Show that $P($ $\limsup $ $\frac{X_n}{\log n}$ $ =$ $ 1$ $)$ $ =$ $ 1$

Question

QUESTION

Let $X_n$ be independent and exponentially distributed with parameter $1$. Show that $P\left(\limsup \frac{X_n}{\log n} = 1\right) = 1$.

ATTEMPT

What I've tried to do is:

$$ P\left( \frac{X_n}{\log n} - 1 \geq \epsilon\right) \begin{aligned}[t] & = P\left(X_n \geq \log n\times(1+\epsilon)\right) \\ & = P(e^{X_n} \geq n^{1+\epsilon}) \\ & \leq \frac{E(e^{X_n})}{n^{1+\epsilon}} \\ & = \frac{1}{n^{1+\epsilon}} \end{aligned} $$

EDIT

Due to the suggestion in the comment, I think that what I wrote before is wrong. The correct one I think is using the fact that: $$ P(X>x)=e^{-\lambda x} \text{ for } \ x\geq 0 \ \text{if } X\sim \text{exp}(\lambda) $$ In my case I have: $$ P\left(X_n \geq \log n\times(1+\epsilon)\right) = e^{1* \log n*(1+\epsilon)} = \frac{1}{n^{1+\epsilon}} $$

The question linked as a duplicate is helpful but does not address the question and so I would be interested in how to explicitly use this result to obtain the result in the question.

What I take advantage of in the link is that : $$ P(X>x)=e^{-\lambda x} \text{ for } \ x\geq 0 \ \text{if } X\sim \text{exp}(\lambda) $$ Thefore, by changing the values, I obtained the result.

Answer 1

It is helpful if we first state the two Borel Cantelli Lemmas and then see how we can apply them to this problem.

Borel Cantelli Lemma I states that for measurable sets $(A_n)_{n \in \mathbb{N}}$ that: $$\sum _{n=1}^{\infty} P(A_n) < \infty \implies P(\limsup _{n \rightarrow \infty} A_n) = 0$$

And we get a (partial) converse if we add in the additional requirement for each of the events to be independent which can be seen in the second Borel Cantelli Lemma below.

Borel Cantelli Lemma II states that for independent events $(A_n)_{n \in \mathbb{N}}$, then: $$\sum _{n=1}^{\infty} P(A_n) = \infty \implies P(\limsup_{n \rightarrow \infty} A_n) = 1 $$

There are a few different approaches we could use to tackle this particular problem. One fairly straightforward approach would be to consider $\sum _{n=1}^{\infty} P(\frac{X_n}{log(n)} \leq 1)$ and $\sum _{n=1}^{\infty} P( \frac{X_n}{log(n)} \ge 1).$ You can use the Borel Cantelli Lemmas to show that both of these sums diverge to infinity (using the CDF for exponentially distributed random variables). This then implies that $P(\limsup _{n \rightarrow \infty} \frac{X_n}{log(n)} = 1)=1 $ almost surely since we have shown that the probability that it is less than or equal to $1$ or greater than or equal to $1$ are both $1$ by the second Borel Cantelli Lemma (using independence).

To show how we can do this explicitly, consider the first sum:

$$\sum _{n=1}^{\infty} P(\frac{X_n}{log(n)}< 1) = \sum _{n=1}^{\infty} P(X_n< log(n)) = \sum _{n=1}^{\infty} 1-e^{-log(n)} = \sum _{n=1}^{\infty}1 - \frac{1}{n} \rightarrow \infty $$

Therefore, using this fact and independence, we can use Borel Cantelli Lemma II to show that $P( \limsup _{n \rightarrow \infty} \frac{X_n}{log(n)} \leq 1) = 1$

Now just use the same method to show $P( \limsup _{n \rightarrow \infty} \frac{X_n}{log(n)} \ge 1) = 1$ and you are done.