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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-37/1/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 37 / 1 / b / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Put

q = 1 - p

. The count has the zero-based geometric distribution, whose probability generating function is

G_{N} (z) = p / (1 - q z)

. Hence the aggregate has a compound geometric distribution, with

κ_{S} (t) = lo g p - lo g (1 - q M_{X_{1}} (t)) .

This transform requires

q M_{X_{1}} (t) < 1

; it holds for every

t \leq 0

. To differentiate explicitly, let

g (t) = M_{X_{1}} (t)

. Then

κ_{S}^{'} (t) = \frac{q g ^{'}}{1 - q g}, κ_{S}^{''} (t) = \frac{q g ^{''}}{1 - q g} + \frac{q ^{2} ( g ^{'} ) ^{2}}{( 1 - q g ) ^{2}},

and

κ_{S}^{'''} (t) = \frac{q g ^{'''}}{1 - q g} + \frac{3 q ^{2} g ^{'} g ^{''}}{( 1 - q g ) ^{2}} + \frac{2 q ^{3} ( g ^{'} ) ^{3}}{( 1 - q g ) ^{3}} .

At zero,

g = 1

and

g^{(j)} = m_{j}

κ_{S, 3} = \frac{q}{p} m_{3} + \frac{3 q ^{2}}{p ^{2}} m_{1} m_{2} + \frac{2 q ^{3}}{p ^{3}} m_{1}^{3} .

Every term is positive for

0 < p < 1

and strictly positive claim sizes. Therefore the unstandardized skewness is positive, even if the claim-size third central moment itself is negative. The zero-based convention matters: there is no factor

M_{X_{1}} (t)

in the numerator of this aggregate transform.

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