Question:
A fair die is rolled twice (independently). Let X 1 and X 2 be the numbers resulting from the first and second rolls,
respectively. Define Y=X 1 +X 2 and Z=4⋅X 1 −X 2 . Find the covariance between Y and Z .
My attempt at an answer:
I tried to find cov(Y,Z) by using their expectations and the equation: cov(Y,Z) = E(Y*Z) – E(Y)*E(Z). I believe my problem is in finding E(Y*Z), but I'll run through how I found E(Y) and E(Z) just in case, because I'm also not sure about that. For E(Y), I said E(Y) = E(X1) + E(X2) = 2*E(die) = 2*3.5 = 7. I said E(X) = 4*E(X1) – E(X2) = (4-1)*E(die) = 3*3.5 = 10.5. In order to find E(Y*Z), I tried multiplying out (X1 + X2)*(4*X1 – X2), which was pretty much just a shot in the dark, then using the same process for how I found E(Y) and E(Z). The problem is, I got that E(Y*Z) = E(Y)*E(Z), which makes covariance zero, and this seems wrong, plus our professors instructions were to enter the answer as a fraction, which makes me think the answer should be nonzero.
Please let me know if you see any of the flaws in my work or if there is an alternate strategy which would be more accurate/efficient. Any help would be greatly appreciated!
Best Answer
$Cov(Y, Z)=Cov(X_1+X_2,\,4X_1-X_2)$
$=4Cov(X_1,X_1) -Cov(X_1, X_2)+4Cov(X_2, X_1)-Cov(X_2, X_2)\,\,by\, linearity$
$=4Var(X_1)+3Cov(X_2, X_1)-Var(X_2)$
$=4Var(X_1)+0-Var(X_2) \,\,\,since\, X_1 \,and\, X_2\, are\, independent$