# How can I find an unbiased estimator for $\frac{1-\theta}{\theta}$ to obtain this quantity's UMVUE?

Let $$X_1,\cdots,X_n$$ be a random sample from $$f(x|\theta)=\theta(1-\theta)^x,x=0,1,\cdots; 0 < \theta <1$$ is unknown. Find the UMVUE of $$\frac{1-\theta}{\theta}$$.

My work:

I know that I should apply the Lehmann-Scheffe Theorem here, so I will need an unbiased estimator of $$\frac{1-\theta}{\theta}$$ and a complete sufficient statistic for $$\theta$$. I know that, since this distribution is an exponential family, the complete sufficient statistic is $$\sum^n_{i=1}x_i$$. However, I am having problems finding an unbiased estimator of $$\frac{1-\theta}{\theta}$$. Is there a systematic approach that I can take to finding an unbiased estimator?

• What is $\mathbb{E}x$ equal to? And what is an unbiased estimator of the population mean? Feb 26 '20 at 3:41
• @jbowman I discovered that this distribution is Negative Binomial with $r=1,p=\theta$, so $E(X)=\frac{1-\theta}{\theta}$, which means any $X_i$ is an unbiased estimator for $\frac{1-\theta}{\theta}$. I just don't see how I can find $E(X|\sum X)$ which is the BUE. Feb 28 '20 at 5:54
• Would $X_1$ have a lower variance than ${1 \over n}\sum x_i$ as an estimate of the quantity of interest? You are making this problem harder than it actually is :) Feb 28 '20 at 15:25
• @jbowman I definitely made this problem much harder than it is. I posted my answer below. Does it check out? Feb 29 '20 at 20:46

We recognize that $$f(x|\theta)$$ follows a Negative Binomial distribution with $$r=1,p=\theta$$.
Following this realization, we notice $$E(X)=\frac{1-\theta}{\theta}$$, so $$W(X)=X$$ is unbiased for our quantity, $$\tau(\theta)$$.
Additionally, we see that $$f(x|\theta)$$ is in the exponential family, so $$T(X)=\sum X_i$$ is complete sufficient for $$\theta$$.
By Lehmann-Scheffe Theorem, $$W^*(X)=E[W(X)|T(X)]$$ is the UMVUE for $$\tau(\theta)$$.
Since $$E(X)=\tau(\theta), E(\bar{X})=\tau(\theta)$$. Since $$\bar{X}$$ is a function of the complete sufficient statistic $$T(X)$$, $$\bar{X}$$ is the UMVUE for $$\tau(\theta)$$.