关于样本方差(统计学)当Y1,…,Yn为独立同分布,且均值μY为和方差σY^2为时,证明:(请详细写出求证过程)

设y = [Y(1)+Y(2)+...+Y(n)]/n为样本均值【就是公式中的Y拔】.
EY(i)=u,E[Y(i)-u]^2 = v^2,i=1,2,...,n.
E[Y(i)-u][Y(j)-u]=0,i不等于j.
E[y]=E[Y(1)+Y(2)+...+Y(n)]/n = [EY(1)+EY(2)+...+EY(n)]/n = u.
cov(y,Y(i) = E[(y-Ey)(Y(i)-EY(i))]
= E{[Y(1)+Y(2)+...+Y(n)]/n-u}{Y(i)-u}
= E{[(Y(1)-u)+(Y(2)-u)+...+(Y(n)-u)]/n}{Y(i)-u}
= E{(Y(1)-u){Y(i)-u}+(Y(2)-u){Y(i)-u}+...+(Y(n)-u){Y(i)-u}}/n
= {E(Y(1)-u){Y(i)-u}+E(Y(2)-u){Y(i)-u}+...+E(Y(n)-u){Y(i)-u}}/n
= v^2/n