 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-30/6/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Apply Q5(b) in each independent coordinate, with

μ = λ

, to change from the Ornstein-Uhlenbeck path law to d-dimensional Wiener measure

Q^{x}

. Its density is

exp (- λ \int_{0}^{t} ω_{s} \cdot d ω_{s} - \frac{λ ^{2}}{2} \int_{0}^{t} ∣ ω_{s} ∣^{2} d s) .

Multiplication by the positive potential factor cancels the second term exactly. Therefore

u (t, x) = \int exp (- λ \int_{0}^{t} ω_{s} \cdot d ω_{s}) Q^{x} (d ω) .

∣ ω ∣^{2}

gives

\int_{0}^{t} ω_{s} \cdot d ω_{s} = \frac{1}{2} (∣ ω_{t} ∣^{2} - ∣ x ∣^{2} - d t) .

Hence an equivalent endpoint-only Wiener integral is

u (t, x) = e^{λ (∣ x ∣^{2} + d t) /2} \int e^{- λ ∣ ω_{t} ∣^{2} /2} Q^{x} (d ω) .

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