Question

Let n X , X ,..., X 1 2 be a random sample of size n > 2 from a distribution with p.d.f

f (x;θ ) =θ x^θ-1 , 0 < x < 1, θ > 0

Find Cramer-Rao lower bound(CRLB)

Answer 1

The joint probability distribution is given by,

$L=\prod_{i=1}^{n}f_\theta \left ( x_i \right )=\theta ^n\left ( \prod_{i=1}^{n} x_i\right )^{\theta - 1}$

Taking logarithm in both sides we get,

$log L=n log\theta+\left ( \theta-1 \right )\sum_{i=1}^{n}logx_i$

Differentiating with respect of $\theta$ we get,

$\frac{\partial logL}{\partial \theta}=\frac{n}{\theta}+\sum_{i=1}^{n}logx_i$

To get maximum likelihood value of $\theta$ we equate right side to zero.

Solving we get,

   $\hat {\theta}=\frac{-n}{\sum_{i=1}^{n}logx_i}$

We know MLE follows invariant property.

So, required MLE of 1/$\theta$ is given by

  $\frac{1}{\hat {\theta}}=-\frac{\sum_{i=1}^{n}logx_i}{n}$

Cramer-Rao lower bound-

Cramer-Rao lower bound is given by

   $\frac{\left [ g'\left ( \theta \right ) \right ]^2}{I\left ( \theta \right )}$

where

   $g\left ( \theta \right )=\frac{1}{\theta}$

and

$I\left ( \theta \right )=-n\left [ \frac{\partial^2 }{\partial \theta^2}logf_\theta(x) \right ]=\frac{1}{\theta^2}$

Also, we get,

$g'\left ( \theta \right )=-\frac{1}{\theta^2}$

So, CRLB is given by $1/\theta^2$.