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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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  • $\begingroup$ Convergence almost surely implies convergence in measure $\endgroup$ – user17762 May 30 '11 at 4:46
  • $\begingroup$ @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. $\endgroup$ – Mike Spivey May 30 '11 at 4:48
  • $\begingroup$ @ Sivaram:The question is to prove the limit is unique. $\endgroup$ – newbie May 30 '11 at 4:56
  • $\begingroup$ @ Mike Spivey: No problem. I put the tag because it serves as a tool for verifying the validity of stochastic integral. $\endgroup$ – newbie May 30 '11 at 5:00
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    $\begingroup$ 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? $\endgroup$ – t.b. May 30 '11 at 5:10
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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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  • $\begingroup$ how did you derive the first one and what does the second one mean(why k>=1 not k>0)? can you explain? 1. $$[|X-Y|\ge 2x]\subseteq[|X_n-X|\geqslant x]\cup[|X_n-Y|\geqslant x],$$ 2. $$ P(|X-Y|\geqslant2x)\leqslant P(|X_n-X|\geqslant x)+P(|X_n-Y|\geqslant x). $$ $\endgroup$ – james black Apr 1 '18 at 8:06
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    $\begingroup$ @jamesblack "(why k>=1 not k>0)?" There is a difference? // Re 1. and 2., please note that 1. implies 2. and that 2. is pure logic: if $|X(\omega)-Y(\omega)|\geqslant2x$ then either $|X_n(\omega)-X(\omega)|\geqslant x$ or $|X_n(\omega)-Y(\omega)|\geqslant x$. $\endgroup$ – Did Apr 2 '18 at 19:50
  • $\begingroup$ i kind of got 1,2 but can you explain the logic a bit more? im sorry but i dont see the reason behind it (like there is four possible combindations of $|Xn-X|\ge x$ and $|Xn-Y|\ge x$ since $|Xn-X|\ge x$ can result in either $Xn-X\ge x$ or $-Xn+X\ge x$; for $$k>=0$$, how do you know the union of 1/k covers all real numbers? because X-Y can be any real number right? or is there any theorem? (I rmb there being some theorem related to ...<1/n, although im not sure) $\endgroup$ – james black Apr 3 '18 at 4:18
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    $\begingroup$ 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 $|w-v|\geqslant x$. $\endgroup$ – Did Apr 3 '18 at 8:02
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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).

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