# Does maximum over n i.i.d. exponential distribution r.v.s. minus ln n coverge almost surely?

Suppose $\{X_i,i\ge 1\}$ is a sequence of i.i.d. random variables of exponential distribution with mean 1. Let $M_n=max_{i=1,\cdots,n}X_i$ and $Z_n=M_n-\ln n$. It is not hard to see $Z_n$ converges to $Z_\infty$ in distribution, where $P(Z_\infty\le x)=e^{-e^{-x}}$. And we need to show whether or not $Z_n$ converges to some limiting r.v. almost surely.

My idea is the following: since the distribution of $Z_\infty$ is continuous, then suppose $Z_n$ converges to some r.v. a.s., then the limiting r.v. should be $Z_\infty$. I want to show actually $Z_n$ does not converge to $Z_\infty$ in probability, then we will get contradiction.
Then pick fixed $x,\epsilon>0$, $P(|Z_n-Z_\infty|>\epsilon)\ge P(Z_\infty>x+\epsilon,Z_n\le x)$. But we don't know $Z_\infty$ is independent of $Z_n$. So is there any other idea to prove it?

• I see... I revised the problem. Jan 28 '17 at 4:59
• In these situations, it's usually easier to show that $Z_n$ is not Cauchy in probability i.e. $P(|Z_n-Z_{m+n}| > \epsilon)$ does not go to zero for all $m\ge 1$ as $n\rightarrow \infty.$ Haven't done this problem before so not sure if it'll work here. (or if no is the right answer) Jan 28 '17 at 5:01
• @spaceisdarkgreen Correct. When m is comparable to n, it will be done. Jan 28 '17 at 5:18
• I think I got it (see answer)... not sure its 100% rigorously correct and maybe there's an easier way. (Is that what you had in mind with $m$ comparable to $n$?) Jan 28 '17 at 6:07
• Yes. Or we can pick a fixed $x$, then $P(|Z_n-Z_{2n}|>\epsilon)\ge P(Z_n\le x, |Z_{2n}|>x+\epsilon)$ and use the fact that $X_{n+1},...,X_{2n}$ is independent of the former ones to show the limit of the second term in the inequality is positive. Jan 31 '17 at 0:56

Let $\epsilon>0$ be anything. Consider $Z_{2n}-Z_n = -\ln(2) + M_{2n}-M_n$ and look at $$P(|Z_{2n}-Z_n|> \epsilon)$$
We have $|Z_{2n}-Z_n| > \epsilon$ in the case that $M_{2n}-M_n > \ln(2) + \epsilon$ so$$P(|Z_{2n}-Z_n|> \epsilon) \ge P(M_{2n}-M_{n}> \ln(2)+\epsilon).$$
Let $$M_{2n,n} \equiv \max\{X_{n+1},\ldots,X_{2n}\}.$$ The only way we can have $M_{2n} -M_n >0$ is if $M_{2n,n}=M_{2n},$ so $M_{2n}-M_n > \ln(2) + \epsilon$ if and only if $M_{2n,n} - M_n > \ln(n)+\epsilon.$ This means $$P(M_{2n}-M_{n}> \ln(2)+\epsilon) = P(M_{2n,n}-M_{n}> \ln(2)+\epsilon).$$
If we then define $Z_{2n,n} \equiv M_{2n,n}-\ln(n)$ and recall $Z_n=M_n-\ln(n),$ we can write $$P(|Z_{2n}-Z_n|> \epsilon) \ge P(M_{2n,n}-M_n>\ln(2)+\epsilon) = P(Z_{2n,n} - Z_n > \ln(2) + \epsilon).$$ But now notice that $Z_{2n,n}$ and $Z_n$ are independent by construction and in the limit of large $n$, they are Gumbel-distributed as you've shown. So we have $$\lim_{n\rightarrow\infty}P(|Z_{2n}-Z_n|> \epsilon) \ge P(A_1-A_2>\ln(2) + \epsilon)$$ where $A_1$ and $A_2$ are independent Gumbels, and it's pretty clear that $P(A_1-A_2>\ln(2) + \epsilon)>0$ since it's a continuous distribution with support on all the reals.
So this shows that $Z_n$ is not Cauchy in probability and therefore does not converge in probability.