1. Conditional Poisson Distibution

Hi,
I have the following problem.

Let X and Y be independent, with X~Poi $(\lambda)$ and Y~Poi $(\mu)$. 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

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

and

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

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

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

2. This should help, since you have the right idea.

$P(X=x|X+Y=z)={P(X=x,X+Y=z)\over P(X+Y=z)}$

$={P(X=x,Y=z-x)\over P(X+Y=z)}={P(X=x)P(Y=z-x)\over P(Z=z)}$

NOW, plug in x, z-x and z into the distribution of the three DIFFERENT Poissons and it should clean up nicely.

3. Thank-you so much matheagle. That was so helpful, I understand conditional probability a lot better now.

Just one thing,
Ive managed to get the solution down to

$\begin{pmatrix} z \\ x \end{pmatrix} \lambda^x \mu^{z-x} (\lambda + \mu)^{-z}$

But I can't work out how to get from my solution down to what it should be.
i.e.

Bin~ $( z, \frac{\lambda}{\lambda + \mu} )$

I bet its something really obvious but I cant see it.

4. Originally Posted by woody198403
Thank-you so much matheagle. That was so helpful, I understand conditional probability a lot better now.

Just one thing,
Ive managed to get the solution down to

$\begin{pmatrix} z \\ x \end{pmatrix} \lambda^x \mu^{z-x} (\lambda + \mu)^{-z}$

But I can't work out how to get from my solution down to what it should be.
i.e.

Bin~ $( z, \frac{\lambda}{\lambda + \mu} )$

I bet its something really obvious but I cant see it.
It's obvious if you work backwards:

${z \choose x} \left( \frac{\lambda}{\lambda + \mu}\right)^x \left( 1 - \frac{\lambda}{\lambda + \mu}\right)^{z-x}$

$= {z \choose x} \left( \frac{\lambda}{\lambda + \mu}\right)^x \left(\frac{\mu}{\lambda + \mu}\right)^{z-x}$