若(x)是取值([0,1])的随机变量，则有(egin{aligned} int_{0}^{1}Pr[lambda=x]xdx&=1-int_{0}^{1}Pr[lambdaleq x]dx end{aligned})。
容易得出(egin{aligned}Pr[lambdaleq x]=int_{0}^{x}Pr[lambda=x]dxend{aligned})
那么由右式可得
(egin{aligned}&=1-int_{0}^{1} left( int_{0}^{x}Pr[lambda=y]dy ight) dx\&=1-int_{0}^{1}dyleft(int_{x}^{1}Pr[lambda=y]dx ight)\&=1-int_{0}^{1}dyPr[lambda=y](1-y)\&=1-int_{0}^{1}Pr[lambda=y]dy+int_{0}^{1}Pr[lambda=y]ydy\&=int_{0}^{1}Pr[lambda=x]xdxend{aligned})
跟换(sum)一样，把两个积分顺序换一下。