 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/29k/c/solution

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

Let

F_{s} = σ (W_{r} : 0 \leq r \leq s)

. For

0 \leq s < t

, the increment

W_{t} - W_{s}

is independent of

F_{s}

and distributed as

t - s Z

. Apply part (a) to the translated function

u \mapsto f (W_{s} + u)

, conditionally on

F_{s}

E [f (W_{t}) ∣ F_{s}] = f (W_{s}) + \frac{1}{2} \int_{0}^{t - s} E [f^{''} (W_{s} + r Z) ∣ F_{s}] d r = f (W_{s}) + \frac{1}{2} E [\int_{s}^{t} f^{''} (W_{r}) d r F_{s}] .

Therefore

E [M_{t} ∣ F_{s}] = E [f (W_{t}) ∣ F_{s}] - \frac{1}{2} \int_{0}^{s} f^{''} (W_{r}) d r - \frac{1}{2} E [\int_{s}^{t} f^{''} (W_{r}) d r F_{s}] = f (W_{s}) - \frac{1}{2} \int_{0}^{s} f^{''} (W_{r}) d r = M_{s} .

The assumed growth conditions supply integrability, so

(M_{t})

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