# Find a function of $\theta$ so that there exists an unbiased estimator and the variance coincides with Cramér-Rao lower bound

Let $$X_1,\dots, X_n$$ be a random sample from the geometric distribution $$P(X=x)=\theta(1-\theta)^x$$ for $$x=0,1,2,\dots$$ where $$0<\theta<1$$.

Find a function of $$\theta$$, say $$\tau=h(\theta)$$ so that there exists an unbiased estimator $$\hat{\tau}$$ of $$\tau$$ and the variance of $$\hat{\tau}$$ coincides with Cramér-Rao lower bound.

My work:

I first obtain the Cramér-Rao lower bound of the variance of $$\hat{\tau}$$: $$\hat{\tau}\ge \frac{(\hat{\tau}')^2}{nI(\theta)}=\frac{(\hat{\tau}'\theta^2(1-\theta))^2}{n}$$ where $$I(\theta)=-E\left[\frac{\partial^2}{\partial \theta^2}\log f(x;\theta)\right]=E\left[\frac{1}{\theta^2}-\frac{x}{(1-\theta)^2}\right]=\frac{1}{\theta^2(1-\theta)}$$

Also, since $$\hat{\tau}$$ is unbiased, then $$E[\hat{\tau}]=\tau$$. But I have no idea how to find such function $$h(\cdot)$$?

• Please add the self-study tag. Mar 15 at 8:57

When $$X_i\overset{\textrm{iid}}{\sim}f(x\mid\theta),$$ and $$\hat\tau(\mathbf x)$$ is an unbiased estimator of $$\tau(\theta),$$ then the unbiased estimator attains the CRLB if and only if there exists some function of $$\theta,~a(\theta)$$(say) such that $$a(\theta)\left[\hat{\tau}(\mathbf x) -\tau(\theta)\right]=\partial_\theta\ln\mathcal L(\theta\mid\mathbf x) \tag 1.\label 1$$
Here $$X_i\sim\textrm{Geom}(\theta).$$ So, \begin{align}\partial_\theta\ln\mathcal L(\theta\mid\mathbf x) &= \partial_\theta[n\ln\theta+\sum x_i\ln(1-\theta)]\\&=\frac n\theta-\frac{\sum x_i}{1-\theta}\\&= \frac{\theta-1}n\left(\frac{\theta-1}{\theta}+\bar x\right).\tag 2\label 2\end{align}
Taking $$a(\theta) :=\frac{\theta-1}n, ~\tau(\theta) :=\frac{1-\theta}\theta$$ in $$\eqref 2,$$ from $$\eqref 1,$$ it can be concluded that $$\bar x$$ (an unbiased estimator of $$\tau(\theta)$$) attains the CRLB.
$$\rm [I]$$ Statistical Inference, George Casella, Roger L. Berger, Wadsworth, $$2002,$$ sec. $$7.3,$$ p. $$341.$$