Vrftsjtryk

The question states ''let $X_{1},...X_{n}$ be random sample from $f(x|theta)=thetacdot x^{theta-1}$ 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 cramer-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 Cramer-Rao lower bound, I calculated as

$E(frac{partial}{partial theta}[Ln(theta cdot x^{theta-1})])^{2} = -n*E(frac{partial^{2}}{partial theta^{2}}(Ln(theta x^{theta-1}) implies frac{theta^{2}}{n}$ taking $frac{1}{I(theta)}$

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

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

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

the attainment theory states
begin{equation}
begin{split}
frac{n}{theta} + sum_{i}Ln(x_{i}) &= a(theta)[W(vec{x})-tau(theta)]\
&=n [frac{ sum Ln(x_{i})}{n} - frac{-1}{theta}]
end{split}
end{equation}
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 $frac{1}{hat{theta_{MLE}}}$ is the best.

consider Y=log(X) now applying the transformation with the jacobian we have
begin{equation}
begin{split}
f_{X}(e^{y})= theta e^{y(theta-1)}cdot e^{y} = theta e^{ytheta} = f_{Y}(y)
end{split}
end{equation}

where EY = $frac{1}{theta}$ (Which I think is an unbiased estimator ? )

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

then
begin{equation}
begin{split}
E[theta^{star}] &= E Big [ E[Y_{i} | sum y_{i} = t] Big ] \
&= E[theta^{star}] \
&= frac{1}{theta^{star}}
end{split}
end{equation}

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

In case you are still interested...

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

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

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^nleft(prod_{i=1}^nx_iright)^{theta-1}mathbf1_{0<x_1,ldots,x_n<1}
\implies ln f_{theta}(x_1,ldots,x_n)&=nlntheta+(theta-1)sum_{i=1}^nln x_i+ln(mathbf1_{0<x_{(1)},x_{(n)}<1})
end{align}

So,

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

This implies that variance of $T=-frac{1}{n}sum_{i=1}^nln X_i$ 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.