Unbiased estimator Let $X_1, X_2, \dots,X_n$ be the random samples from $$f(x,\theta) = \frac{2x}{\theta^2}, \quad 0 < x <  \theta, \; \theta > 0.$$ 
Find the uniformly minimum variance unbiased estimator of $\theta^2$.
 A: The likelihood on the sample is 
$$
   \mathcal{L}(x_1,\ldots,x_n) = \frac{2^n}{\theta^{2n}} \left( \prod_{k=1}^n x_k \right)\left[ \min(x_1,\ldots,x_n) > 0\right] \cdot \left[ \max(x_1,\ldots,x_n) < \theta \right]
$$
where $\left[ \bullet \right]$ is the Iverson bracket. 
The function $\theta^{-2 n}$ is monotonically decreasing function, hence the maximum of the likelihood occurs at $\theta = \max(x_1,x_2, \ldots, x_n)$. Thus $T=\max(X_1,X_2, \ldots, X_n) = X_{n:n}$ is the complete sufficient statistics, as a maximum likelihood estimator. 
It is easy to see that $\delta(x_1,\ldots,x_n) = \frac{2}{n} \left(  x_1^2 + x_2^2 +  \cdots + x_n^2 \right)$ is an unbiased estimator for $\theta^2$ as 
$$
  \mathbb{E}(\delta(X_1,X_2,\ldots,X_n)) = \mathbb{E}(2 X^2) = \theta^2
$$
The MVUE for $\theta^2$ is thus
$$
 \begin{eqnarray}
   \eta\left( X_1,\ldots,X_n\right) &=& \mathbb{E}\left( \delta(X_1,\ldots,X_n) | T \right) \\ &=& \mathbb{E}\left( \frac{2}{n} \sum_{k=1}^n X_{k:n}^2  | T \right)
  \\&=& \frac{2}{n} \sum_{k=1}^{n-1} X_{k:n}^2 + \frac{2}{n} T^2 \\
&=& \delta(X_1,\ldots,X_n) 
\end{eqnarray}
$$
Here $X_{k:n}$ denotes $k$-th out of $n$ order statistics.
