- #1
confused88
- 22
- 0
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:
[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
[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