 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ia/paper-2/3f/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

E (X + Y) = E X + E Y

. Expand the centered product and use linearity of expectation again:

cov (X + Y, Z) = E [((X - E X) + (Y - E Y)) (Z - E Z)] = cov (X, Z) + cov (Y, Z) .

Thus the assertion is true whenever the covariances are defined. Neither independence nor identical distributions are required.

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