0
$\begingroup$

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:

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

$\endgroup$
3
  • $\begingroup$ Looks good to me. What is your question? $\endgroup$ Sep 9, 2021 at 14:22
  • $\begingroup$ @AdrianKeister Was checking that my method was correct as the book I answered this from failed to provide the solution $\endgroup$
    – Stackcans
    Sep 9, 2021 at 14:40
  • 1
    $\begingroup$ I recall providing the answer for all powers $E[X^p]$ in the duplicate thread. $\endgroup$
    – whuber
    Sep 9, 2021 at 14:53

0

Browse other questions tagged or ask your own question.