# 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

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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