0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.