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Math Help - Expectation inequality probelems

  1. #1
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    Expectation inequality probelems

    If E(X) < E(X')
    and E(Y) < E(Y')
    Can we prove that E(XY) < E(X'Y') ?
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  2. #2
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    No, in general E(XY) < E(X'Y') is not true.

    First, if the variables are not independent, here is a counter example:
    X = 0, 1 with equal probability
    Y = X (not independent of X)
    X' = 0, 100 with equal probability
    Y' = 100 - X' (not independent of X')
    E(X) = E(Y) = 0.5
    E(X') = E(Y') = 50
    E(XY) = 0.5
    E(X'Y') = 0

    Second if they were independent E(XY) = E(X)E(Y), and E(X'Y') = E(X')E(Y')
    The expectations still have to be positive.
    Example:
    E(X) = E(Y) = -1
    E(X') = E(Y') = 0
    E(XY) = E(X)E(Y) = 1
    E(X'Y') = E(X')E(Y') = 0
    Last edited by snowtea; December 17th 2010 at 06:44 AM.
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  3. #3
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    Case where X and X', Y and Y' are related

    @snowtea : Nice counter-example !

    Can we still come up with a counter-example if X & X' are related, and Y & Y' are related ?
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  4. #4
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    If X, X' are related and Y, Y' are related. Lets assume that X,X',Y,Y' can only take nonnegative values, and X' = f(X), Y' = g(Y). Otherwise, the counter example will be quite easy to come up with (try it).

    There is still a counter example for the case described.
    X = 0, 10 with equal probability
    X' = 4 when X = 0 and X' = 8 when X = 10
    Y=X and Y'=X'
    E(X) = E(Y) = 5
    E(X') = E(Y') = 6
    E(XY) = E(X^2) = 50
    E(X'Y') = E(X'^2) = 40
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