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Thread: Variance in Inverse Gamma Distribution

  1. #1
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    Variance in Inverse Gamma Distribution

    Can someone help me how to
    show $\displaystyle Var(Y)=\frac{\beta^2}{(\alpha-1)^2(\alpha-2)}$, in Inverse Gamma Distribution.
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  2. #2
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    To compute the expectation value, note that

    $\displaystyle
    EX = \int_0^{\infty} x \frac{\beta^{\alpha}}{\Gamma(\alpha)} x^{-\alpha - 1} e^{-\beta / x} dx
    $

    $\displaystyle
    EX = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_0^{\infty} x^{-(\alpha - 1) - 1} e^{-\beta / x} dx
    $


    Completing the term inside the integral,

    $\displaystyle
    EX = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_0^\infty x^{-(\alpha - 1) - 1} \frac{\beta^{\alpha - 1}}{\Gamma(\alpha - 1)} e^{-\beta/x} dx \frac{\Gamma(\alpha - 1)}{\beta^{\alpha - 1}}
    $

    $\displaystyle
    EX = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \frac{\Gamma(\alpha - 1)}{\beta^{\alpha - 1}}
    $

    Since the term inside the integral is now an inverse Gamma with parameters

    $\displaystyle \alpha - 1 ; \beta $

    So,

    $\displaystyle

    \frac{\beta}{\alpha - 1}
    $

    Do the same for $\displaystyle EX^2 $ and find the variance.
    Last edited by gustavodecastro; Aug 31st 2009 at 07:34 AM.
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  3. #3
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    Thanks Gustavodecastro, can you retype the code, because i can't read it properly?
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  4. #4
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    Sorry for that. Now I think it is okay. You have to remember that

    $\displaystyle
    \Gamma(n) = (n-1)!
    $
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  5. #5
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    I'm not sure how finding $\displaystyle EX^2$, will help me find the variance?
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  6. #6
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    To compute $\displaystyle EX^2 $, note that

    $\displaystyle
    Ex = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \int_0^\infty x^{-(\alpha - 2) - 1} \frac{\beta^{\alpha - 2}}{\Gamma(\alpha - 2)} e^{-\beta/x} dx \frac{\Gamma(\alpha - 2)}{\beta^{\alpha - 2}}
    $

    And so you have an inverse gamma with parameters $\displaystyle \alpha - 2 ; \beta $ inside the integral. Do the same as before and find $\displaystyle EX^2 $.

    $\displaystyle
    EX^2 = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \frac{\Gamma(\alpha - 2)}{\beta^{\alpha - 2}} = \frac{\beta^2}{(\alpha - 2)(\alpha - 1)}
    $

    Then, as $\displaystyle VarX = EX^2 - \left( EX \right)^2 $, find the variance.
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