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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-28/1/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 28 / 1 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-07

Let

G (z) = \sum_{n \geq 0} p_{n} z^{n}

be the probability generating function. For

∣ z ∣ < 1

the series can be differentiated term by term:

\sum n p_{n} ∣ z ∣^{n - 1}

is finite, even without a finite claim-count expected value. Multiply the Panjer claim-count class recurrence by

n

and put

m = n - 1

. Then

G^{'} (z) = \sum_{n \geq 1} n p_{n} z^{n - 1} = \sum_{m \geq 0} {a (m + 1) + b} p_{m} z^{m} = a z G^{'} (z) + (a + b) G (z) .

Hence

(1 - a z) G^{'} (z) = (a + b) G (z) .

This argument is valid throughout the open unit disk. Values at boundary points may be obtained by a limit when the required derivatives exist; one need not assume

E N < \infty

to establish the identity.

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