Answer to Question #104680 in Statistics and Probability for ps

Question

with usual notations, show that
Cov aX( + bY cX, + dY) = acVar(X) + bdVar(Y) + ad( + bc)Cov( X, Y)

is it true or false? Give reasons.

Answer 1

$Cov(aX+bY,cX+dY)=Cov(aX,cX+dY)+Cov(bY,cX +dY)=$

$Cov(aX,cX)+Cov(aX,dY)+Cov(bY,cX)+Cov(bY,dY)=$

$ac\ Cov(X,X)+ad \ Cov(X,Y)+bc\ Cov(Y,X)+bd \ Cov (Y,Y)=$

$ac \ Var(X)+ad\ Cov(X,Y)+bc\ Cov(X,Y)+bd \ Var(Y)=$

$=ac \ Var(X)+bd\ Var(Y)+(ad+bc)Cov(X,Y)$

So, $Cov(aX+bY,cX +dY)=ac \ Var(X)+bd \ Var(Y)+(ad+bc)Cov(X,Y)$

The following properties were used:

$Cov(X,Y)=Cov(Y,X)$

$Cov(aX,Y)=a\ Cov(X,Y)$

$Cov(X+Y,W)=Cov(X,W)+Cov(Y,W)$

$Cov (X,X)=Var(X)$

Answer: this statement is true.