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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.