I am struggling with these questions. Let $X∼\text{Bern}(p)$ and $Y∼\text{Poisson}(\lambda)$ be independent. We are interested in the random variable $Z=XY$.
(a) Find $\mathbb{E}(Z)$ and $\text{Var}(Z)$.
(b) Find the PMF of $Z$.
(c) Compare $f_Z (z)$ to $f_Y (z)$ for all values of $z=0,1,2,\cdots$.
I have tried to solve (a), but I got stuck with (b),(c).
(a) $$\mathbb{E}(Z)=\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)$$
$$\mathbb{E}(Z)=p\lambda \text{Var}(Z)=\text{Var}(XY)=\mathbb{E}((XY)^2)-[\mathbb{E}(XY)]^2$$
so
$$\text{Var}(Z)=-\mathbb{E}(X)^2\mathbb{E}(Y)^2+\mathbb{E}(X^2 )\mathbb{E}(Y^2 )$$
$$\text{Var}(Z)=-(p^2\lambda^2)+p(\lambda^2+\lambda)$$
and thus
$$\text{Var}(Z)=-p^2 \lambda^2+p\lambda+p \lambda^2=p\lambda(-p\lambda+1+\lambda)$$
$$\text{Var}(Z)=pλ(1+λ-pλ)$$
The probability that $Z$ takes value zero we may compute by inclusion-exclusion: \begin{align} \mathbb P(Z=0)&=\mathbb P(X=0)+\mathbb P(Y=0)-\mathbb P(X=0,Y=0)\\ &= 1-p+e^{-\lambda} -(1-p)e^{-\lambda}\\ &= 1-p(1-e^{-\lambda}). \end{align} For $k\geqslant 1$ we have $$ \mathbb P (Z=k)=\mathbb P(X=1,Y=k) = \mathbb P(X=1)\mathbb P(Y=k) = pe^{-\lambda}\frac{\lambda^k}{k!}. $$ It follows from independence that $$ \mathbb E[Z] = \mathbb E[XY]=\mathbb E[X]\mathbb E[Y] = p\lambda. $$ Moreover, we have $$ \mathsf{Var}(XY) = \mathbb E[X^2Y^2] - \mathbb E[XY]^2 = \mathsf{Var}(X)\mathsf{Var}(Y) +\mathsf{Var}(X)\mathbb E[Y]^2 + \mathsf{Var}(Y)\mathbb E[X^2], $$ and as $$ \mathbb E[X^2] = p,\quad \mathbb E[Y^2] = \mathsf{Var}(Y) + \mathbb E[Y^2] = \lambda(1+\lambda) $$ and $$ \mathsf{Var}(X) = p-p^2=p(1-p),\quad \mathsf{Var}(Y) = \lambda, $$ it follows that $$ \mathsf{Var}(Z) = p(1-p)\lambda + p(1-p)\lambda(1-\lambda) + \lambda(1-\lambda)p. $$