题目
设总体X的数学期望EX＝μ，方差DX＝σ2，X1，X2，X3，X4是来自总体X的简单随机样本，则下列μ的估计量中最有效的是（）A. dfrac (1)(6)(X)_(1)+dfrac (1)(6)(X)_(2)+dfrac (1)(3)(X)_(3)+dfrac (1)(3)(X)_(3) B. dfrac (1)(6)(X)_(1)+dfrac (1)(6)(X)_(2)+dfrac (1)(3)(X)_(3)+dfrac (1)(3)(X)_(3) C. dfrac (1)(6)(X)_(1)+dfrac (1)(6)(X)_(2)+dfrac (1)(3)(X)_(3)+dfrac (1)(3)(X)_(3) D. dfrac (1)(6)(X)_(1)+dfrac (1)(6)(X)_(2)+dfrac (1)(3)(X)_(3)+dfrac (1)(3)(X)_(3)
设总体X的数学期望EX＝μ，方差DX＝σ2，X1，X2，X3，X4是来自总体X的简单随机样本，则下列μ的估计量中最有效的是（）
A.
B.
C.
D.
题目解答
答案
D. $\dfrac {1}{4}{X}_{1}+\dfrac {1}{4}{X}_{2}+\dfrac {1}{4}{X}_{3}+\dfrac {1}{4}{X}_{4}$
解析
步骤 1：计算每个选项的期望值
对于选项A，期望值为：
$E(\dfrac {1}{6}{X}_{1}+\dfrac {1}{6}{X}_{2}+\dfrac {1}{3}{X}_{3}+\dfrac {1}{3}{X}_{3}) = \dfrac {1}{6}E(X_{1})+\dfrac {1}{6}E(X_{2})+\dfrac {1}{3}E(X_{3})+\dfrac {1}{3}E(X_{3}) = \dfrac {1}{6}\mu+\dfrac {1}{6}\mu+\dfrac {1}{3}\mu+\dfrac {1}{3}\mu = \mu$
对于选项B，期望值为：
$E(\dfrac {1}{3}{X}_{1}+\dfrac {1}{3}{X}_{2}+\dfrac {1}{3}{X}_{3}) = \dfrac {1}{3}E(X_{1})+\dfrac {1}{3}E(X_{2})+\dfrac {1}{3}E(X_{3}) = \dfrac {1}{3}\mu+\dfrac {1}{3}\mu+\dfrac {1}{3}\mu = \mu$
对于选项C，期望值为：
$E({N}_{1},\dfrac {4}{5}{X}_{3},-\dfrac {1}{5}{X}_{4}) = E({N}_{1})+\dfrac {4}{5}E(X_{3})-\dfrac {1}{5}E(X_{4}) = \mu$
对于选项D，期望值为：
$E(\dfrac {1}{4}{X}_{1}+\dfrac {1}{4}{X}_{2}+\dfrac {1}{4}{X}_{3}+\dfrac {1}{4}{X}_{4}) = \dfrac {1}{4}E(X_{1})+\dfrac {1}{4}E(X_{2})+\dfrac {1}{4}E(X_{3})+\dfrac {1}{4}E(X_{4}) = \dfrac {1}{4}\mu+\dfrac {1}{4}\mu+\dfrac {1}{4}\mu+\dfrac {1}{4}\mu = \mu$
步骤 2：计算每个选项的方差
对于选项A，方差为：
$D(\dfrac {1}{6}{X}_{1}+\dfrac {1}{6}{X}_{2}+\dfrac {1}{3}{X}_{3}+\dfrac {1}{3}{X}_{3}) = \dfrac {1}{36}D(X_{1})+\dfrac {1}{36}D(X_{2})+\dfrac {1}{9}D(X_{3})+\dfrac {1}{9}D(X_{3}) = \dfrac {1}{36}\sigma^{2}+\dfrac {1}{36}\sigma^{2}+\dfrac {1}{9}\sigma^{2}+\dfrac {1}{9}\sigma^{2} = \dfrac {1}{6}\sigma^{2}$
对于选项B，方差为：
$D(\dfrac {1}{3}{X}_{1}+\dfrac {1}{3}{X}_{2}+\dfrac {1}{3}{X}_{3}) = \dfrac {1}{9}D(X_{1})+\dfrac {1}{9}D(X_{2})+\dfrac {1}{9}D(X_{3}) = \dfrac {1}{9}\sigma^{2}+\dfrac {1}{9}\sigma^{2}+\dfrac {1}{9}\sigma^{2} = \dfrac {1}{3}\sigma^{2}$
对于选项C，方差为：
$D({N}_{1},\dfrac {4}{5}{X}_{3},-\dfrac {1}{5}{X}_{4}) = D({N}_{1})+\dfrac {16}{25}D(X_{3})+\dfrac {1}{25}D(X_{4}) = \sigma^{2}$
对于选项D，方差为：
$D(\dfrac {1}{4}{X}_{1}+\dfrac {1}{4}{X}_{2}+\dfrac {1}{4}{X}_{3}+\dfrac {1}{4}{X}_{4}) = \dfrac {1}{16}D(X_{1})+\dfrac {1}{16}D(X_{2})+\dfrac {1}{16}D(X_{3})+\dfrac {1}{16}D(X_{4}) = \dfrac {1}{16}\sigma^{2}+\dfrac {1}{16}\sigma^{2}+\dfrac {1}{16}\sigma^{2}+\dfrac {1}{16}\sigma^{2} = \dfrac {1}{4}\sigma^{2}$
步骤 3：比较方差
根据方差的大小，选项D的方差最小，因此选项D是最有效的估计量。
对于选项A，期望值为：
$E(\dfrac {1}{6}{X}_{1}+\dfrac {1}{6}{X}_{2}+\dfrac {1}{3}{X}_{3}+\dfrac {1}{3}{X}_{3}) = \dfrac {1}{6}E(X_{1})+\dfrac {1}{6}E(X_{2})+\dfrac {1}{3}E(X_{3})+\dfrac {1}{3}E(X_{3}) = \dfrac {1}{6}\mu+\dfrac {1}{6}\mu+\dfrac {1}{3}\mu+\dfrac {1}{3}\mu = \mu$
对于选项B，期望值为：
$E(\dfrac {1}{3}{X}_{1}+\dfrac {1}{3}{X}_{2}+\dfrac {1}{3}{X}_{3}) = \dfrac {1}{3}E(X_{1})+\dfrac {1}{3}E(X_{2})+\dfrac {1}{3}E(X_{3}) = \dfrac {1}{3}\mu+\dfrac {1}{3}\mu+\dfrac {1}{3}\mu = \mu$
对于选项C，期望值为：
$E({N}_{1},\dfrac {4}{5}{X}_{3},-\dfrac {1}{5}{X}_{4}) = E({N}_{1})+\dfrac {4}{5}E(X_{3})-\dfrac {1}{5}E(X_{4}) = \mu$
对于选项D，期望值为：
$E(\dfrac {1}{4}{X}_{1}+\dfrac {1}{4}{X}_{2}+\dfrac {1}{4}{X}_{3}+\dfrac {1}{4}{X}_{4}) = \dfrac {1}{4}E(X_{1})+\dfrac {1}{4}E(X_{2})+\dfrac {1}{4}E(X_{3})+\dfrac {1}{4}E(X_{4}) = \dfrac {1}{4}\mu+\dfrac {1}{4}\mu+\dfrac {1}{4}\mu+\dfrac {1}{4}\mu = \mu$
步骤 2：计算每个选项的方差
对于选项A，方差为：
$D(\dfrac {1}{6}{X}_{1}+\dfrac {1}{6}{X}_{2}+\dfrac {1}{3}{X}_{3}+\dfrac {1}{3}{X}_{3}) = \dfrac {1}{36}D(X_{1})+\dfrac {1}{36}D(X_{2})+\dfrac {1}{9}D(X_{3})+\dfrac {1}{9}D(X_{3}) = \dfrac {1}{36}\sigma^{2}+\dfrac {1}{36}\sigma^{2}+\dfrac {1}{9}\sigma^{2}+\dfrac {1}{9}\sigma^{2} = \dfrac {1}{6}\sigma^{2}$
对于选项B，方差为：
$D(\dfrac {1}{3}{X}_{1}+\dfrac {1}{3}{X}_{2}+\dfrac {1}{3}{X}_{3}) = \dfrac {1}{9}D(X_{1})+\dfrac {1}{9}D(X_{2})+\dfrac {1}{9}D(X_{3}) = \dfrac {1}{9}\sigma^{2}+\dfrac {1}{9}\sigma^{2}+\dfrac {1}{9}\sigma^{2} = \dfrac {1}{3}\sigma^{2}$
对于选项C，方差为：
$D({N}_{1},\dfrac {4}{5}{X}_{3},-\dfrac {1}{5}{X}_{4}) = D({N}_{1})+\dfrac {16}{25}D(X_{3})+\dfrac {1}{25}D(X_{4}) = \sigma^{2}$
对于选项D，方差为：
$D(\dfrac {1}{4}{X}_{1}+\dfrac {1}{4}{X}_{2}+\dfrac {1}{4}{X}_{3}+\dfrac {1}{4}{X}_{4}) = \dfrac {1}{16}D(X_{1})+\dfrac {1}{16}D(X_{2})+\dfrac {1}{16}D(X_{3})+\dfrac {1}{16}D(X_{4}) = \dfrac {1}{16}\sigma^{2}+\dfrac {1}{16}\sigma^{2}+\dfrac {1}{16}\sigma^{2}+\dfrac {1}{16}\sigma^{2} = \dfrac {1}{4}\sigma^{2}$
步骤 3：比较方差
根据方差的大小，选项D的方差最小，因此选项D是最有效的估计量。