UMVUE: either lower bound or variance has problem

Question

For $f(x) = θ(1-θ)^x$; I get $E(x)=(1-θ)/θ$ and variance $= (1-θ)/nθ^2$. For lower bound I get $(θ^2)*(1-θ)/n$ which is not equal to the variance.

Where is the mistake?

Answer 1

I can guess that you meant $g_n = \bar{X}_n$ as an estimator of $g(\theta) = (1-\theta)/\theta$, where $X =0,1,2,...$ As such, for finding the Cramer-Rao lower bound you have to find Fisher's information $$ \ln f(x;\theta)= \ln \theta+x\ln(1-\theta), $$ hence $$ I_X (\theta) = - \mathbb{E} ( \frac{\partial ^2}{\partial\theta^2} \ln f(x;\theta))= \frac{1}{\theta^2} +\frac{1}{(1-\theta)^2}\mathbb{E}X= \frac{1}{\theta^2} +\frac{1}{(1-\theta)\theta} = \frac{1}{\theta^2 ( 1 - \theta)}. $$ Now recall that for $I_{X}(g(\theta))$ you should use $$ I_{X}(g(\theta)) = (g'(\theta)^{-2} I_X(\theta) = \theta^4\frac{1}{\theta^2 ( 1 - \theta)} = \frac{\theta^2}{(1-\theta)}. $$

Thus the CRLB for unbiased estimators of $g(\theta)$ is $$ \frac{1}{nI_X(g( \theta) )} = \frac{ ( 1 - \theta)}{\theta^2n}. $$ And the variance of $\bar{X}_n$ is $$ Var(\bar{X}_n) = \frac{var(X)}{n} = \frac{(1-\theta)}{n \theta^2} = CRLB. $$