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I read this proposition in a book, which was not proved. And I cannot verify it myself. Could anyone help me out here?

If $X_{n}\rightarrow X$ in probability and $X_{n}\rightarrow Y$ almost surely, then $P(X=Y)=1.$

An alternative version is that the p-limit of a sequence is almost surely unique.

Thanks for your time.

Cheers.

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    @Theo Buehler: Yes. I do know. So can it be proved in the following way: take a subsequence which is also convergent almost surely, and since it is also a subsequence of $X_{n}\rightarrow Y$ almost surely, and the limit is unique in the sense of almost sure?2011-05-30

3 Answers 3

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Back to basics: Assume that $X_n\to X$ in probability and that $X_n\to Y$ in probability. Then, for every positive $x$, $P(|X_n-X|\geqslant x)+P(|X_n-Y|\geqslant x)$ converges to zero since both terms do. Now, $[|X-Y|\geqslant 2x]\subseteq[|X_n-X|\geqslant x]\cup[|X_n-Y|\geqslant x],$ hence, for every $n$, $ P(|X-Y|\geqslant2x)\leqslant P(|X_n-X|\geqslant x)+P(|X_n-Y|\geqslant x). $ Considering the limit of the RHS when $n\to+\infty$, this proves that $P(|X-Y|\geqslant2x)=0$. This holds for every positive $x$ and $[X\ne Y]=\bigcup_{k\geqslant1}[|X-Y|\geqslant k^{-1}],$ hence $P(X\ne Y)=0$. This means that $X=Y$ almost surely.

Note: The hypothesis that $X_n\to X$ in probability and $X_n\to Y$ in probability, which we used above, is weaker than the hypothesis that $X_n\to X$ in probability and $X_n\to Y$ almost surely.

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    These are two basic facts from real analysis. First, $u\ne v$ if and only if $|u-v|\ne0$ if and only if there exists some \epsilon>0 such that $|u-v|\geqslant\epsilon$ if and only if there exists a positive integer $k$ such that $|u-v|\geqslant1/k$. Likewise, by the triangular inequality, for every $(u,v,w)$, if $|u-v|\geqslant2x$ then $|w-u|\geqslant x$ or $|w-v|\geqslant x$.2018-04-03
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For the sake of having an answer:

We know the following fact:

If $X_n \to Y$ in probability then there is a subsequence $X_{n_k} \to Y$ almost surely.

So take such a subsequence. As $X_{n} \to X$ a.s. we also have $X_{n_k} \to X$ a.s. and thus $X = Y$ a.s. because the almost sure limit of a sequence is unique a.e. (if it exists).

This is just fleshing out your last comment a bit more formally.

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See, for example, p. 150 in the book An Intermediate Course in Probability by Allan Gut‏ (in particular, the proof of Theorem 2.1(ii) on p. 151).