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 $E[(X-E[X])(aA+……)]$?

**Answer**

**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

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

**Attribution***Source : Link , Question Author : Community , Answer Author : whuber*