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Let $X_i\sim {\rm Poi}(\mu_i)$ independent and $N=\sum_{i=1}^4 X_i$. Show that $$X_1\mid N=n\sim {\rm Bin}\left(n,\frac{\mu_1}{\sum\mu_i}\right)$$

I tried without sucess show that

\begin{align}P(X_1|N=n)&=\frac{P(X_1=x_1,N=n)}{P(N=n)}\\[0.2cm]&=\frac{P(X_1=x_1)P(X_2=n-x_1-x_3-x_4)}{P(N=n)}\\[0.2cm]&=\frac{e^{-\mu_1}\mu_1^{x_1}}{x_1!}\frac{e^{-\mu_2}\mu_2^{n-x_1-x_3-x_4}}{(n-x_1-x_3-x_4)!}\frac{n!}{e^{-\sum \mu_i}\sum \mu_1^n}\end{align}

I make some simplifications but I did not get anywhere, maybe I'm taking a wrong path or there is some binomial transformation that I don't see.

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That should be:$$\mathsf P(X_1=x\mid N=n) ~=~ \dfrac{\mathsf P(X_1=x)~\mathsf P(X_2+X_3+X_4=n-x)}{\mathsf P(X_1+X_2+X_3+X_4=n)}$$

Now use the fact that the sum of independent Poisson random variables is a Poisson random variable whose parameter is the sum of their parameters.

$$\begin{align}X_2+X_3+X_4 ~\sim~& \mathcal {Poiss}(\mu_2+\mu_3+\mu_4)\\[1ex] X_1+X_2+X_3+X_4~\sim~& \mathcal {Poiss}(\mu_1+\mu_2+\mu_3+\mu_4)\end{align}$$

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For simplicity in notation, write $μ:=\sum μ_i$. Then, using that $\sum {\rm Poi}(μ_i)\sim {\rm Poi}(\sum μ_i)$ for independent ${\rm Poi}(μ_i)$ random variables, you get \begin{align}P(X_1=x_1\mid N=n)&=\frac{P(N=n \mid X_1=x_1)P(X_1=x_1)}{P(N=n)}\\[0.2cm]&=\frac{P(X_2+X_3+X_4=n-x_1)P(X_1=x_1)}{P(N=n)}\\[0.2cm]&=\frac{e^{-μ_2-μ_3-μ_4}(μ_2+μ_3+μ_4)^{n-x_1}}{(n-x_1)!}\cdot \frac{e^{-μ_1}μ_1^{x_1}}{x_1!}\cdot\frac{n!}{e^{-μ}\left(μ\right)^n}\\[0.2cm]&=\frac{e^{-μ}}{e^{-μ}}\cdot\frac{n!}{(n-x_1)!x_1!}\cdot \left(\frac{μ_1}{μ}\right)^{x_1}\left(\frac{μ_2+μ_3+μ_4}{μ}\right)^{n-x_1}\\[0.2cm]&=\dbinom{n}{x_1}\left(\frac{μ_1}{μ}\right)^{x_1}\left(1-\frac{μ_1}{μ}\right)^{n-x_1}\end{align} for $0\le x_1 \le n$.

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