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Let $X,A,B,C,D$ be time-series variables and the covariance between any two pairs of these are known.

Suppose we want to find $\textrm{cov}(X,aA + bB + cC + dD)$, where $a,b,c,d$ are constants.

Is there any way of doing this without expanding out $E[(X-E[X])(aA+......)]$?

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Is there any way of doing this without expanding out $E[(X-E[X])(aA+......)]$?

Yes. There is a property of covariance called bilinearity which is that the covariance of a linear combination

$$ {\rm cov}(aX + bY, cW + dZ) $$

(where $a,b,c,d$ are constants and $X,Y,W,Z$ are random variables) can be decomposed as

$$ ac\cdot {\rm cov}(X,W) + ad\cdot {\rm cov}(X,Z) + bc\cdot {\rm cov}(Y,W) + bd\cdot {\rm cov}(Y,Z) $$

In the example you've given, you can use this property to write $\textrm{cov}(X,aA + bB + cC + dD)$ as

$$ a\ {\rm cov}(X, A) +b\ {\rm cov}(X, B) +c\ {\rm cov}(X, C) +d\ {\rm cov}(X, D) $$

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  • $\begingroup$ If I have $cov(X,aA + bX)$, can I still express it in the same way? $\endgroup$ – user14281 Oct 5 '12 at 13:20
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    $\begingroup$ @Harokitty, yes. Keeping in mind that ${\rm cov}(X,X) = {\rm var}(X)$, you can apply this property to find that ${\rm cov}(X,aA+bX) = a \ {\rm cov}(X,A) + b \ {\rm var}(X)$. $\endgroup$ – Macro Oct 5 '12 at 13:23

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