(2) 求 $\theta$ 的极大似然估计 $\hat{\theta}_L ;$ (3) 判断上面所得的矩估计 $\hat{\theta}$ 的无偏性, 说明理由;
(4) 设 $Y=\max \left(X_1, \ldots, X_n\right)$, 求 $E(Y)$
【答案】 (1) $\mu_1=E(X)=\int_0^\theta x \cdot 3 \theta^{-3} x^2 d x=\frac{3}{4} \theta$
$\theta=\frac{4}{3} \mu_1$
$\hat{\theta}=\frac{4}{3} \bar{X}$
(2)
\begin{aligned} & L=\prod_{i=1}^n 3 \theta^{-3} x_i^2=3^n \theta^{-3 n}\left(x_1 x_2 \cdots x_n\right)^2 \\ & \ln L=n \ln 3-3 n \ln \theta+2 \ln \left(x_1 x_2 \cdots x_n\right) \\ & \frac{d \ln L}{d \theta}=-\frac{3 n}{\theta} < 0 \\ & \hat{\theta}_L=\max \left(X_1, \ldots, X_n\right) \end{aligned}
(3) $E(\hat{\theta})=E\left(\frac{4}{3} \bar{X}\right)=\frac{4}{3} E(\bar{X})=\frac{4}{3} E(X)=\theta$
$\hat{\theta}$ 是 $\theta$ 的无偏估计量

(4)
\begin{aligned} & F_X(x)=\int_{-\infty}^x f(t) d t=\left\{\begin{array}{c} 0, x \leq 0 \\ \theta^{-3} x^3, 0 < x \leq \theta \\ 1, x > \theta \end{array}\right. \\ & F_Y(y)=\left[F_X(y)\right]^n=\left\{\begin{array}{c} 0, y \leq 0 \\ \theta^{-3 n} y^{3 n}, 0 < y \leq \theta \\ 1, y > \theta \end{array}\right. \end{aligned}
\begin{aligned} & f_Y(y)=\left\{\begin{array}{c} 3 n \theta^{-3 n} y^{3 n-1}, 0 < y \leq \theta \\ 0, \text { 其他 } \end{array}\right. \\ & E(Y)=\int_0^\theta y \cdot 3 n \theta^{-3 n} y^{3 n-1} d y=\frac{3 n}{3 n+1} \theta \end{aligned}