9
$\begingroup$

I had a homework assignment to express the negative binomial distribution as an exponential family of distributions given that the dispersion parameter was a known constant. This was fairly easy, but I wondered why they would require we held that parameter fixed. I found that I couldn't come up with a way to put it in the right form with the two parameters being unknown.

Looking online, I found claims that it is not possible. However, I have found no proof that this is true. I can't seem to come up with one myself either. Does anybody have a proof of this?

As requested below, I have attached a couple of the claims:

"The family of negative binomial distributions with fixed number of failures (a.k.a. stopping-time parameter) r is an exponential family. However, when any of the above-mentioned fixed parameters are allowed to vary, the resulting family is not an exponential family." http://en.wikipedia.org/wiki/Exponential_family

"The two-parameter negative binomial distribution is not a member of the exponential family. But if we treat the dispersion parameter as a known, fixed constant, then it is a member." http://www.unc.edu/courses/2006spring/ecol/145/001/docs/lectures/lecture21.htm

$\endgroup$
  • 1
    $\begingroup$ I added a couple of the claims above. $\endgroup$ – Larry Sep 30 '14 at 16:34
4
$\begingroup$

If you look at the density of the Negative Binomial distribution against the counting measure over the set of integers, \begin{align*}p(x|N,p)&={x+N-1\choose{N-1}}p^N(1-p)^x\\ &= \frac{(x+N-1)!}{x!(N-1)!}p^N(1-p)^x\\ &= \frac{(x+N-1)\cdots(x+1)}{(N-1)!}\exp\left\{N\log(p)+x\log(1-p) \right\}\\ &= \frac{\exp\left\{N\log(p)\right\}}{(N-1)!}\exp\left\{N\log(p)+x\log(1-p) \right\}(x+N-1)\cdots(x+1)\end{align*} the part $(x+N-1)\cdots(x+1)$ in this density cannot be expressed as $\exp\left\{ A(N)^\text{T}B(x)\right\}$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.