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Function of a random gaussian variable

  1. Apr 25, 2004 #1
    I'm having trouble showing the following relation:

    E(exp(z)) = exp(E(z^2)/2)

    where z is a zero-mean gaussian variable and E() is the avg

    anyone can help?
  2. jcsd
  3. Apr 26, 2004 #2


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    Science Advisor

    Wasn't this posted just recently?

    For the standard Gaussian Normal distribution,
    [tex]E(f(z))= \int_{-\inf}^{\inf}{f(z)e^{-\frac{z^2}{2}}}dz[/tex]

    In this case,
    [tex] f(x)= e^{\frac{z^2}{2}} [/tex]
    so the integral becomes
    [tex]E(e^{\frac{z^2}{2}} )= \int_{-\inf}^{\inf}{e^{\frac{z^2}{2}} e^{-\frac{z^2}{2}}}dz[/tex]

    can you do that?
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