# Thread: Conditional for two Poisson variables is a binomial: derivation help required

1. ## Conditional for two Poisson variables is a binomial: derivation help required

Hi,
I've been struggling with this for the last couple of days. I would appreciate any illumination.

Let $\displaystyle X$ and $\displaystyle Y$ be independent random variables where $\displaystyle p_{X}(k) =\exp(- \lambda) \cdot \frac{\lambda^{k}}{k!}$ and $\displaystyle p_{Y}(k) =\exp(- \mu) \cdot \frac{\mu^{k}}{k!}$ with $\displaystyle k =0, 1, \ldots$

Show that the conditional pdf of $\displaystyle X$ given $\displaystyle X + Y = n$ is binomial with parameters $\displaystyle n$ and $\displaystyle \frac{\lambda}{\lambda + \mu}$.

For notation I say N is the random variable taking a value N =n.

So then as I see it, $\displaystyle p_{X|N}(x, n) = {p_{X, N}(x, n) \over p_{N}(n)} ={ \frac{ \exp{-(\lambda + \mu)} \lambda^{x} \mu^{n-x}}{{x!(n-x)!}} \over \frac{\exp{-(\lambda + \mu)} (\lambda + \mu)^n}{n!}}$

so $\displaystyle p_{X|N}(x, n) = \frac{ \exp{-(\lambda + \mu)} \lambda^{x} \mu^{n-x}}{{x!(n-x)!}} \cdot \frac{n!}{\exp{-(\lambda + \mu)} (\lambda + \mu)^n}$

Which I reckon simplifies to

$\displaystyle p_{X|N}(x, n) ={ \binom{n}{x} \lambda^{x} \mu^{n-x} \over \sum_{k=0}^{n} \binom{n}{k} \lambda^{k} \mu^{n-k}$

Which seems plausible, but I can't massage it into a binomial with parameters $\displaystyle n$ and $\displaystyle \frac{\lambda}{\lambda + \mu}$

Thanks in advance for any light anyone can throw on this. MD

2. ## Re: Conditional for two Poisson variables is a binomial: derivation help required

Originally Posted by Mathsdog

so $\displaystyle p_{X|N}(x, n) = \frac{ \exp{-(\lambda + \mu)} \lambda^{x} \mu^{n-x}}{{x!(n-x)!}} \cdot \frac{n!}{\exp{-(\lambda + \mu)} (\lambda + \mu)^n}$
Simplify your result above to get

$\displaystyle \frac{n! }{{x!(n-x)!}} \cdot \frac{\lambda^{x} \mu^{n-x}}{ (\lambda + \mu)^x ( \lambda+\mu)^{n-x}}$

and simplify a little more .