Correct answer: (B) (min X_i,max X_i) is sufficient.; (D) [2(n+1)/n]max|X_i| is UMVU for θ.
Explanation
The likelihood depends on M=max|X_i| through θ⁻ⁿ1_(θ>2M), so the extreme pair is sufficient but A omits the absolute values. The pair is not complete: min X_i+max X_i has expectation zero by symmetry and is nondegenerate. The absolute observations are iid uniform(0,θ/2); their maximum is complete sufficient, has mean nθ/[2(n+1)], and yields the unbiased estimator in D. Lehmann–Scheffé makes it UMVU. **B,D**.
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