证明：若随机变量X与Y相互独立，则D(X-y)=D(X)+D(Y)．
D(X-Y)=D[X+(-Y)]=D(X)+D(-Y)+2Cov[X，(-Y)] 由于X与Y相互独立，故E(XY)=E(X)E(y)， 所以Cov[ X1(-Y)]=E[X(-Y)]-E(X)E(-Y) =E(XY)+E(X)E(Y)=0． 因此D(X-Y)=D(X)+D(-Y) =D(X)+(-1)2D(Y)=D(X)+D(Y)．
证明：若随机变量X与Y相互独立，则D(X-y)=D(X)+D(Y)．
D(X-Y)=D[X+(-Y)]=D(X)+D(-Y)+2Cov[X，(-Y)] 由于X与Y相互独立，故E(XY)=E(X)E(y)， 所以Cov[ X1(-Y)]=E[X(-Y)]-E(X)E(-Y) =E(XY)+E(X)E(Y)=0． 因此D(X-Y)=D(X)+D(-Y) =D(X)+(-1)2D(Y)=D(X)+D(Y)．
