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Math Help - Convergence question

  1. #1
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    Convergence question

    Suppose for a sequence of random variables X_1,X_2,... is considered as completely convergence to X if

    \displaystyle\sum_{n=1}^{\infty}P(|X_n-X|>\epsilon)<\infty for each \epsilon>0

    How can I show that if the sequence X_n is independent then the complete convergence becomes convergence almost surely?

    Though some searching, the definition of converge almost surely is

    P(\displaystyle\lim_{n\to\infty}X_n=X)=1.

    Then

    P(\displaystyle\lim_{n\to\infty}|X_n-X|<\epsilon)=1

    Do I start from here?
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  2. #2
    Moo
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    Hello,

    By Borel-Cantelli's lemma (part I), we have \displaystyle P(\limsup_n \{|X_n-X|<\epsilon\})=0. By taking the complement, we get that \displaystyle \forall \epsilon>0,P(\liminf_n \{|X_n-X|<\epsilon\})=1

    This means that \displaystyle\forall \epsilon>0,P(\exists N\in\mathbb N,\forall n\geq N,|X_n-X|<\epsilon)=1.

    Then you have to get the epsilon into the probability, and I think you have to assume first that epsilon is rational (so that there is a countable number of epsilon's).
    I don't really remember how to do and I have to go out right now, I'll try to think about it, unless you do before me
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    By Borel-Cantelli's lemma (part I), we have \displaystyle P(\limsup_n \{|X_n-X|<\epsilon\})=0. By taking the complement, we get that \displaystyle \forall \epsilon>0,P(\liminf_n \{|X_n-X|<\epsilon\})=1

    This means that \displaystyle\forall \epsilon>0,P(\exists N\in\mathbb N,\forall n\geq N,|X_n-X|<\epsilon)=1.

    Then you have to get the epsilon into the probability, and I think you have to assume first that epsilon is rational (so that there is a countable number of epsilon's).
    I don't really remember how to do and I have to go out right now, I'll try to think about it, unless you do before me
    I thought u got it correct?

    From Borel-Cantelli's lemma,
    \displaystyle\sum_{n=1}^{\infty}P(|X_n-X|>\epsilon)<\infty becomes

    \displaystyle \forall \epsilon>0,P(\liminf_n \{|X_n-X|<\epsilon\})=1

    which means

    \displaystyle \forall \epsilon>0,P(\lim\bigcup \{|X_n-X|<\epsilon\})=1

    Since it's independent, then

    \displaystyle \forall \epsilon>0,\prod P(\lim \{|X_n-X|<\epsilon\})=1

    which implies that

    P(\displaystyle\lim_{n\to\infty}|X_n-X|<\epsilon)=1
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  4. #4
    Moo
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    It's not independent, because of X
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  5. #5
    MHF Contributor matheagle's Avatar
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    Hsu-Robbins original paper is http://www.ncbi.nlm.nih.gov/pmc/arti...01695-0003.pdf
    there are many others on complete convergence like http://www.math.uregina.ca/~volodin/adv.pdf
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