Product of Poisson Random Variables 
Problem: $X_{1}$ and $X_{2}$ are independent Poisson random variables such that $X_{1}\sim \operatorname{Poisson}(\lambda_{1})$ and $X_{2}\sim \operatorname{Poisson}(\lambda_{2})$. 
Is $Z = X_{1}X_{2}$ also a Poisson random variable? If yes, how do you find the parameter?

I'm thinking either see if the mean and variance are equal, or calculate the pdf. But I'm not sure how to find the variance or the pdf.
 A: Clearly not.  Note that for any Poisson distribution, we have $$\frac{\Pr[X = 2]}{\Pr[X = 1]} = \frac{\lambda^2}{2!} \cdot \frac{1!}{\lambda^1} = \frac{\lambda}{2}.$$
Now $$\Pr[Z = 2] = \Pr[X_1 = 2]\Pr[X_2 = 1] + \Pr[X_1 = 1]\Pr[X_2 = 2],$$ since the only way to make the product $X_1 X_2 = 2$ is if $(X_1, X_2) \in \{(1, 2), (2, 1)\}$.  And $$\Pr[Z = 1] = \Pr[X_1 = 1]\Pr[X_2 = 1].$$  So if $Z \sim \operatorname{Poisson}(\lambda_3)$, we should have $$\frac{\lambda_3}{2} = \frac{\Pr[Z = 2]}{\Pr[Z = 1]} = \frac{\Pr[X_1 = 2]}{\Pr[X_1 = 1]} + \frac{\Pr[X_2 = 2]}{\Pr[X_1 = 1]} = \frac{\lambda_1 + \lambda_2}{2},$$ hence we must have $$\lambda_3 = \lambda_1 + \lambda_2.$$  But $$\begin{align}
e^{-(\lambda_1 + \lambda_2)} &= e^{-\lambda_3} \\ &= \Pr[Z = 0] \\ &= 1 - \Pr[X_1 \ge 1]\Pr[X_2 \ge 1] \\ &= 1 - (1 - e^{-\lambda_1})(1 - e^{-\lambda_2}) \\ &= -e^{-(\lambda_1 + \lambda_2)} + e^{-\lambda_1} + e^{-\lambda_2} \end{align}$$
is not identically true for all $\lambda_1, \lambda_2$, thus $Z$ cannot be Poisson.
A: To test:
$$\begin{align}\mathsf P(Z{=}0)&=\mathsf P(X_1{=}0)+\mathsf P(X_2{=}0)-\mathsf P(X_1{=}0)\mathsf P(X_2{=}0)\\[2ex]\mathsf P(Z{=}1)&=\mathsf P(X_1{=}1)\mathsf P(X_2{=}1)\\[2ex]\mathsf P(Z{=}2)&=\mathsf P(X_1{=}1)\mathsf P(X_2{=}2)+\mathsf P(X_1{=}2)\mathsf P(X_2{=}1)\\[2ex]\mathsf P(Z{=}3)&=\mathsf P(X_1{=}1)\mathsf P(X_2{=}3)+\mathsf P(X_1{=}3)\mathsf P(X_2{=}1)\\[2ex]\mathsf P(Z{=}4)&=\mathsf P(X_1{=}1)\mathsf P(X_2{=}4)+\mathsf P(X_1{=}2)\mathsf P(X_2{=}2)+\mathsf P(X_1{=}4)\mathsf P(X_2{=}1)\\[2ex]&~~\vdots\\[2ex]\mathsf P(Z=z;{z\in\Bbb N^+})&=\sum_{x=1}^z\mathbf 1_{z/x\in\Bbb N^+}\mathsf P(X_1{=}x)\mathsf P(X_2{=}z/x)\end{align}$$
You can evaluate these if you wish, but it does not seem likely to fit a Poisson Distribution.

For the Variance, use the Law of Total Variance, and the favt of independence: $$\begin{align}\mathsf {Var}(Z)&=\mathsf E(\mathsf{Var}(X_1X_2\mid X_1))+\mathsf {Var}(\mathsf{E}(X_1X_2\mid X_1))\\[1ex]&=\mathsf E(X_1^2\,\mathsf{Var}(X_2\mid X_1))+\mathsf {Var}(X_1\,\mathsf{E}(X_2\mid X_1))\\[1ex]&=\mathsf E(X_1^2)\,\mathsf{Var}(X_2)+\mathsf {Var}(X_1)\,\mathsf{E}(X_2)\\&~~\vdots\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\sum_{x = 0}^{\infty}{\lambda_{1}^{x}\expo{-\lambda_{1}} \over x!}
\sum_{y = 0}^{\infty}{\lambda_{2}^{y}\expo{-\lambda_{2}} \over y!}\bracks{z = xy}
\\[5mm] = &\
\expo{-\lambda_{1}}\bracks{z = 0}
\\[2mm] &\ +
\expo{-\lambda_{1} - \lambda_{2}}
\sum_{x = 1}^{z}
{\lambda_{1}^{x}\lambda_{2}^{z/x} \over
x!\pars{z/x}!}\bracks{x\!\mid\! z}
\end{align}
I guess a closed form doesn't seem to be viable at first sight. However, the evaluation involves a finite sum !!!. This link and this one can be helpful.
A: The product is not Poisson. For $Z$ to be Poisson it would need to satisfy
$
E(Z) = Var(Z).
$
But
$$
E(Z) = \lambda_1\lambda_2
$$
and
$$
Var(Z) = \lambda_1\lambda_2 + \lambda_1\lambda_2^2 + \lambda_2\lambda_1^2,
$$
which will not be equal for any $\lambda_1$ or $\lambda_2$ that are greater than zero.
