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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. ?

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  • $\begingroup$ Do you mean $\limsup_{n\to\infty}$? "\limsup_{n\to\infty}"... $\endgroup$
    – d.k.o.
    Mar 7 '21 at 18:52
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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.

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