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Thread: E(Xn) -> E(X) implies E(|Xn - X|) -> 0??

  1. #1
    Senior Member
    Jan 2009

    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 is offline
    A Cute Angle Moo's Avatar
    Mar 2008
    P(I'm here)=1/3, P(I'm there)=t+1/3

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

    And let $\displaystyle X\equiv 0$ almost surely.

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

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