 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/8/a/ii/solution

G_{X}^{'} (1) = \frac{q}{p}, G_{X}^{''} (1) = \frac{2 q ^{2}}{p ^{2}} .

E [X (X - 1)]

. Hence

E X = \frac{1 - p}{p}, Var (X) = G_{X}^{''} (1) + G_{X}^{'} (1) - G_{X}^{'} (1)^{2} = \frac{1 - p}{p ^{2}} .

In particular,

E [X^{2}] = q (2 - p) / p^{2}

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