I have one question about this. I know that if we have X1,X2,…,Xn independent and normally distributed random variables, then the sum X1+X2+…+Xn has the normal distribution with mean M1+M2+..+Mn and variance σ21+…+σ2n.

Why is in this problem the difference W−M the mean obtained by subtraction and variance obtained by addition? Thank you.

**Answer**

Let X,Y be random variables with variances σ2x and σ2y, respectively. It is a fact that var(Z)=cov(Z,Z) for any random variable Z. This can be checked using the definition of covariance and variance. So, the variance of X−Y is

cov(X−Y,X−Y)=cov(X,X)+cov(Y,Y)−2⋅cov(X,Y)

which follows from bilinearity of covariance. Therefore,

var(X−Y)=σ2x+σ2y−2⋅cov(X,Y)

when X,Y are independent the covariance is 0 so this simplifies to σ2x+σ2y. So, the variance of the difference of two independent variables is the sum of the variances.

**Attribution***Source : Link , Question Author : Andrew , Answer Author : Dilip Sarwate*