Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 202 / 1 / b

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- Solution b

Solution

For a simple predictable process

H_{s} = \sum_{i = 0}^{n - 1} H_{i} 1_{(t_{i}, t_{i + 1}]} (s), H_{i} \in L^{\infty} (F_{t_{i}}),

(H \cdot M)_{t} = \sum_{i = 0}^{n - 1} H_{i} (M_{t \land t_{i + 1}} - M_{t \land t_{i}}) .

Each summand is a bounded predictable multiple of a martingale increment, so conditional expectation proves that

H \cdot M

is a martingale. Orthogonality of disjoint martingale increments gives

E ∣ (H \cdot M)_{\infty} ∣^{2} = \sum_{i} E [H_{i}^{2} (M_{t_{i + 1}} - M_{t_{i}})^{2}] < \infty.

It is therefore an

L^{2}

-bounded continuous martingale.

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