Two blends of independent parts can still be correlated

Question

Two independent parts have variances \mathrm{Var}(X) = 4 and \mathrm{Var}(Y) = 1; being independent, \mathrm{Cov}(X, Y) = 0. Two blends are U = X + Y, \qquad V = 2X - Y. What is the covariance of U and V?

Accepted Answer

Even though X and Y are independent, the two blends both contain X, and that shared ingredient links them. Expand to see it: \mathrm{Cov}(X + Y,\; 2X - Y) = \mathrm{Cov}(X, 2X) + \mathrm{Cov}(X, -Y) + \mathrm{Cov}(Y, 2X) + \mathrm{Cov}(Y, -Y). Rewrite each piece, and drop every \mathrm{Cov}(X, Y) because it is zero: = 2\,\mathrm{Var}(X) - \mathrm{Cov}(X, Y) + 2\,\mathrm{Cov}(X, Y) - \mathrm{Var}(Y) = 2\,\mathrm{Var}(X) - \mathrm{Var}(Y). Plug in. = 2(4) - 1 = 8 - 1 = 7. So the two blends have covariance 7 despite being built from independent parts. The shared X term is what creates the link. …

Two blends of independent parts can still be correlated

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