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I'm looking for a counter example of the following statement. The fact that $X_n$ and $X$ are random variables such that $X_n \stackrel{d}{\to} X$ in distribution does not ensure $\mathbb{E}(X_n) \to \mathbb{E}(X)$. But I cannot find one.

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Try $\mathrm P(X_n=n)=1-\mathrm P(X_n=0)=1/n$.

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    +1: to make it even clearer, $\Pr(X_n=0)=\frac{n-1}{n}$ so the convergence in distribution is to $Pr(X=0)=1$ wheich has $E[X]=0$2012-05-07