Hi! Can someone help me prove that the variance for the inverse gamma is:(adsbygoogle = window.adsbygoogle || []).push({});

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

I started with:

[tex]

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:

[tex]

\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:

[tex]

\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

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# Inverse gamma

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