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Math Help - Almost Sure Convergence Criterion

  1. #1
    Guy
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    Almost Sure Convergence Criterion

    Hello all,

    I'm having problems showing this result.

    Problem: Let Z_1, Z_2, ... and Z be simple random variables. Show that Z_n \to Z with probability 1 if and only if for all \epsilon > 0 there exists an n such that P(|Z_k - Z| < \epsilon, n \le k \le m) > 1 - \epsilon for all m > n.

    I haven't made much progress. Hopefully a point in the correct direction will get me going.
    Last edited by Guy; May 14th 2011 at 07:10 AM. Reason: Correcting the typo discussed below
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  2. #2
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    Quote Originally Posted by Guy View Post
    Hello all,

    I'm having problems showing this result.

    Problem: Let Z_1, Z_2, ... and Z be simple random variables. Show that Z_n \to Z with probability 1 if and only if for all \epsilon > 0 there exists an n such that P(|Z_k - Z| < \epsion, n \le k \le m) > 1 - \epsilon for all m > n.

    I haven't made much progress. Hopefully a point in the correct direction will get me going.
    I think you missed something out in the probability (|Z_k-Z|< ????) but it seems like you will need Borel-Cantelli at some point.
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  3. #3
    Guy
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    Yes, sorry, it is supposed to be P(|Z_k - Z| < \epsilon, n \le k \le m) > 1 - \epsilon (which I've now edited so that it is correct). And yeah, I assumed I would need Borel-Cantelli but that is more of a finishing move to one of the directions whereas I can't even get started.
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  4. #4
    Moo
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    Hello,

    I don't understand why you entitled your thread with almost sure convergence ? At first glance, it's not about a.s. convergence !
    What do you call "simple rv's" ? And are the rv independent ?

    Z_n\to Z in probability means that \forall \epsilon>0,\lim_{n\to\infty} P(|Z_m-Z|<\epsilon)=1
    By definition of a limit, we can write :
    \forall \epsilon>0,\forall \tilde{\epsilon}>0,\exists n\in\mathbb{N},\forall m>n,P(|Z_m-Z|<\epsilon)>1-\tilde{\epsilon}

    In particular for \tilde{\epsilon}=\epsilon, we have \forall \epsilon>0, \exists n\in\mathbb{N},\forall m>n, P(|Z_m-Z|<\epsilon)>1-\epsilon

    and hm well, maybe you can finish it off... I can't see well at the moment (they're all equalities/equivalences so it solves the "iff" problem)
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  5. #5
    Guy
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    This problem is 6.1 from Billingsley's Probability and Measure. It is quoted exactly, so no, the random variables are not independent. A simple rv is one with a finite range. And it is, indeed, about a.s. convergence. Billingsley refers to convergence a.s. as convergence with probability 1. It is not a question about convergence in probability.
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  6. #6
    Moo
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    Quote Originally Posted by Guy View Post
    This problem is 6.1 from Billingsley's Probability and Measure. It is quoted exactly, so no, the random variables are not independent. A simple rv is one with a finite range. And it is, indeed, about a.s. convergence. Billingsley refers to convergence a.s. as convergence with probability 1. It is not a question about convergence in probability.
    Yeah sorry, I misread the "with probability 1".
    I'll definitely go to bed
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