# Negative binomial moment estimators

• June 19th 2010, 08:51 AM
losm1
Negative binomial moment estimators
$X_1, ... , X_n$ is random sample from negative binomial distribution $NB(r,p)$ .
I need to find moment estimators $\hat r$ and $\hat p$.
My literature on this topic is very poor, any hints?
• June 19th 2010, 08:58 AM
SpringFan25
My notation:
the sample mean: $\bar{X}$
the sample variance: $\sigma^2$

the population mean: $E(X)$
the population variance: $Var(X)$

i assume you are trying to do a method of moments estimate of the parameters.
background info: http://en.wikipedia.org/wiki/Method_...ts_(statistics)

Quick explanation: equate the sample moments with the theoretical population moments

$\bar{X} = E(X)$
$\sigma^2 = Var(X)$

According to Negative binomial distribution - Wikipedia, the free encyclopedia, the moments for this distribution are:

$E(X) = r \frac{p}{1-p}$
$Var(X) = r \frac{p^2}{(1-p)^2} = \frac{E^2(X)}{r}$

So

$\frac{E^2(X)}{Var(X)}= r$
To obtain the method of moments estimator, replace all the moments in the above equation with their sample analogues.
So your method of moment estimate for r is $\hat{r} = \bar{X}^2 / \sigma^2$

Can you use a similar approach to find the estimate for p?

NBI'd check the theoretical moments are correct if you aren't getting sensible answers. wikipedia is sometimes wrong about those
• June 19th 2010, 12:03 PM
losm1
I have checked NB formulas and used the right ones, is this ok?

$E(X) = \frac{r(1-p)}{p}$
$Var(X) = \frac{r(1-p)}{p^2} = \frac{r(1-p)}{p} \cdot \frac{1}{p} = \frac{E(X)}{p}$

$p= E(X) / Var(X)$

$\hat{p} = \bar{X} / \sigma^2$

$r = \frac{Var(X)}{1-p}$

$\hat{r} = (\sigma^2)^2 / \left(\sigma^2 - \bar{X}\right)$
• June 20th 2010, 03:11 AM
SpringFan25
i hate wikipedia sometimes...