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I read from http://math.arizona.edu/~jwatkins/N_unbiased.pdf

We know the Fisher information is $$I(\theta)=E\bigg[\bigg(\frac{\partial \log f(X) }{\partial \theta}\bigg)^2\bigg]. $$

Cramer-Rao lower bound is $\frac{(h'(\theta))^2}{I(\theta)}$ where $h'(\theta)=E\bigg[d(X) \frac{\partial \log f(X) }{\partial \theta}\bigg]. $

Then in page 214, example 14.15 considered the independent Bernoulli random variables with unknown success probability $\theta$. So density function is $\theta^x (1-\theta)^{1-x}$. In this case $I(\theta)=\frac{1}{\theta\left(1-\theta\right)}$.

Now if we take $d(X)=\bar X,$ then $h'(\theta)=E\bigg[\bar X \frac{\partial \log f(X) }{\partial \theta}\bigg]$. Then how to calculate $h'(\theta)$?

It seems $E\bigg[\bar X \frac{\partial \log f(\bar X) }{\partial \theta}\bigg]= E(\bar X \frac{\bar X -\theta}{\theta (1-\theta)})= \frac{1}{\theta(1-\theta)} E( \bar X ^2 - \theta \bar X)= \frac{1}{\theta (1-\theta)}Var (\bar X)= \frac{1}{n}$.

We know $Var(\bar X)=\frac{\theta(1-\theta)}{n}$. Also $I(\theta)=\frac{1}{\theta (1-\theta)}$. If $h'(\theta)=1$ then $Var(\bar X) < \frac{(h'(\theta))^2}{I(\theta)},$ contradicts the CR bound.

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Your question seems not clear to me. How you get the last equation$E\bigg[\bar X \frac{\partial \ log f(\bar X) }{\partial \theta}\bigg]=\frac{\theta(1-\theta)}{n}$ without knowing the distribution $X$?

From $(14.12)$

$$h(\theta)=E_\theta [d(X)]=\int_{R^n}d(x)f(x|\theta)dx$$ Then

\begin{align*} h'(\theta)=\int_{R^n}d(x)\frac{\partial f(x|\theta)}{\partial \theta}dx&=\int_{R^n}d(x)\frac{\partial f(x|\theta)}{\partial \theta}\frac{f(x|\theta)}{f(x|\theta)}dx\\&=\int_{R^n}d(x)\frac{\partial \log(f(x|\theta))}{\partial \theta}f(x|\theta)dx \tag{1} \end{align*}

i.e. $(1)$ is

$$h'(\theta)=E_\theta\left [d(x)\frac{\partial \log(f(x|\theta))}{\partial \theta}\right] \tag{14.14}$$

Therefore, if you want to calculate $h'(\theta)$ just calculate directly by $(1)$ but you need to know the distribution of $X$


Ok, since you update the question, and $X$ has a Bernoulli distrbituion and your $d(X)$ is the $\bar{X}$

Then why not directly calculte the $h'(\theta)$ by the definiton of $(14.12)$

$$h(\theta)=E_{\theta}[\bar{X}]=n\frac{1}{n}E(X)=\sum_xx(P(X=x))=1*\theta+0*\theta=\theta$$

Therefore,$h'(\theta)=1$

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  • $\begingroup$ Sorry it is coming $\frac{1}{n}$ $\endgroup$ – str Nov 16 '17 at 11:22

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