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 $\{L_1=k\}=\{X_{k-1}<\ldots<X_{1}<X_{k}\},$ 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.

  • $\begingroup$ 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. $\endgroup$ – Mike Earnest May 22 at 0:10
  • $\begingroup$ Thanks @MikeEarnest. You are right; I was misinterpreting the meaning of $\{L_1=k\}.$ $\endgroup$ – 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.

  • $\begingroup$ So clear! Many thanks @sudeep5221. $\endgroup$ – Suiz96 May 22 at 3:56

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