Using the Borel-Cantelli lemmas [closed]

I am really struggling to understand this question any advice would be appreciated.

Let $X_n$ be a sequence of independent random variables, each with the exponential distribution with rate $1/2$.

a) Use the Borel-Cantelli lemmas to show that

$$P(X_n > \alpha \log n\text{ for infinitely many }n) = \begin{cases} 0 \quad& \text{if} \:\:\:\alpha > 2 \\ 1 & \text{if} \:\:\: \alpha \le 2 \\ \end{cases}$$

b) Show that $\limsup_n \dfrac{Xn}{\log n} = 2$ almost surely.

Hint: Consider the events $\{X_n > 2 \log n \:\: \text{i.o.}\}$ and $\{X_n > (2 + 2/k) \log n \:\: \text{i.o.}\}$.

closed as off-topic by zhoraster, Claude Leibovici, user26857, Did, E. JosephDec 1 '16 at 10:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – zhoraster, Claude Leibovici, user26857, Did, E. Joseph
If this question can be reworded to fit the rules in the help center, please edit the question.

a)

Note $P(X_n > \alpha \log n) = \frac{1}{n^{\alpha/2}}$. What can you say about the convergence of $\sum_{n=1}^\infty P(X_n > \alpha\log n)$ for different values of $\alpha$?

Hint: For $\alpha>2$ apply the Borel-Cantelli lemma. For $\alpha \le 2$, apply the "converse" Borel-Cantelli lemma.

b)

By part a), $P\{X_n > 2 \log n \text{ i.o.}\}=1$, so, $\limsup_n \frac{X_n}{\log n} \ge 2$ almost surely.

If we show $\limsup_n \frac{X_n}{\log n} \le 2$ almost surely, or equivalently $P\left\{\limsup_n \frac{X_n}{\log n} > 2\right\}=0$, then we are finished.

For any $k>0$ we have $$P\left\{\limsup_n \frac{X_n}{\log n} > 2+2/k\right\} \le P\{X_n >(2+2/k)\log n \text{ i.o.}\} = 0$$ using part a) again. Taking $k \to \infty$ and noting $\left\{\limsup_n \frac{X_n}{\log n} > 2+2/k\right\}$ is an increasing sequence of sets gives $$P\left\{\limsup_n \frac{X_n}{\log n} > 2\right\} = 0.$$

b)

First show that ${\sf P} (\lim \sup_n \frac{X_n}{\ln n} \ge 2) = 1$:

$\forall \varepsilon > 0 \sum\limits_{n=1}^{\infty} n^{-(1-\varepsilon/2)} = \sum\limits_{n=1}^{\infty} {\sf P} (X_n > (2-\varepsilon)\ln n) = \infty$. Now, using BKL we have:

$$\forall \varepsilon > 0 \sum\limits_{n=1}^{\infty} {\sf P} (X_n > (2-\varepsilon)\ln n) = \infty \Rightarrow \\ \Rightarrow \forall \varepsilon > 0 {\sf P} (\frac{X_n}{\ln n} > 2 - \varepsilon \text{ i. m.} ) = 1 \Rightarrow \\ \Rightarrow \forall \varepsilon > 0 \:\:{\sf P} (\forall n \in \mathbb{N}\:\: \exists m > n: \frac{X_m}{\ln m} > 2 - \varepsilon ) = 1 \Rightarrow \\ \Rightarrow {\sf P}(\forall k \in \mathbb{N} \:\:\forall n \in \mathbb{N}\:\: \exists m > n: \frac{X_m}{\ln m} > 2 - \frac{1}{k}) = 1 \Rightarrow \\ \Rightarrow {\sf P} (\forall \varepsilon > 0\:\:\forall n \in \mathbb{N}\:\: \exists m > n: \frac{X_m}{\ln m} > 2 - \varepsilon) = 1 \Rightarrow \\ \Rightarrow P(\lim \sup_n \frac{X_n}{\ln n} \ge 2 ) = 1$$

Now, similar to this, show that ${\sf P} (\lim \sup_n \frac{X_n}{\ln n} \le 2) = 1$. These inequalities together imply what you want.

• here $\mathbb P(X_n>\alpha \log (n))\le \frac{1}{n^{\frac{\alpha}{2}}}$ – bunny Oct 18 '17 at 5:14