# Inverse gamma

1. Aug 15, 2009

### confused88

Hi! Can someone help me prove that the variance for the inverse gamma is:

$$\frac{\beta^2}{(\alpha - 1)^2 (\alpha -2)}$$

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

then i let $$y = \frac{\beta}{x}; dy = \frac{1}{x}; x = \frac{\beta}{y}$$

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

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

I don't know what to do now, any help would be greatly appreciated

2. Aug 16, 2009

### CompuChip

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 $\Gamma(\alpha + 2)$.
If the $\gamma(\alpha)$ in your denominator is the same function (Euler gamma function) then you can further simplify
$$\frac{ \Gamma(2 + \alpha) }{ \Gamma(\alpha) }$$
using that $\Gamma(z + 1) = z \Gamma(z)$.