Let $g_n$ be given by
$$g_n=\prod_{i=1}^n \left(1+\frac{X_i}{\sqrt{i}}\right)$$
where $X_i$'s are independent random variables with $P(X_i=1)=P(X_i=-1)=0.5$.Then is it true that $P(g_n \rightarrow \infty)=0$
.If yes how can I show it?
My attempt:I think I have to use Borel Cantelli lemma but I am not sure how.
Best Answer
In fact, a much stronger result holds: $P[g_n \to 0]=1$. This is an application of Kakutani's theorem for product Martingales, see Probability with Martingales by David Williams, Section 14.12.
Back to your problem, let $Y_i:= 1+\frac{X_i}{\sqrt{i}}$, and $a_i:=\mathbb{E}\left[\sqrt{Y_i}\right]$. Based on Kakutani's theorem, to show that $P[g_n \to 0]=1$, it suffices to prove that $\sum_{i=1}^\infty(1-a_i)=\infty$. I will now show this.
With an easy calculation, $a_i=\frac{1}{2}\sqrt{1+\frac{1}{\sqrt{i}}}+\frac{1}{2}\sqrt{1-\frac{1}{\sqrt{i}}}$. I will show that $a_i=1-\Theta\left(\frac{1}{i}\right)$, which finishes the proof.
I claim that for any $x,y\in [0,2]$, and for some absolute constant $c>0$, we have $\frac{1}{2}\sqrt{x}+\frac{1}{2}\sqrt{y}+c(x-y)^2\leq \sqrt{\frac{x+y}{2}}$. This is because the function $f(x)=\sqrt{x}$ is strongly-concave inside $[0,2]$. Setting $x=1+\frac{1}{\sqrt{i}}$ and $y=1-\frac{1}{\sqrt{i}}$, we are done.