Finding covariance given variance

Question

Problem: Let $X$ and $Y$ be random variables such that $Var(X)=4$, $Cov(X,Y)=2.$ Find $Cov(3X,X+3Y)$.

We know that: $$Cov(3X,X+3Y)=E(3X^2+9XY)-E(3X)E(X+3Y)$$ $$=3E(X^2)+9E(XY)-9(E(X))^2E(Y).$$ Also: $$Var(X)=E(X^2)-(E(X))^2=4.$$ Also: $$Cov(X,Y)=E(XY)-E(X)E(Y)=2.$$

I'm just not seeing how to keep advancing from here. I could possibly use a linear combination of the last two equations to get the first one. But that seems unlikely since that would give me 4 terms instead of 3. Any help would be appreciated.

Answer 1

Hint: The covariance has the nice property that $\text{Cov}(aU,bV+cW)=ab\text{Cov}(U,V)+ac\text{Cov}(U,W)$.

Remark: Things were going well when you wrote $\text{Cov}(3X,X+3Y)=E(3X^2+9XY)-E(3X)E(X+3Y)$. Then there was a simplification slip. We have $E(X+3Y)=E(X)+3E(Y)$, so that part of the calculation should have gone $3E(X)(E(X)+3E(Y))$ and then $3(E(X))^2 +9E(X)E(Y)$. If it were not for the slip you would have ended up with $$3E(X^2)+9E(XY) -3(E(X))^2-9E(X)E(Y),$$ and then the rest is easy.

Answer 2

Given what André pointed out:

$$3E(X^2)+9E(XY) -3(E(X))^2-9E(X)E(Y)=3Var(X)+9Cov(X,Y) = 30.$$