3、设X1,X2是从正态总体N(A,I)中抽取的一个容量为2的样本,下面4个关于μ的无偏估计-|||-量中最有效的一个是 ()-|||-A、 dfrac (1)(3)(X)_(1)+dfrac (2)(3)(X)_(2) B、 dfrac (1)(4)(X)_(1)+dfrac (3)({A)_(0)}(X)_(3) C、 dfrac (1)(6)(X)_(1)+dfrac (5)(6)(X)_(2) D、 dfrac (1)(2)(X)_(1)+dfrac (1)(2)(X)_(2)

步骤 1：计算每个选项的方差
由于X1和X2是从正态总体N(u,1)中抽取的样本，因此它们的方差为1。我们需要计算每个选项的方差，以确定哪个选项的方差最小，从而是最有效的估计量。
A. $\dfrac {1}{3}{X}_{1}+\dfrac {2}{3}{X}_{2}$
方差为：$D(\dfrac {1}{3}{X}_{1}+\dfrac {2}{3}{X}_{2}) = \dfrac {1}{9}D(X_{1}) + \dfrac {4}{9}D(X_{2}) = \dfrac {1}{9} + \dfrac {4}{9} = \dfrac {5}{9}$
B. $\dfrac {1}{4}{X}_{1}+\dfrac {3}{4}{X}_{3}$
方差为：$D(\dfrac {1}{4}{X}_{1}+\dfrac {3}{4}{X}_{3}) = \dfrac {1}{16}D(X_{1}) + \dfrac {9}{16}D(X_{3}) = \dfrac {1}{16} + \dfrac {9}{16} = \dfrac {5}{8}$
C. $\dfrac {1}{6}{X}_{1}+\dfrac {5}{6}{X}_{2}$
方差为：$D(\dfrac {1}{6}{X}_{1}+\dfrac {5}{6}{X}_{2}) = \dfrac {1}{36}D(X_{1}) + \dfrac {25}{36}D(X_{2}) = \dfrac {1}{36} + \dfrac {25}{36} = \dfrac {13}{18}$
D. $\dfrac {1}{2}{X}_{1}+\dfrac {1}{2}{X}_{2}$
方差为：$D(\dfrac {1}{2}{X}_{1}+\dfrac {1}{2}{X}_{2}) = \dfrac {1}{4}D(X_{1}) + \dfrac {1}{4}D(X_{2}) = \dfrac {1}{4} + \dfrac {1}{4} = \dfrac {1}{2}$
步骤 2：比较方差
比较每个选项的方差，我们可以看到选项D的方差最小，因此是最有效的估计量。