题目

6.设总体Xsim N(mu,sigma^2)，其中mu未知，X_(1),X_(2),X_(3),X_(4)为来自总体X的一个样本，则以下关于mu的4个无偏估计量中，哪一个最有效?hat(mu)_(1)=(1)/(4)(X_(1)+X_(2)+X_(3)+X_(4)),hat(mu)_(2)=(1)/(5)X_(1)+(1)/(5)X_(2)+(1)/(5)X_(3)+(2)/(5)X_(4),hat(mu)_(3)=(1)/(6)X_(1)+(2)/(6)X_(2)+(2)/(6)X_(3)+(1)/(6)X_(4),hat(mu)_(4)=(1)/(7)X_(1)+(2)/(7)X_(2)+(3)/(7)X_(3)+(1)/(7)X_(4).

6.设总体$X\sim N(\mu,\sigma^{2})$，其中$\mu$未知，$X_{1},X_{2},X_{3},X_{4}$为来自总体X的一个样本，则以下关于$\mu$的4个无偏估计量中，哪一个最有效? $\hat{\mu}_{1}=\frac{1}{4}(X_{1}+X_{2}+X_{3}+X_{4}),\hat{\mu}_{2}=\frac{1}{5}X_{1}+\frac{1}{5}X_{2}+\frac{1}{5}X_{3}+\frac{2}{5}X_{4},$ $\hat{\mu}_{3}=\frac{1}{6}X_{1}+\frac{2}{6}X_{2}+\frac{2}{6}X_{3}+\frac{1}{6}X_{4},\hat{\mu}_{4}=\frac{1}{7}X_{1}+\frac{2}{7}X_{2}+\frac{3}{7}X_{3}+\frac{1}{7}X_{4}.$

题目解答

答案

计算各估计量的方差： 1. $\hat{\mu}_1 = \frac{1}{4}\sum_{i=1}^4 X_i$，方差 $D(\hat{\mu}_1) = \frac{\sigma^2}{4}$。 2. $\hat{\mu}_2 = \frac{1}{5}X_1 + \frac{1}{5}X_2 + \frac{1}{5}X_3 + \frac{2}{5}X_4$，方差 $D(\hat{\mu}_2) = \frac{7}{25}\sigma^2$。 3. $\hat{\mu}_3 = \frac{1}{6}X_1 + \frac{2}{6}X_2 + \frac{2}{6}X_3 + \frac{1}{6}X_4$，方差 $D(\hat{\mu}_3) = \frac{5}{18}\sigma^2$。 4. $\hat{\mu}_4 = \frac{1}{7}X_1 + \frac{2}{7}X_2 + \frac{3}{7}X_3 + \frac{1}{7}X_4$，方差 $D(\hat{\mu}_4) = \frac{15}{49}\sigma^2$。 比较方差： - $D(\hat{\mu}_1) = 0.25\sigma^2$ - $D(\hat{\mu}_2) \approx 0.28\sigma^2$ - $D(\hat{\mu}_3) \approx 0.2778\sigma^2$ - $D(\hat{\mu}_4) \approx 0.3061\sigma^2$ **答案：** $\boxed{\hat{\mu}_1}$

解析

本题考查无偏估计量有效性的判断，解题的关键在于明确无偏偏估计量有效性的定义，即方差越小的无偏估计量越有效。因此，我们需要分别计算出四个关于$\mu$的无偏估计量的方差，然后比较它们的大小。

1. 计算$\hat{\mu}_{1}$的方差

已知$\hat{\mu}_{1}=\frac{1}{4}(X_{1}+X_{2}+X_{3}+X_{4})$，根据方差的性质$D(aX)=a^{^{2D(X)$以及$D(X + Y)=D(X)+D(Y)$（$X$与$Y$相互独立），因为$X_{1},X_{2},X_{3},X_{4}$相互独立且都服从$N(\mu,\sigma^{2})$，所以有：
$\begin{align*}D(\hat{\mu}_{1})&=D\left[\frac{1}{4}(X_{1}+X_{2}+X_{3}+X_{4})]\\&=\frac{1}{4^2}{}D(X_{1}+X_{2}+X_{3}+X_{4})\\&=\frac{1}{16}[D(X_{1}) + D(X_{2}) + D(X_{3}) + D(X_{4})]\\&=\frac{1}{1}(\sigma^{2}+\sigma^{2}+\sigma^{2}+\sigma^{2})\\&=\frac{\sigma^{2}}{4}= 0.25\sigma^{2}\end{align*}$

2. 计算$\hat{\mu}_{2}$的方差

已知$\hat{\mu}_{2}=\frac{1}{5}X_{1}+\frac{1}{5}X_{2}+\frac{1}{5}X_{3}+\frac{2}{5}X_{4}$，同样根据方差的性质可得：
$\[\begin{align*}D(\hat{\mu}_{2})&=D(\frac{1}{5}X_{1}+\frac{1{5}X_{2}+\frac{1}{5}X_{3}+\frac{2}{5}X_{4})\\&=(\frac{1}{5})^2D(X_{1})+(\frac{1}{5})^2D(X_{2})+(\frac{1}{5})^2D(X_{3})+(\frac{2}{5})^2D(X_{4})\\&=\frac{1}{25}\sigma^{2}+\frac{1}{25}\sigma^{2}+\frac{1}{25}\sigma^{2}+\frac{4}{25}\sigma^{2}}\\&=\frac{7}{25}\sigma^{2}=0.28\sigma^{2}\end{align*}$

3. 计算$\hat{\mu}_{3}$的方差

已知$\hat{\mu}_{3}=\frac{1}{6}X_{1}+\frac{2}{6}X_{2}+\frac{2}{6}X_{3}+\frac{1}{6}X_{4}$，根据方差性质计算：
$\begin{align*}D(\hat{\mu}_{3})&=D(\frac{1}{6}X_{1}+\frac{2}{6}X_{2}+\frac{2{6}X_{3}+\frac{1}{6}X_{4})\\&=(\frac{1}{6})^2D(X_{1})+(\frac{2}{6})^2D(X_{2})+(\frac{2}{6})^2D(X_{3)+(\frac{1}{6})^2D(X_{4})\\&=\frac{1}{36}\sigma^{2}+\frac{4}{36}\sigma^{2}+\frac{4}{36}\sigma^{2}+\frac{1}{36}\sigma^{2}\\&=\frac{5}{18}\sigma^{2}\approx0.2778\sigma^{2}\end{align*}\)## 4. 计算$\hat{\mu}_{4}$的方差已知$\hat{\mu}_{4}=\frac{1{7}X_{1}+\frac{2}{7}X_{2}+\frac{3}{7}X_{3}+\frac{1}{7}X_{4}$，根据方差性质计算：\[\begin{align*}D(\hat{\mu}_{4})&=D(\frac{1}{7})^2D(X_{1})+(\frac{2}{7})^2D(X_{2})+(\frac{3}{7})^2D(X_{3})+(\frac{1}{7})^2D(X_{4})\\&=\frac{1}{49}\sigma^{2}+\frac{4}{49}\sigma^{2}+\frac{9}{49}\sigma^{2}+\frac{1}{49}\sigma^{2}}\\&=\frac{15}{49}\sigma^{2}\approx0.3061\sigma^{2}\end{align*}$

5. 比较方差大小

比较$D(\\hat{\mu}_{1}) = 0.25\sigma^{2}$，$D(\hat{\mu}_{2}) = 0.28\sigma^{2}$，$D(\hat{\mu}_{3}) \approx 0.2778\sigma^{2}$，$D(\hat{\mu}_{4}) \approx 0.3061\sigma^{2}$的大小，可得$0.25\sigma^{2}<0.2778\sigma^{2}<0.28\sigma^{2}<0.3061\sigma^{2}$，即$D(\hat{\mu}_{1})$最小。