# Convergence in probability of X/(2^i)

Given that $X_1$, $X_2$, ... $X_i$ are independent, identically distributed random variables with $E[X_i]= 2$ and $var(X_i) = 9$, I'm trying to determine to what value the following sequence converges: $$\frac{X_i}{2^i}$$

I've begun by trying to isolate $X_i$: $$P(X_i \leq -\epsilon2^i) + P(X_i \geq \epsilon2^i)$$

But I'm not sure how to treat this or what information this separation offers, in light of the given the expectation and variance. Can anyone offer some guidance on how to proceed?

Hint: use Chebychev's inequality: $$P(|X_i/2^i| > \epsilon) \le \epsilon^{-2} E[(X_i/2^i)^2] = \frac{1}{2^{2i} \epsilon^2} E[X^2_i]$$
• Thank you. As i approaches infinity, it looks like this expression will go to 0 regardless of the form of X, but that doesn't seem right to me because then all we're really looking at is $1/(2^i)$. – Veronica-2016 May 3 '17 at 16:21
• @Veronica-2016 The fact that $E[X_i^2]$ is bounded over all $i$ is essential for the argument. – angryavian May 3 '17 at 17:06