Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Use characteristic function to prove it. – Patrick Li Sep 28 '12 at 16:45
Actually, I suspect that the result above is needed to derive the characteristic function, so it would be circular. – Marco Sep 28 '12 at 17:14

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.