Discussion
Let \(X_{1}\), \(X_{2}\), \(X_{3}\) be iid r.v.s with \(U(\theta, \theta^{2})\); \(\theta \gt 1\). The maximum likelihood estimator (mle) of \(\theta\) is :
where :
\(X_{(1)} = \text{Min}(X_{1}, X_{2}, X_{3})\)
\(X_{(3)} = \text{Max}(X_{1}, X_{2}, X_{3})\).
Answer: B
The MLE is \((X_{(3)})^{1/2}\).
The likelihood is \((\theta^{2} - \theta)^{-3}\) when every observation lies in \([\theta, \theta^{2}]\), and 0 otherwise. That requires \(\theta \le X_{(1)}\) and \(\theta^{2} \ge X_{(3)}\), so
\[ \sqrt{X_{(3)}} \le \theta \le X_{(1)} \]
For \(\theta \gt 1\), \(\theta^{2} - \theta\) is increasing, so the likelihood decreases in \(\theta\) and is largest at the smallest admissible value, \(\sqrt{X_{(3)}}\).
\(X_{(1)}^{2}\) and \(X_{(1)}^{2} + X_{(3)}\) are not even in the admissible interval in general, and the weighted average in D is a different estimator, not the maximiser.
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