MultinomialDistribution 2004-08-26 02:02:04 2006-10-02 22:49:50 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $\textbf{X}=(X_1,\ldots,X_n)$ be a random vector such that \begin{enumerate} \item $X_i\geq 0$ and $X_i\in\mathbb{Z}$ \item $X_1+\cdots+X_n=N$, where $N$ is a fixed integer \end{enumerate} Then $\textbf{X}$ has a \emph{multinomial distribution} if there exists a parameter vector $\boldsymbol{\pi}=(\pi_1,\ldots,\pi_n)$ such that \begin{enumerate} \item $\pi_i\geq 0$ and $\pi_i\in\mathbb{R}$ \item $\pi_1+\cdots+\pi_n=1$ \item $\textbf{X}$ has a discrete probability distribution function $f_{\textbf{X}}(\boldsymbol{x})$ in the form: $$f_{\textbf{X}}(\boldsymbol{x})=\frac{N!}{x_1!\cdots x_n!}\prod_{i=1}^{n}\pi_i^{x_i}$$ \end{enumerate} \par \textbf{Remarks} \begin{itemize} \item $\operatorname{E}[\textbf{X}]=N\boldsymbol{\pi}$ \item $\operatorname{Var}[\textbf{X}]=(v_{ij})$, where $$ v_{ij}= \begin{cases} N\pi_i(1-\pi_i) & \text{if $i=j$;}\\ -N\pi_i\pi_j & \text{if $i\neq j$.} \end{cases} $$ \item When $n=2$, the multinomial distribution is the same as the binomial distribution \item 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 $\lambda=\sum\lambda_i$. \par \textbf{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 $\textbf{X}=(X_1,\ldots,X_n)$, $\boldsymbol{x}= (x_1,\ldots,x_n)$ and $\lambda=\lambda_1+\cdots+\lambda_n$. Next, let $X=X_1+\cdots+X_n$. Then $X$ 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 $\textbf{X}$ 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 $\sum \lambda_i/\lambda=1$. \end{itemize}