Poisson random variable conditioned on sum

Asked Mar 10 '14 at 20:52

Active Mar 10 '14 at 21:16

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If $X_1,X_2...X_N$ are independent Poisson variables with parameters $\lambda_1,\lambda_2...\lambda_3$, then given $\sum_iX_i=N$, we have that

$X_i \sim \mathrm{Binom}(N,\frac{\lambda_i}{\sum_j\lambda_j})$ (straight from wikipedia!)

I am looking for a proof of this -- I haven't really studied the relevant stats to know how to attack this problem. Any help would be great. Cheers!

edited Mar 10 '14 at 21:16

asked Mar 10 '14 at 20:52

Draper

3

I gave a derivation for 2 Poisson variables here. Note that the binomial distribution is conditional on the realized value of the random variable $N$, usually written as $n$. – Scortchi - Reinstate Monica Mar 10 '14 at 21:37
1

@Scortchi A derivation for two is good enough: it immediately implies the result for all $N\ge 2$. – whuber Mar 10 '14 at 21:45

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