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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): http://mathoverflow.net/questions/94140/convergence-of-infinite-moments2012-04-15
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    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.2012-04-15
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    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.2012-04-15
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    **Counterexample**: Let $Y \geq 0$ such that $\mathbb E Y = \infty$. Define $X_n = n^{-1} Y$.2012-04-15
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    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! :)2012-04-15
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    Just to provide some context, I have $X_n \Rightarrow Z$, where all r.v.'s are non-degenerate (there is a density on an interval). Thus, $e^{X_n}\Rightarrow e^Z$ (continuous mapping thm). However, while all moments of $X_n$ & $Z$ are finite, $\mathbb{E}[e^{X_n}] = \infty$. I was hoping $\mathbb{E}[e^Z] = \infty$ too. Note that $e^Z$ is non-degenerate, and so it's a different mold from the counterexamples provided. If someone has a non-degenerate counterexample, it could give me an idea what to avoid so I could still perhaps hope to get my desired result. Thanks!2012-04-16
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    Two thoughts, which are a bit unlikely to be helpful (the information provided is still a little sparse). Define $T_n = \inf_{k \geq n} X_k$. Then, of course, if for some $n$ you can show that $\mathbb E e^{T_n} = \infty$, you'd be done. Second, it's not clear what you mean by "all moments of $X_n$ and $Z$ are finite". If by that you only meant that "for all $n$, $\mathbb E |X_n| < \infty$ and the same for $Z$", then if you can show that $\mathbb E X_n^k = \infty$ for some $k \in \mathbb N$ and all sufficiently large $n$, you'd also be done. :)2012-04-17
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    It is almost as easy to construct counterexamples that converge to any nondegenerate limit you wish. Let $Y \geq 0$ with $\mathbb E Y = \infty$ as before and let $Z$ have an *arbitrary* distribution. Then, take $X_n = (1-n^{-1}) Z + n^{-1} Y$.2012-04-17

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