题目
7.设X_(1),X_(2)是从正态总体N(mu,sigma^2)中抽取的样本.hat(mu)_(1)=(2)/(3)X_(1)+(1)/(3)X_(2); hat(mu)_(2)=(1)/(4)X_(1)+(3)/(4)X_(2); hat(mu)_(3)=(1)/(2)X_(1)+(1)/(2)X_(2);试证hat(mu)_(1),hat(mu)_(2),hat(mu)_(3)都是μ的无偏估计量，并求出每一估计量的方差.
7.设$X_{1},X_{2}$是从正态总体$N(\mu,\sigma^{2})$中抽取的样本.
$\hat{\mu}_{1}=\frac{2}{3}X_{1}+\frac{1}{3}X_{2}$; $\hat{\mu}_{2}=\frac{1}{4}X_{1}+\frac{3}{4}X_{2}$; $\hat{\mu}_{3}=\frac{1}{2}X_{1}+\frac{1}{2}X_{2}$;
试证$\hat{\mu}_{1},\hat{\mu}_{2},\hat{\mu}_{3}$都是μ的无偏估计量，并求出每一估计量的方差.
题目解答
答案
**无偏性证明：**
1. $\hat{\mu}_1 = \frac{2}{3}X_1 + \frac{1}{3}X_2$
$E(\hat{\mu}_1) = \frac{2}{3}\mu + \frac{1}{3}\mu = \mu$
2. $\hat{\mu}_2 = \frac{1}{4}X_1 + \frac{3}{4}X_2$
$E(\hat{\mu}_2) = \frac{1}{4}\mu + \frac{3}{4}\mu = \mu$
3. $\hat{\mu}_3 = \frac{1}{2}X_1 + \frac{1}{2}X_2$
$E(\hat{\mu}_3) = \frac{1}{2}\mu + \frac{1}{2}\mu = \mu$
**方差计算：**
1. $D(\hat{\mu}_1) = \left(\frac{2}{3}\right)^2\sigma^2 + \left(\frac{1}{3}\right)^2\sigma^2 = \frac{5}{9}\sigma^2$
2. $D(\hat{\mu}_2) = \left(\frac{1}{4}\right)^2\sigma^2 + \left(\frac{3}{4}\right)^2\sigma^2 = \frac{5}{8}\sigma^2$
3. $D(\hat{\mu}_3) = \left(\frac{1}{2}\right)^2\sigma^2 + \left(\frac{1}{2}\right)^2\sigma^2 = \frac{1}{2}\sigma^2$
**答案：**
\[
\boxed{
\begin{array}{ll}
\text{无偏性:} & E(\hat{\mu}_1) = E(\hat{\mu}_2) = E(\hat{\mu}_3) = \mu \\
\text{方差:} & D(\hat{\mu}_1) = \frac{5}{9}\sigma^2, \quad D(\hat{\mu}_2) = \frac{5}{8}\sigma^2, \quad D(\hat{\mu}_3) = \frac{1}{2}\sigma^2
\end{array}
}
\]
解析
步骤 1：证明$\hat{\mu}_1$是$\mu$的无偏估计量
根据无偏估计量的定义，需要证明$E(\hat{\mu}_1) = \mu$。由于$X_1$和$X_2$是从正态总体$N(\mu,\sigma^{2})$中抽取的样本，所以$E(X_1) = E(X_2) = \mu$。因此，$E(\hat{\mu}_1) = E(\frac{2}{3}X_1 + \frac{1}{3}X_2) = \frac{2}{3}E(X_1) + \frac{1}{3}E(X_2) = \frac{2}{3}\mu + \frac{1}{3}\mu = \mu$。
步骤 2：证明$\hat{\mu}_2$是$\mu$的无偏估计量
同理，$E(\hat{\mu}_2) = E(\frac{1}{4}X_1 + \frac{3}{4}X_2) = \frac{1}{4}E(X_1) + \frac{3}{4}E(X_2) = \frac{1}{4}\mu + \frac{3}{4}\mu = \mu$。
步骤 3：证明$\hat{\mu}_3$是$\mu$的无偏估计量
同理，$E(\hat{\mu}_3) = E(\frac{1}{2}X_1 + \frac{1}{2}X_2) = \frac{1}{2}E(X_1) + \frac{1}{2}E(X_2) = \frac{1}{2}\mu + \frac{1}{2}\mu = \mu$。
步骤 4：计算$\hat{\mu}_1$的方差
$D(\hat{\mu}_1) = D(\frac{2}{3}X_1 + \frac{1}{3}X_2) = (\frac{2}{3})^2D(X_1) + (\frac{1}{3})^2D(X_2) = \frac{4}{9}\sigma^2 + \frac{1}{9}\sigma^2 = \frac{5}{9}\sigma^2$。
步骤 5：计算$\hat{\mu}_2$的方差
$D(\hat{\mu}_2) = D(\frac{1}{4}X_1 + \frac{3}{4}X_2) = (\frac{1}{4})^2D(X_1) + (\frac{3}{4})^2D(X_2) = \frac{1}{16}\sigma^2 + \frac{9}{16}\sigma^2 = \frac{5}{8}\sigma^2$。
步骤 6：计算$\hat{\mu}_3$的方差
$D(\hat{\mu}_3) = D(\frac{1}{2}X_1 + \frac{1}{2}X_2) = (\frac{1}{2})^2D(X_1) + (\frac{1}{2})^2D(X_2) = \frac{1}{4}\sigma^2 + \frac{1}{4}\sigma^2 = \frac{1}{2}\sigma^2$。
根据无偏估计量的定义，需要证明$E(\hat{\mu}_1) = \mu$。由于$X_1$和$X_2$是从正态总体$N(\mu,\sigma^{2})$中抽取的样本，所以$E(X_1) = E(X_2) = \mu$。因此，$E(\hat{\mu}_1) = E(\frac{2}{3}X_1 + \frac{1}{3}X_2) = \frac{2}{3}E(X_1) + \frac{1}{3}E(X_2) = \frac{2}{3}\mu + \frac{1}{3}\mu = \mu$。
步骤 2：证明$\hat{\mu}_2$是$\mu$的无偏估计量
同理，$E(\hat{\mu}_2) = E(\frac{1}{4}X_1 + \frac{3}{4}X_2) = \frac{1}{4}E(X_1) + \frac{3}{4}E(X_2) = \frac{1}{4}\mu + \frac{3}{4}\mu = \mu$。
步骤 3：证明$\hat{\mu}_3$是$\mu$的无偏估计量
同理，$E(\hat{\mu}_3) = E(\frac{1}{2}X_1 + \frac{1}{2}X_2) = \frac{1}{2}E(X_1) + \frac{1}{2}E(X_2) = \frac{1}{2}\mu + \frac{1}{2}\mu = \mu$。
步骤 4：计算$\hat{\mu}_1$的方差
$D(\hat{\mu}_1) = D(\frac{2}{3}X_1 + \frac{1}{3}X_2) = (\frac{2}{3})^2D(X_1) + (\frac{1}{3})^2D(X_2) = \frac{4}{9}\sigma^2 + \frac{1}{9}\sigma^2 = \frac{5}{9}\sigma^2$。
步骤 5：计算$\hat{\mu}_2$的方差
$D(\hat{\mu}_2) = D(\frac{1}{4}X_1 + \frac{3}{4}X_2) = (\frac{1}{4})^2D(X_1) + (\frac{3}{4})^2D(X_2) = \frac{1}{16}\sigma^2 + \frac{9}{16}\sigma^2 = \frac{5}{8}\sigma^2$。
步骤 6：计算$\hat{\mu}_3$的方差
$D(\hat{\mu}_3) = D(\frac{1}{2}X_1 + \frac{1}{2}X_2) = (\frac{1}{2})^2D(X_1) + (\frac{1}{2})^2D(X_2) = \frac{1}{4}\sigma^2 + \frac{1}{4}\sigma^2 = \frac{1}{2}\sigma^2$。