Discussion
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid \(U(0, \theta)\). Define \(T_{1} = \bar{X}\) and \(T_{2} = (X_{(1)} + X_{(n)})/2\). Where \(X_{(1)} = \min(X_{1}, \ldots X_{n})\), \(X_{(n)} = \max(X_{1} \ldots X_{n})\). Then :
Answer: A
The answer is both \(T_{1}\) and \(T_{2}\) are equally efficient when \(n = 2\).
With two observations, \(X_{(1)} + X_{(2)} = X_{1} + X_{2}\), so \(T_{2} = (X_{(1)} + X_{(2)})/2 = \bar{X} = T_{1}\). They are the same statistic and have the same variance.
For larger \(n\) the midrange is far better: \(\mathrm{Var}(T_{1}) = \theta^{2}/(12n)\), while \(\mathrm{Var}(T_{2}) = \theta^{2}/(2(n + 1)(n + 2))\), which falls like \(1/n^{2}\). So the efficiency of \(T_{1}\) relative to \(T_{2}\) is below 1 for \(n \gt 2\) and tends to 0, which rules out (B) and (D), and (C) fails at \(n = 2\), where the ratio is exactly 1.
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