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Math Help - On Expected Value

  1. #1
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    On Expected Value

    Hi everyone, is the following statement true, if so, why is it true?

    dE[f(x)]/dx=E[df(x)/dx]?

    Thanks

    ---------

    After some thought, I come up with a "heuristic" proof as the following, any idea for a rigorous proof?

    dE[f(x)]/dx={E[f(x+dx)]-E[f(x)]}/dx={E[f(x)+dx*df(x)/dx]-E[f(x)]}/dx={E[f(x)]+E[dx*df(x)/dx]-E[f(x)]}/dx=dx*E[df(x)/dx]/dx=E[df(x)/dx]
    Last edited by zling110; September 16th 2009 at 01:59 PM.
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  2. #2
    Moo
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    Hello,

    dE[f(x)]/dx={E[f(x+dx)]-E[f(x)]}/dx
    this relationship is false, it's an equality when dx goes to 0.

    And how does df/dx appear in the third member (after the second equality) ?



    And... where does this relationship come from ?
    Because E(f(x)) is a constant, so its derivative is 0. And E(df(x)/dx) is not necessarily 0.
    Also, I think taking x instead of X, a random variable, is inconsistent, as it makes the problem "trivial" or "weird" lol
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  3. #3
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    Quote Originally Posted by zling110 View Post
    Hi everyone, is the following statement true, if so, why is it true?

    dE[f(x)]/dx=E[df(x)/dx]?
    I guess you mean \frac{d}{dx}E[f(X,x)]=E[\frac{\partial}{\partial x}f(X,x)], where f:\mathbb{R}^2\to\mathbb{R} is a fixed function and X is a random variable. Don't you?

    If so, the result is not always true, but one can give sufficient condition for that. For instance: this is true if ( x\mapsto f(t,x) is differentiable for all t, and)
    - X takes only finitely many different values (then the expectation is a finite sum)
    - there exists a random variable Y\geq 0 such that E[Y]<\infty and, for all x, \left|\frac{\partial}{\partial x}f(X,x)\right|\leq Y almost surely. (this is the theorem of derivation under the integral symbol)
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