 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/8/b/i/solution

P (X = x, Y = y) = P (X = x) P (Y = y)

. Therefore

E [f (X) g (Y)] = x \sum y \sum f (x) g (y) P (X = x, Y = y) = x \sum y \sum f (x) P (X = x) g (y) P (Y = y) = (x \sum f (x) P (X = x)) (y \sum g (y) P (Y = y)) .

With finite expected values, these rearrangements are justified by absolute convergence: the analogous sum with absolute values equals

E [∣ f (X) ∣] E [∣ g (Y) ∣] < \infty

. Thus

E [f (X) g (Y)] = E [f (X)] E [g (Y)] .

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