Zero Inflated poisson distribution - sum of two random variables

Question

Suppose X,Y are independent variables with $X \sim ZIP(\pi,\lambda)$ and $Y \sim ZIP(\pi,\lambda)$. The extended form of ZIP would be:

$Pr(X=0) = \pi+(1-\pi)e^{-\lambda}$
$Pr(X=x) = (1-\pi)\frac{\lambda^xe^{-\lambda}}{x!}$

(You could find more information about this distribution here)
I assumed the same $\lambda$ and $\pi$ for both, for simplifying the problem. I wonder if we can have a closed-form probability for X+Y as a new ZIP with clear parameters. I tried to write the equations.

$Pr(X+Y=z) = Pr(X=x)Pr(Y=z-x)$

I assumed $z-x$ as y for simplicity. Then we have four different situations.

1. $X=0,Y=0$ $\quad\quad\quad$ $Pr(X=0)Pr(Y=0)= [\pi+(1-\pi)e^{-\lambda}]^2$
2. $X=0,Y>0$ $\quad\quad\quad$ $Pr(X=0)Pr(Y=y)= [\pi+(1-\pi)e^{-\lambda}][(1-\pi)\frac{\lambda^ye^{-\lambda}}{y!}]$
3. $X>0,Y=0$ $\quad\quad\quad$ $Pr(X=x)Pr(Y=0)= [\pi+(1-\pi)e^{-\lambda}][(1-\pi)\frac{\lambda^xe^{-\lambda}}{x!}]$
4. $X>0,Y>0$ $\quad\quad\quad$ $Pr(X=x)Pr(Y=y)= [(1-\pi)\frac{\lambda^ye^{-\lambda}}{y!}][(1-\pi)\frac{\lambda^xe^{-\lambda}}{x!}]$

Now, for assuming ZIP for this probability we would have

$Pr(X+Y=0)=[\pi+(1-\pi)e^{-\lambda}]^2$ $Pr(X+Y=z)=Pr(X=0)Pr(Y=y)+Pr(X=x)Pr(Y=0)+Pr(X=x)Pr(Y=y)$

I wonder if this formulation and assumptions are correct and if I could get a ZIP out of this.

Answer 1

The distribution of the sum of two independent ZIP random variables will be a mixture of Poisson distributions, but not itself a ZIP distribution. In your case, $$P(X+Y=k)=\pi^2\delta_k +2\pi(1-\pi)\frac {\lambda^k}{k!}e^\lambda + (1-\pi)^2\frac{(2\lambda)^k}{k!}e^{-2\lambda},$$ where $\delta_0=1$ and $\delta_k=0$ if $k>0$. This is not a ZIP distribution.

To see this formula, condition on the mixing event. The rv $X$ can be generated by flipping a coin with heads probability $\pi$. If heads comes up, let $X=0$. If tails comes up, pick $X$ from the $Po(\lambda)$ distribution. Flip another coin for $Y$. The result is: with probability $\pi^2$ you got 2 heads, so the result is $0$. With probability $2\pi(1-\pi)$ you got one head and one tail, and the resulting sum is $\sim Po(\lambda)$. The two tails case comes up with probability $(1-\pi)^2$, and the sum is $\sim Po(\lambda+\lambda)$.