Assume [math: X] and [math: Y] are two random variables with [math: Var(X)=4,] [math: Var(Y)=1,] and [math: Cov(X,Y)=2]. What is [math: Cov(2X−1,X−3Y+2)]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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BlackTom AI · Accepted Answer

Let's examine the given covariance problem step by step and evaluate each candidate.
Option a: -4. This matches the standard expansion: Cov(2X−1, X−3Y+2) = Cov(2X, X−3Y) by linearity (constants drop out in covariance), and Cov(2X, X−3Y) = Cov(2X, X) − Cov(2X, 3Y) = 2Var(X) − 6Cov(X,Y). Substituting Var(X)=4 and Cov(X,Y)=2 gives 2(4) − 6(2) = 8 − 12 = −4. Therefore, this option is consistent with the computed result.
Option b: 4. If one misapplies the signs or multiplies incorrectly, they might end up with a positive value, but the correct calculation above yields −4, not +4. This demonstrates a sign error or an incorrect application of linearity.
Option c: −56. This would require a much larger magnitude, implying a mistaken combination like interpreting Cov(2X−1, X−3Y+2) as something involving multiplying variances and covariances without proper scaling, or forgetting that constants vanish in covariance and that the coefficients scale covariances linearly. The correct linear scaling with Var(X)=4 and Cov(X,Y)=2 cannot produce −56 here.
Option d: 20. Achieving +20 would require adding variances or covariances in a way that disregards the negative contribution from −3Y, or miscomputing Cov(2X, X) as 10 (which would come from misusing Var or Cov values). The algebra Cov(2X, X) = 2Var(X) = 8, and Cov(2X, 3Y) = 6Cov(X,Y) = 12, so 8 − 12 = −4, not 20.
Overall, the only option that aligns with the correct arithmetic is option a: −4, since the derivation relies on Cov(aU + b, cV + d) = ac Cov(U,V) when constants b and d drop out, and then substituting Var(X)=4 and Cov(X,Y)=2 leads to −4.

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