Expected value and variance of records Let $X_1,X_2...$ be an i.i.d. continuous random variables. We say a record occurs at time $n$ and we set $Y_n=1$ if $X_n>\max\left\{ X_{1},...,X_{n}\right\} $ and $Y_n=0$ in otherwise. Let $R_n=Y_1+...+Y_n$ be the number of records that have been set by time $n$. Find $\mathbb{E}\left[R_n\right]$ and $\mathbb{\mathbb{V}\mathrm{ar}}\left[R_n\right]$.
My solution:
$\mathbb{E}\left[R_n\right]=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
$\mathbb{\mathbb{V}\mathrm{ar}}\left[R_n\right]=1(1-1)+\frac{1}{2}\left(1-\frac{1}{2}\right)+\frac{1}{3}\left(1-\frac{1}{3}\right)+...++\frac{1}{n}\left(1-\frac{1}{n}\right)$
It's ok?
 A: Note that by continuity we have $\omega \in \Omega$,
$$X_p(\omega) \ne X_q(\omega) \ \forall p \ne q$$
Define:
$$A_n:= (X_n > \max\{X_1, X_2, ... X_n\})$$
Observe that:
$$A_1:= (X_1 > \max\{X_1\}) = (X_1 > \max\{X_1\}) = \emptyset$$
$$A_2:= (X_2 > \max\{X_1, X_2\})$$
$$ = (X_2 > X_1) = \emptyset \ \text{if} \ (X_1 > X_2)$$
$$ = (X_2 > X_2) = \emptyset \ \text{if} \ (X_2 > X_1)$$
$$A_3:= (X_3 > \max\{X_1, X_2, X_3\})$$
$$ = (X_3 > X_1) = \emptyset \ \text{if} \ (X_1 > X_3)$$
$$ = (X_3 > X_2) = \emptyset \ \text{if} \ (X_2 > X_3)$$
$$ = (X_3 > X_3) = \emptyset \ \text{if} \ (X_3 > X_1, X_2)$$
$$\vdots$$
Thus, $$Y_n = 1_{A_n} = 1_{\emptyset} = 0$$
Hence, $$E[Y_n] = Var[Y_n] = Y_n = 0 = E[Y_n^k] \ \forall k \ge 2$$

However, if the inequality is NOT strict i.e.
$$Y_n = 1_{B_n}$$
where
$$B_n:= (X_n \ge \max\{X_1, X_2, ... X_n\}) = (X_n = \max\{X_1, X_2, ... X_n\})$$
Then you are right because


*

*$P(B_n)= 1/n$ (proof here)

*$B_n$'s are independent (needed in computing variance).
Consider
$B_2 = \{X_2 = \max\{X_1, X_2\}\}$
$B_3 = \{X_3 = \max\{X_1, X_2, X_3\}\}$
Since the $X_n$'s are independent, the probability that $X_3$ is the maximum among $X_1, X_2, X_3$ is independent of which is the maximum among $X_1, X_2$
