# Binomial Distribution in the Exponential Family of Distributions

#### rashtastic

##### New member
A pdf is of the exponential family if it can be written $f(x|\theta)=h(x)c(\theta)exp(\sum_{i=1}^{k}{w_{i}(\theta)t_{i}(x))}$ with $\theta$ a finite parameter vector, $c(\theta)>0$, all functions are over the reals, and only $h(x)$ is possibly constant.

I would like to show the binomial distribution with parameters $\theta=(p,n)$ is not in the exponential family.

Actually, if we consider $n$ to be constant, it is an exponential member:

$f(x|\theta)=p^{x}(1-p)^{n-x}\binom{n}{x}=\binom{n}{x}(1-p)^{n}(\frac{p}{1-p})^{x}=\binom{n}{x}(1-p)^{n}exp(x*log(\frac{p}{1-p}))$

Because $n$ is given, $\binom{n}{x}$ is a function of $x$ and will be $h(x)$.

$c(\theta)=(1-p)^{n}$.

$w_{1}(\theta)=log(\frac{p}{1-p})$ and $t_{1}(x)=x$.

If we instead want to consider the full parameter space where $n$ is not given, the binomial distribution is not a member of the exponential family.

Say we wanted to try and fit it into the exponential family model. The $\binom{n}{x}$ term would need to be split into a product of separate functions of $x$ and $n$ to be incorporated into $h(x)c(\theta)$, or split into a sum of products of separate functions to be incorporated into the summation term.

I was able to show that $\binom{n}{x}$ cannot be expressed as a product $u(n)v(x)$, so what is left is showing that it won't work in the summation term either. This means showing that $log(\binom{n}{x})$ is inexpressible as $\sum_{i=1}^{k}{w_{i}(n)t_{i}(x)}$, with $w_{i}$ and $t_{i}$ nonconstant, which I haven't been able to do. Any thoughts are appreciated.

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#### rashtastic

##### New member
So I have read the next section in my text and learned that a characteristic of exponential family distributions is that the values $x$ can take must be the same over the entire parameter space. If we take $\theta=(p,n)$, then x=0,1,2,...,n, which depends on $\theta$, so it cannot be an exponential distribution.

But I'd still like to prove the log(nCr) thing.