Question

求一大学的概率证明题,关于协方差的 若随机变量X,Y不相关,证明D(X+Y)=D(X)+D(Y)

Answer 1

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
此时以x-E(x) y-E(y)为两因子化开平方
=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)]}
此处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)=协方差
因为X Y不相关 所以 他们相关系数为0 进而他们的协方差为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)