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How to show $X_n \overset{\text{P}} \rightarrow X$ and $X_n \overset{\text{P}} \rightarrow Y$, then $X\overset{\text{a.s.}} \rightarrow Y$?

I try to show it below. Though I can not make sure. Can anyone give some suggestions?

$0\leq P(X\neq Y)\leq P(|(X_n-X)-|X_n-Y||>\varepsilon)\leq P(|(X_n-X)|>\varepsilon/2)+P(|(Y_n-X)|>\varepsilon/2) $

Thus $0\leq P(X\neq Y)\leq \lim_{n\rightarrow\infty}P(|(X_n-X)-|Y_n-Y||>\varepsilon)\leq \lim_{n\rightarrow\infty}P(|(X_n-X)|>\varepsilon/2)+P(|(Y_n-X)|>\varepsilon/2) $

Thus $P(X\neq Y)=0$ (since $\lim_{n\rightarrow\infty}P(X\neq Y)=P(X\neq Y)=0$

Thus $X \overset{\text{a.s.}} \rightarrow Y$

Also an question: Why not equal? Can anyone give a counterexample?

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for arbitrary $\varepsilon>0$ Make some change: Thus $0\leq P(|X-Y|>\varepsilon)=P(|(X_n-X)-|Y_n-Y||>\varepsilon)\leq \lim_{n\rightarrow\infty}P(|(X_n-X)|>\varepsilon/2)+P(|(Y_n-X)|>\varepsilon/2) $

Thus $P(|X-Y|>\varepsilon)=0$ (since $\lim_{n\rightarrow\infty}P(|X-Y| > \varepsilon)) = P(|X-Y|>\varepsilon))=0$

Thus $\sum_{n=1}^{+\infty} P(|X-Y|>\varepsilon)<\infty$(?)

According to BC Lemma, we have $P(|X-Y|>\varepsilon, \text{ i.o.})=0$

Thus $X \overset{\text{a.s.}} \rightarrow Y$.

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Make some change: for arbitrary $\varepsilon>0$ $0\leq P(|X-Y|>\varepsilon)=P(|(X_n-X)-|Y_n-Y||>\varepsilon)\leq \lim_{n\rightarrow\infty}P(|(X_n-X)|>\varepsilon/2)+P(|(Y_n-X)|>\varepsilon/2) $

Thus $P(|X-Y|>\varepsilon)=0$ (since $\lim_{n\rightarrow\infty}P(|X-Y|>\varepsilon))=P(|X-Y|>\varepsilon))=0$

Thus $\sum_{n=1}^{+\infty} P(|X-Y|>\varepsilon)<\infty$(?)

According to BC Lemma, we have $P(|X-Y|>\varepsilon, \text{ i.o.})=0$

Thus $X\overset{\text{a.s.}}\rightarrow Y$.

Michael Hardy
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Olivia
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    What does $X\to^{a.s.} Y$ mean here if $X$ is not a sequence? And do you mean $X_n$ everywhere you have written $Y_n$? – kccu Dec 11 '19 at 20:00
  • I do now know. It is from a note of the advanced course which I download online. I try to learn by myself. I thought it means $P(X=Y)=1$. Can $P(X=Y)=1$ mean these two convergence almost surely? Maybe he wants to show $Z_n=X$ and $W_n=Y$ for all n. – Olivia Dec 11 '19 at 20:03

1 Answers1

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The correct conclusion would be "if $X_n \overset p \to X$ and $X_n \overset p \to Y$ then $P(X=Y)=1$ (i.e., $X=Y$ almost surely)." In other words, limits in probability are almost surely unique.

The proof is more or less what you wrote, except the $Y_n$'s should be $X_n$'s (there is no sequence $Y_n$ in this result). Also, what you showed is that $P(X\neq Y) \leq \text{(something going to zero with $n$)}$, but you still need to justify why this implies $P(X \neq Y) = 0$. You can read the whole proof along with that justification here.

Michael Hardy
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kccu
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  • $Y_n $ is a typing mistake. The second suggestion is very helpful! – Olivia Dec 11 '19 at 20:12
  • I have updated it. Though I still can not make sure. – Olivia Dec 11 '19 at 20:34
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    Borel Cantelli applies to sequences of random variables. It doesn't make sense to say "$|X-Y|>\epsilon$ infinitely often". Instead use the fact that $X \neq Y$ implies there exists a positive integer $n$ such that $|X-Y| >\frac{1}{n}$. Therefore $P(X \neq Y) \leq \sum_{n=1}^\infty P(|X-Y| > \frac{1}{n})=0$. – kccu Dec 11 '19 at 20:50
  • Then use the proof as the link show. – Olivia Dec 11 '19 at 20:51