Product of i.i.d. random variables uniformly distributed on $(-1,1)$ converges almost surely to $0$

Question

Let $(\Omega,\mathcal{F},P)$ be a probability space and $(X_n)$ i.i.d. random variables with uniform distribution on $(-1,1)$. If $Y_n=X_1\ldots X_n$, prove that $(Y_n)$ converges to $0$ a.s.

Answer 1

By the Borel-Cantelli lemma, if the series $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} $$ converges for each $\varepsilon>0$, $Y_n\to0$ almost surely as $n\to\infty$. Using Chebyshev's inequality, $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} \le\frac1{\varepsilon^2}\sum_{n=1}^\infty\operatorname E|Y_n|^2 $$ since $\operatorname EY_n=0$. By independence, $$ \operatorname E|Y_n|^2=\operatorname E|X_1|^2\ldots\operatorname E|X_n|^2. $$ We have that $$ E|X_1|^2=\frac{(1-(-1))^2}{12}=\frac13. $$ Hence, $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} \le\frac1{\varepsilon^2}\sum_{n=1}^\infty\biggl(\frac13\biggr)^n<\infty $$ for each $\varepsilon>0$ and $Y_n\to0$ almost surely as $n\to\infty$.

Answer 2

Hint: use the Borel—Cantelli lemma.

In more detail: fix any $\varepsilon > 0$, and let $A_n=A_n(\varepsilon)$ be the event $\{ \lvert Y_n\rvert > \varepsilon \}$. By Borel—Cantelli, to show that $Y_n \xrightarrow[n\to\infty]{\rm a.s.} 0$, it is sufficient to show $$ \sum_{n=1}^\infty \mathbb{P} A_n < \infty. $$

In even more detail: (place your mouse over the gray area to reveal its contents)

Since $\lvert Y_n\rvert \leq \varepsilon $ whenever at least one of the $X_i$ satisfies $\lvert X_i\rvert \leq \varepsilon$ (this is a sufficient condition, clearly not necessary), we have $$\mathbb{P} A_n \leq \mathbb{P}\{ \forall\ 1\leq i \leq n,\ \lvert X_i\rvert > \varepsilon \} = \prod_{i=1}^n \mathbb{P}\{ \lvert X_i\rvert > \varepsilon \} = (1-\varepsilon)^n.$$ It follows that $$ \sum_{n=1}^\infty \mathbb{P} A_n \leq \sum_{n=1}^\infty (1-\varepsilon)^n < \infty. $$