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.

  • $\begingroup$ 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. $\endgroup$ Mar 15, 2018 at 18:00
  • $\begingroup$ @DavidC.Ullrich thanks. does it look okay now? $\endgroup$
    – iceberg
    Mar 15, 2018 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, 2018 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. $\endgroup$ Mar 15, 2018 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... $\endgroup$ Mar 15, 2018 at 18:18

2 Answers 2


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 $\theta.$

  • $\begingroup$ 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, 2018 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. $\endgroup$ Mar 20, 2018 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. $\endgroup$ Mar 20, 2018 at 0:53

Regards to @spaceisdarkgreen's post,

Why $Z_n = Y_n - log(n)$ converges in distribution to a nondegenerate distribution (a Gumbel), so this means $Z_n/log(n)$ converges almost surely to $0$?

I can only see that $Z_n/log(n)$ converges in probability to $0$, not almost surely.


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