# Almost sure convergence of maximum of sequence of random variables

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.

• 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. Mar 15, 2018 at 18:00
• @DavidC.Ullrich thanks. does it look okay now? Mar 15, 2018 at 18:09
• @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 $Y_n = \max(X_1, \ldots, X_n)$." Mar 15, 2018 at 18:11
• 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. Mar 15, 2018 at 18:15
• @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... Mar 15, 2018 at 18:18

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 $\theta.$

• 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. Mar 18, 2018 at 19:29
• @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. Mar 20, 2018 at 0:34
• @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. 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.