Proof that $Cov(aX+b, Y+Z)=aCov(X,Y)+aCov(X,Z)$ 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+...--prophetes.ai

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

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.