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Let $X^n\sim Y^n$ for all $n$ and $X^n\to X$ and $Y^n \to Y$ both in probability. Is $X\sim Y$?

If all variables take values in some measurable space $(S,\mathbb S)$: I'm thinking $P(X\in A)=\lim \, P(X^n\in A) = \lim \, P(Y^n\in A) = P(Y \in A) $ for $A\in \mathbb S$. But suddenly i had doubts that convergence in probability implies that $\lim \, P(X^n\in A)=P(X\in A)$. Is the statement true? Is my argument?


Edit: Sorry. I was a bit fast there. Let's assume they take value in some metric space.

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@StefanHansen: what is the connection between $X$ and $Y$ in your comment? – Ilya Apr 4 '13 at 11:48
@Ilya: Was too fast, misread it :) – Stefan Hansen Apr 4 '13 at 11:49
up vote 3 down vote accepted

Indeed convergence in distribution concerns... well, distributions.

More precisely, call $p_n$ the distribution of $X_n$, $q_n$ the distribution of $Y_n$, $p$ the distribution of $X$ and $q$ the distribution of $Y$. I one weakens your hypotheses in the sense that one replaces the convergence in probability of $(X_n)$ and $(Y_n)$ by their convergence in distribution, one arrives at the following:

$p_n=q_n$ for every $n$, $p_n\to p$, $q_n\to q$.

The conclusion that $p=q$ is direct.

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Your answer made my realize it right away. Thanks! – Henrik Apr 4 '13 at 12:00

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