Proof that $Cov(aX+b, Y+Z)=aCov(X,Y)+aCov(X,Z)$

Question

I'd like to show that $Cov(aX+b, Y+Z)=aCov(X,Y)+aCov(X,Z)$.

Therefore I use:

• $Cov(X,Y)=E(XY)-E(X)\cdot E(Y)$.

So $Cov(aX+b, Y+Z)=$

$=E[(aX+b)(Y+Z)]-E(aX+b)E(Y+Z)$

$=E(aXY+aZ+bEY+bEZ)-[(aEX+b)(EY+EZ)]$

$=aE(XY)+aEZ+bEY+bEZ-[aEXEY+aEXEZ+bEY+bEZ]$

?? I can't use that $X, Y$ are independent; if they were, then $EXEY=E(XY)$.

What I want to show in the end is:

$... = aE(XY)-a(EX)(EY)+aE(XZ)-a(EX)(EY)$, right?

Answer 1

Your line $$E(aXY+aZ+bEY+bEZ)-[(aEX+b)(EY+EZ)]$$ has a mistake: it should be $$E(aXY+a\color{red}{XZ}+bEY+bEZ)-[(aEX+b)(EY+EZ)]\tag1$$ After you fix this, you should be able to collect terms properly: $$ \begin{align} (1)&=aE(XY)+aEXZ+bEY+bEZ-[aEXEY+aEXEZ+bEY+bEZ]\\ &=aE(XY)-aEXEY +aEXZ-aEXEZ \end{align} $$ since the terms $bEY$ and $bEZ$ drop out.

Answer 2

It is much faster to use the fact that $Cov$ is a bi-linear map.

Hence $Cov(aX+b,Y+Z) = a Cov(X,Y+Z) + Cov(b,Y+Z)$ (linearity w.r.t first variable)

$b$ is constant so $Cov(b,Y+Z)=0$.

Then $Cov(aX+b,Y+Z) = aCov(X,Y+Z) = a(Cov(X,Y)+Cov(X,Z))$. (linearity w.r.t second variable)