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Suppose $X_i$'s are i.i.d, with the density distribution $f(x) = e^{-x}$, $x \geq 0$. I was able to show that $P(\limsup X_n/\log{n} =1)=1$ using Borel-Cantelli.

Define $M_n=\max \{X_1,\ldots,X_n\}$, can I claim $M_n/\log{n} \rightarrow 1$ a.s. in this case? Is it still true in general without knowing the distribution of $X_i$?

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    The proof that $M_n/\log n\to1$ (a result which is true) is very similar to the proof that $\limsup X_n/\log n=1$ (a proof you say you know). Hence, you could show where the proof for $\limsup$ fails for $M_n$. And naturally, the result itself depends heavily on the distribution of the random variables $X_i$.2011-09-19

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If $F(x) = 1 - e^{-x}$ for $x > 0$ is the CDF of each $X_i$, the CDF of $M_n$ is $F_{M_n}(x) = F(x)^n = (1 - e^{-x})^n$ for $x > 0$. Note that $\ln(F_{M_n}(x)) = n \ln(1 - e^{-x})$ and since $- t - t^2 < \ln(1-t) < -t$ for $0 < t < .683$, for any $c>0$ we have $-n^{1-c} - n^{1-2c} \le \ln(F_(M_n)(c \ln n)) \le -n^{1-c}$ for $n$ large enough. If $c < 1$, this says $P\left( \frac{M_n}{\ln n} \le c\right) \le e^{-n^{1-c}}$, and $\sum_n e^{-n^{1-c}} < \infty$ so almost surely only finitely many $\frac{M_n}{\ln n} \le c$. If $c > 1$, $P(\left( \frac{M_n}{\ln n} \le c \right) \ge e^{-n^{1-c} - n^{1-2c}} \to 1$ as $n \to \infty$, so almost surely infinitely many $\frac{M_n}{\ln n} \le c$. Thus $\lim \sup_n \frac{M_n}{\ln n} = 1$ almost surely. However, it's not so clear to me that $\lim \inf_n \frac{M_n}{\ln n} = 1$ almost surely.

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    @Robert, forgive the question; I'm sure it's due to a moment of denseness on my part. Reading your argument, it seems that the claim is that since $M_n/\log n \leq c$ only finitely many times for all c < 1 and $M_n/\log n \leq c$ infinitely often for all c > 1, that (somehow) then we can conclude $\limsup_n M_n / \log n = 1 \,\mathrm{a.s.}$. Perhaps you could clarify this, since it would seem we need rather that $M_n / \log n \geq c$ only finitely many times for all c > 1 (which is true!) for the desired result of on the $\limsup$ to follow.2011-09-23
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this is a small observation not an answer:

the distribution is in fact important, for example if the random variables are bounded almost surely the limit is zero a.s.

For the unbounded case, (that more likely you are thinking about), I just got the trivial lower bound $ 1\leq \liminf_{n\to\infty} \frac{M_n}{\log n} $ in the i.i.d case, by the Borel-Cantelli Lemma, supposing that $\sum_{n=1}^{\infty} \mathbb{P}(X_1\leq \ln n)^n<+\infty.$

Edition. I replaced the limsup by liminf, which is better in this case, based on the comments made by Didier Piau.

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    The sum of the probabilities of $M_n\le c\log n$ converges for every c<1. This is all you need to conclude about the liminf.2011-09-19