 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iii/4/b/ii/solution

G_{Y} (s) = \sum_{k \geq 1} \frac{k p _{k}}{μ} s^{k} = \frac{s}{μ} G_{Y}^{'} (s) .

Termwise differentiation of the power series is valid for

∣ s ∣ < 1

. At

s = 1

, the equality extends by the finite mean assumption, and both sides equal one.

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