multinomial distribution

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

1. $X_i\ge 0$ and $X_i\in \mathbb{Z}$

2. $X_1+\mathrm{\cdots }+X_n=N$, where $N$ is a fixed integer

Then X has a multinomial distribution if there exists a parameter vector $𝝅=({\pi }_1,\mathrm{\dots },{\pi }_n)$ such that

1. ${\pi }_i\ge 0$ and ${\pi }_i\in ℝ$

2. ${\pi }_1+\mathrm{\cdots }+{\pi }_n=1$

3. X has a discrete probability distribution function $f_{\text{𝐗}}(𝒙)$ in the form:

   $$f_{\text{𝐗}}(𝒙)=\frac{N!}{x_1!\mathrm{\cdots }{x}_n!}\prod _{i=1}^n{\pi }_i^{x_i}$$

Remarks

• $\mathrm{E}[\text{𝐗}]=N𝝅$

• $\mathrm{Var}[\text{𝐗}]=(v_{ij})$, where

• When $n=2$, the multinomial distribution is the same as the binomial distribution

• If $X_1,\mathrm{\dots },X_n$ are mutually independent Poisson random variables with parameters ${\lambda }_1,\mathrm{\dots },{\lambda }_n$respectively, then the conditional joint distribution of $X_1,\mathrm{\dots },X_n$ given that $X_1+\mathrm{\cdots }+X_n=N$ ismultinomial 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

  $$f_{\text{𝐗}}(𝒙)=\prod _{i=1}^n\frac{e^{-{\lambda }_i}{\lambda }_i^{x_i}}{x_i!}=e^{-\lambda }\prod _{i=1}^n\frac{{\lambda }_i^{x_i}}{x_i!},$$

  where $\text{𝐗}=(X_1,\mathrm{\dots },X_n)$, $𝒙=(x_1,\mathrm{\dots },x_n)$ and $\lambda ={\lambda }_1+\mathrm{\cdots }+{\lambda }_n$.Next, let $X=X_1+\mathrm{\cdots }+X_n$. 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:

  $$f_{\text{𝐗}}(𝒙\mid X=N)=\frac{f_{\text{𝐗}}(𝒙)}{f_X(N)}=(e^{-\lambda }\prod _{i=1}^n\frac{{\lambda }_i^{x_i}}{x_i!})/(\frac{e^{-\lambda }{\lambda }^N}{N!})=\frac{N!}{x_1!\mathrm{\cdots }{x}_n!}\prod _{i=1}^n{(\frac{{\lambda }_i}{\lambda })}^{x_i},$$

  where $\sum x_i=N$ and that $\sum {\lambda }_i/\lambda =1$.