Discussion
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(U(\theta, \theta + 1)\). If \(X_{(1)} \lt X_{(2)} \lt \cdots \lt X_{(n)}\) are corresponding order statistics, then :
Answer: A
The answer is \((X_{(1)}, X_{(n)})\) is jointly sufficient for \(\theta\).
The joint density is
\[ \begin{aligned} &\prod_{i} 1_{\{\theta \lt x_{i} \lt \theta + 1\}} \\ &= 1_{\{x_{(n)} - 1 \lt \theta \lt x_{(1)}\}} \end{aligned} \]
which depends on the data only through the smallest and largest observations. By the factorisation theorem, \((X_{(1)}, X_{(n)})\) is sufficient (indeed minimal sufficient).
Neither order statistic alone is sufficient, because the likelihood is non-zero on an interval whose two ends come from both of them. Adding the constant 1 to a component does not lose information, so the last option is also false.
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