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Math Help - Probabilistic dominated convergence theorem

  1. #1
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    Probabilistic dominated convergence theorem

    Dominated convergence theorem: Suppose Xn->X almost surely and |Xn| ≤ W for all n, where E(W)<∞. Then E(Xn)->E(X).

    1) Suppose Xn->X in probability and |Xn| ≤ W for all n, where E(W)<∞. Show that E(Xn)->E(X).

    2) Suppose Xn->X in probability and (Xn)^2 ≤ W for all n, where E(W)<∞. Show that Xn->X in mean square. (i.e. E[(Xn-X)^2]->0)
    ===================================

    1) Let E(Xnk) be any subsequence of {E(Xn)}.
    Then Xnk->X in probability => there exists a further subsequence Xnk' such that Xnk'->X almost surely.
    By dominated convergence theorem, this implies that E(Xnk')->E(X) and so every subsequence of {E(Xn)} has a further subsequence of it which converges to E(X) => E(Xn)->E(X).

    2) But now I am stuck with question 2. The assumption is replaced by (Xn)^2 ≤ W, and we need to prove more: Xn->X in mean square. I tried to modify the proof in question 1, but it doesn't seem to work here. How can we modify the proof? Or do we need something new?

    Any help is much appreciated!
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  2. #2
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    Quote Originally Posted by kingwinner View Post
    So every subsequence of {E(Xn)} has a further subsequence of it which converges to E(X) => E(Xn)->E(X).
    This implication is true, but not obvious. Can you justify it? (or maybe it is from your lecture notes)


    2) But now I am stuck with question 2. The assumption is replaced by (Xn)^2 ≤ W, and we need to prove more: Xn->X in mean square. I tried to modify the proof in question 1, but it doesn't seem to work here. How can we modify the proof? Or do we need something new?
    Nothing new is needed. Just prove that the sequence Y_n=(X_n-X)^2 satisfies the assumptions of question 1) (with 0 as the limit).

    Hint: using the property about subsequences, you have |X|^2\leq W as well.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    This implication is true, but not obvious. Can you justify it? (or maybe it is from your lecture notes)
    Yes, it's from my notes. I am not sure why it is true though.

    Nothing new is needed. Just prove that the sequence Y_n=(X_n-X)^2 satisfies the assumptions of question 1) (with 0 as the limit).
    Hint: using the property about subsequences, you have |X|^2\leq W as well.
    Xn->X in probability
    =>|Xn-X|->0 in probability
    =>|Xn-X|^2 ->0 in probability
    So the first assumption is satisfied.

    But I don't know how to prove that the other assumptions are satisfied. Here is my attempt:
    |(Xn-X)^2| = (Xn-X)^2 = (Xn)^2 - 2(Xn)(X) + X^2 ≤ |Xn|^2 + 2|Xn| |X| + |X|^2 ≤ 2W +2|Xn| |X|
    But I have no idea how to prove E(2W +2|Xn| |X|) < ∞

    Please help!
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  4. #4
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    Quote Originally Posted by kingwinner View Post
    But I don't know how to prove that the other assumptions are satisfied. Here is my attempt:
    |(Xn-X)^2| = (Xn-X)^2 = (Xn)^2 - 2(Xn)(X) + X^2 ≤ |Xn|^2 + 2|Xn| |X| + |X|^2 ≤ 2W +2|Xn| |X|
    I told you you can prove |X|^2\leq W, so that |X_n||X|\leq \sqrt{W}\sqrt{W}=W. So you just have to prove what I claimed.
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  5. #5
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    Quote Originally Posted by Laurent View Post
    I told you you can prove |X|^2\leq W, so that |X_n||X|\leq \sqrt{W}\sqrt{W}=W. So you just have to prove what I claimed.
    OK, so E(4W)< ∞, so it remains to prove that "if Xn->X in probability and (Xn)^2 ≤ W for all n, then X^2 ≤ W."

    I wrote out all the given information and definitions, and searched for properties and theorems, but I don't see the connection. Which property about subsequence do I need?

    Thanks!
    Last edited by kingwinner; December 14th 2009 at 09:01 AM.
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  6. #6
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    Quote Originally Posted by kingwinner View Post
    Which property about subsequence do I need?
    The fact that there exists a subsequence which converges a.s.. Can you see why?
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