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'multinomial distribution'
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| Title of object: |
multinomial distribution |
| Canonical Name: |
MultinomialDistribution |
| Type: |
Definition |
| Created on: |
2004-08-26 02:02:04 |
| Modified on: |
2006-09-26 17:07:50 |
| Classification: |
msc:60E05 |
Revision comment (for changes between this and next version):
| Changes for correction #10397 ('spelling'). |
Preamble:
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\usepackage{amssymb,amscd}
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%\usepackage{psfrag}
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Content:
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 parameters $\lambda$ (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} |
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