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Math Help - Transforming random variable and finding PDF and Expected Value

  1. #1
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    Transforming random variable and finding PDF and Expected Value

    Suppose X is a continuous random variable with probability density f, and \phi is a continuous increasing differentiable function.

    1) Show that the probability density of g of Y=\phi(X) is given by:

    g(y)=f(\phi^{-1}(y))\psi(y)

    where \phi^{-1} is the inverse of \phi, and

    \psi(y)=\frac{d}{dy}\phi^{-1}(y)

    2) Show that the expected value of Y is:

    E(Y)=\int_{-\infty}^{\infty} \phi(x)f(x) dx
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  2. #2
    MHF Contributor matheagle's Avatar
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    (1) Just differentiate the CDFs...



    g(y)= {d\over dy} F_Y(y)={d\over dy}P(Y\le y)={d\over dy}P(\phi(X)\le y)

     ={d\over dy}P(X\le \phi^{-1}(y))={d\over dy}F_X(\phi^{-1}(y))

    Now use the chain rule..

     =f(\phi^{-1}(y)) \frac{d}{dy}\phi^{-1}(y)=f(\phi^{-1}(y))\psi(y)


    This works as well if \phi is decreasing.

    (2) Is obvious if you just replace Y with \phi(X)
    Last edited by matheagle; June 6th 2010 at 10:43 PM.
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