Cov(2X, Y) is equal to

Question

BlackTom AI · Accepted Answer

Let's examine the given expression Cov(2X, Y) by using a basic property of covariance with a constant multiplier.

- Option 1: '4*Corv(X,Y)' contains a misspelling ('Corv' instead of 'Cov') and incorrectly multiplies by 4. Covariance scales linearly with a constant factor, not by 4, and the expression should be Cov(X, Y) not Corv(X, Y). This option is incorrect due to both the typo and the incorrect scaling.

- Option 2: '4*Var(X)' asserts that Cov(2X, Y) equals four times the variance of X. Covariance between 2X and Y does not depend on Var(X) alone; variance is a different measure and does not appear in the general formula Cov(aX, Y) = a Cov(X, Y). Therefore this option is false.

- Option 3: '4*Cov(X+Y)' suggests Cov(2X, Y) equals four times Cov(X+Y, ?). Even if you tried to expand, Cov of a sum involves Cov(X, Y) and Cov(Y, Y) terms, and multiplying by 4 here does not align with the standard linearity rule Cov(aX, Y) = a Cov(X, Y). This is not a valid transformation for Cov(2X, Y).

- Option 4: '2*Cov(X,Y)' correctly reflects the linearity of covariance with a constant multiplier: Cov(aX, Y) = a Cov(X, Y). Here a = 2, so Cov(2X, Y) = 2 Cov(X, Y). This matches the standard property and is the correct form.

In summary, the reasoning hinges on the linearity of covariance with respect to a constant multiplier, which directly gives the 2 factor in the correct option and rules out the others for errors in form or interpretation.

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