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Math Help - E(Xn) -> E(X) implies E(|Xn - X|) -> 0??

  1. #1
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    E(Xn) -> E(X) implies E(|Xn - X|) -> 0??

    Let X be a random variable. Let Xn be a sequence of random variables.
    Then E(|Xn - X|) -> 0 as n->∞ implies E(Xn) -> E(X) as n->∞.
    [I can prove this by |E(W)|≤E(|W|) and squeeze theorem.]

    Is it also true that E(Xn) -> E(X) implies E(|Xn - X|) -> 0??
    How can we prove it?

    Any help is much appreciated!
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  2. #2
    Moo
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    Hello,

    Let X_n be such that P(X_n=-1)=\frac 12=P(X_n=1)

    And let X\equiv 0 almost surely.

    Then we have E(X_n)=0 and E(X)=0
    So the condition E(X_n)\to E(X) is satisfied.

    But |X_n-X|=1 almost surely. Thus E|X_n-X|=1 and can't have a limit equal to 0
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