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Let $X_1,X_2,...,X_n$ be i.i.d. from a Poisson distribution with mean $\lambda$, so that the common pdf is given by:

$$f_{X_i}(x)=\frac{e^{-\lambda}\lambda^x}{x!}$$

Define the statistic $S=\sum_{i=1}^nX_i$. Clearly $S\sim POISSON(n\lambda)$

Prove that the conditional distribution of $X_1$ given $S=s$ is binomial with parameters $s,n^{-1}$. That is, prove:

$$X_1|_{S=s}\sim B(s,n^{-1})$$

This is really tripping me up since $S$ is actually composed of $X_1$. I have no idea what the joint distribution would be, so I can't take that approach to find the conditional. So I'm really unsure of how to prove this. Any ideas?

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\begin{align} Pr(X_1 = k | \sum_{i=1}^n X_i=s) &= \frac{Pr(X_1=k)Pr(\sum_{i=1}^n X_i=s|X_1=k)}{Pr(\sum_{i=1}^n X_i=s)} \\ &= \frac{Pr(X_1=k)Pr(\sum_{i=2}^n X_i=s-k)}{Pr(\sum_{i=1}^n X_i=s)} \\ \end{align}

Can you proceed?

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You have a stream of point events generated by $n$ independent Poison processes occuring at identical rates.   What is the probability that any particular point event was generated from process 1?   Is that dependent on knowing if any other point event was so generated?   Why so?

Then can it now be argued that the count of point events generated by prossess 1 among $s$ such events is a count of 'successes' among a stream of $s$ independent Bernouli trials with identical 'success' rate?


Alternatively, use the definition of conditional probability, and the nature of a sum of independent poisson events , to identify that $$\mathsf P(X_1=k\mid S_n=s)~{=\dfrac{\mathsf P(X_1=k)\mathsf P(S_{n-1}=s-k)}{\mathsf P(S_n=s)}\\=\dfrac{\lambda^k\mathsf e^{-\lambda}}{k!}\dfrac{((n-1)\lambda)^{s-k}\mathsf e^{-(n-1)\lambda}}{(s-k)!}\dfrac{s!}{(n\lambda)^s\mathsf e^{-n\lambda}}}$$

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