# Borel-Cantelli lemma, exponential distribution problem

Good evening,

I am currently solving an exercise : Let Xn be a sequence of independent random variables, each with the exponential distribution with rate $$\frac{1}{\lambda}$$. I want to prove the part : $$\lim_{n \to \infty} \sup_n (\frac{X_n}{\log(n)}) \leq \frac{1}{\lambda}$$ a.s.

I found $$Pr(\frac{X_n}{\log(n)}\geq \alpha) = \frac{1}{n^{\alpha \lambda}}, \alpha >0$$.

I used the Borel-Cantelli lemma with $$\alpha > \frac{1}{\lambda}$$ and the sum $$\sum_{n \in \mathbb{N}} Pr(X_n \geq \alpha ln(n)) < + \infty$$

In the end, I found $$Pr(\frac{X_n}{\log(n)}\geq \alpha$$ i.o$$) = 0$$ $$=>\lim_{n \to \infty} \sup_n (\frac{X_n}{\log(n)}) \leq \alpha$$ a.s.

How can I demonstrate that $$\lim_{n \to \infty} \sup_n (\frac{X_n}{\log(n)}) \leq \frac{1}{\lambda}$$ a.s. ?

• Do you mean $\limsup_{n\to\infty}$? "\limsup_{n\to\infty}"... Mar 7 '21 at 18:52

First, for any $$\alpha>0$$, $$\mathsf{P}(X_n>\alpha\ln n)=n^{-\alpha\lambda}.$$ The RHS is summable iff $$\alpha>\lambda^{-1}$$. Therefore, by the B-C lemma, $$X_n>\ln n/\lambda$$ i.o., but $$X_n<(1+\epsilon)\ln n/\lambda$$ ev. for any $$\epsilon >0$$. Thus, $$\limsup_{n\to\infty}X_n/\ln n=1/\lambda$$ a.s.