I am having trouble calculating the Cramer-Rao lower bound for an unbiased estimator $\theta$ for the following function:
$$f(x|\theta) = \theta^{2} x e^{-\theta x} , x>0$$
I have already calculated the maximum likelihood estimation which is $$\widehat{\theta} = \frac{2}{\overline{X}}.$$
I have been using the following equation to calculate the Fisher Information:
$$I_{n}(\theta)=E[(\dfrac{\partial }{\partial \theta}\ln f(X|\theta))^{2}]$$
However, I cannot move forward past this point, since whenever I tried to calculate the Fisher Information $I(\theta)$.
$$\ln f(X|\theta) =2\ln (\theta)+\ln(x)-(\theta x)$$
$$\dfrac{\partial}{\partial \theta} \ln f(X|\theta)= \frac{2}{\theta}-x $$
$$(\dfrac{\partial}{\partial \theta} \ln f(X|\theta))^{2}= \frac{4}{\theta^{2}}-\frac{4x}{\theta}+x^{2}$$
When I calculate the expected value of this equation it does not equal to the solution that I was given for the problem, which is $\frac{\theta^{2}}{2n}$.
Best Answer
You are considering this lower bound $$ \frac {1}{n\int_{0}^{\infty}\left( \frac{\partial \ln f(x|\theta)}{\partial \,\theta}\right)^2f(x|\theta)\,dx}.$$
Can you use this lower bound instead? $$ \frac {-1}{n\int_{0}^{\infty}\left( \frac{{\partial}^2 \ln f(x|\theta)}{\partial \,{\theta}^2}\right)f(x|\theta)\,dx}.$$
For $\theta>0$.