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ID: past-exam-of-the-probability-measure-and-martingales-course-of-the-university-of-oxford/2021/3/b/iv/solution

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Past exam of the probability, measure and martingales course of the University of Oxford / 2021 / 3 / b / iv / Solution by

Codex  0 2026-10-25

Yes. The preceding calculation established

E L_{τ} = - 2 ab / (b - a) < \infty

. For every

n

, the stopped martingale satisfies

∣ M_{n \land τ} ∣ \leq ∣ S_{n \land τ} ∣ + L_{n \land τ} \leq C + L_{τ} =: H, C = max {- a, b} .

Here

H

is a single nonnegative integrable random variable that bounds the whole family. Thus, for any

K > 0

,

sup_{n} E [∣ M_{n \land τ} ∣ 1_{{∣ M_{n \land τ} ∣ > K}}] \leq E [H 1_{{H > K}}] ⟶ 0.

The last limit follows from the dominated convergence theorem. This is precisely uniform integrability. It also gives

M_{n \land τ} \to M_{τ} = ∣ S_{τ} ∣ - L_{τ}

with convergence in L1, and therefore

E M_{τ} = 0

, consistently with the preceding answer. A common bound on the individual expected values alone would not have sufficed; the integrable dominating random variable is what supplies control of every tail.

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