Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
I'm sure you might have guessed ... But I'm editing it, thx. –  Student Jun 13 '12 at 13:09
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

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.