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Suppose $X_n  {\buildrel p \over \rightarrow} X$ and $X_n \le Z,\forall n \in \mathbb{N}$. Show $X \le Z$ almost surely.

I've try the following, but I didn't succeed.

By the triangle inequality, $X=X-X_n+X_n \le |X_n-X|+|X_n|$. Hence, $P(X \le Z) \le P(|X_n-X| \le Z) + P(|X_n| \le Z)$. I know that, since $X_n  {\buildrel p \over \rightarrow} X$ then $P(|X_n-X| \le Z) \to 1$, and we have $P( |X_n| \le Z)=1$.
I can't go further.

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1 Answer 1

up vote 4 down vote accepted

$X_n {\buildrel p \over \rightarrow} X$ implies that there is a subsequence $X_{n(k)}$ with $X_{n(k)}\to X$ almost surely.

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I see, thanks. I think, I can finish the proof now. I didn't know the result, you mentionned. Can you give a reference for this result? –  Nicolas Essis-Breton Nov 7 '11 at 18:07
    
I found a reference, I will post it. Our book doesn't mentionned it, so I though, the result was not easy to find. Thanks. –  Nicolas Essis-Breton Nov 7 '11 at 18:15
    
The page looks strange, but it includes a proof of Byron result. Otherwise, graduate book on probability include the proof (Alan Gut, Probability a graduate course). –  Nicolas Essis-Breton Nov 7 '11 at 19:17
1  
It is a pretty standard result in probability. For instance, it is Theorem (13) in section 7.2 of Probability and Random Processes (Third Edition) by Grimmett and Stirzaker (page 314). –  Byron Schmuland Nov 7 '11 at 19:20
    
Thanks for pointing this, I overlooked this classic textbook. –  Nicolas Essis-Breton Nov 7 '11 at 19:49

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