已知随机变量X的期望EX=U,方差DX=&^2,随机变量Y=(x-u)/&,求EY和DY

EY=0 DY=1
EY=E(x-u)/&=(EX-U)/&=0
DY=D[(X-U)^2]/(&^2)
而D[(X-U)^2]=E[(X-U)^2]-[E(X-U)]^2=E[(X-U)^2] (后面项为0)
=E(X^2+U^2-2UX)=EX^2-U^2
EX^2=(EX)^2+DX=U^2+&^2
所以D[(X-U)^2]=&^2 即有DY=1