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Math Help - Pdf of a function of a random variable

  1. #1
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    Pdf of a function of a random variable

    I am supposed to fin the pdf f Y = e^X where X is normal r.v. with parameters \mu, \sigma^2

    The pdf of X is

    \frac{1}{ \sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}

    (Sorry I can't get this to appear in tex for some reason)

    If I integrate this function from negative infinity to ln(y), then differeniate with respect to y, I should get the function I seek.

    However I am not sure how to integrate the normal distribution function, could someone please explain to me some way I way go about doing that?

    Thank you for your help.
    Last edited by CaptainBlack; May 7th 2011 at 09:51 PM.
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  2. #2
    Senior Member Sambit's Avatar
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    Let F_Y(.)be cdf of Y and F_X(.) be cdf of X; f_Y(.) and f_X(.) are pdf of Y and X respectively.

    F_Y(y) = P(Y\leq y) = P(e^X\leq y) = P(X\leq \ln y) = F_X(\ln y)

    Hence,differentiating, f_Y(y) = \frac{d}{dx}F_X(\ln y) = f_X(\ln y)\frac{d}{dx}\ln y = f_X(\ln y)\frac{1}{y}
     = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(\ln y-\mu)^2}{2\sigma^2}}\frac{1}{y} = \frac{1}{y\sqrt{2\pi\sigma^2}}e^{-\frac{(\ln y-\mu)^2}{2\sigma^2}

    This is called Log-Normal distribution.
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  3. #3
    MHF Contributor matheagle's Avatar
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    minor correction, the derivative of F(y) wrt y is f(y)
    you have d/dx, you mean d/dy.
    Otherwise you would need the chain rule

    This is what you meant....

    Hence,differentiating, f_Y(y) = \frac{d}{dy}F_X(\ln y) = f_X(\ln y)\frac{d}{dy}\ln y = f_X(\ln y)\frac{1}{y}
    Last edited by mr fantastic; May 7th 2011 at 11:19 PM. Reason: Merged posts.
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