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Convergence of Random Variables on Discrete Prob Spaces 
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#1
Sep2710, 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! 


#2
Oct110, 03:00 AM

P: 336

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



#3
Oct110, 07:27 AM

P: 523

You've already been given a counterexample http://www.physicsforums.com/showpos...11&postcount=3



#4
Oct110, 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! 


#5
Oct110, 03:05 PM

P: 186




#6
Oct210, 03:46 AM

P: 523

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 nonzero probability (because every nonzero probability contains at least one atom), and this prevents the sequence from weak convergence. HTH 


#7
Oct810, 08:14 PM

P: 186




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