PlanetMath: multinomial distribution

(more info)

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multinomial distribution

(Definition)

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

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

Then $\textbf{X}$ has a multinomial distribution if there exists a parameter vector $\boldsymbol{\pi}=(\pi_1,\ldots,\pi_n)$ such that

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

Remarks

$\operatorname{E}[\textbf{X}]=N\boldsymbol{\pi}$
$\operatorname{Var}[\textbf{X}]=(v_{ij})$ , where
$\begin{displaymath} v_{ij}= \begin{cases} N\pi_i(1-\pi_i) & \text{if $i=j$;}\ -N\pi_i\pi_j & \text{if $i\neq j$.} \end{cases}\end{displaymath}$
When , 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 is distributed as:

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