Suppose X is a random variable X ~ Poisson with mean Y, and Y is a random variable ~ Gamma $(\alpha, \beta)$. Find EY and VarY
For EY and VarY this is what I have: \begin{align*} f(y|\alpha,\beta) &= \frac{1}{\Gamma(\alpha)\beta^{\alpha}}y^{\alpha-1}e^{-\frac{y}{\beta}} \\ E(Y) &= \int_0^\infty y^n \frac{1} \\ {\Gamma(\alpha)\beta^{\alpha}}y^{\alpha-1}e^{-\frac{y}{\beta}} dy &= \frac{1}{\Gamma(\alpha)\beta^{\alpha}}\int_0^\infty y^{\alpha+n-1}e^{-\frac{y}{\beta}}dy = \frac{1} \\ \end{align*} \begin{align*} {\Gamma(\alpha)\beta^{\alpha}}\int_0^\infty (\beta t)^{\alpha+n-1}e^t\beta dt$ **$t &=\frac{y}{\beta}, \frac{dy}{\beta}=dt** \\ &=\frac{\beta^{\alpha+n}}{\Gamma(\alpha)\beta^{\alpha}}\int_0^\infty \\ t^{\alpha+n-1}e^{-t}dt &= \frac{\beta^n}{\Gamma(\alpha)}\int_0^\infty t^{\alpha+n-1}e^{-t}dt = \frac{\beta^n \Gamma(\alpha+n)}{\Gamma^{\alpha}} ** n_1 and n_2 \\ \end{align*} \begin{align*} E(Y) = \frac{\beta \Gamma(\alpha+1)}{\Gamma(\alpha)} = \alpha\beta \\ E(Y^2) &= \frac{\beta^2 \Gamma(\alpha + 2}{\Gamma(\alpha)}= \beta^2(\alpha + 1) \alpha \\ Var(Y) &= (\alpha+1)\alpha\beta^2 - \alpha^2 \beta^2 = \alpha\beta^2 \end{align*} Please tell if this is correct or tips for what am I missing. Thank you.
