Q:

Showing that probability of random variable is greater than another random variable

Let X1, X2, . . . be a sequence of independent identically distributed continuous random variables.
For n ≥ 2, define Xn as a record of the sequence if Xn > max(X1, X2, . . . , Xn−1),
that is, if Xn is larger than each of the values observed so far.
(a)Find the probability that X2 is a record. Use symmetry to obtain the answer without
computation; ans = 1/2
b) Find the probability that Xn is a record, as a function of n. Again use symmetry.
(Answer: 1/n)
c) Find a simple expression for the expected number of records that occur over the first m
trials for any given integer m (X1 is not considered as a record). Show that this number
is infinite in the limit m → +∞.
(Answer: Pm
k=2 1/k → ∞ as m → ∞)
For Question a, I know we have to evaluate P(X2 > X1) and this simplifies to $\int_{- \infty}^{\infty} P(X_2>x)P(X_1=x)dx$. I'm not sure how we can use symmetry. Any hint will help.
For quesiton b, this looks like a generalisation of part a, Except $P(X_n>max(X_1, X_2, .. X_{n-1})) = P(X_n>X_i)^n $
Not sure for part c. 
d) Let $N_1$ be the index of the first record in the sequence. Find P(N1 > n) for each n ≥ 2.
(Answer: 1/n)
My solution: N1$\sim Geo(1/n)$
P(N1 > n) = $(1 - \frac{1}{n})^n(\frac{1}{n}) + (1 - \frac{1}{n})^{n+1}(\frac{1}{n})+..$
= $(1 - \frac{1}{n})^n(\frac{1}{n})( 1+ (1 - \frac{1}{n})+(1 - \frac{1}{n})^2+.. = (1 - \frac{1}{n})^n$ by GP formula.

A:

Let $\mathcal R_n$ be the event that $X_n$ is a record.  That is the event for $X_n=\max\{X_1,..,X_n\}$.
$(a)$ Well, we know $(X_n)_{n\ge 1}$ are i.i.d (independent and identically distributed).   By symmetry each sample thereby has equal probability for being the maxiumum among the first two, so $\mathsf P(X_2=\max(X_1,X_2)) = \frac{1}{2}$.$$\mathsf P(\mathcal R_2)=\tfrac 12$$
$(b)$ Since $(X_n)_{n\ge 1}$ are all i.i.d, then following a similar pattern could find $\mathsf P(\mathcal R_3) = \mathsf P(X_3 {=} \max(X_1,X_2,X_3)) = \frac{1}{3}$.   We find a pattern and see that, as there is no bias as to which sample has the maximum value among the first $n$ samples, the probability will be $\frac{1}{n}$ for $X_n$ being a record. 
$$\mathsf P(\mathcal R_n)=\frac 1n \qquad=\mathsf P(X_n{=}\max{\{X_k\}}_{k=1}^n)$$
$(c)$ So now we want the expected count of records among the first $m$ trials (discounting the first). Let us use indicator random variables for the events, so that count is $\sum_{n=2}^m \mathbf{1}_{\mathcal R_n}$, and we seek its expectation.   We know that $\mathbb{E}(\mathbf 1_{\mathcal R_n}) = \mathsf P(\mathcal R_n).$ So by Linearity of Expectation: $$\mathbb{E}\left(\sum_{n=2}^m \mathbf{1}_{\mathcal R_n}\right) = \sum\limits_{n=2}^m \dfrac 1n$$

