Binomial Distribution in the Exponential Family of Distributions

Click For Summary
SUMMARY

The discussion centers on the binomial distribution with parameters $\theta=(p,n)$ and its classification within the exponential family of distributions. It is established that when $n$ is constant, the binomial distribution can be expressed in the exponential family form, specifically as $f(x|\theta)=\binom{n}{x}(1-p)^{n}exp(x*log(\frac{p}{1-p}))$. However, when considering the full parameter space where $n$ is variable, the binomial distribution does not belong to the exponential family due to the inability to express the term $\binom{n}{x}$ as a product or sum of separate functions of $x$ and $n$. The discussion concludes with the need to prove that $\log(\binom{n}{x})$ cannot be expressed in the required exponential family format.

PREREQUISITES
  • Understanding of probability distributions, specifically the binomial distribution.
  • Familiarity with the concept of the exponential family of distributions.
  • Knowledge of mathematical functions and their properties, particularly logarithmic functions.
  • Basic statistical notation and terminology, including parameters and functions.
NEXT STEPS
  • Research the properties of the exponential family of distributions in detail.
  • Study the derivation and properties of the binomial distribution.
  • Explore methods for proving expressions involving combinatorial terms, such as $\log(\binom{n}{x})$.
  • Investigate the implications of parameter dependence in statistical distributions.
USEFUL FOR

Statisticians, data scientists, and researchers in probability theory who are interested in the classification of distributions and the mathematical foundations of statistical models.

Rashad9607
Messages
14
Reaction score
0
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.
 
Last edited:
Physics news on Phys.org
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.
 

Similar threads

Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K