(I) 求 $Z$ 的概率密度 $f_Z(z)$ 和 $E Z$;
(II) 求二维随机变量 $(X, Y)$ 的分布函数.

\begin{aligned} &X=\max \left\{X_1, X_2\right\}=\left\{\begin{array}{c} X_1, X_1 \geqslant X_2, \\ X_2, X_1 < X_2 \end{array}=\frac{X_1+X_2+\left|X_1-X_2\right|}{2},\right. \\ &Y=\min \left\{X_1, X_2\right\}=\left\{\begin{array}{c} X_1, X_1 < X_2, \\ X_2, X_1 \geqslant X_2 \end{array}=\frac{X_1+X_2-\left|X_1-X_2\right|}{2},\right. \end{aligned}

\begin{aligned} F_Z(z) &=P\left\{\left|X_1-X_2\right| \leqslant z\right\}= \begin{cases}P\left\{-z \leqslant X_1-X_2 \leqslant z\right\}, & z \geqslant 0, \\ 0, & z < 0\end{cases} \\ &= \begin{cases}\Phi\left(\frac{z}{\sqrt{2}}\right)-\Phi\left(-\frac{z}{\sqrt{2}}\right), & z \geqslant 0, \\ 0, & z < 0,\end{cases} \end{aligned}

\begin{aligned}
&f_Z(z)=F_Z^{\prime}(z)= \begin{cases}\frac{1}{\sqrt{\pi}} \mathrm{e}^{-\frac{z^2}{4}}, & z \geqslant 0, \\ 0, & z < 0 .\end{cases}\\
&\text { 从而 } E Z=\int_{-\infty}^{+\infty} z f_Z(z) \mathrm{d} z=\int_0^{+\infty} z \cdot \frac{1}{\sqrt{\pi}} \mathrm{e}^{-\frac{z^2}{4}} \mathrm{~d} z=-\left.\frac{2}{\sqrt{\pi}} \mathrm{e}^{-\frac{z^2}{4}}\right|_0 ^{+\infty}=\frac{2}{\sqrt{\pi}} \text {. }\\
&P(A B) \text {, 其中 }\\
&A=\left\{\max \left\{X_1, X_2\right\} \leqslant x\right\}=\left\{X_1 \leqslant x, X_2 \leqslant x\right\},\\
&B=\left\{\min \left\{X_1, X_2\right\} \leqslant y\right\}=\left\{X_1 \leqslant y\right\} \cup\left\{X_2 \leqslant y\right\},
\end{aligned}

$$\bar{B}=\left\{\min \left\{X_1, X_2\right\} > y\right\}=\left\{X_1 > y, X_2 > y\right\} .$$

\begin{aligned} F(x, y) &=P(A B)=P(A)-P(A \bar{B}) \\ &=P\left\{X_1 \leqslant x, X_2 \leqslant x\right\}-P\left\{X_1 \leqslant x, X_2 \leqslant x, X_1 > y, X_2 > y\right\} \\ &=P\left\{X_1 \leqslant x\right\} P\left\{X_2 \leqslant x\right\}-P\left\{X_1 \leqslant x, X_1 > y\right\} P\left\{X_2 \leqslant x, X_2 > y\right\} \\ &=\Phi^2(x)-P\left\{X_1 \leqslant x, X_1 > y\right\} P\left\{X_2 \leqslant x, X_2 > y\right\} . \end{aligned}

\begin{aligned} F(x, y) &=\Phi^2(x)-P\left\{y < X_1 \leqslant x\right\} P\left\{y < X_2 \leqslant x\right\} \\ &=\Phi^2(x)-[\Phi(x)-\Phi(y)]^2=2 \Phi(x) \Phi(y)-\Phi^2(y) . \end{aligned}

$$F(x, y)= \begin{cases}2 \Phi(x) \Phi(y)-\Phi^2(y), & x > y, \\ \Phi^2(x), & x \leqslant y .\end{cases}$$
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