Hi I'm kind of a newbie of probability and I would like some help/hint with the following exercise:

Let $(Y_n)_{n \geq 1}$ be a sequence of Exponential random variables. Let moreover $\mathbb{E}Y_n = \lambda$

  1. Prove or disprove the following statement:

$$ \limsup_{n\to \infty}\frac{Y_{n}}{\lambda\log n}=1\quad \text{ a.s.}$$

  1. Define now $Z_n=\max\{X_1,...,X_n\} \ \forall n$, prove or disprove that:

$$ \lim_{n\to\infty}\frac{Z_{n}}{\lambda\log n}=1\quad \text{ a.s.}$$

I kind of have the feeling that i should apply in some way the Borel-Cantelli lemma but I'm really confused. Thanks in advance!


1 Answer 1


You wish to prove that $P(\limsup_n \frac{Y_n}{\lambda \log n} = 1) =1$. It suffices to prove that $$P(\limsup_n \frac{Y_n}{\lambda \log n} > 1) = P(\limsup_n \frac{Y_n}{\lambda \log n} < 1)=0$$

Let $\displaystyle X_n = \frac{Y_n}{\lambda \log n}$.

Note that $\displaystyle \limsup_n \frac{Y_n}{\lambda \log n} > 1 = \{w\in \Omega, \limsup_n \frac{Y_n(w)}{\lambda \log n} > 1\} = \bigcup_N \bigcap_n \bigcup_{k\geq n}\left(X_k> 1+\frac 1N \right)$.

It suffices to prove that $\forall N, P\left(\bigcap_n \bigcup_{k\geq n}\left(X_k> 1+\frac 1N \right) \right)=0$.

But $P\left(\bigcap_n \bigcup_{k\geq n}\left(X_k> 1+\frac 1N \right) \right)=P\left(\limsup_n \left( X_n> 1+\frac 1N\right) \right)$.

By Borel-Cantelli, it suffices to prove that $\sum_n P\left( X_n> 1+\frac 1N\right)$ converges. Note that $$P\left( X_n> 1+\frac 1N\right) = P\left(Y_n > \lambda \left( 1 + \frac 1N \right)\log n\right)=\frac{1}{n^{1+\frac 1N}}$$ and convergence follows.

This proves $P(\limsup_n \frac{Y_n}{\lambda \log n} > 1) =0$.

Note that $$ \begin{aligned}[t] \left(\limsup_n \frac{Y_n}{\lambda \log n} \right)< 1 &= \{w\in \Omega, \limsup_n \frac{Y_n(w)}{\lambda \log n} < 1\} \\&\subset \bigcup_n \bigcap_{k\geq n}\left(\frac{Y_k}{\lambda \log k} < 1\right)\\ &= \liminf_n \left( \frac{Y_n}{\lambda \log n} < 1\right) \end{aligned}$$

Since $\liminf_n \left( \frac{Y_n}{\lambda \log n}<1 \right)=\left(\limsup_n \left(\frac{Y_n}{\lambda \log n}>1\right) \right)^c $, it suffices to prove that $$P\left( \limsup_n \left( \frac{Y_n}{\lambda \log n}>1\right)\right)= 1$$

As expected we make use of the second Borel Cantelli lemma (after an independence assumption is added on the $Y_k$) : $$P\left( \frac{Y_n}{\lambda \log n}>1\right) =P\left( Y_n>\lambda \log n\right)= \frac 1n $$ and the sum diverges, proving the claim.

For the second part of your question, I suppose you can proceed similarly.

  • $\begingroup$ Sorry but I have some doubts on the resolution: shouldn't in the first part $\{ \omega \in \Omega : \limsup_n \frac{X_n(\omega)}{\lambda \log n}>1 \} $ be equal to $\bigcup_N \bigcap_n \bigcup_{k \geq n} ( X_k > 1+ \frac{1}{N} )?$ Moreover how can we prove in the second part that $P(\limsup_n \frac{X_n}{\lambda \log n}>1)=1$ when in the first part we showed that it's equal to 0? $\endgroup$ Nov 4, 2017 at 15:20
  • $\begingroup$ @Random-newbie why are you writing $\frac{X_n(\omega)}{\lambda \log n}$ ? You want to prove that $P(\limsup_n \frac{Y_n}{\lambda \log n} = 1) =1$. For notational ease, I have defined $X_n = \frac{Y_n}{\lambda \log n}$. $\endgroup$ Nov 4, 2017 at 15:28
  • $\begingroup$ Yeah sorry $\frac{Y_n(\omega)}{\lambda \log n}$ my mistake, but still i tink that there is something wrong with that set...a limsup should "get smaller" over time, instead it seems to me that that one is getting bigger as $n$ grows $\endgroup$ Nov 4, 2017 at 15:34
  • $\begingroup$ @Random-newbie you're right, sorry for the typo $\endgroup$ Nov 4, 2017 at 15:44
  • $\begingroup$ no problem you are really helping me! Sorry to bother you more but the second part is still unclear to me: in the first step of the proof we showed that $P(\limsup_n \frac{Y_n}{\lambda \log n}>1)=0$ how can we now show in the second part that $P(\limsup_n \frac{Y_n}{\lambda \log n}>1)=1$? It would be a contraddiction $\endgroup$ Nov 4, 2017 at 15:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .