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

**Mathsdog** · July 16th 2012, 12:23 AM

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

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

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

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

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

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

Which I reckon simplifies to

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

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

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

**a tutor** · July 16th 2012, 01:14 AM

Originally Posted by Mathsdog

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

Simplify your result above to get

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

and simplify a little more .

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