Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 5 / 1

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- Solution 1

Solution

A simple predictable process has the form

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

where each bounded

ξ_{i}

is

F_{t_{i}}

-measurable. Define

(H \cdot B)_{t} = \sum_{i} ξ_{i} (B_{t \land t_{i + 1}} - B_{t \land t_{i}}) .

Independent centered Brownian increments show directly by conditioning that this is a martingale. The same conditional expansion, using

E [(B_{v} - B_{u})^{2} ∣ F_{u}] = v - u

, shows that

(H \cdot B)_{t}^{2} - \int_{0}^{t} H_{s}^{2} d s is a martingale .

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