Take the 2-minute tour ×
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.

I'm a little bit confused about what the meaning of multinomial distributions, at least from what i've gleaned from the wikipedia page on Multinomial Distribution. In essence, a multinomial distribution is the generalized form of a binomial distribution. That is, outcomes are independent, however there are k possible outcomes, each with k success, which gives the probability mass function:

$Mult(n, p_1, p_2, p_3, ..., p_n) = {n \choose x_1, x_2, ... x_n} p_1^{x_1} ...p_n^{x_n} $

However, shouldn't evaluating only $X_1$ give a binomial distribution since the multinomial is just a generalization of it? But throwing that in, we get: $Mult(n, p_1) = {n \choose x_1} p_1^{x_1}$ which can't possibly turn into $Bin(n, p) = {n \choose x} p^x(1-p)^{n-x}$

Am I missing something in my understanding of a multinomial distribution? And if so, how do we get the binomial from the multinomial equation then?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Your formula for the multinomial distribution is a little off: there should be $k$ different $p$s and $x$s with $\sum_{i=1}^k x_i=n$, so that the key coefficient is ${n}\choose{x_1,x_2,\ldots,x_k}$. (You can confirm this with Wikipedia.)

To get the binomial distribution, take $k=2$, $x_1=x$, $p_1=p$ and $p_2=1-p$. Then you have ${n}\choose{x_1,x_2}$$p_1^{x_1}p_2^{x_2}=$$n\choose x_1$$p^{x_1}(1-p)^{n-x_1}$, as desired.

share|improve this answer
    
oh, i think i see what i was missing now... since we only have $p_1$ and $p_2$, $p_2 = 1 - p_1$ by definition. Thanks! –  Jaynathan Leung Nov 13 '12 at 2:50

Your Answer

 
discard

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.