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ID: past-exam-of-the-probability-measure-and-martingales-course-of-the-university-of-oxford/2021/2/b/iv/solution

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Past exam of the probability, measure and martingales course of the University of Oxford / 2021 / 2 / b / iv / Solution by

Codex  0 2026-10-25

For

B_{n} = {X_{n} > a_{n}}

, the definition of the common cumulative distribution function gives

P (B_{n}) = 1 - F (a_{n}) .

The strict exceedance matters here: an atom of a measure at

a_{n}

belongs to

{X_{n} \leq a_{n}}

, so it is already included in

F (a_{n})

. The events

B_{n}

are mutually independent events, since each depends on a separate member of the independent random variables

X_{n}

.

If

\sum_{n} (1 - F (a_{n})) < \infty

, countable subadditivity gives, for every

N

,

P (⋃_{n \geq N} B_{n}) \leq \sum_{n \geq N} (1 - F (a_{n})) ⟶ 0.

Continuity of probability then gives

P (lim sup_{n} B_{n}) = 0

. This is the First Borel-Cantelli lemma and does not need independence.

If the series diverges, independence and

1 - u \leq e^{- u}

give, for every fixed

N

,

P (⋂_{n = N}^{M} B_{n}^{c}) = \prod_{n = N}^{M} F (a_{n}) \leq exp (- \sum_{n = N}^{M} (1 - F (a_{n}))) ⟶ 0.

Taking

M \to \infty

shows that the probability of no exceedance after

N

is zero. A countable union over

N

shows that finitely many exceedances has probability zero. This is the Second Borel-Cantelli lemma. Together with the preceding event equality, it proves

P (A) = {0, 1, \sum_{n \geq 1} (1 - F (a_{n})) < \infty, \sum_{n \geq 1} (1 - F (a_{n})) = \infty.

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