- #1

Yagoda

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## Homework Statement

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

## Homework Equations

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

## 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.[itex] \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*}

[/itex]

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?