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If we have two sequence of random variables $X_n$ and $ Y_n$ such that $P(|X_n - Y_n|) > \epsilon \to 0$ for any $ \epsilon > 0$. If $X_n$ converges in distribution to some distribution (for example, normal distribution), does this imply $Y_n$ converges to that distribution as well?

Any help is appreciated.

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Call $F$ the cumulative distribution function of the limit random variable. Let $t$ be a continuity point of this random variable. From the inequality $$\mathbb P\{Y_n\leqslant t\}\leqslant \mathbb P\{X_n\leqslant t+\varepsilon\}+ \mathbb P\{|X_n-Y_n|\geqslant \varepsilon\},$$ we get $\limsup_{n\to\infty}\mathbb P\{Y_n\leqslant t\}\leqslant \limsup_{n\to\infty}\mathbb P\{X_n\leqslant t+\varepsilon\}$.

Since the number of point at which $F$^is discontinuous is at most countable, we can choose a sequence $(\varepsilon_k)_{k\geqslant 1}$ such that $\varepsilon_k\downarrow 0$ and $t+\varepsilon_k$ is a continuity point of $F$. We thus have for each $k$, $$\limsup_{n\to\infty}\mathbb P\{Y_n\leqslant t\}\leqslant F(t+\varepsilon_k),$$ hence $$\limsup_{n\to\infty}\mathbb P\{Y_n\leqslant t\}\leqslant F(t).$$ By a similar argument, we can show that $\liminf_{n\to\infty}\mathbb P\{Y_n\leqslant t\}\geqslant F(t)$.

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  • $\begingroup$ You are welcome. $\endgroup$ – Davide Giraudo Jan 14 '14 at 18:10

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