Prove that the covariance of 2 linear combinations of random variables is the linear combination of the covariances

Question

How would you prove the following property about covariances?

I found it here:

https://www.probabilitycourse.com/chapter5/5_3_1_covariance_correlation.php

Answer 1

It is enough to prove for suitable random variables $X,Y,Z$ and constant $a$ that:

• $\mathsf{Cov}(X+Y,Z)=\mathsf{Cov}(X,Z)+\mathsf{Cov}(Y,Z)$
• $\mathsf{Cov}(aX,Y)=a\mathsf{Cov}(X,Y)$
• $\mathsf{Cov}(X,Y)=\mathsf{Cov}(Y,X)$

which can be done on base of definition of $\mathsf{Cov}$.

The first two bullets show that $\mathsf{Cov}$ is linear on first argument.

The third bullet makes it possible to show that this is also the case for the second argument.

Answer 2

You can prove the result $Cov(aX+bY,cW+dZ)=abCov(X,Y)+acCov(X,W)+bcCov(Y,W)+bdCov(Y,Z)$ as follows.

$$\begin{split}Cov(aX+bY,cW+dZ)&=E[(aX+bY-E(aX+bY))(cW+dZ-E(cW+dZ))]\\ &=E[(a(X-E(X)+b(Y-E(Y))(c(W-E(W)+d(Z-E(Z))]\\ &=E[ac(X-E(X))(W-E(W))]+E[ad(X-E(X))(Z-E(Z))+E[bc(Y-E(Y))(W-E(W))]+E[bd(Y-E(Y))(Z-E(Z))]]\\ &=acCov(X,W)+adCov(X,Z)+bcCov(Y,W)+bdCov(Y,Z)\end{split}$$

Repeated application of this gives the desired result.