# limit in probability is almost surely unique?

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.

Cheers.

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Convergence almost surely implies convergence in measure –  user17762 May 30 '11 at 4:46
@newbie: I didn't see what your question had to do with stochastic integrals, so I removed that tag. If this was a mistake, feel free to add it back. –  Mike Spivey May 30 '11 at 4:48
@ Sivaram:The question is to prove the limit is unique. –  newbie May 30 '11 at 4:56
@ Mike Spivey: No problem. I put the tag because it serves as a tool for verifying the validity of stochastic integral. –  newbie May 30 '11 at 5:00
Hi newbie! Unfortunately, adding a space after the @ sign leads to the users not being notified. Thus @Sivaram and Mike Spivey didn't see your comments. Moreover, only one user per comment gets notified, that's why I didn't add an @ before Mike's name. Concerning your mathematical question: Do you know the following fact: "if a sequence converges in probability then there is a subsequence converging almost surely" or are you asking how to prove that? –  t.b. May 30 '11 at 5:10

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, $$[|X-Y|\ge 2x]\subseteq[|X_n-X|\geqslant x]\cup[|X_n-Y|\geqslant x],$$ 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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Of course, this is much nicer! Sometimes I really wonder what I'm thinking. Thanks for setting this straight. –  t.b. May 30 '11 at 15:15
@Theo: Thanks.  –  Did May 30 '11 at 19:52

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).

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