Show $X_n \overset{\text{P}} \rightarrow X$ and $X_n \overset{\text{P}} \rightarrow Y$, then $X\overset{\text{a.s.}} \rightarrow Y$

measure-theoryprobability theory

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?

#

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

#

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

Best Answer

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.

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