# Prove the mean of $\frac{1}{X}$ is $\frac{\beta}{(\alpha-1)}$ [duplicate]

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:

$$\beta^{\alpha}\beta^{-(\alpha-1)}(\alpha-1)^{-1}\Gamma(\alpha-1)^{-1}\Gamma(\alpha-1) = \frac{\beta}{(\alpha-1)}?$$

• Looks good to me. What is your question? Sep 9, 2021 at 14:22
• @AdrianKeister Was checking that my method was correct as the book I answered this from failed to provide the solution Sep 9, 2021 at 14:40
• I recall providing the answer for all powers $E[X^p]$ in the duplicate thread.
– whuber
Sep 9, 2021 at 14:53