# How to show what law does $P(X_1=i,…,X_m=l|\sum\limits_{i=1}^{m}X_i=n)$ follow,$X_i$~$P(\lambda_i)$

Let say that $X_1,\dots ,X_m$ are independent random variables following Poisson law of parameter $λ_1,\dots, λ_m$.

We write $S=\sum\limits_{i=1}^{m}X_i$ We want the law of $\{X_1,...,X_m\}$ conditionned to $\{S=n\}$ where $n$ is a positive integer.

I know that the sum $X_1+...+X_n$ of $m$ independant variables $X_1,...,X_m$ following Poisson's law $P(\lambda_1),...,P(\lambda_m)$ is a Poisson's variable $P(\lambda_1+...+\lambda_m).$

So:

\begin{align*} P(X_1=i,...,X_m=l|\sum\limits_{i=1}^{m}X_i=n)&=\frac{P(X_1=i,...,X_m=l,\sum\limits_{i=1}^{m}X_i=n)}{P(\sum\limits_{i=1}^{m}X_i=n)}\\ &=\frac{\frac{\lambda_1^i\ *...*\lambda_m^l}{i!...l!}}{\frac{(\lambda_1+\lambda_2+...\lambda_m)^n}{n!}}\\ &=\frac{n!}{i!...l!}\frac{\lambda_1^i\ *...*\lambda_m^l}{(\lambda_1+\ *...*+\lambda_m)^n}\\ &=... \end{align*}

I think a Binomial law of parameter $B(\frac{\lambda_1}{\lambda_1+...+\lambda_m},n)$ should appear but I can't see how. Do you have any hint?

Remark: The parameters of the multinomial are indeed $\frac{\lambda_i}{\lambda_1+\cdots+\lambda_m}$. For as you pointed out, the distribution of the sum is Poisson with parameter $\lambda_1+\cdots+\lambda_m$.
So the denominator is $e^{-(\lambda_1+\cdots+\lambda_m)} \frac{(\lambda_1+\cdots+\lambda_m)^n}{n!}$. The numerator also has $e^{-\lambda_i}$ terms, but they get cancelled, just as in the two random variable case that you have looked at earlier.
• Okay, $\lambda_i$ Stands for the product? This so surprising...! – ThePassenger Nov 3 '15 at 23:01
• $\lambda_i$ stands for the parameter of the $i$-th Poisson. I do not think of it as surprising. As you found out in the case of two Poisson, the conditional distribution given the sum was binomial. The case we are looking at is just a generalization to $m$ Poisson, and the multinomial is a generalization of the binomial, the binomial is the case $m=2$. – André Nicolas Nov 4 '15 at 0:50