Can someone help me how to
show $\displaystyle Var(Y)=\frac{\beta^2}{(\alpha-1)^2(\alpha-2)}$, in Inverse Gamma Distribution.
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.
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.