# 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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### References

#### Bibliography

1. Bosch, Karl: "Elementare Einführung in die Wahrscheinlichkeitsrechnung", vieweg Studium, 1995, 6th Edition