# Sequence of random variable with $\lim_{n \rightarrow \infty}\mathbb{E}[X_n^2] = 0$

Let $$X_n, n \in \mathbb{N}$$ be a sequence of random variables with $$\lim_{n \rightarrow \infty}\mathbb{E}[X_n^2] = 0$$. Show that $$X_N \rightarrow_d X$$ with $$X = 0$$ almost surely, as $$n \rightarrow \infty$$.

I was thinking about showing $$P$$-convergence. This would already imply convergence in distribution. I already noticed $$-\mathbb{E}[X^2] \leq \mathbb{E}[X] \leq \mathbb{E}[X^2]$$. Now I need to show for every $$\epsilon > 0$$:

$$P[|X_n - 0| > \epsilon] \rightarrow 0 \text{ as } n \rightarrow \infty$$

and sure I can somehow show this using the fact above. Thank you.

Note that $$P(|X_n|>\epsilon)\leq P(X_n^2>\epsilon^2)\leq \frac{1}{\epsilon^2}E[X_n^2\times 1_{\{X_n^2>\epsilon^2\}}]\leq \frac{1}{\epsilon^2}E[X_n^2]\to 0$$.
• Can you say a little bit about the second inequality? I know that the expectation of an indicator function is the same as the probability of the set, but I don't see the factor $\frac{1}{\epsilon^2}$. Commented Jul 27, 2019 at 10:18
• Note that $X_n^2\times 1_{\{X_n^2>\epsilon^2\}}\geq \epsilon^2\times 1_{\{X_n^2>\epsilon^2\}}$ pointwise. So now by monotonicity of expectation we get $\frac{1}{\epsilon^2} E[X_n^2\times 1_{\{X_n^2>\epsilon^2\}}]\geq \frac{1}{\epsilon^2}E[\epsilon^2\times 1_{\{X_n^2>\epsilon^2\}}]=\frac{1}{\epsilon^2}\times\epsilon^2E[1_{\{X_n^2>\epsilon^2\}}]=P(X_n^2>\epsilon^2)$.