概率论的期望 协方差的问题
发布网友
发布时间：2024-01-01 20:59
共1个回答
热心网友
时间：2024-01-08 19:22
f(t)=E(Xt-Y)^2
=E(X^2*t^2-2*X*Y*t+Y^2)
=t^2*E(X^2)-2*E(X*Y)*t+E(Y^2)
∵对任意的t，f(t)>=0
∴4*(E(X*Y))^2-4*E(X^2)*E(Y^2)<=0
即(E(X*Y))^2<=E(X^2)*E(Y^2) (1)
2.令X1=X-EX,Y1=Y-EY
则E(X1)=0,E(Y1)=0
[cov(X,Y)]^2=[E(X-EX)(Y-EY)]^2
=[E(X1*Y1)]^2
<=E(X1^2)*E(Y1^2) 由(1)式
=[E(X1^2)-(E(X1)^2)]*[E(Y1^2)-(E(Y1)^2)]
=D(X1)*D(Y1)
=D(X-EX)*D(Y-EY)
=D(X)*D(Y)追问
这两步之间是怎么推导？
∴[E(X1)]^2=0
同理，E(Y1)=0， [E(Y1)]^2=0
E(X1^2)*E(Y1^2) =E(X1^2-0)*E(Y1^2-0)
=[E(X1^2)-(E(X1)^2)]*[E(Y1^2)-(E(Y1)^2)]
