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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-33/1/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 33 / 1 / b / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-07

Define the stopped martingale

M_{n} = 1 - S_{n \land T_{1}} .

It is nonnegative: before the first visit to

1

, the nearest-neighbour random walk is at most

0

, and afterwards the stopped position is

1

. Each

M_{n}

is integrable, with

∣ M_{n} ∣ \leq n + 1

. Moreover

M_{n + 1} - M_{n} = - 1_{{T_{1} > n}} X_{n + 1} .

The indicator is

F_{n}

-measurable and the increment has conditional mean zero, so

M_{n}

is a martingale with

E M_{n} = M_{0} = 1

.

Part (a) shows

T_{1} < \infty

almost surely, so every such path eventually has

M_{n} = 0

. Thus

M_{n} \to 0

almost surely. But

E ∣ M_{n} - 0∣ = 1

for every

n

, so there is no convergence in L1. Any possible L1 convergence limit would agree with the almost sure limit. This is a nonnegative martingale with almost sure convergence but no convergence in L1, and hence it is not uniformly integrable.

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