The probability mass function of a negative binomial random variable $X$ is given by $$P(X=x\mid\theta)= {r+x-1 \choose x}(1-\theta)^x\theta^r$$ and $\Bbb E(X\mid \theta)=\frac{r(1-\theta)}{\theta}$.
Does this model admit a minimum variance unbiased estimator that attains the Cramér-Rao lower bound?
What I have tried:
I have worked out the maximum likelihood estimator $\hat{\theta}$ to be $\frac{r}{r+x}$ and the CRLB to be $\Bbb V(\hat{\theta}) \geq \frac{1}{I(\theta)}= \frac{\theta^2-\theta^3}{r}$.
I know that if the estimator attains the CRLB then we know that we have found a Minimum Variance Unbiased Estimator. Also if the variance of the estimator is equal to the CRLB then it is an MVUE. However I am unsure of how to calculate the variance of $\hat{\theta}$.
Thank you for your help!
An UMVUE (Unformly Minimum Variance Unbiased Estimator) rarely attains the CRLB.
Moreover: if a family of rv's belongs to the exponential family (as it is the case) always exist an estimator that attains CRLB.
To verify that an estimator attains the lower bound there is a IFF condition:
In your case you have
$$\sum_{i=1}^n\frac{\partial}{\partial\theta}\log f(x_i;\theta)=\frac{nr}{\theta}-\frac{\Sigma_i X_i}{1-\theta}=\underbrace{-\frac{n}{(1-\theta)}}_{=b(n,\theta)}\Bigg[\overline{X}_n-\frac{r(1-\theta)}{\theta}\Bigg]$$
Thus $\overline{X}_n$ is UMVUE for $\mathbb{E}[X]$ and it attains CRLB
References taken from Mood Graybill Boes, McGraw Hill