10.(单选题,4.0分)-|||-设随机变量X1,X2,X3是来自总体X的样本,下列关于E(X)的无偏估计中,最-|||-有效的是 ()-|||-A .dfrac (1)(2)((X)_(1)+(X)_(2))-|||-B .dfrac (1)(4)(X)_(1)+dfrac (1)(2)(X)_(2)+dfrac (1)(4)(X)_(3) (-|||-C .dfrac (1)(3)((X)_(1)+(X)_(2)+(X)_(3))-|||-D .dfrac (1)(2)(X)_(1)+dfrac (1)(3)(X)_(2)+dfrac (1)(6)(X)_(3)

步骤 1：无偏估计
无偏估计是指估计量的期望值等于被估计参数的真实值。对于E(X)的无偏估计，需要满足E(估计量) = E(X)。对于选项A、B、C、D，分别计算其期望值。
步骤 2：计算期望值
A. $\dfrac {1}{2}({X}_{1}+{X}_{2})$ 的期望值为 $\dfrac {1}{2}(E(X_{1})+E(X_{2})) = E(X)$
B. $\dfrac {1}{4}{X}_{1}+\dfrac {1}{2}{X}_{2}+\dfrac {1}{4}{X}_{3}$ 的期望值为 $\dfrac {1}{4}E(X_{1})+\dfrac {1}{2}E(X_{2})+\dfrac {1}{4}E(X_{3}) = E(X)$
C. $\dfrac {1}{3}({X}_{1}+{X}_{2}+{X}_{3})$ 的期望值为 $\dfrac {1}{3}(E(X_{1})+E(X_{2})+E(X_{3})) = E(X)$
D. $\dfrac {1}{2}{X}_{1}+\dfrac {1}{3}{X}_{2}+\dfrac {1}{6}{X}_{3}$ 的期望值为 $\dfrac {1}{2}E(X_{1})+\dfrac {1}{3}E(X_{2})+\dfrac {1}{6}E(X_{3}) = E(X)$
步骤 3：计算方差
无偏估计的有效性可以通过估计量的方差来衡量，方差越小，估计量越有效。对于选项A、B、C、D，分别计算其方差。
A. $\dfrac {1}{2}({X}_{1}+{X}_{2})$ 的方差为 $\dfrac {1}{4}(Var(X_{1})+Var(X_{2}))$
B. $\dfrac {1}{4}{X}_{1}+\dfrac {1}{2}{X}_{2}+\dfrac {1}{4}{X}_{3}$ 的方差为 $\dfrac {1}{16}Var(X_{1})+\dfrac {1}{4}Var(X_{2})+\dfrac {1}{16}Var(X_{3})$
C. $\dfrac {1}{3}({X}_{1}+{X}_{2}+{X}_{3})$ 的方差为 $\dfrac {1}{9}(Var(X_{1})+Var(X_{2})+Var(X_{3}))$
D. $\dfrac {1}{2}{X}_{1}+\dfrac {1}{3}{X}_{2}+\dfrac {1}{6}{X}_{3}$ 的方差为 $\dfrac {1}{4}Var(X_{1})+\dfrac {1}{9}Var(X_{2})+\dfrac {1}{36}Var(X_{3})$
步骤 4：比较方差
比较选项A、B、C、D的方差，可以看出选项D的方差最小，因此选项D是最有效的无偏估计。