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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.

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1 Answer 1

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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.

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  • $\begingroup$ How do you change this to \limsup \frac{X_n}{\logn}=1? I am unsure how to formulate this. $\endgroup$
    – K-Q
    Nov 29, 2018 at 15:28

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