Convergence of Random Variables on Discrete Prob Spaces

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Discussion Overview

The discussion revolves around the convergence of random variables in discrete probability spaces, specifically the relationship between weak convergence in probability and almost sure convergence. Participants explore whether weak convergence implies almost sure convergence in this context.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about the distinction between weak convergence and almost sure convergence, citing their professor's claim that both forms of convergence are equivalent in discrete probability spaces.
  • Another participant contests the professor's assertion, stating that weak convergence does not imply almost sure convergence in discrete cases.
  • A third participant references a counterexample provided in a previous post, suggesting that the professor's claim may not hold.
  • Further discussion reveals that some participants are unsure how to reconcile their understanding with the professor's statement, indicating a lack of clarity on the proof of the equivalence in discrete spaces.
  • One participant proposes an approach to demonstrate that a sequence does not almost surely converge by showing it fails on at least one discrete event with non-zero probability, which could prevent weak convergence.

Areas of Agreement / Disagreement

Participants generally disagree on the equivalence of weak convergence and almost sure convergence in discrete probability spaces, with no consensus reached on the validity of the professor's claim.

Contextual Notes

Participants acknowledge the complexity of the topic, noting that the underlying event space may not be discrete despite the individual outcomes being discrete, which could affect the convergence properties discussed.

IniquiTrance
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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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I don't think it's true that weak convergence implies a.s. conv. in the discrete case.
 
@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!
 
Last edited:
Eynstone said:
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...
 
IniquiTrance said:
@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
 
36t25hj.png
 
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