 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-2/12f/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

1

gives, for some

ξ \in [t, 1]

G_{X} (t) = 1 - μ (1 - t) + \frac{1}{2} G_{X}^{''} (ξ) (1 - t)^{2} .

Since

G_{X}^{''}

is nondecreasing and

G_{X}^{''} (1) = E [X (X - 1)] = σ^{2} + μ^{2} - μ =: A,

the stated inequality follows. At the fixed point

d < 1

0 \leq (1 - d) [1 - μ + \frac{A}{2} (1 - d)] .

d \leq d^{*} := 1 - \frac{2 ( μ - 1 )}{σ ^{2} + μ ^{2} - μ} < 1.

The denominator is at least

2 (μ - 1)

because

X

is integer-valued and

E [(X - 1) (X - 2)] \geq 0

, so

d^{*} \geq 0

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