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Suppose we have $X_{n} \overset{p}{\to} X$ Then, $$ \int_{\Omega} \vert X_{n}-X\vert^{2} = \int_{\Omega} \vert X_{n}-X\vert^{2} \mathbb{1}_{\vert X_{n}-X\vert > \epsilon} + \int_{\Omega} \vert X_{n}-X\vert^{2} \mathbb{1}_{\vert X_{n}-X\vert \leq \epsilon} \\ < \int_{\Omega} \vert X_{n}-X\vert^{2} \mathbb{1}_{\vert X_{n}-X\vert > \epsilon}+\epsilon^{2}\int_{\Omega} \mathbb{1}_{\vert X_{n}-X\vert \leq \epsilon} \\ < \int_{\Omega} \vert X_{n}-X\vert^{2} \mathbb{1}_{\vert X_{n}-X\vert > \epsilon} + \epsilon^{2} $$

Does it not then follow that since $\lim_{n} P(\vert X_{n}-X\vert > \epsilon)=0$ that the first integral of the last line converges to 0 in the limit? Since the limit of sets is of probability 0, it follows that the limit of integral is 0 as well?

shem
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    You did not even mention the assumption that $X_n$ have finite second moment. $\int X1_{A_n} \to 0$ if $P(A_n) \to 0$ and $X$ is integeable. You cannot replace $X$ by a squence of r.v.'s. with finite second moment. $X$ has to be independent of $n$. – Kavi Rama Murthy Jul 22 '21 at 05:38
  • Ah yes, I understand now. So if X was independent of n as in say here where X was bounded then it would work. Though could you please explain the relevance of $X_{n}$ having finite second moments? Are you saying that you can bound the $|X_{n} - X|^{2}$ given finite second moments of $X_{n}$? – shem Jul 22 '21 at 05:49
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    $X_{n} \overset{p}{\to} X$ does not imply that $E|X_n-X|^{2}\to 0$ because we may have $E|X_n-X|^{2}=\infty$ for every $n$. Some thing more than just convergence in probabiltiy is needed to say that $E|X_n-X|^{2}\to 0$. – Kavi Rama Murthy Jul 22 '21 at 05:53

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