Let I have a sequence of random variables $X_n$ that converges to random variable $X$ almost surely as $n\to\infty$. How can I proof that $\lim_{n\to\infty}\mathcal{E}[X_n]=\mathcal{E}[X]$ where $\mathcal{E}[\cdot]$ stands for expected value?

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    $\begingroup$ You cannot, because as it stands it is generally not true. $\endgroup$ – Siméon Sep 1 '14 at 10:01
  • $\begingroup$ Really? Why do you think so? $\endgroup$ – Karel Macek Sep 1 '14 at 10:02
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    $\begingroup$ Take $X_n = n^2$ with probability $\frac{1}{n^2}$ and $X_n = 0$ otherwise. The Borel-Cantelli lemma shows that $X_n \to 0$ almost surely. But the expectation of $X_n$ equals $1$ for all $n$. $\endgroup$ – Siméon Sep 1 '14 at 10:08
  • $\begingroup$ Great, Siméon, do you have any idea of assumptions that would make it true? $\endgroup$ – Karel Macek Sep 1 '14 at 10:13
  • $\begingroup$ Monotone converge theorem, dominated convergence theorem, etc. $\endgroup$ – Siméon Sep 1 '14 at 10:14

In general, as pointed out by Siméon, it is not true that $\mathbb E[X_n]\to\mathbb E[X]$.

However, if the family $\{X_n,n\geqslant 1\}$ is uniformly integrable, it is OK, using a truncation and a $2\varepsilon$-argument.

If the $X_n$'s are non-negative the converse holds, namely, if $\mathbb E[X_n]\to\mathbb E[X]$ and $X_n\to X$ almost surely, then the family $\{X_n,n\geqslant 1\}$ is uniformly integrable.


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