# Variance of event counting

I have this question (not homework, review problem for qualifying exam), tried approaching it a couple of ways (unsuccessfully). Any recommendations?

Let $X_1,..,X_n$ be i.i.d continuous rvs. A record is said to occur at time $k$ if $X_k > X_i$ for all $i = 1,...,k-1$. Let $N$ denote the number of records. Find the variance of $N$.

What you need is the theory of record values. $X_1, X_2, \dots, X_n, \dots$ are i.i.d. random variables, with an absolutely continuous distribution with density function $f$. Define the binary variables $$I_i = \begin{cases} 1, \text{i is a record time} \\ 0, \text{i is not a record time}. \end{cases}$$ that is, $I_i=1$ if $X_1<x_i, X_2<X_i, \dots, X_{i-1}<X_i$. Since we have assumed an absolutely continuos distribution, we can disregard the possibility of equality, which have probability zero. Now, $$P(I_i=1)=1/i,$$ since the maximum among the first $i$ observations must have equal probability of occuring at any of the $i$ times $1,2, \dots, i$. Next, somewhat surprisingly, the process $I_1, I_2, \dots$ is independent, since the probability that $j$ will be a new record time cannot in any way depend on the ordering among the first $j-1$ observations! So, letting $N_n=I_1+I_2+\dots+I_n$ be the number of records among the first $n$ observations, we get $$E N_n = \frac 11+\frac12+\frac13+\dots+\frac1n=H_n$$ the $n$th harmonic number, and the variance is $$\text{Var}(N_n) = 1\cdot(1-1)+\frac12(1-\frac12)+\frac13(1-\frac13)+\dots+\frac1n(1-\frac1n).$$