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Let $(X_n)$ a sequence of r.v. s.t. $$\mathbb EX_n^2\longrightarrow \mathbb EX^2.$$

I know that my question looks obvious, but do we have that $X_n\to X$ in $L^2$ ? I would think that yes, but the only think I can have is $$\mathbb |\mathbb EX_n^2-\mathbb EX^2|\leq \mathbb E[(X_n-X)^2]\leq \mathbb EX_n^2+\mathbb EX^2.$$ So if $X_n\to X$ in $L^2$, then $\mathbb E X_n^2\to \mathbb EX$, but I can't get the converse. May be it's wrong ?

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    The converse is not true. This would be similar to saying that two distributions with the same expectation have the same distribution. – 2017-01-14
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    @MichaelBurr: Thank you. And if we would have that $X_n\to X$ a.s., would it be fine ? – 2017-01-14
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    In this case, yes. This is the Vitali convergence theorem. – 2017-01-14

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The converse is not true. For a counter-example, consider $X_n=1, \forall n$, and $X=-1$, such that

$$EX_n^2= EX^2, \:\:\:\:\: E(X_n-X)^2=4.$$