求一大學(xué)的概率證明題,關(guān)于協(xié)方差的 若隨機(jī)變量X,Y不相關(guān),證明D(X+Y)=D(X)+D(Y)

D(X)=E{[x-E(x)]^2}
D(X+Y)=E{[(x+y)-E(x+y)]}^2=E{[x-E(x)+y-E(y)]}^2
此時以x-E(x) y-E(y)為兩因子化開平方
=E{[x-E(x)]^2}+E{[y-E(y)]^2}+2*E{[x-E(x)]*[y-E(y)]}
=D(X)+D(Y)+2*E{[x-E(x)]*[y-E(y)]}
此處2*E{[x-E(x)]*[y-E(y)]}里面因式相乘
得=2*E{E(xy)-E(x)E(y)}=Cov（X,Y)=協(xié)方差
因?yàn)閄 Y不相關(guān) 所以 他們相關(guān)系數(shù)為0 進(jìn)而他們的協(xié)方差為0
即Cov（X,Y)=0 ===》 2*E{[x-E(x)]*[y-E(y)]}=0
====》D(X+Y)==D(X)+D(Y)+2*E{[x-E(x)]*[y-E(y)]}=D(X)+D(Y)