# Issue with derivation of ML parameter estimation for binomial distribution

• May 22nd 2013, 02:48 AM
Mathsdog
Issue with derivation of ML parameter estimation for binomial distribution
Hi, I have an issue with this step in the derivation of the ML solution to the binomial parameter estimation for p. Found here

https://www.projectrhea.org/rhea/ind...utions_OldKiwi

$f(x)=\left(\frac{n!}{x!\left(n-x \right)!} \right){p}^{x}{\left(1-p \right)}^{n-x}$

$L(p)=\prod_{i=1}^{n}f({x}_{i})=\prod_{i=1}^{n} \left( \frac{n!}{{x}_{i}!\left(n-{x}_{i} \right)!} \right){p}^{{x}_{i}}{\left(1-p \right)}^{n-{x}_{i}$

$L(p)=\left( \prod_{i=1}^{n}\left(\frac{n!}{{x}_{i}!\left(n-{x}_{i} \right)!} \right)\right){p}^{\sum_{i=1}^{n}{x}_{i}}{\left(1-p \right)}^{n-\sum_{i=1}^{n}{x}_{i}$

Why is the last term ${\left(1-p \right)}^{n-\sum_{i=1}^{n}{x}_{i}$ and not ${\left(1-p \right)}^{n^{2}-\sum_{i=1}^{n}{x}_{i}$?

Taking a concrete example, suppose $1-p = q$, $n = 5$ and $X = {2, 4, 5, 3, 1}$

then $\prod_{i}^5 (1-p)^{n - x_{i}} = q^{5-2} q^{5-4} q^{5-5} q^{5-3} q^{5-1}=q^{3} q^{1} q^{0} q^{2} q^{4}= q^{10} = q^{5^2 - (2 + 4 + 5 + 3 + 1)} \not= q^{5 - (2 + 4 + 5 + 3 + 1)} = q^{-10}$

Right?

Would really appreciate someone telling me where, if, I am going wrong here.

• May 22nd 2013, 07:42 PM
chiro
Re: Issue with derivation of ML parameter estimation for binomial distribution
Hey Mathsdog.

The reason for n - blah instead of n^2 - blah is that you have a total of n observations. If you have x of those observations corresponding to X = 1 then you have n - x corresponding to X = 0.