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Independence Problem: Please Help

  1. Oct 20, 2009 #1
    Independence Problem: Please Help!!!

    I have been trying to figure out the proof to this problem for the past couple of days and still don't have an answer. The question is as follows:

    Let (Q,F,P) be a probability triple such that Q is countable. Prove that it is impossible for there to exist a sequence A1,A2,A3,... E F which is independent, such that P(Ai) = 1/2 for eash i. [Hint: First prove that for each w E Q, and each n E N, we have P({w}) <= 1/(2^n). Then derive a contradiction.

    Note that Q, represents Omega, F the sigma-algebra/sigma-fiels and E is "an element of or member of"

    Would appreciate any help with this.
     
  2. jcsd
  3. Oct 21, 2009 #2
    Re: Independence Problem: Please Help!!!

    Add this to the hint. If [itex]Q [/itex] is countable, then there is [itex] \omega \in Q[/itex] with [itex]P(\{\omega\}) > 0 [/itex]. So, for large [itex] n [/itex] we have [itex] P(\{\omega\}) > 1/2^n[/itex].
     
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