregardless of whether the annuitant is alive or not. Find a formula for the single premium. Solution. The premium for the guaranteed annuity is 10000 ̈a(c;v), c = (0,1 10). The premium for the non-guaranteed annuity is 10000 ̈a40(f), f = (011,1∞). Thus total premium is just the sum 10000 [̈a(c;v)+ ̈a40(f)]. Letusverifythedurationinthevector f.Thefirstguaranteedpaymentisattime1,sothelast guaranteed payment is at time 10, and the first non-guaranteed payment is at time 11. As we start the indexing with 0, there will be 11 zeros in the non-guaranteed vector. Example4.4 Apersonage40purchasesalifeannuitythatprovides1peryearforlifewith the first payment at age 65. If (40) lives to age 65 he/she will receive at least 10 payments. Nothing is paid if death occurs before age 65. Find a formula for the premium. Solution. The premium for the non-guaranteed annuity is̈a 40(035,1 ∞). One must be careful with the ‘guaranteed’ portion here, which is deferred and therefore not completely guaranteed. The value of this at time 25 is Val25(025,110;v). To get the value at time 0, we multiply this byy40(25). It would be wrong to multiply byv(25), since these payments are not made if (40) dies before age 65, so that accumulation is at interest and survivorship for the first 25 years. The total premium is therefore ̈a 40(035,1∞)+y40(25)Val25(025,110;v). The more traditional way of writing this, incorporating our time shifting notation to handle non-constant interest is v(35) 35p40̈a[40]+35 +v(25)25p40̈a(110;v◦ 25). DEFERRED ANNUITIES WITH ANNUAL PREMIUMS 55 To handle the general case of guaranteed payments, letu denote a vector of benefit paymentsthatareguaranteedprovidedtheannuitantlivestotime g.SoinExample4.3, g= 0 and u = (110), whereas in Example 4.4,g= 25 andu = (025,110). Clearly, we can assume thatthefirst gentriesareequalto0sothat uisoftheform(0 g,w).Then,reasoningasabove, the present value of the guaranteed payments is gpẍa(u;v) = v(g)gpẍa(w; v◦ g), which equals the present value with respect to the discount functionv, multiplied by the probability that the payments will be made. SeeExercise4.15foronevariationontheguaranteed payment concept. Wepresentmore modifications in the next chapter. See Example 5.11 and Exercise 5.19. 4.5 Deferred annuities with annual premiums Deferred annuities are often purchased by a series of annual premiums rather than a single premium. The premium payment period can be any length that does not exceed the deferred period, so that premiums stop when the benefit payments commence. Deferred annuities with annual premiums are frequently used to provide pensions. An individual, together with his/heremployer,willpaypremiumsduringhis/herworkingyearsinordertoprovideincome beginning at retirement. The sequence of annual-premium payments in the contract can be described by means of a premium pattern vector. This is a vector𝜌 with 𝜌 0 = 1. Then,𝜌k denotes the ratio of the premiumpayableattime ktothepremiumpayableattime0.Ifweknowthepremiumpayable at time 0, often referred to as theinitialpremium, the premium pattern vector will determine all premiums. Namely, if𝜋k denotes the premium payable at timek, 𝜋k = 𝜋0𝜌k. (4.12) The vector𝜋 = (𝜋0,𝜋1,…) will be called thepremiumvector. Supposethat( x)purchasesalifeannuitytobeginatage x+n.Themostcommonpatternin practicewouldbe 𝜌 = (1n),signifyingalevelpremiumpayableuntiltheincomecommences. However, many other patterns are encountered. Some annuitants may prefer to pay a higher premium but for a shorter period, adopting a pattern of (1 m)f o rs o m em< n. Others may prefer to pay a lower premium for an initial period, and then more at the end. For example, the pattern𝜌 = (1 k,2n−k) would call for the premium to double afterkyears. Thepremiumpaymentsalsoconstitutealifeannuityon(x )sincetheywillceaseupon( x)’s death. (In this case they are paidby(x) rather than to (x).) To achieve the goal that premiums togetherwithinvestmentearningsaresufficienttoprovidetherequiredannuitypayments,the actuary will set premiums to be actuarially equivalent to benefits, with respect to the interest and survivorship discount function. Example 4.5 (Figure 4.2). An annuity on (40) provides for 1 annually for life, beginning at age 65. Nothing is paid for death before 65. Annual premiums are payable for 25 years 56 LIFE ANNUITIES beginning at age 40. The premium reduces by one-half after 15 years. Find a formula for the initial premium, assuming𝜔 = 110. Benefits 0 00. . . 0 0. . . 1 10. . . 1 Time 2 10 ... ... 151 4 ... ... 26 252 4 ...69 Premiums π 0 π0 π0 π0 π0/ 2 ... π0/2 Figure 4.2 Example 4.6 Solution. Equating present values, we have ̈a40(c) = ̈a40(𝜋) = 𝜋0̈a40(𝜌), where c = (025,145), 𝜌 = (115,0.510). Solving, we obtain 𝜋0 = ̈a40(c) ̈a40(𝜌). 4.6 Some practical considerations 4.6.1 Gross premiums Theannuitypremiums,bothsingleandannual,thatwecalculatedintheprevioussectionsare referred to asnetpremiums orbenefitpremiums. They are the premiums that are required to provide the benefits. The premiums that are actually charged in practice are known asgross premiums, and involve other factors intheircalculation. The insurer must make provision for amounts to cover expenses and profits. In addition, a contingency charge is added to provide for adverse experience. After all, interest earnings may fall short of those predicted by the discount function or people may live longer than predicted by the given life table. Moreover, marketingconsiderationsinevitablyplayarole.Ifaninsurerwantstostayinoperationitmust ensure that the premiums it actually charges are competitive. The many details that go into the calculation of the actual gross premiums are beyond the scope of this book, although we willtouchonthistopicinChapter12.Inanyevent,theinitialbasicstepistocalculatethenet premiums as done in the examples of this chapter. 4.6.2 Gender aspects The role of gender in life annuities is a controversial issue. As noted above, the participants in a ‘female agexroom’ can expect to live longer than those in the ‘male agexroom’. The females will receive less money in forfeiture due to death, and therefore must invest more to STANDARD NOTATION AND TERMINOLOGY 57 receive back the same amounts. If the insurer uses separate life tables for males and females, theresultisthatfemalesmustpayhigherpremiumstoreceivethesameannuitybenefits.This has in particular caused controversy in connection withdefined contributionpension plans. In such a plan, the employee and employer make specified contributions into a fund and the accumulatedamountsarethenpaidoutassomeformoflifeannuitywhentheemployeeretires. (This is as opposed to adefinedbenefit plan, where the annuity income is fixed as a function oflengthofserviceandsalary.)Ifseparatelifetablesareused,thesamelevelofcontributions will purchase smaller pension benefits for a female than for a male of the same age. This has resultedinchargesofdiscriminationbygender.Therehavebeenlengthydebatesonthisissue and many points of view put forward. Some question the fairness of considering only gender while ignoring other factors. For example, a female smoker may actually expect to live a shorterperiodthanamalenonsmokerofthesameage,butwillstillreceiveasmallerpension ifonlygenderistakenintoaccount.Theissueiscomplicatedandwewillnotgointoitfurther here.Thetrendlatelyhasbeentoagreethatthegenderdiscriminationshouldbeavoided,and manypensionplansnowuse‘unisexlifetables’thatareablendofthecorrespondingmaleand femaletables.Theresultisthat,fordefinedcontributionplans,malesreceivesomewhatlower pensionbenefits,andfemalesreceivesomewhathigherpensionbenefits,thantheywouldhave received had separate male and female tables been used. 4.7 Standard notation and terminology Standard notation for life annuities generally follows that for general annuities as introduced in Section 2.14, except that the agexis inserted as a right subscript, as we have done. So, for example, ̈a x: n = ̈ax(1n), ax: n = ̈ax(01,1n); ̈sx: n = yx(n)−1̈ax(1n), sx: n = yx(n)−1̈ax(0,1n); Iax: n = ̈ax(0,1,2,… ,n), Ïax: n = ̈ax(1,2, …,n); Dax: n = ̈ax(0,n,(n−1),(n −2),…,1), D̈ax: n = ̈ax(n,(n−1),(n −2),…,1); k|n̈ax = k−1|nax = ̈ax(0k,1n). The last item is the present value of a deferred annuity where the first payment is at timek, and there arenpayments. In some cases, as here, a duration appears as a left subscript rather than under an ‘angle’. To understand the second symbol, note that the unadornedaindicates a first payment at time 1, and thek−1 before the deferred symbol indicates that payments are k−1 years later than this, that is at timek, which is the same as the first symbol. The present value of ak-year pure endowment of 1 is denoted bykEx, another example where duration is given by a left subscript. Wehavealreadyintroducedthesymbol ̈ax,inwhichtheomitteddurationsymbolindicates a whole life annuity with payments of 1 for life. In the standard notation this carries over to the other symbols as well. That is, ̈a x, ax, Ïax, Iax, k|̈ax = k−1|ax are all defined as the corresponding symbols above, with benefits continuing for as long as (x) is alive. In other words,nis taken to be𝜔−x−j, wherejis the time of the first nonzero payment. 58 LIFE ANNUITIES Traditional actuarial texts include many identities relating these quantities. For the most part,theyallfollowfromthesplittingidentity(4.9).Forexample(assumingconstantinterest), we get the following variation of (4.10), written in standard notation as ̈a x:n+k = ̈ax:k +vk kpẍax+k: n. The termannuitycertain or fixedtermannuity is often used to describe annuities where the payments are certain to be made, as distinguished from life annuities. 4.8 Spreadsheet calculations In order to handle problems involving a life table, we proceed as follows. As in the Chapter 2 spreadsheet we put duration in column A, and interest rates in Column B, starting at row 10.WeinsertthelifetableincolumnN(columnsinbetweenarereservedforotherpurposes, described in later chapters ), withq y entered in cellN(10+y). Our sample table can be input as follows. Enter the two parameters 0.0005 in cell N3, 1.09 in cell N4. This allows us to change parameters when desired. Then enter the formula = 1− EXP(−N$3 ∗ N$4̂$A10) in cell N10 and copy down to cell N128. Enter 1 in cell N129. It is important to also ensure that the remaining cells in Column N are filled with zeros up to at least N250. In cell C1 we insert the agex. We then use the index function in Excel to select the pertinent mortality rates, puttingq x+t in cell C (10+t) This is done by inserting the formula =INDEX(N$10 :N$250,C$1+$A11) in cell C10 and copying down. Entries of zero will appear after duration𝜔−xcoming from the extra zeros in Column N. Wecalculate v(k)incolumnDasinSpreadsheet2,entering1incellD10, =D10/(1+B10) in column D11 and copying down. We calculate the vectoryx in column E, using the recursion (4.5). The entry in E10 is 1, the entry in E11 is = E10 ∗ (1+B10)̂−1 ∗ (1−C10) and we copy down. We insert the cash flow vectorcin column F, and calculate the valuëax(c) in cell F8 with the formula. =SUMPRODUCT ($E10 : $E129,F10 :F129). For example at a constant interest rate of 6%,̈a50(120) = 11.5957. For annual-premium deferred annuities, we enter the premium pattern vector in Column H. We copy the formula in F8–H8 (reserving column G for a later purpose). In I6, we enter =F8/H8, the initial premium. In I10, we enter=$I$6*H10 and copy down, which inserts the premium vector in column I. Copying the formula in H8–I8 provides a check. We should get the same value as in F8. EXERCISES 59 Asacheck,at6%interest,foranannuityon(50)providing1perlifebeginningatage60, the level annual premium payable for 10 years is 0.855. We will leave it to the interested reader to modify this spreadsheet to handle guaranteed periods. Notethatthisspreadsheetcanbeusedtohandleshortdurationproblemswiththemortality rates given individually, such as we have in the examples and exercises. We can ignore the given age, and simply insert the values ofq x+t that we want directly in column C. Remember howevernotto save changes when closing, or alternatively, copy up in column C to restore the formulas when you next use the spreadsheet. Exercises Type A exercises 4.1 Redo Example 4.1, only suppose now that interest rates are 50% for the first2years and 100% after that. 4.2 Agroupofindividualsage40,eachinvest1000inafundearninginterestat5%.Atthe end of 20 years the fund is divided equally among the survivors. If20p40 = 0.8, how much does each get? 4.3 You are givenq60 = 0.20, q61 = 0.25, q62 = 0.40, q63 = 0.50. The interest rate is a constant 5% for the first 2 years, and 7% after that. A 5-year life annuity on (60) provides for payments of 100(1+k) at timek, wherek= 0,1,2,3,4. (a) Find the present value. (b) Supposethatinsteadofbeingastraightlifeannuity,thefirstthreeannuitypayments are guaranteed regardless of whether (60) is alive or not. Find the present value now. 4.4 You are given that interest is constant and that ̈a 40(130) = 21, ̈a60(110) = 6, v20 20p40 = 0.8. Calculate ̈a40(120). 4.5 The present values of a 10-year life annuity on (60), and a 10-year compound inter- est annuity, with annual payments of 1, are given, respectively bÿa60(110) = 5 and ̈a(110;v) = 6. A person age 60 has 100 000 that he plans to pay as a single premium for a life annuity, beginning at age 60. This will provide a level income of 12 500 per year for life. What will the yearly income be if, instead of a straight life annuity, he purchases a life annuity with a 10-year guaranteed period? Type B exercises 4.6 Mortalityisgivenby q52 = 0.1, q53 = 0.2.Theinvestmentdiscountfunction visgiven by i0 = 0.2, i1 = 0.25, i2 = 0.25, i3 = 0.5. Calculatey50 ◦ 2(2) andy52(2), usingvin both cases, to show thatyx◦ kis not equal toyx+k when interest is not constant. 60 LIFE ANNUITIES 4.7 If interest is constant andqx is a constantqfor allx, find an expression for̈ax in terms of vandq. 4.8 Showthat,ataconstantzerorateofinterest, ̈ax(0,1∞) = ex.Giveanintuitiveexplana- tion of this fact. 4.9 A life annuity on (x) provides for annual benefit payments for life beginning at age x. The initial benefit payment is 1000 and each subsequent payment increases by 6% (i.e.,thebenefitpaymentatage x+1is1060,thepaymentatage x+2is1000(1.06) 2, and so on). The first 10 benefit payments areguaranteedand will be made regardless of whether (x) is alive or not. The interest rate is a constant 6%. You are given that 10px = 0.9 andex+10 = 10. Find the present value of this contract. 4.10 Alifeannuitycontracton(80)hasapresentvalueof3.14.Theannuitybenefitsatboth age 80 and 81 are 1. The interest rate in the first year is 25% andp80 = 0.8. Suppose thatthevalueof p81 increasesby10%whileallothermortalityratesremainunchanged. What is the new present value of the contract? 4.11 Given that ̈a40 = 15, ̈a[40]+25 = 10, v(25) = 0.5, 25p40 = 0.4, find the net annual pre- mium, payable for 25 years beginning at age 40, for a deferred annuity, paying 1000 yearly for life, with the first benefit payment at age 65. 4.12 Suppose that for all x,q x+1 ≥ qx. Show thaẗax+1 ≤ ̈ax. Does this remain true if we remove the monotone condition onqx? Spreadsheet exercises The following exercises are to be done using the sample life table of Section 3.7. 4.13 A deferred life annuity on (40) provides for income of 10 000 per year beginning at age 65. Nothing is paid if death occurs before age 65. This is purchased by annual premiums payable for 25 years beginning at age 40. Premiums increase each year by 10%oftheinitialpremium,thatis, 𝜋 k = (1+0.1k)𝜋0.Interestratesare5%forthefirst 10 years and 6% thereafter. Find the initial premium𝜋0. 4.14 Agroupofindividualsage30eachagreetoinvest1000peryearfor40yearsbeginning at age 30. At time 40, the fund is divided up among all the survivors. If interest rates are 5% for the next 10 years, 6% for the following 10 years and 8% after that, how much does each survivor receive? 4.15 A single-premium life annuity on (x) provides 1 unit per year for life, beginning at time 1, with an-year guaranteed period, wheren is the smallest integer greater than or equal to the premium. (This is known as aninstalmentrefund annuity. It guarantees thatatleastthefullamountofthepremium,withoutinterest,willbereturned.)Assume a constant interest rate of 4%. Findn and the single premium if (a)x = 40, (b)x = 70. (Note that there is no direct method of calculation. A trial and error procedure is called for). 4.16 Interest rates are 8% for the first 25 years and 6% thereafter. Comparëa [40]+20 and ̈a[50]+10. Which one of these is smaller? Answer this before any calculation, and then do the actual calculation to verify your answer. 5 Life insurance 5.1 Introduction A life insurance policy is a contract between theinsurer and another party known as the policyholder. In return for a payment of premiums, the insurer will pay a predetermined amount of money, known as adeathbenefit, upon the death of the policyholder. The amount ofthebenefitcanvarywiththetimeofdeath.Inpractice,thismoneywillbepaidimmediately upon death (or more realistically, a short time after, to allow for processing the claim) but for mathematicalconvenienceweassumeinthischapterthatitwillbepaidat theendoftheyear of death. For example, if the policyholder purchases a policy on January 1 and dies a week later,ourassumptionmeansthatthedeathbenefitwillnotbepaiduntilDecember31.Wewill consider the more realistic situation of payment at the moment of death in Chapter 8. The reader should distinguish carefully between life annuity and life insurance contracts. The life annuity provides a sequence of periodic benefit payments. The typical life insurance contract provides only a single benefit payment, paid on the occasion of death. 5.2 Calculating life insurance premiums Considerapolicyon(x ).Let bk betheamountthatwillbepaidattime k+1fordeathbetween timekand timek+1. We will refer to the vector b = (b0,b1,…,b𝜔−x−1) as thedeathbenefitvector. Notation The reader is cautioned that many authors use the subscript onbto refer to the timeofpayment. Whatwecall bk,theywouldcall bk+1,sinceitispaidattheendoftheyear, which is timek+1. We prefer the convention above. All our vectors are then indexed from 0 FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 62 LIFE INSURANCE to 𝜔−x−1. In particular, this facilitates matters when dealing with contracts that combine annuity and insurance benefits. Suppose we have fixed an investment discount functionv and a life table. We want to calculate thenetsinglepremium fortheabove policy, whichwewilldenoteby Ax(b).(The A is a standard symbol that probably came from the word ‘assurance’, an older version of the word ‘insurance’.) The principle to determine this is the same as that used for annuities. The total premiums, together with all investment earnings, must be sufficient to provide all the death benefits. To derive the formulas, we will follow the annuity procedure and begin with the case where b = e k. This is a policy in which 1 is paid at timek+1 providing (x) dies between the ages ofx+k and x+k+1. All of the other death benefits are of zero amount. Suppose that we have𝓁x people agexwho each purchase this same contract. Out of these, there will be dx+k individuals who die between the ages ofx+k and x+k+1, and each of them will receive1attime k+1.Thetotalpresentvalueofallthesedeathbenefitswillbe v(k+1)dx+k. This must be equal to the total amount collected in premiums, which will be𝓁xAx(ek). Therefore Ax(ek) = v(k+1)dx+k 𝓁x . The general policy can be viewed as a sequence of 1-year policies as above, one for each value ofk, where thekth policy paysbk at timek+1 for death in the previous year. The premiumforsucha1-yearpolicyisjust bkv(k+1)dx+k∕𝓁x andthetotalpremiumisobtained by summation. We have Ax(b) = 𝜔−x−1∑ k=0 bkv(k+1)dx+k 𝓁x = 𝜔−x−1∑ k=0 bkv(k+1)(kpx − k+1px) (5.1)‡ = 𝜔−x−1∑ k=0 bkv(k+1)kpxqx+k. The reader should note that the expression above follows the same pattern as the annuity singlepremium.Itisapatternthatwewillencountermanymoretimesinsubsequentmaterial. Namely, we sum up a number of terms, each of which consists ofthreefactors, amount × interest discount factor× probability that payment is made. (5.2) In the case of insurance, formulas [3.5(a)–(c)] give three different expressions for the probabilitythatthepaymentwillbemade,givingusthethreedifferentwaysofwriting A x(b). Each will be useful in certain cases. Aswithannuities,thenotationwillsuppressthedeathbenefitvector1 ∞ = 1𝜔−x.Accord- ingly,Ax will be the net single premium for a policy paying 1 at the end of the year of death, whenever it occurs. CALCULATING LIFE INSURANCE PREMIUMS 63 Example 5.1 Suppose thatq60 = 0.2, q61 = 0.4.q62 = 0.5 andi= 100%. A policy sold to (60) provides for benefits at the end of the year of death of 80 for death in the first year, 75 for death in the second year, and 100 for death in the third year. If the insured lives to age 63, the policy terminates and nothing is paid. Find the net single premium. Benefits 80 75 100 Time 3210 Pattern 0.7521 vs 1/21 /21 /2 qs 0.20 .40 .5 Figure 5.1 Example 5.1 Solution. WeagaincanmakeuseofatimediagramasshowninFigure5.1.Thedeathbenefit amount is put at thebeginningof the year to which it is applicable. That is,bk is inserted above timek. Our convention here is to enclose these amounts in a box, to distinguish them from annuity benefits. The amountbk must be multiplied by the previous interest discount factors,aswellasbytheinterestdiscountfactorfortheyearstartingattime k,sinceitispaid at the end of the year. Moreover it is multiplied by the previouspvalues and as well byqx+k, asindicatedinthethirdformulain(5.1),whichtogether givetheprobabilityoflivingtotime kand then dying in the following year. In this case, the net single premium is ( 80× 1 2 ×0.2 ) + ( 75× 1 4 ×0.8×0.4 ) + ( 100× 1 8 ×0.8×0.6×0.5 ) = 17 As with life annuities, one can as an alternative construct a partial life table, which automatically gives you the multiplication of the probabilities, and then use the first formula in (5.1). So for example in this case, starting with𝓁 60 = 1000, we have in turn,d60 = 200, 𝓁61 = 800, d61 = 320, 𝓁62 = 480, d62 = 240. Normally, the policyholder does not pay for insurance by a single premium but rather by asequenceofperiodicpremiums.Weassumeinthischapterthatpremiumsarepaidannually. The premiums will be given by a premium vector𝜋 = (𝜋 0,𝜋1,…,𝜋𝜔−x−1) as defined for deferred annuities in Section 4.5. Following the principle used there, we want the net single premium to be equal to the present value of the premiums, with respect to the interest and survivorship accumulation function. This means that A x(b) = ̈ax(𝜋) = 𝜋0̈ax(𝜌), so that 𝜋0 = Ax(b) ̈ax(𝜌), (5.3) ‡ where𝜌 is the premium pattern vector. 64 LIFE INSURANCE We follow the terminology used for single premiums and call these premiumsnetannual premiumsor annual benefit premiums. We will later encounter examples of premiums that are different from net premiums, but our convention is that unless otherwise mentioned, all premiums will be net. Example5.2 SupposethattheinsuranceinExample5.1istobepurchasedbythreeannual premiums, beginning at age 60, where the second premium is double the first, and the third premium is three quarters of the first. Find the premiums. Solution. The premium pattern vector is given by𝜌 = (1,2,0 .75). We have ̈a 60(𝜌) = 1+ ( 2× 1 2 ×0.8 ) + ( 0.75× 1 4 ×0.8×0.6 ) = 1.89. so that the initial premium will be 17/1.89= 8.99. The insured will then pay 8.99 in the first year, 17.99 in the second year and 6.75 in the third year. 5.3 Types of life insurance Lifeinsurancepoliciesaretraditionallyclassifiedintodifferenttypes. Terminsuranceprovides death benefits for a fixed number of years (similar to the temporary annuity). After the expiration of the term, coverage ceases and there are no more benefits. Example 5.1 involved such a policy, with a term of 3 years.Wholelife insurance provides death benefits for life, so some payout on the policy is certain to occur. In our mathematical model we will not need to distinguish between the two types. We will assume that all policies will continue to age𝜔. Forterminsurancerunningfor nyears(whichwerefertoas n-yearterm)wesimplywillhave b k = 0f o rk≥ n. Nonetheless, in the next chapter we will see that there are differences inthe natureofthesetwotypesandthatwholelifeinsurancehasa savingscomponentaswellasan insurancecomponent. Another common type of contract isendowmentinsurance, which we will discuss in detail in the next section. A more modern development, known asuniversal lifewill be described in Chapter 13. 5.4 Combined insurance–annuity benefits It is possible to combine both life insurance benefits and life annuity benefits in the same contract. One of the most popular types of such a policy is known asendowmentinsurance. This provides for a payment at some future timenif (x) is then alive, and in addition a death benefit if (x) dies before timen. Such a policy is known asn-year endowment insurance or endowment insuranceat agex+n, since it combines apure endowment(as defined in Section4.2withinsurance.Itisusuallymarketedasasavingsplan,wherebythepolicyholder, by paying premiums each year, will accumulate a certain sum at timen. In addition, the policyholder is protected with life insurance if he/she dies before accumulating the desired amount. It should be noted that a whole life policy is similar to endowment insurance since the policyholderisguaranteedsomedeathbenefit.Mathematicallyitcanbeviewedasendowment COMBINED INSURANCE–ANNUITY BENEFITS 65 insurance at age𝜔, and in practice it actually is interpreted in this way. Since it is to the advantage of the policyholder, the insurer will assume that everybody dies by age𝜔, unlike the case of a life annuity, and pay the death benefit at age𝜔 to all survivors. To calculate the single premium for endowment insurances, we simply view it as two separate contracts and add the premiums. Example 5.3 An insurance policy on (x) provides 1 unit payable at timen if (x) is then alive,plus1unitpayableattheendoftheyearofdeathif( x)diesbeforetime n.Levelannual premiums ofPare payable fornyears. Find a formula forP. Solution. The single premium for the death benefit isA x(1n). The single premium for the pure endowment isyx(n). The premium pattern vector𝜌is (1n). The total single premium for the contract is thenAx(1n)+yx(n) and P= Ax(1n)+yx(n) ̈ax(1n) . Example 5.3 is a very common type of endowment insurance with a level death benefit equal to the pure endowment amount, and level premiums payable for the full term. This is not essential, however, and many other combinations are possible. Example 5.4 Consider a 20-year endowment insurance with the death benefit equal to 1 unit for the first 10 years, and 2 units for the second 10 years. The amount of the pure endowment is 3 units. Level annual premiums ofPare payable for 15 years. Find a formula for P. Solution. Calculating as in the previous example, P= A x(110,2 10)+3yx(20) ̈ax(115) . Example5.5 (Figure5.2).Supposethat q60 = 0.2,q61 = 0.4and i= 100%.A2-yearpolicy provides for benefits at the end of the year of death of 80 for death in the first year, and 75 for death in the second year. In addition, there is a pure endowment of 70 paid at age 62 if thepolicyholderisthenalive.Thisispurchasedbytwo-levelannualpremiums.Calculatethe premium. Benefits 80 70 75 Time 210 Pattern 11 vs 1/21 /2 qs 0.20 .4 Figure 5.2 Example 5.5 66 LIFE INSURANCE Solution. The present value of death benefits is A60(80,75) = ( 80× 1 2 ×0.2 ) + ( 75× 1 4 ×0.8×0.4 ) = 14. The present value of the pure endowment is ̈a60(0,0,70) = 70× 1 4 ×0.8×0.6 = 8.4. The present value for the premium pattern vector is ̈a60(𝜌) = ̈a60(1,1) = 1+ 1 2 ×0.8 = 1.4 By (5.3) 𝜋0 = 𝜋1 = 14+8.4 1.4 = 16. Another common type of combined policy is a deferred annuity that provides for death benefits during the deferred period. This is somewhat similar in nature to endowment insur- ance. The difference is that the accumulated savings are paid out as an annuity rather than as a single payment. Example 5.6 (Figure 5.3). A contract on (40) provides for annuity benefits of 1 per year for life, beginning at age 65. If (40) dies before age 65, a death benefit of 10 will be paid at the end of the year of death. Level annual premiums ofP are payable for 25 years. Find a formula forP. Benefits 10 10 10 ... 1 1 10 ... Time 210 ... 26 25 24 ... Pattern 111 ... 1 Figure 5.3 Example 5.6 Solution. Wecalculatethis,asintheexamplesabove,byaddingthesinglepremiumsforthe death benefits and the annuity benefits and dividing by the annuity for the premium pattern vector. The result is P= 10A 40(125)+ ̈a40(025,1∞) ̈a40(125) . The second term in the numerator can be written alternatively asv(25)25p40̈a[40]+25. An interesting variation on the above involves a popular marketing device, which is to issue policies with a return-of-premium feature. COMBINED INSURANCE–ANNUITY BENEFITS 67 Example 5.7 (Figure 5.4). A contract on (40) provides for annuity payments of 1 per year for life beginning at age 65. Should (40) die before age 65, there will be a return of all premiums (without interest) paid prior to death. Level annual premiums ofPare paid for 25 years. Find a formula forP. Benefits P 2P 3P ... 25P 1 1 ... Time 210 ... 26 25 24 ... Pattern 111 ... 1 Figure 5.4 Example 5.7 Solution. Thisisbasicallythesametypeofproblemasabove,exceptthatthebenefitsdepend on the unknownP. We cannot obtainP directly, but we can set up an equation to solve for it. Suppose (x) dies between timek and k+1, wherek is between 0 and 24 inclusive. The insuredwillhavepaid k+1premiumsofP andthesewillbereturnedattime k+1.Thedeath benefit vector is then b = (P,2P,3P,…,25P) = Pj, wherej = (1,2, …,25). Equating present values of the premiums and benefits, P̈a40(125) = PA40(j)+ ̈a40(025,1∞), from which P= ̈a40(025,1∞) ̈a40(125)−A40(j). (5.4) Example 5.8 Consider the same policy as above, except that the premiums are returned withinterest as determined by the functionv. Find a formula forP. Solution. One method is to simply use (5.4), where the vectorj is changed fromjk = k+1 to jk = Valk+1(1k+1;v). Another approach, which produces the solution in a much easier form for calculation is found through the following reasoning. In the first 25 years the accumulation is according to the discount functionvrather thany 25. To see this clearly, consider the room-box description of Chapter 4. Upon death of a participant before age 65, the person’s box is opened up, but the amount paid back as a death benefit is exactly the same as the amount that was there in the first place. The other participants will receive no survivorship earnings. Therefore, each 68 LIFE INSURANCE survivor will have accumulated at age 65 the amountPVal25(125;v). This amount must be sufficient to provide an annuity of 1 unit per life beginning at age 65. Thus P= ̈a[40]+25 Val25(125;v). We will present an alternative solution to this problem in Section 6.7. A generalization of the above is concerned with partial refunds. Example 5.9 Now consider Example 5.8, except that the payment upon death isk times the accumulated premiums with interest, where 0≤ k≤ 1. For each premiumP, the portionkP accumulates at interest and (1−k)P accumulates with interest and survivorship. Equating values at time 25, kP Val25(125,v)+(1−k)PVal25(125;y40) = ̈a[40]+25 and we can solve to get P= ̈a[40]+25 kVal25(125;v)+(1−k)V a l25(125;y40). For k= 1 this reduces to the solution to Example 5.8 and fork= 0, it is a special case of Example 4.5. Remark For those preferring an alternative to the square bracket notation, there are var- ious other ways in which to write the numerator of the last two examples. For example, Val 25(251∞;y40)or ̈a40(025,1∞)∕y40(25).Ofcoursewithconstantinterest,thesimplestchoice is jusẗa65. The following variation is suggested by provisions in some pension plans where an employee must remain in the plan for a minimum period in order to get credit for premiums paid by the employer. A more realistic version will be given in Section 13.3. Example 5.10 Redo Example 5.8 only with the provision that premiums are returned with interest upon death only if (40) lives to age 45. Nothing is returned if death occurs in the first 5 years. Solution. Theredoesnotappeartobeanyconvenientwaytoapplythesimplifyingapproach of the last two examples. We can however use (5.4) with the vectorj given by j k = {0i f 0 ≤ k< 5, Valk+1(1k+1;v)i f 5 ≤ k< 25. Indeed, (5.4) is a flexible formula which can be adapted to a variety of premium return provisions,includingcaseswheretheinterestrateonthepremiumsmaydifferfromthatused in the discount function. INSURANCES VIEWED AS ANNUITIES 69 We now present an example involving a more complex calculation. Example 5.11 An annuity on (x), purchased by a single premiumS, provides for 1 per year for life, beginning at time 1. If (x) dies before a total income ofShas been paid out, the difference betweenSand the total income received will be refunded at the end of the year of death, so that the policyholder will at least receive total income equal to the single premium. (e.g., ifSis 20 and (x) dies between time 5 and time 6, a death benefit of 15 would be paid at time 6.) Describe a procedure for calculatingS. Solution. The premiumSmust satisfy S= ̈a x(0,1∞)+Ax(b), where bk =m a x{S−k,0}. Sincebdependson S,thereisnodirectwaytosolvethisandaniterativenumericalprocedure must be employed. One guesses at an initial value ofS (it will be close to but greater than ̈ax(0,1∞)), and then continues to adjust the value ofS until the right hand side above also equals this value. This can be carried out automatically in Excel with thegoal-seekfunction. Exercise 5.20 gives some particular examples. ThecontractinExample5.9isusuallytermeda cashrefundannuity anditisavariationon theinstalmentrefundideaintroducedinExercise4.15.Inpractice,bothofthesearesomewhat more complicated than described since the refund is based on the gross premium rather than the net. 5.5 Insurances viewed as annuities A comparison of (4.1) and the third expression in (5.1) shows that Ax(b) = ̈ax(c), where ck = v(k,k+1)bkqx+k. (5.5)‡ Let wx denote the vector with entries (wx)k = v(k,k+1)qx+k. We can then write (5.5) in compact form as Ax(b) = ̈ax(wx ∗ b). (5.6) We can verify this intuitively as follows. Suppose (x) wishes to pay a single premium for a life insurance policy running over several years, based on the death benefit vectorb. The insurer refuses, claiming that it only sells life insurance policies for 1 year at a time. It does, however, sell life annuities. The person therefore purchases a life annuity with benefits of v(k,k+1)b kqx+k at timek if she is alive. She does not keep this annuity payment, but immediately returns it tothe insurer topurchase a 1-year lifeinsurance policy paying adeath benefit at timek+1 if death occurs in the next year. This single premium will purchase a death benefit of exactlyb k. This follows from our previous discussion, but to re-emphasize the point, note that this premium will accumulate tobkdx+k∕𝓁x+k at the end of the year, and assuminglx people engage in this scheme, this amount collected from each of𝓁x+k survivors at timekwill be enough to providebk to each of thedx+k people who die during the year. 70 LIFE INSURANCE Thisillustrationisofcoursefanciful,asallinsurerssellpoliciesforperiodsofmorethan1 year.(Moreover,asapracticalmatteritoverlooksthefactthat1-yearpremiumscouldchange over time, as well as the fact that annuity and insurance premiums are based on different tables.) It does, however, provide a useful point of view. Each policyholder can look upon a life insurance policy as a life annuity, providing what is essentially that person’s share of the deathbenefits.Onecouldsupposethattheseannuitypaymentsarethencollectedinaseparate fundandusedtopayallthebenefitstothosewhodieduringtheyear.Thisisoftenavaluable way of looking at the situation, since all policies can be thought of as life annuities if we wish. The calculation of premiums and other quantities can then be reduced to the general principles outlined in Chapter 2. Thisviewpointallowsustoimmediatelyadaptallresultsobtainedforlifeannuitiestothe insurancesetting.Forexample,the splittingidentity forlifeannuities(4.8)takesthefollowing form for insurances: A x(c) = Ax(kc)+yx(k)A[x]+k(c◦ k), (5.7) where the bracket onxindicates, as with annuities, that the discount function used isv◦ k. We can also maintain the room-box visualization for life insurance policies, which we introduced for annuities in the previous chapter, and which will provide a useful guide in the next chapter. The reader is cautioned that while (5.6) is useful conceptually and also for spreadsheet calculation (as illustrated in Section 5.9), it is not always the best for hand calculation. For suchpurposes,theprocedureusedinExample5.1isusuallylesssubjecttoarithmeticalerrors. 5.6 Summary of formulas Inthissectionwesummarizetheprocedure,developedinthelasttwochapters,forcalculating anannualpremiumonagenerallifeinsurance–annuitycontract.Wefirstidentifyfourpertinent vectors: b, the death benefit vector;c, the life annuity benefit vector;u = (0 g,w), the vector of payments that are guaranteed provided (x) lives to agex+g; and𝜌, the premium pattern vector. In many cases only one or two of the first three vectors will be applicable, and the others will be set equal to the zero vector. We then calculate the initial premium from the equation 𝜋 0̈ax(𝜌) = Ax(b)+ ̈ax(c)+v(g)gpẍa(w;v). This allows for cases where the vectors on the right hand side can themselves depend on𝜋0. 5.7 A general insurance–annuity identity 5.7.1 The general identity There is another useful relationship between insurances and annuities. For the vectorwx introduced in Section 5.5, we can write (wx)k = v(k,k+1)−v(k,k+1)px+k = 1−v(k,k+1)px+k −dk, A GENERAL INSURANCE–ANNUITY IDENTITY 71 and it follows that wx =∇ 1∞ −d in the notation of (2.14) with respect to the discount functionyx. From (5.6) Ax(b) = ̈ax(wx ∗ b) = ̈ax(∇1∞∗ b)− ̈ax(d ∗ b) and from (2.15) we obtain our main identity Ax(b) = ̈ax(Δb)− ̈ax(d ∗ b). (5.8) 5.7.2 The endowment identity We will use (5.8) to derive a well-known actuarial formula, which we call theendowment identity. Assume constant interest. The constant discount ratedthen factors out as a constant multiple and ̈ax(d ∗ b) = d̈ax(b). Supposeb = (1n). ThenΔb = (1,0,0,… ,−1) where the −1 is in position indexed withn(i.e., in the (n+1)th entry since we start with 0). Equation (5.8) says that Ax(1n) = 1−yx(n)−d̈ax(1n). Let Ax:n be the net single premium for ann-year, 1-unit endowment insurance. That is, 1 is paid either at timen, or at the end of the year of death if that occurs before timen. (This is the standard symbol for such a premium.) Addingyx(n) to both sides of this equation, we get the endowment identity, Ax:n = 1−d̈ax(1n). (5.9) This identity is analogous to (2.16) and its derivation as given at the end of Section 2.9. For an interpretation, suppose I lend you 1 unit now, to be repaid in full at the end ofnyears, orattheendoftheyearofyourdeathifthisoccursbefore nyears.Youmustalsopayinterest at the beginning of each year until the loan is paid. The present value of the loan must be equal to the present value of the principal repayments, plus the present value of the interest. The latter is just a temporary life annuity payingdunits yearly fornyears, beginning at time zero. The present value of the loan is just 1 and the present value of the principal repayments is justA x:n, the net single premium for the endowment insurance. We obtain the equation 1 = Ax:n +d̈ax(1n), which gives (5.9). If Pdenotes the level annual premium payable fornyears for the 1-unit,n-year endow- ment insurance, we can also expressP in term of annuities by dividing bÿa(1n) in (5.9) to obtain P= 1 ̈ax(1n) −d. (5.10) This identity can be interpreted as follows. Suppose you invest 1 unit. This will provide you with interest earnings ofd at the beginning of each year, until such time as you wish to 72 LIFE INSURANCE terminate the investment and take back your principal of 1. An alternate scheme is to use the 1 unit to purchase ann-year life annuity, paying 1∕̈a x(1n) at the beginning of each year for nyears, and to use part of these proceeds to purchase a 1-unit,n-year endowment insurance, carrying level annual premiums ofP for n years. The insurance will pay you back your principalattheendof nyears,oratdeathifearlier.Beforerecoveringyourprincipalyouwill havenetannualearningsof1 ∕̈ax(1n)−P,andthismustequaltheincomeof dthatyouwould get from the first alternative. This yields the given identity. Remark This back-to-back annuity–insurance combination has recently become popular as an investment vehicle. At first glance, it appears that in practice it will produce a lower return than the straight investment, since one must pay expenses on both policies and, in addition, the different lifetables used for annuity and insurances will work to the purchaser’s disadvantage.(Oursimplifiedmodelassumesnoexpensesandthatthelifetableisthesamein all cases.) However, the fact that the proceeds of the insurance and annuity contracts receive favourable tax treatment in many jurisdictions often means that the net after-tax return can actually be higher than that of the straight investment. The above formulas are of course true for a 1-unit whole life insurance, which is just endowment insurance at age𝜔.W eh a v e A x = 1−d̈ax, Px = 1 ̈ax −d, (5.11) where Px is the net level annual premium payable for life for a 1-unit whole life contract on (x). 5.8 Standard notation and terminology 5.8.1 Single-premium notation The standard symbol for an insurance single premium isA, as we have given it. As with annuities, thisisembellished withsuperscriptsand subscriptstohandle thecommon types of death benefit vectors. For example: r Ax denotes Ax(1∞) as we have already indicated r A1 x: n denotes Ax(1n), the net single premium for a 1-unit,n-year term insurance. The superscript 1 above thexsignifies that (x) must die before the expiration of then-year period in order to collect. r Ax:n denotes Ax(1n)+yx(n), the net single premium for a 1-unitn-year endowment insurance, as we have indicated above. The subscriptx: n signifies that the death benefit is paid upon the first ‘failure’ of (x)o rt h en-year period. The life (x) fails upon death and then-year period fails at the end ofnyears. Recall that the same subscript is used in the temporary annuity standard symbol to signify that benefits are paid as long as both the life (x) and then-year period are ‘surviving’. STANDARD NOTATION AND TERMINOLOGY 73 r Ax: 1 n isanothersymbolfor nEx (denotedby yx(n)inournotation).Thisisthenetsingle premium for a 1-unit,n-year pure endowment. The superscript 1 is now is above then to signify that then-year period must fail before (x) does in order for the contract to pay. r k|nAx denotes Ax(0k,1n). This isdeferred insurance. A level death benefit of 1 unit begins afterk years and continues forn years. Such a policy would not normally be sold by itself but may be combined with other policies. r k|Ax stands for k|𝜔−xAx in accordance with the usual practice of omitting duration symbols when the contract continues for life. r DA1 x: n denotes Ax(n,n−1,…,1).The Dstands for decreasing. r IA1 x: n denotes Ax(1,2, …,n). TheIstands for increasing. r DAx and IAx are respectively the above two symbols withn= 𝜔−x, in keeping with the general notational principle discussed in Section 4.7. 5.8.2 Annual-premium notation The standard symbol for a level net annual premium isP. This is followed by the single- premiumsymbol.Forthosesinglepremiumsthatbeginwithacapital A,the Aisomitted.The premium payment durationtappears before thePon the lower left. If omitted, it means that premiums are paid for the natural duration of the contract. Examples follow: r Px is the annual premium, payable for life, for a 1-unit whole life policy on (x). r tPx is the annual premium payable fortyears for a 1-unit whole life policy on (x). r P1 x: n is the annual premium payable fornyears for a 1-unit,n-year term policy on (x). r tPx:n istheannualpremiumpayablefor tyearsfora1-unit n-yearendowmentinsurance on (x). r P(IA1 x: n ) is the annual premium payable forn years for ann-year increasing term policy on (x). (In this case we insert the full single-premium symbol since it does not begin with anA.) r P(n|̈ax)isthelevelannualpremiumpayablefor nyears,foradeferredannuityproviding income of 1 unit per year beginning at agex+n. Note here that the missing premium paymentdurationsymbolistakenas n,thenaturalpremiumpaymentduration.Although the contract continues for life, it is not natural to continue paying premiums when the annuity payments begin. Alloftheseannualpremiumsymbolsareevaluatedbytakingthecorrespondingnetsingle premium and dividing bÿa x:t, wheretis the premium paying duration. For example, 10P50:20 = A50:20 ̈a50:10 . 74 LIFE INSURANCE 5.8.3 Identities Thesametypes ofidentitythatwediscussedinSection4.7arisewithinsurances aswell.For example, A x:n = A1 x: k +v(k)kpxA{x}+k: n−k. The derivation will be left to the reader. 5.9 Spreadsheet applications We modify the Chapter 4 Spreadsheet to handle death benefits. The death benefit vector is entered into Column G. The vectorw x is calculated in column L by putting the following formula in L10 and copying down = G10 ∗ (1+B10)̂(−1) ∗ C10 In G8 we then put the same formula as in F8 except with L replacing F, and this will return A x(b). The entry in I6 is changed to=(F8+G8)/H8. The premium vector appears in Column I. Foracheck,theformulainH8canbecopiedtoI8andthiscellshouldreturnatotalofF8 and G8. For a sample problem, compute the initial premium for a 1000-unit, 30-year endowment policyon(40)withpremiumspayablefor20years,andthepremiumtodoubleafter10years. The interest rate is a constant 6%. The answer is 12.68. Exercises Type A exercises 5.1 Given that 𝓁70 = 1000, 𝓁71 = 960, 𝓁72 = 912, and that interest rates are a constant 10%, calculateA70(12). 5.2 A 3-year endowment insurance policy on (60) provides for benefits paid at the end of the year of death of: 500 if death occurs in the first year (i.e., between time 0 and time 1); 800 if death occurs in the second year; and 1000 if death occurs in the third year. In addition, there is a pure endowment of 1000 payable at age 63 if (60) is then alive. Thisispurchasedbythreeannualpremiumsbeginningatage60.Thesecondpremium is double the initial premium and the third premium is three times the initial premium. You are given that thatq 60 = 0.1,q61 = 0.2 andq62 = 0.25. The interest rate is 25% for the first 2 years and 20% after that. Find the initial premium. 5.3 A 2-year term insurance policy on (60) provides for a death benefit of 100 payable at theendoftheyearofdeath.Thisispurchasedbyasinglepremium.If(60)livestoage EXERCISES 75 62, the single premium is returned without interest. Given thatq60 = 0.1,q61 = 0.15, and interest is a constant 10%, find the single premium. 5.4 Youaregiven qx = 0.053, qx+1 = 0.054, qx+2 = 0.055, i0 = 0.06,i1 = 0.08,i2 = 0.10. Ifb = (1,2,3), find the vectorc = b ∗ wx. Type B exercises 5.5 A deferred life annuity on (40) provides for a yearly income of 1000 beginning at age 65 and continuing for life. It is to be purchased by a single premium ofS payable at age 40. If death occurs during the deferred period (i.e., during the first 25 years), then the single premium is refunded without interest at the end of the year of death. (a) Give a formula forSusing the symbols̈aandA. (b) You are now given the following information. The same annuity contract without thepremium-refundfeature(i.e.,nothingispaidduringthedeferredperiod)canbe purchased for a single premium of 2000. In addition, a contract that provides the same annuity benefits, plus a level death benefit of 2000, payable at the end of the year of death, for death during the deferred period, can be purchased for a single premium of 2200. Calculate an exact numerical value forS. 5.6 Acertainelectricalapplianceissoldwitha5-yearguarantee.Thisprovidesthatthefull purchase price is refunded if the product fails within 2 years, and half of the purchase price is refunded if the product fails in the following 3 years. A study shows that out ofatypicalbatchof100items,therewillbe2failuresinthefirstyear,3failuresinthe second year and 4 failures per year after that. Assuming that interest is a constant 5% andthatreimbursementismadeattheendoftheyearoffailure,whatisthecostofthis guarantee to the manufacturer, as a percentage of the purchase price? 5.7 What is A x(1n) if the constant interest ratei= 0? Give both a formal derivation, and a proof by general reasoning. 5.8 Suppose that interest is constant andqy is a constantqfor ally. Find an expression for Ax in terms ofqand y. 5.9 An actuary calculates a single premium for a certain life insurance policy on (40), and then discovers there were two errors made. In the first place, the life table used showedavalueof q 40 thatwasonlyone-halfofwhatthecorrectfigurewas.Second,the first-year death benefit was taken as 20 when it should have been 10. Will the correct premium be the same as, lower, or higher than the one calculated? 5.10 For a certain insurance contract, on (50), the death benefit for the first year of the contract is 1100, payable at the end of the year of death. The single premium for the whole contract is 600. This is based on an interest rate of 10% for the first year and a mortality table withq 50 = 0.20. If the value ofq50 is changed to 0.25, while all other value ofqx are unchanged, what is the new single premium? 5.11 There is a constant interest rate of 20%, and A50 = 0.300, v10 10p50 = 0.10, A61 = 0.400, q60 = 0.20. 76 LIFE INSURANCE Suppose thatq60 is changed to 0.23, while all other values ofqx remain unchanged. What is the new value ofA50? 5.12 The cash flow vectorj = (1,2,3,… ,10). You are given that interest is constant and that ̈ax(j) = 30, Ax(110) = 0.10, ̈ax(110) = 7, v10 10px = 0.48. FindAx(j). 5.13 Consider a whole life policy on (x) with a level death benefit of 1. Suppose that, for someage y> x,thevalueof qy isincreasedwhileallothervaluesof qremainthesame. (a) Show thatAx is increased. (b) Show thatPx is increased, wherePx is the level annual premium payable for life, for this policy. (c) Showbyexamplethattheabovestatementsarenotnecessarilytrueforawholelife policy with a non-constant death benefit. 5.14 (a) Suppose that, for allx, qx+1 ≥ qx. Show thatAx+1 ≥ Ax. (b) Does the above remain true if we remove the monotone condition onqx? 5.15 Show that for the vectorcgiven by formula (5.5), andk= 0,1,2 …,w eh a v e (a) ̈ax(kc) = Ax(kb); (b) ̈a[x]+k(c◦k) = A[x]+k(b◦k). 5.16 A deferred annuity on (40) provides for an income of 1000 per year for life beginning atage60.If(40)livestoage60,thefirst10annuitypaymentsareguaranteedregardless of whether (40) is alive or not. Level annual premiums ofPare payable for 10 years, beginning at age 40. If (40) dies before the annuity begins, all premiums paid prior to death are returned at the end of the year of death, and no annuity payments are made. Find a formula forP, assuming: (a) premiums are returned without interest; (b) premiums are returned with interest. 5.17 One could generalize our definition of an insurance contract by stipulating that for death at timet, the death benefit is paid at time𝜏(t) which is some function oft. (e.g., whenbenefitsarepayableattheendoftheyearofdeath, 𝜏(t) = [t]+1,where[ ⋅]isthe greatest integer function.) Show that a pure endowment contract can be considered as an insurance, in this sense. 5.18 An annuity on (x) provides 1 per year for life beginning at time 0, with the further provision that, n additional payments will be made after (x) dies, beginning at the end of the year of death. Interest is constant. Show that the net single premium is ̈a(1 n;v)+vn̈ax.Derivethisformulaintwoways.(a)Byusing(5.11).(b)Byusing(5.2). 5.19 Redo Example 5.8 only now assuming that for death during the first 5 years, one-half of all premiums paid prior to death are returned at the end of the year of death. EXERCISES 77 5.20 A contract on (70) provides for a payment of 1000 at the end of 3 years if (70) is then alive, and is to be paid for by three-level annual premiums. For death in the first year, nothing is paid. For death in the 2nd or 3rd years, all premiums paid are returned with interestattheendoftheyearofdeath.Youaregiventhat q 70 = 0.1,q71 = 0.2,q73 = 0.3 andi=25%. Find the annual premium. 5.21 Provide an algebraic proof that the two methods given for the solution of Example 5.8 produce the same answer. (Hint: Change the order of summation in a double sum.) 5.22 Provide an algebraic proof and give an intuitive explanation of the following identity. For any vectorc = (c0,c1,…,cn) ̈ax(c) = v(n)npxValn(c;v)+Ax(j) wherejk = Valk+1(kc;v), k= 0,1, …,n−1 andkcis as defined before Equation 2.18. Spreadsheet exercises The following exercises are to be done using the sample life table of Section 3.7. 5.23 A contract on (40) provides for death benefits if death occurs in the next 25 years. The amount of the death benefit is 50 000 for the first 10 years and 100 000 for the next 15 years. If (40) lives to age 65, he/she will receive a life annuity of 10 000 per year for life beginning at age 65. Premiums are payable for 15 years beginning at age 40. The premium doubles after 5 years. Interest rates are 5% for the first 20 years and 6% thereafter. Find the initial premium. 5.24 Interestratesareaconstant6%.Acontracton(50)providesforapaymentof10000at age 70 if then alive. Level premiums ofPare paid only at even-numbered times, that is, at age 50, 52, 54,…. If (50) dies before age 70 there is a return at the end of the year of death of all premiums paid prior to death. FindP. 5.25 Assume a constant interest rate of 4%. Find the premium for the annuity in Example 5.9 if: (a)x= 40 ;(b)x= 70. 6 Insurance and annuity reserves 6.1 Introduction to reserves Givenaninsuranceorannuitycontractandaduration k,thereserve attime kisdefinedexactly as in Definition 2.7. It is the amount that the insurer needs at timek in order to ensure that obligationsunderthecontractcanbemet.Calculatingreservesforeachpolicyisanimportant responsibilityoftheactuary,knownas valuation.Theinsurerwantstobeconfidentthatfunds on hand, together with future premiums and investment earnings, are sufficient to pay the promised future benefits. It is important to thoroughly master the concept of insurance and annuity reserves in order to properly understand and analyze the nature of these contracts. Throughout this chapter we will deal with the following model. As usual we start with a fixed investment discount functionvand a life table. We have a contract issued on (x) with death benefit vectorb, annuity benefit vectorc and premium vector𝜋. (For simplicity we will omit the possibility of guaranteed payments in our discussion, but this feature can easily be incorporated if desired.) Recall from Section 5.5 that we can view the death benefits as a vector of annuity benefitsb ∗ w x, where (wx)k = v(k,k+1)qx+k. We can then form thenet cashflow vector f = 𝝅 −b ∗ wx −c, (6.1) which indeed represents the net cash flow on the contract from the insurer’s viewpoint. The insurerwillcollectpremiumsof 𝜋andpayoutdeathbenefitsintheformof b ∗ w x andannuity benefits ofc. The reserve at timekon the contract is the reserve for the vectorf with respect to the interest and survivorship discount function, as given in Definition 2.6. Denoting the reserve by kV,w eh a v e kV = kV(f;yx) =− Valk (kf;yx ). (6.2)‡ FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial INTRODUCTION TO RESERVES 79 Itisoftenusefultowritethisinanalternatewaythatkeepsbenefitsandpremiumsseparate. Thatis,wenotethatthereserveattime kisthe valueattimekoffuturebenefitslessthevalue attimekoffuturepremiums .Thiscanbewrittenintermsofthestandard Aand ̈asymbolsas kV = 1 yx(k) [Ax(kb)+ ̈ax(kc)− ̈ax(k𝝅)] (6.3) or, equivalently (and the way in which it usually appears in the literature) as kV = A[x]+k(b◦ k)+ ̈a[x]+k(c◦ k)− ̈a[x]+k(𝝅 ◦ k). (6.4) Here we are using (4.8) and the corresponding statement forA. Under the assumption that premiums are actuarially equivalent to benefits, we can also calculate the reserve retrospectively as Bk(f;yx) = 1 yx(k) [̈ax(k𝝅)− ̈ax(kc)−Ax(kb)]. (6.5) Underthisformulationthereserveisthevalueattime kofthepastpremiumslessthevalueat time kof past benefits. The reader is cautioned that the reserve cannot be so calculated when premiums and benefits are not actuarially equivalent. Example 6.1 For the policy of Example 5.5, find the reserves at time 1 and time 2. See Figure 6.1. As in Chapter 2 we use an arrow to mark the point at which values are computed. Benefits 80 70 75 Time 210 Premiums 1616 v s 1/21 /2 qs 0.20 .4 Figure 6.1 Example 6.1 Solution. Value at time 1 of future death benefits=75(1/2)(0.4) =15. Value at time 1 of future annuity benefits=70(1/2)(0.6) =21. Value at time 1 of future premiums=16. 1V = 15+21−16 = 20. Theaboveapproachisrecommendedforhandcalculation.Allreservecalculationscanbe handledinexactlythesameway,althoughtherewillgenerallybemorethanonesummandin each of the three items. 80 INSURANCE AND ANNUITY RESERVES Additional information can be obtained by first computing the net cash flow vector, which also provides the best procedure for spreadsheet calculation. To illustrate, the vector w 60 = (0.1,0.2, −). (Wedon’t have informationtocompute the thirdentrybut itisirrelevant, since it is multiplied by 0.) Then,b ∗ w60 = (8,15,0), c = (0,0,70), 𝝅 = (16,16,0), so that the net cash flow vectorf is (8,1, −70). For present purposes, we can forget about the particular death benefits and premiums. From the insurer’s viewpoint the contract can be viewedassimplyoneofcollecting8attime0,1attime1,andpayingback70attime2.From (6.2), 1V is just the negative of the value at time 1, with respect to interest and survivorship, of the payments at times 1 and 2. This equals−1 +70(0.5)(0.6)=20, as above. Similarly,2V is just the value at time 2 of the payment at time 2 which is 70. In general, foracontractrunningfor nyears,thereserveattime nisjustthepaymentdueattime ntothe survivors. (Recall that from the convention introduced in Chapter 1, reserves are calculated before this payment.) Forn-year term insurance, when there is nothing payable to survivors at the end, thenth year reserve will equal to 0. Since we have used net premiums, we can calculate balances as a check. B 1(f) = 8 y60(1) = 8 0.4 = 20, B2(f) = 1 y60(2)̈a60(2f) = 1 0.12[8+0.4] = 70, which agree with our previous calculations. The following example exhibits a point of interest. Example 6.2 For a 4-year endowment insurance on (60),b2 = 100,b3 = 200 and there is a pure endowment of 200 paid at age 64 if the insured is then alive. You are given that 𝜋 2 = 10,𝜋3 = 20, q62 = 0.1. The interest rate after 2 years is 25%. Find 2V and 3V. See Figure 6.2. Benefits 100 200 200 Time 43210 Premiums 2010 v s 0.80 .8 qs 0.1? Figure 6.2 Example 6.2 Solution. Value at time 2 of death benefits= 100(0.8)(0.1)+200(0.64)(0.9)Value at time 2 of premiums= 10+20(0.8)(0.9) = 24.40. THE GENERAL PATTERN OF RESERVES 81 We don’t knowq63, but it is not needed. Sinceq63 +p63 = 1, the two unspecified terms sum to 200(0.64)(0.9)=115.20 and 2V = 80+115.20−24.40 = 170.80. Similarly,3V = 200(0.8)−20 = 140. Thisexampleshowsthatforanendowmentinsurance(withbenefitspaidattheendofthe yearofdeath)wherethepureendowmentisthesameamountasthefinaldeathbenefit,wedo not need toknow themortalityrate forthe final year. This isin fact evident, sincethe insured gets that amount whether they live or not. Remark There is a subtle point involved with Equation (6.4), which is important to note fortheactualcalulationofreserves.WehavealreadyintroducedtheideainSection2.10.3.In ourmodelwechooseaninvestmentdiscountfunctionandamortalitytable,andtheseremain fixedforthedurationofacontract.Inpractice,whenonereachestime kandactuallywantsto compute kVtheseassumptionsmaywellhavechanged.Thereservecomputedattime kwillbe A′ x+k (b◦ k)+ ̈a′ x+k (c◦ k)− ̈a′ x+k (𝜋◦ k) wheretheprimesindicatequantitiesthatarecalculatedwiththenewinterestandsurvivorship functiony ′ x+k ascomputedunderwhatcouldbechangedconditionsattime k.Ofcoursewhen there is no change in assumptions,y′ x+k will just equalyx◦ k and formula (4.6) shows that bothformulasarethesame.Thepointisthen,thatinpractice,onedoesnotreallyknowwhat kV will be before timek. In this more realistic setting, we can view the original formula (6.4) as the best estimate one could make ofkV, if asked to compute it at time 0. 6.2 The general pattern of reserves Are insurance reserves generally positive or negative? Paradoxically, we will motivate the answer to this question by providing an example where they are neither. Example 6.3 An insurance policy provides a death benefit of 1 paid at the end of the year of death. Level annual premiums are payable for life. The interest rate is constant and the value ofq x is a constantqfor allx. Find the reserves and give an explanation for the answer. Solution. Let p= 1−qbe the constant value ofpx. Thenkpx = pk, which is never 0. There is a positive probability of living to any age, so we have an example in which𝜔 =∞ .O u r vectors will be of infinite length. Let vdenote the constant value ofv(k,k+1). Thenwx is a vector with a constant entry of vqand the premium pattern vector𝜌 is a vector with a constant entry of 1. The net level premium 𝜋 is clearlyvq, since that will make the vectorswx ∗ b and 𝜋 not only actuarially equivalent but actually equal to each other. The net cash flow vectorf is then not just a zero-value vector butactuallyequalto the zero vector. All reserves will be 0. Whatishappeninghereisthatthepremiumof vq,collectedeachyear,willaccumulateto qat the end of the year, and this will be exactly sufficient to pay the death benefits due at that time. There will be nothing left over, so balances, and therefore reserves, equal zero. 82 INSURANCE AND ANNUITY RESERVES Thesituationaboveistypicalformanyformsofinsurance,suchasautomobileorproperty coverage, and insurers of such risks have little in the way of reserves. The given scenario is, however, not realistic for life insurance. The values ofq x are not constant but increase with age. If one paid for the insurance 1 year at a time, the yearly premium per unit ofvqx would rapidly increase and eventually become prohibitively high. As we noted in Chapter 5, the typical life insurance policies will level the premiums out. In most cases, policyholders are paying more in premiums than they need to in the early years, but not enough in the later years. At any point of time after time zero, future premiums will not be sufficient to cover the remaining benefits. The excess collected in the early years is used to cover this deficit. It is expected therefore that reserves are usually positive. This has important implications for the life insurance industry. It means that investing becomes a major activity, as life insurance companiestendtoaccumulatelargeamountsofassets.Somecritics,withlittleunderstanding of insurance, look at these holdings of real estate, stocks and bonds, and claim that they represent unfair profits made at the expense of the policyholder. The truth is, however, that a large portion of these assets represent reserves, which in effect belong to the policyholders, as they will be used to pay the future benefits. Negativereservescanariseonpolicieswherethecostoftheinsurancebenefitsisdecreas- ingeachyear.Anexampleisapolicywitharapidlydecreasingbenefitamount,wheredespite theincreasein q x,thequantity bkv(k,k+1)qx+k decreases.Someexamplesappearintheexer- cises. Similarly, negative reserves can arise in the case where the premiums increase rapidly rather than remaining level. Insurers try to avoid such a situation if possible. A negative reservemeans,viewingthingsprospectively,thatthepolicyholderowesmoneytotheinsurer, whichwillbeprovidedbyfuturepremiums,orequivalently,lookingatthingsretrospectively, that the policyholder has received coverage but not yet paid for it. The problem is that the policyholder may stop paying premiums on the policy, leaving an unpaid debt. 6.3 Recursion In this section, we develop some important recursion formulas. It is convenient to make a slightalterationinnotation.Wewillincorporatetheannuitypaymentswiththepremiumsand let 𝝅 denote 𝝅 −c. In other words, we think of annuity benefits as just negative premiums, which is really what they are, since the policyholder is receiving rather than paying these amounts. Our net cash flow vectorf has entriesf k = 𝜋k −bkv(k,k+1)qx+k, and from our basic recursion formula (2.26), k+1V = ( kV+𝜋k −bkv(k,k+1)qx+k )yx(k+1,k). (6.6) Since yx(k+1,k) = (1+ik)∕px+k this is sometime written as k+1V = (kV+𝜋k)1+ik px+k −bk qx+k px+k . (6.7) Therecursionisstartedwiththeinitialvalue 0V =− ̈ax(f)whichwillbe0underourstandard assumption of net premiums. DETAILED ANALYSIS OF AN INSURANCE OR ANNUITY CONTRACT 83 It is instructive to note that the last term ofqx+k∕px+k is equal todx+k∕𝓁x+k+1. From this, weseeeasilythatitistheamount,perunitofdeathbenefit,thateachsurvivormustpayatthe endoftheyear to provide the benefits paid to those who died during the year. Formula (6.7) takes a retrospective viewpoint, and says that the reserve at timek+1i s obtainedfromthatattime k,byaddingthepremium,accumulatingatinterestandsurvivorship for 1 year, and then subtracting enough to pay the death benefits. It is known as theFackler reserveaccumulationformula,namedafteroneoftheearlyNorthAmericanactuaries,David Parks Fackler. In pre-computer days it was a popular method for calculating reserves. It is used infrequently for calculating purposes now, but is still useful for illustrating how the life insurance reserve changes from one period to the next. Remark Thequantity kV+𝜋k intheaboveformulaisoftencalledthe initialreserve attime kas it represents the reserve at the beginning of the year, after the payment of the premium. In contrast,kV is sometimes referred to as theterminal reserveat timek, reflecting the fact thatitthereserveattheendoftheyear,priortothepaymentofthepremiumforthefollowing year. Alternateversionsofthisformulaprovideinstructiveinformation.Wefirstgiveanimpor- tant definition. Definition 6.1 The quantityb k − k+1V is known asthenetamountatrisk for the (k+1)th yearandwillbedenotedby 𝜂k (thesubscriptischosentocorrespondto bk).(Therearevarious other names in the literature, such asdeathstrainatrisk.) Now, multiplying (6.7) bypx+k = 1−qx+k and rearranging, we get k+1V = (kV+𝜋k)(1+ik)−qx+k𝜂k. (6.8) Formula (6.8) reflects the fact that we can also view the accumulation of funds on an insurance policy as an interest-only investment, rather than as an interest and survivorship investment.Fromthispointofview,thepolicyholder keepsthereservewhenhe/shedies(the amount accumulated in her box), but then the insurer only needs to make up the difference as a death benefit. The insurer is therefore at risk only for the difference between the death benefit and reserve, which is the source of the name. Readers who looked at Section 2.11 will note that this is a special case of the change of discount function that we investigated there. In this casey x and bk are replaced byvand 𝜂k, respectively. 6.4 Detailed analysis of an insurance or annuity contract In this section we use (6.8) to provide a detailed discussion of the workings of an insurance policy. 6.4.1 Gains and losses In practice, interest and mortality rates will not conform exactly to those provided by our model. In a particular year, the insurer may earn more interest than predicted by the given 84 INSURANCE AND ANNUITY RESERVES discount function, which will result in gains. For an insurance contract, there may be fewer deaths than predicted by the given life table, also causing gains. If the insurer earns less interestthanexpected,and/ortherearemoredeathsthanexpected,therewillbelosses.Inany year,theactuarywantstoanalyzethesegainsorlossesandseehowmuchisduetoinvestment earnings and how much is due to mortality. Suppose we wish to measure the gain for a particular policyholder over the period run- ning from timek to timek+1. At the beginning of the year, before premium payment the policyholder’s box will contain the amount kV. At the end of the year the insurer must make certain that the box hask+1V in order that future obligations can be met. Anything in excess of that amount can be considered as a gain, taken out and added to general surplus funds. On the other hand, if there is less than k+1V, the insurer will have to make up the deficit from generalsurplusfundsandtherewillbealoss.Wewillderivesomegeneralformulas.Suppose the actual interest rate earned during this year wasi ∗ k rather thanik, and the actual rate of mortality wasq∗ x+k rather thanqx+k. Then, the actual amount accumulated at timek+1 will be the right hand side of (6.8) with starrediand q. If we subtract the reserve, we obtain the total gainGk from that policy for that year as Gk = (kV+𝜋k)(1+i∗ k ) −q∗ x+k 𝜂k − k+1V. (6.9) If we substitute fork+1V with the left hand side of (6.8) we can write this as Gk = (qx+k −q∗ x+k )𝜂k +(i∗ k −ik )(kV+𝜋k). (6.10) which gives us a decomposition of the gain by source. The first term gives the gain due to mortality, and the second term gives the gain due to interest. That is, the mortality gain is the difference between the expected and actual mortality rates times thenetamountatrisk.T h e interest gain is the difference between actual and expected interest rates times the amount of funds at the beginning of the year, after payment of the premium. Example 6.4 Refer back to the policy of Example 6.1. Suppose that in the first year of the policy, the interest earned was 50% instead of the predicted 100%, and the actual rate of mortality was 0.1 instead of the predicted 0.2. Find the total gain for the year, split into the portion due to interest and the portion due to mortality. Solution. Substituting directly from (6.9), the mortality gain is (0.2−0.1)(80−20) = 6, while the interest gain is (0.5−1)(0+16) =− 8. There is a total gain of−2, or in other words a loss of 2, for each policy. We will verify this by working out an example in the aggregate. Suppose that 10 people age60buythispolicyatacertaintime.Theinsurercollects160,andthisaccumulatesat50% DETAILED ANALYSIS OF AN INSURANCE OR ANNUITY CONTRACT 85 interest to 240 at the end of the year. Out of this, the insurer will pay a death benefit of 80 for the one death that occurred, leaving a total of 160. They have to put aside a total of 180, which is the reserve of 20 for each of the 9 survivors. There is a aggregate shortfall of 20, or 2 from each policy, which has to be drawn from surplus. (Note that the gain given by (6.8) is for each policy in force at thebeginningof the year, not at the end.) Formula (6.9) assumes that the premium paid is the net premium, calculated as in Chap- ter 5. In practice, the premiums actually charged on a policy will normally be different from the valuation premiumswhich are net premiums determined from the interest and mortality assumptions used to compute reserves. (See Section 6.5 for more detail on this.) This neces- sitatesanadjustmentinouranalysis.Supposethepremiumpaidattime kisactually 𝜋 ∗ k rather than valuation premium𝜋k. The first term on the right of (6.8) is then ( kV+𝜋∗ k )(1+i∗ k ) = (kV+𝜋k)(1+i∗ k ) +(𝜋∗ k −𝜋k )(1+i∗ k ), which leads to an extra source of gain or loss. We now have Gk = (qx+k −q∗ x+h )𝜂k +(i∗ k −ik )( kV+𝜋k ) +(𝜋∗ k −𝜋k )(1+i∗ k ). (6.11) Thethirdtermrepresentsthegainorlossarisingfrompremiumsthatdifferfromthevaluation premium. (We will elaborate on this point in Section 12.4.) One application of the formulas in this section is todividend calculation. Insurers fre- quentlyissuewhataretermed participatingpolicies,inwhichgainsresultingfromfavourable investmentandmortalityexperiencearereturnedtothepolicyholderintheformofdividends. We will not go into further detail on this topic, but note that formula (6.10) is a basic tool in computing these amounts. Anotheruseofgainandlossanalysisistodeterminehowchangesinthebasisassumptions will affect the reserves. As an example, suppose that the reserve interest rate decreases. It is clear that valuation premiums will increase. In order to provide the same benefits with decreased earnings from investments, more must be collected from the policyholder. On the otherhand,itisnotclearhowthiswillaffectreserves,sinceatanytime,boththepresentvalue of the future benefits and the present value of the future premiums will increase. Indeed, the effectwilldependonthenatureofthepolicy.Intheusualcasehoweverwithalevelpremium 𝜋 andreservesthatincreasewithtime,theloweringoftheinterestratewillincreasereserves. To see this, suppose that interest rates are constant and there is a change of rate fromi to i ∗ < i. This will cause an interest loss of (kV+𝜋)(i−i∗) for the year running from timekto k+1.Sincethereservesareincreasingwithtime,thelosseswillalsobeincreasingwithtime. Therewillbeofcourseanewlevel valuation premiumof 𝜋 +Δ tocover theselosses.Dueto theincreasinglosses,itmustbethat Δisgreaterthanthelossesinearlyyearsandlessinlater years.Intheearlyyears,thatportionof Δnotusedtocovertheloss,willbesetasideasanextra reserve,tocoverthegreaterlossestocomeinthefuture,andthiswillcausereservestoincrease. 6.4.2 The risk–savings decomposition Wenowlookatausefuldecompositionofthepolicyintoariskportionandasavingsportion. Multiply Equation (6.8) byv(k,k+1) and rearrange to obtain 𝜋 k = v(k,k+1)qx+k𝜂k +[v(k,k+1)k+1V−k V]. 86 INSURANCE AND ANNUITY RESERVES This formula decomposes each premium into two parts. The first term is known as therisk portion of the premium, as it is that amount needed to buy insurance for 1 year for the net amount at risk. The remainder provides for the difference in reserves and is known as the savingsportion of the premium. Example 6.5 Find the decomposition for the premiums of Example 6.1. Solution. In the first year the net amount at risk is 80−20 = 60. The risk portion of the premium is then 1 2 ×0.2×60 = 6. The savings portion is1 2 ×20−0 = 10. (As a check, the two portions add up to the total premium.) In the second year the net amount at risk is 75−70 = 5. The risk portion of the premium is 1 2 ×0.4×5 = 1 and the savings portion is 1 2 ×70−20 = 15. This decomposition shows that any policy can be viewed as being composed of two separate policies, the pure insurance part and the savings part. For the pure insurance part, the premium is the risk premium and the death benefit paid is the net amount at risk. This part of the policy has zero reserves, as the risk premium is just sufficient to purchase coverage for the net amount at risk for 1 year. In the above example, thepolicyholderpays6inthefirstyear,whichisexactlyenoughtopurchasethecoveragefor the net amount at risk of 60. He/she then pays 1 in the second year, which is exactly enough to purchase coverage for the net amount at risk of 5. Thesavingspartofthepolicyoperatesjustlikeabankaccount,withamountsaccumulating at interest only. In the example above, the savings portion of 10 from the first premium accumulatesto40attime2andthesavingportionof15fromthesecondpremiumaccumulates to30attime2.Thetotalsavingsof70arethenpaidoutasthepureendowmenttoallsurvivors at time 2. This is typical of endowment policies, including whole life, which operates as an endowmentatage 𝜔,aswehaveindicated.Theaccumulatedamountsfromthesavingsportion of the premium increase steadily to the pure endowment amount. It is instructive to compare this with term insurance. Example 6.6 Redo Example 6.1 for the corresponding 2-year term policy without the endowment. That is,b still equals (80, 75, 0), butc = 0. Solution. In this case the premium𝜋 will equal (8+15(0.4))∕1.4 = 10, and we havef = (2,−5,0).Then 1V = 5andweknowfromthediscussionfollowingExample6.1that 2V = 0. Inthefirstyear,thenetamountatriskis75,sotheriskportionofthepremiumis 1 2 ×0.2×75 = 7.5, and the savings portion is 2.5. In the second year the net amount at risk is 75, so the risk portion of the premium is 1 2 ×0.4×75 = 15 and the savings portion is−5. This shows the typical savings pattern on term policies. Modest savings are built up in early years, but these must be drawn on in later years when the premium is insufficient to pay for the insurance, resulting in a negative savings portion. As a check, the 2.5 deposited into the savings fund at time 0 will increase to 5 at time 1, which will then be withdrawn to make up for the deficit in the premium payable at that time. Exactlythesameanalysisasaboveholdsforlifeannuityaswellaslifeinsurancecontracts. In this case, the death benefits are all of zero amount, so that the net amount at risk will be negative.Thisisnaturalenoughandreflectsthefactsthatextradeathsinthecaseofannuities BASES FOR RESERVES 87 result in gains. On single-premium annuities, we have𝜋k =− ck < 0, fork> 0. Examples appear in the exercises. 6.5 Bases for reserves Choosingtheappropriatediscountfunctionandlifetableforthepurposeofreservecalculation is another complex subject that we will only discuss briefly. As mentioned, the insurer must ensure that it has sufficient assets to cover their reserves, for if not, it will be in danger of being unable to meet future obligations. It is usually thought that reserve calculations should use conservative assumptions, so there is a built-in safety margin should experience prove to be adverse. In other words, the discount function would use somewhat lower interest rates thanactuallyexpectedandthemortalitytablewouldshowmoredeathsthanactuallyexpected (orfewerinthecaseofannuitycontracts).Inseveraljurisdictions,thebasesusedforreserves are specified by insurance regulatory bodies, whose main goal is to ensure protection for the policyholders. The following example illustrates the resulting effect on profitability. Example 6.7 For the policy in Example 6.1, the company is required by legislation to compute reserves using a 50% interest rate rather than 100%. It actually does achieve the estimated100%returnonitsinvestments,andmortalityfollowsthepredictedratesexactly.It stillchargesthepremiumof16basedontherealisticinterestrateof100%.Analyzetheeffect of the conservative interest rate assumption on the company’s gains and losses. Solution. Redoing the calculations with an interest rate of 50% in place of 100% leads to a valuation premium of 544/23 and a first year reserve of 560/23, as the reader can verify. We willdoanaggregateanalysis.Supposetheinsurersells2300policiesatage60.Itwillcollect totalpremiumsof36800,whichwillaccumulatewithinterestto73600attheendoftheyear. Outofthisitwillpay460peopleadeathbenefitof80units,foratotaldeathbenefitpayment of 36 800. This leaves 36 800. It now must set up a total reserve of 44 800, consisting of 560/23foreachofthe1840survivors.Thereforethelossshownforthefirstyearofthepolicy is 8000. Looking at formula 6.11, the large loss in the third term, more than offsets the gain from the second term. In the second year, it starts with a reserve of 44 800. It collects another 16-unit premium from each of the 1840 survivors for a total of 29 440. This leaves a total of 74 240, which accumulates with interest to 148 480. Out of this it must pay a death benefit of 75 to each of the736 = 1840×0.4peopleremaining1104peoplewhosurvived.Thetotalbenefitpayments are 132 480, which leaves a gain for the year of 16 000. For this group of policies, the insurer will show a loss of 8000 for the first year, and a gain of 16 000 for the second year. If the insurer had used the realistic interest rate of 100% interest, there would have been no gains or losses. In effect the more conservative reserve requires the insurer to borrow 8000 from surplus at the end of the first year, and then repay it with the 16 000 at the end of the second year, which is what the amount should be in view of the 100% interest rate. This example shows that for our present model, with a single discount function, reserve assumptions do not affect the ultimate profitability of the insurer, since this depends solely on what actually happens. It can, however, change the incidence of this profit from year to 88 INSURANCE AND ANNUITY RESERVES year. This means that reserve assumptions can have an effect on profitabiltly when there are different interest rates involved, and we discuss this further in Section 12.4.3. 6.6 Nonforfeiture values Whatshouldhappentopolicyholderswhostoppayingpremiumsbeforethetermstatedinthe contract?Thisisknowntechnicallyas withdrawalorlapse,orsurrender.Inconsiderationfor the premiums they have already paid, they should be entitled to some reduced benefits under the policy. These are known asnonforfeiture benefitssince they are benefits that were not forfeitedbythecessationofpremiums.Infact,inoursimplifiedmodel,theyshouldbeentitled to take the reserve on their policy at any time they wish. Looking at this retrospectively, the reserveistheexcessoftheaccumulatedamountoftheirpremiumsovertheaccumulatedcost oftheinsuranceprotectionthattheyhavereceived.Itwillbetheamountthattheyhaveintheir ‘box’. In practice, insurers pay an amount that is somewhat less than the reserve for several reasons. This is a complex topic that we will only comment on briefly here. One reason is the high incidence of expenses in the early years of the policy (we discuss this more fully in Chapter 12). While these are accounted for by adding an amount to the premiums, the total amount of the initial expenses may not be recovered at the time of withdrawal. Another factor is the phenomenon known asanti-selection. This is a well-established concept in insurance which is simply a recognition of the fact that policyholders will make choices according to their own self-interest, acting on knowledge that they have, but that the insurer may not have. (The prefix ‘anti’ refers to the fact that it is the policyholder doing the selectingagainsttheinsurer.)Onalifeinsurancecontract,theoptiontowithdrawislesslikely to be exercised by an unhealthy policyholder than a healthy one. After all, if someone is told theywilldiewithinafewmonthsfromaterminaldisease,theywouldbefoolishtogiveupthe policy. Consequently, the group that doesnotwithdraw can be expected to experience higher mortalitythannormal.Thereisthereforeananti-selectionexpensetowithdrawal,intheform of these higher mortality rates of the remaining policyholders. The principle followed here is that this expense should be borne equitably by all the policyholders, not just by those who remain. This is done by adjusting the amounts paid out in the case of withdrawal. The cash amount that will be paid to the withdrawing policyholder on a life insurance contract is known as thecashsurrendervalue and is usually guaranteed at the time of issue for all durations. Normally, the policyholder is given the option of taking the nonforfeiture benefitsintheformofareducedlevelofinsuranceratherthanincash.Thereductioncantake the form of either a reduced amount of benefits, or a reduced term for the same benefits. Lifeannuities alsopresent anobvious possibilityforanti-selection. Unhealthy annuitants would find it worthwhile to end the contract, take their reserve, leaving a group who could be expected to live longer than expected, and mortality losses would result. For this reason, nonforfeiturebenefitswouldnotbeofferedonsingle-premiumlifeannuitycontracts,oncethe benefitscommence.Dependingonthecontract,theymightbepresentduringadeferredperiod. 6.7 Policies involving a return of the reserve At the time of death, the policyholder (or, more accurately, the estate of the policyholder) receives the death benefit. Had he/she decided to lapse the policy an instant before death, he/she would have received an amount close to the reserve. Some people, who do not have POLICIES INVOLVING A RETURN OF THE RESERVE 89 a complete understanding of life insurance, have raised the complaint that the company is confiscating the person’s reserve, since it is only returning the death benefit, and not the reserve, at death. The answer to this is that one must adopt one of two points of view. One can view the insurance policy as an interest and survivorship investment. In that case, it is true that the reserve is taken and spread among the survivors, but this is completely fair, as discussed in Chapter 4. Alternatively, one can view the policy as an interest-only investment. In this case the reserve is indeed available at death. But now, one must view the death benefit as the net amount as risk, rather than the originally stated amount. Itispossibletodesignapolicywhereafixedamount,plusthereserve,ispaidatdeath.The policyholder might then be told that the reserve is being paid in addition to the death benefit. This is of course just playing with words. What you really have is a policy with the pattern of death benefits worked out so that when you subtract the reserve you get some prescribed amount. Consider the following example. Example 6.8 A policy on (x) provides for a payment, at the end of the year of death, of 1plusthereserve,shoulddeathoccurwithin nyears.Levelannualpremiumsof Parepayable for nyears. Find a formula forP. Solution. A direct solution of this problem by the method outlined in Chapter 5 will cause difficulties, sinceP depends on the reserves, but the reserves in turn depend onP. While it maybepossibletosolvetheresultingequationsforsmallvaluesof n,itismuchbettertotake the interest-only view for the accumulation of money. That is, we use the discount function vin place ofy x and the net amount at risk𝜂k in place ofbk. Expressing the insurance as an annuity, we wish to solve P̈a(1n;v) = ̈a(𝜂 ∗ wx;v) where 𝜂 is the vector (𝜂0, 𝜂1,…,𝜂n−1). While this could be done on any policy, it would normally be completely impractical, since we would not know the net amounts at risk in advance. In this case, however, it works perfectly. We are given thatb k = 1+k+1V, so that 𝜂k = 1 for allk. This leads to P̈a(1n;v) = n−1∑ k=0 v(k+1)qx+k, and we easily solve forP. For another application of this idea, we can give a more formal solution to Example 5.8. We calculate the balance at time 25 and equate it to the reserve at time 25, which we know is ̈a {40}+25. To calculate the balance, we use the discount functionvand death benefits of𝜂k. In this case, the actual death benefit is just equal to the reserve, so that𝜂k = 0 for allk.T h e balance isjusttheaccumulated value of thepremiums atinterest, whichisPv(25,0) ̈a(125;v), as we had before. Notethatreservecalculationbyrecursionisquitesimpleinthistypeofpolicy.Onesimply uses (6.8) where now𝜂k is just the face amount of the policy. 90 INSURANCE AND ANNUITY RESERVES 6.8 Premium difference and paid-up formulas Consider any policy on (x) with death benefit vectorb and with level annual premiums of P payable forh years, so that the premium vector𝜋 = P(1h). In this case, there are some other formulas that are useful for providing additional insight into the nature of balances and reserves. Throughout this sectionPis arbitrary and not necessarily the net premium. 6.8.1 Premium difference formulas Fixaduration k< h.LetP s bethelevelpremiumthat shouldbechargedforapolicywiththe same remaining benefits, if issued at timek to a person agex+k. That is, the value at time k of these new premiums should equal the value at timekof the death benefits after timek. Equating values at timek, A[x]+k(b◦ k) = Ps̈a[x]+k(1h−k). (6.12) SincekV = A[x]+k(b◦ k)−P̈a[x]+k(1h−k), we substitute in (6.10) to get kV = (Ps −P)̈a[x]+k(1h−k). (6.13) Formula(6.11)isknownasthe premiumdifferenceformula forreserves.Thequantity Ps −P is the difference between what should be charged and what is actually charged after timek. The value at timek of this yearly deficit over the remaining premium payment period gives the reserve. We can also obtain a retrospective premium difference formula. LetPc be the premium thatcouldhave been charged to provide the benefits that were provided up to timek. That is, Ax(kb) = Pc̈ax(1k). (6.14) Since Bk = [P̈ax(1k)−Ax(kb)]∕yx(k), we substitute in (6.14) to get Bk = (P−Pc)̈ax(1k) yx(k) . (6.15) The quantityP−Pc is the difference between what was actually charged and what could have been charged up to timek. The accumulated amount of this yearly excess gives the balance. WhenPis a net premiumBk = kV and (6.15) gives another formula for the reserve. 6.8.2 Paid-up formulas Substituting for the annuity rather the insurance in (6.12) gives kV = ( 1− P Ps ) A{x}+k(b◦ k). (6.16) STANDARD NOTATION AND TERMINOLOGY 91 Similarly, from (6.15), Bk = ( P Pc −1 ) Ax(kb) yx(k) . (6.17) To interpret formula (6.16), note that the fraction of the future benefit that is purchased by the actual premiums isP∕Ps. For example, ifPs = 36 andP= 24 then the policyholder is only paying two-thirds of what they should be for the future benefits. The difference of (1−P∕P s) must be that portion of the future benefits that has already been provided by the excess past premiums, so that multiplying this ratio by the present value of future benefits gives the reserve. In formula (6.17) the ratioP∕P c −1 is that portion of the past benefits that were purchased but not needed. For example, ifP= 24 andPc = 16, then the policyholder has paid 1.5 times what they could have paid to provide those benefits. Multiplying this ratio by the value of these past benefits gives the balance. Formula (6.16) is known as thepaid-upformula. Suppose a policyholder lapses at time k and is given nonforfeiture benefits equal in value to the reserve. If the individual elects to takepaid-upinsuranceforareducedamount,thisformulashowsthattheappropriatefraction is (1−P∕P s). 6.8.3 Level endowment reserves Wecanusethepremiumdifferenceformulatoderiveaverysimpleexpressionforreserveson level endowment insurance where a net level premium is paid for the full period and interest is constant. At durationkof ann-year contract, we know from (5.10) that P= 1 ̈ax(1n) −d, Ps = 1 ̈ax+k(1n−k) −d, and, substituting in (6.13), kV = 1− ̈ax+k(1n−k) ̈ax(1n) , (6.18) reducing the calculation of reserves on such a policy to the calculation of annuity values. 6.9 Standard notation and terminology The standard notation for reserves closely follows the notation for annual premiums as describedinSection5.8.Thebasicsymbolforthereserveattime kis kVaswehaveadopted. This is embellished in exactly the same way as the symbolPwas for annual premiums, with one exception. The premium payment period is moved to the upper left, from the lower left, since the latter is now used for the duration. If the upper left is empty, it signifies again that level premiums are paid for the natural duration of the contract. The following are some examples. 92 INSURANCE AND ANNUITY RESERVES kVx isthereserveattime kfora1-unitwholelifepolicyon( x),withlevelannualpremiums paid for life. The basic prospective formula (6.4) for this policy in standard symbols reads kVx = Ax+k −Pẍax+k. The corresponding retrospective formula is kVx = Pẍsx: k −A1 x: k (1+i)k kpx . To simplify retrospective reserve formulas, a symbol for the last term was introduced. Let tkx = A1 x: t (1+i)t tpx . This is often called theaccumulated cost of insuranceand denotes the single premium that each survivor would pay per unit of death benefit fort years, if this single premium were collected attheend of thek-year period, rather than at the beginning. (This would never be done in practice as it is not feasible to charge people at a time when they have no chance of collecting.) tV1 x: n is the reserve at timet for a 1-unit,n-year term policy on (x) with level annual premiums payable fornyears. Fort≤ n, the prospective formula for this quantity is tV1 x: n = A1 x+t: n−t −P1 x: n ̈ax+t: n−t , while the retrospective formula is tV1 x: n = P1 x: n̈sx: t −t kx. h t Vx: n is the reserve at timetfor a 1-unit,n-year endowment insurance policy on (x), with level annual premiums payable forhyears. Fort≤ n, this is given prospectively as h t Vx: n = { Ax+t: n−t −hPx: n ̈ax+t: h−t if t< h, Ax+t: n−t if t≥ h, or retrospectively as h t Vx: n = {hPx: n ̈sx: t −t kx if t< h, hPx: n ̈sx: h (1+i)t−h t−hpx −t kx if t≥ h. kV(n|̈ax) denotes the reserve at timek for a deferred annuity onx providing 1 unit for life beginning at agex+n and with annual premiums payable forn years. Fork< n,t h e retrospective formula is the easiest and is given by tV(n|̈ax) = P(n|̈ax)̈sx: t . SPREADSHEET APPLICATIONS 93 Fork≥ n, the prospective formula is the easiest and is given by kV(n|̈ax) = ̈ax+k. 6.10 Spreadsheet applications For reserve calculations we add two columns J and K to Chapter 5 spreadsheet. In cell J10, we enter = I10−F10−L10 and copy down to get the net cash flow vector. Then in cell K10 we enter the formula =− SUMPRODUCT(E10 :E$129,J10 :J$129)∕E10, and copy down. This calculates kV in cell K 10+k, as the negative of the value of future net cash flows, divided byyx(k). (Division by 0 error terms will appear for large enough durations, but these occur after age𝜔 and can be ignored If one prefers, the formula in K10 can be suitably modified with an IF statement to replace them with a blank.) As a test calculate15V for the test problem given in Section 5.9. The answer is 333.16. Wenowhaveafinalspreadsheetthatwillcalculatepremiumsandreservesonallinsurance and annuity contracts without guaranteed payments. Here is a complete summary. INPUT FORMULAS Column D: 1 in cell D10, Section 2.14 formula in cell D11. Copy down. Column E: 1 in cell E10, Section 4.8 formula in cell E11. Copy down. Column F: Section 4.8 formula in cell F8. Column G: Section 5.9 formula in cell G8. Column H: Copy cell F8 to cell H8. Column I: Insert the formula=(F8 + G8)/H8 in I6. Copy Cell H8 to I8. Insert the formula =$I$6*H10 in I10 and copy down. Column J: Copy down the formula for cell J10. Column K: Copy down the formula for cell K10. INPUT DATA FOR EACH PARTICULAR PROBLEM Column B: Interest rates. Cell C1: Age at issue. Column F: Annuity benefit vectorc. Column G: Death benefit vectorb. Column H: Premium pattern vector𝜌. Column N: Life table. (For sample table, insert parameters in N3 and N4, Section 4.8 formula in N10. Copy down). 94 INSURANCE AND ANNUITY RESERVES OUTPUT ̈ax(c) in F8. Ax(b)i nG 8 . Present value of all premiums in I8. The premium vector𝜋 in column I. Reserves in column K. Exercises Type A exercises 6.1 A3-yearendowmentinsuranceon(60)providesfordeathbenefitspayableattheendof theyearofdeath.Thedeathbenefitis1000fordeathinthefirstyear,2000fordeathinthe secondyearand3000fordeathinthethirdyear.Inaddition,thereisapureendowment of 4000 paid at time 3 if the insured is then alive. This is purchased by three annual premiums,beginningatage60.Thefirsttwopremiumsareequalandthethirdisdouble the amount of the initial premium. You are givenq 60 = 0.10, q61 = 0.20, q62 = 0.25. The interest rates are 25% for the first 2 years and 100% in the third year. Find the initial premium. Find kV for k= 1,2,3. 6.2 Youaregiven q60 = 0.20,q61 = 0.25.Interestratesare20%inthefirstyear,25%inthe secondyearand50%inthethirdyear.A3-yearendowmentinsurancepolicyissuedto (60)providesfordeathbenefits,attheendoftheyearofdeath,of1000ifdeathoccurs inthefirstyearand2000ifdeathoccurs inthethirdorsecondyears.Inaddition,there is a pure endowment of 2000 paid at age 63 if the insured is then alive. Level annual premiums are payable for 3 years. Find the premium. Find kV,f o rk= 1,2,3. Do this first by the basic prospective reserve formula, and check your answers by using the recursion formula (6.7). 6.3 For a 3-year term insurance policy on (x), level premiums are payable for 3 years. The following data are given. kq x+k ik bk 00 . 1 00 . 2 05 0 0 0 0 10 . 1 50 . 2 52 0 0 0 0 20 . 2 00 . 3 01 5 0 0 0 (a) Find the vectoryx. (b) Find the net cash flow vector. (c) Using your answers to (a) and (b), compute1V and2V. (d) Explain briefly why your answers to (c) are negative. 6.4 A 10-year endowment policy on (40) has death benefits of 900, payable at the end of the year of death, should this occur within 10 years, plus a pure endowment of 900 at age 50 if the insured is then alive. Level annual premiums of 20 are payable for EXERCISES 95 10 years. The interest rate is a constant 50% andq48 = 0.25. You are not givenq49. Find tV for t= 8,9,10. 6.5 An insurance policy has level premiums of 20 payable for the duration of the contract. If the reserve is calculated with a premium of 15, then0V = 100. What is0V if the reserve is calculated with a premium of 16? 6.6 Refer to Exercise 6.2. (a) Decompose each premium into the risk portion and savings portion. (b) Suppose that during the second year of the policy the actual interest rate earned was20%ratherthan25%andtheactualrateofmortalitywas0.20ratherthan0.25. Find the per policy gain during this year from both interest and mortality. 6.7 RefertoExercise6.1.Decomposeeachofthethreepremiumsintotheriskportionand savings portion. 6.8 Foracertaincontractissuedatage60, 5V = 200,thepremiumpayableatage65is40, q65 = 0.20,thedeathbenefitpayableatage66fordeathbetweenage65and66is800, the interest rate for the sixth year of the contract (between age 65 and 66) is 20%. (a) Find 6V. (b) Decompose the premium payable at age 65 into the risk portion and the savings portion. (c) Suppose that during the sixth year, the actual interest rate earned was 25% instead of20%andtheactualrateofmortalityatage65was0.15insteadof0.20.Findthe gain or loss during this year, from interest and from mortality, for each policy in existence at the beginning of the year. Type B exercises 6.9 Refer again to Exercise 6.7. Explain briefly in words why the third premium has a negative risk portion. 6.10 A deferred life annuity on (60) provides for annuity benefits payable for 3 years, beginning at age 63, provided that (60) is alive. The first annuity payment is 1000, the second is 2000 and the third is 3000. Premiums are payable for 3 years, beginning at age 60. The second and third premiums are equal in amount and eachdouble the amount of the initial premium. If (60) dies before age 63, there will be, at the end of the year of death, a return of all premiums paid prior to death, without interest. You are given thatq 60 = 0.1, q61 = 0.2,q62 = 0.25,q63 = 0.3,q64 = 0.4. The interest rates are 25% per year for the first 4 years, and 50% for the fifth year. Find (a) the initial premium, (b) 1V and (c)4V. 6.11 A life insurance policy on (x) has level death benefits of 1000. A life insurance on (y) has exactly the same premiums and reserves as the policy on (x) for the first 10 years, but different death benefits. Suppose thatqy+k = 2qx+k for k= 0,1, …,9.Ifthe commonreserve 8V = 300,whatisthedeathbenefiton(y)’spolicyfortheyearrunning from time 7 to time 8? 96 INSURANCE AND ANNUITY RESERVES 6.12 Use formula (6.8) to derive formula (4.9) withk= 1. (Note that for annuity contracts the death benefits are 0, and the premiums are the negative of the annuity benefits.) 6.13 You are given thaẗa81 = 3.2,q80 = 0.10,i= 0.20. (a) Find ̈a80. (b) A single-premium whole life annuity on (80) provides for a constant benefit of 1 per year. Due to improvements in mortality, the actual mortality rate experienced during the first year of the contract is 0.07 rather than 0.1. On the other hand, the interest earned during that year was 22% rather than 20%. What is the total per contract gain for this year? 6.14 A 2-year term insurance policy on (50), with a constant death benefit of 1000, is purchasedbytwo-levelannualpremiums.Policyholderswhochoosetolapsethepolicy attime1willreceive 1V asacash value. Assumethat mortalityand interestfollowthe projectedpattern,andthatthepremiumchargedisthenetpremium.Suppose,however, that a typical group of people age 51 who have purchased insurance at age 50 can be divided into two groups. Half of them are healthy and half are not. The non-healthy group can expect to have twice as many deaths over the following year as the healthy group. Suppose thatq 51 = 0.03. What is the loss on each remaining policy for the second year, assuming: (a) Allofthehealthypolicyholdersatage51lapsethepolicyatthattime,andnoneof the unhealthy ones do; (b) One-half of the healthy policyholders, and one-quarter of the unhealthy policy- holders, lapse the policy at age 51. 6.15 For a certain whole life policy on (x), given by a death benefit vectorb and premium pattern vector𝜌, reserves are calculated according to two mortality tables, with the 1-yearprobabilitiesofdyingdenotedby q x+k andq∗ x+k ,respectively.Thesamepositive interest rates are used in both calculations, and in each case the premiums are the net premiums as determined by the particular table being used. Suppose the two ‘curves’ cross at one point. That is q x+k ≥ q∗ x+k , k= 0,1, …,n, qx+k ≤ q∗ x+k , k= n,n+1,…,𝜔−x−1. Assumethatdeathbenefitsarenonincreasing.Showthatthereserveattime nishigher for the starred rates (the steeper curve). 6.16 (a) Use (6.8) to derive the formula Ax+1(1n−1)−Ax(1n) = iAx(1n)−qx (1−Ax+1(1n−1)). (b) As an actuary for an insurance company, you receive an angry letter from a pol- icyholder. The person, age 50, has just purchased a single-premium 3-year term insurance policy with a constant death benefit of 1. The complaint is that the per- son’s friend, age 49, purchased a single-premium 4-year term insurance policy, EXERCISES 97 with the same death benefits, for a lower single premium. The letter writer argues that both policies provide coverage for the ages 50–53, but that the friend gets an extra year of coverage, for the year running from age 49 to 50. Therefore, it is claimed,thefriend’spremiumshouldbehigher,notlower.Whatisyourresponse? 6.17 Aspecial2-yearterminsurancepolicyon(70)istobepurchasedbyasinglepremium. Shoulddeathoccurinthefirstyear,theinsuredwillreceive,attheendoftheyear,1000 plusthereserveatthattime.Ifdeathoccursinthesecondyear,theinsuredwillreceive at time 2 only a return of the single premium paid without interest. Supposeq 70 = 0.36, q71 = 0.40, and the interest rate is 100%. Find (a) the single premium, (b)1V. 6.18 A whole life insurance policy on (x) provides for death benefits, paid at the end of the year of death, of 1000 plus the reserve at that time. Level annual premiums ofP are payable for life. Suppose thatqx = 0.2, qx+1 = 0.2 andqy = 0.4f o rally ≥ x+2. The interest rate is a constant 100%. Find (a)P(b)kV, k= 1,2, …. 6.19 Two people, A and B, are both agex. A buys a 20-year endowment policy, with a constant death benefit of 1, and a pure endowment of 1 at time 20, paying level annual premiums for 20 years. B buys a whole life policy with a constant death benefit 1, payinglevelannualpremiumsforlife,andinaddition,eachyear,investsthedifference between his/her premium and A’s premium, in a savings account. At time 20, the reserve on B’s policy, plus the amount in his savings account total 1. All premiums were calculated at a constant interest rate ofi. B earned a constant rate ofj on his investments. Isjgreater than, less than or equal toi?W h y ? 6.20 A whole life policy with a level death benefit of 10 000 carries net annual premium of 100payableforlife.Therateofdiscountisaconstant0.04.Attime n,thepolicyholder wishes to reduce their annual premium payment to 25, and is told they can do so, but the death benefit will be reduced to 7000. Assuming that the entire reserve is available as a nonforfeiture benefit, find nV. 6.21 Supposethatamortalitytableusedtocalculatereservesisalteredbyaddingapositive constant to each value ofqx. Explain how the reserves will be affected, for whole life policies with a constant death benefit and level premiums. Spreadsheet exercise 6.22 SupposethatmortalityfollowsthesamplelifetableofSection3.7andinterestratesare 5% in the first 15 years, 6% for the next 15 years, and 7% thereafter. A contract issued at 40 provides for a life annuity beginning at age 65. The annual annuity payment is 1000 for the first 10 years and 2000 thereafter for life. If death occurs before annuity payments begin, there is a death benefit of 10 000 paid a the end of the year of death. Level annual premiums are payable for 15 years. (a) Find the premium and all reserves. (b) Suppose the interest rate in the first 15 years increases from 5% to 5.5%. Do reserves increase or decrease? (c) If the interest rate in the first 15 years decreases to 4%, what happens to reserves? 7 Fractional durations 7.1 Introduction Up to now, we have considered cash flows where the payments were at integer times. We nowdealwiththecasewherecashflowscanoccuratfractionaldurations.Forexample,ifthe basic time unit was a year, and payments were made monthly, these would be at times that are multiples of 1/12. In practice this is a common occurrence. Purchasers of life annuities oftenwanttheincometobepaidmonthly.Manypurchasersofinsurancepolicieswishtopay premiums monthly, or possibly quarterly or semiannually. One obvious method of handling thisfeatureistochangethetimeunit.Ifwearedealingwithannuitieswithmonthlypayments, we could just take our unit of time as 1 month, and payments would be at integer times. This option was not always feasible in pre-computer days. For cash flows discounted at interest only, calculations were done from tables showing yearly rates of interest, and elaborate formulas were developed to handle the fractional durations. This is no longer necessary, and the change-of-period approach is the modern way to handle the valuation of cash flows at compound interest. For life annuities however it is still common to keep the time unit as a year. There are several reasons for this. In the first place, the year has been so ingrained as a measure of age that it seems hard to break away from this tradition. Another more important reason is that insurers often do not calculate premiums exactly for the different periods, but rely on approximate conversion formulas. For this purpose it is convenient to have some simple relationships established between the yearly annuity values and those with more frequent payments.Stillanotherreasonisthatwealsowanttoconsiderannuitieswithpaymentsmade continuously,whichwillbediscussedinChapter8.Todothis,itisconvenienttofirstdiscuss the case in which payments are madem times per year, and then view continuous annuities as a limiting case asm approaches infinity. We will follow the standard usage of referring to annuities with payments made at times which are multiples of 1∕mas mthly annuities or annuities payablemthly. Fundamentals of Actuarial Mathematics, Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial CASH FLOWS DISCOUNTED WITH INTEREST ONLY 99 We will restrict attention to sequences of cash flows where theperiodic payments during each year are constant,but can vary from year to year. If payments change during the year (e.g., they increase each month), then usually the only feasible approach is to change the period.Thebasicmodelisasfollows.Wearegivenapositiveinteger m,whichisthenumber of payments made each year. We are given a cash flow vector,c = (c 0,c1,…,cN) where, as before,ck isthe total amountpaidinyear k +1,thatis,betweentime k andk +1.Ratherthan being paid as a single cash flow however, there are payments of ck m at timek + j m, j = 0,1, …,m −1. (7.1) Let y denote an arbitrary discount function. We denote the present value of the sequence of cash flows as described bÿa(m)(c;y). The task at hand is to relate this quantity to the case where m = 1, namelÿa(c;y). Our first observation is that it is sufficient to find a formula for ̈a(m)(c;y) in the case thatc = e0, a single payment of 1 at time zero. This follows from the replacementprinciple.Let uk bethevalueattime k ofasequenceofpaymentsof1∕m attime k +(j∕m),j = 0,1, …,m −1. We can write uk = ̈a(m)(e0;y ◦k). The replacement principle tells us that we can replace all the payments between timek and k +1 by a single payment ofckuk at timek. It follows that ̈a(m)(c;y) = ̈a(c ∗ u;y). (7.2) whereu = (u0,u1,…,uk,…). Thekeytoevaluatingfractionalannuitiesistocalculatethequantities uk.Wewilldiscuss particular cases in the following two sections. 7.2 Cash flows discounted with interest only Supposethatwearegivenyearlyinterestrates,andwewanttoconsiderachangeofperiodto 1∕m of a year. We use primed symbols to denote the quantities applicable to the new period, that is,v ′,i′ and d′ in place ofv,i and d, respectively. We will start as usual by supposing that we are given the interest ratesik for each nonnegative integerk. To determine the rates applicabletothenewperiods,thestandardassumptionmadeisthatinterestratesareconstant over each year. To be precise, one postulates is that for all nonnegative integersk, (2.9) holds whenevers,t,s +h,t +h all lie in the interval [k,k +1). Leti ′ k denotetheconstantvalueof v(s +1∕m,s)−1andlet d′ k denotetheconstantvalueof 1−v(s,s +1∕m)f o rk ≤ s < s +1∕m < k +1. An investment of 1 at time 0 will accumulate to 1+i′ k at time 1∕mand then to (1+i′ k )2 at time 2∕m,( 1+i′ k )3 at time 3∕mand so on. Arguing in the same way for the discount rate gives (1+i′ k )m = 1+ik, (7.3)‡ (1−d′ k )m = 1−dk, (7.4)‡ 100 FRACTIONAL DURATIONS since in (7.3) each side of the equation is the value at timek +1 of 1 unit invested at timek, and in (7.4) each side of the equation is the value at timek of 1 unit paid at timek +1. In standard actuarial notation it is common to ‘annualize’ these rates by multiplying by m. That is, one defines i(m) k = mi′ k , d(m) k = md′ k. (7.5) The quantitiesi(m) k and d(m) k are often known asnominal rates, to reflect the fact that they are rates in ‘name only’, and not really applicable to any period. They are sometimes referred to as the yearly interest (respectively discount) ratesconvertible or compounded mthly. The actual rate applicable to eachmthly period in the yeark to k +1 is found by dividing these nominal rates bym. Inthecaseofconstantinterestthesubscript k isnotneededandwesimplyusethesymbols i (m) and d(m) for the nominal rates. Example 7.1 Suppose thati(m) = 0.12. Findi, the yearly rate of interest, if (a)m = 2, (b) m = 12. Solution. (a) We must divide by 2, to deduce that the rate per half-year period is 0.06. From (7.3), (1+i) = 1.062 = 1.1236 so that the corresponding yearly interest rate is 12.36%. (b) Similarly, the corresponding yearly interest rate= (1+0.01)12 −1 = 12.68%. Itisimportanttodistinguishbetweentheactualyearlyratesandnominalratesfordifferent periods, when comparing quoted figures. For example, by custom, the rates on Canadian mortgages are quoted as nominal rates withm = 2, while for US mortgages,m = 12. These provisions would be evident in the actual written contract, but are not always made clear to a prospective borrower. As the above example indicates, the actual annual rates paid on Canadian or US mortgages would be higher than the quoted nominal rates. We now deduce the present value of fractional annuities for the investment discount function. To simplify the notation, we first assume constant interest. A 1-year, 1-unit annuity payable mthly is simply an annuity paying 1∕m units for m periods, discounted with the primed rates. From formulas (2.16) and (7.4) we can write u k = ̈a(m)(e0) = 1 m (1−(1−d′)m d′ ) = d d(m) (7.6) for allk, where the last quantity is taken as 1 ifd = 0. From (7.2), ̈a(m)(c;v) = d d(m) ̈a(c;v). For the general case of non-constant interest we haveuk = dk∕d(m) k , which is taken equal t o1i fd(k) = 0. LIFE ANNUITIES PAIDmTHLY 101 Notethatinthecaseofpositivecashflows(andassumingtheusualcaseofpositiveinterest rates)thepresentvaluewithpaymentspaid mthlyislessthanthecorrespondingyearlypresent value. This can be shown mathematically by noting thatdk is less thand(m) k . For example, if dk = 0.36 then, from(7.4),d′ k = 1−0.641∕2 = 0.20 andd(2) k = 0.40.(The reader is invited to provide a general proof.) This conclusion is also evident from a comparison of payments. In theyearlycase,afullpaymentof c 0 wouldbemadeattime0.Inthe mthlycase,theannuitant receivesc0∕m each period of length 1∕m, and will not have collected the fullc0 amount until one mth of a year before the year end. This is true for each year of the annuity. There is therefore a loss of interest, which is reflected in the lower present value for themthly case. 7.3 Life annuities paidmthly We now consider the case where we have the interest and survivorship discount functionyx. We will usually writëa(m) x (c)f o r̈a(m)(c;yx) as we did whenm = 1. As we are normally given mortality data only for integer ages, we are faced with the necessity of making some assumption in order to extrapolate for the data that we do not have.Manyapproachesarepossible.Inthenextsectionwedescribethemostcommonlyused method. Other possibilities that have been proposed are described in Exercises 7.6, 7.13 and 7.14, as well as later in Chapter 8. 7.3.1 Uniform distribution of deaths This preferred method is known as the assumption of auniform distribution of deaths over each year of age, abbreviated as UDD. It can be viewed as just linear interpolation for𝓁y. Supposeweassumethat 𝓁y,ratherthanbeingdefinedjustforintegervalues,isdefinedforall nonnegativey. The UDD assumption is that for any integerx and 0< t < 1, 𝓁x+t = (1−t)𝓁x +t𝓁x+1. The name comes from the fact that given 0< t < t +h < 1, 𝓁x+t −𝓁x+t+h = h(𝓁x −𝓁x+1) = hdx. Inotherwords,duringanyintervalof h yearswhichliesbetweentwointegralages,thenumber of deaths occurring in that period is justh times the total number of deaths for the yearly interval. We can therefore say that the deaths are spread uniformly over the year. Example7.2 Suppose𝓁 60 = 1000,𝓁61 = 940.AssumingUDD,whatistheprobabilitythat a person age 601 3 will die before reaching age 601 2? Solution. The number of deaths in the1 6-year period between 601 3 and 601 2 equals 10, one-sixth of the total deaths in the year. Out of 1000 lives at age 60, there will 20 who die before reaching age 60 1 3 (one-third of the yearly deaths), leaving 980 alive at age 601 3.T h e probability is therefore 10/980. 102 FRACTIONAL DURATIONS An equivalent form of the UDD assumption is tqx = tqx, for any integer x and 0< t < 1, (7.7) which follows immediately by writingtqx as (𝓁x −𝓁x+t)∕𝓁x. AssumingUDD,wecaneasilycalculate tpy forall t andy (notnecessarilyanintegerage) by the formula tpx+h = t+hpx hpx , which follows from the multiplication rule. This points out an inconsistency in the UDD assumption. Suppose that the life table follows the expected pattern of increasing mortality byage.Thatis,forallintegers x ≤ y wehave q x ≤ qy.ThenitcanbeshownthatunderUDD, tqx ≤t qy for integersx ≤ y and allt ≥ 0, as we would expect. However, this inequality need not hold ifx andy are not integers. See Exercise 7.15. 7.3.2 Present value formulas WenowderiveformulasforfractionaldurationlifeannuitiesbymakingtheUDDassumption for mortality and the normal assumption for interest that we used in Section 7.2. This is the standardprocedure,butthedifferenttreatmentofthetwofactorsleadstosomecomplications in the formulas. We first assume constant interest. For our purpose we require some new interest quantities. Let 𝛽(m) = 1 vm2[v1∕m +2v2∕m +⋯ +(m −1)v(m−1)∕m], 𝛼(m) = d d(m) +d𝛽(m). We will later derive a simplified expression for𝛽(m). We have ̈a(m) x (e0) = 1 m[1+v1∕m 1∕mpx +v2∕m 2∕mpx +⋯ +v(m−1)∕m m−1∕mpx]. From (7.7),j∕mpx = 1−(j∕m)qx, so we can write u0 = ̈a(m) x (e0) = 1 m[1+v1∕m +v2∕m +⋯ +v(m−1)∕m] − qx m2[v1∕m +2v2∕m +⋯ +(m −1)v(m−1)∕m] = ̈a(m)(e0)−𝛽(m)vqx = d d(m) −𝛽(m)vqx, where we refer back to (7.6). Similarly, for any positive integerk, uk = d d(m) −𝛽(m)vqx+k. LIFE ANNUITIES PAIDmTHLY 103 From (7.2) and (5.6) can now write ̈a(m) x (c) = d d(m) ̈ax(c)−𝛽(m)Ax(c). (7.8) Somewhat surprisingly, an insurance term has come into an annuity value formula, but it is easilyexplained.Apartfrominterestloss,areasonforthe mthlyannuitytobeworthlessthan the yearly annuity is the loss of annuity income in the year of death. The annuitant receiving benefits mthly will only receive payments up to the time of death in their final year. For example, if they die shortly after the beginning of the year they will only get 1∕m of the total yearly income. Of course, they could receive the total income by living until the lastmth of the year. On average, annuitants should lose slightly less than half of the income in the year of death, which is reflected by the second term in (7.8). Under UDD the actual proportion is 𝛽(m). This is close to (m−1)∕2m for small interest rates, as shown in Section 7.5 below. To derive an alternate expression to (7.8) that uses only annuity present values we apply the general insurance-annuity identity (5.8) to the second term in (7.8) and obtain ̈a (m) x (c) = 𝛼(m)̈ax(c)−𝛽(m)̈ax(Δc). (7.9)‡ In the general case where the interest rate is constant over each year, but can vary from year to year,𝛼(m) and𝛽(m) will be vectors with the entry corresponding to indexk, equal to the corresponding value for the interest ratei k. The general form of (7.9) will be ̈a(m) x (c) = ̈ax(c ∗ 𝛼(m))− ̈ax(Δ(c ∗ 𝛽(m))). For the particular case of the level payment annuity with constant interest we have ̈a(m) x (1n) = 𝛼(m)̈ax(1n)−𝛽(m)(1−v(n)npx) Example 7.3 A woman age 60, earns a salary of 60 000 per year, payable monthly, and would normally expect to receive a raise of 2000 per year until retirement at age 65. She is injured and unable to work again. She plans to sue the parties responsible for her injury, for anamountequaltothepresentvalueofherlostsalary.Findaformulatocalculatethispresent value. Assume a constant interest rate. Solution. We will assume the accident occurred just before her 60th birthday, and that the first raise is due 1 year later. The present value is 1000 [ ̈a (12) 60 (c) ] = 1000[𝛼(12)̈a60(c)−𝛽(12)̈a60(Δc)], wherec = (60,62,64,66,68), so that Δc = (60,2,2,2,2, −68). 104 FRACTIONAL DURATIONS k… … … … … …c k /4 ck /4 ck /4 ck /4 ck /4 ck /4 ck /4 ck /4 ck+1 /4 ck+1 /4 ck +1 /4 ck +1 /4 ck +1 /4 ck +1 /4ck –1 /4 ck+1 /4ck+1 /4 ck+2 /4 Due payments Time Immediate pa yments k + 1/4 k + 2/4 k + 3/4 k + 5/4 k +6 / 4 k + 7/4 k +2k +1 Figure 7.1 Due and immediate quarterly annuity payments 7.4 Immediate annuities We now discuss an important variation. Refer back to (7.1). Suppose that instead of this scheme we assume payments of c k m at timek + j m, j = 1,2, …,m. (7.10) Wecanviewthisashavingpaymentsmadeatthe end ratherthanthe beginning ofeachperiod of length 1∕m. Annuities satisfying (7.10) are often referred to asimmediate, while those satisfying (7.1) are referred to asdue. The present value of the annuity with benefit vector given by (7.10) and discount functiony is denoted bya(m)(c;y). That is, the two dots are removed for immediate annuities. ThereadermaywishtoreferbacktoSection2.14.1wherewecommentedonthesomewhat unusual terminology. We also pointed out that the distinction between the two types is not needed in the annual case. However, when we postulate a constant rate of payment over the year, we do need to make the distinction in themthly case. See Figure 7.1 for a comparison of the payments on a due and immediate quarterly annuity. To go from the due annuity to the immediate annuity, we must do the following. At time zero, subtract a payment ofc 0∕m. At time 1, subtract a payment ofc1∕m and add a payment of c0∕m. At time 2, subtract a payment ofc2∕m and add a payment ofc1∕m, etc. We then have, forany discount functiony, a(m)(c;y) = ̈a(m)(c;y)− 1 m ̈a(Δc;y). (7.11) Example 7.4 Find a formula for the present value of a life annuity on (40) consisting of payments of $500 at the end of each month for 10 years. Assume a constant interest rate. Solution. In this case, c = 6000(1 10), Δc = 6000(1,09,−1), ̈a40(Δc) = 6000[1−y40(10)]. The present value is 6000 [ a(12) 40 (110) ] = 6000 [ ̈a(12) 40 (110)− 1 12(1−y40(10)) ] = 6000 [ 𝛼(12)̈a40(110)− ( 𝛽(12)+ 1 12 ) (1−y40(10))] . APPROXIMATION AND COMPUTATION 105 7.5 Approximation and computation Intheprevioussectionwedevelopedsomebasicformulas.Wenowconcentrateontheproblem of obtaining numerical values. For this purpose it is convenient to develop a closed-form formula for𝛽(m) in terms of standard interest symbolsi (m) and d(m). Assume for the moment constant interest. From the definition of𝛽(m) we can write, using the discount functionv(k) = (1+i′)−k, 𝛽(m) = 1 m2Valm−1(j), wherej = (1,2, …,m −1). From formulas (2.16) and (2.17), Valm−1(j) = (1+i′)(m−1)̈a(j) = [(1+i′)m−1 −1]∕d′ −(m −1) d′ . We can simplify the numerator since (1+i′)m−1 −1 d′ = (1+i′)m −(1+i′) i′ = i i′ −1. Substituting and noting thatmi′ = i(m),md′ = d(m), we obtain the formula 𝛽(m) = i −i(m) i(m)d(m). (7.12) It then follows that 𝛼(m) = d d(m) +d𝛽(m) = id i(m)d(m). For varying interest we get the same relationships holding on a year-by-year basis. We need merely insert subscripts ofk on all quantities. Exact calculations of𝛼(m) and𝛽(m) are not always done in practice. It is common to use an approximation that assumes a constant interest rate of 0. This implies that 𝛽(m) = 1+2+⋯ +m −1 m2 = m −1 2m . Moreoverd∕d(m) = 1 and𝛼(m) = 1, so we can write the approximate formula ̈a(m) x (c) ≐ ̈ax(c)− m −1 2m ̈ax(Δc). (7.13) In much of the actuarial literature this approximation is not found by setting the interest rate equal to 0 as we have done, but rather by using linear interpolation for the discount function y x. That is, it is assumed that for all nonnegative integersk and 0< s < 1, yx+k(s) = (1−s)+syx+k(1). 106 FRACTIONAL DURATIONS Under this assumption ̈a(m) x (e0) = 1 m [ 1+yx ( 1 m ) +yx ( 2 m ) +⋯ +yx (m −1 m )] = m +1 2m + m −1 2m vpx = 1− m −1 2m [1−vpx], which is the same as̈a(m) x (e0) calculated at 0 interest. By (7.2) both assumptions lead to the same approximation. This reason why this occurs is that the linearity ofyx, in addition to the UDD assumption, necessarily implies that the interest rate must be 0. The advantage of stating the latter condition from the outset is that it makes it clear that the approximation shouldbereasonablewithrelativelylowinterestrates.However,ifratesaretoolargerelative to mortality rates, this approximation can lead to inconsistencies. Example 7.5 Suppose thati =0.21, andq x =0.004. Calculatëa2(e0;v) and̈a(2) x (e0), using (7.13) for the latter. Solution. ̈a (2)(e0;v) = 1 2[1+1.10−1] = 0.955. Using (7.13), ̈a(2) x (e0) = 1− 1 4 [ 1− 0.996 1.21 ] = 0.956. It is, however, inconsistent for an annuity in which payments are contingent upon survival to cost more than an annuity with identical payments that are certain to be made. Equation 7.13 is part of a series of other possible approximation methods known as Woolhouse’s formulas. This one is the two-term formula. The next in the series, the three- term formula, is not used often but appears to give very good results in certain cases. After developing the appropriate tools in the next chapter, we will investigate further the two-term formula inconsistency and derive the three-term formula. See Section 8.11. ∗7.6 Fractional period premiums and reserves Annual premium and reserve formulas remain basically unchanged when premiums are paid mthly. The only difference is that we replacëa x with ̈a(m) x throughout. The purpose of this section is to derive a few useful identities. Consider all policies with a particular issue age (x) and with level premiums payable for h years. The simplest type of such a policy will be a 1-unit,h-year term policy with annual premiums. LetP denote the net annual premium andkV denote the reserve at timek for this term policy. Given any other such policy, let̃P denote the net annual premium if paid yearly, RESERVES AT FRACTIONAL DURATIONS 107 and let̃P(m) denote the net annual premium for this policy if paidmthly (i.e.,̃P(m)∕m is paid at the beginning of eachmth of the year). Letk ̃V denote the reserve at timek on this policy if premiums are paid annually and letk ̃V(m) denote the reserve at timek if premiums are paid mthly. For simplicity, we will assume constant interest. We have ̃P(m)̈a(m) x (1h) = ̃P̈ax(1h), as both sides are equal to the present value of the benefits, so that ̃P ̃P(m) = ̈a(m) x (1h) ̈ax(1h) = d d(m) −𝛽(m)Ax(1h) ̈ax(1h), using (7.8) for the last equality, and we can write ̃P ̃P(m) = d d(m) −𝛽(m)P. (7.14) Since the value of insurance benefits is independent of the premium frequency, we have, for k < h, k ̃V(m) −k ̃V = ̃P̈ax+k(1h−k)− ̃P(m)̈a(m) x+k(1h−k) = ̃P(m) [ ̃P ̃P(m) ̈ax+k(1h−k)− ̈a(m) x+k(1h−k) ] . Now substitute from (7.8) for ̈a(m) x+k(1h) and from (7.14) for ̃P∕̃P(m).T h et e r m(d∕d(m)) ̈ax+k(1h−k) cancels out and we are left with k ̃V(m) −k ̃V = 𝛽(m)̃P(m) kV. (7.15) The term on the right represents the additional reserve that must be held at timek in order to provide for the premiums that will not be collected in the year of death. The significance of (7.14) and (7.15) is as follows. We consider a collection of policies with level premiums, common issue age and premium payment period. The benefits can be anything at all. We know the premium and reserve for the particular case of the level term policy. Then, using only those two quantities, we can easily adjust premiums and reserves from annual mode tomthly mode, for all other policies in the collection. 7.7 Reserves at fractional durations InsurancecompaniesdoacompletecalculationofreservesonDecember31(orwhenevertheir particular fiscal year ends). However, policyholders are not always accommodating enough to purchase their policy on January 1. This means that most of the policy durations at the calendaryearendwillbefractional.Inthissectionwederiveformulasforcalculatingreserves attime k +s wherek isanintegerand0 < s < 1.Considerthegeneralannualpremiumpolicy 108 FRACTIONAL DURATIONS with death benefit vectorb and premium vector𝝅. Suppose first that the death benefit for those dying between timek and k +s is paid at timek +s. The reserve at timek +s would then be, by a calculation analogous to that used in deriving (6.7), (kV +𝜋k)yx(k +s,k)−bk sqx+k spx+k . Thatis,thinkingretrospectively,thebalanceattime k +s isobtainedbytakingthebalanceat time k, adding the premium, accumulating up to timek +s at interest and survivorship, and thensubtractingfromeachsurvivorenoughtopaythedeathbenefitsforthosedyingbetween timek and timek +s. Wemust,however,adjustthistoreflectthefactthatthedeathbenefitwillnotbepaiduntil the end of the year, and to do so we multiply the second term byv(k +s,k +1). Note now that v(k +s,k +1) spx+k = yx(k +s,k +1) (1−spx+k+s)(spx+k) = yx(k +s,k +1) px+k . FromtheUDDassumption, sqx+k = s(qx+k)andsubstitutingtheselasttwoidentitiesgives the desired reserve formula as k+sV = (kV +𝜋k)yx(k +s,k)−sbk qx+k px+k yx(k +s,k +1) = (1−s)(kV +𝜋k)yx(k +s,k)+s [ (kV +𝜋k)yx(k +1,k)−bk qx+k px+k ] yx(k +s,k +1). Finally, substituting from (6.7) yields k+sV = (1−s)[(kV +𝜋k)yx(k +s,k)]+s[k+1Vy x(k +s,k +1)]. In other words, under UDD, the reserve at some intermediate point in the year is obtained by linearly interpolating between the initial reserve, accumulated to that point with interest and survivorship, and the end-of-year terminal reserve, discounted to that point with interest and survivorship. Care must be taken not to use the terminal instead of the initial reserve at time k, as that would involve interpolating over a discontinuity. A common simplification is to assume that interest and mortality rates are sufficiently smallsothatwecantake y x asidenticallyequalto1.Thisresultsintheapproximateformula k+sV = (1−s)(kV +𝜋k)+s(k+1V) = (1−s)kV +s(k+1V)+(1−s)𝜋k. The third term above is known as theunearned premium.It is simply the total premium paid at the beginning of the year multiplied by the fraction of the year remaining. When premiums are paidm times a year rather than annually, the same type of approxi- mation as above is normally used. That is, we calculate the reserve at timek +s by linearly interpolatingbetweentheterminalreservesattime k andk +1andaddtheunearnedpremium. EXERCISES 109 To be precise, suppose thats = h∕m +r where h is an integer and 0< r < 1∕m and that the annual premium𝜋k is paidm times per year. Then a commonly used approximation is k+sV = (1−s)kV +s(k+1V)+ ( 1 m −r ) 𝜋k. Example7.6 Foracertainpolicy 3V = 108, 4V = 180andforthefourthyearofthepolicy the premium is 60, paid quarterly (i.e.,15 is paid at time 3, 31 4,3 1 2 and 33 4. Find the reserve at time 3 years and 7 months. Solution. We haves =7/12=2/4 + 1/12. The unearned premium is ( 1 4 − 1 12 ) 60 = 10. The above formula gives 3+7∕12V = 5 12 ×108+ 7 12 ×180+10 = 160. An alternate way of calculating the unearned premium is to note that the reserve is taken at a time one-third through the quarter-year period, so the unearned premium is two-thirds of the premium payable at the beginning of the quarter, which is 2 3 ×15 = 10. 7.8 Standard notation and terminology Most of the standard notation for fractional payments has been discussed already. The one remaining symbol isA (m) which is used to denote net single premiums for an insurance in which the death payments are paid at the end of themth of the year in which death occurs. For example ifm =12, the death benefit would be made at the end of the month of death. The assumption of Exercise 7.14 below is known as theBalducci hypothesis, named for G. Balducci, an Italian actuary. It arose in conjunction with mortality studies designed to produce life tables. Exercises Assume UDD unless otherwise indicated. Type A exercises 7.1 You are given thaẗa60 = 12,10p50 = 0.8, and interest is constant at 6%. (a) Findthenetsinglepremiumforanannuityon(50)withlevelbenefitpaymentsof 1000 per month, beginning at age 50 and continuing for life, with the provision that the first 120 monthly payments are guaranteed, regardless of whether (50) is alive or not. (b) Redo (a), assuming now that the first payment is made at the end of 1 month. 110 FRACTIONAL DURATIONS 7.2 You are given thatq70 = 0.2,q71 = 0.25,q72 = 0.30. Interest rates are 20% for the first2years and 30% thereafter. A 3-year lifeannuity on(70) provides forbenefits of 1000inthefirstyear,2000inthesecondyearand3000inthethirdyear,providedthat (70) is then alive. Suppose that payments are to be made quarterly. Find the present value if: (a) the first payment is to be made at age 70; (b) the first payment is to be made at age 70 1 4. 7.3 An annuity on (50) provides for yearly payments for 30 years. The amount of the income is 10 000 for the first 10 years, 8000 for the next 10 years and 5000 for the last 10 years. The net single premium for this annuity is 100 000. You are given that theinterestrateisaconstant5%andthat 10p50 = 0.9, 20p50 = 0.8, 30p50 = 0.7.Find the premium if the same income is to be paid monthly instead of annually, assuming payments are made (a) at the beginning of each month, (b) at the end of each month. 7.4 You are given the following figures from a life table.𝓁 60 = 1000,𝓁61 = 700,𝓁62 = 500.Findtheprobabilitythatapersonnow age60 1 3 willdiebetween theagesof60 1 2 and 613 4. 7.5 Given q70 = 0.2, q71 = 0.3, q72 = 0.4,findtheprobabilitythatapersonage70 1 2 will live to age 721 4. 7.6 Instead of assuming UDD, suppose that, for an integerx and 0< t < 1, we assume that tpx = pt x .G i v e nl60 = 100000, l61 = 81000, l62 = 41472, find the probability that (601 2) will die between the ages of 611 3 and 612 3. 7.7 A policy is issued on March 20, 1990. Given that10V = 2000, 11V = 3000 and that the premium is a level 105 per month, find the reserve on December 31, 2000, using the standard approximation. Assume that each month is 30 days. 7.8 For a policy on (x ), 5V = 220, 6V = 80, b5 = 1000, i5 = 0.10, qx+5 = 0.2. Find 5+1∕4V. Type B exercises 7.9 Let r bethepresentvalueofalifeannuityon(x )providing1atthebeginningofeach periodoflength1∕m,withpaymentsguaranteedfor n years.Let s bethepresentvalue ofthesameannuitybutwiththepaymentsatthe end ofeachperiod.If v(n) = 0.9and npx = 0.6, findr −s. You are not given the value ofm. 7.10 Show that under a constant interest rate ofi,a(m)(1n) = (i∕i(m))a(1n). 7.11 You want to evaluate ̈a(4) x (k) at a constant interest rate of 20%, where k = (10,9,8, …,1). Assuming that UDD holds, what error is made if you use the zero- interest approximation? You are given thaẗax(k) = 35 and̈ax(111) = 6. *7.12 A 20-year term insurance policy on (50) has a constant death benefit of 1000, and net level annual premiums of 10, payable for 20 years. The reserve at time 15 is 60. Anotherpolicyon(50)hasthesamepremiumpatternbutdifferentdeathbenefits.On this latter policy, the annual premium is 100, and 15V = 500. The interest rate is a EXERCISES 111 constant 6%. What is the value of the annual premium and the reserve at time 15 on this latter policy, if the premiums are payable monthly instead of annually? 7.13 Suppose we assume that for all integersx and 0< t < 1, 𝓁 x+t = 𝓁1−t x 𝓁t x+1. (a) Show that this is equivalent to the assumption of Exercise 7.6. (b) Show that if (2.9) holds forv, then it also holds for the discount functionyx whenever s,t,s +h,t +h are all in the interval [k,k +1) for some nonnegative integerk. (See Exercise 8.21 for more on this assumption.) 7.14 Suppose we assume that for all integersx and 0< t < 1, 𝓁−1 x+t = (1−t)𝓁−1 x +t𝓁−1 x+1. Show that for all integersx and 0< t < 1,1−tqx+t = (1−t)qx. 7.15 Assume UDD. Suppose that, a life table is such thatqx < qx+1 for all nonnegative integersx. (a) Show that for anyt > 0 and integersx < y we havetqx ≤ tqy. (b) Suppose thatq80 = 0.4, q81 = 0.45.Show that0.5q80.5 >0.5 q81. (c) Show that under the assumption of Exercise 7.6, it is true that for allt > 0 and positive numbersx,y,w eh a v etqx ≤ tqy. Spreadsheet exercise 7.16 ModifytheChapter4spreadsheetsothatbyenteringthefrequency m,thespreadsheet will calculate both due and immediate annuities payablem times per year. 8 Continuous payments 8.1 Introduction to continuous annuities In this section we consider annuities where payments are made continuously. Naturally, this is not physically possible, but we can picture these as a limiting case ofmthly annuities as mgoes to infinity. Suppose, for example, that you are to receive a total of 36 500 units each year. This could be done by paying 100 units per day, or 4 1∕6 units per hour, or 0.001157 units per second and so on. If you can imagine payments coming in every nanosecond, you may get some feeling for what a continuous annuity would be like. This may seem as a somewhat artificial concept, but there are many uses for continu- ous annuities. They can be used to approximatemthly annuity values for large values ofm. Moreover, we will show that insurance contracts with the realistic provision of benefit pay- mentsatthemomentofdeath,canbeviewedascontinuousannuities.Forinsurancecontracts purchased by continuous premium payments we can derive some interesting mathematical relationships that are analogues of those that appeared in Chapters 5 and 6. Inthecontinuouscase,wecannotspecifyanactualpaymentatanypointoftime.Asshown by the figures above, this approaches zero as the frequency of payment increases. Instead we must speak of theperiodicrate of payment. Consider a monthly annuity. If the payment in 1 monthis100,wecoulddescribethisbysayingthattheannual rateofpaymentforthatmonth is1200.Thiswouldmeanthatifthepaymentsremainedatthesamemonthlylevelforayear, then the total payment for that year would be 1200. Each of the annuities described above wouldbepaidattheannual rateof36500.Ofcourseinanannuitypaid mtimesperyear,the actual payment and therefore the periodic rate of payment could change everymth period. In our continuous annuity, it could change from moment to moment. (Picture a person standing under a chute, receiving money that is flowing down continuously. The speed at which it comes down could change at each instant.) Inplaceofacashflowvectorwenowhaveacashflowfunction cdefinedon[0, N],where c(t) is the periodic rate of payment at timet. FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial THE FORCE OF DISCOUNT 113 Wewillassumethatallourcashflowfunctionsare piecewisecontinuous.Thatis,thereis a partition, 0= a1 < a2 < ⋯ < an = N, such thatcis continuous on the intervals (ai−1,ai), i= 1,2, …,n−1. Supposewehaveanarbitrarydiscountfunction yandacontinuousannuitywithcashflow functionc.Thepresentvalueoftheannuitywillbedenotedby a(c;y)orjust a(c).(The‘bar’is astandardactuarialnotationaldevicetodenotecontinuousasopposedtodiscretequantities.) Consider an approximating annuity with payments mademthly. That is, payments are made at timej∕mwhere jis an integer between 0 andmN−1. We assume thatN is an integer and theannual rateofpaymentattime j∕misc(j∕m).Theactualpaymentattime j∕mistherefore (1∕m)c(j∕m), so the present value of thismthly annuity is mN−1∑ j=0 c(j∕m) m y(j∕m). (8.1) Taking the limit asmgoes to∞gives one of the key formulas of this chapter, a(c;y) = ∫ N 0 c(t)y(t)dt, (8.2) ‡ since themthly present values are Riemann sums for this integral. 8.2 The force of discount Throughoutthebookweuse‘log’torefertothenaturallogarithm(oftendenotedby lnorloge.) The evaluation of continuous streams of cash flows leads naturally to a new function, which we will spend some time investigating before returning to the main topic. It can be written in several equivalent forms. Definition 8.1 Given an arbitrary discount functiony, define another function,𝛿 y(t), which we will call theforceofdiscountassociatedwithy,b y 𝛿y(t) = lim h→0 y(t+h,t)−1 h = lim h→0 y(t+h,0) −y(t,0) hy(t,0) = d dty(t,0) y(t,0) = d dtlogy(t,0) =− d dtlogy(t) =− d dty(t) y(t) . (8.3) (To derive the second last equality, recall thaty(t) = y(0,t) = y(t,0)−1.) From the first equality in (8.3) we see that𝛿y represents arelativerate of growth. Each unit invested at timetwill accumulate toy(t+h,t) units over the nexthtime periods, so the numerator gives us the relative growth rate over this time interval. We divide byhto get the relative rate of growth perperiod, and take limits to get the instantaneous periodic relative rate of growth at timet. We are often given𝛿y and wish to recovery. 114 CONTINUOUS PAYMENTS Proposition 8.1 y(t) = e−∫ t 0 𝛿y(r)dr. Proof. From the second last expression in (8.3), we integrate to get ∫ t 0 𝛿y(r)dr=− ∫ t 0 d dr[logy(r)]dr=−l o gy(t)+log y(0). We use the fundamental theorem of calculus for the second equality. Note thatlogy(0) = log1 = 0, multiply by−1, and take exponentials to complete the proof. An immediate consequence of Proposition 8.1 is that fors≤ t, y(s,t) = y(t)∕y(s) = e−∫ t s 𝛿y(r)dr = e−∫ t−s 0 𝛿y(s+r)dr. (8.4) Remark Thistraditionaluseoftheforceofdiscountisfortheinvestmentdiscountfunction v. In this case𝛿v(t) will be denoted by just𝛿(t). The standard actuarial notation is𝛿t,b u tw e find it more convenient to use the bracket in place of a subscript to parallel our treatment for thesymbol v.Thisquantityiscommonlyreferredtoasthe forceofinterest. Theterminologyis justifiedsinceforaninstantaneousperiodthereisnodifferencebetweeninterestanddiscount. (This is shown precisely by the equality of the fourth and last expressions in (8.3)). 8.3 The constant interest case Consider the constant interest discount function given byv(t) = vt. Then𝛿(t) is also constant anddenotedjustby 𝛿.Since log(vt) = tlog(v),thesecondlastexpressionin(8.3)showsthat 𝛿 =−l o g(v) =l o g(1+i), (8.5) and Proposition 8.1 gives v(t) = e−𝛿t,( 1 +i)t = e𝛿t. (8.6) The second expression reflects the fact that under constant interest, invested capital grows exponentially and𝛿 is just the usualrateofexponentialgrowth. Additionalinsightmaybegainedbycomparing 𝛿 withthefractionalinterestquantitiesof Chapter 7. We have that 𝛿 = lim m→∞ i(m) = lim m→∞ d(m). This can be seen by writing i(m) = (1+i)1∕m −1 1∕m , d(m) = 1−v1∕m 1∕m , CONTINUOUS LIFE ANNUITIES 115 so that lim m→∞ i(m) = d dt(1+i)t|t=0 = (1+i)t log(1+i)|t=0 = 𝛿. The calculation ford(m) is similar. The quantity𝛿 is often thought of as an instantaneous rate of interest, but it is important to note that just like the ratesi(m) and d(m), its value depends on the underlying period. The transformationforchangeofperiodthoughismucheasierthanfortheinterestrates,asitjust involves a proportional change. For example, if the annual force of interest is 0.06, then the force of interest for a half year period will be simply 0.03. We now return to continuous annuities, using a constant interest rate, and show that in certain cases we get expressions analogous to discrete annuities, but with𝛿 replacing d.W e will derive the continuous versions of (2.16) and (2.17). Define a function 1 n by 1n(t) = { 1, if t≤ n, 0, if t> n. This is the same notation as for the analogous vector, but the meaning should be clear from the context. We also define the functionI n(t)b y In(t) = { t,i f t≤ n, 0, if t> n. From (8.2) and (8.6), a(1n) = ∫ n 0 e−𝛿tdt= 1−e−𝛿n 𝛿 . (8.7) Integrating by parts, a(In) = ∫ n 0 te−𝛿tdt=− te−𝛿t 𝛿 | | | | n 0 + 1 𝛿 ∫ n 0 e−𝛿tdt= a(1n)−ne−𝛿n 𝛿 . (8.8) 8.4 Continuous life annuities 8.4.1 Basic definition The present value of an annuity issued to (x) that provides benefits paid continuously at the annualrateof c(t)attime t,provided(x )isalive,willbedenotedby ax(c).Formula(8.2)gives the exact formula ax(c) = ∫ 𝜔−x 0 c(t)v(t)tpxdt. (8.9)‡ Note that this formula follows essentially the same pattern that we observed after formula (5.1). Instead of summing we integrate, and for eacht the integrand is the product of three 116 CONTINUOUS PAYMENTS factors: the interest discount factor; the ‘amount paid ’ at timet, (which we can intuitively think of asc(t)dt); and the probability that the payment is made. 8.4.2 Evaluation We will generally only have available the values of𝓁x for integer values of (x), and must interpolatefortheremainingvaluesinordertoevaluatetheintegral.AsinChapter7,wewill assumeUDD,whichwillallowustoexpresscontinuousannuitiesintermsofannualannuities. The procedure closely parallels that formthly annuities and we could simply write down the formulas as limiting cases of those in Chapter 7. It is instructive however to proceed a new. For simplicity of notation, we will at first assume constant interest. The initial step is to define the continuous analogues of the functions𝛽 and𝛼 introduced in Chapter 7. Let 𝛽(∞) = e 𝛿a(I1) = e𝛿 −1−𝛿 𝛿2 = i−𝛿 𝛿2 , invoking (8.7) and (8.8). Let 𝛼(∞) = a(11)+d𝛽(∞) = di 𝛿2. It is clear that𝛽(∞) and𝛼(∞) are the respective limits of𝛽(m) and𝛼(m)a smapproaches∞. WenowusethesameprocedureasweusedinSection7.1andreplaceallthepaymentsin any 1 year by a single payment, equal to the value of these payments at the beginning of the year.Thisreplacementprincipleholdsinthecontinuouscaseaswell.Foraformalverification use the standard integration result ∫ b 0 = ∫ a 0 +∫ b a for a< b. Analogously to (7.2) this leads to ax(c) = ̈ax(z), (8.10) where zk = ∫ 1 0 c(k+s)vs spx+kds= ∫ 1 0 c(k+s)vs(1−s⋅ qx+k)ds. (8.11) This is easily evaluated when the rate of payment is constant over each year. For a nonnegativeinteger k,let ck denotetheconstantvalueof c(k+s),0 ≤ s< 1.Wecanfactor ck outside of the integral sign in (8.11), which gives zk = ck[a(11)−qx+ka(I1)] = ck [d 𝛿 −𝛽(∞)vqx+k ] . Then, from (8.10), ax(c) = d 𝛿 ̈ax(c)−𝛽(∞)Ax(c), CONTINUOUS LIFE ANNUITIES 117 where c = (c0,c1,…). Finally, from Theorem 5.1, ax(c) = 𝛼(∞)̈ax(c)−𝛽(∞)̈ax(Δc). (8.12) As promised, this reduces the calculation to finding the present value of yearly annuities. Asinthe mthlycase,itisusualinpracticetoapproximatethisbycalculatingtheadjusting factorsatzerointerest.Wethenhave 𝛼(∞) = 1,𝛽(∞) = 1 2,andtheapproximatingformulais ax(c) ≐ ̈ax(c)− 1 2̈ax (Δc). In the case wherec = 1∞ so thatΔc = (1,0,0,…), we get the familiar equation ax ≐ ̈ax − 1 2. (8.13) This concludes the development of formulas for the constant interest case. When rates are only constant over each year, but can vary from year to year, we must replace the constants d∕𝛿,𝛼(∞) and𝛽(∞) by vectors, analogously to the treatment in the discrete case. 8.4.3 Life expectancy revisited Comparing the formula for life expectancy (3.8) with formula (4.1), we see thatex is simply thepresentvalueofalifeannuityof1peryear,beginningattime1,ataconstantinterestrate of0.Thismakesenseintuitively,sinceatzerorateofinterest,thepresentvalueoftheannuity will simply be the total amount paid, which in turn is just the whole number of years lived, whichonaverageisjust e x.Wecannowwritedownamorerigorousformulaforthe complete life expectancy, as the present value of a life annuity paying 1 per year for life continuously, at zero interest. This gives ◦ex = ∫ ∞ 0 tpxdt. Assuming UDD, we apply the formulas given above. Withc = 1𝜔−x, and an interest rate of zero, ̈ax = 1+ex, ̈ax(Δc) = 1, and from (8.13), ◦ex = 1+ex − 1 2 = ex + 1 2, recovering the approximation we obtained intuitively in Chapter 3. Similarly,thecomplete n-yeartemporarylifeexpectancyisgivenby ∫ n 0 tpxdt,whichunder UDD, is equal to∑n k=1 k px + 1 2nqx as we obtained previously. Hereisainterestinguseforlifeexpectancy.Supposeyouaregiven ◦ex andtheinvestment discountfunction v,butnottheactuallifetable,andyouwanttoapproximate ax.Onenatural waytodoitissimplytake a(1◦ex ;v)forthisapproximation.Indeed,somemighteventhinkthat 118 CONTINUOUS PAYMENTS this (or an appropriate modification in the discrete case) should be exact, reasoning that the cost of a life annuity on a person should be the cost of providing the individual with income uptotheirpreciselifeexpectancy.Theapproximationshouldbeclose,butinthenormalcase of positive interest rates, this method always overstates the true value. A mathematical proof isoutlinedinExercise22.15butwecaneasilyexplainthisfactbygeneralreasoning.Suppose an insurer sells 1-unit continuous life annuities to a group of people agexand charges them each the price of a fixed period annuity with the period equal to their life expectancy. The insurercanexpecttogainmoneyonthosewhodiebeforelivingtheirlifeexpectancy,butlose moneyonthosewholivelonger,andthesegainsandlossesshouldcancelout.However,since thegainscomeearlier,theywillproduceextrareturnsduetointerestearningsandtheinsurer will end up with more than they need to provide the benefits. The break even premium will then be somewhat less than the premium for the fixed period annuity. 8.5 The force of mortality Beforeproceedingwithinsurances,weintroduceanewquantitythatisanalogoustotheforce of discount. Wefirstmotivatethisintuitively.Whatistheprobabilitythat( x)willdieinthenextinstant of time? We could approximate this by looking at hqx, but if we assume continuity and take thelimitas happroaches0,wewouldget0.Thequestionisanswerablebuttheanswerof0is uninteresting and gives us no information about the mortality of (x) at that point. Instead, we can compute an annualrateof mortality at age (x), by dividing by h before taking the limit. This leads to the following definition. Definition 8.2 The forceofmortality at age (x) is the quantity 𝜇(x) = lim h→0 hqx h = lim h→0 𝓁x −𝓁x+h h𝓁x =− d dx𝓁x 𝓁x =− d dxlog𝓁x. (8.14) The expression after the third equality is useful for obtaining additional insight. We have a group of lives declining over time due to mortality. The quantity𝜇(x) gives us therelative rate of decline in this group at agex. In many cases we are looking at a fixed agexand we want the variable to be the timet. We will accordingly define 𝜇x(t) = 𝜇(x+t). We can view𝜇x(t) as the force of mortality at timetfor an individual agexat time 0. From the fourth expression in (8.14), 𝜇x(t) =− d dttpx tpx , (8.15) and we imitate the proof of Proposition 8.1 to derive tpx = e−∫ t 0 𝜇x(r)dr. (8.16)‡ INSURANCES PAYABLE AT THE MOMENT OF DEATH 119 Example 8.1 Suppose that the force of mortality is given by 𝜇(x) = 1 𝜔−x, x<𝜔 . Find an expression fortpx. Solution. Fort<𝜔 −x, −∫ t 0 1 𝜔−x−rdr=l o g(𝜔−x−t)−log(𝜔−x) =l o g (𝜔−x−t 𝜔−x ) . We then substitute in (8.16) to get tpx = (𝜔−x−t 𝜔−x ) = 1− t 𝜔−x. The following quantity is often useful. Definition8.3 Let𝜆x(t)denotetheforceofdiscountfortheinterestandsurvivorshipfunction at agex Note that 𝜆x(t) =− d dtlogyx(t) =− d dtlogv(t)− d dtlog tpx = 𝛿(t)+𝜇x(t). (8.17) Inotherwords,theforceofdiscountforinterestandsurvivorshipisjustthesumoftheforces forthesetwocomponents.Wecanthenviewthefunction 𝜇 x astheparticularcaseof 𝛿yx when interest rates are zero. 8.6 Insurances payable at the moment of death 8.6.1 Basic definitions Unlikeannuities,thecontinuousversionistherealisticmodelforinsurancepolicies.Itcaptures the idea that, in practice, claims are paid at the moment of death rather than the end of the yearofdeath.Wecanalsoallowforthedeathbenefittovarycontinuouslywithtime.Inplace of our death benefit vector, we have adeathbenefitfunction. This is a piecewise continuous function bdefined on [0,𝜔−x) whereb(t) is the payment made at timetshould death occur at that time. The net single premium for such a policy on (x) will be denoted by Ax(b). To calculate this, we will consider an approximating policy. Letmbe a positive integer. If death occursbetweentime j∕mand( j+1)∕m,where j= 0,1, …,m(𝜔−x)−1,ourapproximating policy will payb(j∕m) at time (j+1)∕m. Up to now we have looked at the case ofm= 1. If m= 365, for example, it would mean that death benefits were paid at midnight on the day of 120 CONTINUOUS PAYMENTS death, clearly getting close to what we want. If we letA(m) x (b) denote the net single premium for this approximating policy, we will have Ax(b) = lim m→∞ A(m)(b). Reasoning as in the middle expression of formula (5.1), we have A(m) x (b) = b(0)v(1∕m)(1− 1∕mpx ) +b(1∕m)v(2∕m)( 1∕mpx − 2∕mpx ) +⋯ . Taking limits asmgoes to∞, Ax(b) =− ∫ 𝜔−x 0 b(t)v(t)dtpx =− ∫ 𝜔−x 0 b(t)v(t) d dttpxdt, and from (8.15) we obtain our final formula Ax(b) = ∫ 𝜔−x 0 b(t)v(t)tpx𝜇x(t)dt. (8.18)‡ Onceagainwehavethefamiliarthree-factorproduct;theamountpaid;theinterestfactor; andtheprobabilitythatpaymentismade.Wecanwritethelatterasbeingitselftheproductof two terms: tpx, the probability of living to timet; and𝜇x(t)dt, which we can intuitively think of as the probability of dying at exact momentt. Aswell,thisformulaillustratesourstatementabovethatinsurancespayableatthemoment of death can be viewed as continuous annuities. Analogously to (5.5), Ax(b) = ax(c), where c(t) = b(t)𝜇x(t). (8.19)‡ 8.6.2 Evaluation Aswithcontinuousannuities,wenormallywillhaveonlythelifetableavailableandwilluse UDD in order to evaluate the integral in (8.18). We again will simplify the notation by first using constant interest. The most useful consequence of UDD for this purpose is tpy𝜇y(t) =− d dttpy =− d dt(1−tqy) = qy. (8.20) whereyis an integer and 0< t< 1. As in the annuity case, we have Ax(b) = ̈ax(z), (8.21) where zk is the value at timek of the benefits paid in the yeark to k+1. Substituting from (8.20) zk = ∫ 1 0 vtb(k+t)tpx+k𝜇x+k(t)dt= vqx+k∫ 1 0 (1+i)1−tb(k+t)dt. INSURANCES PAYABLE AT THE MOMENT OF DEATH 121 This leads to Ax(b) = Ax(̃b), where ̃bk = ∫ 1 0 (1+i)1−tb(k+t)dt, (8.22) thereby reducing the evaluation of moment of death insurances to insurances paid at the end of the year of death. This is further simplified in the case where the benefit functionb is constant over each year.If b k istheconstantvalueof b(k+t)where kisanintegerand0 < t< 1,then bk factors out of the integral in (8.22) and ̃bk = bk∫ 1 0 (1+i)(1−t)dt= bk i 𝛿, leading to the very simple formula Ax(b) = i 𝛿Ax(b), (8.23)‡ whereb = (b0,b1,…). Additional insight can be obtained by observing that i 𝛿 = e𝛿 −1 𝛿 ≐ 1+ 𝛿 2 ≐ 1+ i 2. (Here, ≐ stands for ‘approximately equal to’.) This reflects the fact that in the case where benefits are constant over the year, the difference between paying benefits at the end of the year of death and at the moment of death is that in the latter case the insurer will not earn interest from the time of death to the end of the year, and the premium must be higher to account for this. This time interval will vary with time of death, ranging from 0 for those dying at the end of the year, to 1 year for those dying at the beginning. On average it should be half a year. In the case where interest is only constant over each year, we replace the right hand side above in (8.23) byA x(i∕𝛿 ∗ b) where nowi∕𝛿 is a vector with entries ofik∕𝛿k. Example 8.2 An insurance contract on (x) provides for 2, paid at the moment of death if this occurs withinnyears, plus a pure endowment of 3 ifxis alive at the end ofnyears. Find the present value given that interest rates are a constant 6%,vnnpx = 0.4 andAx(1n) = 0.2. Solution. Care must be taken to apply the adjustment factor only to the term insurance, and not to the pure endowment, which is always paid at timen.A t6 % ,w eh a v ei∕𝛿 = 1.0297 so the present value is equal to 2(0.2)(1.0297)+3(0.4) = 1.612. Whenbenefitpaymentsarenotconstantovereachyear,wehavetoresortdirectlyto(8.22) and try to evaluate the integral. Example8.3 Aninsurancepolicyon(x )providesforadeathbenefitof tpaidatthemoment of death, should this occur at timet. Assuming UDD, find a formula for the present value in terms of end-of-the-year-of-death insurances. 122 CONTINUOUS PAYMENTS Solution. For an integerk and 0< t< 1, we have the death benefitb(k+t) = k+t= (k+ 1)−(1−t).Wecanthinkofthisastwopolicies,onepaying k+1fordeathbetweentime kand timek+1,andtheotherpaying −(1−t)fordeathattime k+t,where k= 0,1, ….(Negati ve deathbenefitsmaybeunrealisticinpractice,buttheyarefinefromamathematicalviewpoint.) From (8.23), the present value of the first policy is (i∕𝛿)A x(j), wherej = (1,2,3 …). From (8.22) the present value of the second policy is the same as that with a constant end of the year death benefit of −∫ 1 0 (1+i)1−t(1−t)dt=− (1+i 𝛿 − i 𝛿2 ) =− i 𝛿 (1 d − 1 𝛿 ) after integration by parts. The total premium is therefore i 𝛿 [ Ax(j)− (1 d − 1 𝛿 ) Ax ] . It may appear more natural to have taken the split of the death benefit ask plus t.T h e split above was done to get an expression involving the death benefit vectorj, representing a linearly increasing benefit starting at 1 unit. This is often encountered in practice. The alternativemethodwouldresultinapolicybasedonavectorwithinitialentry0,whichwould not normally be found in a real-life policy. In a more complicated case the integration could be difficult to carry out, necessitating the use of an approximate integration formula. This can arise in the case of death benefits in pension plans that may depend on salary and length of service and could vary continuously over each year. A frequently used, very simple rule, is simply to assume that the death benefit for the year from timek to k+1 is constant at the value which it takes at time k+1∕2. 8.7 Premiums and reserves The realistic model for life insurance calls for death benefits to be paid at the moment of deathandpremiumstobepaidperiodically(yearly,monthly,quarterly,etc.).Thismeansthat we have to treat death benefits separately when computing premiums and reserves, so we cannot conveniently calculate the net cash flow vectorf as we did in Section 6.1. However, thisusuallycauseslittledifficultyaswecansimplychange A y(b)to Ay(b)inallpremiumand reserve formulas. We also must define the shifted functionsb◦ kanalogous to that for vectors. That is (b◦ k)(t) = b(k+t), 0 ≤ t≤ 𝜔−x−k−t. Example 8.4 Assume a constant interest rate. Consider a policy on (40) with death benefit function b(constant over each year). The total premium for the yearktok+1i s𝜋k, payable monthly. The interest rate is a constant 0.06. Write a formula for the reserve at time 10. Assume UDD. THE GENERAL INSURANCE–ANNUITY IDENTITY IN THE CONTINUOUS CASE 123 Solution. We have 10V = [A50(b◦ 10)− ̈a(12) 50 (𝜋◦ 10)], where𝜋 = (𝜋0,𝜋1,…). Fori= 0.06, i 𝛿 = 1.0297, 𝛼 = 1.00028, 𝛽 = 0.46810, and we can write 10V = 1.0297A50(b◦ 10)−1.00028 ̈a50(𝜋◦ 10)+0.46810 ̈a50 (Δ(𝜋◦ 10)), withb = (b0,b1 …), wherebk is the constant value ofb(t)f o rk≤ t< k+1. Inordertoavoidthisdualtreatment,itissometimesconvenienttoconsiderapolicywhere premiumsarepayablecontinuously.Thisisunrealistic,butareasonableapproximationinthe case of monthly premiums. It often results in simplifying the mathematics. We will look at cases where this is useful in the next two sections. 8.8 The general insurance–annuity identity in the continuous case Inthissectionwedevelopaconnectionbetweenmoment-of-deathinsurancesandcontinuous annuities, which is analogous to that obtained in Section 5.7. We now assume that the associated benefit functionbis differentiable except possibly at the endpoints of an interval. That is, we suppose there exist pointsr and s such thatb(t) is equal to 0 for 0< t< r and s< t<𝜔 −x, and that the derivativeb ′(t)e x i s t sf o rr< t< s. Then Ax(b) = ∫ s r b(t)v(t)tpx𝜇x(t)dt=− ∫ s r b(t)v(t) d dttpxdt. Integration by parts gives Ax(b) = b(t)v(t)tpx| | r s +∫ s r [b′(t)v(t)−b(t)v(t)𝛿(t)]tpxdt. Simplifying, Ax(b) = b(r)v(r)rpx −b(s)v(s)spx +ax(b′)−ax(𝛿b). (8.24) Here 𝛿bis just the function defined by𝛿b(t) = 𝛿(t)b(t). This is in fact a continuous analogue of (5.8). We need the two additional terms at the beginning to handle the possible jumps in the functionb. 124 CONTINUOUS PAYMENTS Wecanusethistoderiveacontinuousversionofthe endowmentidentity ofSection5.7.2. Assume a constant force of interest of𝛿. For a 1-unit,n-year endowment insurance with benefits paid at the moment of death,Ax:n denotes the net single premium, andPx:n denotes the rate of annual premium, payable continuously fornyears. In this case the functionbis constant on the interval (0,n), sob′ = 0. Takingb= (1n), the first two terms in (8.24) are 1−vn npx, which leads to Ax:n = Ax(1n)+v(n)npx = 1−𝛿ax(1n), and Px:n = Ax:n ax(1n) = 1 ax(1n) −𝛿. 8.9 Differential equations for reserves The goal of this section is to develop a differential equation for reserves, which is analogous to the recursion formulas of Sections 2.10.4 and 6.3. Suppose we have a transaction with continuous cash flows defined by the cash flow function cdefined on [0,N], and a general discount functiony.T h ereserveat timet, which wewilldenoteby tV(c;y),isdefinedexactlyasinthediscretecase,namelyasthenegativeof the value at timetof all future cash flows. So tV(c;y) =− ∫ N−t 0 c(t+r)y(t,t+r)dr. It is convenient to make a change of variable fromrtos= r+tand write tV(c;y) =− ∫ N t c(s)y(t,s)ds=− y(t,0) ∫ N t c(s)y(s)ds. Differentiating by the product rule, using (8.3) to differentiate the first factor in the last expression,andthefundamentaltheoremofcalculustodifferentiatethesecondfactor,gives d dttV(c;y) = 𝛿y(t)tV(c;y)+c(t), (8.25) a continuous analogue of (2.26). Considernowaninsurancepolicyon( x)withdeathbenefitsof b(t)paidatthemomentof death,andpremiumspaidcontinuouslyattherateof 𝜋(t)attime t.Let c(t) = 𝜋(t)−b(t)𝜇x(t), the annual rate of continuous cash flow. The reserve at timeton this policy is just−tV(c;yx), which we will denote by justtV. Substituting in (8.25), and using (8.17), d dttV = tV(𝜇x(t)+𝛿(t))+𝜋(t)−b(t)𝜇x(t), SOME EXAMPLES OF EXACT CALCULATION 125 or d dttV = 𝛿(t)(tV)+𝜋(t)−𝜇x(t)(b(t)−tV). These last two formulas are the continuous versions of (6.7) and (6.8), respectively. In the first one, we view accumulation at interest and survivorship. In the second, we view the accumulation at interest only, and take the death benefit as the net amount at risk,b(t)− tV. Thisresultisoftenreferredtoas Thiele’sdifferentialequation namedaftertheDanishactuary T. N. Thiele, who discovered it in the late 19th century. In certain simple cases, for example, constant𝜇 and 𝛿, this equation can be solved analytically.See,forexample,Exercise8.21(b).Ingeneralhowever,theonlyfeasibleapproach is a numerical one. It is of interest to see what the solution will look like when we employ Euler’smethod, one of the main procedures for numerical solutions of differential equations. We describe this first for a general first order equation of the form f ′(t) = g(t,f(t)) given an initial valuef(t0). We pick a step sizeh> 0 and then simply replace the derivative with the difference quotient f(t+h)−f(t)∕hleading to a recursion f(t+h) ≐ f(t) +hg(t,f(t)) (8.26) Wethenworkbackwardsandforwardsfrom t0,startingwithgivenvalueof f(t0).Forexample, from(8.26)wegetanapproximatevalueof f(t0 +h)andwethenagainapply(8.26)withthat value to get an approximate value off(t0 +2h), and so on. Iterating the procedure we can compute approximate values off(t), wheretdiffers fromt0 by an integral multiple ofh.F o r sufficientlysmall h,thisshouldgiveareasonableapproximationstothetruevalues.Applying this to Thiele’s equation we get t+hV = tV(1+h𝛿t)+h𝜋t −h𝜇x(t)(bt −tV). (8.27) Note that what we get is very close to Equation (6.8) applied to time units of lengthh, wherefortheperiodrunningfromtime ttotime t+h,wetaketheinterestrateforthisperiod ash𝛿t,themortalityrateforthisperiodas h𝜇x(t),andthepremiumpaidforthisperiodas h𝜋t. The only difference is that (8.27) does not apply the interest factor to the premium, assuming in effect that it is paid at the end of the period. For very short periods this will make little difference. 8.10 Some examples of exact calculation Asweindicatedabove,thenormalapproachtocalculatingvaluesinthecontinuouscaseisto invoke UDD and reduce to the yearly case. Conceivably, one could have some nice analytic expressions for mortality functions and therefore be able to calculate values exactly. This is rarely done in practice. In the first place, it seems to be extremely difficult to find functions inclosedformthatprovideanaccuratepictureofobservedmortality.Second,evenwithsuch 126 CONTINUOUS PAYMENTS functions, it may be difficult to actually carry out the integration. For illustrative purposes, however, we will look at a few particularly simple cases. These are not meant to be typical of modern human mortality. In Chapter 14, we will introduce some more realistic mortality functions. 8.10.1 Constant force of mortality Supposethat 𝜇(x)isaconstant 𝜇forall x.Thisofcourseisunrealistic,sincewewouldexpect, as verified by observed data, that in general𝜇(x) increases withx. This is precisely what we mean by theaging process. There are exceptions. In the very early years there appears to be a decrease in𝜇(x), reflecting the effect of early childhood diseases. There is also a noted decrease in the early adult ages, around 25 or so. Many attribute this to the effect of a large number of automobile accident deaths for people in their early twenties. From age 30 onwards there is a rapid, exponential growth in𝜇(x), until around age 70, where𝜇(x) continues to increase but the exponential growth dampens out. There are some who predict that with medical and genetic advances, we may be approaching a situation where we can freely transplant or replace any genes or body parts that go awry. They postulate in essence that we will then have eliminated aging and achieved a constant force of mortality. This does not mean that people will no longer die, but only that the 90-year-old will be no more likely to die at any time than the 20-year-old. At the present time, however, we are not at all close to this state and it is still in the realm of science fiction. In any event, we look at some of the mathematical consequences of this assumption. In the first place, note that𝜔 is now∞, since tpx = e−𝜇t > 0, which means that there is at any time, a positive probability of continuing to survive. In our examples we will also assume a constant force of interest𝛿. Our first observation is that with these assumptions, ax = ∫ ∞ 0 e−𝛿te−𝜇tdt= ∫ ∞ 0 e−(𝜇+𝛿)tdt= 1 𝜇 +𝛿. This is easily verified intuitively. As we have seen, the force of discount for the interest and survivorship discount function under our assumptions is the constant𝜇 +𝛿. Therefore, an investment of 1∕(𝜇+𝛿) will provide a continuous return at the annual rate of [1∕(𝜇 +𝛿)]⋅ [𝜇 +𝛿] = 1. Taking𝛿 =0, we see as well that ◦ex = 1 𝜇, which again makes sense, since as𝜇 increases, life expectancy should decrease. Turning to insurance, the constant forces of interest and mortality lead to Ax = ∫ ∞ 0 e−𝛿te−𝜇t𝜇dt= 𝜇 𝜇 +𝛿. (8.28) Note, as a check, that when𝛿 = 0 we getAx = 1. This is obvious, since the insured is sure to die sometime and receive 1 unit, which has a present value of 1 under a zero interest rate. SOME EXAMPLES OF EXACT CALCULATION 127 Thisformulahasaninterestingalternatederivationwhichavoidsanyintegration.Suppose you have wealthAinvested at a constant force of interest𝛿 and each instant of time, you use theinterestearningsof 𝛿Atobuyaninstantofterminsurance,whichwillcostyou 𝜇 perunit. So the amount of insurance purchased at each unit of time will be (𝛿∕𝜇)A. At death you will have the principle ofAplus the insurance benefit, so to make the total wealth at death equal to 1, we want (𝛿∕𝜇)A= 1−A. Solving forAgives precisely the above quantity. For a 1-unit whole-life policy, with level premiums payable continuously for life, the annual rate of premium payment is given by P= Ax ax = 𝜇, and the reserve on such a policy at any timetis given by Ax+t −Pax+t = 0, asweseefromsubstitutingfromabove.Infact,thisisjustthecontinuousanalogueofExample 6.2.Sincethecostofprovidingtheinsurancedoesnotincreasewithtime,theamountcollected atanyinstantisexactlywhatisneededatthatinstant,andthereisnoaccumulationofareserve. 8.10.2 Demoivre’s law Another extremely simple mortality assumption states that for all agesx, tpx = 1− t 𝜔−x,0 ≤ t≤ 𝜔−x. We have already encountered this law in Example 8.1. It is known asDemoivre’s law, named after Abraham Demoivre, an eighteenth-century mathematician, perhaps best known forhistrigonometricidentities.Heisreputedtohaveusedtheaboveformulainmakingsome life table calculations. We can look upon his reasoning as follows. We know that for fixedx, the function tpx decreases from a value of 1 att= 0t oav a l u eo f0a tt= 𝜔. Equipped with nootherinformationatall,Demoivremadethesimplestpossibleassumption,namelythatthe graph of the function was a straight line. It is obviously far too simplisticto be representative of actual mortality, but we will nonetheless investigate its mathematical consequences. We get a very simple expression for life expectancy: ◦ex = ∫ ∞ 0 ( 1− t 𝜔−x ) dt= 1 2(𝜔−x). That is, at any age, one can expect to live one-half of the maximum remaining lifetime. Similarly,takinganupperlimitof nintheaboveintegral,wecalculatethe n-yeartemporary life expectancy asn−n2∕2(𝜔−x) = n( n∕2px ). That is, we multiply the maximum possible future lifetime over the nextnyears by the probability of living to the middle of this period. The force of mortality takes a particularly simple form. 𝜇x(t) =− d dt tpx tpx = 1 𝜔−x−t,0 ≤ t<𝜔 −x 128 CONTINUOUS PAYMENTS whichverifiesExample8.1,wherewedidtheinversecalculation.Thisatleasthasthedesirable feature of being increasing. This leads to tpx𝜇x(t) = 1 𝜔−x = qx,0 ≤ t<𝜔 −x. WenoticedabovethatthisformulaheldunderUDDforaninteger xandtintheinterval(0,1). Under Demoivre’s law, it holds for allxandt. As a consequence, Ax = 1 𝜔−x∫ 𝜔−x 0 v(t)dt= 1 𝜔−xa(1𝜔−x;v). (8.29) Once again, we can check that under zero interest, we getAx = 1. Similarly, Ax(1n) = 1 𝜔−xa(1n;v). A variation of Demoivre’s law, which we will call themodifiedDemoivre’slaw ,i sg i v e n by 𝜇x(t) = 𝛼 𝜔−x−t,0 ≤ t<𝜔 −x−t, for some𝛼> 0. Our original form was with𝛼 = 1. Using (8.16), this means that tpx = ( 1− t 𝜔−x )𝛼 ,0 ≤ t<𝜔 −x, which leads to a simple life expectancy formula, ◦ex = ∫ 𝜔−x 0 ( 1− t 𝜔−x )𝛼 dt= 𝜔−x 𝛼 +1. 8.10.3 An example of the splitting identity The following is an example showing the use of the splitting identity in the continuous case. Example 8.5 Suppose that the force of mortality is given by 𝜇(x) = {0.02, 30 ≤ x≤ 50, 1 100−x,5 0 ≤ x< 100. The force of interest𝛿 is a constant 5 %. CalculateA30. Solution. By the splitting identity, A30 = A30(120)+v(20)20p30A50. FURTHER APPROXIMATIONS FROM THE LIFE TABLE 129 Proceeding as in the previous examples of this section, A30(120) = 𝜇 𝜇 +𝛿(1−e−20(𝜇+𝛿)), A50 = 1 50 ( 1−e−50𝛿 𝛿 ) , v(20)20p30 = e−20(𝜇+𝛿). Substituting for𝜇 and𝛿, we getA30 = 0.3058. 8.11 Further approximations from the life table Suppose you want to approximate the force of mortality given only a life table. For ages that are not integers, the UDD assumption gives an easy answer. For an integerxand 0< t < 1, we obtain from (8.20) that 𝜇(x+t) ≐ q x 1−tqx . Suppose now we want to approximate𝜇(x) wherexis an integer. We could taket= 0i nt h e above, to get an estimated value of𝜇(x)a sj u s tqx, but we could also replacexby x−1 and take t= 1 to get an estimated value of𝜇(x)a sqx−1∕(1−qx−1). These are unlikely to be the same.Itiseasytoseewheretheproblemlies.FromUDD,thefunction tpx ispiecewiselinear asafunctionof tandthederivativewillnotexistatintegervaluesof t.Weareineffectgetting the different right and left hand forces of mortality at the integer points. To get a unique answer we need to consider an interval containingxas a interior point, such as (x−1,x+1). One such approach is to note that p x−1px = 2px−1 = e−∫ 2 0 𝜇x−1(t)dt. We now approximate the integral by one of the simplest methods for doing so, namely the mid-point rule. This simply says that to approximate an integral of a nonnegative function overanintervalyoutakethevalueofthefunctionatthemidpointoftheintervalandmultiply by the length of the interval. (This approximation is exact when the function is linear.) With this method, the right hand side above is approximated bye −2𝜇(x), and solving gives the formula 𝜇(x) ≐ −logpx−1 +log px 2 . (8.30) (See Exercise 8.24 for a check on the accuracy of this approximation.) This can be furthered simplified by using the fact that for small values ofr, −log(1−r) is very close tor, which leads to the approximation 𝜇(x) ≐ q x−1 +qx+1 2 , (8.31) a average of two values ofqat surrounding ages, which has a certain intuitive appeal. We now return to developing the three-term Woolhouse formula introduced in 7.5, which offers an alternative method to the𝛼(m)−𝛽(m) method for fractional annuity computations. 130 CONTINUOUS PAYMENTS FormotivationwerevisitFormula(7.13)whicharosefromtheassumptionthatforallintegers x,and0 ≤ s≤ 1,thequantity y x(s)isoftheform1 −aswherea= 1−vpx.Thisassumption, together with constant interest, implies that spx = e𝛿s(1−as), so that d dsspx = e𝛿s[𝛿(1−as)−a], andat s= 0thisderivativeequals 𝛿 −a.Thiswillbepositivewhen 𝛿> a= 1−vpx whichis equivalent after some manipulation to px > e𝛿(1−𝛿). This explains the inconsistency observed in Example 7.5. If the above condition holds, and we assume the piecewise linearity ofyx(s) as given above, the positivity of the derivative at s= 0 means thatspx actually takes values greater than 1 for sufficiently small values ofs! Writing this in terms of𝓁x we can interpret it fancifully as meaning that people who have died are coming back to life in even greater numbers than were there before, causing the life annuity to be worth more than the one with fixed payments. Suppose we instead make the next simplest assumption. Assume thaty x is a quadratic on [0,1] so that for some constantsa,b, yx(s) = 1−as−bs2,0 ≤ s≤ 1. (8.32) This is the only assumption we make now. We do not need constant interest. Chooseanyinteger mandinvokewell-knownformulasforthesumofconsecutiveintegers (as we did in Chapter 7) as well as the formula for the sum of their squares. These result in m−1∑ i=0 i m = m−1 2 , m−1∑ i=0 ( i m )2 = m−1 2 − m2 −1 6m . We now follow the method introduced in Section 7.1 to compute the relevant vectoru.I t follows from (8.32) that u0 = 1 m m−1∑ j=0 yx(j∕m) = 1−(a+b) (m−1 2m ) +2b ( m2 −1 12m2 ) . Next, calculate the coefficientsa+band 2b. From (8.32) a+b= 1−yx(1) = (∇ 1∞)0. (8.33) using the notation of (2.14) with respect to the discount functionyx. STANDARD ACTUARIAL NOTATION AND TERMINOLOGY 131 Wenowconsidertheinterestandsurvivorshipforceofdiscount 𝜆x(s) =− y′ x (s)∕yx(s),and let 𝝀x denote the vector whose entry at indexkis 𝜆x(k). From (8.32) y′ x (s) =− a−2bs, so that assuming differentiability at integer points, 2b= y′ x (0)−y′ x (1) =− 𝜆x(0)+yx(1)(𝜆x(1) =− (∇𝜆x)0. (8.34) From (8.33) and (8.34) u0 = 1− (m−1 2m ) (∇1∞)0 − ( m2 −1 12m2 ) (∇𝜆x)0. Similarly we can verify that uk = 1− (m−1 2m ) (∇1∞)k − ( m2 −1 12m2 ) (∇𝜆x)k. Then by invoking (7.2) and (2.15) we arrive at the three-term Woolhouse formula. ̈a(m) x (c) ≐ ̈ax(c)− m−1 2m ̈ax(Δc)− m2 −1 12m2 ̈ax(𝝀x ∗Δ c). (8.35) The particular form in the case of level payments is ̈a(m) x (1n) ≐ ̈ax(1n)− m−1 2m (1−v(n)npx) − m2 −1 12m2 (𝜆x(0)−𝜆x(1)v(n)npx ). A direct calculation from (8.35) is not practical, as normally we will be computing from the life table and will not have available the exact values of𝜇(x) needed to compute𝝀x. However we can simply use the approximations given by (8.30) or (8.31) in order to derive numerical results. Formula(8.35)seemstoavoidforthemostparttheinconsistencyofalifeannuityhaving a higher present value than the corresponding fixed interest period annuity, but this can still arise in extreme cases. See Exercise 8.27. 8.12 Standard actuarial notation and terminology We first review some terminology. Policies where benefits are paid at the moment of death, and premiums are paid continuously are sometimes referred to asfully continuous.Policies where benefits are paid at the end of the year of death, and premiums are paid yearly, are sometimesrefereedtoas fullydiscrete.Neitherofthesearerealistic.Thenormallifeinsurance contract involves benefits at the moment of death, and premium payable yearly (or possibly monthly, quarterly, etc). These are referred to assemi-continuous. 132 CONTINUOUS PAYMENTS The single premium notation for standard insurance and annuity contracts is the same as in the yearly case, except, as we have already introduced,A is replaced byA and ̈a is replaced bya. For annual premium contracts, the rules given in 5.8.2 apply. When premiums are payable continuously at a level rate, the same notation is used withPreplacingP. So, for example, r hP(Ax:n)denotesthenetannuallevelpremium,payablefor hyears,fora,1-unit n-year endowmentinsuranceon( x)withdeathbenefitspayableatthemomentofdeath.(Note that we do not omit the single premium symbol, as it starts withArather thanA.) r P(A 1 x:n ) denotes the annual rate of premium payment, when premiums are paid contin- uouslyatalevelratefor nyears,fora1-unit n-yearterminsuranceon( x)withbenefits payable at the moment of death. SymbolsforreservesfollowtherulesinSection6.9,whichsayessentiallythatyousimply takethepremiumsymbolandreplace PbyV,or PbyV,andmovethedurationsymboltothe top rather than bottom left. So, for example, thetth year reserve for the two above policies would be denoted byh t V(Ax:n) andV(A1 x:n ), respectively. TheassumptionofExercise7.13issometimesreferredtoasthe constantforceassumption since it implies that the force of mortality is constant overeach year, as shown by Exercise 8.21 below. It should not be confused with the assumption of a constant force over the entire span of life, as we discussed in Section 8.10. Notes and references The derivation of (8.18) uses a Riemann–Stieljes integral (see Rudin 1976, Theorem 6.17). For an alternate derivation of the general Woolhouse formula (which follows the original derivationofW.Woolhousein1859)seeDickson etal. (2013),AppendixB.2.InSection5.13 of that work, the author’s take a known function for𝜇x and show that the three-term formula gives remarkably close approximations to the true value. See also Exercise 8.26 for more confirmation of this. Exercises Type A exercises 8.1 Alifeannuityon( x)providesforbenefitsmadecontinuouslyfor2years,providedthat (x) is alive. The annual rate of payment at timetis c(t), defined by c(t) = {t,0 < t< 1, 1, 1 < t< 2. Theinterestrateis0, qx = 0.3and qx+1 = 0.4.Findthepresentvalue,assumingUDD. 8.2 Given qx = 0.2, qx+1 = 0.25, i= 0.25 and a vectorc = (100,300), calculate ax(c), assuming UDD. EXERCISES 133 8.3 You are given thatq40 = 0.1,q41 = 0.2, the rate of interest is a constant 25% and that UDD holds. Find the present value of a 2-year endowment insurance on (40), paying 1000 at the moment of death if this occurs before age 42, and 1000 at age 42 if (40) is alive at that time. 8.4 Suppose that q 50 = 0.2. Assume UDD. (a) Calculate𝜇50(0.2). (b) Given that the interest rate is 25%, find the single premium for a 1-year term insurance policy on (50) that pays 1 unit at the moment of death should this occur within 1 year. 8.5 Suppose that the force of mortality is given by 𝜇 x = B(1.09)x and (a) Find a value forB such the resulting mortality table is the illustrative table of Chapter 3. (b) Using this value ofBcompare the true value of𝜇 50 with the estimated value using formula (8.30). 8.6 Alifeannuityon(80)providesforbenefitsmadecontinuouslyfor2years,attheannual rate ofc(t) at timet, provided that (x) is alive. Suppose thatc(t) = t,0 < t< 2, and the interest rate is 0. Find the present value under each of the following assumptions. (a) q 80 = 0.09,q81 = 0.12, and UDD holds. (b) The force of mortality is given by 𝜇80(t) = 1 10−t,0 < t< 10. 8.7 For a certain mortality basis,q50 = 0.2. Suppose that the force of mortality𝜇50(t)i s changed to a new force of mortalitŷ 𝜇given by ̂ 𝜇50(t) = 2𝜇50(t)−0.1, 0 < t<𝜔 −50. Find the new value ofq50. 8.8 The force of mortality is a constant 0.04 and the force of interest is a constant 0.06. A term insurance policy has a level death benefit of 1 payable at the moment of death if this occurs within 40 years. Level annual premiums are payable continuously for 20 years at the annual rate of of𝜋.F i n d( a )𝜋,( b ) 10V. 8.9 Suppose that the force of mortality for ages over 60 is given by 𝜇60(t) = { t,0 < t< 1, 1, 1 < t. Find the probability that (60) will die within 2 years. 134 CONTINUOUS PAYMENTS 8.10 Suppose mortality follows Demoivre’s law with𝜔 = 110 and that the interest rate is 0. A term insurance policy on (60) has a level death benefit of 1000 payable at the momentofdeathprovidedthisoccurswithin40years.Netlevelpremiumsarepayable continuously for 20 years at the annual rate of𝜋. (a) Find𝜋. (b) Find 10V by both the prospective and retrospective methods. 8.11 Two actuaries, A and B, agree that the probability that a female age 60 will die within 10 years is 0.36. (a) ActuaryAdecidesthattheforceofmortalityformalelivesis1.5timestheforceof mortality for females lives, at all ages over 60. What is the probability that a male age 60 will die within 10 years, under A’s assumption? (b) Actuary B disagrees and decides that the force of mortality for males lives is obtained by adding a constant of 0.01 to the force of mortality for female lives, at allagesover60.Whatistheprobabilitythatamaleage60willdiewithin10years under B’s assumption? 8.12 An insurance policy provides for death benefits payable at the moment of death. The amount payable for death at timet is e 0.08t, for allt> 0. The force of mortality is a constant 0.06 and the force of interest is a constant 0.04. Net level premiums are payable at a constant rate for life. Find the rate of premium payment. Type B exercises 8.13 Show that under UDD, for an integerxand 0< t< 1,𝜇x(t) = qx∕(1−tqx). 8.14 The central rate of mortality at agexis defined by mx = dx ∫ 1 0 𝓁x+tdt . (a) Show thatmx can be expressed as a continuous weighted average of values of the force of mortality. (b) Show that under UDD,mx = 𝜇x ( 1 2 ) . 8.15 A whole-life insurance, with net level premiums for life payable continuously at the annual rate of𝜋, provides for death benefits of e𝛾t at the moment of death, for death at timet. Given that the force of mortality is a constant𝜇 and the force of interest is a constant 𝛿, such that𝜇 +𝛿>𝛾 , find𝜋 in terms of𝛾,𝜇 and𝛿. 8.16 A 1-year deferred, 1-year life annuity on (x) provides for continuous payments from time1totime2provided( x)isalive.Therateofpaymentattime1 +tise0.1t,0 < t< 1. This is purchased by a single premium. For death during the first year, the single premium is returned without interest at the moment of death. You are given that q x = 0.1, qx+1 = 0.2, and that the force of interest𝛿 is a constant 0.10. Assuming UDD, find the single premium. EXERCISES 135 8.17 An n-year term insurance policy on (30) has a constant death benefit of 1, payable at themomentofdeath.Premiumsarepayablecontinuouslyataconstantratefor nyears. MortalityfollowsDemoivre’slawwith 𝜔 = 100,andtheinterestrateis i= 0.Findthe rate of premium payment as a function ofn. 8.18 Theforceofinterestisaconstant 𝛿.Theforceofmortalityisaconstant 𝜇1 forthefirst nyears, and a constant𝜇2 after nyears. Derive expressions forax and Ax in terms of 𝜇1,𝜇2,𝛿 andn. 8.19 Suppose Demoivre’s law holds and the force of interest is a constant𝛿.F i n da n expression forax(1n) in two ways. First use formula (8.9) directly, and second apply (8.23) to the expression given forAx(1n) under the same assumptions. Verify that as𝛿 approaches 0, you get the formula for then-year temporary life expectancy. 8.20 A ‘sawtooth’ life insurance policy on (x) provides for a death benefit oftpaid at death if this occurs at timek+t, wherek is an integer and 0< t< 1. Assume UDD and constant interest. Show that the net single premium for this policy is equal to𝛽(∞)Ax. 8.21 Supposethattheforceofinterestisaconstant 𝛿 andtheforceofmortalityisaconstant 𝜇. Consider ann-year endowment insurance with net premiums payable continuously at the constant annual rate of𝜋. Death benefits ofbt are payable at the moment of death for death at timet< n. There is a payment of 1 at timenif the insured is then alive. (a) Supposethat b t = 1forall t< n.Usethereservedifferentialequationtoshowthat tV = (𝜋 −𝜇)et(𝜇+𝛿) −1 𝛿 +𝜇 , 𝜋 = 𝜇 + 𝜇 +𝛿 en(𝜇+𝛿) −1. Give an intuitive explanation of these results. Verify that these agree with the formulas given at the end of Section 8.8. Show that the reserve formula is a continuous version of formula (6.15). (b) Suppose now thatb t = 1+tV for allt< n. Use Thiele’s differential equation to derive formulas fortV and𝜋. 8.22 Show that under the assumption of Exercise 7.13, the force of mortality is constant over each year. That is,𝜇x(t) = 𝜇x(s) wherexis a nonnegative integer andsand tare between 0 and 1. 8.23 Give an alternate derivation of (8.12) by using the continuous endowment identity, as given at the end of Section 8.8, and (8.23). First prove it forc = e0. 8.24 Suppose that modified Demoivre’s law holds with𝛼 = 2. If𝜇x = 30 what is◦ex? 8.25 Suppose that the force of mortality is given by 𝜇(x) = B(1.09)x. (a) Find a value forB such the resulting mortality table is the illustrative table of Chapter 3. 136 CONTINUOUS PAYMENTS (b) Using this value ofB compare the true value of𝜇(50) with the estimated value using formula (8.30). 8.26 Suppose that the force of mortality is a constant 0.2 and the force of interest is a constant 0.1. Computëa(4) x (11) by four methods. (a) formula (7.9), (b) formula (7.13), (c) formula (8.35) and (d) exactly. Compare the results. 8.27 Find an example of some life table values and an interest rate, such that using (8.35) together with ( 8.30), yields an inconsistent result. Hint: Choose life table values so that the estimated value𝜇 x+1 is much higher than the estimated values of𝜇x. 8.28 A life is subject to a constant force of mortality of 0.15. The force of interest is a constant0.10.Aninsurancecontractonthislifeprovidesforadeathbenefitof e.05t for death at timet, with level premiums payable continuously for life. (a) FindtV. (b) Write down Thiele’s differential equation for the reserve, and verify that your answer to (a) satisfies this equation. Spreadsheet exercise 8.29 Modifypreviousspreadsheetstohandlethecaseofdeathbenefitspayableatthemoment of death. 9 Select mortality 9.1 Introduction In this chapter we will discuss a refinement to our basic model. In previous chapters we assumed that for a sufficiently homogeneous group – North American male nonsmokers, for example – mortality rates are a function of attained age only. This may be a reasonably accurate assumption for the totality of people in this group, but it need not apply for certain subsets. An insurer, after all, is not interested in the mortality of the general population, but only the subset of those who buy policies. Observed data show that for purchasers of life insurance, mortality depends not only on age, but onduration since the policy was purchased. To see why this should be true, let us compare two people both age 60, alike in all respects except that A just recently purchased an insurance policy, and B purchased a policy 1 year ago at age 59. During the next year, we would expect that A will have a higher probability of living to age 61 than B will. This arises from the fact that the insurer does not have to accept everyone who applies for a policy. Applicants can be requested to undergo medical exams, or to furnish information regarding their health, lifestyle and other factors, in order to verify that they are reasonable risks. Individual A has just been so certified. Individual B, on the other hand, was certified as a good risk 1 year ago, but this condition could have changed. B may have contracted a fatal disease, or taken up life- threatening habits. Moreover, if we consider a third individual, C, also age 60, similar to A and B, except that he/she was sold a policy at age 58, his/her chance of living over the next year can be expected to be even less that that of B. This person has had 2 years in which to deteriorate. Thestandardactuarialmethodtohandlethissituationistousethesamenotationaldevice that we introduced in Section 4.3.2. We keep the convention that subscripts denote attained age, but put square brackets around the age that the policy was issued, in order to separate age and duration. Therefore, the symbol sq[x]+t denotes the probability that a person now age x +t, who was sold insurance at agex, will die within the nexts years. This is known Fundamentals of Actuarial Mathematics, Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 138 SELECT MORTALITY as aselect mortality rate, as it incorporates the effects of the insurer’s ability to select. We define sp[x]+t = 1−sq[x]+t. As usual we omit left subscripts of 1 forq and p. Our discussion above showed that q[60] ≤ q[59]+1 ≤ q[58]+2 …, and, in general, q[x]+t ≤ q[x−s]+t+s, (9.1) for all nonnegativex andt and 0≤ s ≤ x. 9.2 Select and ultimate tables To fully model this phenomenon we would need a separate mortality table for each agex to cover policies issued at that age. Of course the tables would get shorter with increasing issue age. The agex table would consist of the𝜔−x entries {q [x]+t,t = 0,1, …,𝜔−x −1}. Observationsshow,however,thattheselectioneffectsdecreasewithtimeandcanbeassumed to wear off after a certain duration, known asthe select period. If the select period isr,w e can expect that two individuals of the same age, who have been issued insurancer or more years ago, will exhibit the same mortality rates, even if the issue age is different. This allows for some simplification. We can revert to the symbolq y to mean the probability that a person now agey who was issued insurancer or more years ago will die within a year. In other words, we postulate that equality will hold in (9.1) whent ≥ r, so we can remove the square bracket and denote the common value byqx+t.T h er a t e sqy are known asultimate rates, and the resulting life table is known as aselect and ultimate table. A common choice for select periodis15years,althoughsomerecenttableshaveusedasmuchas25.Thesimpleexample below has a select period of 2, but that is enough to illustrate the basic idea. xq [x] q[x]+1 qx+2 x +2 60 0.10 0.12 0.15 62 61 0.11 0.14 0.17 63 62 0.12 0.15 0.18 64 Tofindthemortalityratesapplicabletoapolicyissuedatage x,westartattheleftcolumn, readacrossuntilwegettotheultimatecolumn,andthenreaddown.So,forexample,therates usedforapolicyissuedatage60wouldbeinorder q [60], q[60]+1, q[60]+2 = q62, q63, q64,…, whichinthisexampleare0.10,0.12,0.15,0.17,0.18,….Wecanpicturetheflowofmortality rates as several streams, all running into the same river, which is the ultimate column. Therefore, we do not need a complete separate table for each age, but onlyr +1 columns. CHANGES IN FORMULAS 139 Tableswithaselectperiodof0years,whichispreciselywhatwehavebeendiscussingin previous chapters, are known asaggregate tables. Some people prefer to construct a select and ultimate table of𝓁x, although this is not really necessary, for as we have seen, one can always work directly with the values ofqx. The process is straightforward, although it requires a little care. One completes the ultimate column first, proceeding as in Chapter 3 to calculate values of𝓁 x starting with an arbitrary value for𝓁r which is the first entry in this column. One then simply works backwards to complete the other entries, using the recursion 𝓁[x]+k = 𝓁[x]+k+1 1−q[x]+k . Forexample,intheabovetable,if 𝓁63 = 5000,then 𝓁[61]+1 = 5000∕0.86 = 5814and 𝓁[61] = 5814∕0.89 = 6353. 9.3 Changes in formulas To change our previous formulas to incorporate select mortality is usually just a matter of inserting a square bracket around the issue age. We will look at some of these in detail. Chapter 3 The multiplication rule (3.6) becomes s+tp[x] = sp[x] tp[x]+s. (9.2) Chapter 4 We will denote the interest survivorship discount function for a policy issued at agex byy[x]. From (9.2) we derive y[x](k,k +n) = y[x](k +n) y[x](k) = v(k,k +n)np[x]+k. (9.3) Takingn = 1, the recursion formula (4.5) now becomes y[x](k +1) = y[x](k)v(k,k +1)p[x]+k. It is important to note that with select mortality, (4.6) and (4.7) no longer hold. As mentioned, we need a separate table for each issue age, and there is no longer any necessary relation between the ages. Note, for example, that y [x+k](n) = v(n)np[x+k], and even ifv(k,k +n) = v(n) this is not the same as the right hand side of (9.3) unlessk is greater than or equal to the select period. 140 SELECT MORTALITY The subscript of [x]+k on annuity symbols is used in the same way as introduced in Section 4.3.2, except that now, modification to the mortality as well as the interest is needed. That is ̈a [x]+k(c) = ̈a(c;y[x] ◦ k) = 𝜔−x−k−1∑ j=0 cjv(k,k +j)jp[x]+k. Notethat ̈a[x]+k(c) = ̈ax+k(c),if k isgreaterthantheselectperiod,andinterestisconstant. With non-constant interest however, these are not necessarily equal even ifk is greater than the select period, as we observed in Chapter 4. Chapter 5 In the third formula in (5.1),qx+k becomes q[x]+k. The subscript [x]+k is used for insurances in the same way to that noted above for annuities. That is A[x]+k(b) = 𝜔−x−k−1∑ j=0 bjv(k,k +j +1)jp[x]+k q[x]+k+j. Chapter 6 Nochangesarenecessaryinthischapterexcepttonotethatthesubscriptsof[ x]+k inreserve formulas are modified as we noted for Chapters 4 and 5 above. Chapter 8 We define 𝜇x(t) = lim h→0 hq[x]+t h = lim h→0 𝓁[x]+t −𝓁[x]+t+h h𝓁[x]+t . (We are following standard notation here. The square bracket is not necessary since thex is already separated from thet.) The basic identity (8.16) can then be written tp[x] = e−∫ t 0 𝜇x(r)dr. Of course, if the select period is 0 then𝜇x(t) is just equal to𝜇(x +t) and agrees with the notation that we have already introduced in Chapter 8. Spreadsheets The interested reader may want to adapt the previous spreadsheets that have been introduced to incorporate a select and ultimate table. However, without doing this fully, one can still do calculations directly on the existing spreadsheet for each particular problem. For a policy issued at age (x) on a table with a select period ofr years, one simply replaces the entries of q x+t withq[x]+t for t = 0,1, …,r −1. PROJECTIONS IN ANNUITY TABLES 141 9.4 Projections in annuity tables There are other situations where a two-dimensional approach to mortality rates is called for. One such example is annuity tables. We need however a different formulation than that presented above. The select tables we have described would not normally be used for annuities. Purchasers of annuities are generally people who expect to live longer than the average or they would be unlikely to enter into such a contract. In any event the insurer is not worried about deteriorating mortality with duration. The problem here is exactly the opposite, namely, people may live longer than the tables predict, requiring that more must be paid out in benefits than the premiums allow for. This has proved to the case in recent years, as advances in medical and public health knowledge, better lifestyles and other factors have led to continually improving mortality, causing issuers of annuities as well as pension plans, to be confronted with what has been termedlongevity risk. In addition to age,year of birth becomes an important variable when pricing annuities. A 60 year old born say in 1950 could beexpectedtohavelesschanceofdyinginthenextyear,thana60yearoldbornin1930,due to improvements in mortality that have come about over the 20 year period. In place of the rates q x, we would like a two variable functionqx,n where x denotes age andn denotes year of birth. We will alter this notation somewhat in order to point out the parallel nature of this situationwithouroriginaldiscussionofselectmortality.Supposewehavepickedabaseyear, say the year a mortality study was completed. We will now letq [x]+t denote the probability of dying within 1 year, for a person now agex +t who was agex in the base year. In other words we have just extended the select notation by allowingt to denote time elapsed since some particular event, which could be anything at all. In our original application, the event wasobservationforinsurancepurposes.Forannuitytables,itistheyearofthemortalitystudy that produced the table. In this case, we have in place of (9.1) q [x]+t ≥ q[x−s]+t+s, since,letting b denotethebaseyear,thetermontheleftreferstosomeoneborninyear(b −x) while that on the right refers to some one born in a later year (b −x +s). There is a however a major problem in constructing a table to show such rates. Suppose for example that a mortality study of annuitants was completed in the year 2000. The result would be a table showingq [x], the one year mortality rates for people agex in the year 2000. Now suppose we want to know whatq[60]+1 is. This would be the rate for a person age 61 who was born in 1940, but in the year of the study no such person would ever have existed. The rateq [60]+1 (as well as all rates witht > 0) could not be based purely by a statistical analysis of past mortality data, but would involve the prediction of future trends in mortality improvement. This is rather difficult task, and present actuarial usage involves rather simple tools. The usual method is to assign a yearly projection factorr y for each age, which represents the reduction in mortality expected at that age for each year. That is, it is assumed that, q [x]+t = q[x+t](1−rx+t)t. Following is an example taken from an actual table known as the Uninsured Pensioner (UP) 1994 Table with projection scale AA. 142 SELECT MORTALITY Example 9.1 The UP 1994 table shows male mortality rates for age 60–62 of 0.008576, 0.009633, 0.010911, respectively and projection factors of 0.016, 0.015, 0.015, respectively. Amaleisage60intheyear 2010. Usingthistablefindtheprobabilitythathewilllivetoage 63.Comparethiswiththeprobabilitythatamalewhowasage60in1994wouldlivetoage63. Solution. The required probability is [(1−(0.008576)(0.984) 16][(1−0.009633)(0.985)17][(1−0.010911)(0.985)18] = 0.9778. For the male age 60 in 1994, the probability would be [(1−(0.008576)][(1−0.009633)(0.985)][(1−0.010911)(0.985)2] = 0.9716. There is no reason to believe that mortality improvement will stop, so the concept of a select and ultimate table does not appear in this context. Therefore it is not usual in the actuarial literature to view the table that we have described as a select table. Rather it is viewed as a collection of separate tables, one for each agex in the base year. Each of these separate tables is often referred to as agenerational table, as it traces the mortality of people all born in the yearb −x. In the construction above each generational table would appear as the row of the select table as described. Each column of this select table can be considered as a version of the original table projected to a future year, and is referred to as thestatic table for that year. It practice, insurers often simplify calculation by making use of the appropriate static table for setting annuity rates in a particular year, rather than using the more accurate generational tables, which would involve a separate table for each age. 9.5 Further remarks There are many other situations in which the concept of select mortality applies. For another example involving rates decreasing with duration, suppose we are studying the effects of a certain treatment for cancer patients. This could depend both on age and the duration since treatment. In this case, if a person was treated a year ago, and is still alive, there is some indication that the treatment is working, and that such person is more likely to live than someone of the same age who has just received the treatment. Exercises 9.1 A person now age 62 was sold insurance 1 year ago at the age of 61. Mortality is given by the table at the beginning of the chapter. Find the probability that this individual will die between the ages of 63 and 65. 9.2 A4-yearterminsurancepolicysoldtoapersonage61providesfor1000paidattheend oftheyearofdeathshouldthisoccurbeforeage65.Levelannualpremiumsarepayable for 4 years. The interest rate is a constant 5% and mortality is given by the table at the beginning of the chapter. (a) Find the premium. (b) Find 2V. EXERCISES 143 9.3 Explain in words what the following indicates: (6p[30]+10)(8q[30]+16). 9.4 For thetablegiven inSection 9.2, construct the corresponding table of𝓁[x]+t, given that 𝓁62 = 5000. 9.5 An annuity mortality table, with projection factors for improvement, constructed in the year 2000, shows q70 = 0.02, q71 = 0.03 q72 = 0.04, r70 = 0.02, r71 = 0.018, r72 = 0.016. Find the probability that a person age 70 in the year 2010 will die between the ages of 72 and 73. Spreadsheet exercises 9.6 Consider a select and ultimate version of the sample table, with a 15-year select period, as follows. Forx ≥ 50, andt = 0,1, …,14wetake q[x]+t = 1−(1.00001)15−te−0.00005(1.09)x+t . Interestratesareaconstant6%.Fora20-yearendowmentinsuranceon(50)withalevel deathbenefitof100000andapureendowmentof100000atage70,purchasedbylevel net premium for 20 years, calculate (a) the premium, (b) 10V. 9.7 Using the same table as in Exercise 9.6, calculate (a)e[60],( b )e[50]+10. 10 Multiple-life contracts 10.1 Introduction Inpreviouschapterswehavestudiedinsuranceandannuitycontractssoldtoasingleindivid- ual. There are many situations when such contracts are sold to a group of several people. We will concentrate on the case of two lives, the most common arrangement, which is sufficient to exhibit most of the complications. The following are a few of the more usual situations. A married couple may wish to purchase an annuity that pays income while either of them is alive. They may also want an insurance policy that pays benefits on the second death, which in some jurisdictions enables themtopassonassetswithoutpayingestatetaxes.Twobusinesspartnersmaydesireapolicy thatpaysadeathbenefituponthefirstdeathofthetwo.Apersonmayarrangeforanannuityto bepaidtoadependent,wherepaymentsbeginonlyonthedeathofthesupportingindividual. Wewilldiscusstheseandmoreinthefollowingsections.Forsimplicityweassumeaggregate mortalityinthischapter.Itisnottoodifficulttomodifytheresultsfortheeffectsofselection. 10.2 The joint-life status Consider a pair of lives, (x) and (y). Suppose that we consider the pair to be in a state of survival when both of them are alive. In other words, the pairfails on the first death. Such a pair is commonly known as ajoint-lifestatus. An annuity sold on a joint-life status would provide income as long as the status survived, that is, as long as both individuals were living. Annuity benefits would stop upon the first death. An insurance policy sold on a joint-life status would pay a death benefit upon the first death. Whileitispossibletoconstructlifetablesinthiscase,itisawkward,anditisbesttowork directly with probabilities. The basic quantity we need is the probability thatbothof (x) and (y) will be alive intyears. We will denote this probability by tpxy. FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial THE JOINT-LIFE STATUS 145 We will now derive an expression fortpxy that can be calculated from the life table for single lives. The method discussed will be well known to the reader who is familiar with the concept of independence in probability theory (see Section A.2). Suppose, for example, that tpx = 0.9, tpy = 0.8. Imagine that we start by observing 100 pairs of lives, where one is agex and the other is agey. Of these 100 pairs, we can expect that in 90 cases, the life agex will be alive int years. Out of these 90 cases we can expect that 80% of the time the life ageywill survivetyears. This leaves on average 72 pairs of the 100 where both lives survive, so we should havetpxy =0.72. It is instructive to derive this in another way by looking at deaths rather than survivors. We expect 20 deaths altogether from the agey individuals. On average, two of these agey deaths should occur among the 10 cases where the agexindividual dies. This leaves 18 age ydeaths among the 90 cases of agexsurvivors, and therefore 90−18 = 72 cases where both survive. The particular numbers 0.9 and 0.8 do not matter and the same arguments show that in general we should have tpxy = tpxt py. (10.1) We define tqxy = 1−tpxy, the probability that the joint-life status will fail withintyears, which is the probability that at least one of (x) and (y) will die withintyears. Thereadershouldcarefullynotethat tqxy isnotequalto tqxt qy.Thelatterexpressionisthe probability thatbothlives will die withintyears, which is certainly less than the probability that at least one will die in this time interval. To expresstqxy in terms of single-life rates, we havetqxy = 1−tpxt py = 1−(1−tqx)(1−tqy), which gives tqxy = tqx +tqy −tqxt qy. (10.2) Wenowconfessthatalthough(10.1)and(10.2)arealmostuniversallyused,theargument wegaveaboveisnotquiteaccurate.Itwouldbeperfectlyvalidifthetwoliveswerecompletely independent of each other. This is unlikely to be true, however, for two people who would choose to buy an insurance or annuity contract together, such as a married couple or two business partners. One can expect that their lives are intertwined to the extent that life- threatening occurrences for one would also affect the other. Forexample,supposethatthetwolivesarebusinesspartnerswhofrequentlyflytogether. If one of them were to die in a plane crash, it is likely that the other one would as well. To illustrate the point, we take an extreme case. Return to the example given above, but suppose that the 100 pairs we start with consist in every case of two people who always fly together. We claimed above that there would be on average 2 agey deaths from the 10 cases where the individual agedx died. But if we expect that the lives are likely to die together, we can certainly expect more than this. For example, if all the agexdeaths were from plane crashes, we would necessarily have 10 agey deaths as well. This is extreme, but suppose that we have even 3 instead of 2 ageydeaths. This leaves 17 instead of 18 deaths from the 90 agex 146 MULTIPLE-LIFE CONTRACTS survivorsandwewouldhave tpxy =0.73.Thisshowsthatratherthan(10.1)wewouldexpect for the typical buyers of an insurance or annuity contract that tpxy > tpx tpy. (10.3) Formarriedcouplesthereisanadditionalfactorthatsupports(10.3).Statisticsbearoutthe fact that upon the death of one partner, the survivor, deprived of the care and companionship of the deceased, often tends to die earlier than otherwise expected. This has been referred to as the ‘broken heart syndrome’. For convenience, we will assume (10.1), or equivalently (10.2), unless otherwise stated. The reader should note, however, that most of the key formulas remain true without this assumption. The main use of (10.1) is to conveniently calculate numerical quantities, given the single-life data. Generalizationsofthejoint-lifestatuscanbefoundinChapters17and19.Onthequestion of dependence, see in particular Sections 17.5 and 17.6. 10.3 Joint-life annuities and insurances Consider a joint-life annuity contract payingck each year provided that both (x) and (y)a r e alive. In order for a payment to be made at timekwe need both lives to have survived to that time, and the probability of this is justkpxy. The present value of the benefits, denoted by ̈axy(c), is given by a formula exactly the same as (4.1), exceptkpxy replaces kpx: ̈axy(c) = N−1∑ k=0 ckv(k)kpxy, (10.4)‡ whereN =m i n{𝜔−x,𝜔−y}. We useN in this way throughout the chapter. To verify that it is the appropriate upper bound, note for example that ifx= 40,y= 70 and𝜔 = 110, the last possible payment would be at time 39. At time 40, (70) would not be living according to our model. As with single lives, we will use 1 ∞ to denote a vector with entries of 1, running to durationN−1, and by convention the omission of a benefit vector or function implies that it is 1∞. In the continuous case we have the formula ̄axy(c) = ∫ N 0 c(t)v(t)tpxydt, (10.5) ‡ for the present value of an annuity, with payments made continuously at the annual rate of c(t) at timet, provided both (x) and (y) are alive. We can also consider ajoint-life insurancepolicy that paysb k at the end of the year of the first death of (x) and (y) when this occurs between timekand timek+1. In other words, paymentwillbemadeattime k+1ifbothlivessurvive kyears,butbothdonotsurvive k+1 years. The probability of this is just kpxy −k+1 pxy. LAST-SURVIVOR ANNUITIES AND INSURANCES 147 This probability can also be expressed as kpxyqx+k:y+k, since for such an event to happen, both lives must survive fork years and then one must die in the subsequent year. (A colon is used to separate the two ages in a joint-life status if neededforclarity.)Thislatterexpressioncanalsobederivedfromthejoint-lifeversionofthe multiplication formula, k+1pxy = kpxypx+k:y+k, sinceforboth(x )and(y)tosurvive k+1years,theymustfirstbothsurvive kyears,andthen, being agex+kandy+krespectively, they must both survive one more year. Thepresentvalueofthejoint-lifeinsurance,denotedby Axy(b),canbewritten,analogously to the second and third expressions in (5.1), as Axy(b) = N−1∑ k=0 bkv(k+1)(kpxy − k+1pxy) = N−1∑ k=0 bkv(k+1)kpxyqx+k:y+k. (10.6)‡ Suppose that such a joint-life insurance were sold with annual premiums, based on the premiumpatternvector 𝜌.Premiumswillceaseuponthefirstdeath,sothepremiumpayments form a joint-life annuity. We determine the initial premium𝜋0 in exactly the same way as with a single-life insurance, namely, 𝜋0 = Axy(b) ̈axy(𝜌). (10.7)‡ 10.4 Last-survivor annuities and insurances 10.4.1 Basic results Consider a pair of lives (x) and (y), which we consider to be in state of survival ifeither of them is alive. In other words, the pair fails upon thesecond death. This is known as a last-survivorstatus. Thestandardsymbolforthisis xy,whichdistinguishesitfromajoint-life status.Theprobabilitythatthisstatuswillbeinastateofsurvivalattime tisdenotedby tpxy. To calculate this, we use (A.3), a basic fact from probability theory, already illustrated above in (10.2), which says that the probability that at least one of two events occurs is given by the sum of the probabilities of each occurring, less the probability that they both occur. This gives tpxy = tpx +tpy −tpxy. (10.8) 148 MULTIPLE-LIFE CONTRACTS We lettqxy denotetheprobabilitythat xywillfailbeforetime t.Clearly tqxy = 1−t pxy,so that if our independence assumption (10.1) holds, we can substitute in (10.8) to obtain tqxy = tqxt qy. (10.9) Alternatively,thiscanbederiveddirectlybynotingthatinorderforthisstatustofailwithin t years, we need to have both lives fail withintyears. The probability that (x) fails between timetand times+tis given either bytpxy −s+t pxy or bys+tqxy −t qxy. Assuming independence, the latter expression simplifies to (s+tqx)(s+tqy)−(tqx)(tqy). Formulasforannuitiesandinsurancesbasedonthelastsurvivorstatusareeasilyobtained. Asinthejoint-lifecasewesimplyreplacethesubscript( x)by( xy).So,foranannuitypaying ck at timek, provided that both (x) and (y) are alive, the present value is just ̈axy(c) = M−1∑ k=0 v(k)ckk pxy = ̈ax(c)+ ̈ay(c)− ̈axy(c), (10.10) where the upper limitMnow denotes the maximum of𝜔−xand𝜔−y.The final expression results directly from (10.8). The analogous formula holds for continuous annuities with̄areplacing ̈a. Similarly, the present value of an insurance contract payingbk at the end of the year of the second death of (x) and (y), if this occurs between timekandk+1 is given by Axy(b) = M−1∑ k=0 v(k+1)bk(kpxy −k+1 pxy) = Ax(b)+Ay(b)−Axy(b), (10.11) whereweagainuse(10.8),aswellasthesecondformulain(5.1)toderivethefinalexpression. Formula(10.11)canbeverifieddirectly.Suppose,forexamplethat( x)diesfirst,atatime betweenjandj+1. Then the first term on the right hand side paysbj and the third term pays −bj, so a net amount of 0 is paid on the first death. When (y) dies later at a time betweenk andk+1 only the second term pays andywill get the required death benefit ofbk.T h es a m e typeofreasoningcanbeappliedtoverify(10.10)aswellasmanyothertypesofmultiple-life contracts. These last survivor annuities and insurances are a special case of more general contracts that we will investigate in Sections 10.6 and 10.7. 10.4.2 Reserves on second-death insurances Reserves on last survivor insurances present a particular problem. For a second-death insur- ance, the true reserve at timek depends on the state of the pair. There are three possible cases. (For simplicity in notation we assume constant interest. In the general case one uses the investment discount functionv◦ kin computingAand ̈a.) kV(0) = Ax+k:y+k(b◦ k)− ̈ax+k:y+k(𝜋◦ k) if both lives survivekyears. kV(1) = Ax+k(b◦ k)− ̈ax+k(𝜋◦ k) if (x) only surviveskyears. kV(2) = Ay+k(b◦ k)− ̈ay+k(𝜋◦ k) if (y) only surviveskyears. MOMENT OF DEATH INSURANCES 149 Theabove formulasalsoassumethatthepremium patternremains unchanged on thefirst death.Ifthecontractcalledforreducedpremiumsinsuchanevent,thensuitableadjustments would be required in the second term of the formulas for kV(1) and kV(2). Itmayseempuzzlingatfirstthatthereserveonthiscontractshouldmakeasuddenchange on the first death, but a moment’s reflection will verify that thisis perfectly logical. The pairs for which the first death has already occurred present a greater liability and a higher reserve is needed. Taking a retrospective, point of view, what happens is that there is a transfer each periodfromthefundsofthepairswithbothindividualsstillliving,tothefundsofthosewhere one has already died. The problem is however that there is no reason for a pair of lives to report the first death to the insurer, unless so stipulated in the contract. This means that the insurer might well not beawareofthestateofthevariouspairs.Inthiscaseanappropriatesolutionwouldbetohold a weighted average of the different reserves. That is, for each contract the reserve at timek would be kpxy kV(0) +(kpx −k pxy) kV(1) +(kpy −k pxy) kV(2) kpxy . If the actual mortality experience is close to that of the weights used above, then this method should produce aggregate reserves which are close to what they should be by the more exact method. Note however that it would not be reasonable to base cash values on such a blended reserve figure, which will clearly be higher than kV(0). This would create a definite anti- selection opportunity as those pairs for which both are still living, could cash in the policy and receive an amount which is more than that which they are entitled to. 10.5 Moment of death insurances To handle moment of death insurances for two-life statuses, we need the joint-life analogue of the force of mortality, Definition 10.1 The forceoffailure for the joint-life status (xy) at timetis the quantity 𝜇 xy(t) =− d∕dttpxy tpxy =− d dtlog(tpxy). Theindependenceassumption(10.1)letsuscalculatethisfromthesinglelifefunctions. 𝜇xy(t) =− d dt(log tpx +log tpy) = 𝜇x(t)+𝜇y(t). (10.12) Consider an insurance policy that paysb(t) at timetif the first death of (x) and (y) occurs atthattime.Thepresentvaluewillbedenotedby ̄Axy(b)andgiven,analogouslyto(8.18),as ̄Axy(b) = ∫ N 0 b(t)v(t)tpxy𝜇xy(t)dt. (10.13)‡ 150 MULTIPLE-LIFE CONTRACTS The present value of an insurance payingb(t) at timetif theseconddeath of (x) and (y) occurs at that time is denoted bȳAxy(b). The same reasoning as used after formula (10.11) establishes the analogous formula ̄Axy(b) = ̄Ax(b)+ ̄Ay(b)− ̄Axy(b). (10.14) Aswiththesingle-lifecase,wearefacedwiththeproblemofevaluating ̄Axy fromthelife table. Assume thatbtakes the constant value ofbk over the year running from timekto time k+1. It is natural to ask if the samei∕𝛿 correction that we used for single-life insurances is also applicable to the joint-life case. Recall that this correction came from (8.20), so the question is whether the corresponding statement is true for joint-life statuses. That is, is it truethat tpxy (𝜇x(t)+𝜇y(t)) = qxy for0 < t< 1?AssumeUDDandtherefore(8.20)forsingle lives. Then, for 0< t< 1, tpxy (𝜇x(t)+𝜇y(t)) = tpyqx +tpxqy = (1−tqy)qx +(1−tqx)qy = qxy +(1−2t)qxqy, where we use (10.2) for the last equality. We see that the required condition does not hold exactly, due to the extra term of (1−2t)q xqy. Arguing as in Section 8.6.2, we deduce that ̄Axy(b) = Axy ( i 𝛿 ∗ b ) +R, (10.15) whereb = (b0,b1,…) and R= N−1∑ k=0 r(k)b(k)v(k+1)kpxyqx+kqy+k, (10.16) withr(k) = ∫ 1 0 v(k+1,k+t)(1−2t)dt.Analysisshowsthat r(k) > 0sothat Rispositive,but likely to be quite small and safely ignored in practice. Using thei∕𝛿 correction will slightly understate the value of the insurance, but it is a reasonable approximation.R is the present valueofacontractthatpaysoffonlyif( x)and( y)dieinthesameyear,aneventthatwillhave smallprobabilityformostages.Moreover, since1 −2tispositivefor0 ≤ t< 1 2 andnegative for 1 2 < t≤ 1, the cancellation in the integral will tend to maker(k) small. 10.6 The general two-life annuity contract There is a demand for two-life annuities that are more flexible than ones we described in previous sections. One popular arrangement is for benefits to continue while either person is alive, but for the amount to reduce when only one of the parties is alive. For example, a married couple may wish to provide for an annuity at retirement that will pay 60 000 yearly whilebotharealive,reducingto40000yearlywhenonlyoneisalive.Thisone-thirdreduction is a common one, reflecting the fact that while one person can live more cheaply than two, he/she will need more than half of the amount, since many expenses, such as housing, will THE GENERAL TWO-LIFE ANNUITY CONTRACT 151 not necessarily reduce. Many variations are possible. Benefits need not be symmetric and could vary according to the particular survivor. A common provision in pension plans is that the income stays level as long as the employee is alive, but will continue to the spouse at a reduced level upon the death of the employee. We present a single formula that covers all possibilities. A general life annuity on the pair (x) and (y) can be described by three annuity benefit vectors:f, wheref k is the amount paid at timekifonly(x)i sa l i v e ;g, wheregk is the amount paid at timekif only(y)i sa l i v e ;a n dh, wherehk is the amount paid at timekif both(x) and (y) are alive. Letj = h−f −g. We can think of the contract as three separate annuities. One isalifeannuityon( x)withbenefitvector f.Anotherisalifeannuityon(y)withbenefitvector g. These two will provide for the required payments when only one of the pair is living. The thirdcontractmustadjustthepaymentwhenbotharealive.Ifbotharelivingattime k,thefirst two annuities will providef k +gk, so the third must provide the difference,jk = hk −fk −gk. Thepresentvalueofthecompletecontractisthenthesumofthethreeseparatepresentvalues, which is just ̈a x(f)+ ̈ay(g)+ ̈axy(j). (10.17)‡ This formula reduces the calculation of present values for the general two-life annuity to calculatingthoseforsingleorjoint-lifeannuities.(SeeSection10.12forspreadsheetmethods to calculate the latter.) Example 10.1 Verify that (10.17) reduces to (10.10), when the yearly benefit is a constant 1 unit. Solution. We havef k = gk = hk = 1 for allk,s ojk =− 1 for allkand (10.17) gives ̈axy = ̈ax + ̈ay − ̈axy. Example 10.2 An annuity on (x) and (y) provides yearly payments as long as either (x)o r (y) are alive. Payments begin at 12, but reduce to 8 if (x)o n l yi sa l i v eo rt o( 6 )i f(y) only is alive. Find the present value Solution. f k = 8, gk = 6, andhk = 12, sojk =− 2 for allk. This gives a present value of 8̈ax +6̈ay −2̈axy. Example10.3 Anannuitypays1uniteachyear,providedthatatleastoneof(50)and(65) is living and is over age 70, but not if (50) is alive and under age 65. Find the present value. Solution. Lettingxrefer to (50) andyto (65), f = (0 20,1∞), g = (05,1∞), h = (015,1∞), 152 MULTIPLE-LIFE CONTRACTS so j =− (05,110,0 5,1∞). The present value is then ̈a50(020,1 ∞)+ ̈a65(05,1 ∞)− ̈a50:65(05,110,05,1∞). Example 10.4 Find a formula for the present value of an annuity that paysck at timek, provided that (y)i sa l i v e ,b u t(x) is not alive. Solution. fk = 0, gk = ck, hk = 0, so thatjk =− ck, for allk. The present value is ̈ay(c)− ̈axy(c). A contract of the type described in Example 10.4 is known as areversionaryannuity.I t provides a life annuity on one life, which does not begin until another life has died. It can be used for a similar purpose as life insurance, except that the proceeds are paid out as a life annuitytoanotherperson,ratherthanasalumpsum.Aswell,reversionaryannuitiesareoften used to put a value on certain inheritances that stipulate that the income from a certain asset will first go to a certain individual and then, upon the death of that party, revert to another person. Remark We comment briefly here on the effect of using assumption (10.1) in which tpxy was approximated by a quantity that is normally too low, as we saw in (10.3). It follows that this approximation will give values for̈a xy(b) that are slightly too low (assuming the entries of b are nonnegative). Since the coefficient of this term is negative in the most common examplesoftwo-lifeannuities,asillustratedabove,thestandardassumptionofindependence meansthatmosttwo-lifeannuitypremiumsaresomewhathigherthantheywouldbeifamore realistic model was used. 10.7 The general two-life insurance contract Themostcommontypeofpolicysoldtotwolivesisthejoint-lifeinsurancedescribedabove, where payment is made on the first death. In some cases, however, people want policies that will pay on the second death. As we indicated in the introduction to this chapter, this can be usedasameansofminimizingestatetaxes.Wewillinfactconsiderageneraltypeofcontract where death benefits can be paid on both deaths. Assuming payment at the moment of death, such a policy will be described by two death benefit functions:b(t), the amount paid at the timeofthefirstdeath;and d(t),theamountpaidatthetimeoftheseconddeath.Wecanview this as two separate insurances, one on the statusxywith benefit functionb, and one on the status xywith benefit vectord. Substituting from (10.14), the present value of benefits is ̄Axy(b)+ ̄Axy(d) = ̄Ax(d)+ ̄Ay(d)+ ̄Axy(b−d). (10.18) Thiscanbeverifieddirectlybythesametypeofreasoningasthatusedafterformula(10.11). CONTINGENT INSURANCES 153 10.8 Contingent insurances In some insurance policies written on two lives, the benefits depend on theorderof death. In the most common cases, a designated individual must die first or second in order to receive benefits. If this does not occur, then no benefits are paid. For this reason these are known as contingentinsurances. We will first consider such contracts with benefits payable at the end of the year of death. It is assumed throughout this section that there is zero probability that two lives will die at exactly the same instant of time. 10.8.1 First-death contingent insurances We first need to compute some new probabilities. Letq1 xy denote the probability that in the next year (x) will die and that (y) will be alive at the time of the death of (x). The symbol of 1 abovexindicatesthat( x)isthefirst todie.Notethat(y)doesnothavetodiewithinthe1-year period in order for the event in question to occur. The first problem is to estimate this probability from the life table. We will derive the answer here intuitively. A more formal derivation appears in Chapter 17. We note thatq1 xy can be approximated by another probability. Consider the following events: eventAconsists of (x) dying within the year, before the death of (y); eventBconsists of (x) dying within the year, and (y) surviving to the middle of the year. Thenq1 xy is the probability ofA. We claim that this is close to the probability of eventB. These events are of coursenotthesame.If( x)and( y)bothdieinthefirsthalfoftheyear,with( x)dyingfirst,then Aoccurs but notB. On the other hand, if (x) and (y) both die in the second half of the year with(y)dyingfirst,then Boccursbutnot A.However,boththeselattersituationsarerelatively rare, involving both lives dying in the same year. In any event, as long as the probabilities of these two death situations are roughly the same, they will tend to cancel each other out, and we can assume that the probability ofAis approximately the same as the probability ofB. The probability ofB can be readily computed from the life table. The probability that (x) dies during the year is justq x. Assuming UDD, the probability that (y) will be alive in the middle of the year is1∕2py = 1−qy∕2. Making the usual independence assumption that we did in deriving (10.1), we deduce that the probability that both of these occurrences will happen is the product of probabilities. In other words, assuming UDD q 1 xy ≐ qx − 1 2qxqy. (10.19)‡ As a check on this, let us compute another expression forqxy, the probability that at least oneofthetwoliveswilldiewithintheyear.Thiscanhappenintwomutuallyexclusiveways. Either (x) dies during the year, with (y) alive at that time, or (y) dies during the year, with (x) aliveatthattime.Bybasicrulesofprobabilitytheorywecansumthetwoprobabilitiestoget the probability that either will happen. We must have that q xy = q1 xy +q1 yx . (10.20) Adding the right hand side of (10.19) to that with (x) and (y) interchanged does indeed give qxy, as shown by (10.2). 154 MULTIPLE-LIFE CONTRACTS Consider now an insurance contract where a designated life must die before another in order to collect. This arises in the following situation. In any insurance policy, a particular individual, known as thebeneficiary, is designated to receive the death benefit upon death of the insured. If the beneficiary dies before the insured, a new beneficiary is normally chosen. Suppose, however, that an insured has no other beneficiary in mind. To handle this, insurers are willing to offer a contract that only pays if a stated beneficiary is alive at the death of the insured. Suppose that a policy sold on the lives (x) and (y) provides for benefits to be paid at the endoftheyearofthedeathof(x )providedthat(y)isaliveatthetimeofsuchdeath.If(y)dies before (x), nothing is paid. For a death benefit vectorb, the present value of the benefits for this contract will be designated byA 1 xy (b). We calculate this by the standard formula for an insurancepresentvalue,asgivenin(5.1)andagainin(10.6).Weneedonlyinsert,foreach k, the probability that the conditions for payment will occur in the yearktok+1. Focus on the secondexpressionintherighthandsideof(10.6)wherewewrotetheprobabilitythatthefirst death of (x) and (y) will occur between timekand timek+1a s kpxyqx+k:y+k. In the present case, in order that the death benefit is paid at timek+1, we still require that both (x) and (y) are alive at time (k) but then we need that (x+k) dies in the next year, and to be the first of the two lives to die. In place ofqx+k:y+k we wantq1 x+k:y+k . This gives A1 xy (b) = N−1∑ k=0 bkv(k+1)kpxyq1 x+k:y+k . (10.21)‡ As a check, we can use (10.20) to see that Axy(b) = A1 xy (b)+A1 yx (b), (10.22) which must be true, since a joint-life insurance can be considered as two separate contingent insurances, each paying benefits if a particular life dies first. 10.8.2 Second-death contingent insurances Suppose we have a policy where death benefits are paid at the end of the year of death only if (x)i st h esecondof the two lives to die. Denote the present value of this byA 2 xy (b). We calculatethisasfollows.Thinkofaregularsingle-lifepolicyon( x)astwoseparatecontracts. The first pays if (x) dies before some other life (y), and the second pays if (x)d i e sa f t e r(y). We must have that Ax(b) = A1 xy (b)+A2 xy (b), so that A2 xy (b) = Ax(b)−A1 xy (b). (10.23)‡ CONTINGENT INSURANCES 155 10.8.3 Moment-of-death contingent insurances Moment-of-deathcontingentinsurancescanbecalculatedbyaformulasimilarto(10.21)but with integrals replacing sums, and𝜇 replacing q.I f̄A 1 xy (b) is the present value of a contract paying b(t) at timetprovided that (x) dies at timetand (y) is alive that time, then ̄A1 xy (b) = ∫ N 0 b(t)v(t)tpxy𝜇x(t)dt. (10.24)‡ Wecanintuitivelyverifythisbynotingthat tpxy𝜇x(t)dtrepresentstheprobabilitythatboth(x ) and (y) survive up to timetand then (x) dies at that time. Note also, from (10.12), that ̄Axy(b) = ̄A1 xy (b)+ ̄A1 yx (b), which is obviously required. For second-death insurance, the same reasoning as in (10.23) shows that ̄A2 xy (b) = ̄Ax(b)− ̄A1 xy (b). (10.25) Example 10.5 Suppose (x) is subject to a constant force of mortality𝜇,( y) is subject to a constant force of mortality𝜈, and the force of interest is a constant𝛿.F i n d̄A1 x:y . Solution. From (10.24), ̄A1 xy = ∫ ∞ 0 e−𝛿te−(𝜇+𝜈)t𝜇dt= 𝜇 𝜇 +𝜈 +𝛿. Normally, we must evaluate contingent insurances directly from the life table. In the case thatbenefitsareconstantovereachyear,theprocedureissimilarto(10.15)exceptthatweget only one-half of the remainder term. This necessarily follows from (10.22) and the fact that Ris symmetric in (x) and (y). That is, assuming UDD for each of the lives (x) and (y), ̄A 1 xy (b) = A1 xy ( i 𝛿 ∗ b ) + R 2, (10.26) whereRis as defined in (10.16). 10.8.4 General contingent probabilities Letsq1 xy denotetheprobabilitythatwithin syears(x)willdieand(y)willbealiveatthetimeof thedeathof(x ).(So q1 xy ,aswehavedefineditaboveisjustthissymbolwith s=1.)Similarly, letsq2 xy denote the probability that withinsyears (x) will die having been predeceased by (y). 156 MULTIPLE-LIFE CONTRACTS Considerfirstthecasewhere sisapositiveinteger n.Wecanevaluatethiseasilybyadding up the probabilities of the required event occurring in each year to obtain nq1 xy = n−1∑ k=0 kpxyq1 x+k:y+k . (10.27)‡ From (10.21) this is justA1 xy (1n) at zero interest. For the general durationswe evaluate the probabilitybytakingthecorrespondingcontinuousinsuranceformula(10.24)atzerointerest: sq1 xy = ∫ s 0 tpxy𝜇x(t)dt. (10.28)‡ Notethatwecantake s=∞ intheabovetogettheprobabilitythat(x )diesbefore(y).(There was no point in doing this in the single life case since∞qx is just equal to 1). Some basic identities are ∞q1 xy +∞ q1 yx = 1, ∞q1 xy +∞ q2 xy = 1. 10.9 Duration problems Certain multiple-life problems involve a duration that runs from the time of death of an individual rather than from time zero. Normally, the best approach to finding present values is to refer to our basic formula where we sum or integrate the product of three factors. Complicationscanarisewhencomputingtheprobabilitythatpaymentwillbemade.Inmany cases a formula for this probability will vary with time, and we will need to break up our total time interval into appropriate subintervals, each with its own formula. Evaluation of the resultingintegralsofteninvolvesthetechniquesofchangingvariablesandreversingtheorder of integration or summation. We will illustrate with several examples involving continuous annuities or moment of death insurances. For simplicity, we assume throughout constant interest and independence of the two lives involved. In all cases we wish to find a formula for either a present value or a probability. Moreover we want the resulting formula to involve either single-life or joint-life statuses, or contingent probabilities and insurances. Example 10.6 An insurance contract provides for 1 to be paid at the moment of the death of(y)providedthat( y)diesatleast nyearsafterthedeathof( x).Findaformulaforthepresent value. Solution. We must consider two separate time periods. For a timetbetween 0 andnthere is no chance of payment being made. For a timet≥ n, payment ismade providedydies at time tandxwas not alive at timet−n. The present value is ∫ ∞ n vt tpy𝜇y(t)(1−t−n px)dt. DURATION PROBLEMS 157 To write this in terms of our standard symbols, we change the variable fromt to s= t−n, which gives us an integral with lower limit 0. Then the integral becomes ∫ ∞ 0 vs+n s+npy𝜇y(s+n)(1−s px)ds= vn npy∫ ∞ 0 vs spy+n 𝜇y+n(s)(1−s px)ds. Inthesecondequalityabove,weusethemultiplicationruleaswellasthefactthatbydefinition, 𝜇 z(r)dependsonlyonthesum z+r.Fromtheaboveformulawecanwritethesolutioninthe required form as vn npy ( ̄Ay+n − ̄A1 y+n:x ) = vn npȳA2 y+n:x . Wecandeducethisintuitivelyifwedosomefancifulreasoning.Wetell( y)toage nyears while we keep (x) at the same age. We then let the two lives, now agey+nand x, ‘race’ to see who dies second. We must multiply the result by an interest factor ofvn to account for the delay in the race, and also bynpy which is the probability thatyw i l ls u r v i v et ob ea b l et o participate in the race in the first place. Example 10.7 What is the probability that (x) and (y) will die at leastnyears apart from each other? Solution. The probability that (y) will dienor more years after the death of (x) is given by the solution to Example 10.6 with 0 interest. This is npy ( 1−∞ q1 y+n:x ) =n py ∞q1 x:y+n . Adding to this the probability that (x)d i e snor more years after the death of (y)g i v e st h e solution of npy ∞q1 x:y+n +n px ∞q1 y:x+n . Example 10.8 A temporary life annuity on (y) provides for payments made continuously at the annual rate of 1 fornyears, but with the payments beginning at the death of (x), rather than at time 0. Payments stop at the death of (y). Find a formula for the present value. Solution. For 0≤ t< n,payment willbemadeprovided(y)isaliveand( x)isnot.For t≥ n, the payment will be made provided (y)i sa l i v e ,(x) is not alive but was alivenyears before. The present value is ∫ n 0 vt tpy(1−t px)dt+∫ ∞ n vt tpy(t−npx −t px)dt= ̄ay(1n)− ̄axy +∫ ∞ n vt tpyt −npxdt, 158 MULTIPLE-LIFE CONTRACTS where we rearrange and combine terms to get the expression on the right. Using the change of variable technique of Example 10.6 on the final term yields the answer of ̄a y(1n)− ̄axy +vn npx̄ax+n:y. Note that this is similar to the usual reversionary annuity formula except (using the same fanciful reasoning as in Example 10.6), afternyears we only award payment to (y)i fl i v i n g jointlywithanindividualwhowasage xandwasthenallowedtoage nyearswhile(y)stayed the same. Another contract of this type, which does not involve a duration directly, but is based on a similar theme, is a reversionary annuity where the benefits can depend on the time of death of the first person as well as the time they are made. Toillustratethetechnique,wefirstrevisitacontinuousversionofthestandardreversionary annuityasgiveninExample10.4.Consideracontractwithcontinuouspaymentsattheannual rate ofc(s) at times, made if (y)i sa l i v eb u t(x) is not. As an alternate approach to that used before, we will view this as a contingent insurance, with a death benefit of̄a y+t(c◦ t) paid at timet,themomentofdeathof(x )ifthisoccursbeforethedeathof(y).Notethetimeshifting necessaryhere.Forexample,if( x)diesattime2,therateofpaymentatatime3periodsafter the annuity begins isc(5) = (c◦ 2)(3). We can write the present value as ∫ ∞ 0 vt tpxt py𝜇x(t)̄ay+t(c◦ t)dt. (10.29) Note now that vt tpȳay+t(c◦ t) = ∫ ∞ t c(s)vs spyds, since each side is the present value of at-year deferred life annuity on (y) where the payment at timesfor s≥ tisc(s). Substituting this into the integral gives us the present value as ∫ ∞ 0 ∫ ∞ t c(s)vs spyt px𝜇x(t)d sdt= ∫ ∞ 0 ∫ s 0 c(s)vs spyt px𝜇x(t)dtds, (10.30) by reversing the order of integration. Now, ∫ s 0 tpx𝜇x(t)ddt= ∫ s 0 − d dttpxdt= 1−s px, the probability thatxdies before times. From this we conclude that the present value of the contract is ∫ ∞ 0 c(s)vs spy(1−s px)ds= ̄ay(c)− ̄axy(c). This appears as a very complicated way of arriving at the same form of answer as that obtained in a much simpler fashion in Section 10.6. However, one benefit of this insurance APPLICATIONS TO ANNUITY CREDIT RISK 159 approach is that it works just as well with the alternative type of reversionary annuities we introducedabove,andprovidesaconvenientmethodforhandlingthose.Suppose,forexample that upon the death of (x), the periodic rate of payment to a surviving (y)i sc(r) at a timeras measuredfromthe startoftheannuity.Theannuitywouldthenbeginwiththeperiodicrateof c(0).In(10.29)wethenhave cinplaceof c◦ t.Fordeathattime ttherateofpaymentatatime s> twould bec(s−t), and that would appear in 10.30 in place ofc(s). This is will normally make for a rather complicated integration, but there are cases where it can be worked out, as in the following example. Example10.9 Supposethattheforceofmortalityfor xisaconstant 𝜇,theforceofmortality for yis a constant𝜈 and the force of interest is a constant𝛿. Upon the death of (x) an annuity will be paid to (y), if surviving, where the rate of payment at a timer after the annuity payments begin is e 𝛾r, with 0≤ 𝛾<𝜈 +𝛿. Find the present value. Solution. Fromtherightsideof10.30,substituting c(s−t) = e𝛾se−𝛾t forc(s)asoutlined,the integral involving the variabletwill be ∫ s 0 e−𝛾te−𝜇t𝜇dt= 𝜇(1−e−(𝜇+𝛾)s) 𝜇 +𝛾 . Substituting and integrating with respect tosgives the present value as 𝜇 𝜇 +𝛾 ∫ ∞ 0 e−𝛿se−𝜈se𝛾s(1−e−(𝜇+𝛾)s)d s= 𝜇 𝜇 +𝛾 [ 1 𝜈 +𝛿 −𝛾 − 1 𝜈 +𝛿 +𝜇 ] . ∗10.10 Applications to annuity credit risk Suppose that you are a prospective annuity purchaser (or an advisor to such), and have a list of various companies and the premiums they charge. The prices will vary, due to different interest and mortality assumptions, as well as a different treatment of many other factors that go into gross premium calculation. A natural tendency would be to select the one with the lowest premium, but that is not necessarily the best choice. This ignorescredit riskwhich is the risk that the company may be unable to meet the promised payments due to financial difficulties. One may want to adjust the prices to reflect the different degrees of this risk among the various issuers. This could be an extensive task, and we will not go into details here. Our goal in this section is to indicate briefly how the theory of multiple-life contracts can be applied towards this endeavour. A first step might be to compare for a fixed interest and mortality basis, the cost of a life annuity calculated in the usual way with one that provides for payments only if a chosen company is viable. For this purpose we just treat the company as another individual, which we’ll denote by𝛾, which is a survival state if it is in position to meet its financial obligations. We need to model tp𝛾, which is the probability that the company is in a survival state at time t. This will produce the associated force of failure,𝜇𝛾(t) =− d∕dtlog tp𝛾. We can then consider two-life contracts based on (x) and𝛾. For simplicity we confine our discussion to a life annuity on (x) that provides continuous payments at a level annual rate ofK. 160 MULTIPLE-LIFE CONTRACTS Takingintoaccountthecreditrisk,theworthofthecontracttothepurchasershouldbethe joint-life annuityK̄ax𝛾 = K∫ ∞ 0 tpxt p𝛾 dtrather thanK̄ax. This assumes that no benefits are paidafterfailureofthecompany.Inpracticehowever,thereisoftenarecoveryrate rwhereby the company, upon financial failure, will continue to pay an income ofr for every 1 unit of promised annuity income. In this case the purchaser will getKrif (x) only is surviving, and K if bothxand𝛾 are surviving so that our general two-life annuity formula gives the value to the purchaser as K[r̄ax +(1−r)̄ax𝛾], which of course reduces toK̄ax when r= 1. Another complication arises in cases where government regulatory bodies provide guar- anteesforannuities.Therearemanypossibleoptionsconcerningtheamountoftheguarantees andhowtheyrelatetotherecoveryratesofthefailingcompany.Wewillconsiderthefollow- ing particular example, which is encountered in some jurisdictions. Upon failure, the insurer continues to pay at their recovery rate. The guarantor makes up the difference, but the value, at interest and survivorship, at the time of failure, of all payments made by the guarantor is limitedtoanamount G.Suppose,forexamplethat K = 30,000, G= 100,000, r= 0.4.Inthe event of default, the issuing company would pay 12,000 each year, and the government body wouldprovidetheannualshortfallof18,000butwouldonlypayitonlyuntilthepresentvalue of100,000wasexhausted,whichshouldbesomewhatover6yearsundernormalinterestand mortality bases. We will arrive at a formula to compute the present value of aK-unit, whole life annuity subject to these default arrangements. We assume that the mortality and interest bases are the same for the issuer and the guaranteeing party. Lets 0 =i n f{s: ̄ax+s ≤ G∕(1−r)K}.Note thats0 would equal 0 if̄ax < G∕(1−r)K.For failure of the issuer at a timet before times0 the guarantee will not cover the full annuity. In that case the annuitant, who, in the absence of default, would have received total future payments with a value at timetofK̄a x+t, will instead receive payments with a reduced value at timet of G+rK̄ax+t. The deficit is (1−r)K̄ax+t −G. We must subtract from the usual present value, the present value of a a contingent insurance that provides this deficit on the failure of𝛾 before times 0, provided that (x) is then alive. Referring to equation (10.24) the credit risk adjusted present value of the annuity is K̄ax −∫ s0 0 vt tpxt p𝛾𝜇𝛾(t)[(1−r)K̄ax+t −G]dt. 10.11 Standard notation and terminology Particularexamplesofthemultiple-lifenotationhavealreadybeenencounteredinthestandard notationfortermandendowment insurancesandtemporaryannuities.Inplaceofthelife( y), we dealt with a period ofn years, denoted by n, which ‘survives’ for exactlyn years and then ‘fails’ at timen. So the standard symbol for the present value of a 1-unit,n-year term insurance premium,A1 x: n, does indeed signify that the policy provides for 1 unit at the end of the year of death provided (x) dies before the ‘failure’ ofn. Similarly, the standard symbol for the present value of a 1-unit,n-year endowment insuranceAx:n indicates that the contract provides for 1 unit to be paid at the death of (x)o ra tt i m en if earlier, that is, on the first failure ofxand n. The 1-unit,n-year, temporary annuity present value symbol̈ax:n indicates EXERCISES 161 thatthecontractprovidesforpaymentstocontinueaslongasboth( x)isaliveand nis‘alive’, meaning that then-year period has not run out. 10.12 Spreadsheet applications WecanadapttheChapter6spreadsheettobeapplicabletojoint-lifeannuitiesandinsurances. In the first place we will make provision to insert two life tables, one for (x), and one for (y). This is common in practice as often one life will be male, and the other female, so different tables apply. Start by inserting a new column D, which we intend show the values ofq y+t. This will move all the existing columns over one to the right. In cell D1 we put the agey. Copy the formula from C10 to D10. It should be the same except with C1 replaced by D1 and a reference to column P. (the reference is column C will be automatically changed from column N to column O). WethencaninsertanewtableinColumnPbyputtingtheparametersinP3andP4,coping the formula from 010 to P10 and copying down. In column F, which now containsy xy, we change the formula in F11 by appending ∗(1−D10) and copy down. In Column M, which now containswxy ∗ b, we change the formula in M10 by replacing C 1 0b y( C 1 0+D 1 0−C10*D10) in order to multiply byqxy rather thatqx. As a test exercise suppose that our example table applies to females and that for the male table the mortality at each age up to 119 is 1.25 times the female rate. Interest rates are a constant 6%. An insurance on a male age 50 and a female age 45 provides 10,000 at the end of the year of the first death. Find the net level premium payable yearly while both lives are living. The answer is 208.75. Notes and references An extensive study of dependence in two-life annuity contracts can be found in Freeset al. (1996). Moreadvancedexamplesofmultiple-lifecontracts,involvinganarbitrarynumberoflives, can be found in Jordan (1967, Part II), or in Bowersetal. (1997, Chapter 18). The result from real analysis that justifies the reversal of integration in formula (10.30) can be found in Royden (1988) Theorem 12.19 (iii). Exercises In all the exercises for this chapter, we assume that (10.1) holds. Type A exercises 10.1 Youaregiventhefollowinginformationfromalifetable: 𝓁80 = 100, 𝓁81 = 80, 𝓁82 = 40, 𝓁83 = 20. You also have the vectorc = (1000,1200,1500) and the vectorb = (1000,2000). Theinterestrateisaconstant25%.Find(a) p80:81 and2p80:81,(b)p 80:81 and 2p80:81,( c )̈a80:81(c), (d)A80:81(b), (e)̈a80:81(c), (f)A80:81(b). 162 MULTIPLE-LIFE CONTRACTS 10.2 Suppose that 5px = 0.8, 6px = 0.7, 5py = 0.6, 6py = 0.4. Find the probability that both (at least one) (exactly one) (neither) of the two lives live 5 years (die within the next 5 years) (die between time 5 and time 6). There are 12 answers required here. 10.3 You are given thatp 70 = 0.9, 2p70 = 0.8, 3p70 = 0.7. The interest rate is 20% for the first year and 25% for the second year. Find the present value of the benefits on a 2-year term insurance policy sold to two lives (70) and (71) that provides benefits upon the first death, payable at the end of the year of death. The amount of the death benefit is 1000 if the first death occurs in the first year, and 2000 if it occurs in the second year. 10.4 An insurance policy pays 1 unit at the moment of the second death of (x) and (y). Level premiums are payable continuously as long as either of the two is alive. Given that (x) is subject to a constant force of mortality of 0.1, (y) is subject to a constant forceofmortalityof(0.3)andtheforceofinterestisaconstant0.1,whatistheannual rate of premium payment? 10.5 For two lives (x) and (y), the forces of mortality are given by𝜇 x(t) = 0.04 for allt, and𝜇y(t) = 1∕(20−t)f o r0≤ t< 20. (a) Find the probability that the joint-life status (xy) will survive to time 10. (b) Find the probability that the last-survivor statusxywill fail before time 10. (c) An insurance contract provides for a death benefit paid at the moment of the first deathof(x )and(y).Theamountofthebenefitise 0.1t fordeathattime t.Theforce of interest is a constant 0.06. Find the present value of the benefits. 10.6 Suppose that Demoivre’s law holds with𝜔 = 100. Consider two lives (80) and (60). (a) Whatistheprobabilitythatthefirstdeathwilloccurbetweentime5andtime10? (b) What is the probability that the second death will occur between time 5 and time 10? 10.7 Thefollowingisaportionofaselectandultimatetablewithaselectperiodof2years. xq [x] q[x]+1 qx+2 x+2 60 0.08 0.14 0.22 62 61 0.09 0.20 0.25 63 62 0.10 0.20 0.30 64 A and B are both age 62. A is a newly selected life, while B was first selected at age 61.WhatistheprobabilitythattheseconddeathofAandBwilloccurbetween2and 3 years from now? EXERCISES 163 10.8 An insurance contract sold to (x) and (y) provides for a death benefit of 1 unit at the moment of the first death and 3 units at the moment of the second death. Annual premiums are payable while at least one of the two is living. The annual rate of premiumpaymentreducestothreequartersoftheinitialrateuponthefirstdeath.You are given that (x) is subject to a constant force of mortality of 0.2, and (y) is subject toaconstant forceofmortalityof0.3.Theforceofinterestisaconstant0.1.Findthe initial annual rate of premium payment. 10.9 A contract on two lives (x) and (y) provides that if (y) dies first, (x) will receive a life annuity of 1 per year, starting at the end of the year of (y)’s death, while if (x)d i e s first,(y)willreceivealifeannuityof2peryear,startingattheendoftheyearof( x)’s death.Levelannualpremiumsof 𝜋 arepayablewhileboth( x)and( y)arealive.Given that ̈a x = 16, ̈ay = 13 and̈axy = 10, find𝜋. 10.10 You are given two independent lives (x) and (y)f o rw h i c h(x) is subject to a constant force of mortality of log(4/3), while (y) is subject to a force of mortality at timetof 1∕(10−t)f o r0≤ t< 10. (a) What is the probability that the first death among these two will occur between time 2 and time 3? (b) Whatistheprobabilitythattheseconddeathamongthesetwowilloccurbetween time 2 and time 3? 10.11 You are given thatq70 = 0.2, q71 = 0.25, q72 = 0.4,q73 = 0.6. The interest rate is a constant25%.Thevector bis(1000,2000,3000).Compute(a) A1 70:71 (b),(b) A1 71:70 (b), (c)A2 70:71 (b), (d)A2 71:70 (b). Type B exercises 10.12 Anannuitycontractissuedtotwolives,(40)and(50),providesforpaymentsof1per year, to be made provided that either or both of the following conditions hold. (i) (40) is alive and under age 60; (ii) (50) is alive and over age 60. Find a formula for the present value of the benefits using terms of the form̈ax(c) or ̈axy(c). 10.13 Repeat Exercise 10.12, but now with (ii) stating that (50) is alive and between age 60 and age 80. 10.14 An insurance policy sold to (x) and (y) provides for a payment of 2 units on the first deathand3unitsontheseconddeath.Levelannualpremiumsarepayableforasalong as either individual is living, with the premiums reducing by one-third upon the first death. Find an expression for the initial premium in terms of̄A x, ̄Ay, ̄Axy, ̈ax, ̈ay, and ̈axy. 10.15 An annuity sold to (x) and (y) provides for annual payments for 20 years, provided at least one of the two is alive. The annual payment begins at 12 units. During the first 164 MULTIPLE-LIFE CONTRACTS 10 years, the payments remain at 12. However, during the second 10-year period, the payment reduces to 6 if (x)o n l yi sa l i v e ,o rt o8i f(y) only is alive. Write the present value in terms of single- and joint-life annuities. 10.16 An annuity contract sold to (40) and (50) provides for yearly payments if either life is living and under age 70. The yearly payment is 6 if both are alive, 4 if (40) only is alive, and 5 if (50) only is alive. Find a formula for the present value of the benefits using terms of the form̈a x(c) and̈axy(c). 10.17 For the policy of Exercise 10.8, calculatetV. 10.18 SupposethatDemoivre’slawholds.Showthat(10.19)holdsexactlyforageneralleft subscriptnin place of 1. That is, nq1 xy = nqx − 1 2nqxn qy. 10.19 For the two lives of Exercise 10.10, find the probability that (x) will die before (y). 10.20 Given two lives (x) and (y) and timess< t, consider the following four events:A,t h e joint-status (xy) fails between timesand timet; B, the last-survivor status (xy) fails between timesand timet;C, at least one of (x) and (y) fails between timesand time t; andD, both (x) and (y) fail between times and timet. Rank the probabilities of these from largest to smallest to the extent that you are able, showing clearly those events that you cannot compare without further information. 10.21 Suppose that interest is a constant 25% and thatq x = 0.2. Assume UDD. Find the ratio of̄Axx(11)t o(i∕𝛿)Axx(11). 10.22 The probability nq1 xy can be generalized tonq1 x 1:x2:…xk , which is the probability that outofagroupof klives,(x1),(x2),…,(xk),(x1)willdiewithin nyearsandbethefirst to die. Express the following in terms of first-death probabilities: (a) theprobabilitythat(x )willdiewithin nyearsandwillbethe secondtodieamong a group of three lives (x),(y), and (z); (b) theprobabilitythat(x )willdiewithin nyearsandbethe thirdtodieamong(x ),(y), and (z). 10.23 Find a formula to calculatēA 2 yx (b) in terms of (a) ̄A1 xy (b), ̄Ay(b), ̄Axy(b), (b) ̄A2 xy (b), ̄Axy(b). 10.24 Suppose that nqx = 0.2, andnqy = 0.3. What isnq1 xy − nq2 yx ? 10.25 An insurance pays 1 at the death of (y) provided that (x) was alivenyears before this death. Assume constant interest. Find a formula for the present value in two ways. Firstbysettingupandevaluatingthenecessaryintegrals.Secondly,byusingExample 10.6 to avoid any integration. EXERCISES 165 10.26 An annuity provides for continuous payments at the annual rate of 1, while both (x) and (y) are alive, and in addition continues the payments forn more years after the firstdeath,aslongasthesurvivorisliving.Paymentsstopcompletelyuponthesecond death. Find a formula for the present value. Verify that your answer gives the correct result in the case thatn= 0. 10.27 In Example 10.9 take𝜈 = 𝛿 = 0 and𝛾< 0. (a) Evaluate the present value and give an intuitive explanation of the answer. (b) Taketheparametersasinpart(a),butnowconsiderthesimplertypeofreversionary annuitywheretherateofpaymentattime r(measuredasusualfrom0)is e 𝛾r.Show that the present value is an increasing function of𝜇 and explain why this is so. 10.28 Inthisproblemalldeathbenefitsarepayableatthemomentofdeath,andallpremiums arepayablecontinuouslyatalevelrate.Alifeinsurancecontracton( x)and( y)provides for death benefits as follows. If (x) dies first at timet, the amount paid is the reserve at timet on a 1-unit whole life insurance on (y) with premiums payable for life. If (y) dies first at timet, the amount paid is the reserve at timetfor a 1-unit whole life insurance on (x) with premiums payable for life. Premiums are payable until the first death. Show that the rate of premium payment is̄Px + ̄Py − ̄Pxy, where ̄Px denotes the rate of premium payment for a 1-unit whole life insurance onx with premiums payable for life, and̄Py and ̄Pxy denote similar quantities foryand xy, respectively. You should demonstrate this mathematically and also give a verbal explanation. 10.29 A contract on ( x) and (y) provides the following death benefits, payable at the moment of death. If (x) dies first, 10 is paid upon the death of (x) and 30 is paid upon the death of (y). If (y) dies first, 20 is paid upon the death of (x) and 40 is paid upon the death of (y). Find the present value given that ̄Ax = 0.30, ̄Ay = 0.45, ̄Axy = 050, ̄A1 xy = 0.20. 10.30 An annuity provides continuous payments at the annual rate of 1 to (y) while living, beginning n years after the death of (x). Assuming constant interest, find a formula for the present value. Spreadsheet exercise 10.31 Suppose that our sample table applies to females and the mortality rate for males is 1.25 times that for females up to age 119. A is a male age 50 and B is a female age 40. The interest rate is a constant 5%. (a) Find the level annual premium for an insurance paying 10 000 at the end of the year of first death of A and B with premiums payable while both are alive. (b) Find the level annual premium for an insurance paying 10 000 at the end of the year of the death of A provided this occurs before the death of B. Premiums are payable while both are alive. (c) Assuming UDD, find the probability that A will die before B. 11 Multiple-decrement theory 11.1 Introduction Our discussion of insurance contracts up to now has been concerned with benefits payable upon the occurrence of death. We now investigate situations when an insured is at the same time subject to several different events that can have financial impact. One example, which we mentioned earlier, is the event ofwithdrawalor lapse, whereby an insured life terminates the contract and receives a cash value. To properly model a life insurance contract, one must consider two causes of termination, death and withdrawal, operating simultaneously. Sometimes the insurer must distinguish between different causes of death. For example, some policies have a feature that provides additional death benefits for accidental death as opposed to death from natural causes. Somepoliciesincludedisabilitybenefits,providingincomeforpeoplewhocannolonger work. For such, the insurer must consider the possibility of disability, as well as death and withdrawal. For employees covered under a pension plan, there are at least four events of interest: disability; death; termination of employment; and retirement. These are just a few of the possibilities, and many more can arise. In this chapter we analyze the general situation. 11.2 The basic model We suppose that we havemdifferent causes offailureoperating simultaneously on a group of lives. These causes are often referred to asdecrements, since they bring about a decrease in the number of lives under observation. The insurance policies introduced in Chapter 5 involved only one decrement, namely that of death. When there are several decrements we are concerned with what is called in the actuarial literature asmultiple-decrementtheory,o r , in biostatistical contexts, the theory ofcompetingrisks. FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial THE BASIC MODEL 167 There is a generalization of these ideas to multi-state insurances and annuities, where individualscantransferfreelybetweenoneofseveralstates.WedealwiththisinChapter19, which requires a basic knowledge of Markov chains, as given in Chapter 18. 11.2.1 The multiple-decrement table It is convenient to make use of a generalized life table, known as amultiple-decrementtable. We will number our causes from 1 tomand use a superscript (j) to refer to causej. We will also use a superscript (𝜏), which the reader can interpret as meaningtotal. We begin with an arbitrary number𝓁(𝜏) 0 of lives age 0. We let𝓁(𝜏) x denote the number of individuals from this group who are still surviving at agex. That is, they have not succumbed to any of them causes. We letd(j) x denote the number of lives who will failfirstfrom cause (j) between the ages ofxandx+1. Let d(𝜏) x = m∑ j=1 d(j) x , (11.1) whichisthenumberofpeoplewhowillfailfrom somecausebetweentheagesof xandx+1. It follows that 𝓁(𝜏) x+1 = 𝓁(𝜏) x −d(𝜏) x . (11.2) That is, the number of survivors at agex+1 is equal to the number of survivors at agex,l e s s thosewhofailedfromsomecausebetweentheagesof xandx+1.Thefollowingisaportion of a sample table with two decrements: x 𝓁(𝜏) x d(1) x d(2) x 0 1000 50 100 1 850 60 105 2 685 70 120 Itisimportanttonotetheword‘first’usedinthedefinitionof d(j) x .Themodelassumesthat any cause of failure results in the individual leaving the group, so they are no longer under observation. For example, if cause 1 denotes death and cause 2 withdrawal, a policyholder whowithdrawsatage60 1 2 andthendiesatage60 3 4 wouldbeincludedin d(2) 60 ,butnotin d(1) 60 . Thisiswhatwewantinthecommonapplications.Inthecaseofwithdrawal,theinsurerwould paythepolicyholdertheircashvalue,thepolicywouldterminate,andtheinsurerwouldhave no further interest in the time of death of this person. Therefore, whenever we refer to failing from a certain causejin the multiple-decrement model, it is understood that this means that this occurs before failure from any other cause. Wedefinedamultiple-decrementtablestartingatage0,butthiscouldbesomeotherage, depending upon the particular application. For example, the multiple-decrement table for an employee pension plan would begin at the first age the employees become eligible for the plan, perhaps age 25 or so. 168 MULTIPLE-DECREMENT THEORY Our model assumes that the multiple-decrement table, as given, can be used for all ages x. That is, it is an aggregate table that does not show selection effects. We could define for eachx, as we did for the single-decrement case, aselectmultiple-decrement table to apply to people first observed at agex. This is important in practice since many causes of decrement, such as withdrawal, will certainly depend more on duration since the policy began than on attained age. However, we will not get into these details in this chapter and will work with the aggregate model. The more general case can be deduced from the material of Chapter 17. 11.2.2 Quantities calculated from the multiple-decrement table We first define probabilities of failure. Let q(j) x = d(j) x 𝓁(𝜏) x be the probability that (x) will fail first from causej within 1 year. In practice, one would start with these probabilities and construct the multiple-decrement table inductively, by calculating d (j) x = 𝓁(𝜏) x q(j) x and then using (11.1) and (11.2) to complete the table. Let kp(𝜏) x = 𝓁(𝜏) x+k 𝓁(𝜏) x (11.3) be the probability that (x) will survive to agex+kwithout succumbing to any cause. Probabilitiesarecalculatedfromthetablewithsimilarformulasasinthesingle-causelife table. One must only remember that the denominators are𝓁(𝜏) x . For example, nq(j) x = ∑n−1 k=0 d(j) x+k 𝓁(𝜏) x is the probability that (x) will fail as a result of causejwithin nyears. Similarly, d(j) x+k 𝓁(𝜏) x =kp(𝜏) x q(j) x+k (11.4) is the probability that (x) will fail from causejin the time intervalktok+1. Another quantity of interest is 𝓁(j) x = 𝜔−x−1∑ k=0 d(j) x+k. INSURANCES 169 This represents the number of people in our group who will fail from causejsometime after age x. We assume in our model that everyone will eventually fail from some cause (which is obvious if death is one of the causes) so we can write 𝓁(𝜏) x = m∑ j=1 𝓁(j) x . Knowing the value of𝓁(j) x for all integral values ofxand alljallows us to complete the table since d(j) x = 𝓁(j) x −𝓁(j) x+1. Note that we can write nq(j) x = 𝓁(j) x −𝓁(j) x+n 𝓁(𝜏) x . (11.5) Another symbol we can define is np(j) x = 𝓁(j) x+n 𝓁(𝜏) x (11.6) which is the probability that (x) will fail first from causej after time n. Here, instead of thinking of the basic symbolnp as denoting survival to timen, we think of the equivalent formulation as failing after timen. 11.3 Insurances Consideraninsurancepolicyon( x)thatprovides,forallrelevantvalues kandj,apaymentof b(j) k attime k+1,providedthatfailureoccursfromcause jintheyear ktok+1.Wecanview this asm separate policies, where thejth policy pays benefits only for failure from causej. The formulas are exactly the same as (5.1), but with the probabilities taken from (11.4). That is, the present value of thejth policy will be ∞∑ k=0 b(j) k v(k+1) d(j) x+k 𝓁(𝜏) x = ∞∑ k=0 b(j) k v(k+1)kp(𝜏) x q(j) x+k, (11.7)‡ and the total present value would be obtained by summing over allj. To handle insurances payable at the moment of failure we will need the following: Definition 11.1 For each causej,l e t 𝜇(j)(x) = lim h→0 hq(j) x h = lim h→0 𝓁(j) x −𝓁(j) x+h h𝓁(𝜏) x (11.8) 170 MULTIPLE-DECREMENT THEORY The forceofdecrementattimet , from cause (j)f o r(x) is the quantity given by 𝜇(j) x (t) = 𝜇(j)(x+t). (11.9) (Inthecaseofselectedmultiple-decrementtableswewouldneedtodefine 𝜇(j) x (t)separately for each agex.) Analogouslyto(8.15)andnotingthatthequantitiesin(11.5)and(11.6)differbyaconstant we can write 𝜇(j) x (t) = −d dt tp(j) x tp(𝜏) x = d dttq(j) x tp(𝜏) x . (11.10) Considerthecontinuousanalogueoftheinsurancewhosepresentvalueisgivenby(11.7). Thisisacontractthatpaysadeathbenefitof b(j)(t)attimet,shouldfailurefromcause joccur at that time. We denote the present value bȳA(j) x (b(j)). It is given by ̄A(j) x (b(j)) = ∫ ∞ 0 b(j)(t)v(t)tp(𝜏) x 𝜇(j) x (t)dt. (11.11)‡ This reflects the fact that in order to collect at timet,(x) must surviveallcauses up to timet, andthensuccumbtocause jattime t.WediscusstheevaluationofthisintegralinSection11.5. For policies which are purchased by annual premiums, payable until the first failure, we determine theinitialpremium as inthe singledecrement case, bysimplydividing thepresent value by ̈a (𝜏) x (𝜌) = ∞∑ k=0 v(k)𝜌kk p(𝜏) x , where𝜌 is the premium pattern vector. 11.4 Determining the model from the forces of decrement The multiple-decrement model is often given from the outset by specifying the forces of decrement, and we must use these to calculate probabilities. Note that in the last expression in(11.8)thenumeratorinvolves 𝓁 (j) whilethedenominatorinvolves 𝓁(𝜏).Thismeansthatwe cannot apply a version of (8.16) directly. We can, however, define the total force 𝜇(𝜏) x (t) = m∑ j=1 𝜇(j) x (t) = lim h→0 𝓁(𝜏) x+t −𝓁(𝜏) x+t+h h𝓁(𝜏) x+t . So𝜇(𝜏) x isthesametypeofquantityasthesingle-lifeforceofmortalityintroducedinChapter8, except based on the total decrement rather than just on failure by death. We can now apply (8.16) to deduce that tp(𝜏) x = e−∫ t 0 𝜇(𝜏) x (r)dr. (11.12)‡ A MACHINE ANALOGY 171 Arguing as we did in deriving (10.28), apply (11.11) with a constant interest rate of 0, and b (j)(t) = 1f o r0≤ t< s, and 0 elsewhere, to conclude that sq(j) x = ∫ s 0 tp(𝜏) x 𝜇(j) x (t)dt. (11.13)‡ Example 11.1 In a model with two decrements, you are given that𝜇(1) x (t) = 0.02 and 𝜇(2) x (t) = 0.03 for allt≥ 0. Find3q(1) x . Solution. 𝜇(𝜏) x = 0.05, so thattp(𝜏) x = e−0.05t and, from (11.13), 3q(1) x = 2 5(1−e−0.15). 11.5 The analogy with joint-life statuses We have already encountered a multiple-decrement model in Chapter 10. The difference is that, rather than single lives, the basic objects were joint-life statuses consisting of a pair of lives (x) and (y). These pairs were subject to two distinct causes of failure, namely the death of (x) and the death of (y). The notation differs somewhat, but the reader should notice that, with this point of view, the contingent insurance present value given in (10.24) is a special caseofthatgivenin(11.11),andtheprobabilityin(10.28)isaspecialcaseofthatin(11.13). It follows that if we assume that the causes of decrement are acting independently, we can evaluate the integral (11.11) from the multiple-decrement table as in Chapter 10. That is, if the benefits are constant over each year, and the particular decrement can be assumed to be uniformly distributed over each year, thei∕𝛿 correction will result in a reasonable approximation. 11.6 A machine analogy Thedifficult partofmultiple-decrement theorydeals withrelationships between thedifferent decrements.Inordertobetterunderstandandmotivatetheideas,wefirstlookatwhatappears tobeadifferentsituation,butwhichwewilleventuallyrelatetoourmodelabove.Tofacilitate this, we will use notation for this new example that parallels that which we have already introduced. Suppose we have a machine with two components, part 1 and part 2, which work com- pletely independently of each other so that the condition of one part does not affect the operation of the other. In order for the machine to work, both parts must be working, so if either part fails, the machine will fail, even though the other part may be in perfect order. We assume also that both parts cannot fail simultaneously, so we can always identify which part caused the machine to fail. Suppose we want to compute probabilities of failure over some time period, say a year. We have four quantities of interest. Forj= 1,2, letq ′(j) be the probability that partj will fail, and letq(j) be the probability that the machine will fail due to the failure of partj, meaning that partjfailed during the year and was the first of the two partstofail.Whataretherelationshipsbetweenthese?Sincethefailureofthemachinedueto 172 MULTIPLE-DECREMENT THEORY the failure of partjobviously implies that partjfailed, it is clear from elementary probability theory that q′(j) ≥ q(j). (11.14) Itisisalsoclearthat,ingeneral,thetwoquantitiesarenotequalandwewouldexpecttheleft handsidetobestrictlygreaterthantheright.Suppose,forexample,thatsometimeduringthe year part 1 fails, causing the machine to fail, and sometime after that but before the end of the year, part 2 fails. Then the event of the failure of part 2 would have occurred, butnotthe event of part 2 causing the failure of the machine. After introducing some additional notation, we will state some further relationships. Let p ′(j) = 1−q′(j) denote the probability that partjwill be working at the end of the year,q(𝜏) denote the probability that the machine will fail within the year, andp(𝜏) = 1−q(𝜏) denote theprobabilitythatthemachineisworkingattheendoftheyear.Themachinecanfailinone of two mutually exclusive ways, namely, failure of part 1 or failure of part 2. By elementary probability theory we have q (𝜏) = q(1) +q(2). Ontheotherhand,inorderthatthemachinebeworkingattheendoftheperiod,bothpartsmust be working. Since the parts work independently, we can multiply probabilities to calculate this (as we illustrated in Chapter 10), which gives p (𝜏) = p′(1)p′(2). From the above two equations we get the fundamental identity q(1) +q(2) = 1−p′(1)p′(2), (11.15) which is often written in the form q(1) +q(2) = q′(1) +q′(2) −q′(1)q′(2). (11.16) Thebasicproblemofinterestisasfollows.Ifwearegiventheunprimedsymbols,canwe calculatetheprimedones,orifwearegiventheprimedsymbolscanwecalculatetheunprimed ones?Ingeneral,wecannotdothisuniquelyandwewillneedadditionalinformationinorder to definitely deduce one set of probabilities from the other. We will, however, present some possible solutions to this problem, and then discuss some conditions that will lead to these solutions. 11.6.1 Method 1 Givenq(j), j= 1,2, we sum to getq(𝜏), then calculatep(𝜏), and then let p′(j) = p(𝜏) ( q(j) q(𝜏) ) , (11.17) A MACHINE ANALOGY 173 which will clearly satisfy (11.15). It is not too difficult to remember this formula. We have to factor the quantityp (𝜏) into two factors, and it is natural to take the factors asp(𝜏) to some exponent where the exponents sum up to one. Two numbers that obviously sum to 1 are q (j)∕q(𝜏), j= 1,2. For the other direction, givenq′(j), j= 1,2, we calculatep′(j), multiply to getp(𝜏), then calculateq(𝜏), and finally take q(j) = logp′(j) logp(𝜏) q(𝜏). (11.18) For this formula we have to split upq(𝜏) as the sum of two terms. We obtain the terms by multiplying by weights that add up to 1. Since we have a multiplicative relationship, we take logs to accomplish this. A straightforward calculation verifies that this approach is consistent. That is, if we are givenq (j), j= 1,2, and use (11.17) to calculateq′(j), j= 1,2, and then apply (11.18) to these numbers, we will end up with the same values ofq(j) as we started with. To see that (11.14) is satisfied we use the inequality (1−x)𝛼 ≤ (1−𝛼x), for 0 ≤ x≤ 1,0 ≤ 𝛼 ≤ 1, which can be demonstrated by looking at the the Taylor polynomial of degree 2 for the left hand side. From (11.17), taking𝛼 = q (j)∕q(𝜏), p′(j) = (1−q𝜏)𝛼 ≤ 1−𝛼q(𝜏) = 1−q(j), and (11.14) follows. The entire discussion above can be generalized to the case ofmindependent parts, where the failure of any one part will cause the machine to fail. Use notation as above, except that jnow takes values 1,2,…,m. Everything is the same with the obvious changes to handlem probabilities instead of two. For example, (11.15) will now read m∑ j=1 q(j) = 1− m∏ j=1 p′(j), (11.19) and the solutions given by (11.17) and (11.18) will satisfy this relationship. 11.6.2 Method 2 We present another method to deduce the unprimed probabilities from the primed, which in certaincasesmaybemorerealisticthanformula(11.18).However,for m> 2itisnoteasyto invert, in order to deduce unprimed from primed. To motivate this method, we look again at the case of the joint-life statuses of Chapter 10. In fact we can directly visualize the machine pointofviewinthissituation.The‘machine’consistsofthejoint-lifestatus( xy)withthetwo distinct components (x) and (y). For our particular problem, suppose we are givenq ′(1) and q′(2),andwanttocomputetheprobabilitythatthefailureofpart1causedthemachinetofail. 174 MULTIPLE-DECREMENT THEORY Reasoning as we did in deriving (10.19), we approximate this by computing the probability that part 1 failed during the year and that at the middle of the year part 2 was still working. Making an assumption of auniform distribution of failures for each part, over each year, which is analogous to our UDD assumption, we have, as in (10.19), q (1) = q′(1) − 1 2q′(1)q′(2), q(2) = q′(2) − 1 2q′(1)q′(2), (11.20) which clearly satisfy (11.14) and (11.16). To calculate the inverse formula, letΔ= q(1) −q(2). From (11.20) q(1) = q′(1) − 1 2 (q′(1))2 + 1 2q′(1)Δ Wehaveaquadraticequationin q′(1),whichcanbesolvedbytheusualformulatoyield q′(1) = (2+Δ)− √ (2+Δ)2 −8q(1) 2 , (11.21) and directly from (11.20) q′(2) = q′(1) −Δ. Example 11.2 Givenq(1) = 0.27, q(2) = 0.17, findq′(1) andq′(2) by Method 2. Solution. q′(1) = 2.1− √ (2.1)2 −2.16 2 = 0.3, q′(2) = 0.3−0.1 = 0.2. As a check we can take these answers and apply (11.20) to get back to where we started. q((1)) = 0.3−(1∕2).06 = 0.27, q((2)) = 0.17. Example 11.3 Givenq′(1) = 0.200,q′(2) = 0.488, estimateq(1) and q(2) by both methods. Solution. ByMethod1: Sincep′(1) = 0.8and p′(2) = 0.512,wehave q(𝜏) = 0.5904and,from (11.18), q(1) = 1 4 ×0.5904 = 0.1476, q(2) = 3 4 ×0.5904 = 0.4428. SolutionbyMethod2: From (11.20), q(1) = 0.2000−0.0488 = 0.1512, q(2) = 0.4880−0.0488 = 0.4392. ASSOCIATED SINGLE-DECREMENT TABLES 175 AswithMethod1,Method2canalsobeextendedtothecaseof mcomponentstodeduce the unprimed from the primed, as we show later in Section 11.7.2. For example, withm= 3 we get q(1) = q′(1) − 1 2 [q′(1)q′(2) +q′(1)q′(3)] + 1 3q′(1)q′(2)q′(3). (11.22) Note that the coefficients of 1/2 and 1/3 are what is needed in order to satisfy (11.19). With m> 2 there is no convenient direct method for the inverse process, and it would have to be done by numerical approximation. 11.7 Associated single-decrement tables 11.7.1 The main methods We now return to the original setting dealing with a group of lives, to which we will apply our ‘machine’ model. For a definite example start with a multiple-decrement table with two decrements; cause 1 is death, and cause 2 is disability. Suppose we wish to use this table to construct a regular single-life table relating only to death, as introduced in Chapter 3. Would webejustifiedinsimplytaking q (1) x asthemortalityrateforage( x),inthesingle-lifetable?The answerisadefinite‘no’.Westressagainthat q(1) x isnottheprobabilitythat(x )willdiewithin the year, but rather the probability that (x) will die within the yearbeforebecoming disabled. Theactualvalueof qxshouldbelargerthan q(1) x .Thesituationisthesameasthemachinemodel intheprevioussection.Apersoncoulddieduringtheyear,afterhavingalreadyleftthegroup by reason of disability. In other words, the mortality rate for age (x) in the single decrement table is analogous to the primed rate in the machine model, and will be denoted byq ′(1) x . We may also want to computeq′(2) x , the probability that (x) will become disabled during the year, assuming that no other causes of failure are operating. This admittedly requires a stretch of the imagination, as the possibility of death would always appear to be present. We have to imagine, however, that we are computing what these probabilities would be if we couldsomehoweliminatethepossibilityofdeath.Thebestapproachissimplytoviewthings in terms of the machine model. That is, we have to think of a person as having a ‘disability component’ that exists independently and could in fact continue to operate even after death. For each causej, the collection of values {q ′(j) x } for various values ofx is known as the associated single-decrement table for cause j. As we have stressed above, these rates give probabilities of failure for the particular causej, assuming thatnoother causes of decrement are operating. Theproblemoftenarisesofgoingfromonesetofratestoanother.Wemight,forexample, construct the multiple-decrement table in the first place by using our knowledge of the associated single-decrement tables. Alternatively, we might have first constructed a multiple tablebyactuallyobservingtheeffectofthevariouscausesactingtogetherandwishtousethat to construct the associated single-decrement tables. In order to do so we can directly apply our machine model and use the twogiven methods, provided we make the key assumption of that model, namely that the various causes are actingindependently.One can certainly argue in many applications that this does not hold, but the standard actuarial model incorporates this independence assumption. The more general theory is discussed in Chapter 17. 176 MULTIPLE-DECREMENT THEORY Consider a numerical example. Suppose that in our death–disability table 𝓁(𝜏) x = 1000, d(1) x = 300, d(2) x = 100. Wewillcomputetheassociatedsingle-decrementratesusingmethod1.Fromformula(11.17), we have q ′(1) x = 1−0.63∕4 = 0.318, q′(2) x = 1−0.61∕4 = 0.120. Supposewearegiventheabovesingle-decrementratesandwanttocomputethemultiple- decrement table. If we use the Method 1 formula (11.18) we necessarily will get back to unprimed rates of 0.300 and 0.100 that we started with, as the reader should verify. If we use the Method 2 formula (11.20) instead, we will get q (1) x = 0.318(1−0.06) = 0.299, q(2) x = 0.120(1−0.159) = 0.101, verifying as we saw in Example 11.3 that the two methods are different, but will generally give answers that are close. 11.7.2 Forces of decrement in the associated single-decrement tables Refer back to the statement made in Section 11.4 that the multiple-decrement model is often constructed from the forces of decrement. A question that naturally arises is, how are these forcesofdecrementobtained?Theansweristhattheycanbetakenastheforcesofdecrement in the associated single-decrement tables, namely 𝜇 ′(j) x (t) =− d dttp′(j) x tp′(j) x . Note that 𝜇′(j) x (t) is defined differently than the quantity𝜇(j) x (t) as given in (11.9) but there is reason to believe that they might be the same. We observed above that we expect that q ′(j) x > q(j) x sincethepersonmayfailwithintheyearfromcause jafterfirstfailingfromanother cause,butthisargumentdoesnotholdwhenwearespeakingof instantaneousratesoffailure rather than failure over a period of positive length. This by itself is not sufficient for equality – see Exercise 11.15 – and we need the independence hypothesis. In Section 17.4.5, we give a formal proof and provide an explanation of the fact that by virtue of independence, 𝜇 ′(j) x (t) = 𝜇(j) x (t). (11.23) We can now summarize the main conclusions which follow from our independence hypothesis. r Given the forces of decrement in the associated single-decrement tables, we can uniquely determine the multiple-decrement model, using (11.23) together with the procedure of Section 11.4. ASSOCIATED SINGLE-DECREMENT TABLES 177 r Givenonlytheratesofdecrementintheassociatedsingle-decrementtables,wecannot uniquely determine the multiple-decrement model, and must choose from different methods, the most notable being the two described in Section 11.7.1. 11.7.3 Conditions justifying the two methods In this section we give conditions to justify the Method 1 and Method 2 introduced above, andalsosomeadditionalproceduresthatapplyinspecialcases.Westartwiththeeasiercase. Method 2 This will hold provided we have auniformdistributionofdecrementsineachof theassociatedsingle-decrementtables. Thatis,foraninteger x,0 < t< 1,and j= 1,2, …,m, tq′(j) x = tq′(j) x . (11.24) To show this, we first note that as in (8.16) tp′(j) x = e−∫ t 0 𝜇′(j) x (r)dr. (11.25) and then, invoking (11.23) and applying (8.20) tp′(j) x 𝜇(j) x (t) =t p′(j) x 𝜇′(j) x (t) = q′(j) x , (11.26) From (11.13) sq(1) x = ∫ 1 0 tp′(1) x (1−t q′(2) x )𝜇(1) x (t)dt. Now substitute from (11.24) and (11.26) and integrate to obtain the Method 2 formula. The sameprocedureworksforthegeneralcaseof mdecrements.Wejustneedtoincludeadditional factors of (1−tq ′(j) x ) in the integrand. Method 1 We need a preliminary definition. We say that theUniform Ratio Hypothesis (UR)holdsataninteger(x )ifthereareconstants Kj forj= 1,2, …msuchthatfor0 < t< 1, 𝜇(j) x (t) 𝜇(𝜏) x (t) = Kj. Asanexampleinthetwodecrementcases,let 𝜇(1) x (t) = 2𝜇(2) x (t)for0 < t< 1.ThenURholds atxwithK1 = 2∕3,K2 = 1∕3. Suppose now that UR holds atx. From (11.13) withs=1, q(j) x = Kj∫ 1 0 tp(𝜏) x 𝜇(𝜏) x dt= Kjq(𝜏) x . (11.27) 178 MULTIPLE-DECREMENT THEORY Finally, (11.23), and (11.27) show that p′(j) x = e−∫ 1 0 𝜇′(j) x (r)dr = e−∫ 1 0 𝜇(j) x (r)dr = e−Kj∫ 1 0 𝜇(𝜏) x (r)dr = p(𝜏) x Kj = p(𝜏) x q(j) x ∕q(𝜏) x , which is the Method 1 formula for going fromq(j) x toq′(j) x . TherearecertainassumptionsthatwillimplyUR.Oneobviousconditionisthattheforces of failure for each decrement are constant over the year from agexto agex+1. That is, for 0 < t< 1 andj= 1,2, …,m, 𝜇x (j)(t) is independent oft. UR also holds at all ages under the assumption of a uniform distribution of deaths in the multiple-decrementtable. That is, for an integer (x), 0< t< 1 andj= 1,2, …,m, tq(j) x = tq(j) x . (11.28) If (11.28) holds, we differentiate (11.13) with respect tosto conclude that for 0< s< 1, q(j) x = sp(𝜏) x 𝜇(j) x (s). Adding this equality for allj q(𝜏) x = sp(𝜏) x 𝜇(𝜏) x (s). Dividing the first equation above by the second shows that UR holds atx. Itshouldbenotedthatcondition(11.28)isasomewhatunnaturalassumptionwhichdiffers from(11.24),andisthereforeinconsistentwiththeusualassumptionsofUDDforeachsingle decrement. To see this, suppose that we assume (11.28) in a situation with two decrements where q ′(j) x = 0.1, for j= 1,2. Then Method 1 applies, so from (11.18),q(j) x = 0.095. Our assumption then gives1∕2q(j) x = 0.0475. Now we can also apply Method 1 to rates over half- yearperiods.From(11.17)wehave 1∕2q′(j) x = 1−0.951∕2 = 0.0487.However,theassumption of UDD for each decrement would give1∕2q′(j) x = 0.05. Special Cases There are decrements that by their nature will not occur uniformly over the year, but rather at definite discrete points. A prime example is withdrawal from an insurance policy.Nobodyislikelytowithdrawinamiddleofapremiumpayingperiodsothatthesecan beassumedtooccuronlyonpremiumduedates.Ifweassumethatalldecrementseitherfollow such a discrete pattern or are uniformly distributed over each year, there is a fairly simple proceduretomovefromtheprimedtounprimedrates.Thisisillustratedbythefollowingtwo examples. Example 11.4 In a double-decrement model with decrements of (d) for death and (w)f o r withdrawal, we are given that deaths are uniformly distributed over each year of age in the single-decrementtable,whileone-thirdofthewithdrawalsinanyyeartakeplaceinthemiddle of the year and two-thirds occur at the end of the year. Given thatq ′(1) x = 0.20,q′(2) x = 0.36 calculateq(1) x andq(2) x . ASSOCIATED SINGLE-DECREMENT TABLES 179 Solution. To calculateq(2) x we add the probability of withdrawal before death in the middle oftheyeartotheprobabilityofwithdrawalbeforedeathattheendofyear.Theprobabilityof a withdrawal in the middle of the year is (1/3)(0.36)= 0.12. In order that this occurs before death, we require that death has not yet occurred by the middle of the year, which by the uniformdistributionassumptionhasaprobabilityof1 −(1∕2)(0.20) = 0.9.Arguingsimilarly for the end of the year we have that q (2) = 0.12(0.9)+0.24(0.8) = 0.300. Since q(𝜏) x = 1−(0.8)(0.64) = 0.488, we know thatq(1) x = 0.188. As a check we can calculate the latter figure directly. The probability of death before withdrawal in the first half of the year is just the probability of death in the first half of the year, since there are no withdrawals during that time. This is equal to 0.10. The probability of death before withdrawal in the second half of the year is the probability of death in the second half, multiplied by the probability that withdrawal didnot take place in the middle of the year. This equals (0.1)(1−0.12) = 0.088. The sum of the two probabilities is indeed equal to 0.188. Note that in both calculations above, we used the independence assumption in order to multiply relevant probabilities. Example 11.5 In a multiple-decrement model with three decrements, failures from causes 1 and 2 both occur uniformly over each year in the single-decrement table. For cause 3, 60% of the failures in any year occur 1/4 of the way through the year and the other 40% occur 3/4 of the way through the year. Find formulas that give the unprimed rates in terms of the primed. Solution. To simply the notation, fix any agex and leta ′,b′,c′ denote q′(j) x for j= 1,2,3, respectively,andlet a,b,cdenoteq(j) x fori= 1,2,3,respectively.ProceedasinExample11.4 only now we need that survival from both decrements 1 and 2 occurred at the two points in question. This gives c= 0.6c ′ ( 1− 1 4a′ )( 1− 1 4b′ ) +0.4c′ ( 1− 3 4a′ )( 1− 3 4b′ ) = c′ − 9 20a′c′ − 9 20b′c′ + 21 80a′b′c′. Formula (11.30) says that a+b+c= 1−(1−a′)(1−b′)(1−c′). Therefore we know by symmetry (since there is nothing to distinguish decrements 1 and 2) that a= a ′ − 1 2a′b′ − 11 20a′c′ + 59 160a′b′c′, b= b′ − 1 2b′a′ − 11 20b′c′ + 59 160a′b′c′. 180 MULTIPLE-DECREMENT THEORY There is a much longer procedure for the first two decrements,which might be done as a check. To illustrate with decrement 1, we use the formula a= ∫ 1 0 a′(1−tb′)tp′(3) x dt, whichfollowsfrom(11.16)aftersubstitutingfrom(11.26).Calculationofthiswillnecessitate integrating over several intervals, since the formula for tp′(3) x changes we get a= a′ [ ∫ 1∕4 0 (1−tb′)dt+∫ 3∕4 1∕4 (1−tb′)(1−0.6c′)dt+∫ 1 3∕4 (1−tb′)(1−c′)dt ] , We leave it to the reader to verify that this gives the same answer as above. 11.7.4 Other approaches We will describe two additional methods, which were popular in pre-computer days for calculating primed quantities from the unprimed. Neither of them can be easily inverted to give the unprimed in terms of the primed. They do satisfy (11.14) but their main drawback is that they donotsatisfy the consistency relation (11.15) which must hold if we assume that the causes operate independently. On the other hand, they give approximate answers quickly and can give additional insights into the difference between the primed and unprimed rates. Looking at the particular data given above in Section 11.7.1, we see that in order to compute q ′(1) x we want to estimate how many of those 100 people who became disabled during the year will die during the year after becoming disabled. Suppose we postulate that, onaverage,peopleleavefromdisabilityinthemiddleoftheyear.Sincethechanceofleaving by death during the year is 0.30, there should be approximately a chance of 0.15 of dying in the second half of the year. In other words, approximately 15 of the 100 disabilities would die during the year after becoming disabled and we could estimateq ′(1) = 0.315, not too far from the result obtained above by method 1. Our final method gives a more plausible argument for decrements other than death, for as we already noted, it is difficult to imagine someone becoming disabled after they have died.Theargumentisasfollows.Theprobabilityofdeathduringtheyearshouldbeobtained by taking the number of people who died during the year and dividing by the number of people who were in the observation period for that year. In the above example we observed 300 deaths. However, we did not really have a full 1000 people under observation for the entire year. Some of the group dropped out due to disability, and were no longer observable as potential deaths. Consider extreme cases. If all the people becoming disabled did so at the beginning of the year we would have only 900 people left. If all of them dropped out at the end of the year, then we would indeed have the full 1000 under observation. We can conclude that on average we would have the equivalent of 950 people to observe and we canestimate q ′(1) x = 300∕950 = 0.316.Similarly,wecouldestimate q′(2) x = 100∕850 = 0.118. Again, we obtain answers close to those of method 1. EXERCISES 181 Both of these alternate methods can be applied to the general case ofmdecrements. To state the formulas it is convenient to define, for each causej, q(−j) x = ∑ i≠j q(i) x , the probability that (x) will leave the group within the next year from some cause other than j. Our first argument above then gives the approximation q′(j) x ≐ q(j) x ( 1+ 1 2q(−j) x ) , (11.29) while the second argument leads to the approximation q′(j) x ≐ q(j) x 1− 1 2q(−j) x . (11.30) Thesegeneralformulasconfirmthatthetwolattermethodsgivedifferentapproximations,but the answers are usually close, since for a small value ofa,1 +ais close to (1−a) −1. The reader may find it instructive to take the case of two decrements and substitute algebraically into the right hand side of (11.16) from both (11.29) and (11.30). In both cases one obtains a result that is somewhat less than the left hand side of (11.16), verifying the inconsistency. However, we see that the difference is likely to be small. It is a sum of terms each of which is close to the product of three or moreq (j)s. Notes and references Promislow (1991) discusses select multiple-decrement tables. Exercises Type A exercises 11.1 You are given the following portion of a double-decrement table. (Blanks indicate data that you must calculate from the given figures if you need them.) x 𝓁(𝜏) x d(1) x d(2) x 50 – 100 300 51 700 50 – 52 470 40 – 53 320 Find the probability that (50) will fail first from cause 2 between the ages of 51 and 53. 182 MULTIPLE-DECREMENT THEORY 11.2 You are given the following portion of a double-decrement table. (Blanks are as in Exercise 11.1.) x 𝓁(𝜏) x d(1) x d(2) x 50 1200 100 300 51 – 200 – 52 300 (a) A 2-year insurance contract on (50) provides for benefits paid at the end of the yearoffailureifthisoccurswithin2years.Thebenefitpayableis1unitiffailure is from cause 1, or 2 units if it is from cause 2. If the interest rate is a constant 50% per year, find the present value of the benefits. (b) Find the associated single-decrement rates,q ′(1) 51 and q′(2) 51 , using the Method 1 approximation. 11.3 In a table with three decrements you are given that, for allx∈ [0,100), 𝓁(1) x = 10(100−x), 𝓁(2) x = 20(100−x), 𝓁(3) x = 30(100−x). Find the probability that (50) will fail first from cause 1 between the ages of 60 and 70. 11.4 Suppose that in a double-decrement model, you are given the associated single- decrement rates q′(1) x = 0.3, q′(2) x = 0.51. Compute q(1) x andq(2) x by Method 1 and Method 2. 11.5 Adisabilityinsurancepolicyprovidesforpaymentsatthemomentofdisabilityshould this occur within 10 years. The amount of the benefit at timetis e0.10t. The policy is purchased by level annual premiums payable continuously for 10 years until either death or disability occurs. Nothing is paid on this policy if the insured dies before becoming disabled. If the force of disability is a constant 0.03, the force of mortality is a constant 0.06, and the force of interest is a constant 0.05, find the annual rate of premium payment. 11.6 Given q (1) x = 0.05,q(2) x = 0.08,q(3) x = 0.10, use Method 1 to findq′(j) x , j= 1,2,3. 11.7 Given q′(1) x = 0.1,q′(2) x = 0.2,q′(3) x = 0.25, use both Method 1 and Method 2 to find q(j) x , j= 1,2,3. Type B exercises 11.8 Suppose that in a double-decrement model,q′(1) x = q′(2) x . Show that Method 1 and Method 2 lead to the same answer for the unprimed rates. EXERCISES 183 11.9 In a double-decrement table, with decrementsdfor death andwwithdrawal, we have that q(d) x is a constant 0.10 for allx and q(w) x is a constant 0.02 for allx. The rate of interest is a constant 0.06. An insurance policy pays 1 at the end of the year of death. Nothing is paid to a withdrawing policyholder. Find the present value of the benefits on this policy. 11.10 Inamultiple-decrementmodelwithtwocausesofdecrement,youaregivenassociated single-decrement rates ofq ′(1) x = 0.20,q′(2) x = 0.36. Calculate the double-decrement rates q(1) x and q(2) x in each of the following three cases: (a) by Method 1; (b) by Method 2; (c) assuming that failures from cause 1 are uniformly distributed over the year, but that the failures from cause 2 all take place at a point three quarters of the way through the year. 11.11 In the case of two decrements, compute the difference between the right hand side and the left hand side of (11.16) in terms ofq (1) x and q(2) x , under each of (11.29) and (11.30). 11.12 Is the following statement true or false? np(j) x +nq(j) x = 1. If false, give a correct version. 11.13 In a model with two decrements, you are given that, for allx, 𝜇(1) x (t) = 0.02 for allt and𝜇(2) x (t) = (10−t)−1 for 0≤ t< 10. Find4q(2) x . 11.14 In a model with two decrements you are given thatq(1) x = 0.1274, q(2) x = 0.1674 and that there is a uniform distribution of decrements in each year for both of the single- decrement tables. Findq ′(1) x andq′(2) x . 11.15 Consider a double-decrement model, where we relax the hypothesis that the causes operate independently. Cause 1 is death, and failure from cause 2 occurs at timetif death occurs at timet−1. Show that 11.23 does not hold. 11.16 In a multiple-decrement model with 4 decrements,q′(i) x = i∕10,i= 1,2,3,4. Each decrement is uniformly distributed over each year in the single-decrement table. Calculate q (j) x , j= 1,2,3,4. 11.17 For an insurance contract with quarterly premiums, withdrawals will occur at times 1/4,1/2,3/4and1.Studiesshowthatataparticularage x,one-thirdofthewithdrawals occur at time 1/4, one-third occur at time 1, one sixth occur at time 1/2 and one sixth occur at time 3/4. Assume that deaths are distributed uniformly over the year. If q ′(d) x = 0.12 andq′(w) x = 0.20, findq(d) x andq(w) x .( H e r e ,ddenotes death andwdenotes withdrawal.) 11.18 Amultiple-decrementmodelhas3decrements.Fordecrement1,60%ofthefailures occurattime1/3,and40%occurattime2/3.Fordecrement2,failuresoccureitherat times1/4or3/4inequalnumbers.Fordecrement3,failuresareuniformlydistributed over each year. Find formulas giving the unprimed rates in terms of the primed rates. 12 Expenses and profits 12.1 Introduction Provision for expenses and profits are two important features that we have ignored in our models up to now. Premiums paid on insurance and annuity contracts must not only provide benefits,butmustalsocontainanextraamounttocovertheexpensesofoperatingthebusiness. Inadditionthepremiumsmustbesufficienttogenerateprofitsasareturntotheinvestorswho provide the capital necessary to start an insurance operation. We will begin with a discussion of expenses, and the first step is to distinguish between three basic categories of regular periodic expenses. These can depend on the premiums, on the amount of the benefits, or be constant per policy. The major expense of the first type is commissions paid to the agents selling the policies. It is traditional for their compensation to take the form of certain percentages of the premiums. There are various expenses that will depend on the amount of the benefits. For example, larger policies will require additional efforts and expenses in the selection procedure to verify that the individual is a sound risk. Finally, there are expenses, such as setting up records for a new policy, that are largely independent of the size of premiums or benefits, and are fixed for each policy. Apeculiarfeatureofexpensesforlifeinsuranceisthatalargeportionoftheseareincurred inthefirstyearofthepolicy.Commissionspaidontheinitialpremiumaretraditionallymuch higher than those paid in subsequent years (known as renewal premiums). The expenses of selectionandofthesettingupofpolicyrecordsalsooccuratthebeginning.Oneconsequence ofthisisthattheamountstobepaiduponwithdrawalbecomeanimportantconsideration.The potential loss on withdrawal must be taken into account, since if a policyholder withdraws early, there may be no chance for the high initial expenses to be recovered through the premiums. Theexpensesmentionedabovewillbepaidatfixedtimes.Buttherearealsoextraexpenses incurredwhenalossoccursandaclaimisreportedtotheinsurer.Theinsurermustinvestigate to ensure that the claim is legitimate and then arrange to disburse the benefits. Some of these FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial INTRODUCTION 185 expenseswoulddependonthepolicyamount,asmorecarewouldbetakenwithalargepolicy. Others would be fixed. These are often referred to assettlementexpenses. For a typical life insurance or annuity policy, the mathematical treatment of expenses is quite straightforward. Suppose the insurer determines that an expensee k will be incurred at each timek= 0,1, …. Then the insurer is simply providing an additional life annuity with benefit vectore = (e0,e1,…). Similarly the expenses of claim settlement would be handled byanincreasetothedeathbenefit.Ofcoursethepolicyholderdoesnotreceivetheseamounts. They are paid to those providing the goods or services involved in the expense, but they still must be provided for from the premiums. Premiums which make provisions for expenses as well as benefits will be referred to as expense-augmentedpremiums. Some authors use the termgrosspremiums. We will however reserve this term for the premiums actually charged, which as pointed out in Section 4.6.1, can take into account factors other than the benefits and expenses. such as profits, which we discuss later on in this chapter. Example 12.1 A whole-life policy issued at agex, with level annual premiums paid for life, provides for a death benefit paid at the moment of death. The expenses are as follows: in the first year, 70% of the initial premium, 1% of the face amount, and 30 per policy; in years 2 to 10, 10% of each premium, 0.5% of the face amount, and 10 per policy; after 10 years, 5% of each premium, 0.2% of the face amount and 5 per policy; the settlement expense is 100 per policy plus 0.5% of the face amount. Assume that the expenses in any year are paid at the beginning of the year. Find a formula to compute the annual expense-augmented premium for a policy with a constant death benefit of 200 000. Solution. Let G be the expense-augmented premium. We simply proceed as we did in Chapter5,equatingthepresentvalueofpremiumswiththepresentvalueofthedeathbenefits plus the present value of the additional expense benefits. Of course the latter depend partly onG, which just means we have to solve a simple equation at the end. It is best to consider two expense vectors. These are e = (2030,1010 9,405∞), which covers the expenses depending on face amount as well as the fixed expenses, and r = (0.07, 0.19,0 .05∞), which gives the proportion of the premium for premium-based expenses. The settlement chargesmeanthattheamountpaidondeathiseffectively201100.Wethenhavetheequation G̈a x = 201100 ̄Ax + ̈ax(e)+G̈ax(r), from which we get G= 201100 ̄Ax + ̈ax(e) ̈ax(0.3, 0.99,0 .95∞). (12.1) Note that the vector in the denominator is just (1∞ −r). 186 EXPENSES AND PROFITS A more realistic model would also involve withdrawal. We would start with a double- decrementtablewithdecrements dfordeathand wforwithdrawal.Wewouldalsoneedacash valuevector cv,where cvk wouldequalthecashvaluegiventoapolicyholderswhowithdraws at timek+1. The subscript is chosen to correspond to the death benefit vector. (Note that we can safely assume that for a policy with premiums paid annually, withdrawal will only occur at an integer time, when a new premium is due.) We need as well a vectore w(standing for theexpensesofwithdrawal ) whereew k would be the expenses incurred for withdrawal at time k, analogous to the settlement expenses added to the death benefit. Formula (12.1) would be modified to G= 201100 ̄A(d) x +A(w) x (cv+ew)+ ̈a(𝜏) x (e) ̈a(𝜏) x (1∞ −r) . 12.2 Effect on reserves Suppose that we compute reserves by adding expenses to the benefits and using expense- augmentedpremiumsinplaceofnet.Theresultingquantitiesareknownas expense-augmented reserves, while the reserves calculated in Chapter 6 are known asnet premium reserves.Is it natural to ask as to how these two quantities compare. In the usual level premium policy, theexpenses constituteadecreasing sequence of benefit payments that arepaid forbyalevel addition to premiums. The extra expense charge included in the premium is less than needed to cover the expenses in early years, and more in later years. This means that typically the expense-augmented reserves will be less than the net premium reserves. To put it another way,thehighinitialexpenses areareceivable totheinsurerthatwillbecollected fromfuture premiums, and this causes a reduction in liabilities. Insurance regulators have usually taken the viewpoint that one should not count on these receivables since the policyholder might withdraw and never pay them. It is traditional that reserves be calculated without taking into accountexpenses,andusingnetinplaceofexpense-augmentedpremiums.Soevenifrealistic mortalityandinterestassumptionswereusedincalculatingreserves,ignoringexpensescauses higher reserves, and will cause losses to be shown in the early years of a policy. Consider a simple example. Suppose that for a certain policy issued to (x) with benefits paid at the end of the year of death,b 0 = 1000,qx = 0.2, i= 0.10, the net premium payable at time 0 is 250, the expense-augmented premium is 300, and the total expenses payable at time 0 are 100. The expense-augmented premium reserve at t time 1 would be (300−100)1.10 0.8 −10000.2 0.8 = 25. while the net premium reserve at time 1 is 2501.10 0.8 −10000.2 0.8 = 93.75. Even if the mortality and interest matched exactly those of the reserve assumptions, the use of net premium reserves will result in a loss of 68.75 per policy in the first year. This arises from the fact that the provision for expenses is made by an extra charge of only 50 to the REALISTIC RESERVE AND BALANCE CALCULATIONS 187 level premium, but this does not cover the actual expense of 100. (Of course, as we showed in Chapter 6, there will be corresponding gains in future years, when the actual expenses are less than the 50 allowed for in the premium.) To alleviate this distortion of the profitability in the first year, many regulatory bodies permit reserves to be calculated by methods that are generally known in North America as modified reserve systems. (Similar procedures are popular in some European countries and are known asZilmerizedreserves.) Recall the termvaluationpremium to refer to premiums used in calculating reserves, which we introduced in Chapter 6. We stress again that these are not necessarily (and in fact almost never) the same as the insurer actually charges. The modifiedsystemsstillcallforexpensestobeignored,buttheyallowforlowerinitialvaluation premiums, followed by higher valuation premiums for the later durations, calculated so that valuationpremiumsremainactuariallyequivalenttobenefits.Thiseffectivelyrecognizesthat the insurer has a smaller premium in the first few years to provide the benefits since a large amount of these premiums must go to pay the high initial expenses. Therearemanysuchsystems.Themostbasicisthe fullpreliminaryterm method.Suppose wehavealifeinsurancepolicywithlevelpremiumspayablefor nyears,andbenefitspayable at the end of the year of death. The initial valuation premium is taken asb 0v(1)qx, the exact amountneededtoprovidethebenefitsforthefirstyear,followedbyalevelvaluationpremium for all subsequent years. This will automatically make 1V = 0. This level valuation premium for years after the first will be A[x]+1(b◦ 1) ̈a[x]+1(1n−1), which will provide for the benefits after the first year. This effectively means that the entire first-yearvaluationpremium,lesstheamountneededforthebenefitsaccordingtothereserve basis, is available to use for expenses. There are several variations to full preliminary term. In some cases there is a recognition that for policies with high initial premiums, such as endowment insurance, or premiums payableoveralimitedperiod,thisprovidesmoreexpenseallowancethanisneededinthefirst year,andthefirst-yearvaluationpremiumistakentobehigherthan b 0vqx.Inothercasesthere may be more than one step. There will be a low premium used in the first year, a somewhat higher one for some duration, for example the next 10 years, and a still higher one after that. We will not discuss the details of the various methods here. Some of these are dealt with in the exercises. 12.3 Realistic reserve and balance calculations As we have indicated, insurers are required to calculate reserves in accordance with the particular regulations that apply in their jurisdiction. These methods are intended mainly to ensure that policyholder will receive their promised benefits. These are often referred to as statutory reserves. There are however many reasons for the insurer, to calculate reserves and balances on a more realistic basis which use expense-augmented or gross premiums and expenses, rather than net premiums with expenses ignored, and which takes into account withdrawals. An acronym that is sometimes used here is that of a ‘GAAP’ reserve, the acronym standing forgenerally accepted accounting principles.Such a reserve would then 188 EXPENSES AND PROFITS be calculated as the present value of all future benefits, including death benefits, expenses, surrender benefits, less the present value of future premiums. It is useful to replace our basic recursion formula (6.8) with the following more general version. For simplicity we assume aggregate mortality, and benefits payable at the end of the year of death. Consider a policy withannualpremiumsof 𝜋 k,deathbenefitsof bk,cashvaluesof cvk,anexpenseof rk perunit of premium paid at timek, fixed expenses at timekof ek, claim settlement expenses ofed k in thecaseofdeathbetweentime kandtime k+1,andsurrenderexpensesof ew k forsurrenderat timek+1.Ourbasicrecursion(6.8),whenmodifiedforexpensesandwithdrawal,becomes k+1V = [kV+𝜋k(1−rk)−ek](1+ik)−q(d) x+k (bk +ed k −k+1 V) −q(w) x+k (cvk +ew k −k+1 V) , (12.2) and solving fork+1V, we obtain the Fackler type equation k+1V = [kV+𝜋k(1−rk)−ek](1+ik)−q(d) x+k (bk +ed k ) −q(w) x+k (cvk +ew k ) 1−q(d) x+k −q(w) x+k . (12.3) We want to discuss three main quantities which arise from the above recursion. They differ in the choice of premiums and initial values. Expense-augmented reserves. These reserves, as defined above, can be calculated from (12.3) with𝜋k equal to the expense-augmented premium. Gross-premium reserves.These are simply the present value of future benefits and expenses less the present value of futuregrosspremiums. They can be calculated from (12.3) with𝜋k equal to the gross premium. AssetShares. Thesearerealisticbalancecalculations.Theygivetheaccumulatedvalueof thepremiumslesstheaccumulatedvalueofthebenefitsandexpenses,Theycanbecalculated from (12.3) using gross premiums, but now with an initial value of 0 at time 0. (Different notation is often used here to distinguish them from reserves. We will write AS k in place of kV.)Thenamecomesfromthefactthattheyrepresentthatshareoftheinsurer’sassetswhich are attributable to the particular contract. We will use the following simplified example to illustrate the difference between these quantities. Example 12.2 Atwo-yearpolicy on( x)provides fordeathbenefits of1000 paidattheend oftheyearofdeath.Premiumsof160arepaidfor2years.Thereareexpenses,incurredatthe beginningofeachyear,of20%ofthepremiumplus18inyear1,and5inyear2.Youaregiven that i= 100%,q x = 0.2, qx+1 = 0.3. There are no withdrawals. Find the expense-augmented reserves, gross-premium reserves, and asset shares. Solution. Let Gdenote the expense-augmented premium. The present value of benefits and expenses is 1000[(0.5)(0.2)+(0.25)(0.8)(0.3)]+18+0.2G+5(0.5)(0.8)] = 180+0.2G. Then G[1+(.5)(.8)] = 180+0.2Gshowing thatG=150. PROFIT MEASUREMENT 189 For the expense-augmented reserves, 0V = 0, 1V = [0.8(150)−18)×2−200]∕0.8 = 5, 2V = 0. For gross-premiums reserves, 0V = 180+0.2(160)−160(1.4) =− 12, 1V = [(−12+0.8(160)−18)×2−200]∕0.8 =− 5, 2V = 0 For the asset shares AS0 = 0, AS1 = [(128−18)×2−200]∕0.8 = 25, 2AS2 = [25+160−5]×2−300]∕0.7 = 85.71. In the following section we will revisit this example, and interpret the figures in terms of profit calculations. Remark For simplicity, some examples involving expenses will ignore withdrawals. This doesnecessarilymeanthatwithdrawalsarenotallowedaswestatedinthepreviousexample. The implicit assumption is that at each duration, the cash value together with any expense of withdrawal is precisely equal to the reserve on the policy. In such a case the withdrawal rates and cash values have no effect on premiums or reserves, and need not be taken into account. 12.4 Profit measurement 12.4.1 Advanced gain and loss analysis WewanttocontinuetheprofitanalysiswebeganinChapter6,butnowincorporatingexpenses and withdrawals, and also by considering a more extensive context. We start with a very general setting. We are given a particular insurance or annuity policy, with a description of the benefits and cash values, and a set of reserves. The latter couldbenetpremiumreserves,expense-augmentedreserves,gross-premiumreserves,oreven somethingelse.Inadditionwearegivenabasisforallrelevantfactors,knownasthe profittest basis. These are interest rates, rates of mortality and withdrawal, expenses and the premiums which will be charged. We want to determine the profit we will have in a specified period, if our actual experience is according to the given profit test basis and we use the specified reserves to measure profit. There are different contexts possible. The profit test basis could be a hypothetical test basis, to see what profits will emerge under our assumptions, and the major goal is to see if the proposed premiums will achieve a desired level of such profits. Alternativelythiscalculationcouldbecarriedoutafterthefactwiththeprofitbasisreflecting the actual observed experience, and the result will be the actual profit that is achieved. 190 EXPENSES AND PROFITS Inanyevent,fortheperiodrunningfromtime ktok+1wehaveaprofitofPr k+1 consisting of the right hand side of (12.2) less the reservek+1V. That is Prk+1 = (kV+𝜋k(1−rk)−ek)(1+ik)−q(d) x+k (bk +ed k −k+1V) −q(w) x+k ( cvk +ew k −k+1V ) −k+1 V, (12.4a) = (kV+𝜋k(1−rk)−ek)(1+ik)−q(d) x+k (bk +ed k ) −q(w) x+k (cvk +ew k ) −p(𝜏) x+k k+1V. (12.4b) The two versions represent two possible points of view, leading to the same result, as we already indicated in Chapter 6. In (12.4a) the insurer sets up the reserve for everyone and then needs to pay only the amount of the benefits above the reserve in the case of death or surrender, while in (12.4b), the insurer pays the benefits in full but then needs to set up a reserve only for the survivors. Version (12.4b) is normally the most efficient for calculation. Yet another version is to write Pr k+1 = (𝜋k(1−rk)−ek)(1+ik)−q(d) x+k (bk +ed k ) −q(w) x+k (cvk +ew k ) −Δk(V), (12.4c) where Δk(V) = p(𝜏) x+k k+1V−k V(1+ik) (12.5) represents the increase in reserves over the year. This version simply says that profit is what is left over after we accumulate the net amounts collected, and provide for benefit payments, expenses and the increase in reserves. Example 12.3 Refer back to Example 12.2 and calculate Pr k when the the underlying reserves are: (a) gross-premium reserves; (b) expense-augmented reserves. In each case, the profit test basis is the same as the reserve basis. Solution. (a) It is clear that Pr 1 = Pr2 = 0. So what happened to the profits? The answer is that with gross-premium reserves, and observed experience the same as the reserve basis, they all occur at time 0. The reserve of−12 at this time, means that the insurer could take 12 out of the company surplus for each such policy and still have enough left to provide for the reserves. This shows that in addition to the definition of Pr k given above we must add the statement that Pr0 =− 0V. (b) Using (12.4(b)), Pr1 = (128−18)2−0.2(1000)−0.8(5) = 16. Pr2 = (5+160−5)2−0.3(1000) = 20. Notethedifferenceintheincidenceofprofitforthedifferenttypesofreserves.Withgross- premium reserves, all the profit is recognized at the beginning. With the expense-augmented PROFIT MEASUREMENT 191 reserves, the profit emerges gradually as the extra amounts above the expense-augmented premiums come in. In this case there is an extra 10 received in the first premium, but this is reduced to 8 after the percentage of premium expense. This accumulates in 1 year to 16. Similarlytheextra10inthesecondpremiumaccumulatesto20attheendofthesecondyear. It will instructive for the reader to verify that in the long run, that the reserve basis will not affect the actual profits received. This example also indicates that we can view asset shares as the sum of the expense- augmented reserves plus an accumulation of the profits at interest and survivorship. For example, AS 2 = 0+8 ( 4 0.56 ) +10 ( 2 0.7 ) = 85.71. as calculated above. 12.4.2 Gains by source We now elaborate on the decomposition of gains by source, as initiated in Section 6.4.1. We willinfactconsiderasomewhatmoregeneralsituation.Supposewehavetwodifferentbases under consideration, one in which quantities are starred and the other unstarred. In the most commontypeofapplicationtheunstarredsymbolsrefertotheexpectedbasis,andthestarred will refer to what actually happened, but mathematically these could be any two bases. We thenhavetwoprofitcalculations,Pr ∗ k andPrk andweareinterestedinthedifferencePr ∗ k −Prk which in the common situation will be the excess of actual profit over the expected profit. In the case where the reserve basis coincides exactly with the unstarred basis, we have that Pr k = 0 for allkand the difference will just be Pr∗ k . This was precisely the point of view we took in Chapter 6. The goal is to decompose this difference into the various gains by source. As well as the mortality and interest gains of the Chapter 6 model, we now have additional sources. Oneisthegainfromwithdrawal,whichiscalculatedanalogouslytothemortalitygainas ( q (w) x+k −q∗(w) x+k ) (cvk −k+1 V). In addition we have gains from the various types of expenses. These present a compli- cation, since they are invariably tied up with other factors. Consider for example the case of death settlement expenses. As a particular example, suppose we haveq (d) x+k = 0.07, q∗(d) x+k = 0.05, ed k = 200 ande∗d k = 100.Soourmortalityexperienceismorefavourablethanexpected and the costs of settling each claim are less than expected. We gain in both ways and in fact the total gain per policy will be 0.07(200)−0.05(100)=9. We would like to divide this gain up into that attributable to mortality, and that attributable to lower settlement expenses. A somewhat surprising fact is that there seems to be no reasonable way to do this uniquely. To illustrate, consider the general situation, which applies as well to withdrawal expenses. (For simplicity in notation we will omit superscripts and subscripts.) We have a total gain of (qe−q ∗e∗).Now we can write this in two distinct ways, namely (q−q∗)e+(e−e∗)q∗,or( q−q∗)e∗ +(e−e∗)q. 192 EXPENSES AND PROFITS In either case it would be reasonable to take the first term to be the mortality gain and the second term to be the expense gain. However, there is nothing to choose between them, and they give different answers. One way to decide is to determine an order in which we will calculate the gains. If we decide to look at mortality first, it would be natural to ignore the expense differential at this point and take (q−q ∗)e as the mortality gain, as in the first expression. Then necessarily we would take (e−e∗)q∗ as the expense gain. If we decide to look at expense first, we would decompose according to the second expression. In short, for the factor considered second, the gain will be expressed with the other factor starred. So in ournumericalexampleabove,wecoulddividethetotalgainof9into4formortalityand5for expenses, or 2 for mortality and 7 for expenses, deciding on which decomposition we used. Anotherwayofviewingthisistonotethatwecanalsowritethetotalgaininasymmetric fashion as qe−q ∗e∗ = (q−q∗)e+q(e−e∗)−(q−q∗)(e−e∗). The first term on the right is strictly mortality gain, the second term is strictly expense gain and the third term, which is a combined effect, can be allocated to one or the other. This seems rather arbitrary and artificial but it is the best we can do if we insist on a decomposition. Indeedwehavealreadyencounteredthisproblembefore,namelyintheexampleinvolving premium difference gain in Chapter 6, a gain which is tied in with interest. Since we had previously talked about interest gain, it seemed natural to do that first, but we could have insteadtakenthepremiumdifferencegainas(𝜋 ∗ −𝜋)(1+i),leavingthetotalinterestgainas (tV+𝜋∗)(i∗ −i). Thesameproblemcomeswithperiodicexpensegainswhichalsoinvolveinterest.Ifwefor example,focusoninterestfirst,thetotalgainof(e k(1+ik)−e∗ k (1+i∗k)willbedecomposed into a part due to interest ofek(ik −i∗ k ) and a part due to expenses of (ek −e∗ k )(1+i∗). (Note thatearningmoreinterestthanexpectedwillresultinthefirstitembeingnegative.Thereader should ensure that they can explain why this is so.) What would happen if there were a source of gain connected withthreedifferent factors? We’ll leave it to the reader to verify that there would now be six possible ways of dividing up the gain. Theoretically this could happen. For example, a percentage of premium expense would be affected by differences in both premiums and interest. In the common situation we have described, this will not occur since the premiums payable on both the actual profit basis and the expected profit basis will be the same, namely the gross premium, and the premium difference gain will be zero. Example 12.4 You are given the following information about a policy issued at age 50. The gross premium payable at age 65 is 2000, and a death benefit of 100 000 is payable at age 66 for death occurring between age 65 and 56. A cash value of 20 000 is paid at age 66 for policyholders surrendering at that time. Moreover 15V = 20 300, and16V = 22 500. The insurer’s best estimate for mortality, withdrawal and interest is given by q(d) 65 = 0.005, q(w) 65 = 0.008, i= 0.04. Anticipated expenses at the beginning of the year are 5% of the gross premium, plus 100. In addition, the expected settlement expenses are 200 for a death claim and 100 for withdrawal. PROFIT MEASUREMENT 193 Nowhereistheactualexperience.Outof1000policiesatthebeginningofthe16thpolicy year, there were 4 deaths and 10 withdrawals. The actual interest rate earned was 0.048. The actual expenses incurred were 230 per policy at the beginning of the year, including both percentage of premium and fixed expenses, 150 to settle each death claim, and 120 for each withdrawal. For the year running from time 15 to time 16, calculate the actual profit less the expected profit, and classify this difference by source. For the latter, assume that all expense gains are calculated afterthe gains from the other sources. Solution. Taking the actual results for the starred quantities, and using (12.3) Pr ∗ 16 = (20 300+2000−230)(1.048)−.004(10 0150)−0.010(20 120)−0.986(22 500) = 342.56 Pr16 = (20 300+2000(1−0.05)−200)(1.04) = 0.005(10 0200)−0.008(20 100) −0.987(22 500) = 114.7 Actual profit−Expected profit = 227.86. This can be decomposed as follows: Gain from mortality (including the portion due to settlement expenses) = (0.005 − 0.004)(10 0200−22 500)=77.70. Gain from withdrawal (including the portion due to withdrawal expenses)= (0.008 − 0.010)(20 100−22 500)=4.80. Gain from interest=(20 300+ 2000−200)(0.048 −0.040) =176.80 Gainfrompercentageofpremiumandperiodicexpenses =(200−230)(1.048)=−31.44 Death settlement expense gain=0.004(200−150)=0.20 Withdrawal expense gain=0.010(100 −120)=− 0.020. Theexpensegainsabove,comingaftertheothercalculations,allusetheactualexperience for the other factors, as we noted in our remarks about order above. The reader should find it instructive to redo the calculations, now assuming that the expense gains are calculated first. Note that since the insurer benefits from a withdrawal, paying out less than the reserve, the higher than expected number of withdrawals caused a positive gain. 12.4.3 Profit testing We now return to the case where we have a proposed profit test basis, and proposed gross premiums. We want to compute the profits over several periods, and see how we can use the resulting figures. Example 12.5 A five-year, 100 000 endowment insurance policy on (60) has benefits payable at the end of the year of death and carries annual premiums of 21 500, payable for 194 EXPENSES AND PROFITS 5 years. Initial expenses are 60% of the premium plus 200, and renewal expenses are 5% of thepremiumplus100.Deathsettlementexpensesforthefirst4yearsare0.2%ofthetheface amount plus 50. The expense of paying any cash value, or the final payment at time 5 is 50. The profit test basis interest rate is 5%. The profit test basis mortality and withdrawal rates, reserves, and cash values are shown in the following table. Find Pr k fork= 0,1, …,5. kq (d) 60+k−1 q(w) 60+k−1 k Vc v k−1 Prk 00 0 0 1 0.009 0.080 18 500 0 −8935.75 2 0.010 0.050 37 500 17 500 3636.25 3 0.012 0.030 57 500 42 500 3151.75 4 0.014 0.020 78 500 70 000 3080.75 5 – – 100 000 0 3716.25 Solution. The values of of Prk are computed directly from (12.3) and are shown in the final column. As an example, here is the computation of Pr2. Pr2 = (18 500+21 500−0.05(21 500)−100)(1.05)−0.01(100 250)−0.05(17 550) −0.94(37 500) = 3636.25 Thevector(Pr 0,Pr1,…)issometimescalledthe profitvector.Itisnothoweverthevector we want in order to accomplish our goal of analyzing profits. Recall that Prk is the profit in the year running from timek−1t ot i m ekfor a policy in force at timek−1. The figure we want is the expected profit for that year. Accordingly we make the following definition. Definition 12.1 The profit signaturefor a policy issued at agex is the vectorΠ, where Π 0 = Pr0, and fork> 0 Πk =k−1 p(𝜏) x Prk. Inotherwordswemultiplytheyearlyprofitsbytheprobabilitythatthepolicyholderwill survive to the beginning of the particular year. Example 12.6 Find the profit signature for Example 12.5. Solution. We first calculate p (𝜏) x = 0.911, 2p(𝜏) x = 0.85634, 3p(𝜏) x = 0.82037, 4p(𝜏) x = 0.79248. A straightforward calculation then yields Π= (0,−8935.75, 3312.62, 2698.97, 2527.37, 2945.06) . PROFIT MEASUREMENT 195 Our problem of analyzing profits is now reduced to the familiar problem, dealt with from the beginning of this book, of analyzing a sequence of cash flows. Forexamplewecansummarizethiscashflowsequencewithasinglenumber,bycomput- ing the present value of the profit signature according to some discount function. This gives the net present value of profits, often abbreviated as NPV. In practice the discounting is at a constant rate of interest which is called therisk discountrate. We can interpret this rate as one which is the desired return by an investor putting up the capital to fund an insurance enterprise.Itwillthereforereflecttheriskinvolvedinsuchaninvestmentandwillaccordingly benormallyhigherthansaytheprofittestrate.Notethatwedonotdiscountbysurvivorshipin calculating the NPV. This has already been considered when calculating the profit signature. RecallnowthatwehavecalculatedthisNPVaccordingtoasinglepolicy,withaspecified amount of benefits and premiums. To interpret it properly and obtain measures which can be used for comparative purposes, we must relate this to some measure of volume. A natural candidate for this is the present value of premiums. Definition 12.2 We define theprofitmarginforthepolicy as the quantity NPV ̈a(𝜏) x (𝜋) where 𝜋 is the premium vector for the policy. The discounting in the denominator is done withbothinterest,atthesameriskdiscountrateusedinthenumerator,andwithsurvivorship as given by the multiple-decrement table. Example 12.7 If the risk discount rate is 0.07, find the profit margin for the policy of Example 12.5. Solution. The denominator is just 21 500̈a 60(15) at 7% interest=83 282.65. The numerator is (1.07−1)(−8935.75)+(1.07)−2(3312.62)+(1.07)−3(2698.97)+(1.07)−4(2527.36) +(1.07)−52984.09 = 773.27 The profit margin is 773.27/ 83292.65=0.0093. This example shows then that for the policy in question, for every dollar in premium received in sales, 0.93 cents would represent profit. This can be used for various decision problems.Ifitdeemedlowerthanthecompanywouldlike,thensomechangesmustbemade in the policy structure, usually by raising the premium charged, or in some case by lowering the surrender benefits if allowed by applicable legislation. The profit margin can also be compared with other policies to see which types of benefit and premium structures are most profitable. Anaturalquestionnowarises.Howistheprofitmarginaffectedbythechoiceofreserves. Our calculations in Section 6.5 suggest that under certain conditions the reserve basis should 196 EXPENSES AND PROFITS notaffecttheprofitmargin,andthisindeedistruewhenthediscountratesusedincomputing the NPV are the same as those of the profit test basis. To see this, look at formulas (12.3) and (12.5). Suppose that the single reserve figure kV is increased by an amountE. ThenΔk−1V is increased bypx+kEso that Prk is decreased bypx+kE. Similarly,Δk(V) is decreased byE(1+ik) so that Prk+1 is increased by (1+ik)E resulting in an decrease inΠk of kpxE and and increase inΠk+1 of kpx(1+ik)E, which will cancel out when we discount, leaving the NPV unchanged. This is not true however in the more typical case where the risk discounting is at higher rates. Suppose thatE> 0. The calculation above shows that discounting the increase inΠk+1 with a rate higher thanik for the year running from timek to k+1 will decrease the NPV. Thisiseasilyexplained.Thehigherreservemeansthatmoremoneymustbesetasideattime k, reducing profits at that time. This money is earning interest and will add to the profits the following year. However under the present scenario, it is not earning sufficient interest to compensate.Theconclusionisthatintheusualcasewhereriskdiscountratesarehigherthan theprofittestrates,increasesinreservescausereducedprofitability,anddecreasesinreserves cause increased profitability. Notes and references See Bowerset al.(1997, Chapter 16) for a detailed discussion of various modified reserve systems. Netpremiumreservesaresometimesreferredtoas benefitreservesintheliteraturesince, they involve only benefits and not expenses. The difference between the expense-augmented reserve and the net premium reserve is sometimes referred to as theexpense reserve.I t represents the present value of the future premiums available for expenses, less the present value of the future expenses. Some authors treat certain expenses incurred right at or just before a policy begins differ- ently than above. They would appear as a negative contribution to Pr 0. In our treatment they are accumulated to the end of the year, and appear as a negative contribution to Pr1,i nt h e same way as the initial premium received at time 0, makes a positive contribution to Pr1. Exercises Type A exercises 12.1 Aninsurancepolicyissuedatage xprovidesforapaymentof100000atthemoment ofdeathprovidedthisoccurswithin20years.Levelannualpremiumsof Garepayable for10years.Theexpensesareasfollows:inthefirstyear,60%oftheinitialpremium, 1%ofthefaceamountand20perpolicy;afterthefirstyear,5%oftheeachpremium, 0.25% of the face amount and 10 per policy; the death benefit settlement expense is EXERCISES 197 50 per policy plus 0.5% of the face amount. Assume that the expenses for any year are paid at the beginning of the year. Find a formula forG. 12.2 Aninsurancepolicyissuedatage xhasdeathbenefitvector bpaidattheendoftheyear of death, and level annual premiums payable fornyears. Assume a constant interest rate. GivenA x(b) = 330, Ax+1(b◦ 1) = 280, ̈ax(1n) = 11, ̈ax+1(1n−1) = 8 and ̈ax+k(1n−k) = 6, find the excess of the reserve at timekcalculated with net premiums over the reserve at timekcalculated by the full preliminary term method. 12.3 A certain policy on (x) has gross-premium reserve of 1000 at time 10. The gross premium payable at time 10 is 100. The death benefit payable at the end of year of death for death between time 10 and time 11 is 1500. The cash value payable at time 11 is 900. Expenses for the year running from time 10 to time 11 are 20, paid at the beginningoftheyear.Theinterestrate i= 10%, q (d) x+10 = 0.08and q(w) x+10 = 0.05.Find the gross premium reserve at time 11. 12.4 A policy on (x) provides death benefits of 1000 paid at the end of the year of death. There is no cash value for withdrawal in the first year. For withdrawal in the second year, an amount of 30 is paid at the end of the year. You are given that q (d) x = 0.1, q(d) x+1 = 0.2, q(w) x = 0.05, q(w) x+1 = 0.06, i= 0.07. Expenses are 120 in the first year and 40 thereafter, paid at the beginning of the year. Thelevelexpense-augmentedpremiumpayableannuallyis250andthecorresponding gross premium is 300. The gross-premium reserve at time 0 is−130. Find (i) the expense-augmentedreserve,(ii)thegross-premiumreserve,(iii)theassetshare,allat the end of 2 years. Type B exercises 12.5 A modified reserve system has a first-year premium ofb0vqx a second-year premium of b1vqx+1, followed by a level renewal premium of𝛽. Assume constant interest and benefits at the end of the year of death. Find an expression for𝛽, when premiums are payable fornyears. 12.6 Considerapolicyissuedatage xwithbenefitspayableattheendoftheyearofdeath. A modified reserve system, used in Canada for tax purposes (sometimes known as the 1 1 2 preliminary term), takes an initial premium tob0vqx, a premium at time 1 of 𝛽,andalevelpremiumof 𝛾 payableaftertime1,suchthat 𝛽 istheaverageof b1vqx+1 and 𝛾. Assume constant interest and benefits paid at the end of the year of death. If premiums are payable for life, show that 𝛾 = Ax+1(b◦ 1)+vpx+1Ax+2(b◦ 2) ̈ax+1 +vpx+1̈ax+2 . 12.7 SupposethatinExample12.3theinsurersells100ofthegivenpolicies.Showthethe total profits accumulated with interest to time 2 are the same, regardless of whether gross premium or expense-augmented reserves are used. 198 EXPENSES AND PROFITS 12.8 Redo Example 12.4 with the expense gains calculatedbeforethose from mortality, withdrawal and interest. 12.9 Refer to Example 12.5 (a) Suppose that1V is changed to 13 000 while all other data remains the same. Calculate the new profit signature and profit margin. (b) Redo part (a) with the risk discount rate equal to 5%. (c) Find the profit margin, for the original problem with a risk discount rate equal to 5%. 12.10 RefertoExample12.5.Supposewedecidetoincreasethepremiumtoachieveaprofit margin of 3%. What should the new premium be? ∗13 Specialized topics FortheconcludingchapterinPartIofthisbook,wecoverafewspecializedtopicsininsurance and annuities that are of current importance. We will not go into any of these topics in depth, but rather provide an introduction to some of the major features. The material here is not needed for any other chapters of the book. 13.1 Universal life 13.1.1 Description of the contract Universallifeisatypeofcontractthatbeganinthe1970s,andnowaccountsforasubstantial portion of life insurance sales. Prior to that time, the mainstay of life insurance was for the most part whole life or endowment contracts. The origin of the change could well have been the advice given by some financial advisors that people should buy term insurance instead of those products, for a much smaller premium, and then invest the difference elsewhere. Now our analysis in Section 6.4.2 shows that this is exactly what happens within the policy itself when one purchases whole life or endowment insurance. The excess premiums, not needed to provide insurance in a particular year, are invested in the savings portion. Of course, the purchaser of endowment or whole life insurance has less flexibility than the term buyer,bothwithregardstotherelativeamountsgoingintotheinsuranceandsavingsportion, and to the types of investments and rates of return. Universal life plans were developed to allow this flexibility within a single contract. The main features involve variable premium payments, and the ability for the policyholder to participate in higher yielding investment opportunities. In the usual type of universal life plan, premiums are not fixed in advanced and may be varied at the option of the purchaser. Each individual has in effect a separate fund that changes over time in the way we describe it by the recursion formula (6.8). Premiums paid FundamentalsofActuarialMathematics , Third Edition. S. David Promislow. © 2015 John Wiley & Sons, Ltd. Published 2015 by John Wiley & Sons, Ltd. Companion website: http://www.wiley.com/go/actuarial 200 SPECIALIZED TOPICS are deposited, the fund is credited with interest or investment earnings, while expenses, and the cost of insurance, based on the net amount at risk are deducted. The policyholder need only pay sufficient premiums to ensure that the fund value will cover the cost of insurance. In this type of plan where there is a more direct relationship between policyholders and their individual accounts, there is a more frequent demand for the type of plan that we discussed in Section 6.7. It is usual to allow two options. The purchaser can fix the amount paid at death, or they can fix the net amount at risk, so at death, the amount in their account (representing the reserve) would be returned in addition to the stipulated face amount, giving an increasing death benefit. In the usual terminology the fixed death benefit is often referred to as a Type A policy, whereas the death benefit plus account value is referred to as type B. Some plans provide the flexibility to alter the face amount as well as the premium, with the stipulation that a request to increase the coverage would normally necessitate the type of information as required by new policyholders as to health and other conditions affecting the risk. There do arise cases however when such changes are mandated. In the United States this occurs due to regulations stipulating that in order to receive the favourable tax treatment accordedtolifeinsurance,thedeathbenefitmustexceedthefundvaluebyacertainminimum percentage. If the fund value gets high enough, the insurer will increase the death benefit in order to maintain this so-calledcorridorrequirement. Policyholderscanlapsethepolicyatanytimeandreceivetheirfundvaluelessaso-called surrender chargethat is intended mainly to account for the initial expenses, as we outlined in Section 6.6. The surrender charge will decrease with time and normally will become zero after a certain number of years, often 20 or so. As well as the flexibility in premium payments a major attraction of universal life for the purchaser is the opportunity to earn a higher yield. The credited interest rate is usually not fixed in advance as it essentially is in traditional plans, but it is allowed to vary. There are differentwaysofaccomplishingthis.Insomecasesthefundsofthepolicyholderareactually invested in certain assets, so that the account is very much like a mutual fund investment togetherwithguaranteedbenefitspayableondeath.Usuallythereisachoiceofvarioustypes of assets or funds to invest in, so the policyholder can control the amount of risk they are willing to take in order to obtain higher returns. This type of contract is often referred to as variableuniversallife. In other cases the account of the policyholder need not be invested in anyparticularassetsbutinterestiscreditedaccordingtoearningsonsomereferenceportfolio. Suchcontractsareknownas equityindexedinsurance. Therearenormallysomequalifications tothecreditingofinterest.Therewillbeastipulatedminimumrateofinterestthatiscredited regardless of what the reference portfolio does. It is also common to have limits on the other end. There may be a ‘participation rate’. So for example if the reference portfolio earned 9% interest over a period and the participation rate was 80%, the the policyholder’s account wouldbecreditedwith7.2%.Inadditionthereisoftenastatedmaximumthatwillbecredited regardless of the actual earnings. Universal life contracts often carry other guarantees that mitigate again unfavourable investment experience. These are referred to assecondary guarantees. A popular option of thistype,isthataslongastheinsuredkeepsupacertainstipulatedminimumrateofpremium payment, the death benefit is guaranteed even though lower than expected rates of return bringtheperson’saccountbelowtheamountneededtomeetthecostsofinsurance.Asimilar provision is commonly used with endowment type contracts that have a maturity date, in which case a minimum guaranteed amount paid at maturity provided that the insured keeps up a minimum schedule of premium payments. UNIVERSAL LIFE 201 13.1.2 Calculating account values For each policyholder, the amount of money in their account is calculated periodically, often monthly. For the most part this is handled easily by the basic recursion formulas that we have seen in previous chapters. There are however some aspects of terminology and special features that will be discussed. Given the account value AV k at timek, the account value AVk+1 is calculated using the following familiar quantities. We assume aggregate mortality and payment of benefits at the end of the period of death. (a) 𝜋k, the premium paid at timek. (b) rk,thepremiumexpenserate.Thisisapercentage,oftenaround5%,takenfromeach premium by the insurer to handle expenses. Only the remaining amount (sometimes referred to as theallocated premium) is credited to the account. In the so-called unit linked policiespopular in the United Kingdom, it is common to incorporate this expense by a device termed abid–offer spread. In these cases the policyholder is deemed to have their own separate fund, distinct from the general assets of the company, and the premium is used to buy a number of units of the fund. These are bought at a certainofferprice, and sold back at a lowerbidprice. So for example if the offer price was 100 per unit and the bid price was 95 per unit, and a premium of 1000 was paid, the purchaser would own 10 units of the fund, which would have a value of 950. So the allocated premium would be effectively 950. (c) e k, an additional flat amount that is charged each period for expenses. (d) ck, the cost of insurance rate. This is just a mortality rate which will apply to the time periodrunningfromtime ktotime k+1.Forayearlycalculationwithanissueageof x, we would haveck = qx+k, for an appropriate life table. For a monthly calculation, assuming UDD, we would haveck = (1∕12)qx+m, where 12m≤ k< 12(m+1). It is oftenexpressedasarateofsomuchper1000.Soforexampleacostofinsurancerate of 15 per 1000 would mean thatc k = 0.015. (e) ik, the credited interest rate. This may be specified directly or tied to the performance of some other investments. (f) bk, the death benefit paid at timek+1 for death between timekandk+1. Then the recursion is just AVk+1 = (AVk +𝜋k(1−rk)−ek)(1+ik)−ck𝜂k. (13.1) FortypeApolicieswehave 𝜂k = bk −AVk+1,ifthisispositive,andwesolvetheequation for AVk+1.For type B policies where the death benefit is paid in addition to the face amount, 𝜂k = bk. Another possible type of arrangement is where the fund itself is returned at death, subject to a guaranteed minimum amountbk. (As mentioned above, this may be ruled out in certain jurisdictions by taxation requirements.) In such case if AVk+1 ≥ bk, we would have 𝜂k =0, while if AVk+1 < bk, we would have𝜂k = bk −AVk+1. It must be kept in mind that the account on a universal life policy should be viewed as belonging strictly to the particular policyholder, so while the account value is similar in 202 SPECIALIZED TOPICS nature to a retrospective reserve, it is not quite the same. For example, withdrawal rates and surrendervaluesarenottakenintoaccountinthiscalculationaswasdoneinsomeChapter12 formulas.Todosowouldmeanthatpolicyholderaccountswouldbeaugmentedwithpositive gainsfromsurrender,butthesebelongstrictlytotheinsurer(asofcoursedosurrenderlosses, should they arise). Example 13.1 A universal life policy has a stated death benefit of 100 000, monthly premiums of 5000, monthly expenses of 100, and a further monthly expense charge of 4% of eachpremium.Theannualcreditedinterestrateis8%,andthemonthlycostofinsurancerate in the 5th year of the policy is 30 per thousand. If the account value at the end of the 52nd monthis50000,findtheaccountvalueattheendofthe53rdmonth,assumingthatthepolicy is (a) type B and (b) type A. Solution. We first must calculate the applicable monthly interest rate which is (1.08) 1∕12 − 1 = .00643. Then (a) AV53 = (50000 +5000(0.96)−100)(1.00643)−0.03(100000) = 52051.72. (b) The100000aboveisreplacedwith100000 −AV53,andsolvingtheequationwejust obtain the Type B amount divided by (1−0.03). AV53 = 52051.72∕(1 −0.03) = 53661.57. Another calculation that will normally be done each period is to compute thecost of insurance(as opposed to the cost of insurancerate). This is sometimes abbreviated as COIk, fortheperiodrunningfromtime ktok+1.(Someauthorspreferadifferentindexingmethod and would refer to this as COIk+1. We prefer to have the index correspond to that of the cost of insurance rateck.) This is normally determined as at thebeginningof each period, so if therearenotsufficientfundstopayit,thepolicywouldlapseintheabsenceofanyguarantees to the contrary. In effect it can be viewed as the net single premium paid each period for the death benefit coverage. That is COI k = (1+ik)−1ck𝜂k. So in the above example we would have COI52 =(1.00643)−1(0.03)100000 = 2980.83, for the type B policy and COI52 =(1.00643)−1(0.03)(100000 −53661.57) =1381.27, for the type A policy. There are some variations which can arise. In some circumstances the insurer may use a discountrate jk incalculatingthecostofinsurancewhichisdifferentfromthecreditedinterest rateik. In this case formula (13.1) would be modified to AVk+1 = [AVk +𝜋k(1−rk)−ek −(1+jk)−1ck𝜂k](1+ik) (13.2) which shows directly the insurance costs being deducted at the start of the period. Clearly when j k = ik, the interest factors cancel and we just get back the same formula as before. VARIABLE ANNUITIES 203 Notice however that in general we still have the same formula as before if we adjust theck. We change this to a new cost of insurance rate given by c′ k = ck 1+ik 1+jk . The following illustrates. Example 13.2 RedoExample13.1andcalculatethecostofinsurance,undertheassumption of a 3% discount rate for the cost of insurance. Solution. We can do all calculations by changingc k from 0.03 to c′ k = 0.03(1.08)1∕12∕ (1.03)1∕12 = 0.03012. In the type B policy, this will change the value of AV 53 to 52039.72 and COI52 to (1.00643)−1(0.03012)100 000= 2992.76. In the type A policy, this will change the value of AV53 to 52039∕(1−0.03012) = 53655.83 and COI52 to (1.00643)−1(0.00301)(100000 −53655.83) = 1386.96. Another modification may be necessary because of the corridor requirement. In a close situation, we must test whether an account value is sufficiently low to meet this restriction. Let cor k denote the minimum allowable ratio ofbk∕AVk+1.I fA Vk+1cork > bk, we must redo the calculations. The death benefit will now be corkFk+1 and𝜂k = (cork −1)AVk+1. Example 13.3 For the policies of Example 13.1, find AV53 and COI52 assuming that cor53 = 1.9. Solution. For the type B policy the corridor requirement is clearly satisfied and the answer will remain as calculated above. For the type A policy, we check the original answer to see that 53 661.57(1.9)=101 957 which is above 100 000, so we must recalculate: AV53 = (50000 +5000(0.96)−100)(1.00643)−0.03(0.9AV53), and solving we get AV53 = 53604.40. The death benefit will increase to 101 848.36 COI52 = 1.00643−10.03(0.9×53604.62) = 1438.07. 13.2 Variable annuities A similar attempt to provide additional flexibility has become popular for certain deferred annuitycontracts,whichhavecometobeknownas variableannuities.Theycorrespondtothe contractdescribedinExample5.8totheextentthatthereisnosurvivorshipaccumulationprior to the annuity payments. In fact they operate in the same basic manner as we have described aboveforUniversalLife,exceptthatnodeductionsaremadeforthecostofinsurance.Infact some purchasers use these as a vehicle to accumulate money in a high yielding investment accountwithoutanyintentionofconvertingthefundstoanannuity.Theannuityaspectenters intothecontracthowever,sinceitcommontoincludeprovisionsprovidingfortheconversion of the funds into annuities at guaranteed rates of interest and mortality. 204 SPECIALIZED TOPICS Asintheinsurancecase, anindividual’s account maybeinvestedinparticular assetsorit may be equity-indexed, with the credited interest tied to some reference portfolio. Similarly to Universal Life, their are several types of possible guarantees that have been designed to reduce the losses under bad investment experience. A common provision is a guaranteed minimum death benefit,whereby the account holder is promised a certain minimumreturnupondeath,regardlessoftheaccountvalue.Thedeathbenefitamountcould be the amount originally invested, or it could be that amount together with a certain fixed accumulation rate. A similar type of provision is theguaranteed minimum accumulation benefit whereby the end of a specified period, the account holder is promised a minimum account value, regardless of theinvestment experience. Stillanother option istheguaranteed minimumwithdrawalbenefit. Under this option the account holder is allowed to withdraw a minimumamount(oraminimumpercentage)fromtheiraccount,eachyear,untiltheoriginal investedvalueisreturned.Evaluationofthesebenefitsrequiresaknowledgeofoptionpricing. An introduction to this topic is found in Chapter 20. 13.3 Pension plans Pensionplansaresetupbycompaniestoprovideretirementincometoagroupofemployees. This is a vast subject and we confine ourself here to providing a survey of principal features and definitions. Pension plans can be classified into two main categories,defined benefit plans,abbreviatedasDB,and definedcontribution plans,abbreviatedasDC,asintroducedin Section 4.6.2. We now provide more detail. 13.3.1 DB plans The usual type of DB plan provides that an employee will receive a life annuity, beginning at a specified normal retirement age, with a periodic paymentK. Rules for calculatingK are specified at the outset. Itwillnormally depend onthe employee’s salaryand years ofservice. The type of annuity is often a whole life annuity, although there may be options to elect a guaranteed period, or a joint-life annuity with another individual, such a spouse. The income of course will be adjusted according to the nature of the annuity selected. A typical formula is thatK will be equal tor times the average of the employee’s last h years of salary times the number of years of service, for some specifiedr and h.A sa n example,supposethatthenormalretirementageis65, r=0.02,h =3andthat Kisanannual payment. Consider an employee who is hired at age 30, retires at age 65, and whose last 3 years of salary were 89 000, 93 000, 100 000. The employee would then receive an annual pension of 0.02×35×94000 = 65800. A measure which is used to compare various plans is known as thereplacement ratio and it is simply the ratio of the pension to the final year’s salary. In the above example the replacement ratio would be 65.8%. The quantityhwill often be between 3 and 5 years. A possible variation involves plans, known ascareer average earnings planwhere h is not fixed, but is equal to the complete number of years of service for each employee. To evaluate benefits on such a plan the actuary will use an investment discount functionv andamultiple-decrementtablewhichwilltypicallyshowdecrementsofdisability,withdrawal PENSION PLANS 205 fromservice,deathandretirement.Anadditionalnecessarytoolisaso-called salaryscaleS x defined as follows. For some minimal agex0,w eh a v eSx0 = 1 and then fory> x0. Sy = the expected ratio of salary at ageyto salary at agex0. The scale is then used to estimate future salaries. For example, if an employee is earning an annual salary ofJ at agex, then an estimate of their annual salary as agex+h will be JS x+h∕Sx. Pension plans must specify what benefits if any will be paid to those who leave the employee group for reasons other than retirement at the normal retirement age. Such reasons include disability, death, withdrawal from company employment and retirement at an age otherthanthenormalone.Wefocusnowonearlyretirement.Plansmayspecifythataperson mayreceivepensionbenefitsiftheyretireearly,withsomeminimumcriteriaspecified,which can depend on age and duration of employment. Early retirement means of course that the pensionispaidforalongerperiod,andalsothatcontributionsintotheplanwillnotbemadefor the remaining time to the normal date, so typically the pension income will be appropriately reduced. Rather than calculate the applicable amount of reduction in each case, it is usual to work out approximate figures, known asactuarial reduction factorsand these are specified as part of the provisions of the plan. In the following simplified example, provision is made for early retirement up to 2 years before the normal date, but not for other forms of decrement. Example 13.4 ConsideraDBplanwhichprovidesayearlypensionof2%ofthefinal3-year average salary, times the number of years of service, for retirement at ages 63–65, but with a reduction of 5% per year should retirement occur before 65. An employee now 45 was hired atage35.His/herpresentsalaryis70000.Thesalaryscaleisgivenby S x = 1.03x−30, x≥ 30. (This simply means that salaries are expected to increase by 3% per year.) An employee may retire at any age from 63 to 65, with a reduction in pension income of 5% per year. Find a formula for the actuarial present value of the pension benefits. Solution. For a person retiring at age 63, the estimated 3-year final average salary is given by 70000[1.03 15 +1.0316 +1.0317]∕3 = 112362 so that the estimated annual pension income, accounting for the actuarial reduction factor of 10%, is given by 0.90×0.02×28×112262 = 56630. For person retiring at age 64 a similar calculation, now using a 5% reduction results in annual pension income of 63 769, and for a person retiring at age 65, the figure is 71 523. The actuarial present value is then given by the following (the subscriptr denoting the retirement decrement). 56630 18p(𝜏) 45q(r) 63 ̈a63 +6376919p(𝜏) 45q(r) 64 ̈a64 +7152320p(𝜏) 45q(r) 65 ̈a65. Our solution here incorporates several simplifying assumptions. We have assumed that thepensionwouldbepaidannually,insteadofthemoreusualmonthlyarrangement.Wehave also assumed that all employees are hired and retired on their birthdays, and that each salary increase occurs on birthdays. It is not too difficult to incorporate more realistic assumptions, 206 SPECIALIZED TOPICS using the fractional duration techniques in Chapter 7. One common provision is to assume hiring, retirements, and salary increases at the middle of each year, that is at agesx+1∕2, wherexis an integer. Fundstoprovidetheincomecouldbeprovidedsolelybytheemployer,ormorecommonly sharedbetweentheemployeesandtheemployeronsomespecifiedbasis.Ausualprovisionis thattheemployeeswillcontributeeachperiodaamountof ctimestheirsalaryforthatperiod.A typicalfiguremightbe c= 0.05.Theemployerwillthencontributeadditionalamountswhich are estimated to be sufficient to provide the promised benefits. There are various methods for doing so. The subject of funding DB pensions is complex, and will not be discussed further. 13.3.2 DC plans AtonetimetheDBarrangementwasthemostcommonone.Howeverperiodsoflowinterest rate earnings and improving mortality meant that many employers needed to put in more money then originally estimated in order to ensure the promised level of benefits. In some casesthisbecameprohibitive,resultinginfirmsswitchingtotheDCmode.Insuchapension, theemployeescontributeacertainpercentageoftheirsalary,theemployeraddsanadditional percentage, and the funds are invested and accumulated as an individual account for each participant until retirement. At the time of retirement the total accumulated contributions made on the employee’s behalf are used to purchase an annuity at the then prevailing interest andmortalityrates.AswiththeDBplanthereisoftenatargetedgoal,butthisisnotguaranteed. Iftheinvestmentexperienceisunfavourable,ormortalityhasimprovedsothatthecostoflife annuities go up, the pension income may be short of the projected amount. Of course, things can go the other way in a DC plan. Very favourable investment returns can result in higher pensions than expected. Funding arrangements are usually much easier for the DC plan than in the DB case. A common practice is to deposit into the account a certain fractionc of a employee’s salary eachyear,whichcouldbesharedinsomewaybetweentheemployeeandemployer.Onedoes not have to worry as much about benefits for withdrawal or early retirement. Withdrawing employees can be given the amount in their account, either in cash or as a deferred annuity. For those retiring early the amount can be used to buy an annuity starting on the normal retirement date. Inthefollowingexamples,wecontinuetomakethesimplifyingassumptionsnotedabove, Example 13.5 For an employee hired at age 35 at a salary of 50 000, it is estimated that an amount of 500000 is needed at at the normal retirement age of 65 to buy an appropriate pension. For exit from the plan before age 65 the accumulated amount of contributions with interest is returned to the employee. Find a formula to calculate the contribution ratec, assumed to be made at the beginning of each year. Solution. TheideaissimilartoExample5.8.Wesolvethefollowingequationtodetermine c. 500000 = 50000 cVal 30(g;v) (13.3) where gk = S35+k∕S35, k= 0,1, …,34. (13.4) EXERCISES 207 There are many variations in practice. Many plans have a so-calledvestingrequirement, whichmeansthatanemployeedoesnotreceivebacktheemployer’scontributionsunlessthey remain in the plan for a minimum period. Example 13.6 Suppose that in the situation described above, the contributions are split, withtheemployeepaying40%andtheemployertheremaining60%.Forparticipantsexiting the plan after 5 years, the total of all contributions with interest is returned at the end of the year of leaving. For those exiting in the first 5 years only the employee’s own contributions with interest are returned. Find an equation to determine the contribution ratec. Solution. We use a variation of formula (5.4). c= 500000v(35) 35p(𝜏) 30 ̈a(𝜏) 35(g)−A(𝜏) 35 (j) . whereg is as in (13.4) and jk = {0.4∑k−1 i=0 v(k,i)S35+i∕S35 if 0≤ k< 5, ∑k−1 i=0 v(k,i)S35+i∕S35 if 5≤ k< 25. Exercises 13.1 A universal life policy provides a death benefit at the end of the period of death equal totheaccountvalue,butsubjecttoaminimumof30000.Monthlypremiumsare2000. There is a monthly expense charge of 50 and a further charge of 3% of each premium. Theannualcreditedinterestrateis6%.Theaccountvalueattheendofthe30thmonth is 27 318. The monthly cost of insurance during the third year of the policy is 15 per 1000. Find AV 31 and COI30. 13.2 A universal life policy provides a death benefit at the end of the period of death of 100 000. The account value at time 20 is 2500. For the time period from time 20 to time 21, the credited interest rate is 0.006, the expense charge is 5% of any premium paid plus 100, and the cost of insurance rate is 30 per thousand. (a) Suppose that at time 20, the policyholder makes the minimum premium payment of 500, which according to the guarantee in the contract ensures that the cost of insurance will be paid regardless of the amount in the account. Find the account value at time 21. (b) If the contract does not carry the secondary guarantee of part (a), what is the minimum premium that the policyholder will have to pay at time 20 to ensure that the cost of insurance can be paid. 13.3 At a certain timek, two universal life policies have exactly the same account value, death benefit, premiums, expenses and cost of insurance rates. In both cases the cost 208 SPECIALIZED TOPICS of insurance discount rate is the same as the credited interest rate. One policy is type A and the other is type B. Show that the difference between the type A and type B account values at the end of the year is the same as the difference between the type B and type A COI for that period, accumulated with interest to the end of the year. 13.4 A universal lifepolicy provides a death benefit equal to the account value plus 50 000. Monthly premiums are 3000. There is a monthly expense charge of 100 and a further charge of 4% of each premium. The annual credited interest rate is 8%. The account value at the end of the 73rd month is 49 200. The monthly cost of insurance for the followingmonthis20per1000andiscomputedatanannualinterestrateof4%.There is a corridor requirement which specifies that the death benefit must be at least twice the account value. Find AV 74 and COI73. 13.5 Redo Example 13.4, only assume salary increases of 4% per year, and an actuarial reduction factor of 3% per year. 13.6 RedoExample13.6,onlyassumingthatanemployeewhodiesduringthefirst5years, receivesattheendoftheyearofdeath,boththeirownandtheemployer’scontribution accumulated with interest. 13.7 An employee starts employment in a firm at age 35 and is offered a choice of either a DC or a DB plan. In the DC plan, contributions of 15% of salary are accumulated withinterestat4%untilage65andthenusedtobuyalifeannuity.UndertheDBplan, the employee is given an annual amount, beginning at age 65 of 1.8% of their 3 years average salary times the number of years of service. If the cost of a 1-unit life annuity atage65is10,andsalariesincreaseby4%peryear,findtheratiooftheannualincome under the DC plan to that of the DB plan, for an employee who remains in the plan until age 65. 13.8 A DB plan provides for an annual pension of a certain percentage of the final 5 years average salary times the number of years of service upon retirement. Retirement is based on the80 factorwhich means that an employee can retire at any time that the sumofageplusyearsofserviceisgreaterthanorequalto80.Anemployeeage55who beganworkatage25,iscomparingtheirincomeforretirementnow,tothatwhichthey would receive if they stay for another year. They can expect a 5% increase in salary for the following year, which is the same that they received in each of the last 4 years. Findtheratioofthepensionincomeforretirementin1yeartothatforretirementnow. Part II THE STOCHASTIC LIFE CONTINGENCIES MODEL