 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/discrete-mathematics-continuous-mathematics-and-probability/7/a/ii/solution

G_{X} (s) = k \geq 1 \sum p (1 - p)^{k - 1} s^{k} = \frac{p s}{1 - ( 1 - p ) s} .

The series converges absolutely for

∣ s ∣ < 1/ (1 - p)

. Differentiating the rational expression yields

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

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