text: # 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
explanation: The document is a detailed mathematical discussion about the conditional distribution of a set of independent Poisson random variables given their sum. It starts with a question about how to show the law that the conditional distribution follows, and then provides a step-by-step derivation of the conditional probability. The document also includes a remark and a follow-up question from a user, along with an answer from another user clarifying the parameters and the generalization to the multinomial distribution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a detailed mathematical discussion about the conditional distribution of a set of independent Poisson random variables given their sum. It starts with a question about how to show the law that the conditional distribution follows, and then provides a step-by-step derivation of the conditional probability. The document also includes a remark and a follow-up question from a user, along with an answer from another user clarifying the parameters and the generalization to the multinomial distribution.
  
  Label: question-and-answer