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

Suppose a sequence of probability measures $\mu_n \Rightarrow \mu$ converges weakly to a limit, and suppose moreover that $$\lim_{n \rightarrow \infty} \int x^k \mu_n (dx) = m_k \in \mathbb{R}$$ for some sequence of numbers $\{m_k\}_{k=1}^{\infty}$. Is it true that $$\int x^k \mu (dx) = m_k?$$ I believe so, but I can't seem to prove it, since the functions $x^k$ are unbounded. If anyone could offer any insight, I'd greatly appreciate it!

share|cite|improve this question
Where does the "II" in the title come from? – Michael Greinecker Jan 30 '13 at 8:37
Homework? $ $ $ $ – Did Jan 30 '13 at 9:23
up vote 2 down vote accepted

Here's one way to show it, though it may be possible to do it with less machinery.

By the Skorohod representation theorem, we can assume that the $\mu_n$ are the laws of a sequence of random variables $\{X_n\}$ on some probability space, and the $X_n$ converge almost surely to some $X$ whose law is $\mu$. Now we have to show that if $E[X_n^k] \to m_k$ for all $k$, then $E[X^k] = m_k$.

Let's do $k=1$ first. Since $E[X_n^2] \to m_2$, in particular we have that $\{X_n\}$ is bounded in $L^2$. There is a fact, sometimes called the "crystal ball condition", that if a sequence of random variables is bounded in $L^p$ for some $p > 1$, then it is uniformly integrable. So we have that $\{X_n\}$ is uniformly integrable and converges to $X$ almost surely. By the Vitali convergence theorem, we have $X_n \to X$ in $L^1$, i.e. $E X_n \to EX$. This shows $EX = m_1$.

For general $k$, choose any even integer $r > k$, and set $p=r/k > 1$. Then we have that $E [|X_n^k|^p] = E[X_n^r]$ is bounded, so $\{X_n^k\}$ is bounded in $L^p$. Since $X_n^k \to X^k$ almost surely, as before we have $E[X_n^k] \to E[X^k]$, which is to say $E[X^k] = m_k$.

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.