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Let $a_1,a_2,b,c_1,c_2,d$ be constants and assume $X_1, X_2, Y_1, Y_2$ are Random Variables.

I am trying to prove $$Cov(a_1X_1+a_2X_2+b, c_1Y_1+c_2Y_2+d)= a_1c_1Cov(X_1,Y_1)+a_1c_2Cov(X_1,Y_2)+a_2c_1Cov(X_2,Y_1)+a_2c_2Cov(X_2,Y_2)$$

I am confused how you would go about proving this formally.

So far I have tried to expand $Cov(X,Y)= E(XY)-E(X)E(Y)$ given $X=a_1X_1+a_2X_2+b$ and $Y=c_1Y_1+c_2Y_2+d$ but I hit a dead end.

Is there an easier way to go about this?

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  • $\begingroup$ How you prove it depends on what definition of covariance you start with (and, optionally, what additional facts you take to be already proven). If we take the expression you are trying to expand as the definition, that's fine and it will work. So: please describe this "dead end" you have reached. $\endgroup$ – whuber♦ Oct 19 '20 at 13:24
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It'd be simplest to prove $\operatorname{cov}(X,Y+Z)=\operatorname{cov}(X,Y)+\operatorname{cov}(X,Z)$ first and go from there.

$$\begin{align}\operatorname{cov}(X,Y+Z)&=\mathbb{E}[X(Y+Z)]-\mathbb E[X]\mathbb E[Y+Z]\\&=\mathbb E[XY]+\mathbb E[XZ]-\mathbb E[X]\mathbb E[Y]-\mathbb E[X]\mathbb E[Z]\\&=(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])+(\mathbb E[XZ]-\mathbb E[X]\mathbb E[Z])\\&=\operatorname{cov}(X,Y)+\operatorname{cov}(X,Z)\end{align}$$

The proof can be followed by assigning temporary variables to the expressions inside the expression, i.e. $X=a_1X_1+a_2X_2+b$ and $Y=c_1Y_1, Z=c_2Y_2+d$, and apply this property repeatedly, together with the property $\operatorname{cov}(X,Y)=\operatorname{cov}(Y,X)$ when necessary.

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