# Show that the maximum of a multinomial distribution is given by N1=N2= =Ns=N/s

1. Mar 23, 2010

### naele

1. The problem statement, all variables and given/known data

So one of the problems in my statistical mechanics textbook (McQuarrie problem 1-50) asks to show that the maximum of a multinomial distribution is given for N1=N2=...=Ns=N/s

In the book, he shows that for large N, the binomial coefficients when taken as a function are maximized at N/2. Then he expands the natural log around that point and shows that it takes on the form of a gaussian function.

The derivation is found at this link

and the exact wording of the problem is here. Problem 50.

The first thing I don't really get where he pulls the N1*=N/2 from. I know that to actually show that isn't too difficult, but I can't just assume the same result for the multinomial case because that's what I'm trying to show.

2. Relevant equations

The multinomial coefficients are given by
$$\frac{N!}{\prod_{j=1}^r N_j!}$$ where $N_1+N_2+\cdots+N_r=N$

3. The attempt at a solution

Just from working with the coefficients expression subject to the above constraint results in the following when maximized with Lagrange multipliers
$$N\ln N - \sum N_j\ln N_j - \lambda(\sum N_j -N)=0=F(N_j)$$
$$(-\sum_{j=1}^s\ln N_j-\lambda)\, dN_j=0$$ which is the total differential
If I write out the partial derivatives with respect to each term I get
$$\frac{\partial F}{\partial N_1}=-\ln N_1-1 - \lambda=0$$

$$\frac{\partial F}{\partial N_2}=-\ln N_2-1 - \lambda=0$$

$$\frac{\partial F}{\partial N_3}=-\ln N_3-1 - \lambda=0$$

etc

So i conclude that $N_1=N_2=N_3=\cdots=N_s$. At the same time, if $sN_j=N$ implies that $N_j=N/s$

Last edited: Mar 23, 2010