Conditional Expectation with 2 distributions

Question

you could help me with this problem I would really appreciate it. This is not a task just an exercise to understand the subject. Let the random variable $X \sim \mathrm{Bernoulli}(p)$. Suppose that given an $X = i$, $Y$ is a random variable of type $\mathrm{Poisson} (3 (i + 1))$. Find $\mathbb{E} \left[(X +1) Y^2\right]$.

Answer 1

\begin{align*} E[(X+1)Y^2] &= E[E[(X+1)Y^2|X]] \\ &= E[(X+1)E[Y^2|X]] & \text{X+1 constant given X} \\ &= E[(X+1)(\mathbb{V}[Y|X]+E[Y|X]^2)] & E[Y^2|X]=\mathbb{V}[Y|X]+E[Y|X]^2 \\ &= E[(X+1)(3(X+1)+(3(X+1))^2)] & Y|X \sim \text{Poisson}(3(X+1)) \\ &= (1-p)((0+1)(3(0+1)+(3(0+1))^2) \\ &+ p((1+1)(3(1+1)+(3(1+1))^2) & X \sim \text{Bernoulli}(p) \\ &= 12(1-p) + 84p = 12+72p \end{align*}

Answer 2

HINT Let $V\sim\mathcal{P}(3)$ and $W\sim \mathcal{P}(6)$. Then, by the Law of Iterated Expectation (aka Law of Total Expectation, Tower Property of Expectation):

$$ \begin{split} \mathbb{E}\left[(X+1)Y^2\right] & = \mathbb P[X=0]\,\Bbb E\left[V^2\right] +\mathbb P[X=1]~\mathbb E\left[2W^2\right] \\ & = (1-p)\,\mathbb{E}\left[V^2\right] + 2p\, \mathbb{E}\left[W^2\right]\\ \end{split} $$

Can you finish this?