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 $h(\cdot)$?

  • $\begingroup$ Please add the self-study tag. $\endgroup$
    – Xi'an
    Mar 15 at 8:57

1 Answer 1


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. $341.$


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