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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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    "Can we apply this property here?" No, because $g(\cdot)$ would be the identity function, which is not bounded.2012-06-03

2 Answers 2

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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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    Could you please give a bit more explanation?2015-10-10
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    @WittawatJ. About what? Please explain your problem.2015-10-11
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    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?2018-11-18
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    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?2018-11-18
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    @JosephGarvin Of course there is, replace $2^n$ by $7n$ in the example of this answer.2018-11-18
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    I see. Am I right to think $E(\lim_{n\to\infty}X_n) = E(X)$ though? Since two variables with the same CDF must have the same expectation.2018-11-18
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    @JosephGarvin Sorry but what is your assumption? That $X_n\to X$ in distribution or that $X_n\to X$ almost surely? 'Cause nobody assumed the latter while you seem to do...2018-11-18
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    @Did the former. Expectation only depends on the CDF, which if $X_n \to X$ in distribution by definition means $X_n$ in the limit becomes a random variable with an equal CDF, right?2018-11-18
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    @JosephGarvin You say you only assume the former -- but then you introduce $\lim\limits_{n\to\infty}X_n$, which assumes de facto the existence of the pointwise limit. So, no, not right at all, precisely for the reason explained in my previous comment.2018-11-18
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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: $$\sup_n \mathbb{E}[|X_n|^{1+\varepsilon}]<\infty,\quad \text{for some }\varepsilon>0.$$