题目

设_(1),(X)_(2),(X)_(3)是来自均值_(1),(X)_(2),(X)_(3)的正态分布总体的样本，现有_(1),(X)_(2),(X)_(3)的3个估计量：_(1),(X)_(2),(X)_(3)，_(1),(X)_(2),(X)_(3)，则（）A._(1),(X)_(2),(X)_(3)是无偏的且_(1),(X)_(2),(X)_(3)比_(1),(X)_(2),(X)_(3)更有效B._(1),(X)_(2),(X)_(3)是无偏的且_(1),(X)_(2),(X)_(3)比_(1),(X)_(2),(X)_(3)更有效C._(1),(X)_(2),(X)_(3)是无偏的且_(1),(X)_(2),(X)_(3)比_(1),(X)_(2),(X)_(3)更有效D._(1),(X)_(2),(X)_(3)是无偏的且_(1),(X)_(2),(X)_(3)比_(1),(X)_(2),(X)_(3)更有效

设 $X_1,X_2,X_3$ 是来自均值 $\mu$ 的正态分布总体的样本，现有 $\mu$ 的3个估计量： $T_1=\frac{1}{4}(X_1+X_2)+\frac{1}{2}X_3,T_2=\frac{1}{5}\left(X_1+2X_2+3X_3\right)$ ， $T_3=\frac{1}{3}(X_1+X_2+X_3)$ ，则（）

A. $T_2,T_3$ 是无偏的且 $T_2$ 比 $T_3$ 更有效

B. $T_2,T_3$ 是无偏的且 $T_3$ 比 $T_2$ 更有效

C. $T_1,T_3$ 是无偏的且 $T_1$ 比 $T_3$ 更有效

D. $T_1,T_3$ 是无偏的且 $T_3$ 比 $T_1$ 更有效

题目解答

答案

$E\left(T_1\right)=E\left[\frac{1}{4}(X_1+X_2)+\frac{1}{2}X_3\right]=\frac{1}{4}\left[E\left(X_1\right)+E\left(X_2\right)\right]+\frac{1}{2}E\left(X_3\right)$ $=\frac{1}{4}\left(\mu+\mu\right)+\frac{1}{2}\mu=\mu$ ， $E\left(T_2\right)=E\left[\frac{1}{5}\left(X_1+2X_2+3X_3\right)\right]=\frac{1}{5}\left[E\left(X_1\right)+2E\left(X_2\right)+3E\left(X_3\right)\right]$ $=\frac{1}{5}\left(\mu+2\mu+3\mu\right)=\frac{6}{5}\mu\ne\mu$ ， $E\left(T_3\right)=E\left[\frac{1}{3}(X_1+X_2+X_3)\right]=\frac{1}{3}\left[E\left(X_1\right)+E\left(X_2\right)+E\left(X_3\right)\right]$ $=\frac{1}{3}\left(\mu+\mu+\mu\right)=\mu$ ，则 $T_1,T_3$ 是无偏估计量， $D\left(T_1\right)=D\left[\frac{1}{4}(X_1+X_2)+\frac{1}{2}X_3\right]=\frac{1}{16}\left[D\left(X_1\right)+D\left(X_2\right)\right]+\frac{1}{4}D\left(X_3\right)$ $=\frac{1}{16}\left(\sigma^2+\sigma^2\right)+\frac{1}{4}\sigma^2=\frac{3}{8}\sigma^2$ ， $D\left(T_3\right)=D\left[\frac{1}{3}(X_1+X_2+X_3)\right]=\frac{1}{9}\left[D\left(X_1\right)+D\left(X_2\right)+D\left(X_3\right)\right]$ $=\frac{1}{9}\left(\sigma^2+\sigma^2+\sigma^2\right)=\frac{1}{3}\sigma^2$ ，则 $D\left(T_3\right)<D\left(T_1\right)$ ，无偏估计量的方差越小越有效，则 $T_3$ 比 $T_1$ 更有效，因此选择D。

解析

步骤 1：计算T1的期望
$E({T}_{1})=E[ \dfrac {1}{4}({X}_{1}+{X}_{2})+\dfrac {1}{2}{X}_{3}] =\dfrac {1}{4}E({X}_{1})+\dfrac {1}{4}E({X}_{2})+\dfrac {1}{2}E({X}_{3})$
由于$E({X}_{1})=E({X}_{2})=E({X}_{3})=3$，所以$E({T}_{1})=\dfrac {1}{4}×3+\dfrac {1}{4}×3+\dfrac {1}{2}×3=3$，因此T1是无偏估计量。

步骤 2：计算T2的期望
$E({T}_{2})=E[ \dfrac {1}{5}({X}_{1}+2{X}_{2}+3{X}_{3})] =\dfrac {1}{5}E({X}_{1})+\dfrac {2}{5}E({X}_{2})+\dfrac {3}{5}E({X}_{3})$
由于$E({X}_{1})=E({X}_{2})=E({X}_{3})=3$，所以$E({T}_{2})=\dfrac {1}{5}×3+\dfrac {2}{5}×3+\dfrac {3}{5}×3=\dfrac {18}{5}\neq 3$，因此T2是有偏估计量。

步骤 3：计算T3的期望
$E({T}_{3})=E[ \dfrac {1}{3}({X}_{1}+{X}_{2}+{X}_{3})] =\dfrac {1}{3}E({X}_{1})+\dfrac {1}{3}E({X}_{2})+\dfrac {1}{3}E({X}_{3})$
由于$E({X}_{1})=E({X}_{2})=E({X}_{3})=3$，所以$E({T}_{3})=\dfrac {1}{3}×3+\dfrac {1}{3}×3+\dfrac {1}{3}×3=3$，因此T3是无偏估计量。

步骤 4：计算T1的方差
$D({T}_{1})=D[ \dfrac {1}{4}({X}_{1}+{X}_{2})+\dfrac {1}{2}{X}_{3}] =\dfrac {1}{16}D({X}_{1})+\dfrac {1}{16}D({X}_{2})+\dfrac {1}{4}D({X}_{3})$
由于$D({X}_{1})=D({X}_{2})=D({X}_{3})={a}^{2}$，所以$D({T}_{1})=\dfrac {1}{16}{a}^{2}+\dfrac {1}{16}{a}^{2}+\dfrac {1}{4}{a}^{2}=\dfrac {3}{8}{a}^{2}$。

步骤 5：计算T3的方差
$D({T}_{3})=D[ \dfrac {1}{3}({X}_{1}+{X}_{2}+{X}_{3})] =\dfrac {1}{9}D({X}_{1})+\dfrac {1}{9}D({X}_{2})+\dfrac {1}{9}D({X}_{3})$
由于$D({X}_{1})=D({X}_{2})=D({X}_{3})={a}^{2}$，所以$D({T}_{3})=\dfrac {1}{9}{a}^{2}+\dfrac {1}{9}{a}^{2}+\dfrac {1}{9}{a}^{2}=\dfrac {1}{3}{a}^{2}$。

步骤 6：比较T1和T3的方差
由于$D({T}_{3})=\dfrac {1}{3}{a}^{2}\lt \dfrac {3}{8}{a}^{2}=D({T}_{1})$，所以T3比T1更有效。