Assume $X$ and $Y$ are two random variables with $Var(X)=4,$ $Var(Y)=1,$ and $Cov(X,Y)=2$. What is $Cov(2X-1, X-3Y+2) $?You may find some of the following screenshots from the lecture notes useful.Answer:Answer:-4Cov(2X-1, X-3Y+2) = Cov(2X, X-3Y) = Cov(2X, X) - Cov(2X, 3Y) = 2Var(X) - 6Cov(X, Y) = 2(4) - 6(2) = -4

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Let's parse the problem step by step and evaluate each option.
Option a. '-4': This is the given result in the worked solution. To see why, use covariance properties: Cov(aX + b, cY + d) = ac Cov(X, Y) when a,b,c,d are constants and you ignore added constants because they do not affect covariance. Here, Cov(2X - 1, X - 3Y + 2) = Cov(2X, X - 3Y) since the constants -1 and +2 drop out. Then Cov(2X, X - 3Y) = 2 Cov(X, X - 3Y) = 2 [Cov(X, X) - 3 Cov(X, Y)] = 2 [Var(X) - 3 Cov(X, Y)]. Given Var(X) = 4 and Cov(X, Y) = 2, this becomes 2 [4 - 3·2] = 2 [4 - 6] = 2(-2) = -4. So this option is consistent with the covariance algebra.
Option b. '4': This would require Cov(2X, X - 3Y) to equal +4, which would imply 2 [Var(X) - 3 Cov(X, Y)] = 4. Solving gives Var(X) - 3 Cov(X, Y) = 2, i.e., 4 - 3·2 = 4 - 6 = -2, which does not equal 2. The given numbers do not support +4, so this option is incorrect.
Option c. '-56': Achieving -56 would demand a much larger negative value from Var(X) and Cov(X, Y). Using the correct formula 2 [Var(X) - 3 Cov(X, Y)] with Var(X) = 4 and Cov(X, Y) = 2 yields -4, not -56. The discrepancy shows this choice is incorrect.
Option d. '20': This would require a positive result far from the actual calculation. Since Cov(X, Y) = 2 contributes a negative term -6·2 = -12 inside the brackets, it drives the total to a negative value rather than a positive 20. Thus this option is incorrect.
In summary, the only option consistent with the covariance properties and the given variances/covariance is -4, while the other choices conflict with the algebra of Cov(2X - 1, X - 3Y + 2).

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