在长为L的线段上独立地任取两点，求两点间距离的数学期望和方差。

解:

$$假设X,Y是线段上的两点,X与Y的距离|X-Y|记作Z$$
$$在线段上任取一点，每一点被取到的概率都相等，X和Y服从均匀分布:$$
$$X\sim U(0，L)，Y\sim U(0，L)$$
$$f_X(x)=\frac{1}{L}=f_Y(y)$$
$$\because X与Y独立$$
$$\therefore f(xy)=f_X(x)\times f_Y(y)=\frac{1}{L^2}$$
$$E(Z)=E(|X-Y|)$$

$$\begin{array}{rl}当X>Y时，E(|X-Y|)=& {\int }_0^{L}dx{\int }_0^X(x-y)\frac{1}{L^2}dy\\ =& \frac{1}{L^2}{\int }_0^L(xy-\frac{1}{2}{y}^2){|}_0^{x}dx\\ =& \frac{1}{L^2}{\int }_0^L\frac{1}{2}{x}^{2}dx\\ =& \frac{1}{6L^2}{x}^3{|}_0^L\\ =& \frac{1}{6}L\end{array}\text{\hspace{1em}(1)}$$

$$\begin{array}{rl}当Y>X时，E(|X-Y|)=& \frac{1}{L^2}{\int }_0^{L}dx{\int }_x^L(y-x)dy\\ =& \frac{1}{L^2}{\int }_0^L(\frac{1}{2}{y}^2-xy){|}_X^{L}dx\\ =& \frac{1}{L^2}{\int }_0^L\frac{1}{2}{L}^2-Lx-(\frac{1}{2}{x}^2-x^2)dx\\ =& \frac{1}{L^2}{\int }_0^L\frac{1}{2}{x}^2-Lx+\frac{1}{2}{L}^{2}dx\\ =& \frac{1}{L^2}\times (\frac{1}{6}{x}^3-\frac{1}{2}L{x}^2+\frac{1}{2}{L}^{2}x){|}_0^L\\ =& \frac{1}{6}L\end{array}\text{\hspace{1em}(2)}$$
$$由(1)(2)可得:E(Z)=\frac{1}{3}L$$

• Note
  $$当Y>X时，还可以先积x后积y:{\int }_0^{L}dy{\int }_0^y(y-x)\frac{1}{L^2}dx$$

$$D(Z)=E(Z^2)-(EZ{)}^2$$
$$\begin{array}{rl}E(Z^2)=& \frac{1}{L^2}{\int }_0^{L}dx{\int }_0^L(x-y{)}^{2}dy\\ =& \frac{1}{L^2}{\int }_0^L-\frac{1}{3}(x-y{)}^3{|}_0^{L}dx\\ =& \frac{1}{L^2}{\int }_0^L-\frac{1}{3}(x-L{)}^3+\frac{1}{3}{x}^{3}dx\\ =& \frac{1}{L^2}\times [(-\frac{1}{12}(x-L{)}^4)+\frac{1}{12}{x}^4]{|}_0^L\\ =& \frac{1}{6}{L}^2\end{array}$$

$$\begin{array}{rl}\therefore D(Z)=& \frac{1}{6}{L}^2-(\frac{L}{3}{)}^2\\ =& L^2(\frac{1}{6}-\frac{1}{9})\\ =& \frac{L^2}{18}\end{array}$$