4
$\begingroup$

I am currently working on the following problem:

Suppose that $X_n$ are independent identically distributed random variables from a unit exponential distribution (so $P(X_n \leq x) = 1 - e^{-x}$, for $x \geq 0$). Let $M_n = \text{max}\{X_1, \ldots, X_n\}$.

Then the question is to find a sequence $u_n \rightarrow \infty$ such that $P(M_n \geq u_n$, infinitely often$) = 0$. I have the result that if $u_n \rightarrow \infty$ is any monotone increasing sequence, then $P(M_n \geq un$, infinitely often$) = P(X_n \geq u_n$, infinitely often$)$.

I am attempting to work through this problem using the Borel-Cantelli lemma, specifically the result that if $\sum_n P(X_n \geq u_n$, infinitely often$) < \infty$ then $P(X_n \geq u_n$, infinitely often$) = 0$. However, I can only calculate $P(X_n \leq x)$ (and therefore also $P(X > x)$), so in the end all I can conclude is that $P(X_n > u_n$, infinitely often$) = 0$.

My question is, how can I use the information I have been given to produce a result for $P(X_n \geq 0$, infinitely often$)$? I suspect that I have missed something about the way the lemma is supposed to be used.

$\endgroup$
  • 1
    $\begingroup$ So... you know that $P(X_n\geqslant u_n)=e^{-u_n}$ and you want $(u_n)$ going to infinity such that $\sum P(X_n\geqslant u_n)$ diverges. No idea at all? $\endgroup$ – Did Jan 9 '17 at 20:03
  • $\begingroup$ Isn't it that $P(X_n > u_n) = e^{-u_n}$? That's where I'm confused, since I don't have an expression for $P(X_n \geq u_n)$. $\endgroup$ – Uzai Jan 9 '17 at 20:43
  • 1
    $\begingroup$ These are random variables $X$ with a PDF hence (you should be well aware that) $P(X=x)=0$ for every $x$. $\endgroup$ – Did Jan 9 '17 at 22:09
  • $\begingroup$ Still stuck? Then addressing my former comment might be a productive option... To be even more explicit: for every $x$, $P(X>x)=P(X\geqslant x)$ hence you are seeing trouble where there is none. $\endgroup$ – Did Feb 7 '17 at 16:56

Your Answer

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

Browse other questions tagged or ask your own question.