$\displaystyle X_1, ... , X_n$ is random sample from negative binomial distribution $\displaystyle NB(r,p)$ .

I need to find moment estimators $\displaystyle \hat r$ and $\displaystyle \hat p$.

My literature on this topic is very poor, any hints?

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- Jun 19th 2010, 07:51 AMlosm1Negative binomial moment estimators
$\displaystyle X_1, ... , X_n$ is random sample from negative binomial distribution $\displaystyle NB(r,p)$ .

I need to find moment estimators $\displaystyle \hat r$ and $\displaystyle \hat p$.

My literature on this topic is very poor, any hints? - Jun 19th 2010, 07:58 AMSpringFan25
My notation:

the sample mean: $\displaystyle \bar{X}$

the sample variance: $\displaystyle \sigma^2 $

the population mean: $\displaystyle E(X)$

the population variance:$\displaystyle Var(X)$

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

$\displaystyle \bar{X} = E(X)$

$\displaystyle \sigma^2 = Var(X)$

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

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

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

So

$\displaystyle \frac{E^2(X)}{Var(X)}= r$

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

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

**NB**I'd check the theoretical moments are correct if you aren't getting sensible answers. wikipedia is sometimes wrong about those - Jun 19th 2010, 11:03 AMlosm1
I have checked NB formulas and used the right ones, is this ok?

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

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

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

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

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

$\displaystyle \hat{r} = (\sigma^2)^2 / \left(\sigma^2 - \bar{X}\right)$ - Jun 20th 2010, 02:11 AMSpringFan25
i hate wikipedia sometimes...

Your answer is perfect though :)