# Thread: Variance in Inverse Gamma Distribution

1. ## Variance in Inverse Gamma Distribution

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

2. 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.

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

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

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

6. 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.