题目
设 X_1, X_2, ..., X_n 是总体 X 的样本，下列估计量是总体均值 mu 的无偏估计量的是(）A. mu_1 = (1)/(2) X_1 - (1)/(4) X_2 + (1)/(3) X_3B. mu_2 = (1)/(6) X_1 + (11)/(12) X_2 + (1)/(12) X_3C. mu_3 = (1)/(3) X_1 + (1)/(2) X_2 + (1)/(6) X_3D. mu_4 = (1)/(2) X_1 + (1)/(2) X_2 + (1)/(2) X_3
设 $X_1, X_2, \cdots, X_n$ 是总体 $X$ 的样本，下列估计量是总体均值 $\mu$ 的无偏估计量的是(）
A. $\mu_1 = \frac{1}{2} X_1 - \frac{1}{4} X_2 + \frac{1}{3} X_3$
B. $\mu_2 = \frac{1}{6} X_1 + \frac{11}{12} X_2 + \frac{1}{12} X_3$
C. $\mu_3 = \frac{1}{3} X_1 + \frac{1}{2} X_2 + \frac{1}{6} X_3$
D. $\mu_4 = \frac{1}{2} X_1 + \frac{1}{2} X_2 + \frac{1}{2} X_3$
题目解答
答案
C. $\mu_3 = \frac{1}{3} X_1 + \frac{1}{2} X_2 + \frac{1}{6} X_3$
解析
步骤 1：计算各选项的期望值
- **选项A**：$E(\hat{\mu}_1) = \left(\frac{1}{2} - \frac{1}{4} + \frac{1}{3}\right)\mu = \frac{7}{12}\mu \neq \mu$
- **选项B**：$E(\hat{\mu}_2) = \left(\frac{1}{6} + \frac{11}{12} + \frac{1}{12}\right)\mu = \frac{7}{6}\mu \neq \mu$
- **选项C**：$E(\hat{\mu}_3) = \left(\frac{1}{3} + \frac{1}{2} + \frac{1}{6}\right)\mu = \mu$
- **选项D**：$E(\hat{\mu}_4) = \left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2}\right)\mu = \frac{3}{2}\mu \neq \mu$
步骤 2：判断无偏估计量
- 无偏估计量的期望值应等于总体均值 $\mu$。根据步骤1的计算结果，只有选项C满足条件。
- **选项A**：$E(\hat{\mu}_1) = \left(\frac{1}{2} - \frac{1}{4} + \frac{1}{3}\right)\mu = \frac{7}{12}\mu \neq \mu$
- **选项B**：$E(\hat{\mu}_2) = \left(\frac{1}{6} + \frac{11}{12} + \frac{1}{12}\right)\mu = \frac{7}{6}\mu \neq \mu$
- **选项C**：$E(\hat{\mu}_3) = \left(\frac{1}{3} + \frac{1}{2} + \frac{1}{6}\right)\mu = \mu$
- **选项D**：$E(\hat{\mu}_4) = \left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2}\right)\mu = \frac{3}{2}\mu \neq \mu$
步骤 2：判断无偏估计量
- 无偏估计量的期望值应等于总体均值 $\mu$。根据步骤1的计算结果，只有选项C满足条件。