# Convergence of Random Variables on Discrete Prob Spaces

1. ### IniquiTrance

190
Well, I thought I understood the difference between (weak) convergence in probability, and almost sure convergence.

My prof stated that when dealing with discrete probability spaces, both forms of convergence are the same.

That is, not only does A.S. convergence imply weak convergence, as it always does, but in the discrete case, weak convergence implies A.S. convergence.

I've been trying to wrap my head around why this is so, but can't seem to "see" it.

Any ideas?

Thanks!

2. ### Eynstone

336
I don't think it's true that weak convergence implies a.s. conv. in the discrete case.

523
4. ### IniquiTrance

190
@bpet:
Thanks for that example in that thread. Like I said, I thought I finally understood the difference.

Yet my professor said one can prove that on a discrete probability space:

$$X_n(\omega)\stackrel{p}{\longrightarrow} X(\omega)\implies X_n(\omega)\stackrel{A.S}{\longrightarrow} X(\omega)$$

This is a totally different question!

Last edited: Oct 1, 2010
5. ### IniquiTrance

190
I don't either, yet my prof said it can be proven that this is true... I can't see how though...

6. ### bpet

523
Ok sorry I didn't take into account that, even though the individual archery outcomes are discrete, there isn't necessarily a discrete event space underlying the joint distribution of the infinite sequence.

An approach for the discrete space could be to assume that a sequence does not a.s. converge and show that this happens on at least one discrete event with non-zero probability (because every non-zero probability contains at least one atom), and this prevents the sequence from weak convergence.

HTH

7. ### IniquiTrance

190

Last edited: Oct 9, 2010