# Cramer-Rao Casella Berger 7.38 for exponential family

The question states ''let $$X_{1}, \dots, X_{n}$$ be random sample from $$f(x \mid \theta) = \theta\cdot x^{\theta-1}$$ for $$0 < x< 1 ; \theta > 0$$. Is there a function of $$\theta, g(\theta)$$ for which there exists an unbiased estimator of $$\theta$$ whose variance $$\textbf{attains}$$ the Cramèr-Rao lower bound ? if so find it!".

so we have an exponential family, and we can interchange differentiation and integration, so the fisher information term, denominator of the Cramér-Rao lower bound, I calculated as

$$E \left(\dfrac{\partial}{\partial \theta}[\ln(\theta \cdot x^{\theta-1})] \right)^{2} = -n E \left(\dfrac{\partial^{2}}{\partial \theta^{2}}(\ln(\theta x^{\theta-1}\right) \implies \dfrac{\theta^{2}}{n}$$ taking $$\dfrac{1}{I(\theta)}$$

Now just taking a stab the statistic I've used is $$W(X)= \overline{X}$$ which I calculate to be UBE since $$E[X] = \dfrac{\theta}{\theta + 1}$$ and thus E$$[\overline{X}] = \frac{\theta}{\theta + 1}$$ from my calculations (hopefully right).

similarly, if I calculate the variance I get $$\dfrac{\theta}{(\theta + 1)^{2}(\theta + 2)} \geq \dfrac{\theta^{2}}{n}$$ satisfies the lower bound.

If I examine the MLE I find: $$\dfrac{n}{\theta} + \sum_{i}\ln(x_{i}) = 0 \implies \hat{\theta_{MLE}} = \dfrac{-n}{\sum \ln(x_{i})}$$.

the attainment theory states $$$$\begin{split} \frac{n}{\theta} + \sum_{i}\ln(x_{i}) &= a(\theta)[W(\vec{x})-\tau(\theta)]\\ &= n [\frac{ \sum \ln(x_{i})}{n} - \dfrac{-1}{\theta}] \end{split}$$$$ where if W(x) satisfies the above, then it is the best estimate for $$\tau$$. We need to use Rao-Blackwell theorem for W in the above equation to show that $$\dfrac{1}{\hat{\theta_{MLE}}}$$ is the best.

Consider Y = \log(X) now applying the transformation with the Jacobian we have $$$$\begin{split} f_{X}(e^{y}) = \theta e^{y(\theta-1)}\cdot e^{y} = \theta e^{y\theta} = f_{Y}(y) \end{split}$$$$
where E[Y] = $$\frac{1}{\theta}$$ (Which I think is an unbiased estimator ?)

Using the Rao-Blackwell theorem, let $$\theta^{\star} = E[Y_{i} \mid \sum y_{i} = t]$$ where since $$f_{Y}(y)$$ is an exponential family then $$\sum y_{i}$$ can be shown to be sufficient statistic.

then $$$$\begin{split} E[\theta^{\star}] &= E \left[ E[Y_{i} \mid \sum y_{i} = t] \right] \\ &= E\left[ \theta^{\star} \right] \\ &= \frac{1}{\theta^{\star}} \end{split}$$$$

which really fits into the attainment equation written above! so my guess is to let $$W(\vec{x}) = \dfrac{1}{\theta_{\text{MLE}}^{\star}}$$ be the best UBE (?)

• Are you sure about the Fisher information? I end up with a Fisher information of 1 variable given by: $\frac{1}{\theta^2}$. Which implies $\frac{\theta^2}{n}$ as a CR-LB. Also, I ended up with a different variance of $X$ and $\overline X$. I got $\text{Var}(X) = \frac{\theta}{(\theta+1)^2(\theta+2)}$. I'm not sure about an answer on the question yet... I'll have to let it sink in :) May 1 '17 at 2:14
• yes you are correct. will update. May 1 '17 at 2:26
• see the related updates, this is very close to the correct answer I reckon , using the MLE, although now it seems like I have it close. is that the function we are looking for $g(\theta) = \frac{1}{\theta}$? which will invert the last relation? May 1 '17 at 2:50

In case you are still interested...

You could have been working with the equality condition of the Cramer-Rao inequality:

$$\frac{\partial}{\partial\theta}\ln f_{\theta}(x_1,\ldots,x_n)=k(\theta)\left(T(x_1,\ldots,x_n)-g(\theta)\right)\tag{*}$$

If $$(*)$$ holds, then variance of the statistic $$T$$ attains the Cramer-Rao lower bound for $$g(\theta)$$.

Moreover, if $$T$$ is unbiased for $$g(\theta)$$, then $$T$$ is the UMVUE of $$g(\theta)$$.

Here joint density of $$(X_1,\ldots,X_n)$$ is

\begin{align} f_{\theta}(x_1,\ldots,x_n)&=\theta^n\left(\prod_{i=1}^nx_i\right)^{\theta-1}\mathbf1_{0

So,

$$\frac{\partial}{\partial\theta}\ln f_{\theta}(x_1,\ldots,x_n)=\frac{n}{\theta}+\sum_{i=1}^n\ln x_i=-n\left(-\frac{1}{n}\sum_{i=1}^n\ln x_i-\frac{1}{\theta}\right)$$

This implies that variance of $$T=-\frac{1}{n}\sum\limits_{i=1}^n\ln X_i$$ attains the Cramer-Rao lower bound for $$1/\theta$$.

Besides,

\begin{align} E_{\theta}(T)&=-\frac{1}{n}\sum_{i=1}^n E_{\theta}(\ln X_i) \\&=-\frac{1}{n}\sum_{i=1}^n \left(-\frac{1}{\theta}\right)=\frac{1}{\theta} \end{align}

So the function you are looking for is indeed $$g(\theta)=\theta^{-1}$$, but you have worked too hard for the answer.