Why is the variance of X−YX-Y equal to the sum of the variances when X,YX,Y are independent?

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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 WM 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 XY is

cov(XY,XY)=cov(X,X)+cov(Y,Y)2cov(X,Y)

which follows from bilinearity of covariance. Therefore,

var(XY)=σ2x+σ2y2cov(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

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