How to find the mean and variance of $Y$ without deriving the distribution of it?

Question

Let $$Y\mid \Lambda \sim \operatorname{Poisson}(\Lambda)$$ and $$\Lambda \sim \operatorname{Gamma}(\alpha, \beta)$$ where $f_{Y\mid\Lambda}(y\mid\lambda)=\frac{\lambda^y}{y!} e^{-\lambda}, y = 0, 1, 2, \ldots$ and $f_\Lambda(\lambda)=\frac 1 {\Gamma(\alpha)\beta^\alpha}\lambda^{\alpha-1} e^{-\frac \lambda \beta}$, $\lambda > 0$.

Answer 1

Now $Y\mid\Lambda\sim\mathcal{Poiss}(\Lambda)$ means we know what the conditional expectation and variance are.

And $\Lambda\sim\Gamma(\alpha, beta)$ means we know what its expectation and variance are.

The law of total expectation: $$\begin{align}\mathsf E(Y) &= \mathsf E(\mathsf E(Y\mid \Lambda)) \\ & = \mathsf E(\Lambda) \\ &= \tfrac \alpha\beta\end{align}$$

The law of total variance: $$\begin{align}\mathsf {Var}(Y) &= \mathsf E(\mathsf {Var}(Y\mid\Lambda))+\mathsf E(\mathsf {Var}(Y\mid \Lambda)) \\ & = \mathsf E(\Lambda)+\mathsf{Var}(\Lambda)\\ &= \tfrac \alpha\beta + \tfrac \alpha{\beta^2}\end{align}$$

Answer 2

We have $\mathbb{E}(Y)=\mathbb{E}(\mathbb{E}(Y|\Lambda))=\mathbb{E}(\Lambda)=\frac {\alpha} {\beta}$. On the other hand $Var(Y)=Var(\mathbb{E}(Y|\Lambda))+\mathbb{E}(Var(Y|\Lambda))=Var(\Lambda)+\mathbb{E}(\Lambda)=\frac {\alpha} {\beta^2}+\frac {\alpha} {\beta}$.