Let X have a gamma distribution with parameters $\alpha > 2$ and $\beta > 0$.
a. Prove that the mean of $\frac{1}{X}$ is $\frac{\beta}{(\alpha -1)}$
My approach:
$$\frac{\beta^\alpha}{\Gamma(\alpha)}\int_0^{\infty} \frac{1}{x}x^{\alpha-1}e^{-\beta x}dx = \frac{\beta^{\alpha}}{\Gamma(\alpha)}\frac{\Gamma(\alpha-1)}{\beta^{\alpha-1}}$$
Using the identity:
1.$\frac{\beta}{\Gamma(\alpha)}\int_0^{\infty} x^{\alpha-1}e^{-\beta x}dx = \frac{\Gamma(\alpha)}{\beta^{\alpha}}$
2. $\Gamma(\alpha) = (\alpha-1)\Gamma(\alpha-1)$
With some simple algebra:
$$\beta^{\alpha}\beta^{-(\alpha-1)}(\alpha-1)^{-1}\Gamma(\alpha-1)^{-1}\Gamma(\alpha-1) = \frac{\beta}{(\alpha-1)}?$$
