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 $X_i$?

  • $\begingroup$ 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 $X_i$. $\endgroup$ – Did Sep 19 '11 at 5:11

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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    $\begingroup$ You know that almost surely finitely many terms of a sequence are $\le c$, for every $c<1$. This fact alone implies that the liminf of the sequence is $\ge1$. $\endgroup$ – Did Sep 19 '11 at 6:01
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    $\begingroup$ @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. $\endgroup$ – cardinal Sep 23 '11 at 21:40
  • $\begingroup$ The last part of this answer seems to be arguing that because $$ \sum_{n \in \mathbb{N}} P\left(\frac{M_n}{\log n} \leq c \right) = \infty $$ for c>1, we can then conclude that $$ P\left(\left\{ \frac{M_n}{\log n} \leq c \text{ i.o.}\right\} \right) = 1 $$ But I don't see how, since Borel-Cantelli II needs also that the sequence of events be independent, which clearly fails here since the $M_n$ "look" at increasing subsets of the $X_n$'s. $\endgroup$ – mbiron Oct 1 '19 at 2:01
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    $\begingroup$ @mbiron It's not that the sum of probabilities is infinite, it's that the probabilities go to $1$ as $n \to \infty$. If $A_n = \{M_n/\log n \le c\}$, $P\left(\bigcap_{m>n} A_m^c\right) \le \inf_{m > n} P(A_m^c) = 0$. $\endgroup$ – Robert Israel Oct 2 '19 at 4:02

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.

  • $\begingroup$ The series you wrote is divergent. $\endgroup$ – Did Sep 19 '11 at 5:13
  • $\begingroup$ Hi Piau, I guess that is convergent in some cases, for example, if $\mathbb{P}(X_1\leq n)^n= \left(1-\frac{1}{\ln n}\right)^n$. $\endgroup$ – Leandro Sep 19 '11 at 5:39
  • $\begingroup$ The series you wrote is divergent in the case at hand (exponential with parameter $1$). I understand now that you are not answering this question (exponential case) but rather, considering a general setting where you could prove something about $M_n$. Note that the convergence of the series you write implies that $\liminf M_n/n\ge1$ (liminf instead of limsup and $n$ instead of $\log n$). $\endgroup$ – Did Sep 19 '11 at 5:51
  • $\begingroup$ Still have to replace limsup by liminf... $\endgroup$ – Did Sep 19 '11 at 6:02
  • $\begingroup$ Hi Piau, thanks for point the corrections needed. But I still think that first Borel-Cantelli Lemma should give me the lower bound for limsup. I considered the events $E_n=\{\max\{X_1,\ldots,X_n\}\leq \ln n\}$. Using the Hypothesis I posted in my comment the limsup of this sequence has probability zero. Do you agree ? $\endgroup$ – Leandro Sep 19 '11 at 6:04

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