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Convergence of Random Variables on Discrete Prob Spaces

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IniquiTrance
#1
Sep27-10, 03:16 PM
P: 186
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!
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Eynstone
#2
Oct1-10, 03:00 AM
P: 336
I don't think it's true that weak convergence implies a.s. conv. in the discrete case.
bpet
#3
Oct1-10, 07:27 AM
P: 523
You've already been given a counterexample http://www.physicsforums.com/showpos...11&postcount=3

IniquiTrance
#4
Oct1-10, 03:03 PM
P: 186
Convergence of Random Variables on Discrete Prob Spaces

@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:

[tex]X_n(\omega)\stackrel{p}{\longrightarrow} X(\omega)\implies X_n(\omega)\stackrel{A.S}{\longrightarrow} X(\omega)[/tex]

This is a totally different question!
IniquiTrance
#5
Oct1-10, 03:05 PM
P: 186
Quote Quote by Eynstone View Post
I don't think it's true that weak convergence implies a.s. conv. in the discrete case.
I don't either, yet my prof said it can be proven that this is true... I can't see how though...
bpet
#6
Oct2-10, 03:46 AM
P: 523
Quote Quote by IniquiTrance View Post
@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:

[tex]X_n(\omega)\stackrel{p}{\longrightarrow} X(\omega)\implies X_n(\omega)\stackrel{A.S}{\longrightarrow} X(\omega)[/tex]

This is a totally different question!
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
IniquiTrance
#7
Oct8-10, 08:14 PM
P: 186


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