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Let $\mathbb EX$ exist and $X_n$ converges in probability to $X$. How to prove that $\mathbb EX_n → \mathbb EX$ then and only when $\mathbb E| X_n - X | → 0$?

Lisa
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  • If $X_n \to X$ almost surely, then this follows from Scheffe's lemma, see e.g. this question. To get the result for convergence in probability you can use a similiar reasoning as here. – saz Mar 10 '19 at 11:50
  • @saz Scheffe's Lemma does not apply here. In the post you have quoted it is given that $E|X_n| \to E|X|$. – Kavi Rama Murthy Mar 10 '19 at 11:56
  • @KaviRamaMurthy Ah right, I missed that the OP is not assuming convergence of $L^1$ norms, thanks. – saz Mar 10 '19 at 12:04

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This is not true. Consider $[-1,1]$ with the measure $P(A)=\frac {\lambda(A)} 2$ where $\lambda$ is Lebesgue measure. Let $X_n(x)=n^{2} xI_{(-\frac 1 n, \frac 1 n)}(x)$, and $X\equiv0$. Then $EX_n =EX=0$ for all $n$ and $X_n \to X$ almost surely, hence in probability. But $E|X_n|=1$ for all $n$.

Mars Plastic
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Kavi Rama Murthy
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