Instruction: what is the most likely time of failure?

Response: 14 Survival distributions and failure times 14.1 Introduction to survival distributions Our goal in this part of the book is to introduce a stochastic model for mortality to replace the deterministic model used in Part I. This will not only provide us with a more realistic description of human mortality, but it will also have more general applications. The basic information a prospective issuer of an insurance or annuity contract wants to know is how long the life in question will live. The insurer obviously cannot hope to answer this question exactly, since the actual future lifetime lived is random. Some people age 50, for example, will live another 40 years or more, while others will die very soon. In the deterministicmodel,wecircumventedthisissuebyassumingthatwhilewecouldnotidentify how long a particular individual would live, we could identify how many individuals of a given age would live to some other age. Clearly, however, the number of such individuals is also random. In the stochastic model we will face this randomness directly. Thisandsubsequentchapterswillrequireamoreadvancedknowledgeofprobabilitythan we have assumed so far. We follow the notation and terminology of Appendix A. For the present chapter, see in particular Sections A.4–A.8 and note thatP will denote probability. We do not need to confine ourselves to looking at the time of death of an individual. Suppose we are interested in some event that will occur once and only once at some random future time. We will refer to this event as ‘failure’. The random variableT, the time of occurrence of such an event, is known as afailure time, and its distribution is often referred to as asurvival distribution. At any time before ‘failure’ we will say we are in a state of ‘survival’. Our motivating example is the case where the event in question is the death of (x). In this case we will denoteT by T(x). 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 212 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES There are, however, many other examples of interest. Suppose a manufacturer sells a product with a guarantee that it will be replaced it if fails before a certain time. In order to assess the cost of the guarantee the manufacturer wants to know the distribution of the product’s failure times. We see that a guarantee can be viewed as type of insurance policy. WehavealreadyencounteredmoregeneralfailuretimesinChapter10.Joint-lifeinsurance and annuities were seen to be the same type of contracts as those for single lives except that the failure time was defined to be the time of first death. Similarly, for last-survivor annuities orsecond-deathinsurances,thefailuretimewasthetimeofseconddeath.Multiple-decrement theory provided many more examples of failure times. 14.2 The discrete case Consider the case where failure can occur only at integer times 1,2,3,…,s oT is a discrete randomvariablewithpositiveintegersasvalues.ReferinparticulartoSectionA.4.Inplaceof thecumulativedistributionfunction F,itisoftenmoreconvenienttousethe survival function s defined by s(k) = 1−F(k) = P(T > k). This gives the probability that failure has not yet occurred by timek, or in other words that we are still in a state of survival at timek.I ff is the probability function ofT, it is clear from (A.5) that f (k) = s(k −1)−s(k), s(k) = ∞∑ i=k+1 f (i) = 1− k∑ i=1 f (i). (14.1) There is another important method of describing the distribution ofT. Definition 14.1 The hazard function ofT at timek, denoted by𝜆(k) is the condition proba- bility of failure at timek given survival up to timek −1. That is, 𝜆(k) = f (k) s(k −1). This of course is defined only for those integersk such thats(k −1) > 0. Onces(k −1) equals0,thereisnopossibilityofsurvivaluptothatpoint.(Thehazardfunctionisalsoknown as the hazardrate or intensity function orfailure function.) Readersshouldmakesurethattheyunderstandthedifferencebetween f (k)and 𝜆(k).Both quantities give the probability of failure at timek, but from different perspectives. If, at time 0,weareaskedtoassessthelikelihoodthatfailurewilloccurattime k,theanswerisjust f (k). As time goes on, our assessment of this likelihood must change. For example, if failure takes place before timek, then we know that the probability of failure at timek is zero. Suppose that at timek −1, failure has not yet occurred, and we ask ourselves the same question. The THE CONTINUOUS CASE 213 answer now is𝜆(k). The difference between the two functions is further clarified by writing the definition of𝜆in the form f (k) = s(k −1)𝜆(k), (14.2) expressing the fact that we view failure at timek as arising from two events. First, there must be survival up to timek −1, and then, given this survival, failure must occur at timek. In deriving an appropriate survival distribution, it is often easier to model the hazard rate rather than s or f .G i v e n𝜆, we can then easily determine the other functions as follows. Equating the two different expressions forf (k) given in (14.1) and (14.2), s(k) = s(k −1)[1−𝜆(k)]. (14.3) Beginning withs(0) = P(T > 0) = 1, we know thats(1) = 1−𝜆(1),s(2) = s(1)[1−𝜆(2)] = [1−𝜆(1)][1−𝜆(2)], and proceeding inductively, s(k) = [1−𝜆(1)][1−𝜆(2)]⋯[1−𝜆(k)]. (14.4) Example 14.1 A bag contains 3 green balls and 1 red ball. A ball is drawn at random. If green, it is replaced and the draw is repeated. Failure occurs when a red ball is drawn. If this is on thekth draw we say that failure occurs at timek.F i n ds(k),f (k) and𝜆(k). Solution. From the conditions of the problem we can immediately deduce that 𝜆(k) = 1 4, k = 1,2, …. From (14.4), s(k) = (3 4 )k , k = 1,2, …. and from (14.2), f (k) = 3k−1 4k , k = 1,2, …. We have the well-known geometric distribution. See Section A.11.3. As a check we can also compute f (k) from the first formula in (14.1), which of course yields the same answer. 14.3 The continuous case In most applications, the time of failure is not restricted to the integers but can be arbitrary. We model this by assuming thatT is a continuous random variable with values in [0,∞). 214 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES 14.3.1 The basic functions We define thesurvival function sas in the discrete case, s(t) = 1−F(t) = P(T > t), the probability of survival up to timet. It is related to the probability density function (p.d.f) by f (t) =− s′(t), s(t) = ∫ ∞ t f (r)dr = 1−∫ t 0 f (r)dr. (14.5) Definition 14.2 The hazard function of a continuous failure timeT at timet, denoted by 𝜇(t), is the continuous density function for failure at timet given survival up to that point. It is given by 𝜇(t) = f (t) s(t) for allt such thats(t) > 0. For small Δt,𝜇(t)Δt approximates the probability thatT takes a value in the interval [t,t +Δt] given survival up timet. Analogously to (14.2), we have the expression f (t) = s(t)𝜇(t). (14.6) To determine the other quantities from𝜇 we note from (14.5) and (14.6) that 𝜇(t) = −s′(t) s(t) =− d dt[logs(t)]. Following the proof of Proposition 8.1, we deduce that s(t) = e−∫ t 0 𝜇(r)dr. (14.7)‡ Formula (14.7) seems quite different from its discrete counterpart (14.4), but if you look at it in the right way, it really is a natural continuous version. Recall that e−∑ai = ∏e−ai . So if we think of an integral as a type of generalized sum, then e to the integral is a type of generalized product of terms of the form e −𝜇, which for ‘small’ values of𝜇 are close to 1−𝜇. 14.3.2 Properties of 𝝁 Weknowthatthedensityfunction f mustbenonnegativeandsatisfy ∫ ∞ 0 f (t)dt = 1,andthatthe distributionfunction F mustbenondecreasingandsatisfy F(0) = 0,limt→∞ F(t) = 1.Thelat- terimpliesthatthesurvivalfunction s isnonincreasingandsatisfies s(0) = 1,limt→∞ s(t) = 0. EXAMPLES 215 What are the corresponding properties for the hazard function? The hazard function𝜇 must be nonnegative on its domain and in addition satisfy ∫ ∞ 0 𝜇(t)dt =∞ . To see this, suppose that the above integral had a finite valuea. From (14.7) we would deduce thatlimt→∞ s(t) = e−a which is not equal to 0, contradicting the fact that failure must occur at some time. In other words, if the integral is finite, the hazard function is not large enough to guarantee failure and there would be some chance of surviving forever. Of course, we may want to model situations where there is a chance that failure will never occur, and in this case we would want the above integral to be finite. 14.3.3 Modes The general shape of the density function ofT can often be inferred by looking at𝜇.I n particular,wemaybeinterestedin modes,whicharepointswherethedensityfunctionassumes a local maximum. We assume thatf is differentiable, and note from (14.5) and (14.6) that f ′ = (s𝜇)′ = s′𝜇 +s𝜇′ =− f 𝜇 +s𝜇′ =− s𝜇𝜇 +s𝜇′ = s(𝜇′ −𝜇2). (14.8) Weseethenthat f isincreasingordecreasingaccordingas 𝜇′ isgreaterthanorlessthan 𝜇2. Points at which modes can occur are restricted toor those values oft for which𝜇′(t) = 𝜇(t)2, or possibly to endpoints of the domain. We will look at some particular examples later. 14.4 Examples In this section we survey some familiar distributions that can be used to model failure times in certain situations. The family of exponential distributions (see Section A.11.6) can be easily described as those with aconstant hazard function. If𝜇(t) = 𝜇 for all t, then from (14.6) and (14.7), s(t) = e −𝜇t.Suchadistributionissuitableformodellingsituationswherethechanceoffailure in the next instant remains constant regardless of the time. This is shown by the constancy of thehazardfunctionaswellasthepropertygiveninA.57.Itisnotsuitableformodellinghuman mortality,wheretheagingprocessmeansthatwewouldexpectthehazardfunctiontoincrease with time. See Section 8.10.1 where we discussed the same point in the deterministic model. Another overly simplistic approach is to take a continuous uniform distribution (see Sec- tionA.11.4).Formostapplications,thisisnotrealistic,butithasbeenusedasaroughapprox- imation to model human mortality. This is simply the stochastic version of Demoivre’s law, introducedpreviouslyinChapter8.Notethatitdoesgiveanincreasinghazardfunctionsince 𝜇(t) = f (t) s(t) = 1 N −t,0 ≤ t < N. The first serious attempt to capture some of the relevant features for mortality was the Gompertz distribution given by 𝜇(t) = Bct, for some parameters B > 0 andc > 1. 216 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES This was introduced by Benjamin Gompertz in 1825. His idea was that the hazard function for human mortality should increase with age at a rate proportional to itself, that is, at an exponential rate. It has been used extensively for modelling mortality in both humans and other species. For human mortality, a good fit can usually be found over the middle span of ages by choosing the parameters appropriately. One feature of this distribution is that𝜇 ′(t)−𝜇(t)2 = 𝜇(t)[logc −𝜇(t)] is positive, zero, ornegative,preciselywhen 𝜇(t)islessthan,equalto,orgreaterthan logc.Then,(14.8)shows thatif 𝜇(0) ≥ logc,thedensityfunctionisdecreasing,whileif 𝜇(0) < logc,thereisaunique mode at the pointt where 𝜇(t) =l o gc. The density function increases up to this point and then decreases. This does indeed seem to fit the pattern of human mortality over the middle range of ages. Starting about age 30, the probability of dying at agex will increase withx up to a certain point. There is clearly more chance of dying at age 70 than at age 30. However, atsufficientlyhighagesthechanceofdyingstartstodecreasewithage.Forexample,thereis a very small chance of dying at age 110, for the simple reason that very few people will live that long in the first place. It may be enlightening to look again at (14.6), which expressesf as the product of two functions, one of which is decreasing, and one increasing. The Gompertz distribution does not fit well with observed mortality at ages below 30 or above 70 or so, for the reasons explained at the beginning of Section 8.10.1. In 1860, a modification was advocated by Makeham. He suggested that a constant be added to the hazard function to cover causes of death that were age-independent, such as accidents. The Makehamdistribution,oftenreferredtoastheGompertz–Makehamdistribution,isthengiven by 𝜇(t) = A +Bc t,f o rs o m eA > 0. The extra parameter allows for more flexibility in shape (see Exercise 14.12) but still does not capture observed mortality behaviour at very young or very old ages. 14.5 Shifted distributions Suppose we are given a failure timeT. (For convenience we will deal with the continuous case,buttheconclusionsforthediscretecasearesimilar.)Asremarkedabove,havingreached apoint u atwhichfailurehasnotyetoccurred,wemustalterourassessmentofthelikelihood of failure at different times. To formalize this, we define a new random variable. The random variableT ◦u is equal to the time until failure occurs, as measured from time u, given survival up to timeu. Therefore, it takes a value ofT −u when T takes a value greaterthan u,withprobabilitiesconditionedontheeventthat T isgreaterthan u.Thesurvival function, density function, and hazard rate function, respectively, forT ◦u are expressed in terms of the corresponding functions forT as follows. (Since we are dealing with more than one distribution here, we will insert appropriate subscripts ons,f and𝜇.) s T ◦u(t) = sT (u +t) sT (u) . (14.9) fT ◦u(t) =− d dtsT ◦u(t) = fT (u +t) sT (u) . (14.10) 𝜇T ◦u(t) = fT ◦u(t) sT ◦u(t) = 𝜇T (u +t). (14.11) THE STANDARD APPROXIMATION 217 Formula (14.10) provides a nice way to visualize this concept. To get the graph of the density function of the shifted random variable, we take the graph of the original density function, ignore everything to the left ofu, and scale all values upward to get the total area equal to 1. Forsomedistributions,theshiftedrandomvariablesareofthesamefamilyastheoriginal. This makes these distributions convenient for modelling purposes. Example 14.2 Describe the random variablesT ◦u when (a)T is an exponential random variable with constant hazard𝜇,( b )T is uniform on the interval [0,N], (c)T is Gompertz– Makeham with parametersA, B andc. Solution. In all cases it is convenient to use the hazard function and (14.11). (a) 𝜇 T ◦u(t) = 𝜇(u +t) = 𝜇, showing thatT ◦u has the same distribution asT. This indi- cates the so-calledmemory-less property of exponential random variables. The time until the event in question occurs will not be affected by how long you have already waited. (b) 𝜇 T ◦u(t) = 𝜇(u +t) = 1∕(N −u −t), showing that T ◦u is uniform on the interval [0,N −u]. (c) 𝜇T ◦u(t) = A +Bcuct, showing that T ◦u is also Gompertz–Makeham with the B parameter changed toBcu. 14.6 The standard approximation Associatedwitheverycontinuousfailuretime T isadiscretefailuretime ̃T.Failureaccording to ̃T will occur at the integer timek, if failure underT has occurred in the time interval (k −1,k]. To be precise, ̃T = [T]+1, where [⋅] denotes the greatest integer function. Letf ands denotethedensityfunctionandsurvivalfunction,respectively,of T.Let ̃f and ̃s denote the corresponding functions for̃T. Then, for all positive integersk, ̃f (k) = ∫ 1 0 f (k −1+t)dt = s(k −1)−s(k), (14.12) ̃s(k) = s(k). (14.13) Suppose we want to deduce information aboutT by observing when failure occurs. It could be difficult or impossible to observe continuously and we may be only able to view the situation at certain discrete times, say 1, 2, 3, etc. If we observe at timek, and see that failure has occurred, we know only that failure occurred between timek −1 andk. In other words, we are observing values of̃T. We would like therefore to infer the distribution ofT from that 218 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES of ̃T.Clearly,wecannotdothisexactlyandmustmakesometypeofapproximation.Asimple method, consistent with (14.12), is just to assume that for all nonnegative integersk, and any t in the interval (0,1), f (k +t) = ̃f (k +1). (14.14) We will refer to this asthe standard approximation.In fact, we have already used this approximationinourdeterministicmodel,intheformoftheUDDassumptionfortherandom variable T(x), as indicated by the equivalent formulation in terms of survival functions, namely, s(k +t) = (1−t)s(k)+ts(k +1), (14.15) for an integerk and 0< t < 1. To see this equivalence, note that, given (14.15), we obtain (14.14) by differentiating with respect tot. Conversely, given (14.14), we have s(k +t) = s(k)−∫ t 0 f (k +r)dr = s(k)−∫ t 0 ̃f (k +1)dr = s(k)−t̃f (k +1) = s(k)−t[s(k)−s(k +1)]. For the purpose of calculating moments, it is useful to introduce the random variable R = ̃T −T, thedurationfromthetimeoffailureuntiltheendoftheyearoffailure.Givenavalue r inthe interval(0,1),considertheprobabilitythat R ≤ r given that̃T =k.Thisistheprobabilitythat T takesavaluebetween k −r andk,giventhat ̃T = k.Usingthestandardapproximation,this probability is given by 1 ̃f (k) ∫ k k−r f (t)dt = 1 ̃f (k) ∫ k k−r ̃f (k)dt = r. This shows that underR is independent of̃T and has a uniform distribution on the interval (0,1). Using standard results about the uniform distribution, we can calculate that under the standard approximation E(T) = E(̃T)−E(R) = E(̃T)− 1 2, (14.16) Var(T) = Var(̃T)+Var(R) = Var(̃T)+ 1 12, (14.17) whereE denotes expectation and Var denotes variance. THE STOCHASTIC LIFE TABLE 219 14.7 The stochastic life table Thegoalinthissectionistoestablishalinkbetweenthestochasticmodelandthedeterministic model for mortality. In this and subsequent chapters we will use this link to show that the questions that we answered in the deterministic model can be answered in the stochastic model in thevery sameway. In addition, we will be able to answer more questions. The tool for doing so will be to return to the fundamental concept of a life table but look at it in a different way. As we did before, we start out with an arbitrary number𝓁 0 of newly born lives. We now letℒx be the number of these still alive at agex, and𝒟x be the number of these who will die between the age ofx and x +1. These quantities look like the𝓁x and dx of Chapter 3, but we now recognize that these are not numbers butrandom variables.We obtain the life table by letting 𝓁x = E(ℒx), dx = E(𝒟x). Since it is clear thatℒx+1 = ℒx −𝒟x, we have the same formula (3.1) as we had before. In other words, in our stochastic model we still can introduce a life table, but we now view the numbersasexpectedvaluesratherthanasanexactaccountofthenumberswhowillliveordie. Consider now the random variableT(x), the time to the death of (x). We will write s x(t),fx(t),𝜇x(t) for the survival function, density function, and hazard function respectively of T(x). Forage0,thereissomewhatadifferentstandardnotation,whichcancausesomeconfusion if care is not taken. The random variableT(0) is traditionally denoted byX, and the variable denoted byx rather thant, sinceT(0) is really theage at death. It is common to suppress the 0 and just writes(x), f (x) and𝜇(x)f o rs0(x), f0(x) and𝜇0(x), respectively. A key observation is that 𝓁x = 𝓁0s(x). (14.18) The argument is the same as showing that if you flipN coins, each with a probabilityp of comingupheads,youcanexpecttoget Np heads.Inthiscase,ifyoutake 𝓁0 people,eachwith a probabilitys(x) of surviving to age (x), then you can expect to get𝓁0s(x) such survivors. We are now ready for the main purpose of this section, which is to establish a correspon- dence between the survival, hazard, and density functions ofT(x) and the life table functions for the stochastic life table. At first, we assume aggregate mortality, as in Chapter 3, where we used a single-life table for the future mortality of all agesx, by just starting at agex.T h i s corresponds stochastically to the statement T(x) has the distribution ofT(0)◦x, which simply says that the future lifetime of (x) is the future lifetime after timex of a person whowasage0,giventhattheyhavelivedtoage x.Therearethereforenoeffectsofselection. Suppose also that we calculate the quantities tpx,tqx,𝜇(x) starting from the life table, exactly as we did in the deterministic model. We then have from (14.10) and (14.23) that sx(t) = s(x +t) s(x) = 𝓁x+t 𝓁x = tpx. (14.19) 220 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES This is certainly reasonable. Although looked at from different points of view, both of these are probabilities that a person agex will survivet years. It follows immediately that F x(t) = 1−tpx = tqx, (14.20) and that the hazard rate function ofT(x)i sg i v e nb y 𝜇x(t) =− d dt logtpx = 𝜇(x +t), (14.21) where the last quantity is the force of mortality of Chapter 8. (Note that in the case of hazard rates,thenotationinthedeterministicmodelwaschosenfromtheoutsettocorrespondtothat in the stochastic setting.) Finally, f x(t) = tpx𝜇x(t), (14.22) the familiar expression that we used for insurance premiums in Chapter 8. IfweassumethestrictlyselectmortalityofChapter9,thenwecannotassumethat T(x)is distributed asX ◦x. The above correspondences would be modified as follows: sx(t) = tp[x] and fx(t) = tp[x]𝜇x(t). It is also instructive to compute the distribution of multiple-life failure times in terms of the quantities introduced in the discrete model. For example, considerT(xy), the time of failure of the joint-life status. For this random variable, the survival function is justtpxy,t h e hazard function is𝜇xy(t), and the density function istpxy𝜇xy(t). 14.8 Life expectancy in the stochastic model Let ̃T(x) be the discrete random variable associated withT(x) as defined in Section 14.7. It will have survival, probability and hazard functions given by ̃sx(k) = kpx, (14.23) ̃fx(k) = k−1px − kpx = k−1px qx+k−1, (14.24) 𝜆x(k) = qx+k−1. (14.25) Traditional actuarial terminology defines a discrete random variableK(x) known as the curtate future lifetime,whichmeasuresthewholenumberoffutureyearstobelivedby( x).It is clear that K(x) = ̃T(x)−1. STOCHASTIC INTEREST RATES 221 From (14.24), (3.8), (A.12), and the fact thatsx(0) = 1, we can write ex = 𝜔−x−1∑ k=1 sx(k) = E[̃T(x)]−sx(0) = E[K(x)]. Therefore life expectancy is indeed an expectation in the sense of probability theory. The formal definition ofcomplete life expectancyin the stochastic model is ◦ex = E[T(x)]. From(A.15)weseethatthisagreeswithourChapter8definition.Moreover,usingthestandard approximation and (14.16), we verify the fact that E[T(x)] = e x + 1 2. We can already see one advantage of the stochastic model. As well as computing the expected future lifetime, we can compute the variance of future lifetime. Example14.3 Given𝓁 90 = 100, 𝓁91 = 90, 𝓁92 = 70, 𝓁93 = 40, 𝓁94 = 10, 𝓁95 = 0,com- pute the expectation and variance ofT(90), assuming UDD. Solution. Using either (A.7) or (A.16) we find thatE[̃T(90)] = 3.1, E[̃T(90)]2 = 10.9, so that Var[̃T(90)] = 1.29. Then, from (14.16) and (14.22), ◦e90 = 2.6, Var[T (90)] = 1.373. 14.9 Stochastic interest rates Stochastic interest is a large topic which we will only allude to briefly. As well as the life table, the other main ingredient of the actuaries tool-kit, the discount function, should be treated stochastically. For the most part, the method of doing so in the pricing and valuation of insurance and annuity contracts has been by a simulation technique. Various scenarios of possible future interest rates are selected, together with some weighting as to how likelihood their occurrence is. Premiums or reserves are computed using each scenario, exactly as we havedescribed,andthetotalityofresultsareanalyzed.Inthisway,onecanarriveatestimates of the expected value of the quantities, or in some cases, particularly for reserve purposes, an idea of the worst case scenarios. Amoresophisticatedapproachistomodel v(s,t)asarandomvariablesratherthandefinite numbers,aswementionedinChapter2.SomeaspectsofthisideaaredealtwithinChapter20. Numerous stochastic interest rate models have been proposed. The most popular approach is to express the force of interest𝛿(t) as a stochastic process rather than a function oft. 222 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES Notes and references Standardactuarialterminology,whichisbasedondeterministicthinking,referstoGompertz’s law,ratherthantoaGompertzdistribution.Thisisstatedas 𝜇(x) = Bc x.Inourframeworkthis meansthat T(0)hasaGompertzdistributionand T(x) = T(0)◦x,sothat T(x)hasaGompertz distribution for allx. Similar remarks apply to the Makeham modification. It was common practice at one point to use life tables that satisfied Gompertz’s or Makeham’s law. A major motivation was to simplify joint-life calculations, by reducing joint-life statuses to a single-life or a joint-life status with equal ages. See Exercise 14.10. There is little need for this with modern computing methods. AdditionalmaterialonGompertz,Makehamandmoregeneralmortalitydistributionscan be found in Brillinger (1961), Carri`ere (1994a) and Tennenbein and Vanderhoof (1980). Exercises Type A exercises 14.1 Redo Example 14.1, but now assume that each time a green ball is drawn, it is replaced with a red ball. In this case 14.2 RedoExample14.1,butnowassumethateachtimeagreenballisdrawn,anewred ball is added. 14.3 A failure time T has the hazard function𝜇(t) = (1+t)−2. What is the probability that failure never occurs? 14.4 Suppose that q97 = 0.5, q98 = 0.6, q99 = 0.8, q100 = 1, and that UDD holds. Find (a)E[T(97)] and (b) Var[T(97)]. 14.5 If Demoivre’s law holds with𝜔 = 100, find the variance ofT(60). Type B exercises 14.6 Consider two copying machines. Machine 1 will make 20 000 copies per month. Whennew,itwilllast T months,where T isarandomvariablewithhazardfunction 𝜇T (t) = 1 40−t,0 ≤ t < 40. Machine 2 will make 15 000 copies per month. When new, it will last forS months, whereS is a random variable with hazard function 𝜇S(t) = 2 96−t,0 ≤ t < 96. Aprospective purchaser istryingtodecide between buyinganew machine1,or a 2-year-old machine 2. Which of these two choices will produce, on average, the largest total number of copies during its lifetime? EXERCISES 223 14.7 Suppose that T has an exponential distribution with constant hazard 0.2. (a) What is the probability that̃T = 3? (b) Whatistheerrormadeifweapproximatetheprobabilitythat2 .5 ≤ T ≤ 2.75by using the standard approximation? 14.8 Consider the distribution with density function f (t) = 𝛽2te−𝛽t, t ≥ 0. (This is a gamma distribution with first parameter 2.) (a) Show thats(t) = (1+𝛽t)e −𝛽t. (b) Find the hazard function𝜇(t). (c) Describe precisely the shifted distributionT ◦u.( Hint:I ti sam i x t u r eo ft w o well-known distributions.) 14.9 For two independent lives (x) and (y), (x) is subject to a force of a mortality of 𝜇x(t) = 2∕(10−t) while (y) is subject to the survival functionsy(t) = 1−(t∕10), in both cases for 0≤ t < 10. Find the p.d.f. for the following random variables. (a) T(xy), the time of failure of the first death; (b) T(xy), the time of failure of the second death. 14.10 (The law of uniform seniority) (a) SupposethatGompertz’slawholds(seeSection14.9).Showthatthereisfunction g, defined on the nonnegative reals, such that tpx:x+n = tpx+g(n), for allt > 0. (b) Suppose that Makeham’s law holds. Show that there is a functionh, defined on the nonnegative reals, such that tpx:x+n = tpx+h(n):x+h(n), for allt > 0. 14.11 Show that the sample table of Section 3.7 satisfies Gompertz’s law (that is up to age 119 where a modification was made to get a finite value of𝜔). *14.12 Suppose that T has a Gompertz–Makeham distribution withA > 0. Show thatfT is either (i) decreasing, (ii) increasing then decreasing or (iii) bimodal with two local maximums. Identify the conditions on the parameters that will result in each case. 14.13 For each of the following two families of distributions, decide whether or not the shifted distributions are from the same family, with changed parameters. (a) 𝜇(t) = 𝛼(t +𝜃) −1 for positive𝛼 and𝜃 (Pareto); (b) 𝜇(t) = ktr for positivek andr (Weibull). 15 The stochastic approach to insurance and annuities 15.1 Introduction In this chapter we deal with a perfectly general failure timeT, and develop the stochastic approach for calculating premiums and reserves for insurance and annuity contracts based on T. These will all be calculated as expected values of appropriate random variables. In the particular case whereT = T(x), we will show, using the correspondence established in Chapter 15, that these agree with the results that we obtained in the deterministic model. The advantageofthestochasticapproachisthatwecanaugmenttheseexpectedvalueswithother quantities, such as variances. Throughout the chapter,f, sand 𝜇 will denote the density, survival and hazard functions respectivelyof T.W elet̃Tbetheassociateddiscretefailuretime, ⌊T⌋ +1.Wewilllet ̃f and𝜆 denotetheprobabilityfunctionandhazardfunctionrespectivelyof ̃T.Wedenotethesurvival function of ̃T with the same symbols. This will not cause confusion since the value at any integer is the same in both cases. Insomecasesthevaluesof Tarebounded(e.g.,thevaluesof T(x)areboundedby 𝜔−x), but this is not necessary. The upper limit on integrals or sums will be written as∞ to cover all possibilities. The approach differs somewhat from the deterministic case and it is important to under- stand the distinction. We suppose, as we did before, that we have a fixed deterministic investment discount functionv. We will consider contracts that pay certain benefits provided that failure has not occurred, and/or benefits at the time of failure. We will want to determine present values of these benefits, and these will all be computed with respect to the discount function v. In this model, we do not have the interest and survivorship discount function that we had before. However, the present values that we obtain will not be definite numbers but ratherrandom variables, since they will depend on the unknown value thatT assumes. 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 STOCHASTIC APPROACH TO INSURANCE BENEFITS 225 Premiums and reserves will be calculated asexpectations of these random variables, and it is in calculating these expectations that the probabilities of life and death are taken into account. These expectations are commonly known asactuarial present values, abbreviated as APV. Throughout this chapter and the next, we adopt the following notation, which will ensure that our theory is applicable both to the net premium situation, as discussed in Chapters 4–6 and to the gross premium situation as discussed in Chapter 12. The premium symbol𝜋 k will denote thetotal inflowreceived at timek, which will be the net premium in a model which ignores expenses, or in the expense situation will be the gross- or expense-augmented premium less the expenses paid at timek. Similarly the benefitb k paid for failure at timek will include any costs involved with making the benefit payment. However to keep things simple we confine our attention to a single decrement case where we have only one cause of failure. In particular then, in the context of life insurance we are ignoring withdrawals, implicitly making the assumption discussed in the second remark after Example 12.2. 15.2 The stochastic approach to insurance benefits We deal here with contracts that provide benefits upon failure, and consider both the discrete and continuous cases. 15.2.1 The discrete case We consider a contract with failure benefit vectorbpayingbk−1 at timekfor failure between timek−1andtime k.LetZ denotethepresentvalueofthebenefitswithrespectto v.Then Z is a function of the random variablẽT, since wheñT = k,t h ev a l u eo fZis bk−1v(k). We can therefore write Z = b̃T−1v(̃T). LettingÃT(b;v) denoteE(Z), we have ÃT(b;v) = ∞∑ k=1 bk−1v(k)̃f(k) = ∞∑ k=0 bkv(k+1)̃f(k+1). (15.1)‡ As before, we will often suppress thev and write this asÃT(b). Note that whenT = T(x), formula (15.29) shows that we obtainAx(b) as calculated in (5.1). For many applications we want more information aboutZthan just its expected value; at the very least, we would want to calculate Var(Z). This is easily done, sinceZ2 = b2 ̃T−1v(̃T)2. In other words, to calculate the second moment, we simply square the benefits, and square the discount function. This leads to Var(Z) = A ̃T(b2;v2)−ÃT(b;v)2. (15.2)‡ 226 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES 15.2.2 The continuous case Consider a contract that paysb(t) at the moment of failure, should failure occur at timet. We will letZ denote the present value of the benefits and denote the expected value ofZ by AT(b;v), often shortened toAT(b). WhenT = t,b(t)i sp a i da tt i m et,s o Z = b(T)v(T) and AT(b;v) = ∫ ∞ 0 b(t)v(t)f(t)dt. (15.3)‡ When T = T(x), we see from (15.27) that this is the same quantity as we obtained in (8.18). The variance is computed as we did in the discrete case: Var(Z) = AT(b2;v2)−AT(b;v)2. (15.4)‡ Example 15.1 Suppose that𝜇(t) is a constant 0.04 and the force of interest is a constant 0.07. An insurance contract pays a benefit of e0.03t at timet, should failure occur at that time. Find the expectation and variance ofZ, the present value of the benefits. Solution. This isanexponential failuretime,so f(t) = 0.04e−0.04t.W eh a v eb(t) = e0.03t and v(t) = e−0.07t. Substituting into (15.3), E(Z) = ∫ ∞ 0 0.04e0.03te−0.07te−0.04tdt= 0.04 0.04+0.07−0.03 = 1 2. Nowb(t)2 = e0.06t and v(t)2 = e−0.14t, so calculating as above, E(Z2) = 0.04 0.04+0.14−0.06 = 1 3,V a r ( Z) = 1 3 − 1 4 = 1 12. 15.2.3 Approximation Suppose now that we know only the distribution of̃T, and wish to approximateAT(b). If interest and benefits are constant over each year, we can use the standard approximation to obtain the samei∕𝛿 adjustment as we obtained before forT(x). We will give an alternate stochasticderivationofthisresult.Wefirstassumeconstantinterest,sothat v(t) = v t forsome constant v.L e tbk denote the constant value ofb(t) over the time intervalk to k+1, and let b = (b0,b1,…). Recall the random variableR= ̃T−T introduced in Section 15.7. Then E(Z) = E[b(T)vT] = E[b(T)ṽT−R] = E[b(T)ṽT]E[v−R], THE STOCHASTIC APPROACH TO INSURANCE BENEFITS 227 where we invoke independence for the last equality. When ̃T = j,b(T) = bj−1, so that E[b(T)ṽT]i sj u s tÃT(b). Moreover,v−R = (1+i)R, and sinceRis uniform on [0,1]. E[v−R] = ∫ 1 0 (1+i)udu= i 𝛿, leading toAT(b) = (i∕𝛿)ÃT(b). The general formula, where interest can vary from year to year is then given by AT(b) = ÃT ( i 𝜹 ∗ b ) . (15.5) wherei∕𝜹 is now a vector. 15.2.4 Endowment insurances Inthedeterministicmodel,endowmentinsuranceswereviewedasakindofhybrid,consisting ofbothinsuranceandannuitycomponents.Inourstochasticmodel,wecanviewthesestrictly as insurances. Define the failure time T =m i n{T(x),n}, so that failure occurs at the death of (x), or at timen if earlier. Insurances based onT are precisely n-year endowment insurances. When benefits are paid at the moment of death,T is an example of a random variable that has both a continuous part and a discrete part (see Section A.5). It is continuous over the interval [0,n), on which it has a density functionf x satisfying P(a≤ T ≤ b) = ∫ b a fx(t)dt, provided b< n. Thisdensitywilljustbetherestrictionofthedensityfunctionfor T(x)totheinterval[0, n), and it integrates toFx(n) over this interval. The remaining probability is all concentrated on the pointn, sinceT will take this value wheneverT(x) takes a value ofnor greater, an event with probabilitysx(n). Expectations of a general functiongof T must be calculated as a sum of two terms, E[g(T)] = ∫ n 0 g(t)fx(t)dt+g(n)sx(n). In particular, consider ann-year endowment insurance with death benefit functionbdefined on[0, n].Thatis, b(t)ispaidatthemomentofdeathifthisoccursbefore nandb(n)ispaidat timenif the insured is then alive (so that failure occurs at timen). Then AT(b) = E[v(T)b(T)] = ∫ n 0 b(t)v(t)fx(t)dt+b(n)v(n)sx(n), which is easily seen to agree with the present value obtained in the deterministic model. 228 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES From (15.4) we calculate the variance of the benefits as ∫ n 0 [b(t)v(t)]2fx(t)dt+[b(n)v(n)]2sx(n)− [ AT(b) ]2 . Some care is needed when applying the approximation (15.5). This is only valid for the continuouspartoftherandomvariable T.Inthepresentexampleoftheendowmentinsurance, Equation (15.5) would be modified to AT(b) = AT ( i 𝜹 ∗ b ) +b(n)v(n)s(n). The last term does not get multiplied by the interest adjustment, since in both the end-of-the- year-of-deathandmoment-of-deathcasesthefinalamountof b(n)ispaidtosurvivorsatexact timen. Example 15.2 A two-year-endowment insurance on (x) has benefits of 100 if death occurs in the first year or 80 if death occurs in the second year, and a pure endowment of 80 if the insured is alive at time two. You are given thatq x = 0.2, qx+1 = 0.3, and the interest rate is a constant100%.Findtheexpectationandvarianceofthebenefitswhen(a)benefitsarepayable at the end of the year of death; (b) benefits are payable at the moment of death. Solution. (a) LetZ denote the present value of the benefits in the end of the year of death case.AswenotedinExample6.2,wedonotevenneedthevalueof q x+1.Infactwecouldjust notice thatZ takes the value 50 with probability 0.2 and 20 with probability 0.8. It therefore has a mean of 26 and a variance of 302(0.2)(0.8)=144. (b) LetZ denote the present value of the benefits in the moment of death case. Now we do need the value ofqx+1 since it makes a difference if the person dies in the second year, in which case 80 is paid at the moment of death, or if they survive, in which case 80 is paid at the end of the year. Begin again with the calculation ofE(Z) andE(Z 2) but now split into the insurance and pure endowment amounts. E(Z) = 14.8+80(1∕4)(0.56) = 14.8+11.2. E(Z 2 ) = 596+6400(1∕16)(0.56) = 596+224. Now fori= 1,𝛿 =l o g(2) so E(Z) = ( 1 log2 ) 14.8+11.2 = 32.55 Squaringvmeans that𝛿 is doubled and theichanges to (1+i)2 −1 = 2i+i2. THE STOCHASTIC APPROACH TO ANNUITY BENEFITS 229 which equals 3 when the original value is 1. So, for the second moment calculation we take i= 3,𝛿 =l o g(4), to get E( Z 2 ) = ( 3 log4 ) 596+224 = 1513.77 Var(Z) = 1513.77−32.552 = 454.27 The variance is much larger in (b), due to the variation in the present value of the benefits, depending on when the person dies during the year, which is significant in view of the high interest rate. 15.3 The stochastic approach to annuity benefits 15.3.1 Discrete annuities Consideracontractwithannuitybenefitvector cthatpays ck attime kprovidedthatfailurehas notyetoccurred.Let Y denotethepresentvalueofthebenefitswithrespecttotheinvestment discount functionv, and leẗãT(c;v) (usually shortened töãT(c)) denoteE(Y). (The subscript ̃T reflects the fact that in view of the yearly payments, this quantity will depend only on the value of̃T, rather than the exact value ofT). There are two methods for computing expectations and variances of annuities in the stochastic setting. This first has the advantage of simplifying variance calculations. Recall that in the deterministic model we often viewed insurances as annuities. It is now helpful to view annuities as insurances in a certain sense. (Readers who attempted Exercise 5.22 will have encountered this idea previously.) To be precise, we want to expressY as a function of ̃T.L e tgbe the function defined on the positive integers by g(k) = c 0 +c1v(1)+c2v(2)+⋯ +ck−1v(k−1). We can also writeg(k) = ̈a(kc;v) using the notation introduced in Section 2.10.1. Suppose that̃T takes the valuek. Then payments will have been made at all integer times from 0 tok−1, soY will take the valueg(k). We can therefore write Y = g(̃T), and it follows that ̈ãT(c) = ∞∑ k=1 g(k)̃f(k) (15.6)‡ and Var(Y) = ∞∑ k=1 g(k)2̃f(k)− ̈ãT(c)2. (15.7)‡ 230 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES Expression(14.6)isknownasthe aggregatepaymentformula,sinceweconsiderthefunction g(k) which is the aggregate present value received for failure at timek. Example 15.3 For a 4-year life annuity on (60). The sequence of annuity benefits is 1, 2, 3, 4 beginning at age 60. You are givenq60 = 0.1, q61 = 0.2, q62 = 0.3 andi= 100%.F i n d E(Y) and Var(Y). Solution. Calculate recursively g(1) = 1,g(2) = 1+2× 1 2 = 2,g(3) = 2+3× 1 4 = 2.75, g(k) = 2.75+4× 1 8 = 3.25 for allk> 3. The distribution of̃T is given bỹf60(1) = 0.1, ̃f60(2) = 0.180, ̃f60(3) = 0.216, s60(3) = 0.504. We can now calculate E(Y) = 0.100+2(0.180)+2.75(0.216)+3.25(0.504) = 2.692, Var(Y) = 0.100+22 ×0.180+(2.75)2 ×0.216+(3.25)2 ×0.504−2.6922 = 0.530. Remark Note carefully that when the final payment of an annuity is at timek0, thenY takes the value ofg(k0 +1) with probabilitys(k0) sinceg(k) = g(k0 +1) for allk> k0. Remark There is an important point here which is often overlooked. Equation (15.6) applied to life annuities says that we can view the present value of a set of cash flows with respect to the interest and survivorship function as the expected value of the present value of payments received with respect to an interest only function. Some people have mistakenly assumed that similarly, the accumulated value of a life annuity is the same as the expected value of the payments received, accumulated with interest. This is not true and in fact the equality does not hold for values at any time other than 0. We have Val k(c,yx) = yx(k)−1̈ax(c), while the expected value at timek with respect tov, of the payments received fromc is v(k)−1̈ax(c), which will be less than the first quantity for allk> 0. Thesecondmethodtocalculateannuitypresentvaluesismorecloselyrelatedtowhatwe did in the deterministic model. We define random variables Ik = { 1, if ̃T > k, 0, if ̃T ≤ k. The contract pays a present value ofckv(k) for each integerksuch that̃T > k, or equiva- lently, such thatIk = 1. We can therefore write Y = ∞∑ k=0 ckv(k)Ik. (15.8) THE STOCHASTIC APPROACH TO ANNUITY BENEFITS 231 This is analogous to viewing the annuity as consisting of several pure endowments. Since E(I k) = s(k), taking expectations, gives ̈ãT(c) = ∞∑ k=0 ckv(k)s(k). (15.9)‡ Expression(15.9)isknownasthe currentpaymentformula,sinceitlooksateachpayment received, discounts it with interest, and multiplies it by the probability of receiving it. In the case whereT = T(x), this is precisely the formula obtained in (4.1). The equality of the expressions in (14.6) and (15.9) can be seen from (A.16). (Condition (A.14) will almost always hold. It is automatic when benefits are positive sinceg will be increasing,orwhentherearefinitelymanybenefitpayments,evenifnegative,since gwillbe bounded.) Note that theI k are not independent. To calculate variances from the current payment approach we use the following facts. Var(Ik) = s(k)(1−s(k)). (15.10) Forj< k,w eh a v eIjIk = Ik, so that Cov(Ij,Ik) = E(IjIk)−E(Ij)E(Ik) = s(k)(1−s(j)). (15.11) To simplify notation let rk = ckv(k)s(k), uk = ckv(k)(1−s(k)). Then from (A.24), (15.10) and (15.11), Var(Y) = ∑ k rkuk +2 ∑ j<k rkuj. (15.12) This will normally be a much more involved calculation than that given by the aggregate payment approach. 15.3.2 Continuous annuities Consider a contract that makes payments continuously at the annual rate ofc(t) at timet, providedT has not yet occurred. We will denote the actuarial present value byaT(c;v)( o f t e n shortened toaT(c)). Define the functiongby g(t) = ∫ t 0 c(r)v(r)dr. IfT fails at times, the present value of the benefits received will be preciselyg(s), so that Y = g(T). 232 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES It follows that aT(c) = E(Y) = ∫ ∞ 0 g(t)f(t)dt (15.13)‡ and Var(Y) = ∫ ∞ 0 g(t)2f(t)dt−[aT(c)]2. (15.14)‡ For the alternate current payment formula we use (A.15). From the fundamental theorem of calculus,g′(t) = c(t)v(t) and, moreover,g(0) = 0. If (A.14) holds (which as in the discrete case will almost always be the case) we can write the current payment formula aT(c) = ∫ ∞ 0 c(t)v(t)s(t)dt, (15.15)‡ which agrees with (8.9) whenT = T(x). There is no convenient variance formula analogous to (15.12) for continuous annuities, since we do have the representation as a sum of indicator random variables that was present in the discrete case. Example15.4 Anannuityprovidescontinuousbenefitsattherateof1perperiod,provided failure has not occurred. The failure timeT has an exponential distribution with constant hazard 𝜇. The force of interest is a constant𝛿.I f Y is the present value of the benefits, find Var(Y). Firstsolution. Since g(t) = ∫ t 0 e−𝛿rdr= 1−e−𝛿t 𝛿 , We have E(Y 2 ) = ∫ ∞ 0 [ 1−e−𝛿r 𝛿 ]2 𝜇e−𝜇rdr = 𝜇 𝛿2 ∫ ∞ 0 [1−2e−𝛿r +e−2𝛿r]e−𝜇rdr = 𝜇 𝛿2 [ 1 𝜇 − 2 𝜇 +𝛿 + 1 𝜇 +2𝛿 ] = 2 (𝜇 +𝛿)(𝜇 +2𝛿). From(15.13),orsimplyreferringbacktoSection8.10,weseethat E(Y) = 1∕(𝜇 +𝛿)sothat Var(Y) = 2 (𝜇 +𝛿)(𝜇 +2𝛿) − 1 (𝜇 +𝛿)2 = 𝜇 (𝜇 +𝛿)2(𝜇 +2𝛿). Note that, as a check, for𝛿 = 0, this reduces to𝜇−2, the variance of the given exponential distribution (see Section A.11.6) which must be true sinceY = T at zero interest. THE STOCHASTIC APPROACH TO RESERVES 233 Asimplification. We can simplify and also obtain a more general formula by the following trick: Y 2 = 1−2e−𝛿T +e−2𝛿T 𝛿2 = 2 𝛿 [( 1−e−𝛿T 𝛿 ) − ( 1−e−2𝛿T 2𝛿 )] . Upontakingexpectations andrecallingthat E[e−𝛿T] = 𝜇∕(𝜇 +𝛿),wegetthesameansweras above. ∗15.4 Deferred contracts Consider any contract based on the failure timeT, in which no payments are to be made in the firstmyears regardless of whether failure occurs or not. A typical example is a deferred annuity, but this can be a perfectly general contract, continuous or discrete, including death benefits, annuity benefits or both. LetY be the present value of the benefits, and letY ′ be the value of the benefits at the starting timem. As mentioned previously, there is no real need for any special mathematical treatment of such, as we simply take zero entries or zero values in thebenefitvectororfunction.Itissometimesconvenient,however,toexpressVar( Y)interms of Var(Y ′). The purpose of this section is to derive such a formula. The idea is to notice thatY takes the value 0 with probability 1−s(m) andv(m)Y′ with probability s(m). It follows from basic probability theory that E(Y) = s(m)E(v(m)Y′) = v(m)s(m)E(Y′), a formula that we are familiar with from the deterministic model. It similarly follows that E(Y2) = v(m)2s(m)E(Y′2) = v(m)2s(m)[Var(Y′)+E(Y′)2]. Substituting forE(Y) in terms ofE(Y′) in the expression Var(Y) = E(Y2)−E(Y)2 gives Var(Y) = v(m)2s(m)Var(Y′)+v(m)2s(m)(1−s(m))E(Y′)2. (15.16) This gives a familiar decomposition of the variance ofY. The first term measures the uncertainty arising from actual benefits once they start at timem, and the second term measures the uncertainty arising from the fact that failure may or may not occur prior to timem. 15.5 The stochastic approach to reserves We will illustrate the stochastic approach to reserves with the discrete model. The results are easily adapted to the continuous case. We have vectorsb and 𝜋, whereb k is paid at time k+1 if failure occurs between timekandk+1, and𝜋k is paid at timekif failure has not yet 234 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES occurred. We are adopting the point of view (used before in Chapter 6) of treating annuity benefits as negative premiums, so that𝜋 k represents a net inflow at timek, consisting of the premiumreceived,lessexpensespaidattime kifthesearebeingconsidered,andalsolessany annuity payments made at timek. We will therefore not need to refer to a separate annuity benefit vector. Fix an integer durationrand suppose that failure has not yet occurred by that time. Definition 15.1 The prospective loss at time r,denoted byrL, is the value at timer of the future net cash flows to bepaidout. These net cash flows are the future benefits to be paid minus the future net inflows to be received. The definition here is different than that used in the deterministic model. The benefits are the actual benefits paid on the contract, and not the individual share of the total deathbenefits.AswestressedinSection14.1,thediscountingiswithregardtotheinvestment discount functiononly. The value willdepend ofcourseonwhen failureoccurs, so rLwillbe arandomvariable.Itisafunctionoftheshiftedrandomvariable ̃T◦ r,sincewehaveassumed survival up to timer. We next determine this function. Suppose that failure occurs at timer+twhere k≤ t< k+1,forsomeinteger k.Then ̃T◦ rwillassumeavalueof k+1.Thecontractwillpayouta failure benefit ofbk at timek+1. Offsetting this, the net inflows would have been collected from timerto timer+k, so that in this case the value ofrLwill be br+kv(r,r+k+1)−[𝜋r +𝜋r+1v(r,r+1)+⋯ +𝜋r+kv(r,r+k)]. In terms of random variables, we can write rL= br+̃T◦ r−1v(r,r+ ̃T◦ r)−[𝜋r +𝜋r+1v(r,r+1)+⋯ +𝜋r+̃T◦ r−1v(r,r+ ̃T◦ r−1)]. Remark We have used the standard terminology here, but the reader should be aware that the word ‘loss’ in the above definition has a completely different connotation than that discussed in Section 6.4.1. It is not a definite number, measuring the loss brought about by deviationfromtheexpectedpatternofmortalityorinterest,butratheritisarandomvariable, giving the difference between what is collected and what is paid out under the contract at the differentrandomtimesoffailure.Itshouldalsobenotedthat rLisofthefewrandomvariables we encounter that can take negative values, which occur when there is a ‘gain’. Definition 15.2 In this stochastic model, we define thereserveattimer, denoted as before by rV,s i m p l ya s rV = E(rL). SorVistheexpectedvalueoftheprospectiveloss,whichistheexpectedpresentvalueof the future benefits less the expected present value of the future inflows. It is clear that in the case ofT = T(x) we obtain the same value as in the deterministic model. THE STOCHASTIC APPROACH TO PREMIUMS 235 Example 15.5 Let T = T(60). We are givenq60 = 0.1, q61 = 0.2, q62 = 0.3, i= 100%. Suppose we have a 3-year term insurance policy on (60) withb0 = 200, b1 = 200, b2 = 100, 𝜋0 = 20, 𝜋1 = 20, 𝜋2 = 10. (a) Find the distribution of0L. (b) Find the distribution of1L. (c) From your answer to (b), calculate1V. Solution. (a) The distribution of0Lis as follows: k Value of0Lwhen ̃T = kP (̃T = k) 1 200 × 1 2 −20 = 80 0.100 2 200 × 1 4 − ( 20+20× 1 2 ) = 20 0.180 3 100 × 1 8 − ( 20+20× 1 2 +10× 1 4 ) =− 20 0.216 ≥ 40 − ( 20+20× 1 2 +10× 1 4 ) =− 32.5 0.504 (b) k Value of1Lwhen ̃T◦ 1 = kP (̃T◦ 1 = k) 1 200× 1 2 −20 = 80 0.20 2 100× 1 4 − ( 20+10× 1 2 ) = 00 . 2 4 ≥ 30 − ( 20+10× 1 2 ) =− 25 0.56 (c) 1V = 80×0.20+(−25×0.56) = 2. The reader should verify this agrees with the answer that we would get from the methods of Chapter 6. 15.6 The stochastic approach to premiums For this section only, it is convenient to adopt a different convention. Assume now that expenses and annuity payments are included as benefits, so that the inflows𝜋 k are just the premiums collected. 15.6.1 The equivalence principle The random variable0L is usually denoted just asL. It is the present value of all future benefits less the present value of all future premiums. Recall that in the deterministic model 236 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES wecalculatednetpremiumsbysettingthepresentvalueofbenefitsequaltothepresentvalue of premiums. The stochastic version of this concept is to set premiums so that E(L) = 0. This is known as theequivalenceprinciple. 15.6.2 Percentile premiums The stochastic viewpoint allows us to incorporate other features of the random variableL rather than just its expectation, when computing premiums. For example, it would be highly desirable for an insurer to avoid a positive loss. Of course, it is impossible to ensure that this willalwayshold.Iftheinsureddiesshortlyafterpurchasingalifeinsurancepolicy,then Lwill almost certainly be positive. Suppose, however, we set a threshold𝛼, and then set premiums as small as we can, so that the probability will be at least𝛼 that the loss will be nonpositive. Typically𝛼 willbeanumber closeto1suchas0.95or 0.99. Thesearesometimesreferredto as percentilepremiums. This sounds reasonable, but there are major problems to such an approach. In the first place, the method does not take into account the importantright tailof the distribution, consisting of those values ofL which are greater than the specified percentile. If there are very large claims present, the method may not provide sufficient premiums to cover the risk. Thisisevidencedbythefactthatthemethodcanproducepremiumsthatareactuallylessthan the equivalence principle premium. For an extreme example of this, suppose we have ann year term policy and the probability of (x) dying withinnyears is less than 1−𝛼. We could achieve the stated goal by charging premiums of 0, which is absurd. On the other hand, for high values of𝛼 the method can produce premiums that seem inordinatelyhighwhencomparedwiththeequivalenceprinciplepremium.SeeExample15.7 below. Still another instance of the pathological behaviour produced by percentile premiums is illustrated by the following example. Example 15.6 On a single premium contract, the present value of the benefits takes the value 0 with probability 0.94, 1 with probability 0.01 and 3 with probability 0.05. (a) Find the percentile premium for𝛼 = 0.95. (b) Suppose the insurer plans to sell 2 contracts. Find the smallest possible premium for each,suchthatthereisaprobabilityofatleast0.95thatthetotalpremiumswillcover the total benefits on both contracts. Solution. (a) This is obviously 1, which gives exactly a 0.95 probability of covering the benefits. (b) Using independence, direct calculation shows that the probability of the total present value of benefits being less than or equal tozis 0.9025 forz= 2 or 0.9965 forz= 3. The smallest total premium needed is then 3 resulting in a charge of 1.5 per policy. THE STOCHASTIC APPROACH TO PREMIUMS 237 Theconclusionofthisexampleisexactlycontrarytowhatonewouldexpect,andindicates a major flaw in the percentile premium approach. As the number of policies increases, the risk should decrease due to diversification. Unfavourable occurrences on one contract can be offset by favourable ones on another. The premium per policy should therefore decrease rather than increase. Despitethedrawbackinherentinpercentilepremiums,theremaybetimeswhenonewant to compute these, possibly for comparison with premiums produced by other methods. Here is a procedure for the case where benefits are paid at the end of the year of failure. We want to find the smallest possible level premium𝜋 payable forhyears, such that the probability of a nonpositiveLis greater than or equal to some given number𝛼. We consider only the case where L decreases as the value ofT increases. This is fairly typical. As the time of failure increases, the discount factor reduces the present value of the benefit paid. In addition, the later the occurrence of failure, the more premiums will be collected. Both of these factors tendtodecrease L.(Ofcourse,therequiredconditionmaynotholdinthecasewherebenefits are increasing rapidly.) The procedure is as follows: 1. Letk 0 be the largest positive integerksuch thats(k) ≥ (𝛼). 2. Solve for𝜋 so thatLis 0 wheñT = k0 +1. That is, bk0v(k0 +1) = 𝜋̈a(1r;v), where ris the minimum ofk0 +1 andh. SinceLis decreasing withk, L≤ 0 implies that ̃T ≥ k0. Therefore, 𝛼 ≤ s(k0) ≤ P(L≤ 0). Moreover,ifwetakeanysmallerpremium 𝜋′,thenfor ̃T = k0 +1,Lwillbepositive,sothat P(L≤ 0) ≤ sk0+1 <𝛼 . by our choice ofk0. Example 15.7 Refer back to Example 15.5. Find the level percentile premium𝜋 payable for 3 years if𝛼 = 0.8. Solution. We takek0 = 1 sinceP(̃T > 1) = 0.90 > 0.8, whileP(̃T > 2) = 0.72 < 0.8. When ̃T = 2,L= 200× 1 4 −𝜋(1+ 1 2), so that𝜋 = 33.33. This is much higher than the equivalence principle level premium of 13.31. 15.6.3 Aggregate premiums An analysis of the examples above indicate that the percentile premium method can give unreasonable results with highly asymmetric distributions, where there is a large probability of a value in one of the tails. This approach however is a reasonable one to use in order 238 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES to compute the total premiums for a large group of policies, where the distribution of total benefits is likely to be more heavily concentrated about the mean. Suppose the insurer issues severalcontractsofthesametypeandwanttobefairlycertain,thatintheaggregate,thetotal premiums collected will cover the total benefits. There are two major questions that could be considered. First, the premiums may be fixed and we want to determine the number of contracts to sell to achieve the desired confidence. Second, the number of contracts may be fixed and the problem is to determine the premium to charge. In general, exact calculations areprohibitive,butwecangetapproximateanswersbyassumingthatthetotalityofalllosses has a normal distribution (see Section A.11.5). It is convenient to introduce the following quantity, which we will discuss in some generality. Definition15.3 Forarandomvariable Xwithapositivemean,the coefficientofvariation of Xis the quantity CV(X) = √ Var(X) E(X) . It is clear from (A.8) and (A.10) that for any constantc, CV(cX) = CV(X), showing that, unlike variance or standard deviation, this is a measure of variation that is independentoftheparticularunits.Soforexample,ifwemeasurelossinUSdollarsandthen change the units to pounds sterling, the variance will be quite different, but the coefficient of variation will remain the same. There are some other facts about this quantity that we want to derive. SupposeS= X 1 +X2 +⋯ +XN where these are independent random variables each distributed asX. Then Var(S) = N Var(X), E(X) = NE(X), showing that CV(S) = CV(X)√ N , (15.17) illustrating the expected diversification effect that occurs when we take an independent sum of random variables. We expect to reduce variation, as some high values are offset by low values. Next, suppose thatX is a normal random variable with positive mean, and we want the probability thatX takes a value greater than or equal to 0. This is clearly independent of any units, so we should be able to express it in terms of the coefficient of variation. Indeed, suppose Xhas mean𝜇 and standard deviation𝜎, so thatX = 𝜇 +𝜎Z where Zis the standard THE STOCHASTIC APPROACH TO PREMIUMS 239 normal.Then, P(X ≥ 0) = P(Z ≥ −𝜇∕𝜎)whichbysymmetryisequalto P(Z ≤ 𝜇∕𝜎).Wecan write P(X ≥ 0) =Φ [CV(X)−1], (15.18) whereΦ is the cumulative distribution function of the standard normal. Considernowthefirstquestiongivenabove.Supposewehaveacontractwithaprospective loss at time 0 ofL, whereE(L) < 0 and we want to determine the smallest value ofN,s o that ifNindependent contracts are sold, the probability is at least𝛼 that premiums will cover benefits. Let S denotes the aggregate loss on all contracts and apply our results above to−S,t h e aggregategain.Wewantaprobabilityof 𝛼that−Swillbepositive.Assumethat −Sisnormal. Invoke (15.18) withX =− S, and takeΦ−1 of each side, resulting in Φ−1(𝛼) = CV(−S)−1. Then apply (15.17) withX =− Lto getΦ−1(𝛼) = √ N∕CV(−L), so that √ N =Φ −1(𝛼)CV(−L). (15.19)‡ We see then that the required number of contracts increases as the confidence level goes up, and also as the uncertainty inLgoes up, which is just what we expect to happen. For the second question above whereN is fixed, consider a single premium contract with presentvalueofbenefits Z.Lettherequiredpremiumbeoftheform(1 +𝜃)E(Z).Thequantity 𝜃 is known as therelativesecurityloading –t h ew o r d sriskloadingand contingencyloading a r ea l s ou s e d .( S oi f𝜃= 0.20 for example, it means that the premium is calculated by adding 20% to the equivalence principle premium.) Now in this case −L= (1+𝜃)E(Z)−Z, so that Var(−L) = Var(Z), andE(−L) = 𝜃E(Z). Therefore CV(−L) = CV(Z) 𝜃 . (15.20) Substituting from (15.19) we can write 𝜃 = Φ−1(𝛼)CV(Z)√ N . (15.21)‡ Thisiscertainlyreasonableasitshowsthattheloadingincreaseswiththetolerancelevel, and the uncertainty in the benefits, but decreases as more contracts are sold. Example 15.8 Suppose the force of interest is a constant 0.06 and that𝜇(t) = 0.04 for all t. An insurer sells 100 contracts, each of which pays 1000 upon failure. Assuming a normal approximation, how much should be charged as a single premium on each contract, so that 240 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES there is a 95% chance that premiums will be sufficient to cover the aggregate claims for this group of 100 contracts. Solution. If Z=the benefit for a contract paying one unit at failure. Then E(Z) = 𝜇 𝜇 +𝛿 = 0.4, E(Z 2 ) = 𝜇 𝜇 +2𝛿 = 0.25. The formula for the second moment reflects the fact that squaring the discount function is equivalent to doubling the force of interest. It follows that Var( Z) = 0.09 andCV(Z) = 0.75. From(15.21),with 𝛼 = 0.95,𝜃 = 1.645(0.75)∕10 = 0.123375.Foreach100-unitcontractthe equivalenceprinciplepremiumis400andsothepremiumchargedshouldbe400(1.123375) = 449.35. Example 15.9 Suppose the contract in Example 15.8 is sold with premiums paid con- tinuously for life at a level rate, where the security loading𝜃 = 0.10. Assuming a normal approximation, how many contracts must be sold, so there is a 95% chance that total premi- ums will cover total benefits? Solution. Sinceourformulasareindependentofunitswecanignorethe1000andconsidera 1-unit contract. The main problem is to determineCV(−L). This is not always so easy when wenolongerhaveasinglepremium,butitcanbedoneforlevelbenefitsandpremiums.Thisis carriedoutinChapter16,andwereferthereadertoformulas(16.5)and(16.6)with t= 0.The equivalence principle premium is 0.04, so the premium𝜋 = 0.044 and𝜋∕𝛿 = 11∕15. From formula(16.6) √ Var(L) = (26∕15))0.3 = 0.52.Fromformula(16.5) E(L) = (26∕15)(0.04)− 11∕15 =− 0.04. So CV(−L) = 0.52 0.04 = 13. From(15.19) √ N = 1.645×13,so N = 457.32.Thismeansthat458contractsmustbesold. 15.6.4 General premium principles Therearemanyotherpossibilitiesforcomputingpremiumstoallowfortherandomnatureof theloss.Theseareknowningeneralas premiumprinciples.Formally,apremiumprincipleis a functionH, which assigns to each random variableX representing a loss, a numberH(X), which is the premium to be charged for accepting the riskX. This definition is applicable to contracts purchased by a single premiums, rather than periodic premiums. The equivalence principle premium is given byH(X) = E(X). The standard deviation principle is given byH(X) = E(X)+𝛽 √ Var(X), for some positive𝛽. Dividing this byE(X), we can see from (15.21) that the standard deviation principle can be interpreted (assuming a normal approximation) as setting premiums so that there is a certain probability, which will THE VARIANCE OFr L 241 dependon N,thatpremiumswillcoverlosses.Anotherexampleisthe varianceprinciplegiven by E(X)+𝛽Var(X), for some positive𝛽. (Exercise 15.25 points out a flaw in this principle.) There is extensive treatment in the actuarial literature comparing the properties of these and other premium principles. We will not elaborate further here, but more comments can be found in Chapter 22. 15.7 The variance ofr L We have seen that in our stochastic model, the reserve represents the expected value of the amountthatweneedinordertomeetfutureobligations.Itisusefultohavemoreinformation about the distribution of this random variable. In this section we present a formula for calculatingthevarianceof rLunderthesetupofSection14.2.Ofcourse,ifwehavecomputed the exact distribution ofrL, as we did in Example 15.5 we could calculate the variance directly. The present method will allow us to calculate the variance without knowing the exact distribution, as long as we know all the reserves after timer. Moreover, it gives us a decompositionofthisvarianceintothatattributabletoeachofthefutureyears.Intheactuarial literature, this result is known asHattendorf’stheorem. We start by takingr= 0, so we compute the variance ofL= 0L. To simplify the task we look at the net cash outflow year by year. LetCk denote the value at timekof the net amount paid out in the year fromk to k+1. That is,Ck is equal to the value at timekof the failure benefit paid out at the end of the year, less the inflow at timek.S oCk is a random variable taking three possible values depending on whether failure has occurred before timek, during the year (k,k+1) or at timek+1 or later. If failure occurs before timek, nothing is paid and nothing is received. If failure occurs during the year (k,k+1), thenbk is paid at time k+1 and𝜋k is received at timek. If failure occurs after timek+1, nothing is paid out and a premium of𝜋k is received at timek. The following table summarizes the possible values of Ck and the respective probabilities: Time of failure Value ofCk Probability Before timek 01 −s(k) Between timekand k+1 v(k,k+1)bk −𝜋k ̃f(k+1) = s(k)𝜆(k+1) At timek+1 or later −𝜋k s(k+1) = s(k)(1−𝜆(k+1)) The random variableLis related to the values ofCk by L= ∞∑ k=0 v(k)Ck. (15.22) A fundamental property is that CjCk =− 𝜋jCk, j< k. (15.23) 242 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES To verify (15.23), observe that it is obvious ifCk = 0, since then both sides are 0. IfCk is not equal to 0, it means that failure did not occur before timekand therefore must have occurred at timej+1 or later. We see from the table above thatCj is equal to−𝜋j. We will first derive a variance formula for the case of a contract with zero reserves. This means that the premium paid at any timek is just enough to cover the death benefit paid at the end of that year, so that 𝜋k = v(k,k+1)bk𝜆(k+1), for all k. (15.24) The significance of this case is that, as we show below, the random variablesCh for different values ofhare uncorrelated (although they are not independent). Moreover, we have a fairly easy expression for the variance of eachCh, so that (15.22) gives us a formula for Var(L). Substitutingfrom(15.24),weseethatthesecondandthirdvaluesof Ck aslistedaboveare respectively v(k,k+1)bk[1−𝜆(k+1)] and −v(k,k+1)bk𝜆(k+1). It follows immediately that under assumption (15.24), E[Ck] = 0, for all k, (15.25) so that from (15.23), ifj< k, Cov(Cj,Ck) =− 𝜋jE(Ck) = 0. (15.26) Moreover, after some algebraic manipulation, Var[Ck] = E[C2 k ] = v(k,k+1)2b2 k s(k)𝜆(k+1)(1−𝜆(k+1)). (15.27) Invoking the fact the variance of a sum of uncorrelated random variables is the sum of the variances, and noticing thats(k)(1−𝜆(k+1)) = s(k+1), we deduce from (15.22) that Var(L) = ∞∑ k=0 v(k+1)2b2 k s(k+1)𝜆(k+1). We next consider the case of a general durationr, still keeping the assumption of zero reserves.Inthiscase, rLwilljustbe 0Lforthepolicyon(x +r)withthebenefitsandpremiums starting at ager, and with probabilities conditioned on the survival of (x) to agex+r.T h es and 𝜆 in the formula above will be those associated with the distribution of̃T◦ r. The result, using (15.10) and the discrete counterpart of (15.8), is Var(rL) = 1 s(r) ∞∑ k=0 v(r,r+k+1)2b2 r+k s(r+k+1)𝜆(r+k+1). Finally, we consider the general case where we remove the restriction given by (15.24). We split the policy into two parts, the risk portion and the savings portion, in exactly the same way as we did in Section 6.4.2. This leads to a corresponding split of the prospective loss into that portion of this loss attributable to the risk portion and that attributable to the STANDARD NOTATION AND TERMINOLOGY 243 savingsportion.Buttheprospectivelossforthesavingsportionmustbeaconstant,sinceitis completely independent of the mortality experience. (In fact this constant is just the present value, with respect to interest, of the pure endowment paid out at the end, less the savings portions of the premiums. Since the fund is reduced to zero, this present value must be what youstartedwith,whichis rV.)Themeansthatthevarianceoftheprospectivelossisthesame as the variance of the prospective loss on the risk portion. Therefore, the only change from thepreviousformulaistoreplacetheoriginaldeathbenefitsbythebenefitsattributabletothe risk portion, namely the net amount at risk. Our final formula is Var( rL) = 1 s(r) ∞∑ k=0 v(r,r+k+1)2𝜂2 r+ks(r+k+1)𝜆(r+k+1). (15.28)‡ This leads to a backwards recursion formula. Splitting off the first term, Var(rL) = v(r,r+1)2s(r+1) s(r) [𝜂2 r𝜆(r+1)+Var(r+1L)]. (15.29) This is useful ifT is bounded, as is true withT(x). We can start the recursion with a value of 0 for the variance of the loss at the final duration. Example 15.10 Refer back to Example 15.5. Use (14.23) to find the variance of 1L. Solution. We will first calculate the product of the three mortality factors in (14.23). In the case whereT = T(x), this product for indexk is k+1px+r qx+r+k.F o rk = 0w eg e t0.8×0.2, and fork= 1 we get 0.8×0.7×0.3. We also calculate that2V = 5, so that𝜂1 = 195. Then Var(1L) = 1 4 ×1952 ×0.16+ 1 16 ×1002 ×0.168 = 1626. We can verify this answer directly from the solution to Example 15.5, where we have the complete distribution of 1L. 15.8 Standard notation and terminology Theonlymajoritemwehavenotyetdiscussedistheuseofanupperleftsubscript2todenote a quantity that is computed at a force of interest that is double that of the standard one or, equivalently,atadiscountfunctionthatisthesquareofthestandardone.Thiswasdeveloped for variance formulas. Its use is, however, restricted to cases where the failure benefits are either 0 or 1. In that case, squaring the benefits leaves them unchanged, and the second moments are calculated by just squaring the discount function. For example, the variance for the benefits paid on a 1-unit, end-of-the-year-of-death,n-year term insurance would be written as 2A1 x:n ⌝ − ( A1 x:n ⌝ )2 . 244 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES Notes and references To be precise, one should specify what convergence means in Equation (15.8) in the case that the sum is infinite. We did not worry about that here, since the important result for our purposes is (15.9) which, as we noted, is valid in almost all cases. Young (2004) discusses a variety of premium principles and their properties. Exercises Type A exercises 15.1 Afailuretime T isuniformlydistributedontheinterval[0,10].Aninsurancecontract pays e0.06t at the moment of failure if this occurs at timet. The force of interest is a constant 0.04. Find the expectation and variance of the present value of the benefits. 15.2 A 3-year life annuity on (x) and (y) provides for annuity benefits of 1 at time 0, 2 at time 1 and 4 at time 2, provided that both individuals are alive. You are given that p x = 0.8, 2px = 0.6, py = 0.7, 2py = 0.5 and the interest rate is a constant 100%. (a) Find the expectation and variance of the present value of the benefits. (b) A single premium for the contract is found by adding a 20% relative risk loading tothepresentvalue.Whatistheprobabilitythatthepremiumwillbesufficientto provide for the benefit payments on a single contract? 15.3 A life annuity contract provides for continuous payments at the annual rate of e −0.05t attime t.Theforceofmortalityisaconstant0.1,andtheforceofinterestisaconstant 0.05. IfY is the present value of the benefits, findE(Y) and Var(Y). 15.4 A 10-year term insurance policy provides for a payment of 10−t at the moment of the first death of (x) and (y), should this occur at timet. The mortality of (x) and (y) is as given in Exercise 14.9. Assuming an interest rate of 0, find the expectation and variance of the present value of the benefits. 15.5 A failure time T is uniformly distributed on the interval [0,10]. There is a constant interest rate of zero. An annuity provides for continuous payments at the rate oftat time t, provided that failure has not yet occurred. Find the expectation and variance of the present value of the benefits. 15.6 A 4-year pure endowment contract on (60) provides for 1000 paid at age 64 if (60) is thenalive.Nothingispaidifdeathoccursbeforeage64.Thisispurchasedbyfourlevel annual premiums of 100. You are given thatq 60 = 0.1,q61 = 0.2,q62 = 0.25,q63 = 0.30,andtheinterestrate iisaconstant25%.Writedowntheexactdistributionof 1L. Use this to find1V. 15.7 Acertainproducthasafailuretimeof T.Theapproximatingdiscreterandomvariable ̃T has a constant hazard function𝜆(k) = 0.3. (In other words, there is a 30% chance that the product will fail each year, given that it was still working at the beginning of the year.) An insurer agrees to pay 100 at the end of the year of failure should the product fail within 2 years. In return it collects a premium of 40 now and a second EXERCISES 245 premium of 20 at time 1, if failure did not occur in the first year. The interest rate is 25%. (a) FindE(L), Var(L), Var( 1L). (b) Find the probability that, for a given contract, the premiums collected will be sufficient to pay the benefits. 15.8 A 3-year endowment insurance on (70) provides death benefits of 100 payable at the endoftheyearofdeath,ifthisoccurswithin3years,plusapureendowmentof100at age 63 if the insured is then alive. This is purchased by two premiums of 30 payable at age 60 and 61. You are given thatq 60 = 0.2,q61 = 0.4, and the interest rate is a constant25%.Givethecompletedistributionoftherandomvariable Landusethisto find 0V. 15.9 Aninsurerchargesasinglepremiumof E(X)+0.1 √ Var(X)foracontractwithpresent value of benefits equal toX. Use a normal approximation. (a) How many contracts must be sold so that the probability is 95% that premiums cover benefits? (b) What is the probability that premiums cover benefits if 225 contracts are sold? 15.10 A failure time has a hazard rate of 𝜇(t) = 2 20−t. A contract provides for a benefit of 1 at time 10 provided that failure occurs before time 10 (so, for example, if failure occurs at time 6, the benefit is not paid until 4 years later). You are given thatv(10) = 0.6. Find the expectation and variance of the present value of the benefits. 15.11 Consider a 3-year life annuity on (60). The sequence of payments (beginning at time 0) is 4, 2, 1. You are given thatq 60 = 0.2, q61 = 0.4, q62 = 0.5. The interest rate is a constant 25%. The insurer sells 100 such contracts, charging (1+r) times the equivalenceprinciplesinglepremium.Whatshould rbesothatthereisatleasta95% chance that premiums will cover the benefits? (Use a normal approximation.) 15.12 A2-yearterminsurancecontractsoldtotwolives,(70)and(71),providesforbenefits payable at the end of the year of the first death. The benefit paid is 40 in the first year and 50 in the second year. You are given thatq 70 = 0.4,q71 = 0.5,q72 = 0.6, and the interest rate is 100%. LetZdenotes the present value of the benefits. (a) FindE(Z) and Var(Z). (b) Suppose the contract is to be paid for by level annual premiums ofPpayable for 2 years. What shouldP be if the equivalence principle is used, and what is the resulting probability thatL≤ 0? (c) What is the smallest amount thatPcould be if we want a probability of at least 0.25 that the lossLwill be less than or equal to 0? 246 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES 15.13 A 3-year endowment insurance on (70) provides death benefits of 400 payable at the end of the year of death, if this occurs within 3 years, plus a pure endowment of 200 atage73,iftheinsuredisthenalive.Thisispurchasedbythreelevelpremiumsof40. You are given thatq 70 = 0.2,q71 = 0.4,q72 = 0.5, and the interest rate is a constant 100%. (a) Givethecompletedistributionsfortherandomvariable L,andtherandomvariable 1L. Use the latter to find the reserve at time 1. (b) Whatshouldthelevelpremiumbeiftheinsurerwantsthesmallestpremiumsuch that the probability of a positive loss is less than 25%? 15.14 Use Equation (15.12) to calculate the variance in Example 15.3. 15.15 You are given thatq x = 0.1, qx+1 = 0.2, qy = 0.2, qy+1 = 0.3, and the rate of interest is a constant 25%. Find the variance of the present value of the benefits in each of the following contracts (a) A payment of 100 is made at the end of the year of the second death, provided this occurs within 2 years. (b) Three annual payments of 100 are made, the first at time 0, provided that at least one of (x)o r(y) are alive. 15.16 Afailuretime Thasauniformdistributionon[0,30].Theforceofinterestisaconstant 5%. A single premium insurance contract pays 1 unit at the moment of failure. It is desiredtohaveatleasta95%probabilitythatpremiumswillcoverbenefitsonalarge group of contracts. Answer the following assuming a normal approximation. (a) Suppose the security loading is 20%. How many contracts should be sold? (b) It is estimated that 100 contracts will be sold. What should the security loading be? 15.17 RedoExample15.2(b)onlynowassumingthattheinterestrateis25%inthefirstyear and 50% in the second year. Type B exercises 15.18 A failure time has a constant hazard rate of𝜇 and the force of interest is a constant 𝛿. A contract pays e𝛾t at the moment of failure for failure at timet, where𝛾<𝜇 +𝛿. Find the expectation and variance of the present value of the benefits as a function of 𝜇,𝛿 and𝛾. 15.19 Suppose that the force of mortality𝜇 is a constant 0.04 and the force of interest is a constant 0.06. (a) An insurer sells to one individual a single-premium, whole-life policy, with 100 payable at the moment of death, and also sells, to an independent life, a life annuity with benefits payable continuously at the rate ofcper year. IfW denotes the present value of the benefits on the two contracts combined, find Var(W)a sa function ofc. EXERCISES 247 (b) Supposenowthatthetwocontractsin(a)aresoldtothesameindividual.Whatis thevarianceof Winthiscase?Isitlessthan,greaterthanorequaltothevariance in (a)? Explain why. For what values ofc, if any, will the variance be equal to 0? Explain. 15.20 Afailuretime T hasap.d.f.of f(t) = (12−t)∕72for0 ≤ t< 12.Anannuitycontract provides for continuous payments at the annual rate of 1, which stop at failure, or at time 6 if earlier. The interest rate is 0. Let Y denote the present value of the benefits. (a) ExpressY as a function ofT. (b) CalculateE(Y) and Var(Y). (c) The single premium charged for this contract is the expected value plus 20%. What is the probability that the premium will cover the benefits? 15.21 Foracertainfailuretime T,aninsurancecontractpaysbenefitsattheendoftheyearof failure. You are given thatsT(9) = 0.8, sT(10) = 0.7 E(10L) = 70,Var(10L) = 1000. The amount payable at time 10 for failure between time 9 and time 10 is 100. The interest rate is a constant 20 %. Find Var( 9L). 15.22 Adiscretefailuretime T hasaconstanthazardrateof 𝜆(k) = 1∕2.Thereisaconstant interest rate of 100%. An insurance contract pays a benefit of 1 unit at the end of the year of failure. Level annual equivalence principle premiums are payable prior to failure. Using formula (15.28), show that for any nonnegative integerr, Var( rL) = 1∕15. 15.23 Consider two independent lives (x),(y) , whereT(x) is uniform on [0,1] andT(y)i s uniform on [0,2]. (a) Calculate the p.d.f. of the last survivor failure timeT( xy). (b) An annuity provides continuous payments for as long as either (x)o r(y)i sa l i v e . The rate of payment at timetis 3t2. The rate of interest is zero. IfY denotes the present value of the benefits for this annuity, findE(Y) and Var(Y). 15.24 An endowment insurance on (x) provides for 1000 at the end of the year of death, provided this occurs within 4 years, and 1000 at time 4 if (x) is still alive. Net level annual premiums are payable for 4 years. You are given thatqx = 0.05,qx+1 = 0.08,qx+2 = 0.10. Interest rates are 5% for the first 2 years and 6% thereafter. Find Var(1L) in two ways: first, by finding the exact distribution of1L; and second, by finding 2V and 3V and then using Equation (15.28). 15.25 A failure time takes the values 1 or 2 each with probability 1/2. The interest rate is 0. Insurance contract A pays 5 at failure. Insurance contract B pays 1 if failure occurs at time 1, and 5 if failure occurs at time 2. Show that using the variance premium principle with𝛽 = 1, the premium for contract B is higher than that for contract A. Spreadsheet exercise 15.26 Modify the spreadsheet of Chapter 6 to calculate Var(rL). 16 Simplifications under level benefit contracts 16.1 Introduction The calculation of variances and other distributional features simplifies considerably when we have level benefits and constant interest. In fact, we can write down formulas for exact distributions of the major random variables of interest. Throughout this chapter, we consider the following setup. We have a general failure timeT. We will consider insurances paying a level amount upon failure, either at the end of the year of failure or at the moment of such, and we will consider annuities paid prior to the failure ofT with either a level pay- ment or continuous payments at a level rate. In addition, we assume a constant force of interest 𝛿. By takingT to beT(x), this will apply to level benefit, whole-life insurances and to level benefit whole-life annuities. By takingT = min{T(x),n}, this will apply to level benefitn- yearendowmentpoliciesandtolevelbenefit n-yeartemporarylifeannuities.Ourassumption does not apply to term insurance, even when there is a level benefit during the term, since the benefit drops to zero after the expiration of the contract. However, in Section 16.5 we do illustrate that the calculation of exact distributions is possible for term or deferred insurances with a level death benefit paid over the benefit period. 16.2 Variance calculations in the continuous case It is convenient to begin with a continuous failure timeT. 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 VARIANCE CALCULATIONS IN THE CONTINUOUS CASE 249 16.2.1 Insurances Consider an insurance policy paying 1 at the moment of failure. The discount function is given byv(t) = v t = e−𝛿t.L e t̄Z be the present value of the benefits. In this case there is little simplification, and we know from the previous chapter that ̄AT = E(̄Z) = E(vT ), Var( ̄Z) = E(v2T)−(̄AT )2. (16.1) 16.2.2 Annuities Consider an annuity with continuous payments at the rate of 1 per year, made prior to the occurrence of failure. If̄Y is the present value of the benefits, then ̄Y = ̄a(1 T ) = 1−vT 𝛿 = 1− ̄Z 𝛿 , (16.2)‡ ̄aT = E(̄Y) = 1−E(̄Z) 𝛿 ,V a r ( ̄Y) = Var(̄Z) 𝛿2 . (16.3)‡ For the case whereT = min{T(x),n}, we have already seen the first part of (16.3). This was the continuous ‘endowment identity’ given at the end of Section 8.8. 16.2.3 Prospective losses Consideracontractthatpays1unitatthemomentoffailureandhascontinuouslevelpremiums at an annual rate of𝜋 payable prior to failure. (As a practical matter, this means that we are dealingwithanetpremiummodelthatignoresexpenses,whichareunlikelytobelevel.)The prospective loss at timet is tL = ̄Z ◦ t −𝜋(̄Y ◦ t), (16.4) where ̄Z ◦ t = vT ◦ t and ̄Y ◦ t = (1− ̄Z ◦ t)∕𝛿. This leads to tL = ( 1+ 𝜋 𝛿 ) vT ◦ t − 𝜋 𝛿. (16.5) This greatly simplifies the calculation of the variance of the prospective loss. Compare the following with (14.23). Var( tL) = ( 1+ 𝜋 𝛿 )2 Var(vT ◦ t). (16.6)‡ 16.2.4 Using equivalence principle premiums Supposethat 𝜋isanequivalenceprinciplepremium.Then 𝜋E(̄Y) = E(̄Z),sothatfrom(16.3), ̄ 𝜋= 1 ̄aT −𝛿. (16.7) 250 SIMPLIFICATIONS UNDER LEVEL BENEFIT CONTRACTS Takingt = 0 in (16.6), Var(L) = Var(vT ) (𝛿̄aT )2 = Var(̄Z) (1− ̄AT )2, (16.8) showing that Var(̄Z) < Var (L). This reflects the fact that there is more risk involved in selling an insurance contract where premiums are payable over the entire life of the contract as opposed to the single-premium case. For failure occurring early, the insurer not only loses interest but will have collected relatively small amounts in premiums. For a final formula, express̄Z ◦ t in terms of̄Y ◦ t in (16.4). Then tL = (1−(𝜋 +𝛿)(̄Y ◦ t)) If𝜋 isanequivalenceprinciplepremium,wecansubstitutefrom(16.7)andtakeexpectations to give a simple formula for the reserve at timet. t ̄V = 1− ̄aT ◦ t aT . (16.9) (ForT(x), we obtained the discrete version of this in (6.18).) 16.3 Variance calculations in the discrete case Wenow consider thecasewhere failurebenefits arepaidat theend oftheyear of failure,and annuity benefits and premiums are payable yearly. All of the formulas in Section 16.2 have discrete counterparts, which for the most part are obtained by replacingT by ̃T, ̄A by A,a by ̈a and 𝛿 by d. We will leave the formal derivations to the reader, but will list the formulas with the corresponding equation numbers as in Section 16.2, only with a prime to denote the discrete case. If Z denotes the present value of the benefits for an insurance paying 1 unit at the end of the year of failure, A ̃T = E(Z) = E(ṽT ), Var(Z ) = E(v2̃T)−(ÃT )2. (16.1′) IfY denotes the present value of an annuity paying 1 unit yearly provided that failure has not occurred, Y = ̈a(1T ) = 1−ṽT d = 1−Z d , (16.2′)‡ E(Y) = 1−E(Z) d ,V a r ( Y) = Var(Z) d2 . (16.3′)‡ VARIANCE CALCULATIONS IN THE DISCRETE CASE 251 Now consider a contract which pays 1 unit at the end of the year of failure and has level annual premiums of𝜋 payable prior to failure. Then for any positive integerk, kL = ( 1+ 𝜋 d ) ṽT ◦ k − 𝜋 d , (16.5 ′) Var(kL) = ( 1+ 𝜋 d )2 Var(ṽT ◦ k). (16.6′)‡ Suppose that𝜋 is an equivalence principle premium. Then 𝜋 = 1 ̈ãT −d, (16.7 ′) Var(L) = Var(ṽT ) (d̈ãT )2 = Var(ṽT ) (1−ÃT )2, (16.8′) kV = 1− ̈ãT ◦ k ̈ãT . (16.9′) One could possibly consider expenses when using the above simplified prospective loss formulas if the difference is only in the first year, as the following example shows. Example16.1 Aninsurancepolicyprovides1000attheendoftheyearofdeathwithlevel premiums payable for life. There are expenses in the first year of 60% of the premium plus 50, and in the subsequent years of 15% of the premium plus 20. In addition, there is a death benefitsettlementexpenseof50.Youaregiventhat E(Z) = 0.4andVar(Z ) = 0.10where Z is the present value of 1 unit of death benefit. The rate of discount is a constant 0.06. Find Var (L) whereL is calculated using expense-augmented premiums, and including all expenses. Solution. We first compute the expense-augmented premiumG as G̈a x = 0.45G +0.15G̈ax +30+20̈ax +1050Ax so that G = 1050Ax +30+20̈ax .85̈ax −0.45 . We know thaẗax = (1−Ax)∕d = (1−0.4)∕6 = 10, and substituting in the above,G = 80.75.This means that the total inflow after the first year is 0.85(80.75)−20 =48.64. Nowconsiderapolicywherethetotalinflowwas48.64ineveryyear.Thevalueof L from thatcontractwoulddifferfromthatontheoneinquestiononlybyaconstantamount,namely the extra amount in the first year due to higher expenses. Therefore, the variance ofL would bethesame,andwecanuseformula(16.6 ′),with 𝜋 = 48.64,andthe1replacedbythedeath benefit of 1050. From (16.6′), Var(L) = ( 1050+ 48.64 0.06 )2 0.10 = 346208. 252 SIMPLIFICATIONS UNDER LEVEL BENEFIT CONTRACTS 16.4 Exact distributions Inthissection,wecalculatetheexactdistributionsfor ̄Z, ̄Y andL.Ineachcase,wewillderive distribution functions. These will be given for values between the greatest lower bound and leastupperboundofthevalues.(Weknowthat F takesthevalue0forargumentslessthanthe greatestlowerbound,and1forargumentsgreaterthantheleastupperbound.)Itisconvenient here to introduce some new notation. For any random variableX,l e t ̂F X(x) = P(X < x), ̂sX(x) = P(X ≥ x). Of course, whenX is continuous,̂F = F and ̂s = s. The distribution functions we want are all easily expressed in terms of the distribution of T.L e tN denote the least upper bound of the values ofT. In the case thatN =∞ (as for example whenT is exponential), the termvN in the formulas below will be equal to 0. 16.4.1 The distribution of ̄Z The distribution of̄Z is given by F̄Z(z) = P(e−𝛿T ≤ z) = P ( T ≥ −logz 𝛿 ) = ̂sT ( −logz 𝛿 ) , vN < z < 1. (16.10) 16.4.2 The distribution of ̄Y First note that arguing as in (16.10) gives ̂FZ(z) = sT (−log(z) 𝛿 ) . Using (16.2) F̄Y (y) = P(̄Z ≥ 1−𝛿y) = 1− ̂F̄Z(1−𝛿y). Substituting from above, F̄Y (y) = 1−sT ( −log(1−𝛿y) 𝛿 ) = FT ( −log(1−𝛿y) 𝛿 ) ,0 ≤ y < 1−vN 𝛿 . (16.11) 16.4.3 The distribution ofL The minimum value ofL will be −𝜋 𝛿 + ( 1+ 𝜋 𝛿 ) vN, EXACT DISTRIBUTIONS 253 occurringforfailureattime N,anditsmaximumvaluewillbe1,occurringforfailureattime 0. Using (16.5), FL(u) = P [( 1+ 𝜋 𝛿 ) e−𝛿T ] ≤ u + 𝜋 𝛿 = P [ e−𝛿T ≤ (𝛿u +𝜋 𝛿 +𝜋 )] (16.12) = ̂sT [ −1 𝛿 log (𝛿u +𝜋 𝛿 +𝜋 )] , −𝜋 𝛿 + ( 1+ 𝜋 𝛿 ) vN ≤ u ≤ 1. For the more general case,FtL(u) is given by the same formula, but witĥsT ◦ t replacing ̂sT and vN−t replacing vN. 16.4.4 The case whereT is exponentially distributed In the particular case whereT is exponential with constant hazard function𝜇, we know that ̂sT (z) = sT (z) = e−𝜇z andN =∞ . The above formulas simplify to F̄Z(z) = z𝜇∕𝛿,0 < z < 1, (16.13) F̄Y (y) = 1−(1−𝛿y)𝜇∕𝛿,0 ≤ y < 1 𝛿, (16.14) FL(u) = (𝛿u +𝜋 𝛿 +𝜋 )𝜇∕𝛿 , −𝜋 𝛿 ≤ u ≤ 1. (16.15) Itisofinteresttoobservethatif 𝜇 = 𝛿intheexponentialcase,theexponentintheformulas above equals 1 so that̄Z, ̄Y andL are alluniform random variables. Example 16.2 A company decides to add 20% to its equivalence principle premiums as a protection against unfavourable experience. In each of the following cases, find the probabilitythatpremiumswillcoverclaims.Supposethat T isexponentialwith 𝜇 = 0.04and that𝛿 = 0.06. (a) Asingle-premiumannuityprovidingcontinuouspaymentsattheannualrateof1prior to failure. (b) Acontractpaying1unitatfailure,withlevelpremiumspayablecontinuouslypriorto failure. Solution. (a) The equivalence principle premium is 1∕(𝜇 +𝛿) = 10. The actual premium charged will be 12. From (16.14), P(Y ≤ 12) = 1−0.28 2∕3 = 0.57. 254 SIMPLIFICATIONS UNDER LEVEL BENEFIT CONTRACTS (b)Theequivalenceprincipleannualpremiumrateisjust 𝜇 = 0.04,sotheactualpremium rate charged is 0.048. The probability that premiums cover claims is just P(L ≤ 0) = (0.048 0.108 )2∕3 = 0.58. 16.5 Some non-level benefit examples It is also possible to obtain exact distributions in some simple cases involving non-level benefits such as term or deferred insurance. 16.5.1 Term insurance Consider a contract that pays 1 at failure, provided failure occurs withinn years. In order to handlefailuretimesthatarenotcontinuous,weadopttheconventionthatabenefitispaidfor failureatexacttime n.Wewillcomputethedistributionof ̄Z,thepresentvalueofthebenefits. The minimumpositive value of̄Z is e −𝛿n occurring for death at timen. Since nothing is paid for death strictly after timen, which occurs with probabilitysT (n), we can easily obtain the term distribution from the whole life distribution. Pick up the probability mass to theleft of z = e−𝛿n andsetitdownasapointmassatthepoint0,asillustratedinFigure16.1.Fromthis, utilizing formula (16.10), we can read off this distribution of̄Z. F̄Z(z) = ⎧ ⎪ ⎨ ⎪ ⎩ s T (n), 0 ≤ z < e−𝛿n, ̂sT (−log z 𝛿 ) , e−𝛿n ≤ z ≤ 1. (16.16) Example16.3 Redopart(b)ofExample16.2,butnowassuming n-yearterminsurancefor n = 15 and 10. Level premiums are payable continuously forn years. Solution. Theequivalenceprinciplepremiumrateisstill 𝜇 = 0.04sotheactualpremiumrate charged is still 0.048 and𝜋∕𝛿 = 0.8. For the contract in Example 16.2, the value ofL when T = n is 1.8e−0.06n −0.8. Forn = 15, this is negative. This means thatL becomes negative at some point prior to expiration of the term contract, so the probability that premiums cover claims is 0.58, exactly the same as it was in constant benefit case. Whenn = 10, the value of L for the contract in Example 16.1 is positive. Therefore, in order thatL in this example be negative, it is necessary thatT ≥ 10, so that no benefits are paid. The probability of this is e −0.6 = 0.67. 16.5.2 Deferred insurance Asimilarexampleisprovidedby deferred insurance.Consideracontractthatpays1atfailure, provided failure occurs after timen. We now adopt the convention that nothing is paid for failure at exact timen. In view of this convention, there need not be a maximum value of̄Z but e−𝛿n is certainly an upper bound, since any benefits will be paid at a time later thann. SOME NON-LEVEL BENEFIT EXAMPLES 255 0 z e–δn e–δn e–δn 1 Whole life ST (n) FT (n) 11 z z 00 n-Year deferred n-Year term ST (n)FT (n) ST (n) FT (n) Figure 16.1 Graph of f̄Z(z) for various types of insurance Refer again to Figure 16.1. Since nothing is paid for death at or before timen, which occurs with probabilityFT (n), the probability mass to theright of e−𝛿n is picked up and set down as a point mass at 0. Utilizing (16.10), we can write F̄Z(z) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ F T (n) z = 0, FT (n)+ ̂sT (−log z 𝛿 ) 0 < z < e−𝛿n, 1 e−𝛿n ≤ z ≤ 1. 16.5.3 An annual premium policy We next investigate a more complicated case where we compute the exact distribution ofL in an annual premium policy. Consider a failure timeT that is unbounded (i.e.N =∞ ). We consider insurances which have premiums payable continuously at a level annual rate𝜋 for the duration of the contract. We will compare the distribution ofL for a contract that pays 1 unit on failure or at timen if earlier, and a contract that pays 1 unit at failure provided this occurs withinn years.These thencorrespond respectively toendowment andterminsurance. To simplify the notation, let b0 =− 𝜋 𝛿 + ( 1+ 𝜋 𝛿 ) vn, c0 =− 𝜋 𝛿 + 𝜋 𝛿vn. In both cases, the value ofL is given by (16.5) (witht =0), provided thatT takes a value less than n, which corresponds toL taking a value greater thanb0. So for any interval (a,b) with a > b0, we have thatP(a < L ≤ b) is the same in both cases and this value can be calculated directly from (16.12). 256 SIMPLIFICATIONS UNDER LEVEL BENEFIT CONTRACTS b0–π/δ c0 0 b0–π/δ c0 0 b0–π/δ c0 0 u uu FT (n) FT (n) FT (n) ST (n) ST (n) ST (n) Whole life n-Year term n-Year endowment Figure 16.2 Graph offL(u) for various types of insurance Consider the remaining probability ofsT (n). For the endowment contract, this will all be concentrated at the single pointb0. For the term contract, it will all be concentrated at the single pointc0, which lies in the interval (−𝜋∕𝛿,b0). Asn increases to∞ both b0 and c0 approach −𝜋∕𝛿 and the distribution approaches the whole-life case as given in (16.12). The various density functions are compared in Figure 16.2. The probability mass to left ofthepoint b0 onthewhole-lifegraphispickedupandsetdownasapointmassat b0 forthe endowment policy, or atc0 for the term policy. Exercises Type A exercises 16.1 Foracertainfailuretime T,aninsurancecontractpays1atthemomentoffailure,and has level premiums payable continuously prior to failure at the annual rate of 0.06. You are given that Var(̄Z) = 0.064, Var( ̄Y) = 10, where ̄Z is the present value of the failure benefits, and̄Y is the present value of an annuity contract with continuous payments at the annual rate of 1, payable prior to failure. Find Var(L). 16.2 For a certain failure timeT, an insurance contract pays 1 at the moment of failure. Level equivalence principle premiums are payable prior to failure. IfE(̄Z) = 0.6 and Var(L) = 2, find Var(̄Z). 16.3 A whole-life insurance contract provides for 1 unit payable at the moment of death. Level premiums are payable continuously for life at the annual rate of 0.08. The force of mortality is a constant 0.05, and the force of interest is a constant 0.10. How many contracts must be sold in order that there is at least a 95% chance that the total premiums on all these contracts will cover the total benefits? Use a normal approximation. EXERCISES 257 16.4 Aninsurersells100whole-lifeinsurancepolicieseachprovidingfor1unitpayableat the moment of death. Level premiums are payable continuously for life. The force of interest is a constant 0.06, and the force of mortality is a constant 0.04. What should the rate of premium payment be on each policy, in order that there is a 95% chance that total premiums on all 100 policies will cover the total benefits on all policies? Use a normal approximation. Type B exercises 16.5 (See Exercise 13.8.) A continuous failure time has a density function,fT (t) = 𝛽2te−𝛽t (a gamma distribution with first parameter 2). The force of interest is a constant𝛿. Find expressions in terms of𝛽 and 𝛿 for: (a) ̄AT ;( b )̄aT ; (c) the net annual rate of premium payment when premiums are payable continuously prior to failure, for an insurance paying 1 at failure and (d) the reserve at timek for the contract in (c). 16.6 An insurance contract, based on the failure timeT, pays 1 unit at the moment of failure provided this occurs within 5 years. Nothing is paid for failure after that time. The force of interest is a constant 0.1. IfT has the hazard function 𝜇 T (t) = 2 10−t, find the probability that the present value of the benefits is strictly positive, but less than or equal to e −0.3. 16.7 Suppose T isuniformon[0,10],and 𝛿 = 0.05.Acontractpays1unitatfailure.Level premiums are payable continuously for 10 years. The premiums charged are the net premiums plus 20%. (a) Find the probability thatL ≤ 0. (b) Supposenowthat,insteadof1.2timesequivalenceprinciplepremiums,thecom- pany charges premiums at the rate of 0.05 per year. What is the probability that L ≤ 0? 16.8 A failure timeT is uniformly distributed on the interval [0, 20]. The force of interest is a constant 0.05. (a) Adeferredinsurancecontractprovidesforapaymentof1atthemomentoffailure providedthatthisoccursaftertime5.If ̄Z isthepresentvalueofthebenefits,find the 80th percentile of̄Z. That is, find the pointz such thatP(̄Z ≤ z) = 0.80. (b) Find the 80th percentile of̄Z when the contract is a term insurance policy that pays 1 unit if failure occurs in the first 5 years. 16.9 Consider a deferred term insurance contract. It provides for 1 unit payable at the moment of failure provided that this occurs betweenN and 2N years from now. Nothing is paid if failure occurs in the firstN years or after 2Nyears. You are given that the force of failure𝜇 and the force of interest𝛿 are constants such that𝜇 = 𝛿∕2. Moreover, you are given that e −N𝛿 = 0.36. If ̄Z is the present value of the benefits, calculateF̄Z(z) for all nonnegative values ofz. 258 SIMPLIFICATIONS UNDER LEVEL BENEFIT CONTRACTS 16.10 An insurance contract provides for e𝛿t payable at failure, should this occur at timet. Net level premiums are payable continuously prior to failure. The force of interest is a constant𝛿. Show that L = v T − ̄AT 1− ̄AT . 16.11 An insurance policy provides a benefit of 1 at the moment of failure provided this occurs after 10 years. The hazard rate of the failure time is a constant 0.10 and the force of interest is a constant 0.05. If̄Z is the present value of the benefits, find the probability that: (a) 0< ̄Z ≤ 0.5, (b) 0≤ ̄Z ≤ 0.5 and (c) 0≤ ̄Z ≤ 0.7. 16.12 An insurance policy provides a benefit of 1 at the moment of failure plus a pure endowmentof1attime10.Thisispurchasedbylevelpremiumspayablecontinuously for 10 years. The premium is the net premium plus 10%. There is a constant force or mortalityof0.05andaconstantforceofinterestof0.05.Find(a) P(−1∕3 < L ≤ 1∕2) and (b)P(−1∕4 ≤ L ≤ 1∕2). 16.13 Let T be any failure time, and assume constant interest. (a) Show that E[̄a(1 T )]2 = 2 𝛿(̄aT −2̄aT ), where the superscript 2 on the right-hand side indicates calculation at a force of interest equal to 2𝛿or equivalently withv replaced byv 2. (b) Is the following formula true? If not, give a correct version. E[̈a(1̃T )]2 = 2 d (̈ãT −2 ̈ãT ). 16.14 ModifytheformulainSection16.5.2inthecasethatthereisabenefitpaidforfailure at exact timen. 17 The minimum failure time 17.1 Introduction Supposethat T1,T2,…,Tm arefailuretimesdefinedonthesamesamplespace.Inthischapter, we investigate the random variable T =m i n{T1,T2,…,Tm}. Inotherwords, Tisthetimeofthefirstfailuretooccuramongthe mdifferentfailuretimesthat arepossible.Inparticular,thiswillrecastthematerialofChapters10and11intoastochastic framework, and show that both joint-life theory and multiple-decrement theory are special cases of this general problem. In addition, it will provide more rigorous arguments for some of the results of those chapters that were obtained in an intuitive fashion. Finally it will deal with the important cases where the failure times need not be independent. Inthejoint-lifecase,wherewehaveagroupof mlivesnumbered1,2, …,m,wecantake T i tobethefuturelifetimeofthe ithlife,sothat T isthefailuretimeofthejoint m-lifestatus. In the multiple-decrement context, we can takeTi to be the time of failure from causei in theassociatedsingle-decrement setting, so it is the failure time of theith cause, assuming no other causes of failure are operating. Then, the random variableT is the time of failure in the multiple-decrement model. (In the machine analogy of Section 11.6,Ti would be the failure time of theith part.) 17.2 Joint distributions Wewishtoexpandsomewhatonourbriefdescriptionforthecase m= 2giveninAppendixA. Supposethateach Ti iscontinuous.Therearevariouswaysofdescribingthejointdistribution. 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 260 THE MINIMUM FAILURE TIME We can do so by the joint density functionfT1T2,…,Tm(t1,t2,…,tm), or alternatively by the joint distribution function FT1,T2,…,Tm(t1,t2,…,tm) = P[T1 ≤ t1,T2 ≤ t2,…,Tm ≤ tm], or by the joint survival function sT1,T2,…,Tm(t1,t2,…,tm) = P[T1 > t1,T2 > t2,…,Tm > tm]. To simplify the notation, we will often omit the subscripts and just writef,F or s when no confusion arises. The reader is cautioned thats(t1,t2,…,tm) ≠ 1−F(t1,t2,…,tm) whenm> 1. Asintheone-dimensionalcase,weintegratetoobtainthedistributionorsurvivalfunctions from the density function, and differentiate to go in the other direction. We must, however, use multiple integrals and partial derivatives. For example, withm= 3, F(t 1,t2,t3) = ∫ t1 0 ∫ t2 0 ∫ t3 0 f(u,v,w)dwdvdu, and similarly s(t1,t2,t3) = ∫ ∞ t1 ∫ ∞ t2 ∫ ∞ t3 f(u,v,w)dwdvdu. Nowdifferentiatethelatterexpressionwithrespectto t1.Thefundamentaltheoremofcalculus tells us to replace the variableuin the integrand witht1 and affix a minus sign sincet1 is a lower limit. The integrand consists of the second two integrals. The result is 𝜕 𝜕t1 s(t1,t2,t3) =− ∫ ∞ t2 ∫ ∞ t3 f(t1,v,w)dwdv (17.1) After two more iterations of the procedure, we have f(t1,t2,t3) = (−1)3 𝜕 𝜕t1 𝜕 𝜕t2 𝜕 𝜕t3 s(t1,t2,t3). Similarly, we can derive f(t1,t2,t3) = 𝜕 𝜕t1 𝜕 𝜕t2 𝜕 𝜕t3 F(t1,t2,t3). Analogous expressions hold for generalm, which appears as the exponent of−1i nt h efi r s t formula. Note that the individual distributions are easily obtained from the the joint distribution or survival functions by sTi(t) = sT1,T2,…,Tm(0,0, …,t,…,0), FTi(t) = FT1,T2,…,Tm(∞,∞,…,t,…,∞), where thetis in theith position, and∞indicates that you take limits. THE JOINT DISTRIBUTION OF (T,J) 261 17.3 The distribution ofT 17.3.1 The general case It is a simple matter to deduce the distribution ofT from the joint distribution. Clearly, the minimum will take a value greater thant if and only if eachTi takes a value greater thant. Therefore, sT(t) = s(t,t,…,t). (17.2) (The reader should note thatFT(t) ≠ F(t,t,…,t).) 17.3.2 The independent case Let the density function, survival function, and hazard rate function ofTi be denoted respec- tively byfi,si,𝜇i. Things become much easier to deal with when the random variablesTi are independent. We can then readily write down the relevant functions forT in terms of the corresponding functions forTi. For example, from (17.2) we obtain sT(t) = s1(t)s2(t)⋯sm(t). (17.3) By taking logs and differentiating, we can find a similar relationship involving hazard func- tions. 𝜇 T(t) = m∑ i=i 𝜇i(t). (17.4) We have already encountered a particular case of (17.4) in (10.12). Example 17.1 Suppose thatT 1 andT2 are independent and both have the hazard function 𝜇(t) = 2 1−t,0 ≤ t< 1. Find the probability that the minimum value of these two random variables will be less than or equal to 1/2. Solution. We note thats T(t) = (1−t)4,f o r0≤ t< 1, either by first noting that, fori= 1,2, si(t) = (1−t)2 or by noting that𝜇T(t) = 4∕(1−t). The desired probability isFT(1∕2) = 1−sT(1∕2) = 15∕16. 17.4 The joint distribution of (T,J) 17.4.1 The distribution function for (T,J) Suppose that there is zero probability of the simultaneous occurrence of two or more failure times T 1,T2,…,Tm. We can then define the random variableJ as the index of the random 262 THE MINIMUM FAILURE TIME variable giving the minimum. For an example withm= 3, suppose T1 = 7, T2 = 5, and T3 = 10. ThenT would take the value 5, andJ would take the value 2, since the minimum occurs forT2. In many applications, we are interested in the joint distribution ofT and J. This can be described in various ways. One method is by the joint distribution function, FT,J(t,j) = P(T ≤ tandJ = j). Thisjointdistributionhasthesomewhatunusualfeaturethattherandomvariable Tisnormally continuous,while Jisdiscrete.Therefore,unliketheusualnotationfordistributionfunctions, the second variable inFT,J(t,j) is not cumulative, but refers to one specific index. We now consider the problem of deducing the joint distribution (T,J) from the joint distribution ofT1,T2,…,Tm. If we are given the joint density function, thenFT,J(t, j) is calculated by integrating over a suitable region – see (A.17). Takem= 2. ThenFT,J(t,1) is the probability thatT1 takes a value less than or equal tot, and thatT2 takes any value that is greater than that taken byT1. This is given by the double integral FT,J(t,1) = ∫ t 0 ∫ ∞ u fT1,T2(u,v)dvdu, (17.5) and similarly FT,J(t,2) = ∫ t 0 ∫ ∞ v fT1,T2(u,v)dudv. (17.6) Example 17.2 The joint distribution ofT1 andT2 is given by fT1,T2(s,t) = {6(s−t)2,0 ≤ s≤ 1,0 ≤ t≤ 1, 0, elsewhere . FindFT,J(t,j)f o rj= 1,2. Solution. FT,J(t,1) = ∫ t 0 ∫ 1 u 6(u−v)2dvdu= 2∫ t 0 (1−u)3du= 1−(1−t)4 2 ,f o r 0≤ t≤ 1. By symmetry, we must have FT,J(t,2) = 1−(1−t)4 2 ,0 ≤ t≤ 1. Since Ti is bounded above by 1 fori= 1,2, we necessarily have FT,J(t,i) = FT,J(1,i) = 1∕2, i= 1,2, t> 1. THE JOINT DISTRIBUTION OF (T,J) 263 The sum ofFT,J(t,1) and FT,J(t,2) must of course equalFT(t) = 1−sT(t). It will be instructive for the reader to verify this by drawing a picture in the plane, showing that the unionoftheregionsofintegrationin(17.5)and(17.6),andtheregioncorrespondingto s T(t), is the entire positive quadrant. In the general case, we will needm-dimensional integrals to computeF(t,j)f r o mt h e densityfunction.However,ifwealreadyhavethejointsurvivalfunction,thecomputationcan be simplified. To illustrate, consider the case withm= 3. Then, reasoning as above, F T,J(t,1) = ∫ t 0 ∫ ∞ u ∫ ∞ u f(u,v,w)dwdvdu. From(17.1)theinnertwointegralscanbewrittencompactlyasapartialderivative.Wehave FT,J(t,1) = ∫ t 0 𝜎(u)du, where 𝜎(u) =− 𝜕 𝜕t1 s(t1,t2,t3), evaluated at t1 = t2 = t3 = u. In the general case, FT,J(t,j) = ∫ t 0 𝜎j(u)du, (17.7) where 𝜎j(u) =− 𝜕 𝜕tj s(t1,t2,…tm), evaluated at t1 = t2 = ⋯ = tm = u. Example 17.3 Suppose thatm= 2 and the joint survivor function is given by sT1,T2(u,v) = 1 2[(1−u)4 +(1−v)4 −(u−v)4], 0 ≤ u≤ 1, 0≤ v≤ 1. FindFT,J(t,1). Solution. We could take two derivatives to calculatef(u,v) = 6(u−v)2, and apply (17.5). (ThisisinfactthesamedistributionasinExample16.2.)Note,however,that(17.5)wouldjust ‘undo’thecalculationofthesecondderivativebyintegrating.Forthisformofthedistribution, it is easier to apply (17.7). On the given region − 𝜕 𝜕us(u,v) = 2[(1−u)3 +(u−v)3], 𝜎1(u) = 2(1−u)3, 264 THE MINIMUM FAILURE TIME so FT,J(t,1) = ∫ t 0 2(1−u)3du= 1−(1−t)4 2 ,0 ≤ t≤ 1, verifying the previous example. 17.4.2 Density and survival functions for (T,J) We can also define the distribution of (T,J) by the joint density functionfT,J(t,j). This is the function satisfying FT,J(t,j) = ∫ t 0 fT,J(s,j)ds, fT,J(t,j) = d dtFT,J(t,j). (17.8) The functionf is interpreted in the normal way. Namely, for ‘small’Δt,fT,J(t,j)Δtis approx- imately the probability that the first failure will be from causejand that it will take place in the time interval fromtto t+Δt. The precise statement, following from the first expression in (17.8), is that the probability that the first failure will be from causejand will take place between timeaand timebis given by FT,J(b,j)−FT,J(a,j) = ∫ b a fT,J(t,j)dt. Wedefinethejointsurvivalfunctionfor( T,J)bythinkingofsurvivalaswedidattheend of Section 11.2.2. Let sT,J(t,j) = P(T > tandJ = j) = ∫ ∞ t fT,J(s,j)ds. This is the probability that failure will occur after timetdue to causej. Figure 17.1 shows a typical graph off(t,j)f o rm= 2. Note that the mass is concentrated on parallel sheets. If we cut thejth sheet by the planeT = t, the area of the left portion will be F(t,j) and the area of the right portion will bes(t,j). Allfunctionspertainingtotherandomvariable T aloneareobtainedbysummingoverall j, as we noted above forFin the casem= 2. That is, FT(t) = m∑ j=1 FT,J(t,j), (17.9) fT(t) = m∑ j=1 fT,J(t,j), (17.10) sT(t) = m∑ j=1 sT,J(t,j). (17.11) THE JOINT DISTRIBUTION OF (T,J) 265 fT,J (t, 2) fT,J (t, 1) j =2 j =1 t j Figure 17.1 The graph offT,J(t,j) 17.4.3 The distribution of J To obtain the distribution ofJ from the joint distribution, we compute the other marginal, which can be expressed in various ways. IffJ(j) denotes the probability thatJ = j, then fJ(j) = ∫ ∞ 0 fT,J(t,j)dt= sT,J(0,j) = limt→∞FT,J(t,j) (17.12) In Figure 17.1,fJ(j) is the area of thejth sheet. Note thatFT,J(t,j)+sT,J(t,j) is not equal to 1, but rather tofJ(j). Example 17.4 Takem= 2.If T1 isuniformon[0,1], T2 isuniformon[0,2],and T1 andT2 are independent, find the distribution ofJ. Solution. The joint density function takes a constant value of 1/2 on the rectangle 0≤ s≤ 1,0 ≤ t≤ 2, and is 0 elsewhere. The maximum value ofT is the minimum of the respective maximums of theTi, which in this case is 1. Therefore, fJ(1) = FT,J(1,1) = ∫ 1 0 ∫ 2 u (1∕2)dvdu= 3 4, fJ(2) = FT,J(1,2) = ∫ 1 0 ∫ 1 v (1∕2)dudv= 1 4. 266 THE MINIMUM FAILURE TIME 17.4.4 Hazard functions for (T,J) Definition 17.1 The hazard function for (T,J) is given by 𝜇T,J(t,j) = fT,J(t,j) sT(t) . This is a conditional density. For smallΔt, 𝜇T,J(t,j)Δt is approximately the probability that failure will occur first from causejin the time interval fromttot+Δt, given that failure fromanycause has not yet taken place before timet. In the Chapter 11 multiple-decrement model for a life agex, 𝜇T,J(t,j) corresponds to 𝜇(j) x (t). Given the hazard rates, we can obtain the joint distribution of (T,J) by the same method as employed in Chapter 11. From (17.10) and the definition of𝜇(t,j), 𝜇T(t) = m∑ j=1 𝜇T,J(t,j), (17.13) and then, from (15.7), sT(t) = e−∫ t 0 𝜇T(r)dr. We know thatfT,J(s,j) = sT(s)𝜇T,J(s,j) and, from the first expression in (17.5), FT,J(t,j) = ∫ t 0 sT(s)𝜇T,J(s,j)ds. (17.14) The last formula is easily explained intuitively. For the event in question to occur, there must be some points, beforet, for which failure from any cause has not yet occurred, and then failure will occur from thejth cause at timesAlthough (17.14) has this intuitive appeal, it is not necessarily useful for computingFT,J(t,j) as we may not know the joint hazard rates untilwehavealreadycomputed FT,J(t,j)and fT,J(t,j).Itis,however, animportantformulain the independent case to which we now turn. 17.4.5 The independent case As in Section 17.2, we can simplify calculations when theTi are independent, and deduce the distribution of (T,J) directly from the individual distributions of eachTi. Sinces′ j (t) = −sj(t)𝜇j(t), (17.3) shows that 𝜎j(t) = ∏ i≠j si(t)s′ j (t) = sT(t)𝜇j(t) THE JOINT DISTRIBUTION OF (T,J) 267 and then from (17.7), FT,J(t,j) = ∫ t 0 sT(u)𝜇j(u)du, for independent Ti. (17.15) Note, as a comparison to (17.14.), that (17.15) can be used directly to computeFT,J(t,j) in the independent case, when we know𝜇i from the individual distributions. Moreover, differentiating and dividing bysT(t) verifies that in the independent case 𝜇T,J(t,j) = 𝜇j(t) (17.16) forj= 1,2, …mandall tforwhich sT(t) ≥ 0,Thisprovidesthepromisedprooffortheresult stated in in formula (11.23) . The result is easily explained intuitively. Looking at the machine model of Section 11.6, forexample,bothquantitiesin(17.16)giveaconditionaldensityforfailureofpart jattime t. Inthecaseof 𝜇T,J(t,j),theconditionisthatallpartshavesurviveduptotime tandingeneral this may give information regarding the failure time of partj. Suppose, for example, that the parts are connected so that part 2 cannot fail until part 1 does, and then it fails five seconds later.(Thesameideainthemultipledecrementmodelprovidedthesimplecounter-exampleof Exercise11.15.)Intheindependentcase,however,weobtainexactlythesameinformationas ifwetoldonlythatpart jhassurviveduptotime t,whichispreciselytheconditionapplicable to𝜇 j(t). Wehavealreadyencounteredspecialcasesof(17.15).Oneexampleis(10.28).Thehazard rate 𝜇x(t) corresponds to𝜇1(t) which equals𝜇T,J(t,1) since we postulated independence. Another example is (11.13). Example 17.5 Suppose that theT i are independent and exponential with constant hazard 𝜇i. (a) FindFT,J(t,j). (b) FindfJ(j). (c) Show thatT andJare independent. Solution. (a) Let𝜇 = 𝜇1 +𝜇2 +⋯+𝜇m.Formula(17.4)showsthat Tisexponentialwithconstant hazard𝜇, and, from (17.15), FT,J(t,j) = ∫ t 0 e−s𝜇𝜇jds= 𝜇j 𝜇 (1−e−𝜇t). (b) fJ(j) = lim t→∞ FT,J(t,j) = 𝜇j 𝜇 . (c) We just note that FT,J(t,j) = FT(t)fJ(j). The solution to (b) gives an important result that has many applications. It says that for independent exponential failure times, the probabilities of first failure are proportional to the hazardrates.Thismakessense,sincethehigherthehazardrate,thelowerthemean,whichis the reciprocal of the hazard rate, and therefore more likelihood of occurring first. 268 THE MINIMUM FAILURE TIME Example 17.6 Takem= 2. SupposeT1 and T2 are independent,T1 is uniform on [0,a], andT2 is uniform on [0,b], where 0< a≤ b.F i n dFT,J(t,1) andFT,J(t,2). Solution. Since sT(t)𝜇1(t) = s2(t)s1(t)𝜇1(t) = s2(t)f1(t), we can write FT,J(t,1) = ∫ t 0 ( 1− s b ) 1 ads= t a − t2 2ab = F1(t)− 1 2F1(t)F2(t), 0 < t< a. Similarly, FT,J(t,2) = t b − t2 2ab = F2(t)− 1 2F1(t)F2(t), 0 < t< a. Since failure must take place before timea, for anyt≥ a, FT,J(t,1) = FT,J(a,1) = 1− a 2b, FT,J(t,2) = FT,J(a,2) = a 2b As a check, note thatFT,J(t,1) +FT,J(t,2) = F1(t)+F2(t)−F1(t)F2(t), which must be true, since forT to take a value less thantmeans that at least one ofT1 and T2 takes a value less thant. Also note that the answer here could have been written down immediately, since thedistributionssatisfythestochasticversionofthecondition givenin(11.24),andtherefore we can apply Method 2 of Chapter 11. 17.4.6 Nonidentifiability Wemotivatetheideaofthissectionbyanexample.Firstnotethatthejointdistributionsgiven in Examples 17.1 and 17.2 are easily seen to be different. One way is to note thatT 1 and T2 are not independent in the latter. Example 17.7 Calculate F T,J(t,j) for the distribution given in Example 17.1 Solution. We could do this by (17.15), but it is easier to note that, by symmetry,FT,J(t,1) = FT,J(t,2)andthetwomustsumto FT(t) = 1−sT(t).WecanconcludedirectlyfromExample 17.1 that FT,j(t,1) = FT,J(t,2) = 1−(1−t)4 2 ,0 ≤ t≤ 1, and necessarily FT,J(t,1) = FT,J(t,2) = 1 2,1 ≤ t. ComparetheaboveresultwithExample17.2.Thesomewhatsurprisingconclusionisthat two completely different distributions for (T1,T2) have led to exactly the same distribution for (T,J). This is known as thenonidentifiabilityproblem and it has statistical implications. Supposewewanttomakeinferencesaboutthejointdistributionofthe Ti byobservingfailure THE JOINT DISTRIBUTION OF (T,J) 269 times. In many cases, all we can possibly observe is the joint distribution (T,J). An example is when the random variables represent the time of death from various causes. Once death occurs, we know the time and the cause, but no further observation of the subject is possible. Our example above shows that it is impossible to uniquely determine the joint distribution of the random variables that give rise to a given (T,J). We need additional information in order to obtain a unique solution. One instance when this occurs is in the independent case. Theorem 17.1 Givenanyjointdistributionfor (T,J) thereisauniquejointdistributionof (T 1,T2,…,Tm)suchthatthe (Ti)aremutuallyindependentandinducethegivendistribution of(T,J). Proof. Uniqueness follows immediately, since, given independence, we know the joint dis- tributionifweknowthedistributionofeach Ti,and(17.16)impliesthateach Ti isnecessarily a random variable with hazard function𝜇T,J(t,i), which is determined uniquely from (T,J). For the existence, given any joint distribution functionFfor (T,J), we letTi be a random variable with hazard function𝜇T,J(t,i). This collection of independentTi in turn generates a joint distribution function̂FT,J. From (17.15), ̂FT,J(t,j) = ∫ t 0 sT(s)𝜇T,J(s,j)ds, which equalsFT,J(t,j) as shown by (17.14). To illustrate the use of this theorem, suppose you are told that FT,J(t,i) = 1−(1−t)4 2 ,0 ≤ t≤ 1, FT,J(t,1) = 1, t> 1 fori= 1,2 and asked to identify the joint distribution of (T1,T2). You cannot do this without further information, for it could be either the joint distribution of Example 16.1 or that of Example 16.2, or indeed several other possibilities. However, if you are given the additional information thatT 1 andT2 are independent, then you know that it must be the distribution of Example 16.1. 17.4.7 Conditions for the independence ofT andJ Another question of interest is to determine whenT andJare independent. In Example 17.5, we saw that this occurred with constant hazard functions. We present here a more general criterion. Define the ratios K(t,j) = 𝜇(t,j) 𝜇T(t) for allj, and allt, such thatsT(t) > 0. Theorem 17.2 K(t,j) = P(J = j|T = t).Therefore, T and J areindependent ifand onlyif K(t,j) isindependentoft. 270 THE MINIMUM FAILURE TIME Proof. We have fT,J(t,j) = sT(t)𝜇(t,j) = K(t,j)sT(t)𝜇T(t) = K(t,j)fT(t). So P(J = j|T = t) = fT,J(t,j) fT(t) = K(t,j). The condition of this theorem is sometimes expressed by saying that the hazards for the individual causes are fixed proportions of the total hazard. 17.5 Other problems Thereareseveralotherquestionsregardingthejointdistributionof( T1,T2,…,Tm)thatcanbe answeredbysimilartechniquestothoseinSection17.3.Thatis,wefindacertainprobability byintegratingfailuretimesoverasuitableregionof m-dimensionalspace.Asanexample,we illustrate the method for a problem analogous to Example 10.7 of Section 10.9. Takem= 2, and consider the probability that both causes of failure will occur within a specified duration of each other. That is, for some fixedn, we want the probability that (|T 1 −T2| ≤ n). It will normally be easier to compute this as 1−P(|T1 −T2| > n) = 1−[P(T2 > T1 +n)+P(T1 > T1 +n)]. Each term is found by integrating the joint density function over a suitable region in the plane. For example, P(T2 > T1 +n) = ∫ ∞ 0 ∫ ∞ u+n fT1,T2(u,v)dvdu. Example 17.8 Find P(|T1 −T2| ≤ n), whenT1, T2 are independent, and both are expo- nential with hazard functions𝜇1 and𝜇2, respectively. Solution. The integral above reduces to 𝜇1 𝜇1 +𝜇2 e−n𝜇2, so the final answer is 1− 𝜇1 𝜇1 +𝜇2 e−n𝜇2 − 𝜇2 𝜇1 +𝜇2 e−n𝜇1. THE COMMON SHOCK MODEL 271 17.6 The common shock model In many applications, we have a group of objects whose future lifetimes are generally inde- pendent, except that they are all subject to a common hazard, which will result in the failure of all, should it occur. In the case of human lives, this could be a natural disaster such as a hurricane. In the case of machine parts, it could be something like an electrical problem that affects all components at once. The presence of the common shock introduces dependence into what would otherwise be independent future lifetimes. Tomodelthegeneralsituation,wehave m+1independent,continuousrandomvariables, (T ∗ 1,T∗ 2,…,T∗ m,Z), and for eachiwe let Ti =m i n(T∗ i ,Z). Theinterpretationisthat T∗ i isthetimeuntilfailureofthe ithobjectforreasonsotherthanthe common shock, andZis the time until the common shock occurs. It follows then thatTi will be simply the time until failure of theith object, since such failure will occur at either time T∗ i or timeZ, whichever is earlier. In the remainder of this section, we will confine ourselves to the case wherem= 2. Quantities referring toT∗ i will have a superscript∗. We are interested in questions about the joint distribution (T1,T2), which involves depen- dentrandomvariables.However,inmanycases,wecananswerthesequestionsbyconsidering theindependent collection (T ∗ 1,T∗ 2,Z).Wewillillustratewithseveral examples. Akey factto note is that T =m i n(T1,T2) =m i n(T∗ 1,T∗ 2,Z), since both give the time of first failure. Example 17.9 Find a formula for the probability that both objects will survive to timet. Solution. This is justs ∗ 1 (t)s∗ 2 (t)sZ(t). Example 17.10 What is the probability that failure will occur as a result of the common shock? Solution. This is justP(J = 3) in the joint distribution of (T,J), whereJ takes the values 1, 2, 3 andT is the minimum ofT ∗ 1,T∗ 2 andZ. Example 17.11 What is the probability that the second failure will occur before timet? Solution. We divide this up into two mutually exclusive cases. It will always occur ifZ ≤ t. IfZ > t, we need bothT∗ 1 andT∗ 2 less than or equal tot. The probability is FZ(t)+sZ(t)F∗ 1(t)F∗ 2(t) = 1−sZ(t)[s∗ 1 (t)+s∗ 2 (t)−s∗ 1 (t)s∗ 2 (t)]. 272 THE MINIMUM FAILURE TIME Other problems are not so straightforward and require special attention. The joint distri- bution of (T1,T2) is quite different from the typical two-dimensional continuous distribution. Itstilliscontinuous,butithasamassofpositiveprobabilityallconcentratedonasingleline, namely the diagonal, since the occurrence of the common shock will cause failure from both causes, leading toafailurepoint of theform (t,t).Indetermining theprobability that (T 1,T2) lies in some regionA, we will in general have to breakAup into three pieces, the part that is above the diagonal, the part that is below the diagonal, and the part that is on the diagonal. We adopt the convention that the value ofT1 is on the horizontal axis. For the part of the plane above the diagonal, {(u,v): u< v}, we use the joint density function f∗ 1 (u)f2(v), since the only wayT1 can take a valueu< vis ifT∗ 1 took the valueu. In other words, failure fromcause1attime udidnotoccurfromthecommonshock,sinceifitdid,thenfailurefrom cause 2 would also have occurred at timeuand could not have occurred at the later datev. Similarly,forthepartofthepositivequadrantbelowthediagonal,{( u,v): v< u},weuse the joint density function f1(u)f∗ 2 (v). Since Ti =m i n(T∗ i ,Z), the densitiesfi,i= 1,2, are easily calculated as fi(t) =− (s∗ i sz)′(t) = f∗ i (t)sz(t)+s∗ i (t)fz(t). (17.17) Failure on the diagonal arises if and only if the occurrence of the common shock occurs before the other two causes. We use the one-dimensional density function, fT,J(t,3) = s∗ 1 (t)s∗ 2 (t)fZ(t), andprojectthediagonalontotheline.Thatis,tofindtheprobabilitythatfailuretookplaceat a point (t,t), wherea≤ t≤ b, we integrate this density fromatob. Example 17.12 Suppose T ∗ 1, T∗ 2, andZ are exponential with hazard functions𝜇1, 𝜇2, and 𝜌, respectively. Consider the event thatT1 and T2 are both less than or equal ton. This can be subdivided into three cases according as (a)T1 < T2,( b) T2 > T1,( c) T1 = T2.F i n dt h e probability of each case. Solution. (a) Note first that, from (17.17), we can calculate f 2(t) = (𝜇2 +𝜌)e−(𝜇2+𝜌)t, so that, using the above-diagonal joint density, the required probability is 𝜇1(𝜇2 +𝜌)∫ n 0 ∫ n u e−𝜇1ue−(𝜇2+𝜌)vdvdu, COPULAS 273 which equals −e−n(𝜇2+𝜌) + 𝜇1 𝜇1 +𝜇2 +𝜌 + 𝜇2 +𝜌 𝜇1 +𝜇2 +𝜌e−n(𝜇1+𝜇2+𝜌). (b) Similarly, the required probability in this case is −e−n(𝜇1+𝜌) + 𝜇2 𝜇1 +𝜇2 +𝜌 + 𝜇1 +𝜌 𝜇1 +𝜇2 +𝜌e−n(𝜇1+𝜇2+𝜌). (c) The required probability is ∫ n 0 𝜌e−𝜌te−𝜇1te−𝜇2tdt= 𝜌(1−e−n(𝜇1+𝜇2+𝜌)) 𝜇1 +𝜇2 +𝜌 . The sum of these three cases is 1−e−n(𝜇1+𝜌) −e−n(𝜇2+𝜌) +e−n(𝜇1+𝜇2+𝜌), as we can verify from the general formula given in Example 16.12. 17.7 Copulas This section, like the previous one, is concerned with situations where there is a lack of independence. We present a general method that is often used to deal with this. Attention is confined to the casem= 2. A joint distribution (T 1,T2) can be thought of as having two ingredients. One is the distributionsofthetwo-componentrandomvariables,andtheotheristhewayinwhichthese are linked together. The latter can be described by a device known as acopula, which can then be applied to an arbitrary pair of individual distributions. The copula provides a means of dealing with these two ingredients separately. To elaborate, we start with the observation that wheneverT 1 andT2 are independent, we immediately recover the joint distribution from the individual distributions by the rule FT1,T2(t1,t2) = FT1(t1)FT2(t2). (17.18) We can then ask whether we can replace the multiplication on the right-hand-side of (17.18) by other transformations, and still obtain a joint distribution – that is, ifIdenotes the unit interval [0,1], whether we can find a functionC from I×I to itself, so that we obtain a legitimate joint distribution by the rule FT1,T2(t1,t2) = C(FT1(t1),FT2(t2)). (17.19) 274 THE MINIMUM FAILURE TIME We need some restrictions on the functionC. Take any points in I.I fT1 > s, so that FT1(s) = 0,then,forall t,FT1,T2(s,t) = 0.Thesameholdsfor T2,leadingtotheconditionthat for allu,vinI, C(0,v) = C(u,0) = 0. (17.20) If T1 ≤ s, so thatFT1(s) = 1, then for allt, FT1,T2(s,t) = FT2(t), leading to the condition that for allu,vinI, C(1,v) = v, C(u,1) = u. (17.21) Another requirement stems from the fact that probabilities cannot be negative. For any sub-rectangle R⊆ I×I, the probability that (T1,T2) lies inRis just the sum of the values of FT1,T2 on the northeast and southwest corners, minus the sum of the values on the other two corners. Since this is nonnegative, it follows that C(u2,v2)+C(u1,v1)−C(u1,v2)−C(u2,v1) ≥ 0. (17.22) wheneveru1 ≤ u2 andv1 ≤ v2. These are the only conditions we need and we can now state the formal definition. Definition 17.2 Acopulais a functionCfromI×ItoIsatisfying (17.20)–(17.22). Itcanbeshownthatif Cisacopula,then(17.19)givesavalidjointprobabilitydistribution forany T1 andT2.Conversely(andhardertoshow),foranyjointdistribution(T 1,T2)thereis a copulaCsuch thatFT1,T2 is given by (17.19). IfT1 andT2 arebothuniformdistributionson I,then FTi(u) = ufori= 1,2,andall u∈ I, from which it follows that FT1,T2(u,v) = C(u,v). This shows that as an alternate definition, we can simply define a copula as a distribution function of a joint distribution involving two random variables that are uniform onI. The following are three simple examples of copulas: 1. C(u,v) = uv. This is just the copula for an independent distribution, as mentioned; 2. C(u,v) =m i n(u,v); 3. C(u,v) =m a x(u+v−1,0). Copulas 2 and 3 are extreme in the sense that for any copulaCand for allu,v,∈ I, max(u+v−1,0) ≤ C(u,v) ≤ min(u,v). They are also extreme in the following sense. Consider all possible joint distributions for a givenT 1 and T2. In many cases, one is interested in the sumT1 +T2. For example, an insurer sells two insurance contracts andTi denotes the claim on theith policy, or a person COPULAS 275 buys two stocks andTi is the value of theith stock at some future date. We may want to compare all possible joint distributions as to their degree of risk. We will not go into the details of comparing joint distributions as to risk here. (In Section 22.4, we do introduce this idea for single variable distributions). However, it seems clear that risker possibilities arise when when large values of one random variable tend to go with large values of the other so there is a tendency for either both values to be large or both to be small. The less riskier possibilities arise when large values of one tend to go with small values of another, so there is a possibility for bad results in one case to be balanced by good results in the other. It can be shown, that under some natural risk comparing criteria, copula 2 will give the most risky jointdistributionandcopula3theleastriskyjointdistribution.Thispointisillustratedfurther in Exercise 17.14. We deal only briefly with problems of choosing a copula to model a given situation. In many cases, the modeler likes to choose a copula from a parametric family, and select the parameter to suit certain conditions. A popular choice for this is Frank’s family of copulas given by C Θ(u,v) = 1 Θ log ( 1+ (eΘu −1)(eΘv −1) eΘ −1 ) , whereΘcan be any nonzero real number. It can be shown, using L’Hˆopital’s rule, that for all u,v,∈ I, lim Θ→0 CΘ(u,v) = uv, so that the smaller the parameter is in absolute value, the greater the extent of independence between the two random variables, with full independence occurring forΘ=0. An interesting feature which some copulas, but not all, have is that C(u,v) = u+v−1+C(1−u,1 −v). (17.23) It is straightforward to verify this property for copulas 1–3 above. It is true, but harder to verify, that this holds for Frank’s family. The significance of (17.23) is that s T1,T2(t1,t2) = 1−FT1(t1)−FT2(t2)+FT1,T2(t1,t2) = C(sT1(t1),sT2(t2)). In other words, the same transformation rule can be applied to either distribution or survival functions. Example17.13 SupposethatDemoivre’slawholdswith 𝜔 = 100.Considertwolives(60) and (70). Find the probability that both will be alive at the end of 10 years, assuming each of the three basic copulas given above, for the joint distribution ofT(60) andT(70). Solution. Theindividualsurvivalprobabilitiesare3/4and2/3,sothesurvivalofthejoint-life status is in the respective cases: 1. 1/2 as we known already from Chapter 10; 276 THE MINIMUM FAILURE TIME 2. min{3∕4,2∕3} = 2∕3. This copula applied to a joint-life status just means that the younger life will die at exactly the same time as the older; 3. (3∕4+2∕3−1) = 5∕12. For a particular application of copulas, refer back to the nonidentifiability problem of Section17.4.6.Insteadofassumingindependence,wemightpostulateacertaincopula Cand thenaskforajointdistributionwiththechosencopulathatgivesrisetothegivendistribution (T,J). In many cases this will be unique. Notes and references Nelsen (1999) is a good general source for additional material on copulas, including full derivationsoftheunprovedresultsthatwehavegiven.FreesandValdez(1998)discussvarious actuarial applications of copulas. Carri´erre (1994b) discusses the application of copulas to the nonidentifiability problem in multiple-decrement theory. For methods of comparing two- variable distributions for risk, see Shaked and Shantikumar (2007), Chapter 6. Exercises Type A exercises 17.1 A joint distribution is given by fT1,T2(s,t) = ⎧ ⎪ ⎨ ⎪ ⎩ 3s,0 ≤ s≤ t≤ 1, 3t,0 ≤ t≤ s≤ 1, 0, elsewhere . FindF T,J(t,1) andFT,J(t,2). 17.2 A joint survival function is given by s(u,v) = {1−(3∕2)(u2 +v2)+(1∕2)u3 +(3∕2)uv2,0 ≤ v≤ u≤ 1, 1−(3∕2)(u2 +v2)+(1∕2)v3 +(3∕2)vu2,0 ≤ u≤ v≤ 1. FindFT,J(t,1). 17.3 Suppose that T1 andT2 are independent with p.d.f.’s f1(r) = 2e−2r, f2(r) = 3e−3r. (a) FindFT,J(t,1) andFT,J(t,2). (b) Find the distribution of J. 17.4 Suppose that T1 and T2 are independent. T1 has an exponential distribution with constant hazard rate𝜇. T2 is uniform on [0,a]. Find (a)FT,J(t,2),(b) P(J = 2). EXERCISES 277 17.5 The failure timesT1 andT2 are independent, and have respective hazard functions 𝜇1(t) = 3 2−t,0 ≤ t< 2, 𝜇2(t) =l o g(2). Find the probability that the minimum value of these two random variables will be less than or equal to 1. 17.6 A machine is subject to two independent causes of failure. The time of the first cause of failure is uniformly distributed on the interval [0, 4]. The time of the second cause of failure is uniformly distributed on the interval [0, 5]. Find the probability that: (a) the machine will fail from cause 1 before time 3; (b) the machine will eventually fail from cause 2. 17.7 Twofailuretimes T 1 andT2 haveajointdistributiongivenbythejointdensityfunction fT1,T2(u,v) = ⎧ ⎪ ⎨ ⎪ ⎩ 4(v−u) 2,0 ≤ u≤ v≤ 1, 4(u−v), 0 ≤ v< u≤ 1, 0, elsewhere . Find: (a)FT,J(t,1) andFT,J(t,2); (b) the distribution ofJ;( c )𝜇(t,1) and𝜇(t,2). 17.8 Twofailuretimes T1 andT2 haveajointdistributiongivenbythejointdensityfunction fT1,T2(u,v) = ⎧ ⎪ ⎨ ⎪ ⎩ (8∕3)v 2,0 ≤ u≤ v≤ 1, (8∕3)uv,0 ≤ v< u≤ 1, 0, elsewhere . (a) FindFT,J(t,1) andF−T,J(t,2). (b) Find the distribution ofJ. (c) Suppose that ̂T1 and ̂T2 are two independent variables whose joint distribution leads to the same distribution of (T,J) as you found in part (a). What are the hazard functions of̂T1 and ̂T2? Type B exercises 17.9 ForthejointdistributiongiveninExample17.2,find 𝜇(t,j)and 𝜇j(t)for j= 1,2.Show that these are not the same. 17.10 ForthejointdistributiongiveninExercise17.2,find 𝜇(t,j)and 𝜇j(t)for j= 1,2.Show that these are not the same. 17.11 Consider independent failure timesT1 and T2 where the hazard rate ofT1 is 𝛼∕(1− t),0 ≤ t< 1, and the hazard rate ofT2 is 𝛽∕(1−t),0 ≤ t< 1. Answer the following in terms of𝛼 and𝛽. (a) FindFT,J(t,1). 278 THE MINIMUM FAILURE TIME (b) FindP(J = 2). (c) An insurance contract provides for a benefit at the moment of the first failure, provided that this is due to ‘cause 1’ (i.e. provided thatT1 < T2). The amount of the benefit for failure at timetis (1−t)e0.1t. The force of interest𝛿 is a constant 0.10. Find the expected present value of the benefits. 17.12 Twolivesage( x)and( y)aresubjecttothecommonshockmodel. T∗(x)hasaconstant hazard rate of 0.06,T∗(y) has a constant hazard rate of 0.04, andZ has a constant hazard rate of 0.02. The force of interest is a constant 0.05. (a) Calculate ̄A x(110)andwriteitasasumofthreeterms,namely,theexpectedpresent value of benefits when: (i) (x) dies at a time strictly before the death of (y); (ii) (x)diesattimestrictlyafterthedeathof( y);(iii)(x )diesasresultofthecommon shock. (b) Calculate ̄Axy. 17.13 You are given a common shock model (T∗ 1,T∗ 2,Z) whereT∗ 1 has the survival function s(t) = 1−0.1t,T∗ 2 hasaconstanthazardfunctionof0.04,and Zhasaconstanthazard function of 0.02. Find the probabilities that (a)T1 < T2,( b )T2 < T1,( c )T1 = T2. 17.14 Suppose that T1 and T2 each take the values 0 with probability 1/2, and 1 with probability1/2.Calculatetheprobabilityfunctionforthejointdistributionof( T1,T2) under each of the following copulas: (a)C(u,v) =m i n(u,v), (b)C(u,v) =m a x(u+ v−1,0), (c)C(u,v) = uv, (d) Frank’s copula with values of𝜃 = 0.01,50, −50. What happens when𝜃 approaches∞or−∞? 17.15 A joint life insurance on (x) and (y) has death benefits which are constant over each year.Assumethateithercopulas2or3ofSection17.7applytothejointdistributionof T(x) andT(y). Show that unlike the independent case, the termRin equation (10.15) is equal to 0. PART III ADVANCED STOCHASTIC MODELS

Human: what is the most likely time of failure?

Actuarial AI: 14 Survival distributions and failure times 14.1 Introduction to survival distributions Our goal in this part of the book is to introduce a stochastic model for mortality to replace the deterministic model used in Part I. This will not only provide us with a more realistic description of human mortality, but it will also have more general applications. The basic information a prospective issuer of an insurance or annuity contract wants to know is how long the life in question will live. The insurer obviously cannot hope to answer this question exactly, since the actual future lifetime lived is random. Some people age 50, for example, will live another 40 years or more, while others will die very soon. In the deterministicmodel,wecircumventedthisissuebyassumingthatwhilewecouldnotidentify how long a particular individual would live, we could identify how many individuals of a given age would live to some other age. Clearly, however, the number of such individuals is also random. In the stochastic model we will face this randomness directly. Thisandsubsequentchapterswillrequireamoreadvancedknowledgeofprobabilitythan we have assumed so far. We follow the notation and terminology of Appendix A. For the present chapter, see in particular Sections A.4–A.8 and note thatP will denote probability. We do not need to confine ourselves to looking at the time of death of an individual. Suppose we are interested in some event that will occur once and only once at some random future time. We will refer to this event as ‘failure’. The random variableT, the time of occurrence of such an event, is known as afailure time, and its distribution is often referred to as asurvival distribution. At any time before ‘failure’ we will say we are in a state of ‘survival’. Our motivating example is the case where the event in question is the death of (x). In this case we will denoteT by T(x). 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 212 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES There are, however, many other examples of interest. Suppose a manufacturer sells a product with a guarantee that it will be replaced it if fails before a certain time. In order to assess the cost of the guarantee the manufacturer wants to know the distribution of the product’s failure times. We see that a guarantee can be viewed as type of insurance policy. WehavealreadyencounteredmoregeneralfailuretimesinChapter10.Joint-lifeinsurance and annuities were seen to be the same type of contracts as those for single lives except that the failure time was defined to be the time of first death. Similarly, for last-survivor annuities orsecond-deathinsurances,thefailuretimewasthetimeofseconddeath.Multiple-decrement theory provided many more examples of failure times. 14.2 The discrete case Consider the case where failure can occur only at integer times 1,2,3,…,s oT is a discrete randomvariablewithpositiveintegersasvalues.ReferinparticulartoSectionA.4.Inplaceof thecumulativedistributionfunction F,itisoftenmoreconvenienttousethe survival function s defined by s(k) = 1−F(k) = P(T > k). This gives the probability that failure has not yet occurred by timek, or in other words that we are still in a state of survival at timek.I ff is the probability function ofT, it is clear from (A.5) that f (k) = s(k −1)−s(k), s(k) = ∞∑ i=k+1 f (i) = 1− k∑ i=1 f (i). (14.1) There is another important method of describing the distribution ofT. Definition 14.1 The hazard function ofT at timek, denoted by𝜆(k) is the condition proba- bility of failure at timek given survival up to timek −1. That is, 𝜆(k) = f (k) s(k −1). This of course is defined only for those integersk such thats(k −1) > 0. Onces(k −1) equals0,thereisnopossibilityofsurvivaluptothatpoint.(Thehazardfunctionisalsoknown as the hazardrate or intensity function orfailure function.) Readersshouldmakesurethattheyunderstandthedifferencebetween f (k)and 𝜆(k).Both quantities give the probability of failure at timek, but from different perspectives. If, at time 0,weareaskedtoassessthelikelihoodthatfailurewilloccurattime k,theanswerisjust f (k). As time goes on, our assessment of this likelihood must change. For example, if failure takes place before timek, then we know that the probability of failure at timek is zero. Suppose that at timek −1, failure has not yet occurred, and we ask ourselves the same question. The THE CONTINUOUS CASE 213 answer now is𝜆(k). The difference between the two functions is further clarified by writing the definition of𝜆in the form f (k) = s(k −1)𝜆(k), (14.2) expressing the fact that we view failure at timek as arising from two events. First, there must be survival up to timek −1, and then, given this survival, failure must occur at timek. In deriving an appropriate survival distribution, it is often easier to model the hazard rate rather than s or f .G i v e n𝜆, we can then easily determine the other functions as follows. Equating the two different expressions forf (k) given in (14.1) and (14.2), s(k) = s(k −1)[1−𝜆(k)]. (14.3) Beginning withs(0) = P(T > 0) = 1, we know thats(1) = 1−𝜆(1),s(2) = s(1)[1−𝜆(2)] = [1−𝜆(1)][1−𝜆(2)], and proceeding inductively, s(k) = [1−𝜆(1)][1−𝜆(2)]⋯[1−𝜆(k)]. (14.4) Example 14.1 A bag contains 3 green balls and 1 red ball. A ball is drawn at random. If green, it is replaced and the draw is repeated. Failure occurs when a red ball is drawn. If this is on thekth draw we say that failure occurs at timek.F i n ds(k),f (k) and𝜆(k). Solution. From the conditions of the problem we can immediately deduce that 𝜆(k) = 1 4, k = 1,2, …. From (14.4), s(k) = (3 4 )k , k = 1,2, …. and from (14.2), f (k) = 3k−1 4k , k = 1,2, …. We have the well-known geometric distribution. See Section A.11.3. As a check we can also compute f (k) from the first formula in (14.1), which of course yields the same answer. 14.3 The continuous case In most applications, the time of failure is not restricted to the integers but can be arbitrary. We model this by assuming thatT is a continuous random variable with values in [0,∞). 214 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES 14.3.1 The basic functions We define thesurvival function sas in the discrete case, s(t) = 1−F(t) = P(T > t), the probability of survival up to timet. It is related to the probability density function (p.d.f) by f (t) =− s′(t), s(t) = ∫ ∞ t f (r)dr = 1−∫ t 0 f (r)dr. (14.5) Definition 14.2 The hazard function of a continuous failure timeT at timet, denoted by 𝜇(t), is the continuous density function for failure at timet given survival up to that point. It is given by 𝜇(t) = f (t) s(t) for allt such thats(t) > 0. For small Δt,𝜇(t)Δt approximates the probability thatT takes a value in the interval [t,t +Δt] given survival up timet. Analogously to (14.2), we have the expression f (t) = s(t)𝜇(t). (14.6) To determine the other quantities from𝜇 we note from (14.5) and (14.6) that 𝜇(t) = −s′(t) s(t) =− d dt[logs(t)]. Following the proof of Proposition 8.1, we deduce that s(t) = e−∫ t 0 𝜇(r)dr. (14.7)‡ Formula (14.7) seems quite different from its discrete counterpart (14.4), but if you look at it in the right way, it really is a natural continuous version. Recall that e−∑ai = ∏e−ai . So if we think of an integral as a type of generalized sum, then e to the integral is a type of generalized product of terms of the form e −𝜇, which for ‘small’ values of𝜇 are close to 1−𝜇. 14.3.2 Properties of 𝝁 Weknowthatthedensityfunction f mustbenonnegativeandsatisfy ∫ ∞ 0 f (t)dt = 1,andthatthe distributionfunction F mustbenondecreasingandsatisfy F(0) = 0,limt→∞ F(t) = 1.Thelat- terimpliesthatthesurvivalfunction s isnonincreasingandsatisfies s(0) = 1,limt→∞ s(t) = 0. EXAMPLES 215 What are the corresponding properties for the hazard function? The hazard function𝜇 must be nonnegative on its domain and in addition satisfy ∫ ∞ 0 𝜇(t)dt =∞ . To see this, suppose that the above integral had a finite valuea. From (14.7) we would deduce thatlimt→∞ s(t) = e−a which is not equal to 0, contradicting the fact that failure must occur at some time. In other words, if the integral is finite, the hazard function is not large enough to guarantee failure and there would be some chance of surviving forever. Of course, we may want to model situations where there is a chance that failure will never occur, and in this case we would want the above integral to be finite. 14.3.3 Modes The general shape of the density function ofT can often be inferred by looking at𝜇.I n particular,wemaybeinterestedin modes,whicharepointswherethedensityfunctionassumes a local maximum. We assume thatf is differentiable, and note from (14.5) and (14.6) that f ′ = (s𝜇)′ = s′𝜇 +s𝜇′ =− f 𝜇 +s𝜇′ =− s𝜇𝜇 +s𝜇′ = s(𝜇′ −𝜇2). (14.8) Weseethenthat f isincreasingordecreasingaccordingas 𝜇′ isgreaterthanorlessthan 𝜇2. Points at which modes can occur are restricted toor those values oft for which𝜇′(t) = 𝜇(t)2, or possibly to endpoints of the domain. We will look at some particular examples later. 14.4 Examples In this section we survey some familiar distributions that can be used to model failure times in certain situations. The family of exponential distributions (see Section A.11.6) can be easily described as those with aconstant hazard function. If𝜇(t) = 𝜇 for all t, then from (14.6) and (14.7), s(t) = e −𝜇t.Suchadistributionissuitableformodellingsituationswherethechanceoffailure in the next instant remains constant regardless of the time. This is shown by the constancy of thehazardfunctionaswellasthepropertygiveninA.57.Itisnotsuitableformodellinghuman mortality,wheretheagingprocessmeansthatwewouldexpectthehazardfunctiontoincrease with time. See Section 8.10.1 where we discussed the same point in the deterministic model. Another overly simplistic approach is to take a continuous uniform distribution (see Sec- tionA.11.4).Formostapplications,thisisnotrealistic,butithasbeenusedasaroughapprox- imation to model human mortality. This is simply the stochastic version of Demoivre’s law, introducedpreviouslyinChapter8.Notethatitdoesgiveanincreasinghazardfunctionsince 𝜇(t) = f (t) s(t) = 1 N −t,0 ≤ t < N. The first serious attempt to capture some of the relevant features for mortality was the Gompertz distribution given by 𝜇(t) = Bct, for some parameters B > 0 andc > 1. 216 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES This was introduced by Benjamin Gompertz in 1825. His idea was that the hazard function for human mortality should increase with age at a rate proportional to itself, that is, at an exponential rate. It has been used extensively for modelling mortality in both humans and other species. For human mortality, a good fit can usually be found over the middle span of ages by choosing the parameters appropriately. One feature of this distribution is that𝜇 ′(t)−𝜇(t)2 = 𝜇(t)[logc −𝜇(t)] is positive, zero, ornegative,preciselywhen 𝜇(t)islessthan,equalto,orgreaterthan logc.Then,(14.8)shows thatif 𝜇(0) ≥ logc,thedensityfunctionisdecreasing,whileif 𝜇(0) < logc,thereisaunique mode at the pointt where 𝜇(t) =l o gc. The density function increases up to this point and then decreases. This does indeed seem to fit the pattern of human mortality over the middle range of ages. Starting about age 30, the probability of dying at agex will increase withx up to a certain point. There is clearly more chance of dying at age 70 than at age 30. However, atsufficientlyhighagesthechanceofdyingstartstodecreasewithage.Forexample,thereis a very small chance of dying at age 110, for the simple reason that very few people will live that long in the first place. It may be enlightening to look again at (14.6), which expressesf as the product of two functions, one of which is decreasing, and one increasing. The Gompertz distribution does not fit well with observed mortality at ages below 30 or above 70 or so, for the reasons explained at the beginning of Section 8.10.1. In 1860, a modification was advocated by Makeham. He suggested that a constant be added to the hazard function to cover causes of death that were age-independent, such as accidents. The Makehamdistribution,oftenreferredtoastheGompertz–Makehamdistribution,isthengiven by 𝜇(t) = A +Bc t,f o rs o m eA > 0. The extra parameter allows for more flexibility in shape (see Exercise 14.12) but still does not capture observed mortality behaviour at very young or very old ages. 14.5 Shifted distributions Suppose we are given a failure timeT. (For convenience we will deal with the continuous case,buttheconclusionsforthediscretecasearesimilar.)Asremarkedabove,havingreached apoint u atwhichfailurehasnotyetoccurred,wemustalterourassessmentofthelikelihood of failure at different times. To formalize this, we define a new random variable. The random variableT ◦u is equal to the time until failure occurs, as measured from time u, given survival up to timeu. Therefore, it takes a value ofT −u when T takes a value greaterthan u,withprobabilitiesconditionedontheeventthat T isgreaterthan u.Thesurvival function, density function, and hazard rate function, respectively, forT ◦u are expressed in terms of the corresponding functions forT as follows. (Since we are dealing with more than one distribution here, we will insert appropriate subscripts ons,f and𝜇.) s T ◦u(t) = sT (u +t) sT (u) . (14.9) fT ◦u(t) =− d dtsT ◦u(t) = fT (u +t) sT (u) . (14.10) 𝜇T ◦u(t) = fT ◦u(t) sT ◦u(t) = 𝜇T (u +t). (14.11) THE STANDARD APPROXIMATION 217 Formula (14.10) provides a nice way to visualize this concept. To get the graph of the density function of the shifted random variable, we take the graph of the original density function, ignore everything to the left ofu, and scale all values upward to get the total area equal to 1. Forsomedistributions,theshiftedrandomvariablesareofthesamefamilyastheoriginal. This makes these distributions convenient for modelling purposes. Example 14.2 Describe the random variablesT ◦u when (a)T is an exponential random variable with constant hazard𝜇,( b )T is uniform on the interval [0,N], (c)T is Gompertz– Makeham with parametersA, B andc. Solution. In all cases it is convenient to use the hazard function and (14.11). (a) 𝜇 T ◦u(t) = 𝜇(u +t) = 𝜇, showing thatT ◦u has the same distribution asT. This indi- cates the so-calledmemory-less property of exponential random variables. The time until the event in question occurs will not be affected by how long you have already waited. (b) 𝜇 T ◦u(t) = 𝜇(u +t) = 1∕(N −u −t), showing that T ◦u is uniform on the interval [0,N −u]. (c) 𝜇T ◦u(t) = A +Bcuct, showing that T ◦u is also Gompertz–Makeham with the B parameter changed toBcu. 14.6 The standard approximation Associatedwitheverycontinuousfailuretime T isadiscretefailuretime ̃T.Failureaccording to ̃T will occur at the integer timek, if failure underT has occurred in the time interval (k −1,k]. To be precise, ̃T = [T]+1, where [⋅] denotes the greatest integer function. Letf ands denotethedensityfunctionandsurvivalfunction,respectively,of T.Let ̃f and ̃s denote the corresponding functions for̃T. Then, for all positive integersk, ̃f (k) = ∫ 1 0 f (k −1+t)dt = s(k −1)−s(k), (14.12) ̃s(k) = s(k). (14.13) Suppose we want to deduce information aboutT by observing when failure occurs. It could be difficult or impossible to observe continuously and we may be only able to view the situation at certain discrete times, say 1, 2, 3, etc. If we observe at timek, and see that failure has occurred, we know only that failure occurred between timek −1 andk. In other words, we are observing values of̃T. We would like therefore to infer the distribution ofT from that 218 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES of ̃T.Clearly,wecannotdothisexactlyandmustmakesometypeofapproximation.Asimple method, consistent with (14.12), is just to assume that for all nonnegative integersk, and any t in the interval (0,1), f (k +t) = ̃f (k +1). (14.14) We will refer to this asthe standard approximation.In fact, we have already used this approximationinourdeterministicmodel,intheformoftheUDDassumptionfortherandom variable T(x), as indicated by the equivalent formulation in terms of survival functions, namely, s(k +t) = (1−t)s(k)+ts(k +1), (14.15) for an integerk and 0< t < 1. To see this equivalence, note that, given (14.15), we obtain (14.14) by differentiating with respect tot. Conversely, given (14.14), we have s(k +t) = s(k)−∫ t 0 f (k +r)dr = s(k)−∫ t 0 ̃f (k +1)dr = s(k)−t̃f (k +1) = s(k)−t[s(k)−s(k +1)]. For the purpose of calculating moments, it is useful to introduce the random variable R = ̃T −T, thedurationfromthetimeoffailureuntiltheendoftheyearoffailure.Givenavalue r inthe interval(0,1),considertheprobabilitythat R ≤ r given that̃T =k.Thisistheprobabilitythat T takesavaluebetween k −r andk,giventhat ̃T = k.Usingthestandardapproximation,this probability is given by 1 ̃f (k) ∫ k k−r f (t)dt = 1 ̃f (k) ∫ k k−r ̃f (k)dt = r. This shows that underR is independent of̃T and has a uniform distribution on the interval (0,1). Using standard results about the uniform distribution, we can calculate that under the standard approximation E(T) = E(̃T)−E(R) = E(̃T)− 1 2, (14.16) Var(T) = Var(̃T)+Var(R) = Var(̃T)+ 1 12, (14.17) whereE denotes expectation and Var denotes variance. THE STOCHASTIC LIFE TABLE 219 14.7 The stochastic life table Thegoalinthissectionistoestablishalinkbetweenthestochasticmodelandthedeterministic model for mortality. In this and subsequent chapters we will use this link to show that the questions that we answered in the deterministic model can be answered in the stochastic model in thevery sameway. In addition, we will be able to answer more questions. The tool for doing so will be to return to the fundamental concept of a life table but look at it in a different way. As we did before, we start out with an arbitrary number𝓁 0 of newly born lives. We now letℒx be the number of these still alive at agex, and𝒟x be the number of these who will die between the age ofx and x +1. These quantities look like the𝓁x and dx of Chapter 3, but we now recognize that these are not numbers butrandom variables.We obtain the life table by letting 𝓁x = E(ℒx), dx = E(𝒟x). Since it is clear thatℒx+1 = ℒx −𝒟x, we have the same formula (3.1) as we had before. In other words, in our stochastic model we still can introduce a life table, but we now view the numbersasexpectedvaluesratherthanasanexactaccountofthenumberswhowillliveordie. Consider now the random variableT(x), the time to the death of (x). We will write s x(t),fx(t),𝜇x(t) for the survival function, density function, and hazard function respectively of T(x). Forage0,thereissomewhatadifferentstandardnotation,whichcancausesomeconfusion if care is not taken. The random variableT(0) is traditionally denoted byX, and the variable denoted byx rather thant, sinceT(0) is really theage at death. It is common to suppress the 0 and just writes(x), f (x) and𝜇(x)f o rs0(x), f0(x) and𝜇0(x), respectively. A key observation is that 𝓁x = 𝓁0s(x). (14.18) The argument is the same as showing that if you flipN coins, each with a probabilityp of comingupheads,youcanexpecttoget Np heads.Inthiscase,ifyoutake 𝓁0 people,eachwith a probabilitys(x) of surviving to age (x), then you can expect to get𝓁0s(x) such survivors. We are now ready for the main purpose of this section, which is to establish a correspon- dence between the survival, hazard, and density functions ofT(x) and the life table functions for the stochastic life table. At first, we assume aggregate mortality, as in Chapter 3, where we used a single-life table for the future mortality of all agesx, by just starting at agex.T h i s corresponds stochastically to the statement T(x) has the distribution ofT(0)◦x, which simply says that the future lifetime of (x) is the future lifetime after timex of a person whowasage0,giventhattheyhavelivedtoage x.Therearethereforenoeffectsofselection. Suppose also that we calculate the quantities tpx,tqx,𝜇(x) starting from the life table, exactly as we did in the deterministic model. We then have from (14.10) and (14.23) that sx(t) = s(x +t) s(x) = 𝓁x+t 𝓁x = tpx. (14.19) 220 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES This is certainly reasonable. Although looked at from different points of view, both of these are probabilities that a person agex will survivet years. It follows immediately that F x(t) = 1−tpx = tqx, (14.20) and that the hazard rate function ofT(x)i sg i v e nb y 𝜇x(t) =− d dt logtpx = 𝜇(x +t), (14.21) where the last quantity is the force of mortality of Chapter 8. (Note that in the case of hazard rates,thenotationinthedeterministicmodelwaschosenfromtheoutsettocorrespondtothat in the stochastic setting.) Finally, f x(t) = tpx𝜇x(t), (14.22) the familiar expression that we used for insurance premiums in Chapter 8. IfweassumethestrictlyselectmortalityofChapter9,thenwecannotassumethat T(x)is distributed asX ◦x. The above correspondences would be modified as follows: sx(t) = tp[x] and fx(t) = tp[x]𝜇x(t). It is also instructive to compute the distribution of multiple-life failure times in terms of the quantities introduced in the discrete model. For example, considerT(xy), the time of failure of the joint-life status. For this random variable, the survival function is justtpxy,t h e hazard function is𝜇xy(t), and the density function istpxy𝜇xy(t). 14.8 Life expectancy in the stochastic model Let ̃T(x) be the discrete random variable associated withT(x) as defined in Section 14.7. It will have survival, probability and hazard functions given by ̃sx(k) = kpx, (14.23) ̃fx(k) = k−1px − kpx = k−1px qx+k−1, (14.24) 𝜆x(k) = qx+k−1. (14.25) Traditional actuarial terminology defines a discrete random variableK(x) known as the curtate future lifetime,whichmeasuresthewholenumberoffutureyearstobelivedby( x).It is clear that K(x) = ̃T(x)−1. STOCHASTIC INTEREST RATES 221 From (14.24), (3.8), (A.12), and the fact thatsx(0) = 1, we can write ex = 𝜔−x−1∑ k=1 sx(k) = E[̃T(x)]−sx(0) = E[K(x)]. Therefore life expectancy is indeed an expectation in the sense of probability theory. The formal definition ofcomplete life expectancyin the stochastic model is ◦ex = E[T(x)]. From(A.15)weseethatthisagreeswithourChapter8definition.Moreover,usingthestandard approximation and (14.16), we verify the fact that E[T(x)] = e x + 1 2. We can already see one advantage of the stochastic model. As well as computing the expected future lifetime, we can compute the variance of future lifetime. Example14.3 Given𝓁 90 = 100, 𝓁91 = 90, 𝓁92 = 70, 𝓁93 = 40, 𝓁94 = 10, 𝓁95 = 0,com- pute the expectation and variance ofT(90), assuming UDD. Solution. Using either (A.7) or (A.16) we find thatE[̃T(90)] = 3.1, E[̃T(90)]2 = 10.9, so that Var[̃T(90)] = 1.29. Then, from (14.16) and (14.22), ◦e90 = 2.6, Var[T (90)] = 1.373. 14.9 Stochastic interest rates Stochastic interest is a large topic which we will only allude to briefly. As well as the life table, the other main ingredient of the actuaries tool-kit, the discount function, should be treated stochastically. For the most part, the method of doing so in the pricing and valuation of insurance and annuity contracts has been by a simulation technique. Various scenarios of possible future interest rates are selected, together with some weighting as to how likelihood their occurrence is. Premiums or reserves are computed using each scenario, exactly as we havedescribed,andthetotalityofresultsareanalyzed.Inthisway,onecanarriveatestimates of the expected value of the quantities, or in some cases, particularly for reserve purposes, an idea of the worst case scenarios. Amoresophisticatedapproachistomodel v(s,t)asarandomvariablesratherthandefinite numbers,aswementionedinChapter2.SomeaspectsofthisideaaredealtwithinChapter20. Numerous stochastic interest rate models have been proposed. The most popular approach is to express the force of interest𝛿(t) as a stochastic process rather than a function oft. 222 SURVIVAL DISTRIBUTIONS AND FAILURE TIMES Notes and references Standardactuarialterminology,whichisbasedondeterministicthinking,referstoGompertz’s law,ratherthantoaGompertzdistribution.Thisisstatedas 𝜇(x) = Bc x.Inourframeworkthis meansthat T(0)hasaGompertzdistributionand T(x) = T(0)◦x,sothat T(x)hasaGompertz distribution for allx. Similar remarks apply to the Makeham modification. It was common practice at one point to use life tables that satisfied Gompertz’s or Makeham’s law. A major motivation was to simplify joint-life calculations, by reducing joint-life statuses to a single-life or a joint-life status with equal ages. See Exercise 14.10. There is little need for this with modern computing methods. AdditionalmaterialonGompertz,Makehamandmoregeneralmortalitydistributionscan be found in Brillinger (1961), Carri`ere (1994a) and Tennenbein and Vanderhoof (1980). Exercises Type A exercises 14.1 Redo Example 14.1, but now assume that each time a green ball is drawn, it is replaced with a red ball. In this case 14.2 RedoExample14.1,butnowassumethateachtimeagreenballisdrawn,anewred ball is added. 14.3 A failure time T has the hazard function𝜇(t) = (1+t)−2. What is the probability that failure never occurs? 14.4 Suppose that q97 = 0.5, q98 = 0.6, q99 = 0.8, q100 = 1, and that UDD holds. Find (a)E[T(97)] and (b) Var[T(97)]. 14.5 If Demoivre’s law holds with𝜔 = 100, find the variance ofT(60). Type B exercises 14.6 Consider two copying machines. Machine 1 will make 20 000 copies per month. Whennew,itwilllast T months,where T isarandomvariablewithhazardfunction 𝜇T (t) = 1 40−t,0 ≤ t < 40. Machine 2 will make 15 000 copies per month. When new, it will last forS months, whereS is a random variable with hazard function 𝜇S(t) = 2 96−t,0 ≤ t < 96. Aprospective purchaser istryingtodecide between buyinganew machine1,or a 2-year-old machine 2. Which of these two choices will produce, on average, the largest total number of copies during its lifetime? EXERCISES 223 14.7 Suppose that T has an exponential distribution with constant hazard 0.2. (a) What is the probability that̃T = 3? (b) Whatistheerrormadeifweapproximatetheprobabilitythat2 .5 ≤ T ≤ 2.75by using the standard approximation? 14.8 Consider the distribution with density function f (t) = 𝛽2te−𝛽t, t ≥ 0. (This is a gamma distribution with first parameter 2.) (a) Show thats(t) = (1+𝛽t)e −𝛽t. (b) Find the hazard function𝜇(t). (c) Describe precisely the shifted distributionT ◦u.( Hint:I ti sam i x t u r eo ft w o well-known distributions.) 14.9 For two independent lives (x) and (y), (x) is subject to a force of a mortality of 𝜇x(t) = 2∕(10−t) while (y) is subject to the survival functionsy(t) = 1−(t∕10), in both cases for 0≤ t < 10. Find the p.d.f. for the following random variables. (a) T(xy), the time of failure of the first death; (b) T(xy), the time of failure of the second death. 14.10 (The law of uniform seniority) (a) SupposethatGompertz’slawholds(seeSection14.9).Showthatthereisfunction g, defined on the nonnegative reals, such that tpx:x+n = tpx+g(n), for allt > 0. (b) Suppose that Makeham’s law holds. Show that there is a functionh, defined on the nonnegative reals, such that tpx:x+n = tpx+h(n):x+h(n), for allt > 0. 14.11 Show that the sample table of Section 3.7 satisfies Gompertz’s law (that is up to age 119 where a modification was made to get a finite value of𝜔). *14.12 Suppose that T has a Gompertz–Makeham distribution withA > 0. Show thatfT is either (i) decreasing, (ii) increasing then decreasing or (iii) bimodal with two local maximums. Identify the conditions on the parameters that will result in each case. 14.13 For each of the following two families of distributions, decide whether or not the shifted distributions are from the same family, with changed parameters. (a) 𝜇(t) = 𝛼(t +𝜃) −1 for positive𝛼 and𝜃 (Pareto); (b) 𝜇(t) = ktr for positivek andr (Weibull). 15 The stochastic approach to insurance and annuities 15.1 Introduction In this chapter we deal with a perfectly general failure timeT, and develop the stochastic approach for calculating premiums and reserves for insurance and annuity contracts based on T. These will all be calculated as expected values of appropriate random variables. In the particular case whereT = T(x), we will show, using the correspondence established in Chapter 15, that these agree with the results that we obtained in the deterministic model. The advantageofthestochasticapproachisthatwecanaugmenttheseexpectedvalueswithother quantities, such as variances. Throughout the chapter,f, sand 𝜇 will denote the density, survival and hazard functions respectivelyof T.W elet̃Tbetheassociateddiscretefailuretime, ⌊T⌋ +1.Wewilllet ̃f and𝜆 denotetheprobabilityfunctionandhazardfunctionrespectivelyof ̃T.Wedenotethesurvival function of ̃T with the same symbols. This will not cause confusion since the value at any integer is the same in both cases. Insomecasesthevaluesof Tarebounded(e.g.,thevaluesof T(x)areboundedby 𝜔−x), but this is not necessary. The upper limit on integrals or sums will be written as∞ to cover all possibilities. The approach differs somewhat from the deterministic case and it is important to under- stand the distinction. We suppose, as we did before, that we have a fixed deterministic investment discount functionv. We will consider contracts that pay certain benefits provided that failure has not occurred, and/or benefits at the time of failure. We will want to determine present values of these benefits, and these will all be computed with respect to the discount function v. In this model, we do not have the interest and survivorship discount function that we had before. However, the present values that we obtain will not be definite numbers but ratherrandom variables, since they will depend on the unknown value thatT assumes. 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 STOCHASTIC APPROACH TO INSURANCE BENEFITS 225 Premiums and reserves will be calculated asexpectations of these random variables, and it is in calculating these expectations that the probabilities of life and death are taken into account. These expectations are commonly known asactuarial present values, abbreviated as APV. Throughout this chapter and the next, we adopt the following notation, which will ensure that our theory is applicable both to the net premium situation, as discussed in Chapters 4–6 and to the gross premium situation as discussed in Chapter 12. The premium symbol𝜋 k will denote thetotal inflowreceived at timek, which will be the net premium in a model which ignores expenses, or in the expense situation will be the gross- or expense-augmented premium less the expenses paid at timek. Similarly the benefitb k paid for failure at timek will include any costs involved with making the benefit payment. However to keep things simple we confine our attention to a single decrement case where we have only one cause of failure. In particular then, in the context of life insurance we are ignoring withdrawals, implicitly making the assumption discussed in the second remark after Example 12.2. 15.2 The stochastic approach to insurance benefits We deal here with contracts that provide benefits upon failure, and consider both the discrete and continuous cases. 15.2.1 The discrete case We consider a contract with failure benefit vectorbpayingbk−1 at timekfor failure between timek−1andtime k.LetZ denotethepresentvalueofthebenefitswithrespectto v.Then Z is a function of the random variablẽT, since wheñT = k,t h ev a l u eo fZis bk−1v(k). We can therefore write Z = b̃T−1v(̃T). LettingÃT(b;v) denoteE(Z), we have ÃT(b;v) = ∞∑ k=1 bk−1v(k)̃f(k) = ∞∑ k=0 bkv(k+1)̃f(k+1). (15.1)‡ As before, we will often suppress thev and write this asÃT(b). Note that whenT = T(x), formula (15.29) shows that we obtainAx(b) as calculated in (5.1). For many applications we want more information aboutZthan just its expected value; at the very least, we would want to calculate Var(Z). This is easily done, sinceZ2 = b2 ̃T−1v(̃T)2. In other words, to calculate the second moment, we simply square the benefits, and square the discount function. This leads to Var(Z) = A ̃T(b2;v2)−ÃT(b;v)2. (15.2)‡ 226 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES 15.2.2 The continuous case Consider a contract that paysb(t) at the moment of failure, should failure occur at timet. We will letZ denote the present value of the benefits and denote the expected value ofZ by AT(b;v), often shortened toAT(b). WhenT = t,b(t)i sp a i da tt i m et,s o Z = b(T)v(T) and AT(b;v) = ∫ ∞ 0 b(t)v(t)f(t)dt. (15.3)‡ When T = T(x), we see from (15.27) that this is the same quantity as we obtained in (8.18). The variance is computed as we did in the discrete case: Var(Z) = AT(b2;v2)−AT(b;v)2. (15.4)‡ Example 15.1 Suppose that𝜇(t) is a constant 0.04 and the force of interest is a constant 0.07. An insurance contract pays a benefit of e0.03t at timet, should failure occur at that time. Find the expectation and variance ofZ, the present value of the benefits. Solution. This isanexponential failuretime,so f(t) = 0.04e−0.04t.W eh a v eb(t) = e0.03t and v(t) = e−0.07t. Substituting into (15.3), E(Z) = ∫ ∞ 0 0.04e0.03te−0.07te−0.04tdt= 0.04 0.04+0.07−0.03 = 1 2. Nowb(t)2 = e0.06t and v(t)2 = e−0.14t, so calculating as above, E(Z2) = 0.04 0.04+0.14−0.06 = 1 3,V a r ( Z) = 1 3 − 1 4 = 1 12. 15.2.3 Approximation Suppose now that we know only the distribution of̃T, and wish to approximateAT(b). If interest and benefits are constant over each year, we can use the standard approximation to obtain the samei∕𝛿 adjustment as we obtained before forT(x). We will give an alternate stochasticderivationofthisresult.Wefirstassumeconstantinterest,sothat v(t) = v t forsome constant v.L e tbk denote the constant value ofb(t) over the time intervalk to k+1, and let b = (b0,b1,…). Recall the random variableR= ̃T−T introduced in Section 15.7. Then E(Z) = E[b(T)vT] = E[b(T)ṽT−R] = E[b(T)ṽT]E[v−R], THE STOCHASTIC APPROACH TO INSURANCE BENEFITS 227 where we invoke independence for the last equality. When ̃T = j,b(T) = bj−1, so that E[b(T)ṽT]i sj u s tÃT(b). Moreover,v−R = (1+i)R, and sinceRis uniform on [0,1]. E[v−R] = ∫ 1 0 (1+i)udu= i 𝛿, leading toAT(b) = (i∕𝛿)ÃT(b). The general formula, where interest can vary from year to year is then given by AT(b) = ÃT ( i 𝜹 ∗ b ) . (15.5) wherei∕𝜹 is now a vector. 15.2.4 Endowment insurances Inthedeterministicmodel,endowmentinsuranceswereviewedasakindofhybrid,consisting ofbothinsuranceandannuitycomponents.Inourstochasticmodel,wecanviewthesestrictly as insurances. Define the failure time T =m i n{T(x),n}, so that failure occurs at the death of (x), or at timen if earlier. Insurances based onT are precisely n-year endowment insurances. When benefits are paid at the moment of death,T is an example of a random variable that has both a continuous part and a discrete part (see Section A.5). It is continuous over the interval [0,n), on which it has a density functionf x satisfying P(a≤ T ≤ b) = ∫ b a fx(t)dt, provided b< n. Thisdensitywilljustbetherestrictionofthedensityfunctionfor T(x)totheinterval[0, n), and it integrates toFx(n) over this interval. The remaining probability is all concentrated on the pointn, sinceT will take this value wheneverT(x) takes a value ofnor greater, an event with probabilitysx(n). Expectations of a general functiongof T must be calculated as a sum of two terms, E[g(T)] = ∫ n 0 g(t)fx(t)dt+g(n)sx(n). In particular, consider ann-year endowment insurance with death benefit functionbdefined on[0, n].Thatis, b(t)ispaidatthemomentofdeathifthisoccursbefore nandb(n)ispaidat timenif the insured is then alive (so that failure occurs at timen). Then AT(b) = E[v(T)b(T)] = ∫ n 0 b(t)v(t)fx(t)dt+b(n)v(n)sx(n), which is easily seen to agree with the present value obtained in the deterministic model. 228 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES From (15.4) we calculate the variance of the benefits as ∫ n 0 [b(t)v(t)]2fx(t)dt+[b(n)v(n)]2sx(n)− [ AT(b) ]2 . Some care is needed when applying the approximation (15.5). This is only valid for the continuouspartoftherandomvariable T.Inthepresentexampleoftheendowmentinsurance, Equation (15.5) would be modified to AT(b) = AT ( i 𝜹 ∗ b ) +b(n)v(n)s(n). The last term does not get multiplied by the interest adjustment, since in both the end-of-the- year-of-deathandmoment-of-deathcasesthefinalamountof b(n)ispaidtosurvivorsatexact timen. Example 15.2 A two-year-endowment insurance on (x) has benefits of 100 if death occurs in the first year or 80 if death occurs in the second year, and a pure endowment of 80 if the insured is alive at time two. You are given thatq x = 0.2, qx+1 = 0.3, and the interest rate is a constant100%.Findtheexpectationandvarianceofthebenefitswhen(a)benefitsarepayable at the end of the year of death; (b) benefits are payable at the moment of death. Solution. (a) LetZ denote the present value of the benefits in the end of the year of death case.AswenotedinExample6.2,wedonotevenneedthevalueof q x+1.Infactwecouldjust notice thatZ takes the value 50 with probability 0.2 and 20 with probability 0.8. It therefore has a mean of 26 and a variance of 302(0.2)(0.8)=144. (b) LetZ denote the present value of the benefits in the moment of death case. Now we do need the value ofqx+1 since it makes a difference if the person dies in the second year, in which case 80 is paid at the moment of death, or if they survive, in which case 80 is paid at the end of the year. Begin again with the calculation ofE(Z) andE(Z 2) but now split into the insurance and pure endowment amounts. E(Z) = 14.8+80(1∕4)(0.56) = 14.8+11.2. E(Z 2 ) = 596+6400(1∕16)(0.56) = 596+224. Now fori= 1,𝛿 =l o g(2) so E(Z) = ( 1 log2 ) 14.8+11.2 = 32.55 Squaringvmeans that𝛿 is doubled and theichanges to (1+i)2 −1 = 2i+i2. THE STOCHASTIC APPROACH TO ANNUITY BENEFITS 229 which equals 3 when the original value is 1. So, for the second moment calculation we take i= 3,𝛿 =l o g(4), to get E( Z 2 ) = ( 3 log4 ) 596+224 = 1513.77 Var(Z) = 1513.77−32.552 = 454.27 The variance is much larger in (b), due to the variation in the present value of the benefits, depending on when the person dies during the year, which is significant in view of the high interest rate. 15.3 The stochastic approach to annuity benefits 15.3.1 Discrete annuities Consideracontractwithannuitybenefitvector cthatpays ck attime kprovidedthatfailurehas notyetoccurred.Let Y denotethepresentvalueofthebenefitswithrespecttotheinvestment discount functionv, and leẗãT(c;v) (usually shortened töãT(c)) denoteE(Y). (The subscript ̃T reflects the fact that in view of the yearly payments, this quantity will depend only on the value of̃T, rather than the exact value ofT). There are two methods for computing expectations and variances of annuities in the stochastic setting. This first has the advantage of simplifying variance calculations. Recall that in the deterministic model we often viewed insurances as annuities. It is now helpful to view annuities as insurances in a certain sense. (Readers who attempted Exercise 5.22 will have encountered this idea previously.) To be precise, we want to expressY as a function of ̃T.L e tgbe the function defined on the positive integers by g(k) = c 0 +c1v(1)+c2v(2)+⋯ +ck−1v(k−1). We can also writeg(k) = ̈a(kc;v) using the notation introduced in Section 2.10.1. Suppose that̃T takes the valuek. Then payments will have been made at all integer times from 0 tok−1, soY will take the valueg(k). We can therefore write Y = g(̃T), and it follows that ̈ãT(c) = ∞∑ k=1 g(k)̃f(k) (15.6)‡ and Var(Y) = ∞∑ k=1 g(k)2̃f(k)− ̈ãT(c)2. (15.7)‡ 230 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES Expression(14.6)isknownasthe aggregatepaymentformula,sinceweconsiderthefunction g(k) which is the aggregate present value received for failure at timek. Example 15.3 For a 4-year life annuity on (60). The sequence of annuity benefits is 1, 2, 3, 4 beginning at age 60. You are givenq60 = 0.1, q61 = 0.2, q62 = 0.3 andi= 100%.F i n d E(Y) and Var(Y). Solution. Calculate recursively g(1) = 1,g(2) = 1+2× 1 2 = 2,g(3) = 2+3× 1 4 = 2.75, g(k) = 2.75+4× 1 8 = 3.25 for allk> 3. The distribution of̃T is given bỹf60(1) = 0.1, ̃f60(2) = 0.180, ̃f60(3) = 0.216, s60(3) = 0.504. We can now calculate E(Y) = 0.100+2(0.180)+2.75(0.216)+3.25(0.504) = 2.692, Var(Y) = 0.100+22 ×0.180+(2.75)2 ×0.216+(3.25)2 ×0.504−2.6922 = 0.530. Remark Note carefully that when the final payment of an annuity is at timek0, thenY takes the value ofg(k0 +1) with probabilitys(k0) sinceg(k) = g(k0 +1) for allk> k0. Remark There is an important point here which is often overlooked. Equation (15.6) applied to life annuities says that we can view the present value of a set of cash flows with respect to the interest and survivorship function as the expected value of the present value of payments received with respect to an interest only function. Some people have mistakenly assumed that similarly, the accumulated value of a life annuity is the same as the expected value of the payments received, accumulated with interest. This is not true and in fact the equality does not hold for values at any time other than 0. We have Val k(c,yx) = yx(k)−1̈ax(c), while the expected value at timek with respect tov, of the payments received fromc is v(k)−1̈ax(c), which will be less than the first quantity for allk> 0. Thesecondmethodtocalculateannuitypresentvaluesismorecloselyrelatedtowhatwe did in the deterministic model. We define random variables Ik = { 1, if ̃T > k, 0, if ̃T ≤ k. The contract pays a present value ofckv(k) for each integerksuch that̃T > k, or equiva- lently, such thatIk = 1. We can therefore write Y = ∞∑ k=0 ckv(k)Ik. (15.8) THE STOCHASTIC APPROACH TO ANNUITY BENEFITS 231 This is analogous to viewing the annuity as consisting of several pure endowments. Since E(I k) = s(k), taking expectations, gives ̈ãT(c) = ∞∑ k=0 ckv(k)s(k). (15.9)‡ Expression(15.9)isknownasthe currentpaymentformula,sinceitlooksateachpayment received, discounts it with interest, and multiplies it by the probability of receiving it. In the case whereT = T(x), this is precisely the formula obtained in (4.1). The equality of the expressions in (14.6) and (15.9) can be seen from (A.16). (Condition (A.14) will almost always hold. It is automatic when benefits are positive sinceg will be increasing,orwhentherearefinitelymanybenefitpayments,evenifnegative,since gwillbe bounded.) Note that theI k are not independent. To calculate variances from the current payment approach we use the following facts. Var(Ik) = s(k)(1−s(k)). (15.10) Forj< k,w eh a v eIjIk = Ik, so that Cov(Ij,Ik) = E(IjIk)−E(Ij)E(Ik) = s(k)(1−s(j)). (15.11) To simplify notation let rk = ckv(k)s(k), uk = ckv(k)(1−s(k)). Then from (A.24), (15.10) and (15.11), Var(Y) = ∑ k rkuk +2 ∑ j<k rkuj. (15.12) This will normally be a much more involved calculation than that given by the aggregate payment approach. 15.3.2 Continuous annuities Consider a contract that makes payments continuously at the annual rate ofc(t) at timet, providedT has not yet occurred. We will denote the actuarial present value byaT(c;v)( o f t e n shortened toaT(c)). Define the functiongby g(t) = ∫ t 0 c(r)v(r)dr. IfT fails at times, the present value of the benefits received will be preciselyg(s), so that Y = g(T). 232 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES It follows that aT(c) = E(Y) = ∫ ∞ 0 g(t)f(t)dt (15.13)‡ and Var(Y) = ∫ ∞ 0 g(t)2f(t)dt−[aT(c)]2. (15.14)‡ For the alternate current payment formula we use (A.15). From the fundamental theorem of calculus,g′(t) = c(t)v(t) and, moreover,g(0) = 0. If (A.14) holds (which as in the discrete case will almost always be the case) we can write the current payment formula aT(c) = ∫ ∞ 0 c(t)v(t)s(t)dt, (15.15)‡ which agrees with (8.9) whenT = T(x). There is no convenient variance formula analogous to (15.12) for continuous annuities, since we do have the representation as a sum of indicator random variables that was present in the discrete case. Example15.4 Anannuityprovidescontinuousbenefitsattherateof1perperiod,provided failure has not occurred. The failure timeT has an exponential distribution with constant hazard 𝜇. The force of interest is a constant𝛿.I f Y is the present value of the benefits, find Var(Y). Firstsolution. Since g(t) = ∫ t 0 e−𝛿rdr= 1−e−𝛿t 𝛿 , We have E(Y 2 ) = ∫ ∞ 0 [ 1−e−𝛿r 𝛿 ]2 𝜇e−𝜇rdr = 𝜇 𝛿2 ∫ ∞ 0 [1−2e−𝛿r +e−2𝛿r]e−𝜇rdr = 𝜇 𝛿2 [ 1 𝜇 − 2 𝜇 +𝛿 + 1 𝜇 +2𝛿 ] = 2 (𝜇 +𝛿)(𝜇 +2𝛿). From(15.13),orsimplyreferringbacktoSection8.10,weseethat E(Y) = 1∕(𝜇 +𝛿)sothat Var(Y) = 2 (𝜇 +𝛿)(𝜇 +2𝛿) − 1 (𝜇 +𝛿)2 = 𝜇 (𝜇 +𝛿)2(𝜇 +2𝛿). Note that, as a check, for𝛿 = 0, this reduces to𝜇−2, the variance of the given exponential distribution (see Section A.11.6) which must be true sinceY = T at zero interest. THE STOCHASTIC APPROACH TO RESERVES 233 Asimplification. We can simplify and also obtain a more general formula by the following trick: Y 2 = 1−2e−𝛿T +e−2𝛿T 𝛿2 = 2 𝛿 [( 1−e−𝛿T 𝛿 ) − ( 1−e−2𝛿T 2𝛿 )] . Upontakingexpectations andrecallingthat E[e−𝛿T] = 𝜇∕(𝜇 +𝛿),wegetthesameansweras above. ∗15.4 Deferred contracts Consider any contract based on the failure timeT, in which no payments are to be made in the firstmyears regardless of whether failure occurs or not. A typical example is a deferred annuity, but this can be a perfectly general contract, continuous or discrete, including death benefits, annuity benefits or both. LetY be the present value of the benefits, and letY ′ be the value of the benefits at the starting timem. As mentioned previously, there is no real need for any special mathematical treatment of such, as we simply take zero entries or zero values in thebenefitvectororfunction.Itissometimesconvenient,however,toexpressVar( Y)interms of Var(Y ′). The purpose of this section is to derive such a formula. The idea is to notice thatY takes the value 0 with probability 1−s(m) andv(m)Y′ with probability s(m). It follows from basic probability theory that E(Y) = s(m)E(v(m)Y′) = v(m)s(m)E(Y′), a formula that we are familiar with from the deterministic model. It similarly follows that E(Y2) = v(m)2s(m)E(Y′2) = v(m)2s(m)[Var(Y′)+E(Y′)2]. Substituting forE(Y) in terms ofE(Y′) in the expression Var(Y) = E(Y2)−E(Y)2 gives Var(Y) = v(m)2s(m)Var(Y′)+v(m)2s(m)(1−s(m))E(Y′)2. (15.16) This gives a familiar decomposition of the variance ofY. The first term measures the uncertainty arising from actual benefits once they start at timem, and the second term measures the uncertainty arising from the fact that failure may or may not occur prior to timem. 15.5 The stochastic approach to reserves We will illustrate the stochastic approach to reserves with the discrete model. The results are easily adapted to the continuous case. We have vectorsb and 𝜋, whereb k is paid at time k+1 if failure occurs between timekandk+1, and𝜋k is paid at timekif failure has not yet 234 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES occurred. We are adopting the point of view (used before in Chapter 6) of treating annuity benefits as negative premiums, so that𝜋 k represents a net inflow at timek, consisting of the premiumreceived,lessexpensespaidattime kifthesearebeingconsidered,andalsolessany annuity payments made at timek. We will therefore not need to refer to a separate annuity benefit vector. Fix an integer durationrand suppose that failure has not yet occurred by that time. Definition 15.1 The prospective loss at time r,denoted byrL, is the value at timer of the future net cash flows to bepaidout. These net cash flows are the future benefits to be paid minus the future net inflows to be received. The definition here is different than that used in the deterministic model. The benefits are the actual benefits paid on the contract, and not the individual share of the total deathbenefits.AswestressedinSection14.1,thediscountingiswithregardtotheinvestment discount functiononly. The value willdepend ofcourseonwhen failureoccurs, so rLwillbe arandomvariable.Itisafunctionoftheshiftedrandomvariable ̃T◦ r,sincewehaveassumed survival up to timer. We next determine this function. Suppose that failure occurs at timer+twhere k≤ t< k+1,forsomeinteger k.Then ̃T◦ rwillassumeavalueof k+1.Thecontractwillpayouta failure benefit ofbk at timek+1. Offsetting this, the net inflows would have been collected from timerto timer+k, so that in this case the value ofrLwill be br+kv(r,r+k+1)−[𝜋r +𝜋r+1v(r,r+1)+⋯ +𝜋r+kv(r,r+k)]. In terms of random variables, we can write rL= br+̃T◦ r−1v(r,r+ ̃T◦ r)−[𝜋r +𝜋r+1v(r,r+1)+⋯ +𝜋r+̃T◦ r−1v(r,r+ ̃T◦ r−1)]. Remark We have used the standard terminology here, but the reader should be aware that the word ‘loss’ in the above definition has a completely different connotation than that discussed in Section 6.4.1. It is not a definite number, measuring the loss brought about by deviationfromtheexpectedpatternofmortalityorinterest,butratheritisarandomvariable, giving the difference between what is collected and what is paid out under the contract at the differentrandomtimesoffailure.Itshouldalsobenotedthat rLisofthefewrandomvariables we encounter that can take negative values, which occur when there is a ‘gain’. Definition 15.2 In this stochastic model, we define thereserveattimer, denoted as before by rV,s i m p l ya s rV = E(rL). SorVistheexpectedvalueoftheprospectiveloss,whichistheexpectedpresentvalueof the future benefits less the expected present value of the future inflows. It is clear that in the case ofT = T(x) we obtain the same value as in the deterministic model. THE STOCHASTIC APPROACH TO PREMIUMS 235 Example 15.5 Let T = T(60). We are givenq60 = 0.1, q61 = 0.2, q62 = 0.3, i= 100%. Suppose we have a 3-year term insurance policy on (60) withb0 = 200, b1 = 200, b2 = 100, 𝜋0 = 20, 𝜋1 = 20, 𝜋2 = 10. (a) Find the distribution of0L. (b) Find the distribution of1L. (c) From your answer to (b), calculate1V. Solution. (a) The distribution of0Lis as follows: k Value of0Lwhen ̃T = kP (̃T = k) 1 200 × 1 2 −20 = 80 0.100 2 200 × 1 4 − ( 20+20× 1 2 ) = 20 0.180 3 100 × 1 8 − ( 20+20× 1 2 +10× 1 4 ) =− 20 0.216 ≥ 40 − ( 20+20× 1 2 +10× 1 4 ) =− 32.5 0.504 (b) k Value of1Lwhen ̃T◦ 1 = kP (̃T◦ 1 = k) 1 200× 1 2 −20 = 80 0.20 2 100× 1 4 − ( 20+10× 1 2 ) = 00 . 2 4 ≥ 30 − ( 20+10× 1 2 ) =− 25 0.56 (c) 1V = 80×0.20+(−25×0.56) = 2. The reader should verify this agrees with the answer that we would get from the methods of Chapter 6. 15.6 The stochastic approach to premiums For this section only, it is convenient to adopt a different convention. Assume now that expenses and annuity payments are included as benefits, so that the inflows𝜋 k are just the premiums collected. 15.6.1 The equivalence principle The random variable0L is usually denoted just asL. It is the present value of all future benefits less the present value of all future premiums. Recall that in the deterministic model 236 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES wecalculatednetpremiumsbysettingthepresentvalueofbenefitsequaltothepresentvalue of premiums. The stochastic version of this concept is to set premiums so that E(L) = 0. This is known as theequivalenceprinciple. 15.6.2 Percentile premiums The stochastic viewpoint allows us to incorporate other features of the random variableL rather than just its expectation, when computing premiums. For example, it would be highly desirable for an insurer to avoid a positive loss. Of course, it is impossible to ensure that this willalwayshold.Iftheinsureddiesshortlyafterpurchasingalifeinsurancepolicy,then Lwill almost certainly be positive. Suppose, however, we set a threshold𝛼, and then set premiums as small as we can, so that the probability will be at least𝛼 that the loss will be nonpositive. Typically𝛼 willbeanumber closeto1suchas0.95or 0.99. Thesearesometimesreferredto as percentilepremiums. This sounds reasonable, but there are major problems to such an approach. In the first place, the method does not take into account the importantright tailof the distribution, consisting of those values ofL which are greater than the specified percentile. If there are very large claims present, the method may not provide sufficient premiums to cover the risk. Thisisevidencedbythefactthatthemethodcanproducepremiumsthatareactuallylessthan the equivalence principle premium. For an extreme example of this, suppose we have ann year term policy and the probability of (x) dying withinnyears is less than 1−𝛼. We could achieve the stated goal by charging premiums of 0, which is absurd. On the other hand, for high values of𝛼 the method can produce premiums that seem inordinatelyhighwhencomparedwiththeequivalenceprinciplepremium.SeeExample15.7 below. Still another instance of the pathological behaviour produced by percentile premiums is illustrated by the following example. Example 15.6 On a single premium contract, the present value of the benefits takes the value 0 with probability 0.94, 1 with probability 0.01 and 3 with probability 0.05. (a) Find the percentile premium for𝛼 = 0.95. (b) Suppose the insurer plans to sell 2 contracts. Find the smallest possible premium for each,suchthatthereisaprobabilityofatleast0.95thatthetotalpremiumswillcover the total benefits on both contracts. Solution. (a) This is obviously 1, which gives exactly a 0.95 probability of covering the benefits. (b) Using independence, direct calculation shows that the probability of the total present value of benefits being less than or equal tozis 0.9025 forz= 2 or 0.9965 forz= 3. The smallest total premium needed is then 3 resulting in a charge of 1.5 per policy. THE STOCHASTIC APPROACH TO PREMIUMS 237 Theconclusionofthisexampleisexactlycontrarytowhatonewouldexpect,andindicates a major flaw in the percentile premium approach. As the number of policies increases, the risk should decrease due to diversification. Unfavourable occurrences on one contract can be offset by favourable ones on another. The premium per policy should therefore decrease rather than increase. Despitethedrawbackinherentinpercentilepremiums,theremaybetimeswhenonewant to compute these, possibly for comparison with premiums produced by other methods. Here is a procedure for the case where benefits are paid at the end of the year of failure. We want to find the smallest possible level premium𝜋 payable forhyears, such that the probability of a nonpositiveLis greater than or equal to some given number𝛼. We consider only the case where L decreases as the value ofT increases. This is fairly typical. As the time of failure increases, the discount factor reduces the present value of the benefit paid. In addition, the later the occurrence of failure, the more premiums will be collected. Both of these factors tendtodecrease L.(Ofcourse,therequiredconditionmaynotholdinthecasewherebenefits are increasing rapidly.) The procedure is as follows: 1. Letk 0 be the largest positive integerksuch thats(k) ≥ (𝛼). 2. Solve for𝜋 so thatLis 0 wheñT = k0 +1. That is, bk0v(k0 +1) = 𝜋̈a(1r;v), where ris the minimum ofk0 +1 andh. SinceLis decreasing withk, L≤ 0 implies that ̃T ≥ k0. Therefore, 𝛼 ≤ s(k0) ≤ P(L≤ 0). Moreover,ifwetakeanysmallerpremium 𝜋′,thenfor ̃T = k0 +1,Lwillbepositive,sothat P(L≤ 0) ≤ sk0+1 <𝛼 . by our choice ofk0. Example 15.7 Refer back to Example 15.5. Find the level percentile premium𝜋 payable for 3 years if𝛼 = 0.8. Solution. We takek0 = 1 sinceP(̃T > 1) = 0.90 > 0.8, whileP(̃T > 2) = 0.72 < 0.8. When ̃T = 2,L= 200× 1 4 −𝜋(1+ 1 2), so that𝜋 = 33.33. This is much higher than the equivalence principle level premium of 13.31. 15.6.3 Aggregate premiums An analysis of the examples above indicate that the percentile premium method can give unreasonable results with highly asymmetric distributions, where there is a large probability of a value in one of the tails. This approach however is a reasonable one to use in order 238 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES to compute the total premiums for a large group of policies, where the distribution of total benefits is likely to be more heavily concentrated about the mean. Suppose the insurer issues severalcontractsofthesametypeandwanttobefairlycertain,thatintheaggregate,thetotal premiums collected will cover the total benefits. There are two major questions that could be considered. First, the premiums may be fixed and we want to determine the number of contracts to sell to achieve the desired confidence. Second, the number of contracts may be fixed and the problem is to determine the premium to charge. In general, exact calculations areprohibitive,butwecangetapproximateanswersbyassumingthatthetotalityofalllosses has a normal distribution (see Section A.11.5). It is convenient to introduce the following quantity, which we will discuss in some generality. Definition15.3 Forarandomvariable Xwithapositivemean,the coefficientofvariation of Xis the quantity CV(X) = √ Var(X) E(X) . It is clear from (A.8) and (A.10) that for any constantc, CV(cX) = CV(X), showing that, unlike variance or standard deviation, this is a measure of variation that is independentoftheparticularunits.Soforexample,ifwemeasurelossinUSdollarsandthen change the units to pounds sterling, the variance will be quite different, but the coefficient of variation will remain the same. There are some other facts about this quantity that we want to derive. SupposeS= X 1 +X2 +⋯ +XN where these are independent random variables each distributed asX. Then Var(S) = N Var(X), E(X) = NE(X), showing that CV(S) = CV(X)√ N , (15.17) illustrating the expected diversification effect that occurs when we take an independent sum of random variables. We expect to reduce variation, as some high values are offset by low values. Next, suppose thatX is a normal random variable with positive mean, and we want the probability thatX takes a value greater than or equal to 0. This is clearly independent of any units, so we should be able to express it in terms of the coefficient of variation. Indeed, suppose Xhas mean𝜇 and standard deviation𝜎, so thatX = 𝜇 +𝜎Z where Zis the standard THE STOCHASTIC APPROACH TO PREMIUMS 239 normal.Then, P(X ≥ 0) = P(Z ≥ −𝜇∕𝜎)whichbysymmetryisequalto P(Z ≤ 𝜇∕𝜎).Wecan write P(X ≥ 0) =Φ [CV(X)−1], (15.18) whereΦ is the cumulative distribution function of the standard normal. Considernowthefirstquestiongivenabove.Supposewehaveacontractwithaprospective loss at time 0 ofL, whereE(L) < 0 and we want to determine the smallest value ofN,s o that ifNindependent contracts are sold, the probability is at least𝛼 that premiums will cover benefits. Let S denotes the aggregate loss on all contracts and apply our results above to−S,t h e aggregategain.Wewantaprobabilityof 𝛼that−Swillbepositive.Assumethat −Sisnormal. Invoke (15.18) withX =− S, and takeΦ−1 of each side, resulting in Φ−1(𝛼) = CV(−S)−1. Then apply (15.17) withX =− Lto getΦ−1(𝛼) = √ N∕CV(−L), so that √ N =Φ −1(𝛼)CV(−L). (15.19)‡ We see then that the required number of contracts increases as the confidence level goes up, and also as the uncertainty inLgoes up, which is just what we expect to happen. For the second question above whereN is fixed, consider a single premium contract with presentvalueofbenefits Z.Lettherequiredpremiumbeoftheform(1 +𝜃)E(Z).Thequantity 𝜃 is known as therelativesecurityloading –t h ew o r d sriskloadingand contingencyloading a r ea l s ou s e d .( S oi f𝜃= 0.20 for example, it means that the premium is calculated by adding 20% to the equivalence principle premium.) Now in this case −L= (1+𝜃)E(Z)−Z, so that Var(−L) = Var(Z), andE(−L) = 𝜃E(Z). Therefore CV(−L) = CV(Z) 𝜃 . (15.20) Substituting from (15.19) we can write 𝜃 = Φ−1(𝛼)CV(Z)√ N . (15.21)‡ Thisiscertainlyreasonableasitshowsthattheloadingincreaseswiththetolerancelevel, and the uncertainty in the benefits, but decreases as more contracts are sold. Example 15.8 Suppose the force of interest is a constant 0.06 and that𝜇(t) = 0.04 for all t. An insurer sells 100 contracts, each of which pays 1000 upon failure. Assuming a normal approximation, how much should be charged as a single premium on each contract, so that 240 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES there is a 95% chance that premiums will be sufficient to cover the aggregate claims for this group of 100 contracts. Solution. If Z=the benefit for a contract paying one unit at failure. Then E(Z) = 𝜇 𝜇 +𝛿 = 0.4, E(Z 2 ) = 𝜇 𝜇 +2𝛿 = 0.25. The formula for the second moment reflects the fact that squaring the discount function is equivalent to doubling the force of interest. It follows that Var( Z) = 0.09 andCV(Z) = 0.75. From(15.21),with 𝛼 = 0.95,𝜃 = 1.645(0.75)∕10 = 0.123375.Foreach100-unitcontractthe equivalenceprinciplepremiumis400andsothepremiumchargedshouldbe400(1.123375) = 449.35. Example 15.9 Suppose the contract in Example 15.8 is sold with premiums paid con- tinuously for life at a level rate, where the security loading𝜃 = 0.10. Assuming a normal approximation, how many contracts must be sold, so there is a 95% chance that total premi- ums will cover total benefits? Solution. Sinceourformulasareindependentofunitswecanignorethe1000andconsidera 1-unit contract. The main problem is to determineCV(−L). This is not always so easy when wenolongerhaveasinglepremium,butitcanbedoneforlevelbenefitsandpremiums.Thisis carriedoutinChapter16,andwereferthereadertoformulas(16.5)and(16.6)with t= 0.The equivalence principle premium is 0.04, so the premium𝜋 = 0.044 and𝜋∕𝛿 = 11∕15. From formula(16.6) √ Var(L) = (26∕15))0.3 = 0.52.Fromformula(16.5) E(L) = (26∕15)(0.04)− 11∕15 =− 0.04. So CV(−L) = 0.52 0.04 = 13. From(15.19) √ N = 1.645×13,so N = 457.32.Thismeansthat458contractsmustbesold. 15.6.4 General premium principles Therearemanyotherpossibilitiesforcomputingpremiumstoallowfortherandomnatureof theloss.Theseareknowningeneralas premiumprinciples.Formally,apremiumprincipleis a functionH, which assigns to each random variableX representing a loss, a numberH(X), which is the premium to be charged for accepting the riskX. This definition is applicable to contracts purchased by a single premiums, rather than periodic premiums. The equivalence principle premium is given byH(X) = E(X). The standard deviation principle is given byH(X) = E(X)+𝛽 √ Var(X), for some positive𝛽. Dividing this byE(X), we can see from (15.21) that the standard deviation principle can be interpreted (assuming a normal approximation) as setting premiums so that there is a certain probability, which will THE VARIANCE OFr L 241 dependon N,thatpremiumswillcoverlosses.Anotherexampleisthe varianceprinciplegiven by E(X)+𝛽Var(X), for some positive𝛽. (Exercise 15.25 points out a flaw in this principle.) There is extensive treatment in the actuarial literature comparing the properties of these and other premium principles. We will not elaborate further here, but more comments can be found in Chapter 22. 15.7 The variance ofr L We have seen that in our stochastic model, the reserve represents the expected value of the amountthatweneedinordertomeetfutureobligations.Itisusefultohavemoreinformation about the distribution of this random variable. In this section we present a formula for calculatingthevarianceof rLunderthesetupofSection14.2.Ofcourse,ifwehavecomputed the exact distribution ofrL, as we did in Example 15.5 we could calculate the variance directly. The present method will allow us to calculate the variance without knowing the exact distribution, as long as we know all the reserves after timer. Moreover, it gives us a decompositionofthisvarianceintothatattributabletoeachofthefutureyears.Intheactuarial literature, this result is known asHattendorf’stheorem. We start by takingr= 0, so we compute the variance ofL= 0L. To simplify the task we look at the net cash outflow year by year. LetCk denote the value at timekof the net amount paid out in the year fromk to k+1. That is,Ck is equal to the value at timekof the failure benefit paid out at the end of the year, less the inflow at timek.S oCk is a random variable taking three possible values depending on whether failure has occurred before timek, during the year (k,k+1) or at timek+1 or later. If failure occurs before timek, nothing is paid and nothing is received. If failure occurs during the year (k,k+1), thenbk is paid at time k+1 and𝜋k is received at timek. If failure occurs after timek+1, nothing is paid out and a premium of𝜋k is received at timek. The following table summarizes the possible values of Ck and the respective probabilities: Time of failure Value ofCk Probability Before timek 01 −s(k) Between timekand k+1 v(k,k+1)bk −𝜋k ̃f(k+1) = s(k)𝜆(k+1) At timek+1 or later −𝜋k s(k+1) = s(k)(1−𝜆(k+1)) The random variableLis related to the values ofCk by L= ∞∑ k=0 v(k)Ck. (15.22) A fundamental property is that CjCk =− 𝜋jCk, j< k. (15.23) 242 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES To verify (15.23), observe that it is obvious ifCk = 0, since then both sides are 0. IfCk is not equal to 0, it means that failure did not occur before timekand therefore must have occurred at timej+1 or later. We see from the table above thatCj is equal to−𝜋j. We will first derive a variance formula for the case of a contract with zero reserves. This means that the premium paid at any timek is just enough to cover the death benefit paid at the end of that year, so that 𝜋k = v(k,k+1)bk𝜆(k+1), for all k. (15.24) The significance of this case is that, as we show below, the random variablesCh for different values ofhare uncorrelated (although they are not independent). Moreover, we have a fairly easy expression for the variance of eachCh, so that (15.22) gives us a formula for Var(L). Substitutingfrom(15.24),weseethatthesecondandthirdvaluesof Ck aslistedaboveare respectively v(k,k+1)bk[1−𝜆(k+1)] and −v(k,k+1)bk𝜆(k+1). It follows immediately that under assumption (15.24), E[Ck] = 0, for all k, (15.25) so that from (15.23), ifj< k, Cov(Cj,Ck) =− 𝜋jE(Ck) = 0. (15.26) Moreover, after some algebraic manipulation, Var[Ck] = E[C2 k ] = v(k,k+1)2b2 k s(k)𝜆(k+1)(1−𝜆(k+1)). (15.27) Invoking the fact the variance of a sum of uncorrelated random variables is the sum of the variances, and noticing thats(k)(1−𝜆(k+1)) = s(k+1), we deduce from (15.22) that Var(L) = ∞∑ k=0 v(k+1)2b2 k s(k+1)𝜆(k+1). We next consider the case of a general durationr, still keeping the assumption of zero reserves.Inthiscase, rLwilljustbe 0Lforthepolicyon(x +r)withthebenefitsandpremiums starting at ager, and with probabilities conditioned on the survival of (x) to agex+r.T h es and 𝜆 in the formula above will be those associated with the distribution of̃T◦ r. The result, using (15.10) and the discrete counterpart of (15.8), is Var(rL) = 1 s(r) ∞∑ k=0 v(r,r+k+1)2b2 r+k s(r+k+1)𝜆(r+k+1). Finally, we consider the general case where we remove the restriction given by (15.24). We split the policy into two parts, the risk portion and the savings portion, in exactly the same way as we did in Section 6.4.2. This leads to a corresponding split of the prospective loss into that portion of this loss attributable to the risk portion and that attributable to the STANDARD NOTATION AND TERMINOLOGY 243 savingsportion.Buttheprospectivelossforthesavingsportionmustbeaconstant,sinceitis completely independent of the mortality experience. (In fact this constant is just the present value, with respect to interest, of the pure endowment paid out at the end, less the savings portions of the premiums. Since the fund is reduced to zero, this present value must be what youstartedwith,whichis rV.)Themeansthatthevarianceoftheprospectivelossisthesame as the variance of the prospective loss on the risk portion. Therefore, the only change from thepreviousformulaistoreplacetheoriginaldeathbenefitsbythebenefitsattributabletothe risk portion, namely the net amount at risk. Our final formula is Var( rL) = 1 s(r) ∞∑ k=0 v(r,r+k+1)2𝜂2 r+ks(r+k+1)𝜆(r+k+1). (15.28)‡ This leads to a backwards recursion formula. Splitting off the first term, Var(rL) = v(r,r+1)2s(r+1) s(r) [𝜂2 r𝜆(r+1)+Var(r+1L)]. (15.29) This is useful ifT is bounded, as is true withT(x). We can start the recursion with a value of 0 for the variance of the loss at the final duration. Example 15.10 Refer back to Example 15.5. Use (14.23) to find the variance of 1L. Solution. We will first calculate the product of the three mortality factors in (14.23). In the case whereT = T(x), this product for indexk is k+1px+r qx+r+k.F o rk = 0w eg e t0.8×0.2, and fork= 1 we get 0.8×0.7×0.3. We also calculate that2V = 5, so that𝜂1 = 195. Then Var(1L) = 1 4 ×1952 ×0.16+ 1 16 ×1002 ×0.168 = 1626. We can verify this answer directly from the solution to Example 15.5, where we have the complete distribution of 1L. 15.8 Standard notation and terminology Theonlymajoritemwehavenotyetdiscussedistheuseofanupperleftsubscript2todenote a quantity that is computed at a force of interest that is double that of the standard one or, equivalently,atadiscountfunctionthatisthesquareofthestandardone.Thiswasdeveloped for variance formulas. Its use is, however, restricted to cases where the failure benefits are either 0 or 1. In that case, squaring the benefits leaves them unchanged, and the second moments are calculated by just squaring the discount function. For example, the variance for the benefits paid on a 1-unit, end-of-the-year-of-death,n-year term insurance would be written as 2A1 x:n ⌝ − ( A1 x:n ⌝ )2 . 244 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES Notes and references To be precise, one should specify what convergence means in Equation (15.8) in the case that the sum is infinite. We did not worry about that here, since the important result for our purposes is (15.9) which, as we noted, is valid in almost all cases. Young (2004) discusses a variety of premium principles and their properties. Exercises Type A exercises 15.1 Afailuretime T isuniformlydistributedontheinterval[0,10].Aninsurancecontract pays e0.06t at the moment of failure if this occurs at timet. The force of interest is a constant 0.04. Find the expectation and variance of the present value of the benefits. 15.2 A 3-year life annuity on (x) and (y) provides for annuity benefits of 1 at time 0, 2 at time 1 and 4 at time 2, provided that both individuals are alive. You are given that p x = 0.8, 2px = 0.6, py = 0.7, 2py = 0.5 and the interest rate is a constant 100%. (a) Find the expectation and variance of the present value of the benefits. (b) A single premium for the contract is found by adding a 20% relative risk loading tothepresentvalue.Whatistheprobabilitythatthepremiumwillbesufficientto provide for the benefit payments on a single contract? 15.3 A life annuity contract provides for continuous payments at the annual rate of e −0.05t attime t.Theforceofmortalityisaconstant0.1,andtheforceofinterestisaconstant 0.05. IfY is the present value of the benefits, findE(Y) and Var(Y). 15.4 A 10-year term insurance policy provides for a payment of 10−t at the moment of the first death of (x) and (y), should this occur at timet. The mortality of (x) and (y) is as given in Exercise 14.9. Assuming an interest rate of 0, find the expectation and variance of the present value of the benefits. 15.5 A failure time T is uniformly distributed on the interval [0,10]. There is a constant interest rate of zero. An annuity provides for continuous payments at the rate oftat time t, provided that failure has not yet occurred. Find the expectation and variance of the present value of the benefits. 15.6 A 4-year pure endowment contract on (60) provides for 1000 paid at age 64 if (60) is thenalive.Nothingispaidifdeathoccursbeforeage64.Thisispurchasedbyfourlevel annual premiums of 100. You are given thatq 60 = 0.1,q61 = 0.2,q62 = 0.25,q63 = 0.30,andtheinterestrate iisaconstant25%.Writedowntheexactdistributionof 1L. Use this to find1V. 15.7 Acertainproducthasafailuretimeof T.Theapproximatingdiscreterandomvariable ̃T has a constant hazard function𝜆(k) = 0.3. (In other words, there is a 30% chance that the product will fail each year, given that it was still working at the beginning of the year.) An insurer agrees to pay 100 at the end of the year of failure should the product fail within 2 years. In return it collects a premium of 40 now and a second EXERCISES 245 premium of 20 at time 1, if failure did not occur in the first year. The interest rate is 25%. (a) FindE(L), Var(L), Var( 1L). (b) Find the probability that, for a given contract, the premiums collected will be sufficient to pay the benefits. 15.8 A 3-year endowment insurance on (70) provides death benefits of 100 payable at the endoftheyearofdeath,ifthisoccurswithin3years,plusapureendowmentof100at age 63 if the insured is then alive. This is purchased by two premiums of 30 payable at age 60 and 61. You are given thatq 60 = 0.2,q61 = 0.4, and the interest rate is a constant25%.Givethecompletedistributionoftherandomvariable Landusethisto find 0V. 15.9 Aninsurerchargesasinglepremiumof E(X)+0.1 √ Var(X)foracontractwithpresent value of benefits equal toX. Use a normal approximation. (a) How many contracts must be sold so that the probability is 95% that premiums cover benefits? (b) What is the probability that premiums cover benefits if 225 contracts are sold? 15.10 A failure time has a hazard rate of 𝜇(t) = 2 20−t. A contract provides for a benefit of 1 at time 10 provided that failure occurs before time 10 (so, for example, if failure occurs at time 6, the benefit is not paid until 4 years later). You are given thatv(10) = 0.6. Find the expectation and variance of the present value of the benefits. 15.11 Consider a 3-year life annuity on (60). The sequence of payments (beginning at time 0) is 4, 2, 1. You are given thatq 60 = 0.2, q61 = 0.4, q62 = 0.5. The interest rate is a constant 25%. The insurer sells 100 such contracts, charging (1+r) times the equivalenceprinciplesinglepremium.Whatshould rbesothatthereisatleasta95% chance that premiums will cover the benefits? (Use a normal approximation.) 15.12 A2-yearterminsurancecontractsoldtotwolives,(70)and(71),providesforbenefits payable at the end of the year of the first death. The benefit paid is 40 in the first year and 50 in the second year. You are given thatq 70 = 0.4,q71 = 0.5,q72 = 0.6, and the interest rate is 100%. LetZdenotes the present value of the benefits. (a) FindE(Z) and Var(Z). (b) Suppose the contract is to be paid for by level annual premiums ofPpayable for 2 years. What shouldP be if the equivalence principle is used, and what is the resulting probability thatL≤ 0? (c) What is the smallest amount thatPcould be if we want a probability of at least 0.25 that the lossLwill be less than or equal to 0? 246 THE STOCHASTIC APPROACH TO INSURANCE AND ANNUITIES 15.13 A 3-year endowment insurance on (70) provides death benefits of 400 payable at the end of the year of death, if this occurs within 3 years, plus a pure endowment of 200 atage73,iftheinsuredisthenalive.Thisispurchasedbythreelevelpremiumsof40. You are given thatq 70 = 0.2,q71 = 0.4,q72 = 0.5, and the interest rate is a constant 100%. (a) Givethecompletedistributionsfortherandomvariable L,andtherandomvariable 1L. Use the latter to find the reserve at time 1. (b) Whatshouldthelevelpremiumbeiftheinsurerwantsthesmallestpremiumsuch that the probability of a positive loss is less than 25%? 15.14 Use Equation (15.12) to calculate the variance in Example 15.3. 15.15 You are given thatq x = 0.1, qx+1 = 0.2, qy = 0.2, qy+1 = 0.3, and the rate of interest is a constant 25%. Find the variance of the present value of the benefits in each of the following contracts (a) A payment of 100 is made at the end of the year of the second death, provided this occurs within 2 years. (b) Three annual payments of 100 are made, the first at time 0, provided that at least one of (x)o r(y) are alive. 15.16 Afailuretime Thasauniformdistributionon[0,30].Theforceofinterestisaconstant 5%. A single premium insurance contract pays 1 unit at the moment of failure. It is desiredtohaveatleasta95%probabilitythatpremiumswillcoverbenefitsonalarge group of contracts. Answer the following assuming a normal approximation. (a) Suppose the security loading is 20%. How many contracts should be sold? (b) It is estimated that 100 contracts will be sold. What should the security loading be? 15.17 RedoExample15.2(b)onlynowassumingthattheinterestrateis25%inthefirstyear and 50% in the second year. Type B exercises 15.18 A failure time has a constant hazard rate of𝜇 and the force of interest is a constant 𝛿. A contract pays e𝛾t at the moment of failure for failure at timet, where𝛾<𝜇 +𝛿. Find the expectation and variance of the present value of the benefits as a function of 𝜇,𝛿 and𝛾. 15.19 Suppose that the force of mortality𝜇 is a constant 0.04 and the force of interest is a constant 0.06. (a) An insurer sells to one individual a single-premium, whole-life policy, with 100 payable at the moment of death, and also sells, to an independent life, a life annuity with benefits payable continuously at the rate ofcper year. IfW denotes the present value of the benefits on the two contracts combined, find Var(W)a sa function ofc. EXERCISES 247 (b) Supposenowthatthetwocontractsin(a)aresoldtothesameindividual.Whatis thevarianceof Winthiscase?Isitlessthan,greaterthanorequaltothevariance in (a)? Explain why. For what values ofc, if any, will the variance be equal to 0? Explain. 15.20 Afailuretime T hasap.d.f.of f(t) = (12−t)∕72for0 ≤ t< 12.Anannuitycontract provides for continuous payments at the annual rate of 1, which stop at failure, or at time 6 if earlier. The interest rate is 0. Let Y denote the present value of the benefits. (a) ExpressY as a function ofT. (b) CalculateE(Y) and Var(Y). (c) The single premium charged for this contract is the expected value plus 20%. What is the probability that the premium will cover the benefits? 15.21 Foracertainfailuretime T,aninsurancecontractpaysbenefitsattheendoftheyearof failure. You are given thatsT(9) = 0.8, sT(10) = 0.7 E(10L) = 70,Var(10L) = 1000. The amount payable at time 10 for failure between time 9 and time 10 is 100. The interest rate is a constant 20 %. Find Var( 9L). 15.22 Adiscretefailuretime T hasaconstanthazardrateof 𝜆(k) = 1∕2.Thereisaconstant interest rate of 100%. An insurance contract pays a benefit of 1 unit at the end of the year of failure. Level annual equivalence principle premiums are payable prior to failure. Using formula (15.28), show that for any nonnegative integerr, Var( rL) = 1∕15. 15.23 Consider two independent lives (x),(y) , whereT(x) is uniform on [0,1] andT(y)i s uniform on [0,2]. (a) Calculate the p.d.f. of the last survivor failure timeT( xy). (b) An annuity provides continuous payments for as long as either (x)o r(y)i sa l i v e . The rate of payment at timetis 3t2. The rate of interest is zero. IfY denotes the present value of the benefits for this annuity, findE(Y) and Var(Y). 15.24 An endowment insurance on (x) provides for 1000 at the end of the year of death, provided this occurs within 4 years, and 1000 at time 4 if (x) is still alive. Net level annual premiums are payable for 4 years. You are given thatqx = 0.05,qx+1 = 0.08,qx+2 = 0.10. Interest rates are 5% for the first 2 years and 6% thereafter. Find Var(1L) in two ways: first, by finding the exact distribution of1L; and second, by finding 2V and 3V and then using Equation (15.28). 15.25 A failure time takes the values 1 or 2 each with probability 1/2. The interest rate is 0. Insurance contract A pays 5 at failure. Insurance contract B pays 1 if failure occurs at time 1, and 5 if failure occurs at time 2. Show that using the variance premium principle with𝛽 = 1, the premium for contract B is higher than that for contract A. Spreadsheet exercise 15.26 Modify the spreadsheet of Chapter 6 to calculate Var(rL). 16 Simplifications under level benefit contracts 16.1 Introduction The calculation of variances and other distributional features simplifies considerably when we have level benefits and constant interest. In fact, we can write down formulas for exact distributions of the major random variables of interest. Throughout this chapter, we consider the following setup. We have a general failure timeT. We will consider insurances paying a level amount upon failure, either at the end of the year of failure or at the moment of such, and we will consider annuities paid prior to the failure ofT with either a level pay- ment or continuous payments at a level rate. In addition, we assume a constant force of interest 𝛿. By takingT to beT(x), this will apply to level benefit, whole-life insurances and to level benefit whole-life annuities. By takingT = min{T(x),n}, this will apply to level benefitn- yearendowmentpoliciesandtolevelbenefit n-yeartemporarylifeannuities.Ourassumption does not apply to term insurance, even when there is a level benefit during the term, since the benefit drops to zero after the expiration of the contract. However, in Section 16.5 we do illustrate that the calculation of exact distributions is possible for term or deferred insurances with a level death benefit paid over the benefit period. 16.2 Variance calculations in the continuous case It is convenient to begin with a continuous failure timeT. 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 VARIANCE CALCULATIONS IN THE CONTINUOUS CASE 249 16.2.1 Insurances Consider an insurance policy paying 1 at the moment of failure. The discount function is given byv(t) = v t = e−𝛿t.L e t̄Z be the present value of the benefits. In this case there is little simplification, and we know from the previous chapter that ̄AT = E(̄Z) = E(vT ), Var( ̄Z) = E(v2T)−(̄AT )2. (16.1) 16.2.2 Annuities Consider an annuity with continuous payments at the rate of 1 per year, made prior to the occurrence of failure. If̄Y is the present value of the benefits, then ̄Y = ̄a(1 T ) = 1−vT 𝛿 = 1− ̄Z 𝛿 , (16.2)‡ ̄aT = E(̄Y) = 1−E(̄Z) 𝛿 ,V a r ( ̄Y) = Var(̄Z) 𝛿2 . (16.3)‡ For the case whereT = min{T(x),n}, we have already seen the first part of (16.3). This was the continuous ‘endowment identity’ given at the end of Section 8.8. 16.2.3 Prospective losses Consideracontractthatpays1unitatthemomentoffailureandhascontinuouslevelpremiums at an annual rate of𝜋 payable prior to failure. (As a practical matter, this means that we are dealingwithanetpremiummodelthatignoresexpenses,whichareunlikelytobelevel.)The prospective loss at timet is tL = ̄Z ◦ t −𝜋(̄Y ◦ t), (16.4) where ̄Z ◦ t = vT ◦ t and ̄Y ◦ t = (1− ̄Z ◦ t)∕𝛿. This leads to tL = ( 1+ 𝜋 𝛿 ) vT ◦ t − 𝜋 𝛿. (16.5) This greatly simplifies the calculation of the variance of the prospective loss. Compare the following with (14.23). Var( tL) = ( 1+ 𝜋 𝛿 )2 Var(vT ◦ t). (16.6)‡ 16.2.4 Using equivalence principle premiums Supposethat 𝜋isanequivalenceprinciplepremium.Then 𝜋E(̄Y) = E(̄Z),sothatfrom(16.3), ̄ 𝜋= 1 ̄aT −𝛿. (16.7) 250 SIMPLIFICATIONS UNDER LEVEL BENEFIT CONTRACTS Takingt = 0 in (16.6), Var(L) = Var(vT ) (𝛿̄aT )2 = Var(̄Z) (1− ̄AT )2, (16.8) showing that Var(̄Z) < Var (L). This reflects the fact that there is more risk involved in selling an insurance contract where premiums are payable over the entire life of the contract as opposed to the single-premium case. For failure occurring early, the insurer not only loses interest but will have collected relatively small amounts in premiums. For a final formula, express̄Z ◦ t in terms of̄Y ◦ t in (16.4). Then tL = (1−(𝜋 +𝛿)(̄Y ◦ t)) If𝜋 isanequivalenceprinciplepremium,wecansubstitutefrom(16.7)andtakeexpectations to give a simple formula for the reserve at timet. t ̄V = 1− ̄aT ◦ t aT . (16.9) (ForT(x), we obtained the discrete version of this in (6.18).) 16.3 Variance calculations in the discrete case Wenow consider thecasewhere failurebenefits arepaidat theend oftheyear of failure,and annuity benefits and premiums are payable yearly. All of the formulas in Section 16.2 have discrete counterparts, which for the most part are obtained by replacingT by ̃T, ̄A by A,a by ̈a and 𝛿 by d. We will leave the formal derivations to the reader, but will list the formulas with the corresponding equation numbers as in Section 16.2, only with a prime to denote the discrete case. If Z denotes the present value of the benefits for an insurance paying 1 unit at the end of the year of failure, A ̃T = E(Z) = E(ṽT ), Var(Z ) = E(v2̃T)−(ÃT )2. (16.1′) IfY denotes the present value of an annuity paying 1 unit yearly provided that failure has not occurred, Y = ̈a(1T ) = 1−ṽT d = 1−Z d , (16.2′)‡ E(Y) = 1−E(Z) d ,V a r ( Y) = Var(Z) d2 . (16.3′)‡ VARIANCE CALCULATIONS IN THE DISCRETE CASE 251 Now consider a contract which pays 1 unit at the end of the year of failure and has level annual premiums of𝜋 payable prior to failure. Then for any positive integerk, kL = ( 1+ 𝜋 d ) ṽT ◦ k − 𝜋 d , (16.5 ′) Var(kL) = ( 1+ 𝜋 d )2 Var(ṽT ◦ k). (16.6′)‡ Suppose that𝜋 is an equivalence principle premium. Then 𝜋 = 1 ̈ãT −d, (16.7 ′) Var(L) = Var(ṽT ) (d̈ãT )2 = Var(ṽT ) (1−ÃT )2, (16.8′) kV = 1− ̈ãT ◦ k ̈ãT . (16.9′) One could possibly consider expenses when using the above simplified prospective loss formulas if the difference is only in the first year, as the following example shows. Example16.1 Aninsurancepolicyprovides1000attheendoftheyearofdeathwithlevel premiums payable for life. There are expenses in the first year of 60% of the premium plus 50, and in the subsequent years of 15% of the premium plus 20. In addition, there is a death benefitsettlementexpenseof50.Youaregiventhat E(Z) = 0.4andVar(Z ) = 0.10where Z is the present value of 1 unit of death benefit. The rate of discount is a constant 0.06. Find Var (L) whereL is calculated using expense-augmented premiums, and including all expenses. Solution. We first compute the expense-augmented premiumG as G̈a x = 0.45G +0.15G̈ax +30+20̈ax +1050Ax so that G = 1050Ax +30+20̈ax .85̈ax −0.45 . We know thaẗax = (1−Ax)∕d = (1−0.4)∕6 = 10, and substituting in the above,G = 80.75.This means that the total inflow after the first year is 0.85(80.75)−20 =48.64. Nowconsiderapolicywherethetotalinflowwas48.64ineveryyear.Thevalueof L from thatcontractwoulddifferfromthatontheoneinquestiononlybyaconstantamount,namely the extra amount in the first year due to higher expenses. Therefore, the variance ofL would bethesame,andwecanuseformula(16.6 ′),with 𝜋 = 48.64,andthe1replacedbythedeath benefit of 1050. From (16.6′), Var(L) = ( 1050+ 48.64 0.06 )2 0.10 = 346208. 252 SIMPLIFICATIONS UNDER LEVEL BENEFIT CONTRACTS 16.4 Exact distributions Inthissection,wecalculatetheexactdistributionsfor ̄Z, ̄Y andL.Ineachcase,wewillderive distribution functions. These will be given for values between the greatest lower bound and leastupperboundofthevalues.(Weknowthat F takesthevalue0forargumentslessthanthe greatestlowerbound,and1forargumentsgreaterthantheleastupperbound.)Itisconvenient here to introduce some new notation. For any random variableX,l e t ̂F X(x) = P(X < x), ̂sX(x) = P(X ≥ x). Of course, whenX is continuous,̂F = F and ̂s = s. The distribution functions we want are all easily expressed in terms of the distribution of T.L e tN denote the least upper bound of the values ofT. In the case thatN =∞ (as for example whenT is exponential), the termvN in the formulas below will be equal to 0. 16.4.1 The distribution of ̄Z The distribution of̄Z is given by F̄Z(z) = P(e−𝛿T ≤ z) = P ( T ≥ −logz 𝛿 ) = ̂sT ( −logz 𝛿 ) , vN < z < 1. (16.10) 16.4.2 The distribution of ̄Y First note that arguing as in (16.10) gives ̂FZ(z) = sT (−log(z) 𝛿 ) . Using (16.2) F̄Y (y) = P(̄Z ≥ 1−𝛿y) = 1− ̂F̄Z(1−𝛿y). Substituting from above, F̄Y (y) = 1−sT ( −log(1−𝛿y) 𝛿 ) = FT ( −log(1−𝛿y) 𝛿 ) ,0 ≤ y < 1−vN 𝛿 . (16.11) 16.4.3 The distribution ofL The minimum value ofL will be −𝜋 𝛿 + ( 1+ 𝜋 𝛿 ) vN, EXACT DISTRIBUTIONS 253 occurringforfailureattime N,anditsmaximumvaluewillbe1,occurringforfailureattime 0. Using (16.5), FL(u) = P [( 1+ 𝜋 𝛿 ) e−𝛿T ] ≤ u + 𝜋 𝛿 = P [ e−𝛿T ≤ (𝛿u +𝜋 𝛿 +𝜋 )] (16.12) = ̂sT [ −1 𝛿 log (𝛿u +𝜋 𝛿 +𝜋 )] , −𝜋 𝛿 + ( 1+ 𝜋 𝛿 ) vN ≤ u ≤ 1. For the more general case,FtL(u) is given by the same formula, but witĥsT ◦ t replacing ̂sT and vN−t replacing vN. 16.4.4 The case whereT is exponentially distributed In the particular case whereT is exponential with constant hazard function𝜇, we know that ̂sT (z) = sT (z) = e−𝜇z andN =∞ . The above formulas simplify to F̄Z(z) = z𝜇∕𝛿,0 < z < 1, (16.13) F̄Y (y) = 1−(1−𝛿y)𝜇∕𝛿,0 ≤ y < 1 𝛿, (16.14) FL(u) = (𝛿u +𝜋 𝛿 +𝜋 )𝜇∕𝛿 , −𝜋 𝛿 ≤ u ≤ 1. (16.15) Itisofinteresttoobservethatif 𝜇 = 𝛿intheexponentialcase,theexponentintheformulas above equals 1 so that̄Z, ̄Y andL are alluniform random variables. Example 16.2 A company decides to add 20% to its equivalence principle premiums as a protection against unfavourable experience. In each of the following cases, find the probabilitythatpremiumswillcoverclaims.Supposethat T isexponentialwith 𝜇 = 0.04and that𝛿 = 0.06. (a) Asingle-premiumannuityprovidingcontinuouspaymentsattheannualrateof1prior to failure. (b) Acontractpaying1unitatfailure,withlevelpremiumspayablecontinuouslypriorto failure. Solution. (a) The equivalence principle premium is 1∕(𝜇 +𝛿) = 10. The actual premium charged will be 12. From (16.14), P(Y ≤ 12) = 1−0.28 2∕3 = 0.57. 254 SIMPLIFICATIONS UNDER LEVEL BENEFIT CONTRACTS (b)Theequivalenceprincipleannualpremiumrateisjust 𝜇 = 0.04,sotheactualpremium rate charged is 0.048. The probability that premiums cover claims is just P(L ≤ 0) = (0.048 0.108 )2∕3 = 0.58. 16.5 Some non-level benefit examples It is also possible to obtain exact distributions in some simple cases involving non-level benefits such as term or deferred insurance. 16.5.1 Term insurance Consider a contract that pays 1 at failure, provided failure occurs withinn years. In order to handlefailuretimesthatarenotcontinuous,weadopttheconventionthatabenefitispaidfor failureatexacttime n.Wewillcomputethedistributionof ̄Z,thepresentvalueofthebenefits. The minimumpositive value of̄Z is e −𝛿n occurring for death at timen. Since nothing is paid for death strictly after timen, which occurs with probabilitysT (n), we can easily obtain the term distribution from the whole life distribution. Pick up the probability mass to theleft of z = e−𝛿n andsetitdownasapointmassatthepoint0,asillustratedinFigure16.1.Fromthis, utilizing formula (16.10), we can read off this distribution of̄Z. F̄Z(z) = ⎧ ⎪ ⎨ ⎪ ⎩ s T (n), 0 ≤ z < e−𝛿n, ̂sT (−log z 𝛿 ) , e−𝛿n ≤ z ≤ 1. (16.16) Example16.3 Redopart(b)ofExample16.2,butnowassuming n-yearterminsurancefor n = 15 and 10. Level premiums are payable continuously forn years. Solution. Theequivalenceprinciplepremiumrateisstill 𝜇 = 0.04sotheactualpremiumrate charged is still 0.048 and𝜋∕𝛿 = 0.8. For the contract in Example 16.2, the value ofL when T = n is 1.8e−0.06n −0.8. Forn = 15, this is negative. This means thatL becomes negative at some point prior to expiration of the term contract, so the probability that premiums cover claims is 0.58, exactly the same as it was in constant benefit case. Whenn = 10, the value of L for the contract in Example 16.1 is positive. Therefore, in order thatL in this example be negative, it is necessary thatT ≥ 10, so that no benefits are paid. The probability of this is e −0.6 = 0.67. 16.5.2 Deferred insurance Asimilarexampleisprovidedby deferred insurance.Consideracontractthatpays1atfailure, provided failure occurs after timen. We now adopt the convention that nothing is paid for failure at exact timen. In view of this convention, there need not be a maximum value of̄Z but e−𝛿n is certainly an upper bound, since any benefits will be paid at a time later thann. SOME NON-LEVEL BENEFIT EXAMPLES 255 0 z e–δn e–δn e–δn 1 Whole life ST (n) FT (n) 11 z z 00 n-Year deferred n-Year term ST (n)FT (n) ST (n) FT (n) Figure 16.1 Graph of f̄Z(z) for various types of insurance Refer again to Figure 16.1. Since nothing is paid for death at or before timen, which occurs with probabilityFT (n), the probability mass to theright of e−𝛿n is picked up and set down as a point mass at 0. Utilizing (16.10), we can write F̄Z(z) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ F T (n) z = 0, FT (n)+ ̂sT (−log z 𝛿 ) 0 < z < e−𝛿n, 1 e−𝛿n ≤ z ≤ 1. 16.5.3 An annual premium policy We next investigate a more complicated case where we compute the exact distribution ofL in an annual premium policy. Consider a failure timeT that is unbounded (i.e.N =∞ ). We consider insurances which have premiums payable continuously at a level annual rate𝜋 for the duration of the contract. We will compare the distribution ofL for a contract that pays 1 unit on failure or at timen if earlier, and a contract that pays 1 unit at failure provided this occurs withinn years.These thencorrespond respectively toendowment andterminsurance. To simplify the notation, let b0 =− 𝜋 𝛿 + ( 1+ 𝜋 𝛿 ) vn, c0 =− 𝜋 𝛿 + 𝜋 𝛿vn. In both cases, the value ofL is given by (16.5) (witht =0), provided thatT takes a value less than n, which corresponds toL taking a value greater thanb0. So for any interval (a,b) with a > b0, we have thatP(a < L ≤ b) is the same in both cases and this value can be calculated directly from (16.12). 256 SIMPLIFICATIONS UNDER LEVEL BENEFIT CONTRACTS b0–π/δ c0 0 b0–π/δ c0 0 b0–π/δ c0 0 u uu FT (n) FT (n) FT (n) ST (n) ST (n) ST (n) Whole life n-Year term n-Year endowment Figure 16.2 Graph offL(u) for various types of insurance Consider the remaining probability ofsT (n). For the endowment contract, this will all be concentrated at the single pointb0. For the term contract, it will all be concentrated at the single pointc0, which lies in the interval (−𝜋∕𝛿,b0). Asn increases to∞ both b0 and c0 approach −𝜋∕𝛿 and the distribution approaches the whole-life case as given in (16.12). The various density functions are compared in Figure 16.2. The probability mass to left ofthepoint b0 onthewhole-lifegraphispickedupandsetdownasapointmassat b0 forthe endowment policy, or atc0 for the term policy. Exercises Type A exercises 16.1 Foracertainfailuretime T,aninsurancecontractpays1atthemomentoffailure,and has level premiums payable continuously prior to failure at the annual rate of 0.06. You are given that Var(̄Z) = 0.064, Var( ̄Y) = 10, where ̄Z is the present value of the failure benefits, and̄Y is the present value of an annuity contract with continuous payments at the annual rate of 1, payable prior to failure. Find Var(L). 16.2 For a certain failure timeT, an insurance contract pays 1 at the moment of failure. Level equivalence principle premiums are payable prior to failure. IfE(̄Z) = 0.6 and Var(L) = 2, find Var(̄Z). 16.3 A whole-life insurance contract provides for 1 unit payable at the moment of death. Level premiums are payable continuously for life at the annual rate of 0.08. The force of mortality is a constant 0.05, and the force of interest is a constant 0.10. How many contracts must be sold in order that there is at least a 95% chance that the total premiums on all these contracts will cover the total benefits? Use a normal approximation. EXERCISES 257 16.4 Aninsurersells100whole-lifeinsurancepolicieseachprovidingfor1unitpayableat the moment of death. Level premiums are payable continuously for life. The force of interest is a constant 0.06, and the force of mortality is a constant 0.04. What should the rate of premium payment be on each policy, in order that there is a 95% chance that total premiums on all 100 policies will cover the total benefits on all policies? Use a normal approximation. Type B exercises 16.5 (See Exercise 13.8.) A continuous failure time has a density function,fT (t) = 𝛽2te−𝛽t (a gamma distribution with first parameter 2). The force of interest is a constant𝛿. Find expressions in terms of𝛽 and 𝛿 for: (a) ̄AT ;( b )̄aT ; (c) the net annual rate of premium payment when premiums are payable continuously prior to failure, for an insurance paying 1 at failure and (d) the reserve at timek for the contract in (c). 16.6 An insurance contract, based on the failure timeT, pays 1 unit at the moment of failure provided this occurs within 5 years. Nothing is paid for failure after that time. The force of interest is a constant 0.1. IfT has the hazard function 𝜇 T (t) = 2 10−t, find the probability that the present value of the benefits is strictly positive, but less than or equal to e −0.3. 16.7 Suppose T isuniformon[0,10],and 𝛿 = 0.05.Acontractpays1unitatfailure.Level premiums are payable continuously for 10 years. The premiums charged are the net premiums plus 20%. (a) Find the probability thatL ≤ 0. (b) Supposenowthat,insteadof1.2timesequivalenceprinciplepremiums,thecom- pany charges premiums at the rate of 0.05 per year. What is the probability that L ≤ 0? 16.8 A failure timeT is uniformly distributed on the interval [0, 20]. The force of interest is a constant 0.05. (a) Adeferredinsurancecontractprovidesforapaymentof1atthemomentoffailure providedthatthisoccursaftertime5.If ̄Z isthepresentvalueofthebenefits,find the 80th percentile of̄Z. That is, find the pointz such thatP(̄Z ≤ z) = 0.80. (b) Find the 80th percentile of̄Z when the contract is a term insurance policy that pays 1 unit if failure occurs in the first 5 years. 16.9 Consider a deferred term insurance contract. It provides for 1 unit payable at the moment of failure provided that this occurs betweenN and 2N years from now. Nothing is paid if failure occurs in the firstN years or after 2Nyears. You are given that the force of failure𝜇 and the force of interest𝛿 are constants such that𝜇 = 𝛿∕2. Moreover, you are given that e −N𝛿 = 0.36. If ̄Z is the present value of the benefits, calculateF̄Z(z) for all nonnegative values ofz. 258 SIMPLIFICATIONS UNDER LEVEL BENEFIT CONTRACTS 16.10 An insurance contract provides for e𝛿t payable at failure, should this occur at timet. Net level premiums are payable continuously prior to failure. The force of interest is a constant𝛿. Show that L = v T − ̄AT 1− ̄AT . 16.11 An insurance policy provides a benefit of 1 at the moment of failure provided this occurs after 10 years. The hazard rate of the failure time is a constant 0.10 and the force of interest is a constant 0.05. If̄Z is the present value of the benefits, find the probability that: (a) 0< ̄Z ≤ 0.5, (b) 0≤ ̄Z ≤ 0.5 and (c) 0≤ ̄Z ≤ 0.7. 16.12 An insurance policy provides a benefit of 1 at the moment of failure plus a pure endowmentof1attime10.Thisispurchasedbylevelpremiumspayablecontinuously for 10 years. The premium is the net premium plus 10%. There is a constant force or mortalityof0.05andaconstantforceofinterestof0.05.Find(a) P(−1∕3 < L ≤ 1∕2) and (b)P(−1∕4 ≤ L ≤ 1∕2). 16.13 Let T be any failure time, and assume constant interest. (a) Show that E[̄a(1 T )]2 = 2 𝛿(̄aT −2̄aT ), where the superscript 2 on the right-hand side indicates calculation at a force of interest equal to 2𝛿or equivalently withv replaced byv 2. (b) Is the following formula true? If not, give a correct version. E[̈a(1̃T )]2 = 2 d (̈ãT −2 ̈ãT ). 16.14 ModifytheformulainSection16.5.2inthecasethatthereisabenefitpaidforfailure at exact timen. 17 The minimum failure time 17.1 Introduction Supposethat T1,T2,…,Tm arefailuretimesdefinedonthesamesamplespace.Inthischapter, we investigate the random variable T =m i n{T1,T2,…,Tm}. Inotherwords, Tisthetimeofthefirstfailuretooccuramongthe mdifferentfailuretimesthat arepossible.Inparticular,thiswillrecastthematerialofChapters10and11intoastochastic framework, and show that both joint-life theory and multiple-decrement theory are special cases of this general problem. In addition, it will provide more rigorous arguments for some of the results of those chapters that were obtained in an intuitive fashion. Finally it will deal with the important cases where the failure times need not be independent. Inthejoint-lifecase,wherewehaveagroupof mlivesnumbered1,2, …,m,wecantake T i tobethefuturelifetimeofthe ithlife,sothat T isthefailuretimeofthejoint m-lifestatus. In the multiple-decrement context, we can takeTi to be the time of failure from causei in theassociatedsingle-decrement setting, so it is the failure time of theith cause, assuming no other causes of failure are operating. Then, the random variableT is the time of failure in the multiple-decrement model. (In the machine analogy of Section 11.6,Ti would be the failure time of theith part.) 17.2 Joint distributions Wewishtoexpandsomewhatonourbriefdescriptionforthecase m= 2giveninAppendixA. Supposethateach Ti iscontinuous.Therearevariouswaysofdescribingthejointdistribution. 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 260 THE MINIMUM FAILURE TIME We can do so by the joint density functionfT1T2,…,Tm(t1,t2,…,tm), or alternatively by the joint distribution function FT1,T2,…,Tm(t1,t2,…,tm) = P[T1 ≤ t1,T2 ≤ t2,…,Tm ≤ tm], or by the joint survival function sT1,T2,…,Tm(t1,t2,…,tm) = P[T1 > t1,T2 > t2,…,Tm > tm]. To simplify the notation, we will often omit the subscripts and just writef,F or s when no confusion arises. The reader is cautioned thats(t1,t2,…,tm) ≠ 1−F(t1,t2,…,tm) whenm> 1. Asintheone-dimensionalcase,weintegratetoobtainthedistributionorsurvivalfunctions from the density function, and differentiate to go in the other direction. We must, however, use multiple integrals and partial derivatives. For example, withm= 3, F(t 1,t2,t3) = ∫ t1 0 ∫ t2 0 ∫ t3 0 f(u,v,w)dwdvdu, and similarly s(t1,t2,t3) = ∫ ∞ t1 ∫ ∞ t2 ∫ ∞ t3 f(u,v,w)dwdvdu. Nowdifferentiatethelatterexpressionwithrespectto t1.Thefundamentaltheoremofcalculus tells us to replace the variableuin the integrand witht1 and affix a minus sign sincet1 is a lower limit. The integrand consists of the second two integrals. The result is 𝜕 𝜕t1 s(t1,t2,t3) =− ∫ ∞ t2 ∫ ∞ t3 f(t1,v,w)dwdv (17.1) After two more iterations of the procedure, we have f(t1,t2,t3) = (−1)3 𝜕 𝜕t1 𝜕 𝜕t2 𝜕 𝜕t3 s(t1,t2,t3). Similarly, we can derive f(t1,t2,t3) = 𝜕 𝜕t1 𝜕 𝜕t2 𝜕 𝜕t3 F(t1,t2,t3). Analogous expressions hold for generalm, which appears as the exponent of−1i nt h efi r s t formula. Note that the individual distributions are easily obtained from the the joint distribution or survival functions by sTi(t) = sT1,T2,…,Tm(0,0, …,t,…,0), FTi(t) = FT1,T2,…,Tm(∞,∞,…,t,…,∞), where thetis in theith position, and∞indicates that you take limits. THE JOINT DISTRIBUTION OF (T,J) 261 17.3 The distribution ofT 17.3.1 The general case It is a simple matter to deduce the distribution ofT from the joint distribution. Clearly, the minimum will take a value greater thant if and only if eachTi takes a value greater thant. Therefore, sT(t) = s(t,t,…,t). (17.2) (The reader should note thatFT(t) ≠ F(t,t,…,t).) 17.3.2 The independent case Let the density function, survival function, and hazard rate function ofTi be denoted respec- tively byfi,si,𝜇i. Things become much easier to deal with when the random variablesTi are independent. We can then readily write down the relevant functions forT in terms of the corresponding functions forTi. For example, from (17.2) we obtain sT(t) = s1(t)s2(t)⋯sm(t). (17.3) By taking logs and differentiating, we can find a similar relationship involving hazard func- tions. 𝜇 T(t) = m∑ i=i 𝜇i(t). (17.4) We have already encountered a particular case of (17.4) in (10.12). Example 17.1 Suppose thatT 1 andT2 are independent and both have the hazard function 𝜇(t) = 2 1−t,0 ≤ t< 1. Find the probability that the minimum value of these two random variables will be less than or equal to 1/2. Solution. We note thats T(t) = (1−t)4,f o r0≤ t< 1, either by first noting that, fori= 1,2, si(t) = (1−t)2 or by noting that𝜇T(t) = 4∕(1−t). The desired probability isFT(1∕2) = 1−sT(1∕2) = 15∕16. 17.4 The joint distribution of (T,J) 17.4.1 The distribution function for (T,J) Suppose that there is zero probability of the simultaneous occurrence of two or more failure times T 1,T2,…,Tm. We can then define the random variableJ as the index of the random 262 THE MINIMUM FAILURE TIME variable giving the minimum. For an example withm= 3, suppose T1 = 7, T2 = 5, and T3 = 10. ThenT would take the value 5, andJ would take the value 2, since the minimum occurs forT2. In many applications, we are interested in the joint distribution ofT and J. This can be described in various ways. One method is by the joint distribution function, FT,J(t,j) = P(T ≤ tandJ = j). Thisjointdistributionhasthesomewhatunusualfeaturethattherandomvariable Tisnormally continuous,while Jisdiscrete.Therefore,unliketheusualnotationfordistributionfunctions, the second variable inFT,J(t,j) is not cumulative, but refers to one specific index. We now consider the problem of deducing the joint distribution (T,J) from the joint distribution ofT1,T2,…,Tm. If we are given the joint density function, thenFT,J(t, j) is calculated by integrating over a suitable region – see (A.17). Takem= 2. ThenFT,J(t,1) is the probability thatT1 takes a value less than or equal tot, and thatT2 takes any value that is greater than that taken byT1. This is given by the double integral FT,J(t,1) = ∫ t 0 ∫ ∞ u fT1,T2(u,v)dvdu, (17.5) and similarly FT,J(t,2) = ∫ t 0 ∫ ∞ v fT1,T2(u,v)dudv. (17.6) Example 17.2 The joint distribution ofT1 andT2 is given by fT1,T2(s,t) = {6(s−t)2,0 ≤ s≤ 1,0 ≤ t≤ 1, 0, elsewhere . FindFT,J(t,j)f o rj= 1,2. Solution. FT,J(t,1) = ∫ t 0 ∫ 1 u 6(u−v)2dvdu= 2∫ t 0 (1−u)3du= 1−(1−t)4 2 ,f o r 0≤ t≤ 1. By symmetry, we must have FT,J(t,2) = 1−(1−t)4 2 ,0 ≤ t≤ 1. Since Ti is bounded above by 1 fori= 1,2, we necessarily have FT,J(t,i) = FT,J(1,i) = 1∕2, i= 1,2, t> 1. THE JOINT DISTRIBUTION OF (T,J) 263 The sum ofFT,J(t,1) and FT,J(t,2) must of course equalFT(t) = 1−sT(t). It will be instructive for the reader to verify this by drawing a picture in the plane, showing that the unionoftheregionsofintegrationin(17.5)and(17.6),andtheregioncorrespondingto s T(t), is the entire positive quadrant. In the general case, we will needm-dimensional integrals to computeF(t,j)f r o mt h e densityfunction.However,ifwealreadyhavethejointsurvivalfunction,thecomputationcan be simplified. To illustrate, consider the case withm= 3. Then, reasoning as above, F T,J(t,1) = ∫ t 0 ∫ ∞ u ∫ ∞ u f(u,v,w)dwdvdu. From(17.1)theinnertwointegralscanbewrittencompactlyasapartialderivative.Wehave FT,J(t,1) = ∫ t 0 𝜎(u)du, where 𝜎(u) =− 𝜕 𝜕t1 s(t1,t2,t3), evaluated at t1 = t2 = t3 = u. In the general case, FT,J(t,j) = ∫ t 0 𝜎j(u)du, (17.7) where 𝜎j(u) =− 𝜕 𝜕tj s(t1,t2,…tm), evaluated at t1 = t2 = ⋯ = tm = u. Example 17.3 Suppose thatm= 2 and the joint survivor function is given by sT1,T2(u,v) = 1 2[(1−u)4 +(1−v)4 −(u−v)4], 0 ≤ u≤ 1, 0≤ v≤ 1. FindFT,J(t,1). Solution. We could take two derivatives to calculatef(u,v) = 6(u−v)2, and apply (17.5). (ThisisinfactthesamedistributionasinExample16.2.)Note,however,that(17.5)wouldjust ‘undo’thecalculationofthesecondderivativebyintegrating.Forthisformofthedistribution, it is easier to apply (17.7). On the given region − 𝜕 𝜕us(u,v) = 2[(1−u)3 +(u−v)3], 𝜎1(u) = 2(1−u)3, 264 THE MINIMUM FAILURE TIME so FT,J(t,1) = ∫ t 0 2(1−u)3du= 1−(1−t)4 2 ,0 ≤ t≤ 1, verifying the previous example. 17.4.2 Density and survival functions for (T,J) We can also define the distribution of (T,J) by the joint density functionfT,J(t,j). This is the function satisfying FT,J(t,j) = ∫ t 0 fT,J(s,j)ds, fT,J(t,j) = d dtFT,J(t,j). (17.8) The functionf is interpreted in the normal way. Namely, for ‘small’Δt,fT,J(t,j)Δtis approx- imately the probability that the first failure will be from causejand that it will take place in the time interval fromtto t+Δt. The precise statement, following from the first expression in (17.8), is that the probability that the first failure will be from causejand will take place between timeaand timebis given by FT,J(b,j)−FT,J(a,j) = ∫ b a fT,J(t,j)dt. Wedefinethejointsurvivalfunctionfor( T,J)bythinkingofsurvivalaswedidattheend of Section 11.2.2. Let sT,J(t,j) = P(T > tandJ = j) = ∫ ∞ t fT,J(s,j)ds. This is the probability that failure will occur after timetdue to causej. Figure 17.1 shows a typical graph off(t,j)f o rm= 2. Note that the mass is concentrated on parallel sheets. If we cut thejth sheet by the planeT = t, the area of the left portion will be F(t,j) and the area of the right portion will bes(t,j). Allfunctionspertainingtotherandomvariable T aloneareobtainedbysummingoverall j, as we noted above forFin the casem= 2. That is, FT(t) = m∑ j=1 FT,J(t,j), (17.9) fT(t) = m∑ j=1 fT,J(t,j), (17.10) sT(t) = m∑ j=1 sT,J(t,j). (17.11) THE JOINT DISTRIBUTION OF (T,J) 265 fT,J (t, 2) fT,J (t, 1) j =2 j =1 t j Figure 17.1 The graph offT,J(t,j) 17.4.3 The distribution of J To obtain the distribution ofJ from the joint distribution, we compute the other marginal, which can be expressed in various ways. IffJ(j) denotes the probability thatJ = j, then fJ(j) = ∫ ∞ 0 fT,J(t,j)dt= sT,J(0,j) = limt→∞FT,J(t,j) (17.12) In Figure 17.1,fJ(j) is the area of thejth sheet. Note thatFT,J(t,j)+sT,J(t,j) is not equal to 1, but rather tofJ(j). Example 17.4 Takem= 2.If T1 isuniformon[0,1], T2 isuniformon[0,2],and T1 andT2 are independent, find the distribution ofJ. Solution. The joint density function takes a constant value of 1/2 on the rectangle 0≤ s≤ 1,0 ≤ t≤ 2, and is 0 elsewhere. The maximum value ofT is the minimum of the respective maximums of theTi, which in this case is 1. Therefore, fJ(1) = FT,J(1,1) = ∫ 1 0 ∫ 2 u (1∕2)dvdu= 3 4, fJ(2) = FT,J(1,2) = ∫ 1 0 ∫ 1 v (1∕2)dudv= 1 4. 266 THE MINIMUM FAILURE TIME 17.4.4 Hazard functions for (T,J) Definition 17.1 The hazard function for (T,J) is given by 𝜇T,J(t,j) = fT,J(t,j) sT(t) . This is a conditional density. For smallΔt, 𝜇T,J(t,j)Δt is approximately the probability that failure will occur first from causejin the time interval fromttot+Δt, given that failure fromanycause has not yet taken place before timet. In the Chapter 11 multiple-decrement model for a life agex, 𝜇T,J(t,j) corresponds to 𝜇(j) x (t). Given the hazard rates, we can obtain the joint distribution of (T,J) by the same method as employed in Chapter 11. From (17.10) and the definition of𝜇(t,j), 𝜇T(t) = m∑ j=1 𝜇T,J(t,j), (17.13) and then, from (15.7), sT(t) = e−∫ t 0 𝜇T(r)dr. We know thatfT,J(s,j) = sT(s)𝜇T,J(s,j) and, from the first expression in (17.5), FT,J(t,j) = ∫ t 0 sT(s)𝜇T,J(s,j)ds. (17.14) The last formula is easily explained intuitively. For the event in question to occur, there must be some points, beforet, for which failure from any cause has not yet occurred, and then failure will occur from thejth cause at timesAlthough (17.14) has this intuitive appeal, it is not necessarily useful for computingFT,J(t,j) as we may not know the joint hazard rates untilwehavealreadycomputed FT,J(t,j)and fT,J(t,j).Itis,however, animportantformulain the independent case to which we now turn. 17.4.5 The independent case As in Section 17.2, we can simplify calculations when theTi are independent, and deduce the distribution of (T,J) directly from the individual distributions of eachTi. Sinces′ j (t) = −sj(t)𝜇j(t), (17.3) shows that 𝜎j(t) = ∏ i≠j si(t)s′ j (t) = sT(t)𝜇j(t) THE JOINT DISTRIBUTION OF (T,J) 267 and then from (17.7), FT,J(t,j) = ∫ t 0 sT(u)𝜇j(u)du, for independent Ti. (17.15) Note, as a comparison to (17.14.), that (17.15) can be used directly to computeFT,J(t,j) in the independent case, when we know𝜇i from the individual distributions. Moreover, differentiating and dividing bysT(t) verifies that in the independent case 𝜇T,J(t,j) = 𝜇j(t) (17.16) forj= 1,2, …mandall tforwhich sT(t) ≥ 0,Thisprovidesthepromisedprooffortheresult stated in in formula (11.23) . The result is easily explained intuitively. Looking at the machine model of Section 11.6, forexample,bothquantitiesin(17.16)giveaconditionaldensityforfailureofpart jattime t. Inthecaseof 𝜇T,J(t,j),theconditionisthatallpartshavesurviveduptotime tandingeneral this may give information regarding the failure time of partj. Suppose, for example, that the parts are connected so that part 2 cannot fail until part 1 does, and then it fails five seconds later.(Thesameideainthemultipledecrementmodelprovidedthesimplecounter-exampleof Exercise11.15.)Intheindependentcase,however,weobtainexactlythesameinformationas ifwetoldonlythatpart jhassurviveduptotime t,whichispreciselytheconditionapplicable to𝜇 j(t). Wehavealreadyencounteredspecialcasesof(17.15).Oneexampleis(10.28).Thehazard rate 𝜇x(t) corresponds to𝜇1(t) which equals𝜇T,J(t,1) since we postulated independence. Another example is (11.13). Example 17.5 Suppose that theT i are independent and exponential with constant hazard 𝜇i. (a) FindFT,J(t,j). (b) FindfJ(j). (c) Show thatT andJare independent. Solution. (a) Let𝜇 = 𝜇1 +𝜇2 +⋯+𝜇m.Formula(17.4)showsthat Tisexponentialwithconstant hazard𝜇, and, from (17.15), FT,J(t,j) = ∫ t 0 e−s𝜇𝜇jds= 𝜇j 𝜇 (1−e−𝜇t). (b) fJ(j) = lim t→∞ FT,J(t,j) = 𝜇j 𝜇 . (c) We just note that FT,J(t,j) = FT(t)fJ(j). The solution to (b) gives an important result that has many applications. It says that for independent exponential failure times, the probabilities of first failure are proportional to the hazardrates.Thismakessense,sincethehigherthehazardrate,thelowerthemean,whichis the reciprocal of the hazard rate, and therefore more likelihood of occurring first. 268 THE MINIMUM FAILURE TIME Example 17.6 Takem= 2. SupposeT1 and T2 are independent,T1 is uniform on [0,a], andT2 is uniform on [0,b], where 0< a≤ b.F i n dFT,J(t,1) andFT,J(t,2). Solution. Since sT(t)𝜇1(t) = s2(t)s1(t)𝜇1(t) = s2(t)f1(t), we can write FT,J(t,1) = ∫ t 0 ( 1− s b ) 1 ads= t a − t2 2ab = F1(t)− 1 2F1(t)F2(t), 0 < t< a. Similarly, FT,J(t,2) = t b − t2 2ab = F2(t)− 1 2F1(t)F2(t), 0 < t< a. Since failure must take place before timea, for anyt≥ a, FT,J(t,1) = FT,J(a,1) = 1− a 2b, FT,J(t,2) = FT,J(a,2) = a 2b As a check, note thatFT,J(t,1) +FT,J(t,2) = F1(t)+F2(t)−F1(t)F2(t), which must be true, since forT to take a value less thantmeans that at least one ofT1 and T2 takes a value less thant. Also note that the answer here could have been written down immediately, since thedistributionssatisfythestochasticversionofthecondition givenin(11.24),andtherefore we can apply Method 2 of Chapter 11. 17.4.6 Nonidentifiability Wemotivatetheideaofthissectionbyanexample.Firstnotethatthejointdistributionsgiven in Examples 17.1 and 17.2 are easily seen to be different. One way is to note thatT 1 and T2 are not independent in the latter. Example 17.7 Calculate F T,J(t,j) for the distribution given in Example 17.1 Solution. We could do this by (17.15), but it is easier to note that, by symmetry,FT,J(t,1) = FT,J(t,2)andthetwomustsumto FT(t) = 1−sT(t).WecanconcludedirectlyfromExample 17.1 that FT,j(t,1) = FT,J(t,2) = 1−(1−t)4 2 ,0 ≤ t≤ 1, and necessarily FT,J(t,1) = FT,J(t,2) = 1 2,1 ≤ t. ComparetheaboveresultwithExample17.2.Thesomewhatsurprisingconclusionisthat two completely different distributions for (T1,T2) have led to exactly the same distribution for (T,J). This is known as thenonidentifiabilityproblem and it has statistical implications. Supposewewanttomakeinferencesaboutthejointdistributionofthe Ti byobservingfailure THE JOINT DISTRIBUTION OF (T,J) 269 times. In many cases, all we can possibly observe is the joint distribution (T,J). An example is when the random variables represent the time of death from various causes. Once death occurs, we know the time and the cause, but no further observation of the subject is possible. Our example above shows that it is impossible to uniquely determine the joint distribution of the random variables that give rise to a given (T,J). We need additional information in order to obtain a unique solution. One instance when this occurs is in the independent case. Theorem 17.1 Givenanyjointdistributionfor (T,J) thereisauniquejointdistributionof (T 1,T2,…,Tm)suchthatthe (Ti)aremutuallyindependentandinducethegivendistribution of(T,J). Proof. Uniqueness follows immediately, since, given independence, we know the joint dis- tributionifweknowthedistributionofeach Ti,and(17.16)impliesthateach Ti isnecessarily a random variable with hazard function𝜇T,J(t,i), which is determined uniquely from (T,J). For the existence, given any joint distribution functionFfor (T,J), we letTi be a random variable with hazard function𝜇T,J(t,i). This collection of independentTi in turn generates a joint distribution function̂FT,J. From (17.15), ̂FT,J(t,j) = ∫ t 0 sT(s)𝜇T,J(s,j)ds, which equalsFT,J(t,j) as shown by (17.14). To illustrate the use of this theorem, suppose you are told that FT,J(t,i) = 1−(1−t)4 2 ,0 ≤ t≤ 1, FT,J(t,1) = 1, t> 1 fori= 1,2 and asked to identify the joint distribution of (T1,T2). You cannot do this without further information, for it could be either the joint distribution of Example 16.1 or that of Example 16.2, or indeed several other possibilities. However, if you are given the additional information thatT 1 andT2 are independent, then you know that it must be the distribution of Example 16.1. 17.4.7 Conditions for the independence ofT andJ Another question of interest is to determine whenT andJare independent. In Example 17.5, we saw that this occurred with constant hazard functions. We present here a more general criterion. Define the ratios K(t,j) = 𝜇(t,j) 𝜇T(t) for allj, and allt, such thatsT(t) > 0. Theorem 17.2 K(t,j) = P(J = j|T = t).Therefore, T and J areindependent ifand onlyif K(t,j) isindependentoft. 270 THE MINIMUM FAILURE TIME Proof. We have fT,J(t,j) = sT(t)𝜇(t,j) = K(t,j)sT(t)𝜇T(t) = K(t,j)fT(t). So P(J = j|T = t) = fT,J(t,j) fT(t) = K(t,j). The condition of this theorem is sometimes expressed by saying that the hazards for the individual causes are fixed proportions of the total hazard. 17.5 Other problems Thereareseveralotherquestionsregardingthejointdistributionof( T1,T2,…,Tm)thatcanbe answeredbysimilartechniquestothoseinSection17.3.Thatis,wefindacertainprobability byintegratingfailuretimesoverasuitableregionof m-dimensionalspace.Asanexample,we illustrate the method for a problem analogous to Example 10.7 of Section 10.9. Takem= 2, and consider the probability that both causes of failure will occur within a specified duration of each other. That is, for some fixedn, we want the probability that (|T 1 −T2| ≤ n). It will normally be easier to compute this as 1−P(|T1 −T2| > n) = 1−[P(T2 > T1 +n)+P(T1 > T1 +n)]. Each term is found by integrating the joint density function over a suitable region in the plane. For example, P(T2 > T1 +n) = ∫ ∞ 0 ∫ ∞ u+n fT1,T2(u,v)dvdu. Example 17.8 Find P(|T1 −T2| ≤ n), whenT1, T2 are independent, and both are expo- nential with hazard functions𝜇1 and𝜇2, respectively. Solution. The integral above reduces to 𝜇1 𝜇1 +𝜇2 e−n𝜇2, so the final answer is 1− 𝜇1 𝜇1 +𝜇2 e−n𝜇2 − 𝜇2 𝜇1 +𝜇2 e−n𝜇1. THE COMMON SHOCK MODEL 271 17.6 The common shock model In many applications, we have a group of objects whose future lifetimes are generally inde- pendent, except that they are all subject to a common hazard, which will result in the failure of all, should it occur. In the case of human lives, this could be a natural disaster such as a hurricane. In the case of machine parts, it could be something like an electrical problem that affects all components at once. The presence of the common shock introduces dependence into what would otherwise be independent future lifetimes. Tomodelthegeneralsituation,wehave m+1independent,continuousrandomvariables, (T ∗ 1,T∗ 2,…,T∗ m,Z), and for eachiwe let Ti =m i n(T∗ i ,Z). Theinterpretationisthat T∗ i isthetimeuntilfailureofthe ithobjectforreasonsotherthanthe common shock, andZis the time until the common shock occurs. It follows then thatTi will be simply the time until failure of theith object, since such failure will occur at either time T∗ i or timeZ, whichever is earlier. In the remainder of this section, we will confine ourselves to the case wherem= 2. Quantities referring toT∗ i will have a superscript∗. We are interested in questions about the joint distribution (T1,T2), which involves depen- dentrandomvariables.However,inmanycases,wecananswerthesequestionsbyconsidering theindependent collection (T ∗ 1,T∗ 2,Z).Wewillillustratewithseveral examples. Akey factto note is that T =m i n(T1,T2) =m i n(T∗ 1,T∗ 2,Z), since both give the time of first failure. Example 17.9 Find a formula for the probability that both objects will survive to timet. Solution. This is justs ∗ 1 (t)s∗ 2 (t)sZ(t). Example 17.10 What is the probability that failure will occur as a result of the common shock? Solution. This is justP(J = 3) in the joint distribution of (T,J), whereJ takes the values 1, 2, 3 andT is the minimum ofT ∗ 1,T∗ 2 andZ. Example 17.11 What is the probability that the second failure will occur before timet? Solution. We divide this up into two mutually exclusive cases. It will always occur ifZ ≤ t. IfZ > t, we need bothT∗ 1 andT∗ 2 less than or equal tot. The probability is FZ(t)+sZ(t)F∗ 1(t)F∗ 2(t) = 1−sZ(t)[s∗ 1 (t)+s∗ 2 (t)−s∗ 1 (t)s∗ 2 (t)]. 272 THE MINIMUM FAILURE TIME Other problems are not so straightforward and require special attention. The joint distri- bution of (T1,T2) is quite different from the typical two-dimensional continuous distribution. Itstilliscontinuous,butithasamassofpositiveprobabilityallconcentratedonasingleline, namely the diagonal, since the occurrence of the common shock will cause failure from both causes, leading toafailurepoint of theform (t,t).Indetermining theprobability that (T 1,T2) lies in some regionA, we will in general have to breakAup into three pieces, the part that is above the diagonal, the part that is below the diagonal, and the part that is on the diagonal. We adopt the convention that the value ofT1 is on the horizontal axis. For the part of the plane above the diagonal, {(u,v): u< v}, we use the joint density function f∗ 1 (u)f2(v), since the only wayT1 can take a valueu< vis ifT∗ 1 took the valueu. In other words, failure fromcause1attime udidnotoccurfromthecommonshock,sinceifitdid,thenfailurefrom cause 2 would also have occurred at timeuand could not have occurred at the later datev. Similarly,forthepartofthepositivequadrantbelowthediagonal,{( u,v): v< u},weuse the joint density function f1(u)f∗ 2 (v). Since Ti =m i n(T∗ i ,Z), the densitiesfi,i= 1,2, are easily calculated as fi(t) =− (s∗ i sz)′(t) = f∗ i (t)sz(t)+s∗ i (t)fz(t). (17.17) Failure on the diagonal arises if and only if the occurrence of the common shock occurs before the other two causes. We use the one-dimensional density function, fT,J(t,3) = s∗ 1 (t)s∗ 2 (t)fZ(t), andprojectthediagonalontotheline.Thatis,tofindtheprobabilitythatfailuretookplaceat a point (t,t), wherea≤ t≤ b, we integrate this density fromatob. Example 17.12 Suppose T ∗ 1, T∗ 2, andZ are exponential with hazard functions𝜇1, 𝜇2, and 𝜌, respectively. Consider the event thatT1 and T2 are both less than or equal ton. This can be subdivided into three cases according as (a)T1 < T2,( b) T2 > T1,( c) T1 = T2.F i n dt h e probability of each case. Solution. (a) Note first that, from (17.17), we can calculate f 2(t) = (𝜇2 +𝜌)e−(𝜇2+𝜌)t, so that, using the above-diagonal joint density, the required probability is 𝜇1(𝜇2 +𝜌)∫ n 0 ∫ n u e−𝜇1ue−(𝜇2+𝜌)vdvdu, COPULAS 273 which equals −e−n(𝜇2+𝜌) + 𝜇1 𝜇1 +𝜇2 +𝜌 + 𝜇2 +𝜌 𝜇1 +𝜇2 +𝜌e−n(𝜇1+𝜇2+𝜌). (b) Similarly, the required probability in this case is −e−n(𝜇1+𝜌) + 𝜇2 𝜇1 +𝜇2 +𝜌 + 𝜇1 +𝜌 𝜇1 +𝜇2 +𝜌e−n(𝜇1+𝜇2+𝜌). (c) The required probability is ∫ n 0 𝜌e−𝜌te−𝜇1te−𝜇2tdt= 𝜌(1−e−n(𝜇1+𝜇2+𝜌)) 𝜇1 +𝜇2 +𝜌 . The sum of these three cases is 1−e−n(𝜇1+𝜌) −e−n(𝜇2+𝜌) +e−n(𝜇1+𝜇2+𝜌), as we can verify from the general formula given in Example 16.12. 17.7 Copulas This section, like the previous one, is concerned with situations where there is a lack of independence. We present a general method that is often used to deal with this. Attention is confined to the casem= 2. A joint distribution (T 1,T2) can be thought of as having two ingredients. One is the distributionsofthetwo-componentrandomvariables,andtheotheristhewayinwhichthese are linked together. The latter can be described by a device known as acopula, which can then be applied to an arbitrary pair of individual distributions. The copula provides a means of dealing with these two ingredients separately. To elaborate, we start with the observation that wheneverT 1 andT2 are independent, we immediately recover the joint distribution from the individual distributions by the rule FT1,T2(t1,t2) = FT1(t1)FT2(t2). (17.18) We can then ask whether we can replace the multiplication on the right-hand-side of (17.18) by other transformations, and still obtain a joint distribution – that is, ifIdenotes the unit interval [0,1], whether we can find a functionC from I×I to itself, so that we obtain a legitimate joint distribution by the rule FT1,T2(t1,t2) = C(FT1(t1),FT2(t2)). (17.19) 274 THE MINIMUM FAILURE TIME We need some restrictions on the functionC. Take any points in I.I fT1 > s, so that FT1(s) = 0,then,forall t,FT1,T2(s,t) = 0.Thesameholdsfor T2,leadingtotheconditionthat for allu,vinI, C(0,v) = C(u,0) = 0. (17.20) If T1 ≤ s, so thatFT1(s) = 1, then for allt, FT1,T2(s,t) = FT2(t), leading to the condition that for allu,vinI, C(1,v) = v, C(u,1) = u. (17.21) Another requirement stems from the fact that probabilities cannot be negative. For any sub-rectangle R⊆ I×I, the probability that (T1,T2) lies inRis just the sum of the values of FT1,T2 on the northeast and southwest corners, minus the sum of the values on the other two corners. Since this is nonnegative, it follows that C(u2,v2)+C(u1,v1)−C(u1,v2)−C(u2,v1) ≥ 0. (17.22) wheneveru1 ≤ u2 andv1 ≤ v2. These are the only conditions we need and we can now state the formal definition. Definition 17.2 Acopulais a functionCfromI×ItoIsatisfying (17.20)–(17.22). Itcanbeshownthatif Cisacopula,then(17.19)givesavalidjointprobabilitydistribution forany T1 andT2.Conversely(andhardertoshow),foranyjointdistribution(T 1,T2)thereis a copulaCsuch thatFT1,T2 is given by (17.19). IfT1 andT2 arebothuniformdistributionson I,then FTi(u) = ufori= 1,2,andall u∈ I, from which it follows that FT1,T2(u,v) = C(u,v). This shows that as an alternate definition, we can simply define a copula as a distribution function of a joint distribution involving two random variables that are uniform onI. The following are three simple examples of copulas: 1. C(u,v) = uv. This is just the copula for an independent distribution, as mentioned; 2. C(u,v) =m i n(u,v); 3. C(u,v) =m a x(u+v−1,0). Copulas 2 and 3 are extreme in the sense that for any copulaCand for allu,v,∈ I, max(u+v−1,0) ≤ C(u,v) ≤ min(u,v). They are also extreme in the following sense. Consider all possible joint distributions for a givenT 1 and T2. In many cases, one is interested in the sumT1 +T2. For example, an insurer sells two insurance contracts andTi denotes the claim on theith policy, or a person COPULAS 275 buys two stocks andTi is the value of theith stock at some future date. We may want to compare all possible joint distributions as to their degree of risk. We will not go into the details of comparing joint distributions as to risk here. (In Section 22.4, we do introduce this idea for single variable distributions). However, it seems clear that risker possibilities arise when when large values of one random variable tend to go with large values of the other so there is a tendency for either both values to be large or both to be small. The less riskier possibilities arise when large values of one tend to go with small values of another, so there is a possibility for bad results in one case to be balanced by good results in the other. It can be shown, that under some natural risk comparing criteria, copula 2 will give the most risky jointdistributionandcopula3theleastriskyjointdistribution.Thispointisillustratedfurther in Exercise 17.14. We deal only briefly with problems of choosing a copula to model a given situation. In many cases, the modeler likes to choose a copula from a parametric family, and select the parameter to suit certain conditions. A popular choice for this is Frank’s family of copulas given by C Θ(u,v) = 1 Θ log ( 1+ (eΘu −1)(eΘv −1) eΘ −1 ) , whereΘcan be any nonzero real number. It can be shown, using L’Hˆopital’s rule, that for all u,v,∈ I, lim Θ→0 CΘ(u,v) = uv, so that the smaller the parameter is in absolute value, the greater the extent of independence between the two random variables, with full independence occurring forΘ=0. An interesting feature which some copulas, but not all, have is that C(u,v) = u+v−1+C(1−u,1 −v). (17.23) It is straightforward to verify this property for copulas 1–3 above. It is true, but harder to verify, that this holds for Frank’s family. The significance of (17.23) is that s T1,T2(t1,t2) = 1−FT1(t1)−FT2(t2)+FT1,T2(t1,t2) = C(sT1(t1),sT2(t2)). In other words, the same transformation rule can be applied to either distribution or survival functions. Example17.13 SupposethatDemoivre’slawholdswith 𝜔 = 100.Considertwolives(60) and (70). Find the probability that both will be alive at the end of 10 years, assuming each of the three basic copulas given above, for the joint distribution ofT(60) andT(70). Solution. Theindividualsurvivalprobabilitiesare3/4and2/3,sothesurvivalofthejoint-life status is in the respective cases: 1. 1/2 as we known already from Chapter 10; 276 THE MINIMUM FAILURE TIME 2. min{3∕4,2∕3} = 2∕3. This copula applied to a joint-life status just means that the younger life will die at exactly the same time as the older; 3. (3∕4+2∕3−1) = 5∕12. For a particular application of copulas, refer back to the nonidentifiability problem of Section17.4.6.Insteadofassumingindependence,wemightpostulateacertaincopula Cand thenaskforajointdistributionwiththechosencopulathatgivesrisetothegivendistribution (T,J). In many cases this will be unique. Notes and references Nelsen (1999) is a good general source for additional material on copulas, including full derivationsoftheunprovedresultsthatwehavegiven.FreesandValdez(1998)discussvarious actuarial applications of copulas. Carri´erre (1994b) discusses the application of copulas to the nonidentifiability problem in multiple-decrement theory. For methods of comparing two- variable distributions for risk, see Shaked and Shantikumar (2007), Chapter 6. Exercises Type A exercises 17.1 A joint distribution is given by fT1,T2(s,t) = ⎧ ⎪ ⎨ ⎪ ⎩ 3s,0 ≤ s≤ t≤ 1, 3t,0 ≤ t≤ s≤ 1, 0, elsewhere . FindF T,J(t,1) andFT,J(t,2). 17.2 A joint survival function is given by s(u,v) = {1−(3∕2)(u2 +v2)+(1∕2)u3 +(3∕2)uv2,0 ≤ v≤ u≤ 1, 1−(3∕2)(u2 +v2)+(1∕2)v3 +(3∕2)vu2,0 ≤ u≤ v≤ 1. FindFT,J(t,1). 17.3 Suppose that T1 andT2 are independent with p.d.f.’s f1(r) = 2e−2r, f2(r) = 3e−3r. (a) FindFT,J(t,1) andFT,J(t,2). (b) Find the distribution of J. 17.4 Suppose that T1 and T2 are independent. T1 has an exponential distribution with constant hazard rate𝜇. T2 is uniform on [0,a]. Find (a)FT,J(t,2),(b) P(J = 2). EXERCISES 277 17.5 The failure timesT1 andT2 are independent, and have respective hazard functions 𝜇1(t) = 3 2−t,0 ≤ t< 2, 𝜇2(t) =l o g(2). Find the probability that the minimum value of these two random variables will be less than or equal to 1. 17.6 A machine is subject to two independent causes of failure. The time of the first cause of failure is uniformly distributed on the interval [0, 4]. The time of the second cause of failure is uniformly distributed on the interval [0, 5]. Find the probability that: (a) the machine will fail from cause 1 before time 3; (b) the machine will eventually fail from cause 2. 17.7 Twofailuretimes T 1 andT2 haveajointdistributiongivenbythejointdensityfunction fT1,T2(u,v) = ⎧ ⎪ ⎨ ⎪ ⎩ 4(v−u) 2,0 ≤ u≤ v≤ 1, 4(u−v), 0 ≤ v< u≤ 1, 0, elsewhere . Find: (a)FT,J(t,1) andFT,J(t,2); (b) the distribution ofJ;( c )𝜇(t,1) and𝜇(t,2). 17.8 Twofailuretimes T1 andT2 haveajointdistributiongivenbythejointdensityfunction fT1,T2(u,v) = ⎧ ⎪ ⎨ ⎪ ⎩ (8∕3)v 2,0 ≤ u≤ v≤ 1, (8∕3)uv,0 ≤ v< u≤ 1, 0, elsewhere . (a) FindFT,J(t,1) andF−T,J(t,2). (b) Find the distribution ofJ. (c) Suppose that ̂T1 and ̂T2 are two independent variables whose joint distribution leads to the same distribution of (T,J) as you found in part (a). What are the hazard functions of̂T1 and ̂T2? Type B exercises 17.9 ForthejointdistributiongiveninExample17.2,find 𝜇(t,j)and 𝜇j(t)for j= 1,2.Show that these are not the same. 17.10 ForthejointdistributiongiveninExercise17.2,find 𝜇(t,j)and 𝜇j(t)for j= 1,2.Show that these are not the same. 17.11 Consider independent failure timesT1 and T2 where the hazard rate ofT1 is 𝛼∕(1− t),0 ≤ t< 1, and the hazard rate ofT2 is 𝛽∕(1−t),0 ≤ t< 1. Answer the following in terms of𝛼 and𝛽. (a) FindFT,J(t,1). 278 THE MINIMUM FAILURE TIME (b) FindP(J = 2). (c) An insurance contract provides for a benefit at the moment of the first failure, provided that this is due to ‘cause 1’ (i.e. provided thatT1 < T2). The amount of the benefit for failure at timetis (1−t)e0.1t. The force of interest𝛿 is a constant 0.10. Find the expected present value of the benefits. 17.12 Twolivesage( x)and( y)aresubjecttothecommonshockmodel. T∗(x)hasaconstant hazard rate of 0.06,T∗(y) has a constant hazard rate of 0.04, andZ has a constant hazard rate of 0.02. The force of interest is a constant 0.05. (a) Calculate ̄A x(110)andwriteitasasumofthreeterms,namely,theexpectedpresent value of benefits when: (i) (x) dies at a time strictly before the death of (y); (ii) (x)diesattimestrictlyafterthedeathof( y);(iii)(x )diesasresultofthecommon shock. (b) Calculate ̄Axy. 17.13 You are given a common shock model (T∗ 1,T∗ 2,Z) whereT∗ 1 has the survival function s(t) = 1−0.1t,T∗ 2 hasaconstanthazardfunctionof0.04,and Zhasaconstanthazard function of 0.02. Find the probabilities that (a)T1 < T2,( b )T2 < T1,( c )T1 = T2. 17.14 Suppose that T1 and T2 each take the values 0 with probability 1/2, and 1 with probability1/2.Calculatetheprobabilityfunctionforthejointdistributionof( T1,T2) under each of the following copulas: (a)C(u,v) =m i n(u,v), (b)C(u,v) =m a x(u+ v−1,0), (c)C(u,v) = uv, (d) Frank’s copula with values of𝜃 = 0.01,50, −50. What happens when𝜃 approaches∞or−∞? 17.15 A joint life insurance on (x) and (y) has death benefits which are constant over each year.Assumethateithercopulas2or3ofSection17.7applytothejointdistributionof T(x) andT(y). Show that unlike the independent case, the termRin equation (10.15) is equal to 0. PART III ADVANCED STOCHASTIC MODELS