 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/ii/solution

Assume the variances are finite. Expand the square and apply linearity of expectation:

var (X + Y) = E [X^{2}] + 2 E [X Y] + E [Y^{2}] - (E [X] + E [Y])^{2} = var (X) + var (Y) + 2 (E [X Y] - E [X] E [Y]) .

Part (i), applied to the identity functions, makes the bracketed covariance zero by independence of random variables. Hence

var (X + Y) = var (X) + var (Y) .

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