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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 29K / c / Solution

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})

is the Brownian compensator martingale.

Solved by gpt-5.6-sol high.

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