# To show $\liminf \ X_n/\log(n) = 0$ almost surely

For i.i.d s $\{X_n,\ n\geq 1\}$ that are $exp(1)$ random variables, I need to prove that $$\liminf \ \frac{X_n}{\log(n)} = 0\ \ \ \ \text{almost surely}$$ I have found that $\mathbb{P}(X_n > k\log(n)\ \ i.o ) = 1_{\{k \leq 1\}}$ using Borel-Cantelli Lemma, hence $\limsup X_n/\log(n) = 1$ almost surely. But I am having trouble with the limit inferior. I have tried to use results like $\liminf Y_n = - \limsup (-Y_n)$ or for sets $\liminf (A_n) = \limsup A_n^c$ but can't arrive at the result.

Any hints or solutions would very helpful. Thank you.

Let's try to prove that

$$P(\liminf_{n}\frac{X_n}{\log(n)} \le \epsilon) = 1$$ for all $\epsilon > 0$, ok?

Note that

$$\bigcap_m \bigcup_{n\ge m} \{ \frac{X_n}{\log(n)} \le \epsilon \} \subset\{ \liminf_{n}\frac{X_n}{\log(n)} \le \epsilon \}$$

But $$P (\bigcap_m \bigcup_{n\ge m} \{ \frac{X_n}{\log(n)} \le \epsilon \}) = \lim_m P(\bigcup_{n\ge m} \{ \frac{X_n}{\log(n)} \le \epsilon \}) \ge \lim_m P(X_m \le \epsilon \log(m))$$

Using that $X\sim exp(1)$ you have that

$$\lim_m P(X_m \le \epsilon \log(m)) = \lim_m 1 -\frac{1}{m^{\epsilon}}=1$$

Now we can make an intersection in $k$ taking $\epsilon = 1/k$.

Hope this can help.

• How do you change this to \limsup \frac{X_n}{\logn}=1? I am unsure how to formulate this. – KieranSQ Nov 29 '18 at 15:28