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. ?

  • $\begingroup$ Do you mean $\limsup_{n\to\infty}$? "\limsup_{n\to\infty}"... $\endgroup$
    – d.k.o.
    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.


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.