Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 201 / 2 / a / ii

Solution

n \geq 0

and

A \in F_{n}

T_{A} = {n + 1, n, ω \in A, ω \in / A .

Indeed, its only nontrivial sublevel event is

{T_{A} \leq n} = A^{c} \in F_{n}

. Applying the assumed stopped-expectation equality to

T_{A}

and to the deterministic stopping time

n

gives

0 = E (M_{T_{A}} - M_{n}) = E [1_{A} (M_{n + 1} - M_{n})] .

This holds for every

A \in F_{n}

. Since

M_{n}

is adapted and both variables are integrable, the defining test-event property of conditional expectation yields

E [M_{n + 1} ∣ F_{n}] = M_{n} .

Together with the given adaptation and integrability, this is exactly the martingale property.

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