What is Cov(X-Y,Z-W)

Question

I know the equality of the covariance $$ \operatorname{Cov}(X+Y, Z+W) = \operatorname{Cov}(X,Z) + \operatorname{Cov}(X,W) + \operatorname{Cov}(Y,Z) + \operatorname{Cov}(Y,W), $$ But I have the doubt that if it had negative signs then the equality would be $$ \operatorname{Cov}(X-Y, Z-W) = \operatorname{Cov}(X,Z) - \operatorname{Cov}(X,W) - \operatorname{Cov}(Y,Z) + \operatorname{Cov}(Y,W), $$ I am not sure if this equality is true

Answer 1

It's true.

This is one of those cases where the definition is simple enough that you can trace it all through without getting hopelessly lost in the algebra. Using $\operatorname{Cov}(X, Y) = \mathbb E[XY] - (\mathbb EX)(\mathbb E Y)$, which isn't quite the definition but a theorem you've probably encountered (and can probably prove if you haven't):

\begin{align*} \operatorname{Cov}(X-Y, Z-W) &= \mathbb E[(X-Y)(Z-W)] - \mathbb E[X-Y] \mathbb E[Z-W] \\ &= \mathbb E[XZ - XW - YZ + YW] - \left( \mathbb E X - \mathbb E Y \right) \left(\mathbb E Z - \mathbb E W \right)\\ &= \color{blue}{\mathbb E[XZ]} - \color{green}{\mathbb E[XW]} - \color{red}{\mathbb E[YZ]} + \mathbb E [YW]\\ & \qquad - \color{blue}{(\mathbb E X) (\mathbb E Z)} + \color{green}{(\mathbb E X)(\mathbb E W)} + \color{red}{(\mathbb E Y) (\mathbb E Z)} - (\mathbb E Y) (\mathbb E W) \end{align*} and grouping the terms in the last line by color shows that your last equality is true.