If $X_n \xrightarrow{d} X$ what are the minimal hypothesis to have $E[X_n]\rightarrow E[X]$ ? For example I think that if all the second moments are bounded it's true, but I'm not sure if is true if the first moments are bounded.
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2It's true when ${X_n\mid n\geq 1}$ is uniformly integrable, see this. In particular it holds when the second moments are bounded. – Stefan Hansen Mar 21 '14 at 18:14
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But not when only the first moments are bounded. – Did Mar 21 '14 at 18:15
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Why second moments bounded $\rightarrow$ uniformly integrable ? – user136725 Mar 21 '14 at 18:25
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@Febo This follows from Markov's inequality: $$\int_{|X_n|>R} |X_n| , d\mathbb{P} \leq \frac{1}{R} \int X_n^2 , d\mathbb{P}.$$ – saz Mar 22 '14 at 07:24
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If $(X_n)_{n\in\mathbb{N}}$ is uniformly integrable and $X_n\xrightarrow{d}X$ then $E[X_n]\rightarrow E[X]$ ( See this ).
Now if $\exists\ M>0\ $ and $\ \exists\ p>0$ s.t. $E[|X_n|^p]<M$, $\ \ (X_n)_{n\in\mathbb{N}}$ is uniformly integrable by Markov inequality: $$R^{p-1}\int_{|X_n|>R} |X_n| \, d\mathbb{P} \leq \int |X_n|^p \, d\mathbb{P}\leq M.$$
If the first moment is bounded this is not true, in fact if you take $X_n:=n\mathbb{1}_{[0,\frac{1}{n}]}$, $X:=0$ in $((0,1),\mathcal B, \mathbb P)$, with $\mathbb P$ the Lebesgue measure, you have a counterexample. The first moments are constantly $1$, but the expectations don't converge to $0$.
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