The gamma function (not to be confused with the gamma density function; see below) is given by
\[ \Gamma(\alpha)=\int_0^\infty x^{\alpha - 1}e^{-x} \, dx, \alpha > 0. \]
Show that \(\int_0^\infty x^{\alpha - 1}e^{-\beta x} \, dx=\frac{\Gamma(\alpha)}{\beta^\alpha}\) for \(\alpha,\beta > 0\). Hint: Use the substitution \(u=\beta x\).
Show that \(\Gamma(\alpha + 1)=\alpha\Gamma(\alpha)\). Hint: Use integration by parts.
Let \(Z\sim\text{Gamma}(\alpha, \beta)\). The density function is given by
\[ [z|\alpha, \beta]=\frac{\beta^\alpha}{\Gamma(\alpha)}z^{\alpha-1}e^{-\beta z}. \]
Use the properties of the gamma function from the previous exercise to show the following:
\(E(Z)=\frac{\alpha}{\beta}\)
\(Var(Z)=\frac{\alpha}{\beta^2}\). Hint: Use the fact that \(Var(Z)=E(Z^2) - (E(Z))^2\). You can prove this fact for bonus points.
Let \(\mu=E(Z)\) and \(\sigma^2=Var(Z)\). Show the following:
\(\alpha=\frac{\mu^2}{\sigma^2}\)
\(\beta=\frac{\mu}{\sigma^2}\)