题目

4、设X_(1),X_(2),X_(3),X_(4)是来自总体N(μ,1)的样本，则以下估计量最有效的是（） (A.）hat(mu)_(1)=(1)/(4)X_(1)+(1)/(3)X_(2)+(1)/(4)X_(3)+(1)/(6)X_(4) (B.）hat(mu)_(2)=(1)/(4)X_(1)+(1)/(4)X_(2)+(1)/(4)X_(3)+(1)/(4)X_(4) (C.）hat(mu)_(3)=(1)/(3)X_(1)+(1)/(3)X_(2)+(1)/(3)X_(3) (D.）hat(mu)_(4)=(1)/(2)X_(1)+(1)/(2)X_(2)

4、设$X_{1},X_{2},X_{3},X_{4}$是来自总体N(μ,1)的样本，则以下估计量最有效的是（） (
A.）$\hat{\mu}_{1}=\frac{1}{4}X_{1}+\frac{1}{3}X_{2}+\frac{1}{4}X_{3}+\frac{1}{6}X_{4}$ (
B.）$\hat{\mu}_{2}=\frac{1}{4}X_{1}+\frac{1}{4}X_{2}+\frac{1}{4}X_{3}+\frac{1}{4}X_{4}$ (
C.）$\hat{\mu}_{3}=\frac{1}{3}X_{1}+\frac{1}{3}X_{2}+\frac{1}{3}X_{3}$ (
D.）$\hat{\mu}_{4}=\frac{1}{2}X_{1}+\frac{1}{2}X_{2}$

题目解答

答案

为了确定哪个估计量最有效，我们需要比较它们的方差。一个估计量的方差越小，它就越有效。给定的估计量都是总体均值$\mu$的线性组合，由于$X_1, X_2, X_3, X_4$是来自总体$N(\mu, 1)$的样本，每个$X_i$的方差为1。因此，估计量的方差是$X_i$的系数平方的和。 让我们计算每个估计量的方差： 1. 对于$\hat{\mu}_1 = \frac{1}{4}X_1 + \frac{1}{3}X_2 + \frac{1}{4}X_3 + \frac{1}{6}X_4$： \[ \text{Var}(\hat{\mu}_1) = \left(\frac{1}{4}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{6}\right)^2 = \frac{1}{16} + \frac{1}{9} + \frac{1}{16} + \frac{1}{36} = \frac{9}{144} + \frac{16}{144} + \frac{9}{144} + \frac{4}{144} = \frac{38}{144} = \frac{19}{72} \] 2. 对于$\hat{\mu}_2 = \frac{1}{4}X_1 + \frac{1}{4}X_2 + \frac{1}{4}X_3 + \frac{1}{4}X_4$： \[ \text{Var}(\hat{\mu}_2) = \left(\frac{1}{4}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{4}\right)^2 = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \] 3. 对于$\hat{\mu}_3 = \frac{1}{3}X_1 + \frac{1}{3}X_2 + \frac{1}{3}X_3$： \[ \text{Var}(\hat{\mu}_3) = \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 = 3 \times \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \] 4. 对于$\hat{\mu}_4 = \frac{1}{2}X_1 + \frac{1}{2}X_2$： \[ \text{Var}(\hat{\mu}_4) = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = 2 \times \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \] 现在，我们比较方差： - $\text{Var}(\hat{\mu}_1) = \frac{19}{72} \approx 0.2639$ - $\text{Var}(\hat{\mu}_2) = \frac{1}{4} = 0.25$ - $\text{Var}(\hat{\mu}_3) = \frac{1}{3} \approx 0.3333$ - $\text{Var}(\hat{\mu}_4) = \frac{1}{2} = 0.5$ 最小的方差是$\frac{1}{4}$，这对应于估计量$\hat{\mu}_2$。 因此，最有效的估计量是$\boxed{B}$。

解析

本题考查的知识点是估计量有效性的判断，解题思路是通过比较各估计量的方差大小来确定最有效的估计量，方差越小的估计量越有效。

已知$X_{1},X_{2},X_{3},X_{4}$是来自总体$N(\mu,1)$的样本，即$Var(X_i)=1$，$i = 1,2,3,4$。对于线性估计量$\hat{\theta}=\sum_{i = 1}^{n}a_iX_i$，其方差$Var(\hat{\theta})=\sum_{i = 1}^{n}a_i^2Var(X_i)$，由于$Var(X_i)=1$，所以$Var(\hat{\theta})=\sum_{i = 1}^{n}a_i^2$。下面分别计算各估计量的方差：

  1. 对于$\hat{\mu}_{1}=\frac{1}{4}X_{1}+\frac{1}{3}X_{2}+\frac{1}{4}X_{3}+\frac{1}{6}X_{4}$：
    根据上述方差计算公式可得：
    $\begin{align*}Var(\hat{\mu}_{1})&=(\frac{1}{4})^2+(\frac{1}{3})^2+(\frac{1}{4})^2+(\frac{1}{6})^2\\&=\frac{1}{16}+\frac{1}{9}+\frac{1}{16}+\frac{1}{36}\\&=\frac{9}{144}+\frac{16}{144}+\frac{9}{144}+\frac{4}{144}\\&=\frac{9 + 16+9 + 4}{144}\\&=\frac{38}{144}\\&=\frac{19}{72}\end{align*}$
  2. 对于$\hat{\mu}_{2}=\frac{1}{4}X_{1}+\frac{1}{4}X_{2}+\frac{1}{4}X_{3}+\frac{1}{4}X_{4}$：
    $\begin{align*}Var(\hat{\mu}_{2})&=(\frac{1}{4})^2+(\frac{1}{4})^2+(\frac{1}{4})^2+(\frac{1}{4})^2\\&=4\times\frac{1}{16}\\&=\frac{4}{16}\\&=\frac{1}{4}\end{align*}$
  3. 对于$\hat{\mu}_{3}=\frac{1}{3}X_{1}+\frac{1}{3}X_{2}+\frac{1}{3}X_{3}$：
    $\begin{align*}Var(\hat{\mu}_{3})&=(\frac{1}{3})^2+(\frac{1}{3})^2+(\frac{1}{3})^2\\&=3\times\frac{1}{9}\\&=\frac{3}{9}\\&=\frac{1}{3}\end{align*}$
  4. 对于$\hat{\mu}_{4}=\frac{1}{2}X_{1}+\frac{1}{2}X_{2}$：
    $\begin{align*}Var(\hat{\mu}_{4})&=(\frac{1}{2})^2+(\frac{1}{2})^2\\&=2\times\frac{1}{4}\\&=\frac{2}{4}\\&=\frac{1}{2}\end{align*}$
    为了方便比较大小，将各方差值化为小数：
  • $Var(\hat{\mu}_{1})=\frac{19}{72}\approx0.2639$
  • $Var(\hat{\mu}_{2})=\frac{1}{4}=0.25$
  • $Var(\hat{\mu}_{3})=\frac{1}{3}\approx0.3333$
  • $Var(\hat{\mu}_{4})=\frac{1}{2}=0.5$

比较可得$0.25<0.2639<0.3333<0.5$，即$Var(\hat{\mu}_{2})$最小。