题目

五、证明题（共10分）1.设X_(1)，X_(2)，X_(3)是来自正态总体N(mu,sigma^2)的3个样本，证明：hat(mu)_(1)=(1)/(2)X_(1)+(1)/(6)X_(2)+(1)/(3)X_(3)，hat(mu)_(2)=(1)/(4)X_(1)+(1)/(2)X_(2)+(1)/(4)X_(3)是mu的无偏估计，并判断哪一个估计更有效.

五、证明题（共10分） 1.设$X_{1}$，$X_{2}$，$X_{3}$是来自正态总体$N(\mu,\sigma^{2})$的3个样本，证明： $\hat{\mu}_{1}=\frac{1}{2}X_{1}+\frac{1}{6}X_{2}+\frac{1}{3}X_{3}$，$\hat{\mu}_{2}=\frac{1}{4}X_{1}+\frac{1}{2}X_{2}+\frac{1}{4}X_{3}$是$\mu$的无偏估计，并判断哪一个估计更有效.

题目解答

答案

证明无偏性：
对于 $\hat{\mu}_1 = \frac{1}{2}X_1 + \frac{1}{6}X_2 + \frac{1}{3}X_3$，
$E(\hat{\mu}_1) = \frac{1}{2}\mu + \frac{1}{6}\mu + \frac{1}{3}\mu = \mu$
对于 $\hat{\mu}_2 = \frac{1}{4}X_1 + \frac{1}{2}X_2 + \frac{1}{4}X_3$，
$E(\hat{\mu}_2) = \frac{1}{4}\mu + \frac{1}{2}\mu + \frac{1}{4}\mu = \mu$
两者均为无偏估计。

比较有效性：
计算方差，
$D(\hat{\mu}_1) = \frac{1}{4}\sigma^2 + \frac{1}{36}\sigma^2 + \frac{1}{9}\sigma^2 = \frac{7}{18}\sigma^2$
$D(\hat{\mu}_2) = \frac{1}{16}\sigma^2 + \frac{1}{4}\sigma^2 + \frac{1}{16}\sigma^2 = \frac{3}{8}\sigma^2$
由于 $D(\hat{\mu}_1) > D(\hat{\mu}_2)$，$\hat{\mu}_2$ 更有效。

结论：
$\hat{\mu}_1$ 和 $\hat{\mu}_2$ 均为 $\mu$ 的无偏估计，且 $\hat{\mu}_2$ 更有效。

$\boxed{\hat{\mu}_2 \text{ 更有效}}$

解析

本题主要考察正态总体样本的无偏估计和有效性的判断。解题思路如下：

  1. 证明无偏性：根据无偏估计的定义，若一个估计量的数学期望等于被估计的参数，则该估计量为无偏估计。对于正态总体$N(\mu,\sigma^{2})$的样本$X_{1}$，$X_{2}$，$X_{3}$，有$E(X_{i}) = \mu$，$i = 1,2,3$。我们需要分别计算$\hat{\mu}_{1}$和$\hat{\mu}_{2}$的数学期望，看是否等于$\mu$。
    • 计算$E(\hat{\mu}_{1})$：
      已知$\hat{\mu}_{1}=\frac{1}{2}X_{1}+\frac{1}{6}X_{2}+\frac{1}{3}X_{3}$，根据数学期望的线性性质$E(aX + bY)=aE(X)+bE(Y)$，可得：
      $E(\hat{\mu}_{1}) = E(\frac{1}{2}X_{1}+\frac{1}{6}X_{2}+\frac{1}{3}X_{3})=\frac{1}{2}E(X_{1})+\frac{1}{6}E(X_{2})+\frac{1}{3}E(X_{3})$
      因为$E(X_{1}) = E(X_{2}) = E(X_{3}) = \mu$，所以$E(\hat{\mu}_{1})=\frac{1}{2}\mu+\frac{1}{6}\mu+\frac{1}{3}\mu$
      通分可得：$\frac{1}{2}\mu+\frac{1}{6}\mu+\frac{1}{3}\mu=\frac{3}{6}\mu+\frac{1}{6}\mu+\frac{2}{6}\mu=\frac{3 + 1+2}{6}\mu=\mu$
    • 计算$E(\hat{\mu}_{2})$：
      已知$\hat{\mu}_{2}=\frac{1}{4}X_{1}+\frac{1}{2}X_{2}+\frac{1}{4}X_{3}$，同理可得：
      $E(\hat{\mu}_{2}) = E(\frac{1}{4}X_{1}+\frac{1}{2}X_{2}+\frac{1}{4}X_{3})=\frac{1}{4}E(X_{1})+\frac{1}{2}E(X_{2})+\frac{1}{4}E(X_{3})$
      因为$E(X_{1}) = E(X_{2}) = E(X_{3}) = \mu$，所以$E(\hat{\mu}_{2})=\frac{1}{4}\mu+\frac{1}{2}\mu+\frac{1}{4}\mu$
      通分可得：$\frac{1}{4}\mu+\frac{1}{2}\mu+\frac{1}{4}\mu=\frac{1}{4}\mu+\frac{2}{4}\mu+\frac{1}{4}\mu=\frac{1 + 2+1}{4}\mu=\mu$
      由于$E(\hat{\mu}_{1}) = E(\hat{\mu}_{2}) = \mu$，所以$\hat{\mu}_{1}$和$\hat{\mu}_{2}$都是$\mu$的无偏估计。
  2. 比较有效性：在无偏估计中，方差越小的估计量越有效。对于正态总体$N(\mu,\sigma^{2})$的样本$X_{1}$，$X_{2}$，$X_{3}$，有$D(X_{i}) = \sigma^{2}$，$i = 1,2,3$，且样本之间相互独立，根据方差的性质$D(aX + bY)=a^{2}D(X)+b^{2}D(Y)$（$X$与$Y$相互独立），分别计算$\hat{\mu}_{1}$和$\hat{\mu}_{2}$的方差。
    • 计算$D(\hat{\mu}_{1})$：
      已知$\hat{\mu}_{1}=\frac{1}{2}X_{1}+\frac{1}{6}X_{2}+\frac{1}{3}X_{3}$，可得：
      $D(\hat{\mu}_{1}) = D(\frac{1}{2}X_{1}+\frac{1}{6}X_{2}+\frac{1}{3}X_{3})=\frac{1}{4}D(X_{1})+\frac{1}{36}D(X_{2})+\frac{1}{9}D(X_{3})$
      因为$D(X_{1}) = D(X_{2}) = D(X_{3}) = \sigma^{2}$，所以$D(\hat{\mu}_{1})=\frac{1}{4}\sigma^{2}+\frac{1}{36}\sigma^{2}+\frac{1}{9}\sigma^{2}$
      通分可得：$\frac{1}{4}\sigma^{2}+\frac{1}{36}\sigma^{2}+\frac{1}{9}\sigma^{2}=\frac{9}{36}\sigma^{2}+\frac{1}{36}\sigma^{2}+\frac{4}{36}\sigma^{2}=\frac{9 + 1+4}{36}\sigma^{2}=\frac{14}{36}\sigma^{2}=\frac{7}{18}\sigma^{2}$
    • 计算$D(\hat{\mu}_{2})$：
      已知$\hat{\mu}_{2}=\frac{1}{4}X_{1}+\frac{1}{2}X_{2}+\frac{1}{4}X_{3}$，可得：
      $D(\hat{\mu}_{2}) = D(\frac{1}{4}X_{1}+\frac{1}{2}X_{2}+\frac{1}{4}X_{3})=\frac{1}{16}D(X_{1})+\frac{1}{4}D(X_{2})+\frac{1}{16}D(X_{3})$
      因为$D(X_{1}) = D(X_{2}) = D(X_{3}) = \sigma^{2}$，所以$D(\hat{\mu}_{2})=\frac{1}{16}\sigma^{2}+\frac{1}{4}\sigma^{2}+\frac{1}{16}\sigma^{2}$
      通分可得：$\frac{1}{16}\sigma^{2}+\frac{1}{4}\sigma^{2}+\frac{1}{16}\sigma^{2}=\frac{1}{16}\sigma^{2}+\frac{4}{16}\sigma^{2}+\frac{1}{16}\sigma^{2}=\frac{1 + 4+1}{16}\sigma^{2}=\frac{6}{16}\sigma^{2}=\frac{3}{8}\sigma^{2}$
      为了比较$D(\hat{\mu}_{1})$和$D(\hat{\mu}_{2})$的大小，对$\frac{7}{18}\sigma^{2}$和$\frac{3}{8}\sigma^{2}$进行通分，$\frac{7}{18}\sigma^{2}=\frac{28}{72}\sigma^{2}$，$\frac{3}{8}\sigma^{2}=\frac{27}{72}\sigma^{2}$
      因为$\frac{28}{72}\sigma^{2}>\frac{27}{72}\sigma^{2}$，即$D(\hat{\mu}_{1}) > D(\hat{\mu}_{2})$，所以$\hat{\mu}_{2}$更有效。