# Parameter space for the negative binomial distribution

1. Aug 13, 2014

### Mogarrr

1. The problem statement, all variables and given/known data
For the negative binomial distribution, with r known, describe the natural parameter space

2. Relevant equations
the pmf for the negative binomial distribution with parameters r and p can be
1) $P(X=x|r,p)= \binom {x-1}{r-1}p^{r}(1-p)^{x-r}$ where $x=r,r+1,...$, or
2) $P(Y=y|r,p)= \binom {y+r-1}{y}p^{r}(1-p)^{y}$ where $y=0,1,...$.

A distribution, like the one above where r is known, is a member of the exponential family of distributions. An exponential distribution is one that can be expressed as...

$h(x)c^{*}(\eta) exp(\sum_{i=1}^{k} \eta_i t_i(x))$

The parameter space are the values of $\eta$ such that $\sum_A h(x) exp(\sum_{i=1}^{k} \eta_i t_i(x)) < \infty$ where $A$ is the support of the pmf.
3. The attempt at a solution
Rewriting the 2nd pmf for the negative binomial distribution, as an exponential distribution, I have

$h(y) = \binom {y+r-1}{y}$, $c(p) = p^{r} \cdot I_(0,1)(p)$, $t_1(y)=y$, and $w_1(p) = ln(1-p)$.

Then I let $\eta = w_1(p)$, and find the values for $\eta$ where the sum converges.

I have $\sum_{y=0}^{\infty} \binom{r+y-1}{y}(e^{\eta})^{y}$, and I don't recognize this sum as anything that converges.

Any help would be appreciated.

2. Aug 13, 2014

### Ray Vickson

$$\frac{1}{(1-z)^r} \equiv (1-z)^{-r} = \sum_{k=0}^{\infty} \binom{r+k-1}{k} z^k.$$
The reason for the name "negative binomial distribution" is that it comes from "negative binomial coefficients" $\binom{-r}{k} = (-1)^k \binom{r+k-1}{k}$ associated with the "negative binomial" series $(1-z)^{-r}$. You ought to be able to show that the series converges for $|z| < 1$ because it just generalizes the series for $1/(1-z)$.

3. Aug 13, 2014

### Mogarrr

That look's like a winner.

I had thought of using the 1st pmf for the negative binomial distribution. Then I'd have the exponential distribution as:

$h(x) = \binom {x-1}{r-1} \cdot I_{(r,r+1,...)}(x)$, $c(p) = p^{r} \cdot I_{(0,1)}$, $t_1(x) = x-r$, and $w_1(p) = ln(1-p)$.

Then letting $w_1(p) = \eta$ and solving the series for $\eta$:

$\sum_X \binom {x-r}{r-1} e^{(x-r)\eta} = \sum_X \binom {x-r}{r-1} (e^{\eta})^{x-r}$,

then I recognized this series: $\sum_{x=r}^{\infty} \binom {x-1}{r-1}w^{x-r} = (1-w)^{r}$.

So I have $(1-e^{\eta})^r$.

I'm not sure, but I think the constraint is that $|e^{\eta}| < 1$.

Is this right?

4. Aug 13, 2014

### Ray Vickson

I cannot make any sense out of what you write above.

Your previous post had a simple, explicit question, and I answered it. If you put $z = e^{\eta}$ you will get the summation you had in your first post, and I already pointed out what restrictions apply to $z$.

5. Aug 13, 2014

### Mogarrr

Maybe my reply was lengthy and ambiguous. I asked you about the constraints for w in this series,

$\sum_{x=r}^{\infty} \binom {x-1}{r-1} w^{x-r} = (1-w)^{r}$, which I found at this webpage.

I think it's the same series you gave, with a change in the index, letting $x=y+r$, I have

$\sum_{y=0}^{\infty} \binom {r+y-1}{y} z^{y} = \sum_{x=r}^{\infty} \binom {x-1}{x-r}z^{x-r} = \sum_{x=r}^{\infty} \binom {x-1}{r-1}z^{x-r}$

My question is this: Is the webpage wrong?

6. Aug 13, 2014

### Mogarrr

Looking at the negative binomial distribution as a particular case of the general binomial series, I see the formula on the webpage was wrong.

Note to self: do not trust all websites.

Last edited: Aug 13, 2014