 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-1/30k/b/solution

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

V (n, x) = A_{n} + B_{n} x + C_{n} x^{2}

. The terminal values are

(A_{N}, B_{N}, C_{N}) = (0, 0, 1)

. Since the noise is centred, backward induction keeps

B_{n} = 0

u^{*} = - \frac{C _{n}}{1 + C _{n}} x, qq u a d C_{n - 1} = \frac{C _{n}}{1 + C _{n}}, qq u a d A_{n - 1} = A_{n} + C_{n} σ^{2} .

Hence

C_{n} = \frac{1}{N - n + 1}, B_{n} = 0, A_{n} = σ^{2} j = n + 1 \sum N \frac{1}{N - j + 1}

(with the empty sum equal to zero). This proves the asserted quadratic form of the value function.

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