Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
"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
up vote 12 down vote accepted

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

share|cite|improve this answer
Could you please give a bit more explanation? – wij Oct 10 '15 at 14:25
@WittawatJ. About what? Please explain your problem. – Did Oct 11 '15 at 23:57

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.