 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/8/a/ii/solution

Let

q = 1 - p

G_{N} (s) = E [s^{N}] = \sum_{n = 0}^{\infty} p q^{n} s^{n} = \frac{p}{1 - q s} .

The series converges absolutely when

∣ q s ∣ < 1

, equivalently

∣ s ∣ < 1/ (1 - p)

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