Unbiased estimator

Question

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 $\theta^2$.

Answer 1

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.