Proof concerning the multinomial distribution Despite a long search I was not able to find a rigorous proof of the fact that a random vector having a multinomial distribution with parameters p (the vector of probabilities) and n (the number of trials) can be written as the sum of n independent random vectors all having a multinomial distribution with parameters p and 1. Can anyone suggest where to look?
 A: Suppose $X_1,\ldots,X_n$ are independent identically distributed random variables and
$$
\Pr(X_1 = (0,0,0,\ldots0,0,\underset{\uparrow}{1},0,0,\ldots,0,0,0)) = p_i
$$
where there are $k$ components and the single "$1$" is the $i$th component, for $i=1,\ldots,k$.
Suppose $c_1+\cdots+c_n = n$, and ask what is
$$
\Pr((X_1+\cdots+X_n)=(c_1,\ldots,c_n)).
$$
The vector $(c_1,\ldots,c_n)$ is a sum of $c_1$ terms equal to $(1,0,0,0,\ldots,0)$, then $c_2$ terms equal to $(0,1,0,0,\ldots,0)$, and so on.  The probability of getting any particular sequence of $c_1$ terms equal to $(1,0,0,0,\ldots,0)$, then $c_2$ terms equal to $(0,1,0,0,\ldots,0)$, and so on, is $p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k}$.  So the probability we seek is
$$
(p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k}) + (p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k}) + \cdots + (p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k}),
$$
where the number of terms is the number of distinguishable orders in which we can list $c_1$ copies of $(1,0,0,0,\ldots,0)$, $c_2$ copies of $(0,1,0,0,\ldots,0)$, and so on.  That is a combinatorial problem, whose solution is $\dbinom{n}{c_1,c_2,\ldots,c_k}$.  Hence the probability we seek is
$$
\binom{n}{c_1,c_2,\ldots,c_k} p_1^{c_1}p_2^{c_2}\cdots p_k^{c_k},
$$
so there we have the multinomial distribution.
