Expected Value of Inverse Gamma Distribution

[Yagoda]

Homework Statement

I have that X is distributed with Gamma(a,b) and that $Y = \frac{1}{X}$. I found the pdf of Y to be $\frac{1}{\Gamma(a)b^a} \left(\frac{1}{y}\right)^{a+1} e^{-1/yb}$ for y > 0. I need to use this to find the expected value.

Homework Equations

The gamma function is defined as $\Gamma(a) = \int_{0}^{\infty} t^{a-1} e^{-t} \, dt$

The Attempt at a Solution

I know how to approach the problem, but I am having trouble actually getting the correct answer. Here is what I've done.
$\begin{align*} E[Y] &=\frac{1}{\Gamma(a)b^a} \int_{0}^{\infty} y \left(\frac{1}{y}\right)^{a+1} e^{-1/yb} \, dy \\ &=\frac{1}{\Gamma(a)b^a} \int_{0}^{\infty}\left(\frac{1}{y}\right)^{a} e^{-1/yb} \, dy \\ \text{Let t = 1/y so dt = -1/y^2.} \\ &= \frac{-1}{\Gamma(a)b^a} \int_{\infty}^{0} t^{a-2} e^{-t/b} \, dt \\ &= \frac{1}{\Gamma(a)b^a} \int_{0}^{\infty} \left(\frac{t}{b}\right)^{a-2} b^{a-2} e^{-t/b} \, dt \\ &= \frac{b^{a-2}\Gamma(a-1)}{\Gamma(a)b^a} \\ &= \frac{1}{\Gamma(a-1) b^2} \end{align*}$

But I think I'm not supposed to have that extra b in the denominator. I've gone over the substitution a few times and it seems like raising t to the a-2 power is correct since I've removed two of the (1/y)'s to substitute in for dt. What am I missing?

 

[Ray Vickson]

Homework Statement

I have that X is distributed with Gamma(a,b) and that $Y = \frac{1}{X}$. I found the pdf of Y to be $\frac{1}{\Gamma(a)b^a} \left(\frac{1}{y}\right)^{a+1} e^{-1/yb}$ for y > 0. I need to use this to find the expected value.

Homework Equations

The gamma function is defined as $\Gamma(a) = \int_{0}^{\infty} t^{a-1} e^{-t} \, dt$

The Attempt at a Solution

I know how to approach the problem, but I am having trouble actually getting the correct answer. Here is what I've done.
$\begin{align*} E[Y] &=\frac{1}{\Gamma(a)b^a} \int_{0}^{\infty} y \left(\frac{1}{y}\right)^{a+1} e^{-1/yb} \, dy \\ &=\frac{1}{\Gamma(a)b^a} \int_{0}^{\infty}\left(\frac{1}{y}\right)^{a} e^{-1/yb} \, dy \\ \text{Let t = 1/y so dt = -1/y^2.} \\ &= \frac{-1}{\Gamma(a)b^a} \int_{\infty}^{0} t^{a-2} e^{-t/b} \, dt \\ &= \frac{1}{\Gamma(a)b^a} \int_{0}^{\infty} \left(\frac{t}{b}\right)^{a-2} b^{a-2} e^{-t/b} \, dt \\ &= \frac{b^{a-2}\Gamma(a-1)}{\Gamma(a)b^a} \\ &= \frac{1}{\Gamma(a-1) b^2} \end{align*}$

But I think I'm not supposed to have that extra b in the denominator. I've gone over the substitution a few times and it seems like raising t to the a-2 power is correct since I've removed two of the (1/y)'s to substitute in for dt. What am I missing?

You have changed variables incorrectly: when you set $t = 1/y$ you will have $e^{-t^b}$ in the integrand, so you cannot 'do' the integral as it stands. You need a different change of variables.

 

[Yagoda]

Do you have a suggestion for what to use for the substitution? I also tried substituting in $t = \frac{1}{by}$ so that I would just have $e^{-t}$ in the integrand, but I again ran into the problem of having $b^{a-2}$ pulled outside the integral rather than the $b^{a-1}$ that I am supposed to have.

 

[Yagoda]

Oh, never mind I think I see where I made my mistake. Forgot about that extra b when substituting dt.