When weak convergence implies moment convergence?

Question

Given a sequence $(\mu_n)_n$ of probability measures on $\mathbb R$, which converges weakly to a probability measure $\mu$, when do we have $$ \tag{1} \lim_{n}\int x^kd\mu_n(x)=\int x^k d\mu(x) \qquad \forall k\geq 0\;? $$ Is "$\mu$ has compact support" a sufficient condition?

Note that $\mu_n$ converges to $\mu$ weakly if $$ \int \varphi d\mu_n \to \int \varphi d\mu$$ for all $\varphi$ which is continuous and has compact support. Note that $x^k$ are continuous but not of compact support, so (1) is not immediately obvious.

Answer 1

A condition on the limit measure will never be enough. The sequence $\left(1-{1\over n}\right)\delta_0+{1\over n}\delta_{x(n)}$ converges to $\delta_0$ weakly, but we can make its moments behave horribly by choosing $x(n)$ to be very large.

A sufficient condition for your moments to converge is if all the $\mu_n$s have the same compact support.

Answer 2

This is almost 10 years late. Possibly by know you are Student no longer. In any event, if $\mu_n\stackrel{n}{\Longrightarrow\infty}\mu$, a sufficient condition for $E_{\mu_n}[X]\rightarrow E_\mu[X]$ is that $$\lim_{a\rightarrow\infty}\sup_n\int_{\{|x|>a\}}|x|\,\mu_n(dx)=0$$ This is equivalent to say that a sequence of random variable $X_n\sim \mu_n$, defined on a common domain $(\Omega,\mathscr{F},\mathbb{P})$ is uniformly integrable. This can be seen for instance in Billingsley, P. Convergence in Probability Measures, John Wiley & Sons, 1968, pp. 32.

From this result, is not difficult to derived sufficient conditions for $E_{\mu_n}[X^r]\xrightarrow{n\rightarrow\infty}E_{\mu}[X^r]$ (again, uniform integrability is the key).

In the counterexample given by the responder (user940), $\mu_n=(1-\tfrac1n)\delta_0+\frac1n\delta_{x_n}$ with $x_n\xrightarrow{n\rightarrow\infty}\infty$ Notice that given $a>0$, for all $n$ large enough $x_n>a$ and so, $$\int\mathbb{1}_{\{|x|>a\}}|x|\mu_n(dx)=\frac{1}{n}x_n$$ If $x_n\geq n$ for all $n$, then $$\sup_n\int\mathbb{1}_{\{|x|>a\}}|x|\mu_n(dx)\geq1$$ This means that in this case, the measures do not satisfy the uniformintegrability condition.