Let $x,y,z$ be three random variables. How can you show that: $$\operatorname{cov}(x+y,z) = \operatorname{cov}(x,z) + \operatorname{cov}(y,z)$$ by using the definition of covariance.

-

Use the fact that $cov(X,Y)=E(X-EX)(Y-EY)$ and rearrange the terms. I've included the full solution below, just move the mouse on the grey area.

\begin{align*} cov(x+y,z)&=E[(x+y-E(x+y))(z-Ez)]\\ &=E[((x-Ex)+(y-Ey))(z-Ez)]\\ &=E(x-Ex)(z-Ez)+E(y-Ey)(z-EZ)=cov(x,z)+cov(y,z) \end{align*}

-