Prove that $\text{Cov}(aX + bY, Z) = a\text{Cov}(X, Z) + b\text{Cov}(Y, Z)$.

Question

So far, I have: $$\mathbb{E}[(aX+bY-\mathbb{E}(aX+bY))(Z-\mathbb{E}(Z))] = \mathbb{E}[Z(aX + bY)-Z\mathbb{E}(aX +bY) - \mathbb{E}(Z)(aX + bY)-\mathbb{E}(Z)E(aX + bY)] = \mathbb{E}(Z(aX + bY - \mathbb{E}(aX + bY)) - \mathbb{E}(Z)(aX + bY - \mathbb{E}(aX + bY))$$

I don't really know what to do next.

Answer 1

Can you use this formula instead? $$ \text{Cov}(X,Y) =\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y).$$

Thus \begin{align*} \text{Cov}(aX + bY,Z) &=\mathbb{E}[(aX + bY)Z]-\mathbb{E}(aX + bY)\mathbb{E}(Z)\\ &=\mathbb{E}(aXZ + bYZ) - [a \mathbb{E}(X) + b \mathbb{E}(Y)]\mathbb{E}(Z)\\ &=... \end{align*} Can you proceed from here?