# Does convergence in distribution implies convergence of expectation?

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?

• "Can we apply this property here?" No, because $g(\cdot)$ would be the identity function, which is not bounded. – leonbloy Jun 3 '12 at 15:59

Try $\mathrm P(X_n=2^n)=1/n$, $\mathrm P(X_n=0)=1-1/n$.

• Could you please give a bit more explanation? – wij Oct 10 '15 at 14:25
• 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? – Joseph Garvin Nov 18 '18 at 2:46
• 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? – Joseph Garvin Nov 18 '18 at 2:53
• @JosephGarvin Of course there is, replace $2^n$ by $7n$ in the example of this answer. – Did Nov 18 '18 at 10:13
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: $$\sup_n \mathbb{E}[|X_n|^{1+\varepsilon}]<\infty,\quad \text{for some }\varepsilon>0.$$