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: $$\mathbb{E}\left(\sum_{n=2}^m \mathbf{1}_{\mathcal R_n}\right) = \sum\limits_{n=2}^m \dfrac 1n$$
A: For (a) and (b) use symmetry.  They are equally distributed and random therefore any one of them is equally likely to be the largest, so the probability that it is the $n^{th}\ is\ \frac{1}{n}$.
For (c) limit you may need a condition that $ F(x) \lt 1,\ for\ finite \ x$.  Otherwise a variable $X_n$ could be at the maximum and none $(k\gt n)$ could be greater.
A: For (a), $X_2$ is a record if and only if $X_2$ is the greatest of $X_1$ and $X_2$.  By symmetry, each of those has equal probability of being the greatest.
Similarly for (b), each of $X_1, \ldots, X_n$ has equal probability of being the greatest of those $n$, and $X_n$ is a record iff it is the greatest.
