【答案】 (1) 设 $X$ 的密度函数为 $f_X(x)$, 则 $Y$ 的密度函数为 $f_X(y) . X$ 和 $Y$ 相互独立, 因此 $(X, Y)$ 的密度函数为 $f_X(x) f_X(y)$. 利用二重积分的轮换对称性, 得
\begin{aligned} P\{Y-X \geqslant x\} & =\iint_{v \geqslant u} f_X(u) f_X(v) \mathrm{d} u \mathrm{~d} v=\iint_{u \geqslant v} f_X(v) f_X(u) \mathrm{d} u \mathrm{~d} v \\ & =\iint_{v-u \leqslant-x} f_X(u) f_X(v) \mathrm{d} u \mathrm{~d} v=P\{Y-X \leqslant-x\}, \end{aligned}

(2) 由 $\int_{-1}^1 y \mathrm{~d} y=0, \int_{-1}^1\left(|x|-\frac{1}{2}\right) \mathrm{d} x=0$ 得 $X$ 和 $Y$ 的边缘密度函数分别为

\begin{aligned} & f_X(x)=\left\{\begin{array}{lc} \int_{-1}^1\left[\frac{1}{4}-\frac{1}{4}\left(|x|-\frac{1}{2}\right) y\right] \mathrm{d} y=\frac{1}{2},|x| < 1, \\ 0, & \text { 其他, } \end{array}\right. \\ & f_Y(y)=\left\{\begin{array}{lc} \int_{-1}^1\left[\frac{1}{4}-\frac{1}{4}\left(|x|-\frac{1}{2}\right) y\right] \mathrm{d} x=\frac{1}{2},|y| < 1, \\ 0, & \text { 其他, } \end{array}\right. \\ & \end{aligned}

\begin{aligned} P\{Y-X \geqslant 0\} & =\iint_{y \geqslant 2} f(x, y) \mathrm{d} x \mathrm{~d} y=\int_{-1}^1 \mathrm{~d} x \int_{-}^1\left[\frac{1}{4}-\frac{1}{4}\left(|x|-\frac{1}{2}\right) y\right] \mathrm{d} y \\ & =\frac{1}{2}-\frac{1}{8} \int_{-1}^1\left(|x|-\frac{1}{2}\right)\left(1-x^2\right) \mathrm{d} x \\ & =\frac{1}{2}-\frac{1}{4} \int_0^1\left(x-\frac{1}{2}\right)\left(1-x^2\right) \mathrm{d} x \\ & =\frac{1}{2}-\frac{1}{4} \cdot\left(-\frac{1}{12}\right)=\frac{1}{2}+\frac{1}{48} > \frac{1}{2}, \end{aligned}

$$\text { 所以 } P\{Y-X \leqslant 0\} < \frac{1}{2} \text {, 因此 } P\{Y-X \geqslant 0\} \neq P\{Y-X \leqslant 0\} \text {, 所以 } Y-X \text { 不是对称的. }$$