题目
设某种产品的寿命X服从指数分布，其概率密度为f(x)=}(1)/(theta)e^-(x)/(theta)&x>0,0&xleq0其中theta为未知参数，X_(1),X_(2),X_(3),X_(4)是来自总体的样本设有theta的估计量hat(theta)_(1)=(1)/(6)(X_(1)+X_(2))+(1)/(3)(X_(3)+X_(4))hat(theta)_(2)=(1)/(5)(X_(1)+2X_(2)+3X_(3)+4X_(4))hat(theta)_(3)=(1)/(4)(X_(1)+X_(2)+X_(3)+X_(4))问哪一个最优？
设某种产品的寿命X服从指数分布，其概率密度为
$f(x)=\begin{cases}\frac{1}{\theta}e^{-\frac{x}{\theta}}&x>0,\\0&x\leq0\end{cases}$
其中$\theta$为未知参数，$X_{1},X_{2},X_{3},X_{4}$是来自总体的样本
设有$\theta$的估计量
$\hat{\theta}_{1}=\frac{1}{6}(X_{1}+X_{2})+\frac{1}{3}(X_{3}+X_{4})$
$\hat{\theta}_{2}=\frac{1}{5}(X_{1}+2X_{2}+3X_{3}+4X_{4})$
$\hat{\theta}_{3}=\frac{1}{4}(X_{1}+X_{2}+X_{3}+X_{4})$
问哪一个最优？
题目解答
答案
首先，判断无偏性：
- $\hat{\theta}_1 = \frac{1}{6}(X_1 + X_2) + \frac{1}{3}(X_3 + X_4)$，期望为 $\theta$，无偏。
- $\hat{\theta}_2 = \frac{1}{5}(X_1 + 2X_2 + 3X_3 + 4X_4)$，期望为 $2\theta$，有偏。
- $\hat{\theta}_3 = \frac{1}{4}(X_1 + X_2 + X_3 + X_4)$，期望为 $\theta$，无偏。
排除有偏估计量 $\hat{\theta}_2$，比较 $\hat{\theta}_1$ 和 $\hat{\theta}_3$ 的方差：
- $Var(\hat{\theta}_1) = \frac{5}{18}\theta^2 \approx 0.2778\theta^2$，
- $Var(\hat{\theta}_3) = \frac{1}{4}\theta^2 = 0.25\theta^2$。
由于 $\frac{1}{4} < \frac{5}{18}$，$\hat{\theta}_3$ 的方差更小，更优。
**答案：** $\boxed{\hat{\theta}_3}$
解析
步骤 1：判断无偏性
- 对于 $\hat{\theta}_1 = \frac{1}{6}(X_1 + X_2) + \frac{1}{3}(X_3 + X_4)$，期望为 $E[\hat{\theta}_1] = \frac{1}{6}E[X_1] + \frac{1}{6}E[X_2] + \frac{1}{3}E[X_3] + \frac{1}{3}E[X_4] = \frac{1}{6}\theta + \frac{1}{6}\theta + \frac{1}{3}\theta + \frac{1}{3}\theta = \theta$，无偏。
- 对于 $\hat{\theta}_2 = \frac{1}{5}(X_1 + 2X_2 + 3X_3 + 4X_4)$，期望为 $E[\hat{\theta}_2] = \frac{1}{5}E[X_1] + \frac{2}{5}E[X_2] + \frac{3}{5}E[X_3] + \frac{4}{5}E[X_4] = \frac{1}{5}\theta + \frac{2}{5}\theta + \frac{3}{5}\theta + \frac{4}{5}\theta = 2\theta$，有偏。
- 对于 $\hat{\theta}_3 = \frac{1}{4}(X_1 + X_2 + X_3 + X_4)$，期望为 $E[\hat{\theta}_3] = \frac{1}{4}E[X_1] + \frac{1}{4}E[X_2] + \frac{1}{4}E[X_3] + \frac{1}{4}E[X_4] = \frac{1}{4}\theta + \frac{1}{4}\theta + \frac{1}{4}\theta + \frac{1}{4}\theta = \theta$，无偏。
步骤 2：计算方差
- 对于 $\hat{\theta}_1$，方差为 $Var(\hat{\theta}_1) = Var\left(\frac{1}{6}(X_1 + X_2) + \frac{1}{3}(X_3 + X_4)\right) = \frac{1}{36}Var(X_1) + \frac{1}{36}Var(X_2) + \frac{1}{9}Var(X_3) + \frac{1}{9}Var(X_4) = \frac{1}{36}\theta^2 + \frac{1}{36}\theta^2 + \frac{1}{9}\theta^2 + \frac{1}{9}\theta^2 = \frac{5}{18}\theta^2$。
- 对于 $\hat{\theta}_3$，方差为 $Var(\hat{\theta}_3) = Var\left(\frac{1}{4}(X_1 + X_2 + X_3 + X_4)\right) = \frac{1}{16}Var(X_1) + \frac{1}{16}Var(X_2) + \frac{1}{16}Var(X_3) + \frac{1}{16}Var(X_4) = \frac{1}{16}\theta^2 + \frac{1}{16}\theta^2 + \frac{1}{16}\theta^2 + \frac{1}{16}\theta^2 = \frac{1}{4}\theta^2$。
步骤 3：比较方差
- 由于 $\frac{1}{4} < \frac{5}{18}$，$\hat{\theta}_3$ 的方差更小，更优。
- 对于 $\hat{\theta}_1 = \frac{1}{6}(X_1 + X_2) + \frac{1}{3}(X_3 + X_4)$，期望为 $E[\hat{\theta}_1] = \frac{1}{6}E[X_1] + \frac{1}{6}E[X_2] + \frac{1}{3}E[X_3] + \frac{1}{3}E[X_4] = \frac{1}{6}\theta + \frac{1}{6}\theta + \frac{1}{3}\theta + \frac{1}{3}\theta = \theta$，无偏。
- 对于 $\hat{\theta}_2 = \frac{1}{5}(X_1 + 2X_2 + 3X_3 + 4X_4)$，期望为 $E[\hat{\theta}_2] = \frac{1}{5}E[X_1] + \frac{2}{5}E[X_2] + \frac{3}{5}E[X_3] + \frac{4}{5}E[X_4] = \frac{1}{5}\theta + \frac{2}{5}\theta + \frac{3}{5}\theta + \frac{4}{5}\theta = 2\theta$，有偏。
- 对于 $\hat{\theta}_3 = \frac{1}{4}(X_1 + X_2 + X_3 + X_4)$，期望为 $E[\hat{\theta}_3] = \frac{1}{4}E[X_1] + \frac{1}{4}E[X_2] + \frac{1}{4}E[X_3] + \frac{1}{4}E[X_4] = \frac{1}{4}\theta + \frac{1}{4}\theta + \frac{1}{4}\theta + \frac{1}{4}\theta = \theta$，无偏。
步骤 2：计算方差
- 对于 $\hat{\theta}_1$，方差为 $Var(\hat{\theta}_1) = Var\left(\frac{1}{6}(X_1 + X_2) + \frac{1}{3}(X_3 + X_4)\right) = \frac{1}{36}Var(X_1) + \frac{1}{36}Var(X_2) + \frac{1}{9}Var(X_3) + \frac{1}{9}Var(X_4) = \frac{1}{36}\theta^2 + \frac{1}{36}\theta^2 + \frac{1}{9}\theta^2 + \frac{1}{9}\theta^2 = \frac{5}{18}\theta^2$。
- 对于 $\hat{\theta}_3$，方差为 $Var(\hat{\theta}_3) = Var\left(\frac{1}{4}(X_1 + X_2 + X_3 + X_4)\right) = \frac{1}{16}Var(X_1) + \frac{1}{16}Var(X_2) + \frac{1}{16}Var(X_3) + \frac{1}{16}Var(X_4) = \frac{1}{16}\theta^2 + \frac{1}{16}\theta^2 + \frac{1}{16}\theta^2 + \frac{1}{16}\theta^2 = \frac{1}{4}\theta^2$。
步骤 3：比较方差
- 由于 $\frac{1}{4} < \frac{5}{18}$，$\hat{\theta}_3$ 的方差更小，更优。