问题 设总体$X\sim \operatorname{Exp}(1/\theta)$, $x_1,\cdots,x_n$是其样本, $\theta$的矩估计和最大似然估计都是$\bar{x}$, 它也是$\theta$的相合估计和无偏估计, 试证明在均方误差准则$\left(\operatorname{MSE}(\hat{\theta})=E(\hat{\theta}-\theta)^2\right)$下存在优于$\bar{x}$的估计. ## 答案 由于总体 $X \sim \operatorname{Exp}(1/\theta)$，其概率密度函数为

$$f(x) = \frac{1}{\theta} e^{-x/\theta}, \quad x > 0.$$

因此，$X$ 的期望和方差为

$$E(X) = \theta, \quad \operatorname{Var}(X) = \theta^2.$$

样本均值 $\bar{x}$ 的期望和方差为

$$E(\bar{x}) = E\left( \frac{1}{n} \sum_{i=1}^n X_i \right) = \theta,$$

$$\operatorname{Var}(\bar{x}) = \operatorname{Var}\left( \frac{1}{n} \sum_{i=1}^n X_i \right) = \frac{\operatorname{Var}(X)}{n} = \frac{\theta^2}{n}.$$

考虑估计类 $\hat{\theta}_a

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