Decompose Poisson random variable as sum of Poisson random variables If $X,Y$ are independent Poisson random variables with parameter $\lambda_1, \lambda_2$, then $X+Y$ is Poisson random variable with parameter $\lambda_1+\lambda_2$. I am wondering whether the converse if true, given a poisson random variable on a probability space $(\Omega, \mathcal{F},P)$, can we always decompose it into independent poisson random variables with parameter $\lambda_1,\lambda_2$ such that there sum is the given random variable? 

Suppose $W\sim\operatorname{Poisson}(\lambda).$
Suppose $0<\lambda_1 <\lambda,$ and let $\lambda_2 = \lambda - \lambda_1.$
Let $p = \dfrac{\lambda_1}\lambda = \dfrac{\lambda_1}{\lambda_1+\lambda_2}.$
Let $X\mid W \sim\operatorname{Binomial}(W,p),$ i.e. this is the number of successes in $W$ independent trials with probability $p$ of success on each trial. Let $Y= W-X.$ Then $Y\mid W \sim\operatorname{Binomial}(W,1-p).$
Given all of this, one can conclude that


*

*$W=X+Y.$

*$X\sim\operatorname{Poisson}(\lambda_1).$

*$Y \sim\operatorname{Poisson}(\lambda_2).$

*$X,Y$ are independent.


Proving this is a standard exercise. How to do it is a question that has been posted here a number of times.
