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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We start by restating the problem to keep all given information in view: Var(X) = 4, Var(Y) = 1, Cov(X, Y) = 2. The target is Cov(2X − 1, X − 3Y + 2).

Option a: -4
- This option matches the standard covariance algebra: constants disappear, and Cov(aX + b, cU + dV) behaves linearly in the random variables. Specifically, Cov(2X − 1, X − 3Y + 2) = Cov(2X, X − 3Y) since constants have zero covariance with any variable. Then Cov(2X, X − 3Y) = Cov(2X, X) − Cov(2X, 3Y) by linearity of Cov in its second argument. Cov(2X, X) = 2 Cov(X, X) = 2 Var(X) = 8. Cov(2X, 3Y) = 6 Cov(X, Y) = 6·2 = 12. Therefore, Cov(2X − 1, X − 3Y + 2) = 8 − 12 = −4.

Option b: 4
- If someone computed Cov(2X − 1, X − 3Y + 2) and somehow obtained a positive 4, they likely misused the linearity or sign. The key step is Cov(2X, X − 3Y) = Cov(2X, X) − Cov(2X, 3Y); with Cov(X, Y) positive (2), the subtraction term reduces the result, not increases it to +4.

Option c: −56
- This would require a much larger magnitude from the cross-covariances. Since Var(X) = 4 and Cov(X, Y) = 2 are relatively modest, the combination 2 Var(X) − 6 Cov(X, Y) = 8 − 12 yields −4, not −56. Any calculation yielding −56 would imply an incorrect scaling or missing factors in the Cov(2X, X) or Cov(2X, 3Y) terms.

Option d: 20
- Achieving +20 would demand Cov(2X, X − 3Y) to be 20, which again contradicts Cov(2X, X) = 8 and Cov(2X, 3Y) = 12. The signs and the magnitudes in the standard expansion lead to 8 − 12 = −4, not +20. A misapplication of the linearity or a wrongCov(X, Y) value could produce unrelated numbers like 20.

In summary, the calculation hinges on applying linearity of covariance and the fact that constants drop out: Cov(2X − 1, X − 3Y + 2) = Cov(2X, X) − Cov(2X, 3Y) = 2 Var(X) − 6 Cov(X, Y) = 2(4) − 6(2) = 8 − 12 = −4.

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