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Consider two sequences of random variables $X_n, Y_n$ and suppose $X_n\to X$ in distribution. Does the following hold: $\lim_{n\to\infty} E[|X_n-Y_n|]=0 \implies Y_n\to X$ in distribution?

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What do you mean by convergence in distribution? maybe you meant in probability? –  Mercy King Jun 22 '12 at 12:11
@Mercy… –  Byron Schmuland Jun 22 '12 at 12:26

1 Answer 1

Let $Z_n = Y_n - X_n$. From the assumption, we have $Y_n = X_n + Z_n$, where $X_n \to X$ in distribution and $Z_n \to 0$ in $L^1$.

Then $Z_n \to 0$ in probability, and therefore by Slutsky's theorem we have the desired conclusion.

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