Application of Borel-Cantelli Lemma

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Let $\{X_n\}_{n\geq1}$ be i.i.d. exponential random variables with parameter $\lambda$, then $$\mathscr{P}\left\{\underset{n\rightarrow\infty}{\limsup}\frac{X_n}{\log{n}}=\frac{1}{\lambda}\right\}=1.$$

 $\bullet$ Proof.
Recall Berel-Cantelli Lemma,

 [Lemma]
Let $\mathscr{F}$ be a Borel field and $\{E_n\}_{n\geq1}\in\mathscr{F}$ are events.  We have
(1) $\sum_{n=1}^\infty \mathscr{P}\{E_n\} < \infty \implies \mathscr{P}\{E_n\mbox{ i.o.}\}=0;$
(2) If $\sum_{n=1}^\infty \mathscr{P}\{E_n\} = \infty$ and $E_n$'s are independent.  Then $\mathscr{P}\{E_n\mbox{ i.o.}\}=1.$

Back to the proof.  $X_n\overset{i.i.d.}{\sim}\mbox{exp}(\lambda)$, we have $\mathscr{P}\{X_n>x\}=e^{-\lambda x}.$

(Part I)
Given $\varepsilon>0$,$$\sum_{n=1}^\infty\mathscr{P}\left\{\frac{X_n}{\log{n}}>\frac{1}{\lambda}+\varepsilon\right\}=\sum_{n=1}^\infty\exp{\{-\log{n^{(1+\lambda\varepsilon)}}\}}=\sum_{n=1}^\infty n^{-(1+\lambda\varepsilon)}<\infty.$$Thus by the first Borel-Cantelli Lemma, we have $$\mathscr{P}\left\{\underset{n\rightarrow\infty}{\limsup}\frac{X_n}{\log{n}}>\frac{1}{\lambda}+\varepsilon\right\}=0\implies\mathscr{P}\left\{\underset{n\rightarrow\infty}{\limsup}\frac{X_n}{\log{n}}\leq\frac{1}{\lambda}+\varepsilon\right\}=1.$$
(Part II)
Consider that $\{X_n\}_{n\geq1}$ are i.i.d. and $$\sum_{n=1}^\infty\mathscr{P}\left\{\frac{X_n}{\log{n}}\geq\frac{1}{\lambda}\right\}=\sum_{n=1}^\infty e^{-\log{n}}=\sum_{n=1}^\infty\frac{1}{n}=+\infty.$$Thus by the second Borel-Cantelli Lemma, we have $$\mathscr{P}\left\{\underset{n\rightarrow\infty}{\limsup}\frac{X_n}{\log{n}}\geq\frac{1}{\lambda}\right\}=1.$$
Let's combine (Part I) and (Part II) together.  The event $$\left\{\underset{n\rightarrow\infty}{\limsup}\frac{X_n}{\log{n}}=\frac{1}{\lambda}\right\}=\bigcap_{k=1}^\infty\left\{\frac{1}{\lambda}\leq\underset{n\rightarrow\infty}{\limsup}\frac{X_n}{\log{n}}\leq\frac{1}{\lambda}+\frac{1}{k}\right\},$$by setting $\varepsilon=\frac{1}{k}$ in (Part I).  Then, since $\varepsilon$ (or $k$) is arbitrary, we have $$\mathscr{P}\left\{\underset{n\rightarrow\infty}{\limsup}\frac{X_n}{\log{n}}=\frac{1}{\lambda}\right\}=1.$$

$\Box$