Hi,

I have the following problem.

Let X and Y be independent, with X~Poi $\displaystyle (\lambda)$ and Y~Poi$\displaystyle (\mu) $. Let Z = X + Y. Find the distribution X | Z = z.

Now I know that the sum of two independent random variables which both have a Poisson distribution, also has a Poisson distribution.

so the pmf of Z is

$\displaystyle \mathbb{P}(Z=z) = \frac{(\lambda + \mu)^z}{z!} e^{-(\lambda + \mu)} $

and

$\displaystyle \mathbb{P}(X|Z = z) = \frac{\mathbb{P}(X \cap Z = z)}{\mathbb{P}(Z = z)} $

but Im having trouble calculating $\displaystyle \mathbb{P}(X \cap Z = z)$. This is where Im stuck.

I know that the conditional probability $\displaystyle X|Z=z $ will return a binomial distribution. Can anyone please help me understand how to calculate $\displaystyle \mathbb{P}(X \cap Z = z)$ and consequently obtain a binomial distribution?