Is the negative binomial not expressible as in the exponential family if there are 2 unknowns?

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

Answer

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\}.

Attribution
Source : Link , Question Author : Larry , Answer Author : Xi’an

Leave a Comment