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If a sequence of random variables $X_n$ converges in distribution to some r.v. $X$, the convergence of moments doesn't immediately follow. However, if the sequence is uniformly integrable, then we have the convergence of moments.

Thus, for example, if $X_n\Rightarrow X$ and $\sup \mathbb{E}[|X_n|^{1+\varepsilon}]<\infty$ for some $\varepsilon >0$ (a sufficient condition for uniform integrability), then $\mathbb{E}[|X|]<\infty$ and $\mathbb{E}[X_n]→\mathbb{E}[X]$. (See for example Theorem 25.12 and Corollary in Billingsley's Probability and Measure).

My situation however is this: $X_n\Rightarrow X$, and $\mathbb{E}[X_n]=\infty$.

QUESTION: Does it follow that $\mathbb{E}[X]=\infty$ too?

Let me add that all the $X_n$ and $X$ are nonnegative so their moments are defined ($\mathbb{R}_+ \cup \infty$).

The moment convergence results I've seen all invoke uniform integrability and finiteness of moments, which doesn't apply here. Is it even possible to have $\mathbb{E}[X]<\infty$ (a counterexample would be instructive)? Or might anyone be able to suggest other additional conditions so that $\mathbb{E}[X]=\infty$?

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Crossposted to MO (please don't do this): mathoverflow.net/questions/94140/… –  cardinal Apr 15 '12 at 20:21
    
For example you can have that $X=0$ a.s.. Try to construct a counterexample: start with a r.v. that is not integrable, and use the indicator function. –  Kolmo Apr 15 '12 at 20:22
    
Re: crossposting Thanks for the advice. Rest assured I didn't intend to violate written and unwritten rules of etiquette. I didn't see anything against crossposting in the FAQ so I didn't think much of it when I posted in both mathoverflow and stackexchange; I just wanted to reach a wider audience. I'll take this into account next time. –  Chuck Eden Apr 15 '12 at 20:26
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Counterexample: Let $Y \geq 0$ such that $\mathbb E Y = \infty$. Define $X_n = n^{-1} Y$. –  cardinal Apr 15 '12 at 20:26
    
No worries, Richard (or Charles?). Quasi-official policy on this is on the meta site, but it's hard to expect new users to know this. Consider the previous comments as a courtesy; welcome to the site! :) –  cardinal Apr 15 '12 at 20:30

1 Answer 1

Let $X_n$ of density $f_n(x):=\frac{2n}{\pi(1+nx^2)}\chi_{(0,+\infty)}$; if $g$ is a continuous bounded function then $$\int_0^{+\infty}(f_n(x)g(x)-f_n(x)g(0))dx\leq 2\int_0^{+\infty}\frac{|g(tn^{-1})-g(0)|}{\pi(1+t^2)}dt$$ and by the dominated convergence theorem it converges to $0$. So $X_n$ converges in distribution to $\delta_0$ which have moments of each orders.

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I don't think it needs to be this complicated. Any sequence that converges a.s. to zero but has infinite mean for all $n$ works. –  cardinal Apr 15 '12 at 20:28
    
@cardinal Indeed (at least, if the OP adds "continuous" we have an example). –  Davide Giraudo Apr 15 '12 at 20:30
    
Ah wonderful! The counterexamples are instructive, and cardinal's was really simple too. –  Chuck Eden Apr 16 '12 at 22:29

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