题目
X_(1), X_(2), X_(3), X_(4)为参数为theta的指数分布总体的样本,设theta的估计量 T_(1) = (X_(1) + X_(2))/ 6 + (X_(3) + X_(4))/ 3, T_(2) = (X_(1) + X_(2) + X_(3) + X_(4))/(4), T_(3) = (X_(1) + 2X_(2) + 3X_(3) + 4X_(4))/ 5, 则theta的无偏估计量为A. T_(1), T_(2)B. T_(1), T_(3)C. T_(2), T_(3)D. T_(1), T_(2), T_(3)
$X_{1}, X_{2}, X_{3}, X_{4}$为参数为$\theta$的指数分布总体的样本,设$\theta$的估计量 $T_{1} = (X_{1} + X_{2})/ 6 + (X_{3} + X_{4})/ 3$, $T_{2} = \frac{X_{1} + X_{2} + X_{3} + X_{4}}{4}$, $T_{3} = (X_{1} + 2X_{2} + 3X_{3} + 4X_{4})/ 5$, 则$\theta$的无偏估计量为
A. $T_{1}, T_{2}$
B. $T_{1}, T_{3}$
C. $T_{2}, T_{3}$
D. $T_{1}, T_{2}, T_{3}$
题目解答
答案
A. $T_{1}, T_{2}$
解析
步骤 1:计算估计量 $T_1$ 的期望值
\[ E(T_1) = E\left(\frac{X_1 + X_2}{6} + \frac{X_3 + X_4}{3}\right) = \frac{E(X_1) + E(X_2)}{6} + \frac{E(X_3) + E(X_4)}{3} \]
由于 $X_1, X_2, X_3, X_4$ 是参数为 $\theta$ 的指数分布总体的样本,所以 $E(X_i) = \theta$ 对于 $i = 1, 2, 3, 4$。因此,
\[ E(T_1) = \frac{\theta + \theta}{6} + \frac{\theta + \theta}{3} = \frac{2\theta}{6} + \frac{2\theta}{3} = \frac{\theta}{3} + \frac{2\theta}{3} = \theta \]
步骤 2:计算估计量 $T_2$ 的期望值
\[ E(T_2) = E\left(\frac{X_1 + X_2 + X_3 + X_4}{4}\right) = \frac{E(X_1) + E(X_2) + E(X_3) + E(X_4)}{4} \]
同样地,由于 $E(X_i) = \theta$ 对于 $i = 1, 2, 3, 4$,所以
\[ E(T_2) = \frac{\theta + \theta + \theta + \theta}{4} = \frac{4\theta}{4} = \theta \]
步骤 3:计算估计量 $T_3$ 的期望值
\[ E(T_3) = E\left(\frac{X_1 + 2X_2 + 3X_3 + 4X_4}{5}\right) = \frac{E(X_1) + 2E(X_2) + 3E(X_3) + 4E(X_4)}{5} \]
由于 $E(X_i) = \theta$ 对于 $i = 1, 2, 3, 4$,所以
\[ E(T_3) = \frac{\theta + 2\theta + 3\theta + 4\theta}{5} = \frac{10\theta}{5} = 2\theta \]
\[ E(T_1) = E\left(\frac{X_1 + X_2}{6} + \frac{X_3 + X_4}{3}\right) = \frac{E(X_1) + E(X_2)}{6} + \frac{E(X_3) + E(X_4)}{3} \]
由于 $X_1, X_2, X_3, X_4$ 是参数为 $\theta$ 的指数分布总体的样本,所以 $E(X_i) = \theta$ 对于 $i = 1, 2, 3, 4$。因此,
\[ E(T_1) = \frac{\theta + \theta}{6} + \frac{\theta + \theta}{3} = \frac{2\theta}{6} + \frac{2\theta}{3} = \frac{\theta}{3} + \frac{2\theta}{3} = \theta \]
步骤 2:计算估计量 $T_2$ 的期望值
\[ E(T_2) = E\left(\frac{X_1 + X_2 + X_3 + X_4}{4}\right) = \frac{E(X_1) + E(X_2) + E(X_3) + E(X_4)}{4} \]
同样地,由于 $E(X_i) = \theta$ 对于 $i = 1, 2, 3, 4$,所以
\[ E(T_2) = \frac{\theta + \theta + \theta + \theta}{4} = \frac{4\theta}{4} = \theta \]
步骤 3:计算估计量 $T_3$ 的期望值
\[ E(T_3) = E\left(\frac{X_1 + 2X_2 + 3X_3 + 4X_4}{5}\right) = \frac{E(X_1) + 2E(X_2) + 3E(X_3) + 4E(X_4)}{5} \]
由于 $E(X_i) = \theta$ 对于 $i = 1, 2, 3, 4$,所以
\[ E(T_3) = \frac{\theta + 2\theta + 3\theta + 4\theta}{5} = \frac{10\theta}{5} = 2\theta \]