Moment Matching for Gamma Distribution

The gamma function (not to be confused with the gamma density function; see below) is given by

$$\Gamma \left(\alpha \right)={\int }_0^{\infty }{x}^{\alpha -1}{e}^{-x}dx,\alpha >0.$$

1. Show that ${\int }_0^{\infty }{x}^{\alpha -1}{e}^{-\beta x}dx=\frac{\Gamma \left(\alpha \right)}{{\beta }^{\alpha }}$ for $\alpha ,\beta >0$. Hint: Use the substitution $u=\beta x$.

2. Show that $\Gamma (\alpha +1)=\alpha \Gamma \left(\alpha \right)$. Hint: Use integration by parts.

Let $Z\sim \text{Gamma}(\alpha ,\beta )$. The density function is given by

$$[z|\alpha ,\beta ]=\frac{{\beta }^{\alpha }}{\Gamma \left(\alpha \right)}{z}^{\alpha -1}{e}^{-\beta z}.$$

Use the properties of the gamma function from the previous exercise to show the following:

1. $E\left(Z\right)=\frac{\alpha }{\beta }$

2. $Var\left(Z\right)=\frac{\alpha }{{\beta }^2}$. Hint: Use the fact that $Var\left(Z\right)=E\left(Z^2\right)-\left(E\right(Z){)}^2$. You can prove this fact for bonus points.

Let $\mu =E\left(Z\right)$ and ${\sigma }^2=Var\left(Z\right)$. Show the following:

1. $\alpha =\frac{{\mu }^2}{{\sigma }^2}$

2. $\beta =\frac{\mu }{{\sigma }^2}$