Our professor for Introduction to Probability 1 course gave us the following question as an exercise:
Let $X_i\sim Uniform[0,1.1]$ iid and let $Y_n=\prod_{i=1}^n X_i$. Does $Y_n\overset{\text{a.s}}{\to}0?$
My attempt:
$ Y_n = \prod_{i=1}^n X_i = e^{\frac{n}{n}\sum_{i=1}^n\ln{X_i}} $ Then we can show $\mathbb{E}[\ln(X_i)]=\int_0^{1.1}\ln{x}\frac{1}{1.1}\ dx \approx -0.904$
By the strong law of large numbers: $\frac{1}{n}\sum_{i=1}^n\ln(X_i)\overset{\text{a.s}}{\to} \mathbb{E}[\ln(X_i)] \approx -0.904 $
$e^x$ is a continuous function, and thus $ Y_n = \prod_{i=1}^n X_i \overset{\text{a.s} } \approx e^{-0.904n} \overset{n\to\infty}{\longrightarrow} 0 $
but I'm more interested in showing this using Borel-Cantelli lemmas using the definition of a.s convergence.
How would one go about this?
Any help would be appreciated,
thanks!
Best Answer
Observe that for any $\epsilon>0,$
$$\sum_{n=1}^\infty P(Y_n>\epsilon)\underbrace{\leq}_{\text{Markov's ineq.}} \sum_{n=1}^\infty \frac{E[Y_n]}{\epsilon}\underbrace{=}_{\text{iid}}\sum_{n=1}^\infty \frac{(E[X_1])^n}{\epsilon}=\sum_{n=1}^\infty \frac{(.55)^n}{\epsilon}=\frac{0.55}{\epsilon(1-0.55) }<\infty,$$
so by the first Borel Cantelli lemma $Y_n\overset{\text{a.s.}}{\rightarrow}0.$