Question

Recall the definition of covariance for two random variables, X and W: Cov(X,W) =

E(X − E(X))(W − E(W)).

Prove or disprove

Cov(aX + b, cW) = ac Cov(X,W)

where a, b, and c are constants. Start with applying the covariance definition to Cov(aX +
b, cW), proceed by applying appropriate properties of expectation operator, and see if your
simplified expression is the same as ac Cov(X,W).

Answer 1

The statement Cov(aX + b, cW) = ac Cov(X,W) is true.

Explanation:

Starting with the definition of covariance:

Cov(aX + b, cW) = E((aX + b) - E(aX + b))(cW - E(cW))

Using linearity of expectation:

Cov(aX + b, cW) = E(a(X - E(X)) + (b - E(b)))(c(W - E(W)))

Expanding and rearranging the product:

Cov(aX + b, cW) = acE(X - E(X))(W - E(W)) + E(b - E(b))E(c(W - E(W)))

Since the expectation of a constant is the constant itself and E(b - E(b)) = 0, the above expression simplifies to:

Cov(aX + b, cW) = ac Cov(X,W)