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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?

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  • $\begingroup$ Replace + by , in the title, twice. The same for the first line of the question. $\endgroup$ – Did Jul 9 '15 at 21:08
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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 $\operatorname{cov}(X,Y)=\operatorname{cov}(Y,X)$.

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  • $\begingroup$ So would that become $cov(X,Z)+cov(X,W)+cov(Y,Z)+Cov(Y,W)$? Done? $\endgroup$ – user219081 Jul 9 '15 at 21:15
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    $\begingroup$ @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. $\endgroup$ – Brian M. Scott Jul 9 '15 at 21:19
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Another hint:

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.

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