 Solution

ID: past-exam-of-the-continuous-martingales-and-stochastic-calculus-course-of-the-university-of-oxford/2023/2/b/ii/solution

Apply the exact discrete product identity

t_{j + 1} B_{j + 1} - t_{j} B_{j} = t_{j} Δ B_{j} + B_{j + 1} Δ t_{j}

and sum over the deterministic partition of an interval. Since

B_{0} = 0

\sum_{j} t_{j} Δ B_{j} = t B_{t} - \sum_{j} B_{j + 1} Δ t_{j} .

The left side converges in L2 space to

\int_{0}^{t} s d B_{s}

\int_{0}^{t} ∣ s - η (s) ∣^{2} d s = \frac{1}{3} \sum_{j} (Δ t_{j})^{3} \leq \frac{t ∣ π ∣ ^{2}}{3} .

The right-endpoint Riemann sum converges in L2 space to

\int_{0}^{t} B_{s} d s

: its squared error is at most

t \sum_{j} \int_{t_{j}}^{t_{j + 1}} E ∣ B_{j + 1} - B_{s} ∣^{2} d s = \frac{t}{2} \sum_{j} (Δ t_{j})^{2} ⟶ 0.

Consequently,

\int_{0}^{t} s d B_{s} = t B_{t} - \int_{0}^{t} B_{s} d s .

As in (a)(iii), continuity extends fixed-time equality to equality of the continuous processes.

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