1
$\begingroup$

Let $X_1, X_2, \dots$ be a sequence of i.i.d random variables from distribution $F$ with exponential tails. Denote $Y_n = \max (X_1, \dots , X_n)$. How can we prove the following: $$ \lim_{n \rightarrow \infty} \frac{Y_n}{\log n} = c $$ almost surely for some constant $c$.

And also, how can we determine what that value of $c$ is? What if we knew that the distribution $F$ is, say, a Gamma distribution (or another common distribution)?

This result seems standard, as indicated in the question here, but I could not discover how to prove it.

$\endgroup$
  • $\begingroup$ No, you didn't really fix the notation much. First, your $(X_j)$ appreas to be a finite sequence. And saying "Let $Y_n$ be the maximum of the sequence" means that $Y_1=Y_2=Y_5$, because the sequence has only one maximum. You could try simply copying the problem carefully. $\endgroup$ – David C. Ullrich Mar 15 '18 at 18:00
  • $\begingroup$ @DavidC.Ullrich thanks. does it look okay now? $\endgroup$ – iceberg Mar 15 '18 at 18:09
  • $\begingroup$ @shoeburg I think it would be better to write "Let $X_1, X_2, \ldots$ be a sequence of i.i.d. random variables and let $Y_n = \max(X_1, \ldots, X_n)$." $\endgroup$ – angryavian Mar 15 '18 at 18:11
  • $\begingroup$ It looks like you didn't read my last comment. You're still talking about a finite sequence $(X_j)$, which is clearly not what you intended. And your $X_{max}$ is still just one variable. Since there's no "$n$" in $X_{max}$ that limit is still obviously zero, whhich is clearly nnot what you intended. Why not just copy the problem exactly as it appears wherever you found it? I mean it's not hard to guess what you actually mean, but if you can't be bothered to state the problem correctly you shouldn't expect people to help. $\endgroup$ – David C. Ullrich Mar 15 '18 at 18:15
  • $\begingroup$ @angryavian That's clearly what was intended. In my opinion we shouldn't help people state their problem coherently - if they're unable or uunwilling to do that then what's the point? Here all he has to do is copy the statement carefully from whatever the source was... $\endgroup$ – David C. Ullrich Mar 15 '18 at 18:18
4
$\begingroup$

For a standard exponential, we know that $Z_n=Y_n-\log(n)$ converges in distribution to a nondegenerate distribution (a Gumbel), so this means $\frac{Z_n}{\log(n)}$ converges almost surely to zero, which in turn means that $\frac{Y_n}{\log(n)}$ converges almost surely to $1.$

So the almost-sure convergence of something like this follows from the extreme value distribution. In general, for a distribution with an infinite tail that decays faster than a power law, we have that $\frac{Y_n-b_n}{a_n}$ converges in distribution to a Gumbel. For something like a Gamma, with a pure exponential tail $\sim e^{-x/\theta}$, we can work out that we have $a_n=\theta$ and $b_n$ to leading order in $n$ is $\theta\log(n).$ So for a Gamma with PDF $\frac{1}{\Gamma(\alpha) \theta^\alpha}x^{\alpha-1}e^{-x/\theta},$ $\frac{Y_n}{\log(n)}$ converges almost surely to $\theta.$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ To work out those $a_n$ and $b_n$ terms, one needs a good grasp of extreme value theory? I was hoping there is a "direct" way to prove such a result about a.s. convergence, maybe using union bound and Borel-Cantelli lemma for example. $\endgroup$ – iceberg Mar 18 '18 at 19:29
  • $\begingroup$ @shoeburg BC doesn't seem to work. For instance, for a std exponential, we have $P(Y_n/\log(n) >2)=1-(1-n^{-2})^n \sim 1/n,$ so $\sum_n P(Y_n/\log(n)>2)=\infty.$ So the dependency of the $Y_n$ is important here. $\endgroup$ – spaceisdarkgreen Mar 20 '18 at 0:34
  • 1
    $\begingroup$ @shoeburg That said, perhaps writing things down in terms of EVT mystified things too much here... everything can be done on a case-by case basis. The way it works is to pick $a_n$ and $b_n$ so that $P((Y_n-b_n)/a_n<x) =P(Y_n<a_nx+b_n)=(1-F(a_nx +b_n))^n$ tends to a nondegenerate function, which just involves some asymptotics on the tail of $F$ (though it helps to know at that outset that the function it will tend to is $e^{-e^{-x}}$). I don't doubt there's a way like you're alluding to, though, but the 'stickiness' of the maximum will have to be accounted for as the previous example shows. $\endgroup$ – spaceisdarkgreen Mar 20 '18 at 0:53

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.