 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-2/12f/ii/solution

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

The sample mean

S_{n} / n

has expectation

p

\frac{p ( 1 - p )}{n} \leq \frac{1}{4 n} .

The sum in the question is exactly

P (\frac{S _{n}}{n} - p > δ) .

Chebyshev's inequality therefore gives, uniformly in

p

k \in K_{δ} \sum (k n) p^{k} (1 - p)^{n - k} \leq \frac{1}{4 n δ ^{2}} .

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