题目

设随机变量的分布函数为F(x)=0.3Φ(x)+0.7Φ(0.5x-0.5),其中Φ（x）为标准正态分布的分布函数,则E(X)=

提问时间：2020-12-24

答案

E(X)=∫xdF(x)=∫xd[0.3Φ(x)+0.7Φ(0.5x-0.5)]=∫x[0.3Φ'(x)+0.35Φ'(0.5x-0.5)]dx
=∫x[0.3Φ'(x)]dx+∫x[0.35Φ'(0.5x-0.5)]dx
∫x[0.3Φ'(x)]dx=0
对∫x[0.35Φ'(0.5x-0.5)]dx令0.5x-0.5=t得
∫x[0.35Φ'(0.5x-0.5)]dx=∫(2t+1)[0.35Φ'(t)]*2dt=∫(2t)[0.35Φ'(t)]*2dt+∫[0.35Φ'(t)]*2dt
=0+0.7=0.7
注:以上积分上下限都是+∞,-∞
所以E(X)=0.7