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Inverse gamma

  1. Aug 15, 2009 #1
    Hi! Can someone help me prove that the variance for the inverse gamma is:

    [tex]\frac{\beta^2}{(\alpha - 1)^2 (\alpha -2)}[/tex]

    I started with:
    E(x) = \int_0^\infty x \frac{\beta^{\alpha}}{\gamma(\alpha)}(\frac{1}{x})^{\alpha + 1} exp (\frac{-\beta}{x}) dx[/tex]

    then i let [tex] y = \frac{\beta}{x}; dy = \frac{1}{x}; x = \frac{\beta}{y}[/tex]

    so then i get:
    \frac{\beta^{\alpha}}{\gamma(\alpha)} \int_0^\infty (\frac{y}{\beta})^{\alpha + 1} e^{-y} dy [/tex]

    I'm really not sure if I'm doing this right..but i'll keep going just in case, so then i get:
    \frac{\beta^{\alpha}}{\gamma(\alpha)} \frac{1}{\beta^{\alpha + 1}}\int_0^\infty y^{\alpha + 1} e^{-y} dy [/tex]

    I don't know what to do now, any help would be greatly appreciated
  2. jcsd
  3. Aug 16, 2009 #2


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    Homework Helper

    I can't quite follow what you are doing, but by the looks of it you are close... the integral you have on your last line is [itex]\Gamma(\alpha + 2)[/itex].
    If the [itex]\gamma(\alpha)[/itex] in your denominator is the same function (Euler gamma function) then you can further simplify
    [tex]\frac{ \Gamma(2 + \alpha) }{ \Gamma(\alpha) }[/tex]
    using that [itex]\Gamma(z + 1) = z \Gamma(z)[/itex].
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