How to show that Pr(limsup A_n) = 1?

Question

Let $ \{X_n \}_{n \in N} $ be a sequence of iid r.v. having exponential distribution with parameter $\lambda>0$.

How can I show that $$ Pr\left(\limsup_n \cfrac{X_n}{\log n}=\cfrac{1}{\lambda} \right)=1 $$

?

Answer 1

Let $\varepsilon>0$ and consider the events $$E_n = \left\{\frac{X_n}{\log n} > \frac1\lambda + \varepsilon\right\}.$$ Since $\mathbb P(X_n>t)= e^{-\lambda t}$, we have \begin{align}\sum_{n=1}^\infty \mathbb P(E_n)&=\sum_{n=1}^\infty e^{-\lambda \log n\left(\frac1\lambda + \varepsilon \right)}\\&=\sum_{n=1}^\infty \left(e^{\log n}\right)^{-(1+\lambda\varepsilon)}\\ &=\sum_{n=1}^\infty n^{-(1+\lambda\varepsilon)}<\infty, \end{align} so by the first Borel-Cantelli lemma, $$\limsup_{n\to\infty} \frac{X_n}{\log n}\leqslant\frac1\lambda+\varepsilon $$ almost surely. Since the $X_n$ are independent, applying the second Borel-Cantelli lemma to the events $$F_n = \left\{\frac{X_n}{\log n}>\frac1\lambda\right\} $$ yields $$\limsup_{n\to\infty} \frac{X_n}{\log n}\geqslant\frac1\lambda $$ almost surely. Since for any $k$ $$\left\{\frac1\lambda \leqslant\limsup_{n\to\infty} \frac{X_n}{\log n}\leqslant\frac1\lambda+\frac1k \right\}\supset \left\{\frac1\lambda \leqslant\limsup_{n\to\infty} \frac{X_n}{\log n}\leqslant\frac1\lambda+\frac1{k+1} \right\}, $$ it follows that

\begin{align} \mathbb P\left(\limsup_{n\to\infty} \frac{X_n}{\log n}=\frac1\lambda \right)&= \mathbb P\left(\bigcap_{k=1}^\infty \left\{\frac1\lambda \leqslant\limsup_{n\to\infty} \frac{X_n}{\log n}\leqslant\frac1\lambda+\frac1k \right\} \right)\\ &=\lim_{k\to\infty} \mathbb P\left(\frac1\lambda \leqslant\limsup_{n\to\infty} \frac{X_n}{\log n}\leqslant\frac1\lambda+\frac1k \right)\\ &=1. \end{align}