4
$\begingroup$

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.

1 Answers 1

5

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

  • 0
    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?2011-11-07
  • 0
    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.2011-11-07
  • 0
    The [page](http://www.opentradingsystem.com/quantNotes/Convergence_in_probability_.html) looks strange, but it includes a proof of Byron result. Otherwise, graduate book on probability include the proof (Alan Gut, Probability a graduate course).2011-11-07
  • 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).2011-11-07
  • 0
    Thanks for pointing this, I overlooked this classic textbook.2011-11-07