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If we have a sequence of random variables $X_1,X_2,\ldots,X_n$ converges in distribution to $X$, i.e. $X_n \rightarrow_d X$, then is $$ \lim_{n \to \infty} E(X_n) = E(X) $$ correct?

I know that converge in distribution implies $E(g(X_n)) \to E(g(X))$ when $g$ is a bounded continuous function. Can we apply this property here?

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    $\begingroup$ "Can we apply this property here?" No, because $g(\cdot)$ would be the identity function, which is not bounded. $\endgroup$ Commented Jun 3, 2012 at 15:59

2 Answers 2

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With your assumptions the best you can get is via Fatou's Lemma: $$\mathbb{E}[|X|]\leq \liminf_{n\to\infty}\mathbb{E}[|X_n|]$$ (where you used the continuous mapping theorem to get that $|X_n|\Rightarrow |X|$).

For a "positive" answer to your question: you need the sequence $(X_n)$ to be uniformly integrable: $$\lim_{\alpha\to\infty} \sup_n \int_{|X_n|>\alpha}|X_n|d\mathbb{P}= \lim_{\alpha\to\infty} \sup_n \mathbb{E} [|X_n|1_{|X_n|>\alpha}]=0.$$ Then, one gets that $X$ is integrable and $\lim_{n\to\infty}\mathbb{E}[X_n]=\mathbb{E}[X]$.

As a remark, to get uniform integrability of $(X_n)_n$ it suffices to have for example: $$\sup_n \mathbb{E}[|X_n|^{1+\varepsilon}]<\infty,\quad \text{for some }\varepsilon>0.$$

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  • $\begingroup$ Clarification: my comment doesn't really answer the original question since Fatou is employed -- I've additionally assumed convergence of random variables. The original question only assumes convergence in distribution for which the random variables need not be defined on a common probability space. Thank you, Andrei V. Zorine (Lobachevsky University of Nizhni Novgorod, Russia) for pointing this out. $\endgroup$ Commented Jan 31, 2023 at 17:25
  • $\begingroup$ It looks to me that the "convergence in distribution" doesn't really put an obstacle. There are a sequence of r.v.s $Y_n, 1\leq n\leq +\infty$ satisfying that $Y_n$ and $X_n$ have the same distribution and $\lim_{n\to +\infty} Y_n=Y$ a.s. So you can apply Fatou to $Y_n$. The inequality passes to $X_n$ as $E[|X_n|]=E[|Y_n|]$. $\endgroup$ Commented Nov 9, 2024 at 5:13
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Try $\mathrm P(X_n=2^n)=1/n$, $\mathrm P(X_n=0)=1-1/n$.

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  • $\begingroup$ Could you please give a bit more explanation? $\endgroup$ Commented Oct 10, 2015 at 14:25
  • $\begingroup$ @WittawatJ. About what? Please explain your problem. $\endgroup$ Commented Oct 11, 2015 at 23:57
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    $\begingroup$ So in the limit $X_n$ becomes a point mass at 0, so $\lim_{n\to\infty} E(X_n) = 0$. Then $E(X) = 0$. I don't see a problem? $\endgroup$ Commented Nov 18, 2018 at 2:46
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    $\begingroup$ Answering my own question: $E(X_n) = (1/n)2^n + (1-1/n)0 = (1/n)2^n$. Then taking the limit the numerator clearly grows faster, so the expectation doesn't exist. This begs the question though if there is example where it does exist but still isn't equal? $\endgroup$ Commented Nov 18, 2018 at 2:53
  • $\begingroup$ @JosephGarvin Of course there is, replace $2^n$ by $7n$ in the example of this answer. $\endgroup$ Commented Nov 18, 2018 at 10:13

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