Discussion
Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) are independent and identically distributed random variables each having \(U(-\theta, \theta)\) r.vs. Then the maximum likelihood estimator of \(\theta\) is :
Answer: D
The MLE of \(\theta\) is \(\max\{|X_{1}|, \ldots, |X_{n}|\}\).
The likelihood \(L(\theta) = (2\theta)^{-n}\) for \(\theta \ge \max|x_{i}|\), and 0 otherwise. It decreases in \(\theta\), so it is maximised at the smallest admissible value, \(\hat{\theta} = \max|X_{i}|\).
\(X_{(n)}\) or \(-X_{(1)}\) alone can be smaller than \(\max|X_{i}|\), and then the likelihood at that value is 0 because some observation lies outside \((-\theta, \theta)\). \((X_{(1)} + X_{(n)})/4\) is not even guaranteed to be positive.
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