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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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    @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: $\sup_n \mathbb{E}[|X_n|^{1+\varepsilon}]<\infty,\quad \text{for some }\varepsilon>0.$