I'm self-studying probability, and I'm unable to solve this problem:
You have a sequence of (possibly dependent) random variables $\{X_n\}$ defined on the probability space $(\Omega,\mathcal{F},P)$, and you are given that $$ EX_n^2 \leq 1 \quad \text{ for all } n = 1,2,…$$
Show that if the $X_n$ all have the same distribution, then $X_n/\sqrt{n}$ converges to zero with probability one. Hint: You will need the First Borel Cantelli Lemma.
My instinct was to use Chebyshev's inequality, i.e. $$a^2 P(|X_n| \geq a) \leq EX_n^2$$
For each $n$, we can set $a = n$ and use the fact that $EX_n^2 \leq 1$ to get $$P(|X_n| \geq n) \leq \frac{1}{n^2}$$
I really don't know how to continue here. I got to the $1/n^2$ in hopes that the application of Borel Cantelli will jump out at me, but it isn't (if I apply it now, it just tells me that the $X_n$ are bounded a.s.). I'm not sure where the equidistributed hypothesis comes in, also. Any help would be greatly appreciated.
Best Answer
Hint: Applying the inequality $$\sum_{n \geq 1} \mathbb{P}(|Y| \geq n) \leq \mathbb{E}(|Y|)$$ for $Y := \epsilon^{-1} X_1^2$ yields $$\sum_{n \geq 1} \mathbb{P}\left(|X_1|^2 \geq n \epsilon \right)<\infty$$ for any $\epsilon>0$.
Now use the Borel-Cantelli theorem and the fact that $$\mathbb{P}(|X_1|^2 \geq n \epsilon) = \mathbb{P}\left(|X_n|^2 \geq n \epsilon \right).$$