Past exam of the continuous martingales and stochastic calculus course of the University of Oxford / 2023 / 2 / b / i

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- Solution i

Solution

To establish the Brownian weighted quadratic variation in L2 for the deterministic partition of an interval, decompose

\sum_{j} B_{j} (Δ B_{j})^{2} - \int_{0}^{t} B_{s} d s = U_{π} + V_{π},

where

U_{π} = \sum_{j} B_{j} ((Δ B_{j})^{2} - Δ t_{j})

and

V_{π} = \sum_{j} B_{j} Δ t_{j} - \int_{0}^{t} B_{s} d s

.

The summands in

U_{π}

are martingale differences:

B_{j}

is

F_{t_{j}}

-measurable, and the following Brownian increment is independent of

F_{t_{j}}

. Their conditional means are zero, so distinct summands are orthogonal in L2 space. Since

E B_{j}^{2} = t_{j}

and

E [((Δ B_{j})^{2} - Δ t_{j})^{2}] = 2 (Δ t_{j})^{2}

,

E U_{π}^{2} = 2 \sum_{j} t_{j} (Δ t_{j})^{2} \leq 2 t^{2} ∣ π ∣ ⟶ 0.

For the Riemann sum error, the Cauchy-Schwarz inequality gives

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

Thus the triangle inequality in L2 space proves

j \sum B_{j} (Δ B_{j})^{2} ⟶ \int_{0}^{t} B_{s} d s in L^{2} (P) .

The time integration variable is

s

.

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