Variance in Inverse Gamma Distribution

**ynotidas** · August 31st 2009, 05:28 AM

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

**gustavodecastro** · August 31st 2009, 06:36 AM

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.

**ynotidas** · August 31st 2009, 06:44 AM

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

**gustavodecastro** · August 31st 2009, 07:00 AM

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

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

**ynotidas** · August 31st 2009, 07:09 AM

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

**gustavodecastro** · August 31st 2009, 07:42 AM

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.

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