题目

4、设总体Xsim N(mu,sigma^2)，X_(1),X_(2),X_(3)是来自X的样本，试证：估计量hat(mu)_(1)=(1)/(5)X_(1)+(3)/(10)X_(2)+(1)/(2)X_(3)；hat(mu)_(2)=(1)/(3)X_(1)+(1)/(4)X_(2)+(5)/(12)X_(3)；hat(mu)_(3)=(1)/(3)X_(1)+(1)/(6)X_(2)+(1)/(2)X_(3)都是μ的无偏估计，并指出它们中哪一个最有效.

4、设总体$X\sim N(\mu,\sigma^{2})$，$X_{1},X_{2},X_{3}$是来自X的样本，试证：估计量 $\hat{\mu}_{1}=\frac{1}{5}X_{1}+\frac{3}{10}X_{2}+\frac{1}{2}X_{3}$；$\hat{\mu}_{2}=\frac{1}{3}X_{1}+\frac{1}{4}X_{2}+\frac{5}{12}X_{3}$；$\hat{\mu}_{3}=\frac{1}{3}X_{1}+\frac{1}{6}X_{2}+\frac{1}{2}X_{3}$ 都是μ的无偏估计，并指出它们中哪一个最有效.

题目解答

答案

**答案：** 1. **无偏性证明：** - $\hat{\mu}_1 = \frac{1}{5}X_1 + \frac{3}{10}X_2 + \frac{1}{2}X_3$，系数和为1，故无偏。 - $\hat{\mu}_2 = \frac{1}{3}X_1 + \frac{1}{4}X_2 + \frac{5}{12}X_3$，系数和为1，故无偏。 - $\hat{\mu}_3 = \frac{1}{3}X_1 + \frac{1}{6}X_2 + \frac{1}{2}X_3$，系数和为1，故无偏。 2. **方差计算：** - $D(\hat{\mu}_1) = 0.38\sigma^2$ - $D(\hat{\mu}_2) = \frac{25}{72}\sigma^2 \approx 0.3472\sigma^2$（最小） - $D(\hat{\mu}_3) = \frac{7}{18}\sigma^2 \approx 0.3889\sigma^2$ **结论：** $\boxed{ \begin{array}{ll} \text{无偏估计：} & \hat{\mu}_1, \hat{\mu}_2, \hat{\mu}_3 \\ \text{最有效估计：} & \hat{\mu}_2 \end{array} }$

解析

本题考查正态总体参数的无偏估计和有效性的判断。解题思路是先根据无偏估计的定义证明三个估计量都是 $\mu$ 的无偏估计，然后通过计算它们的方差，比较方差大小来确定最有效的估计量。

无偏性证明

根据无偏估计的定义，若 $E(\hat{\theta}) = \theta$，则称 $\hat{\theta}$ 是 $\theta$ 的无偏估计。

  • 对于 $\hat{\mu}_1$：
    已知总体 $X\sim N(\mu,\sigma^{2})$，样本 $X_{1},X_{2},X_{3}$ 来自总体 $X$，则 $E(X_{i}) = \mu$，$i = 1,2,3$。
    $E(\hat{\mu}_1) = E(\frac{1}{5}X_{1}+\frac{3}{10}X_{2}+\frac{1}{2}X_{3}) = \frac{1}{5}E(X_{1})+\frac{3}{10}E(X_{2})+\frac{1}{2}E(X_{3})$
    $=\frac{1}{5}\mu+\frac{3}{10}\mu+\frac{1}{2}\mu = (\frac{1}{5}+\frac{3}{10}+\frac{1}{2})\mu = (\frac{2 + 3 + 5}{10})\mu = \mu$
    所以 $\hat{\mu}_1$ 是 $\mu$ 的无偏估计。
  • 对于 $\hat{\mu}_2$：
    $E(\hat{\mu}_2) = E(\frac{1}{3}X_{1}+\frac{1}{4}X_{2}+\frac{5}{12}X_{3}) = \frac{1}{3}E(X_{1})+\frac{1}{4}E(X_{2})+\frac{5}{12}E(X_{3})$
    $=\frac{1}{3}\mu+\frac{1}{4}\mu+\frac{5}{12}\mu = (\frac{4 + 3 + 5}{12})\mu = \mu$
    所以 $\hat{\mu}_2$ 是 $\mu$ 的无偏估计。
  • 对于 $\hat{\mu}_3$：
    $E(\hat{\mu}_3) = E(\frac{1}{3}X_{1}+\frac{1}{6}X_{2}+\frac{1}{2}X_{3}) = \frac{1}{3}E(X_{1})+\frac{1}{6}E(X_{2})+\frac{1}{2}E(X_{3})$
    $=\frac{1}{3}\mu+\frac{1}{6}\mu+\frac{1}{2}\mu = (\frac{2 + 1 + 3}{6})\mu = \mu$
    所以 $\hat{\mu}_3$ 是 $\mu$ 的无偏估计。

方差计算

根据方差的性质，若 $X_{1},X_{2},X_{3}$ 相互独立，则 $D(aX_{1}+bX_{2}+cX_{3}) = a^{2}D(X_{1})+b^{2}D(X_{2})+c^{2}D(X_{3})$。
已知 $D(X_{i}) = \sigma^{2}$，$i = 1,2,3$。

  • 对于 $\hat{\mu}_1$：
    $D(\hat{\mu}_1) = D(\frac{1}{5}X_{1}+\frac{3}{10}X_{2}+\frac{1}{2}X_{3}) = (\frac{1}{5})^{2}D(X_{1})+(\frac{3}{10})^{2}D(X_{2})+(\frac{1}{2})^{2}D(X_{3})$
    $=\frac{1}{25}\sigma^{2}+\frac{9}{100}\sigma^{2}+\frac{1}{4}\sigma^{2} = (\frac{4 + 9 + 25}{100})\sigma^{2} = \frac{38}{100}\sigma^{2} = 0.38\sigma^{2}$
  • 对于 $\hat{\mu}_2$：
    $D(\hat{\mu}_2) = D(\frac{1}{3}X_{1}+\frac{1}{4}X_{2}+\frac{5}{12}X_{3}) = (\frac{1}{3})^{2}D(X_{1})+(\frac{1}{4})^{2}D(X_{2})+(\frac{5}{12})^{2}D(X_{3})$
    $=\frac{1}{9}\sigma^{2}+\frac{1}{16}\sigma^{2}+\frac{25}{144}\sigma^{2} = (\frac{16 + 9 + 25}{144})\sigma^{2} = \frac{25}{72}\sigma^{2} \approx 0.3472\sigma^{2}$
  • 对于 $\hat{\mu}_3$：
    $D(\hat{\mu}_3) = D(\frac{1}{3}X_{1}+\frac{1}{6}X_{2}+\frac{1}{2}X_{3}) = (\frac{1}{3})^{2}D(X_{1})+(\frac{1}{6})^{2}D(X_{2})+(\frac{1}{2})^{2}D(X_{3})$
    $=\frac{1}{9}\sigma^{2}+\frac{1}{36}\sigma^{2}+\frac{1}{4}\sigma^{2} = (\frac{4 + 1 + 9}{36})\sigma^{2} = \frac{7}{18}\sigma^{2} \approx 0.3889\sigma^{2}$

比较三个估计量的方差大小：$D(\hat{\mu}_2) < D(\hat{\mu}_1) < D(\hat{\mu}_3)$，方差越小越有效，所以 $\hat{\mu}_2$ 最有效。