Let ${X}=(X_1,\ldots,X_n)$ be a random vector such that

1. $X_i\geq 0$ and $X_i\in\mathbb{Z}$
2. $X_1+\cdots+X_n=N$ , where $N$ is a fixed integer

1. $\pi_i\geq 0$ and $\pi_i\in\mathbb{R}$
2. $\pi_1+\cdots+\pi_n=1$
3. ${X}$ has a discrete probability distribution function $f_{{X}}(\boldsymbol{x})$ in the form: $$f_{\textbf{X}}(\boldsymbol{x})=\frac{N!}{x_1!\cdots x_n!}\prod_{i=1}^{n}\pi_i^{x_i}$$

Remarks

• $\operatorname{E}[{X}]=N\boldsymbol{\pi}$
• $\operatorname{Var}[{X}]=(v_{ij})$ , where
  
• When $n=2$ , the multinomial distribution is the same as the binomial distribution
• If $X_1,\ldots,X_n$ are mutually independent Poisson random variables with parameters $\lambda_1,\ldots,\lambda_n$ respectively, then the conditional joint distribution of $X_1,\ldots,X_n$ given that $X_1+\cdots+X_n=N$ is multinomial with parameters $\lambda_i/\lambda$ , where $\lambda=\sum\lambda_i$ .

  Sketch of proof. Each $X_i$ is distributed as: $$f_{X_i}(x_i) = \frac{e^{-\lambda_i} \lambda_i^{x_i}}{x_i!}$$ The mutual independence of the $X_i$ 's shows that the joint probability distribution of the $X_i$ 's is given by $$f_{\textbf{X}}(\boldsymbol{x})=\prod_{i=1}^{n}\frac{e^{-\lambda_i} \lambda_i^{x_i}}{x_i!}= e^{-\lambda}\prod_{i=1}^{n}\frac{\lambda_i^{x_i}}{x_i!},$$ where ${X}=(X_1,\ldots,X_n)$ , $\boldsymbol{x}= (x_1,\ldots,x_n)$ and $\lambda=\lambda_1+\cdots+\lambda_n$ . Next, let $X=X_1+\cdots+X_n$ . Then $X$ is Poisson distributed with parameter $\lambda$ (which can be shown by using induction and the mutual independence of the $X_i$ 's): $$f_X(x)=\frac{e^{-\lambda} \lambda^{x}}{x!}.$$ The conditional probability distribution of ${X}$ given that $X=N$ is thus given by: $$f_{\textbf{X}}(\boldsymbol{x}\mid X=N)=\frac{f_{\textbf{X}}(\boldsymbol{x})}{f_X(N)}=(e^{-\lambda}\prod_{i=1}^{n}\frac{\lambda_i^{x_i}}{x_i!})/(\frac{e^{-\lambda} \lambda^{N}}{N!})=\frac{N!}{x_1!\cdots x_n!}\prod_{i=1}^{n}(\frac{\lambda_i}{\lambda})^{x_i},$$ where $\sum x_i=N$ and that $\sum \lambda_i/\lambda=1$ .