$$f(x ; \theta)= \begin{cases}1, & \theta-\frac{1}{2} \leqslant x \leqslant \theta+\frac{1}{2}, \\ 0, & \text { 其他, }\end{cases}$$

$$X_{(1)}=\min \left\{X_1, X_2, \cdots, X_n\right\}, X_{(n)}=\max \left\{X_1, X_2, \cdots, X_n\right\} .$$
(1) 求参数 $\theta$ 的矩估计量 $\hat{\theta}_M$ 和最大似然估计量 $\hat{\theta}_L$;
(2) 判断 $\frac{X_{(1)}+X_{(n)}}{2}$ 是否为 $\theta$ 的无偏估计量, 并说明理由.

【解】(1) 由 $E(X)=\int_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} x \mathrm{~d} x=\left.\frac{1}{2} x^2\right|_{\theta-\frac{1}{2}} ^{\theta+\frac{1}{2}}=\theta, \bar{X}=\frac{1}{n} \sum_{i=1}^n X_i$, 令 $E(X)=\bar{X}$, 得 $\hat{\theta}_M=\bar{X}$

\begin{aligned} L(\theta) & =\prod_{i=1}^n f\left(x_i ; \theta\right)= \begin{cases}1, & \theta-\frac{1}{2} \leqslant x_1, x_2, \cdots, x_n \leqslant \theta+\frac{1}{2}, \\ 0, \quad \text { 其他 }\end{cases} \\ & = \begin{cases}1, & \theta-\frac{1}{2} \leqslant \min \left\{x_1, x_2, \cdots, x_n\right\} \leqslant x_1, x_2, \cdots, x_n \leqslant \max \left\{x_1, x_2, \cdots, x_n\right\} \leqslant \theta+\frac{1}{2}, \\ 0, & \text { 其他, }\end{cases} \end{aligned}

$$\left\{\begin{array}{l} \hat{\theta}_L-\frac{1}{2} \leqslant X_{(1)}, \\ \hat{\theta}_L+\frac{1}{2} \geqslant X_{(n)}, \end{array}\right.$$

(2) $\frac{X_{(1)}+X_{(n)}}{2}$ 为区间 $\left[X_{(n)}-\frac{1}{2}, X_{(1)}+\frac{1}{2}\right]$ 的中点, 即其为 $\theta$ 的一个最大似然估计量, 由

$$F(x ; \theta)= \begin{cases}0, & x < \theta-\frac{1}{2}, \\ x-\theta+\frac{1}{2}, & \theta-\frac{1}{2} \leqslant x < \theta+\frac{1}{2}, \\ 1, & x \geqslant \theta+\frac{1}{2},\end{cases}$$

$$\begin{gathered} f_{(1)}(x)=n[1-F(x ; \theta)]^{n-1} f(x ; \theta)= \begin{cases}n\left(\frac{1}{2}+\theta-x\right)^{n-1}, & \theta-\frac{1}{2} \leqslant x \leqslant \theta+\frac{1}{2}, \\ 0, & \text { 其他, }\end{cases} \\ f_{(n)}(x)=n[F(x ; \theta)]^{n-1} f(x ; \theta)= \begin{cases}n\left(x+\frac{1}{2}-\theta\right)^{n-1}, & \theta-\frac{1}{2} \leqslant x \leqslant \theta+\frac{1}{2}, \\ 0, & \text { 其他, }\end{cases} \end{gathered}$$

\begin{aligned} \text { 故 } E\left[X_{(1)}\right] & =\int_{-\infty}^{+\infty} x f_{(1)}(x) \mathrm{d} x=\int_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} x n\left(\frac{1}{2}+\theta-x\right)^{n-1} \mathrm{~d} x \stackrel{\frac{1}{2}+\theta-x=t}{=} \int_0^1\left(\frac{1}{2}+\theta-t\right) t^{n-1} \mathrm{~d} t \\ & =\theta+\frac{1}{2}-\frac{n}{n+1}, \\ & E\left[X_{(n)}\right]=\int_{-\infty}^{+\infty} x f_{(n)}(x) \mathrm{d} x=\int_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} x n\left(x+\frac{1}{2}-\theta\right)^{n-1} \mathrm{~d} x=\theta-\frac{1}{2}+\frac{n}{n+1}, \end{aligned}

$$E\left[\frac{X_{(1)}+X_{(n)}}{2}\right]=\frac{1}{2}\left\{E\left[X_{(1)}\right]+E\left[X_{(n)}\right]\right\}=\frac{1}{2}\left(\theta+\frac{1}{2}-\frac{n}{n+1}+\theta-\frac{1}{2}+\frac{n}{n+1}\right)=\theta,$$