Can I write $\text{Cov}[x,y+z]=\text{Cov}[x,y]+\text{Cov}[x,z]$,
where $\text{Cov}(.)$ is referring to the population covariance? $x,y$ and $z$ are random variables.
(very fundamental question... :)
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$\begingroup$ Hint: $E[x(y+z)]−ExE(y+z)=Exy+Exz−ExEy−ExEz=Exy−ExEy+Exz−ExEz=Cov(x,y)+Cov(x,z)$ $\endgroup$– user31766Commented Feb 5, 2014 at 22:40
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Yes, covariance is linear in either of its arguments, which is to say:
$\text{Cov}(aX,Y) = \text{Cov}(X,aY) = a\text{Cov}(X,Y)$ (in the univariate case; an analogous formula holds for the multivariate case) and
$\text{Cov}(A+B,C) = \text{Cov}(A,C) + \text{Cov}(B,C)\,$.
This linearity follows from the definition of covariance and the basic properties of expectation - in particular, linearity of expectation.
Let $\mu_X = E(X)$ and similarly for the other variables:
\begin{eqnarray} \text{Cov}(X,Y+Z) &=& \text{E}[(X-\text{E}(X))(Y+Z-\text{E}(Y+Z)]\\ &=& \text{E}[(X-\mu_X)(Y+Z-\{\mu_Y+\mu_Z\})]\\ &=& \text{E}[(X-\mu_X)(Y-\mu_Y+Z-\mu_Z)]\\ &=& \text{E}[(X-\mu_X)(Y-\mu_Y)]+\text{E}[(X-\mu_X)(Z-\mu_Z)]\\ &=& \text{Cov}(X,Y)+\text{Cov}(X,Z) \end{eqnarray}
(which is pretty similar what you have in your own hint in comments there)
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1$\begingroup$ I wouldn't proceed quite that way, but I can edit something similar in. $\endgroup$– Glen_bCommented Feb 5, 2014 at 22:57
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$\begingroup$ You didn't do anything wrong in your hint, but it simply takes an unnecessary detour which I sought to avoid. As for assumptions, the only assumptions are those necessary to use the definition of covariance and the linearity of expectation (which to my understanding is "the covariances all exist, the expectation exists"). $\endgroup$– Glen_bCommented Feb 5, 2014 at 23:14
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$\begingroup$ Pretty much; it's correct, but it adds steps where you split things up then a couple of steps later put it all back together and to use it requires an additional step of justification. But mostly, it simply seems less intuitive to me than to just apply the definition plus the linearity of expectation property (twice). $\endgroup$– Glen_bCommented Feb 5, 2014 at 23:19