# Show $Cov(X+Y, Z+W)=Cov(X,Z)+Cov(X,W)+Cov(Y,Z)+Cov(Y,W)$

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?

• Replace + by , in the title, twice. The same for the first line of the question. – Did Jul 9 '15 at 21:08

$$\operatorname{cov}(X+Y,Z+W)=\operatorname{cov}(X,Z+W)+\operatorname{cov}(Y,Z+W)\;.$$
You also need to remember that $\operatorname{cov}(X,Y)=\operatorname{cov}(Y,X)$.
• So would that become $cov(X,Z)+cov(X,W)+cov(Y,Z)+Cov(Y,W)$? Done? – user219081 Jul 9 '15 at 21:15
• @Alyssa: That’s right: reverse the order, apply the previous result to each covariance, and reverse the orders again as necessary. It’s like using the distributive law to show that $(a+b)(c+d)=ac+ad+bc+bd$: first $(a+b)(c+d)=a(c+d)+b(c+d)$, then you use it again to expand each of those terms. – Brian M. Scott Jul 9 '15 at 21:19
You only need to show that $\text{Cov}(X + Y, Z) = \text{Cov}(X, Z) + \text{Cov}(Y, Z)$, because the statement you have follows from this. This saves some unnecessary effort.