# Expectation of infimum of records

I'm trying to prove the next:

Let $$L_{1}=\inf\{j\geq 2: X_j\space\text{is a record}\}.$$

Prove that $$E(L_{1})=\infty.$$

Here, we say $$X_n$$ is a record if $$X_n>\max\{X_2,\ldots,X_{n-1}\}$$ and $$\{X_n\}$$ is a sequence of i.i.d. with continuous distribution.

I'm having prblems proving this; this is my attempt:

For $$k\geq 2$$ we have $$\{L_1=k\}=\{X_{k-1}<\ldots so we have $$P(\{L_1=k\})=\frac{1}{k!},$$ but this is not the density of random variable $$L_1$$ because $$\sum_{k\geq 2}P(\{L_1=k\})=e^{1}-2.$$

In fact $$E(L_1)<\infty$$ because of the above.

How to prove the expectation is infinity? What's wrong with the previous?

Any kind of help is thanked in advanced.

• There are more ways that $\{L_1=k\}$ can occur other than $\{X_{k-1}<X_{k-2}<\dots<X_1<X_k\}$. There is also, for example, $\{X_{k-2}<X_{k-1}<X_{k-3}<\dots<X_1<X_k\}$. All that matters is that among the first $k$ observations, $X_k$ is the biggest and $X_1$ is the second biggest. – Mike Earnest May 22 at 0:10
• Thanks @MikeEarnest. You are right; I was misinterpreting the meaning of $\{L_1=k\}.$ – Suiz96 May 22 at 3:39

Let $$f(x)$$ be the probability distribution function of $$X_n$$ for all $$n$$ and let $$F(x)$$ be the corresponding cdf. Let us compute the probability that $$L_1 = k$$. If $$X_1 = x$$, then we have that $$\displaystyle P(L_1 = k| X_1 = x) = \prod_{i = 2}^{k-1} P(X_i < x) Pr(X_k > x) = F(x)^{k-2}(1 - F(x))$$. To obtain $$P(L_1 = k)$$, we can integrate over $$X_1$$, which gives us $$\displaystyle P(L_1 = k) = \int_{\mathbb{R}}F(x)^{k-2}(1 - F(x)) \ f(x)\ dx = \frac{1}{k -1} - \frac{1}{k} = \frac{1}{k(k-1)}$$. Thus, we can now write, $$\displaystyle \mathbb{E}(L_1) = \sum_{k = 2}^{\infty} k P(L_1 = k) = \sum_{k = 2}^{\infty} k \frac{1}{k(k-1)} = \sum_{k = 1}^{\infty} \frac{1}{k}$$, which we knows diverges.