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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.

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Note that $P(|X_n|>\epsilon)\leq P(X_n^2>\epsilon^2)\leq \frac{1}{\epsilon^2}E[X_n^2\times 1_{\{X_n^2>\epsilon^2\}}]\leq \frac{1}{\epsilon^2}E[X_n^2]\to 0$.

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  • $\begingroup$ 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 $\frac{1}{\epsilon^2}$. $\endgroup$
    – user7802048
    Commented Jul 27, 2019 at 10:18
  • $\begingroup$ 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 $\frac{1}{\epsilon^2} E[X_n^2\times 1_{\{X_n^2>\epsilon^2\}}]\geq \frac{1}{\epsilon^2}E[\epsilon^2\times 1_{\{X_n^2>\epsilon^2\}}]=\frac{1}{\epsilon^2}\times\epsilon^2E[1_{\{X_n^2>\epsilon^2\}}]=P(X_n^2>\epsilon^2)$. $\endgroup$
    – Mark
    Commented Jul 27, 2019 at 10:22
  • $\begingroup$ Thank you so much, it's clear now. $\endgroup$
    – user7802048
    Commented Jul 27, 2019 at 10:53

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