[Math] Show $Cov(X+Y, Z+W)=Cov(X,Z)+Cov(X,W)+Cov(Y,Z)+Cov(Y,W)$

probabilityprobability distributions

Show $cov(X+Y,Z+W)=cov(X,Z)+cov(X,W)+cov(Y,Z)+cov(Y,W)$

I want to use this theorem I just proved to show this: $cov(X\pm Y, Z)= cov(X,Z) \pm cov(Y,Z)$

For the previous proof I re-wrote both sides in terms of the expected value and show that they were equivalent.

So for this one I would re-write the left hand side to be
E((X+Y)(Z+W))-E(X+Y)E(Z+W) Now do I have to re-write the right hand side and manipulate as well? I know there has to be a way to use the theorem I had just proved. Any suggestions?

Best Answer

HINT: Just use the previous result three times. Here’s the first step:

$$\operatorname{cov}(X+Y,Z+W)=\operatorname{cov}(X,Z+W)+\operatorname{cov}(Y,Z+W)\;.$$

You also need to remember that $\operatorname{cov}(X,Y)=\operatorname{cov}(Y,X)$.