Suppose that X has a Poisson distribution with the rate $\lambda$ and suppose the conditional distribution of Y, given$ X=x$, is binomial with parameters x and p.
Find E(Y). The law of total expectation is \begin{align} &E(Y)= E(E(Y|X)) \\ =&E(Xp)\\=& p E(X) \end{align}
Question: Can someone justify the problem solving process please?
By the law of total probability, for any nonegative integer $k$ we have \begin{align} \mathbb P(Y=k) &= \sum_{n=k}^\infty \mathbb P(Y=k\mid X=n)\mathbb P(X=n)\\ &= \sum_{n=k}^\infty \binom nk p^k(1-p)^{n-k} e^{-\lambda}\frac{\lambda^n}{n!}\\ &= e^{-\lambda p}\frac{(\lambda p)^k}{k!}, \end{align} so that $Y\sim\mathrm{Pois}(\lambda p)$. It follows immediately that $\mathbb E[Y] = \lambda p$.