Making a sequence of random variables converge almost surely to $0$

Prove: for every sequence of random variables $X_1,X_2,X_3,\dots$ there exists a sequence $a_1,a_2,a_3,\dots$ of real numbers ($a_i \neq 0$) such that the sequence $\frac{X_1}{a_1},\frac{X_2}{a_2},\frac{X_3}{a_3},\dots$ converges almost surely to the constant $0$. Note: $X_n$ doesn't neccisarily have finite expectation.

What I tried: I wanted to use Borel Cantelli lemma, proving that: $$\sum_{i=0}^n P(|\frac{X_n}{a_n}|\geq \epsilon) <\infty \; \;, \; \forall \epsilon >0$$

Then I got: $$P(|\frac{X_n}{a_n}| \geq \epsilon)=1-F_{X_n}(a_n \epsilon)+F_{X_n}(-a_n \epsilon)$$

When $F_{X_n}$ is the cumulatie distribution function of $X_n$.

My ideas was to show that $1-F_{X_n}(a_n \epsilon)+F_{X_n}(-a_n \epsilon) < \frac{C}{n^2}$ for some constant, so the series will converge. To show that I tried using the fact that $\lim \limits_{t \to \infty} F_{X_n}(t) = 1, \lim \limits_{t \to -\infty}F_{X_n}(t)=0$, that is choosing big enough $a_n$ such that $F_{X_n}(-a_n \epsilon)< \frac{1}{n^2} \;$ $, \; F_{X_n}(a_n \epsilon) > 1- \frac{1}{n^2}$.

My problem is that the sequence I get is dependant with $\epsilon$, which is not good (I want one sequence for all $\epsilon$).

Is there a way to fix my proof ? or another way to solve? Thanks.

You understood perfectly the problem with your construction, but luckily there is a way to solve the dependence on $\varepsilon.$ Indeed you can choose $a_n$ so that the sequence diverges and this does the trick. For example choose $b_n$ such that: $$\mathbb{P}(|X_n| \ge b_n) \le \frac{1}{2^n}$$ (such a sequence can always be chosen). Then choose through an iteration $a_0 = 1$ and $$a_n = b_n^2 \vee (a_{n-1}+1)$$ and notice that in this way $a_n \to {+}\infty.$ Hence if we go back to the original sum we can estimate: $$\sum_{n} P \Big(\Big|\frac{X_n}{a_n}\Big|\geq \varepsilon \Big) \le \sum_{n} P \Big(\Big|\frac{X_n}{b_n} \Big|\geq \varepsilon \sqrt{a_n} \Big)< {+} \infty$$ since for any $\varepsilon > 0$ there exists an $n_0 \ge \varepsilon$ such that $\sqrt{a_n} \varepsilon \ge 1$ for all $n \ge n_0$ and thus we can simply sum up the geometric series after $n_0$.
• I'm not sure I understand how $a_n$ is constructed, how do we choose if $a_n = b_n^2$ or $a_n = a_{n-1} + 1$ ? and why $P(| \frac{X_n}{a_n}| \geq \epsilon ) \leq P(| \frac{X_n}{b_n}| \geq \epsilon \sqrt a_n )$ ? – user401516 Jan 4 '18 at 20:33
• Sorry, the symbol $\vee$ might be ambiguous in general. In this case I do not mean the logical symbol or, but the $\max$ function: $$a \vee b : = \max\{a,b\}$$ – Kore-N Jan 4 '18 at 21:27