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

  1. #1
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    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 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
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  2. #2
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    Re: Conditional for two Poisson variables is a binomial: derivation help required

    Quote Originally Posted by Mathsdog View Post


    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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