 Solution

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

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

G_{N} (z) = (1 - p + p z)^{n}

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

h (t) = 1 - p + p M_{X_{1}} (t)

. Since

h (0) = 1

and

h^{(j)} (0) = p m_{j}

, differentiating

n lo g h

three times gives

κ_{S, 3} = n (p m_{3} - 3 p^{2} m_{1} m_{2} + 2 p^{3} m_{1}^{3}) .

The first two cumulants, for comparison, are

n p m_{1}

and

n (p m_{2} - p^{2} m_{1}^{2})

Choose deterministic claim sizes

X_{i} = d > 0

. Then

S = d N

, so

m_{j} = d^{j}

and

κ_{S, 3} = n d^{3} p (1 - p) (1 - 2 p) .

This is strictly negative for

p > 1/2

and

n \geq 1

. A concrete negative-skewness example is

n = 1

p = 3/4

d = 1

, for which

κ_{S, 3} = - 3/32 < 0.

There is no contradiction with positive claims: bounded Bernoulli-type totals can have more mass near their upper value and a longer left tail.

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