3
$\begingroup$

Suppose $X_{n}$ is any sequence of (real-valued) random variables defined on a common probability space $(\Omega,\mathcal{F},\mathbb{P})$. Show that there exists a sequence $c_{n}\rightarrow \infty$ such that $\frac{X_{n}}{c_{n}}\rightarrow 0$ almost surely. The key is to use the Borel Cantelli lemma, but I am not able to formulate pairwise independent events $A_{n}$ such that $\sum\limits_ {n=1}^{\infty}\mathbb{P}(A_{n})=\infty$ implies $\mathbb{P}(A_{n}~\text{i.o.})=1$, where i.o. stands for infinitely often.

Just a few small hints would be greatly appreciated.

$\endgroup$

2 Answers 2

4
$\begingroup$

Use (or reprove) the extended form of Borel Cantelli, which says that if there exists a decreasing sequence $\epsilon_i\rightarrow 0$ such that $\sum_i P(|X_i|>\epsilon_i)<\infty$, then $X_i$ converges to 0 almost surely. You can pick any such sequence $\epsilon_i$, such as $\epsilon_n=1/n$, and then pick a corresponding $c_n$ sequence, that decreases $P(|X_n|>c_n/n)$ sufficiently fast.

$\endgroup$
2
  • $\begingroup$ Thank you so much!! Will try the problem with this. $\endgroup$ Sep 3, 2016 at 18:32
  • 1
    $\begingroup$ Can you post a link to this extended form pls, I can't find it. $\endgroup$
    – Alex
    Jun 15, 2020 at 10:40
1
$\begingroup$

First note that if $X_n$ is real valued then \begin{equation} \mathbb{P}[X_n=\infty] = 0. \hspace{2cm}(1) \end{equation} Let $\{c_n\}$ be an increasing and unbounded sequence $(c_n \nearrow \infty)$. Define $E_n = \Big \{\omega\in \Omega : |X_n(\omega)| > \frac{c_n}{n}\Big\}$. Then, in order to get a contradiction, Suppose that there is no sequence of positive numbers ${c_n}$ such that $$\mathbb{P}[E_n] ≤ 2^{−n},$$ i.e. for every $N\in\mathbb{N}$, setting for a fixed $n$ $$A_N:= \Big \{\omega\in \Omega : |X_n(\omega)| > \frac{N}{n}\Big\}, $$ and $$A_\infty:= \Big \{\omega\in \Omega : |X_n(\omega)| = \infty \Big\},$$
it follows that $$\mathbb{P}[A_{N}] > 2^{-n}.$$ From there, proving that the sequence $\{A_N\}$ is decreasing and $A_\infty = \bigcap_N A_N$, you can show that $\mathbb{P}[A_\infty]>0$ $\hspace{0.3cm}(\longrightarrow\longleftarrow )$ since $(1)$ implies $\mathbb{P}[A_\infty]=0$. Thus, there exists a sequence with characteristics as mention above such that $\mathbb{P}[E_n] \leq 2^{-n}$.

So, using comparison criterion for series shows that $\sum_{n=1}^\infty \mathbb{P}[E_n] < \infty$, which is all necesary to use Borel - Cantelli lemma to get $\mathbb{P}[\limsup E_n] = 0$. This implies that $0 \leq \frac{|X_n|}{c_n} \leq \frac{1}{n} \longrightarrow 0$ almost everywhere $(\omega \notin \limsup E_n)$ having in mind that $\mathbb{P}\Big[\Big(\limsup E_n\Big)^c\ \Big] = 1$, this complete the proof.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .