Almost Sure Convergence Criterion

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May 13th 2011, 04:37 PM
Guy

Almost Sure Convergence Criterion

Hello all,

I'm having problems showing this result.

Problem: Let $Z_1, Z_2, ...$ and $Z$ be simple random variables. Show that $Z_n \to Z$ with probability 1 if and only if for all $\epsilon > 0$ there exists an n such that $P(|Z_k - Z| < \epsilon, n \le k \le m) > 1 - \epsilon$ for all m > n.

I haven't made much progress. Hopefully a point in the correct direction will get me going.
May 14th 2011, 06:06 AM
Focus

Quote:

Originally Posted by Guy

Hello all,

I'm having problems showing this result.

Problem: Let $Z_1, Z_2, ...$ and $Z$ be simple random variables. Show that $Z_n \to Z$ with probability 1 if and only if for all $\epsilon > 0$ there exists an n such that $P(|Z_k - Z| < \epsion, n \le k \le m) > 1 - \epsilon$ for all m > n.

I haven't made much progress. Hopefully a point in the correct direction will get me going.

I think you missed something out in the probability (|Z_k-Z|< ????) but it seems like you will need Borel-Cantelli at some point.
May 14th 2011, 07:09 AM
Guy

Yes, sorry, it is supposed to be $P(|Z_k - Z| < \epsilon, n \le k \le m) > 1 - \epsilon$ (which I've now edited so that it is correct). And yeah, I assumed I would need Borel-Cantelli but that is more of a finishing move to one of the directions whereas I can't even get started.
May 14th 2011, 11:59 AM
Moo

Hello,

I don't understand why you entitled your thread with almost sure convergence ? At first glance, it's not about a.s. convergence !
What do you call "simple rv's" ? And are the rv independent ?

$Z_n\to Z$ in probability means that $\forall \epsilon>0,\lim_{n\to\infty} P(|Z_m-Z|<\epsilon)=1$
By definition of a limit, we can write :
$\forall \epsilon>0,\forall \tilde{\epsilon}>0,\exists n\in\mathbb{N},\forall m>n,P(|Z_m-Z|<\epsilon)>1-\tilde{\epsilon}$

In particular for $\tilde{\epsilon}=\epsilon$ , we have $\forall \epsilon>0, \exists n\in\mathbb{N},\forall m>n, P(|Z_m-Z|<\epsilon)>1-\epsilon$

and hm well, maybe you can finish it off... I can't see well at the moment (they're all equalities/equivalences so it solves the "iff" problem)
May 14th 2011, 12:05 PM
Guy

This problem is 6.1 from Billingsley's Probability and Measure. It is quoted exactly, so no, the random variables are not independent. A simple rv is one with a finite range. And it is, indeed, about a.s. convergence. Billingsley refers to convergence a.s. as convergence with probability 1. It is not a question about convergence in probability.
May 14th 2011, 12:35 PM
Moo

Quote:

Originally Posted by Guy

This problem is 6.1 from Billingsley's Probability and Measure. It is quoted exactly, so no, the random variables are not independent. A simple rv is one with a finite range. And it is, indeed, about a.s. convergence. Billingsley refers to convergence a.s. as convergence with probability 1. It is not a question about convergence in probability.

Yeah sorry, I misread the "with probability 1".
I'll definitely go to bed :D