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Assume that for sequence $X_n \in L^2(\Omega,F,P)$ which converges in distribution to CDF $F_X$ ($F_n(t)\rightarrow F_X(t)$ for every point of continuity of $F_X$), we have also that $X_n$ converges weakly to $Z$ in $L^2(\Omega,F,P)$ $(\mathbb{E}X_nY \rightarrow \mathbb{E}ZY$ for every $Y \in L^2(\Omega,F,P))$. Does it mean that CDF of $Z$ is equal to $F_X$?

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  • $\begingroup$ You're using $F$ for both the $\sigma$-algebra and the CDF. Might wanna change that. $\endgroup$
    – Michael
    Jun 2 '13 at 16:42
  • $\begingroup$ @Michael : I once took a course in which a professor used the letter j for two different things in different parts of a long proof. When I mentioned it, she said "Pretend one of them is a Greek j." $\endgroup$ Jun 2 '13 at 16:48
  • $\begingroup$ Michael Hardy, I trust you don't do that to your students. $\endgroup$
    – Michael
    Jun 3 '13 at 19:00
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That is not true. Take $\{ X_n \}$ to be a white-noise process that is i.i.d. normal with mean $0$ and variance $1$. Then the sequence converges in distribution trivially. But $\{X_n \}$ is an orthonormal sequence in $L^2$. So it converges to $0 \in L^2$ weakly. The CDF of $0$ is evidently not the normal distribution.

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