Thread: Variance in Inverse Gamma Distribution

1. Variance in Inverse Gamma Distribution

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

2. To compute the expectation value, note that

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

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

Completing the term inside the integral,

$
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}}
$

$
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

$\alpha - 1 ; \beta$

So,

$

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

Do the same for $EX^2$ and find the variance.

3. Thanks Gustavodecastro, can you retype the code, because i can't read it properly?

4. Sorry for that. Now I think it is okay. You have to remember that

$
\Gamma(n) = (n-1)!
$

5. I'm not sure how finding $EX^2$, will help me find the variance?

6. To compute $EX^2$, note that

$
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 $\alpha - 2 ; \beta$ inside the integral. Do the same as before and find $EX^2$.

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

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