 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/28k/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Substituting part (b) gives

g_{N} - g^{'} (x_{0}) = \frac{h}{2 N} \sum_{i = 1}^{N} Z_{i} g^{''} (ξ_{i}) + \frac{1}{N h} \sum_{i = 1}^{N} Z_{i} ε_{i} = A + B .

The curvature bound gives

∣ A ∣ \leq h M /2

. Conditional on the

Z_{i}

and the evaluation points,

B

has mean zero, so the cross term has expectation zero. Independence of the normal random variables

ε_{i}

gives

E B^{2} = \frac{1}{N ^{2} h ^{2}} \sum_{i = 1}^{N} E ε_{i}^{2} = \frac{σ ^{2}}{N h ^{2}} .

Consequently

E ∣ g_{N} - g^{'} (x_{0}) ∣^{2} \leq \frac{h ^{2} M ^{2}}{4} + \frac{σ ^{2}}{N h ^{2}} .

The two terms are balanced by

h_{N}^{2} = \frac{2 σ}{M N}, h_{N} = (\frac{2 σ}{M})^{1/2} N^{- 1/4} .

At this value both terms equal

σ M / (2 N)

, and hence

E ∣ g_{N} - g^{'} (x_{0}) ∣^{2} \leq \frac{σ M}{N} .

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