 Solution

ID: past-exam-of-the-stochastic-differential-equations-course-of-the-university-of-oxford/2025/2/a/iii/solution

B_{t}^{2}

yields

\int_{0}^{t} B_{u} d B_{u} = (B_{t}^{2} - t) /2

. Substituting into the expression for

Z_{t}

gives the pathwise identity

Z_{t} = exp [- \frac{β}{2} B_{t}^{2} + \frac{βt}{2} - \frac{β ^{2}}{2} \int_{0}^{t} B_{u}^{2} d u] .

Consequently

Z_{t} exp [(\frac{β}{2} - α) B_{t}^{2} - \frac{βt}{2}] = exp [- α B_{t}^{2} - \frac{β ^{2}}{2} \int_{0}^{t} B_{u}^{2} d u] .

L (t, α, β) = E [Z_{t} exp ((\frac{β}{2} - α) B_{t}^{2} - \frac{βt}{2})] .

The random variable on the preceding line is bounded by one for

α \geq 0

, so its integrability is immediate.

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