For $n$ random variables with PDF $f(x;\theta)=\frac{2x}{\theta^2}, 0<x\leq\theta, \theta > 0$ Is $Var(\theta^*)$ equal to $CRLB(\theta)$?

Question

Let $X_1,...,X_n$ be iid random varaibles from the following PDF $$f(x;\theta)=\frac{2x}{\theta^2}, 0<x\leq\theta, \theta > 0$$ Let $\theta^*$ be an unbiased estimator for $\theta$ based on $\hat{\theta}_{ML}$.

Is $Var(\theta^*)$ equal to $CRLB(\theta)$?

My try:

I'm mainly struggling to find $\theta^*$.

First I got that the $\hat{\theta}_{ML}$ is $X_{(n)}$

Then $\theta^*$ must be a function of $X_{(n)}$.

But I get that $E[X_{(n)}]=\frac{x^2}{\theta}$ and I'm not sure on how to calculate $\theta^*$ from it. Any suggestions would be great!

Answer 1

There's a mistake in your computation of $E(X_{(n)})$. Observe that the CDF of $X_i$ is : $$F_X(x) = \int_0^x \frac{2t}{\theta^2}dt = \frac{x^2}{\theta^2}$$ So that the CDF of $X_{(n)}$ is given by $$F_{X_{(n)}} (x) = F_{X}(x)^n = \frac{x^{2n}}{\theta^2}$$ And its density is simply $f_{X_{(n)}}(x) = \frac{2n x^{2n-1}}{\theta^2}$. From there, the expectation is: $$E(X_{(n)}) = \int_0^\theta 2n \cdot x \cdot \frac{x^{2n-1}}{\theta^2} = \frac{2n}{2n+1} \theta$$ You should be able to compute $\theta^*$ from this and determine whether it achieves the Cramér-Rao lower bound.