Proposition: Multinomial Distribution

Consider a random experiment repeated \(n\) times, in which only one of the collectively exhaustive and mutually exclusive events \(A_1,A_2,\ldots,A_r\) can occur. Assume that all repetitions are mutually independent. Furthermore, assume, in each repetition the event \(A_i\) can occur with the constant probability \(p_i:=p(A_i)\) for \(i=1,\ldots,r\). Let \(X_1,\ldots,X_r\) be random variables counting the numbers \(k_i\) of the realizations of each event \(A_i\), \(i=1,\ldots,r\). Clearly, we have \(k_i\ge 0\) for and \(k_1+k_2+\ldots+k_r=n\).

The probability of the event that we will observe the event \(S_{k_1k_2\ldots k_r}\) defined by \[S_{k_1k_2\ldots k_r}:=\cases{A_1&\text{occurred }k_1\text{ times}\\\vdots\\A_r&\text{occurred }k_r\text{ times}\\}\]

is given by

\[p(S_{k_1k_2\ldots k_r})=\binom {n}{k_1,k_2,\ldots,k_r}p_1^{k_1}p_2^{k_2}\ldots p_r^{k_r}.\quad\quad ( * )\]

Therefore, the probability mass function of all random variables is given by

\[p(X_1=k_1,\ldots,X_r=k_r)=\cases{\binom {n}{k_1,k_2,\ldots,k_r}p_1^{k_1}p_2^{k_2}\ldots p_r^{k_r}&\text{if } k_1+k_2+\ldots+k_r=n\\ 0&\text{else}.}\]

Because the probabilities \(( * )\) can be retrieved from the expansion of the multinomial theorem. \[1=(p_1 + p_2 + \ldots + p_r)^n=\sum_{\substack{k_1+\ldots+k_r=n \\ k_1,\ldots,k_r}}\binom{n}{k_1,k_2\ldots,k_r}p_1^{k_1} p_2^{k_2}\ldots p_r^{k_r}.\]

this kind of a distribution is called a multinomial distribution.

Proofs: 1

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  1. Bosch, Karl: "Elementare Einf├╝hrung in die Wahrscheinlichkeitsrechnung", vieweg Studium, 1995, 6th Edition