Let$ X_1, ...,X_n$ be a random sample feom a distribution $Geometric(\theta)$ for $0<\theta<1$. I.e,
$$p_{\theta}(x)=\theta(1-\theta)^{x-1} I_{\{1,2,...\}}(x)$$
Find the unbiased estimator with minimum variance for $g(\theta)=\frac{1}{\theta}$
My attempt:
Since the geometric distribution is from the exponential family, the statistics $$\sum X_i $$ is complete and sufficient for $ \theta$. Also, if $$T(X)=X_1$$ is an estimator for $g(\theta)$, it is unbiased. Therefore, by the Rao-Blackwell theorem and Lehmann-Scheffé Theorem, $$W(X) = E[X_1|\sum X_i]$$ is the estimator we are looking for.
We have the following:
$W(X) = \sum_{i=1}^t i\, P(X_1=i|\sum X_i =t) = \sum_{i=1}^t i\, \frac{P(\sum_{i \geq 2} X_i =t-i)P(X_1=i)}{P(\sum_{i \geq 1}X_i =t)}$
Since the variables are iid geometric, the sums distributions are both negative binomials. But i am having troubles tosimplify the binomial coefficients and give a final answer with a better form, if it is possible.I wpuld be glad if I could get some help.
Thanks!
Edit: I dont think you guys understand my doubt: Ithink I made all the correct steps, maybe only forgot some indicator function. Here is what I did:
$$...=\sum_{i=1}^ti\frac{\binom{t-i-1}{n-2}\theta^{n-i}(1-\theta)^{t-i-n+1} \theta(1-\theta)^{i-1}}{\binom{t-1}{n-1}\theta^n(1-\theta)^{t-n}}=\sum_{i=1}^t i \frac{\binom{t-i-1}{n-2}}{\binom{t-1}{n-1}}$$
As i said, I am having troubles to simplify this and with the somatory index