 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/29l/b/solution

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

Let

c = q lo g (1 + b) + (1 - q) lo g (1 + a), m = N - n .

Risk-neutral valuation or direct backward induction gives

V (n, s) = (1 + r)^{- m} (lo g s + m c) .

Thus

A_{n} = (1 + r)^{- (N - n)}, B_{n} = (N - n) c (1 + r)^{- (N - n)} .

Substitution into the recursion verifies the terminal condition and the induction step.

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