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ID: past-exam-of-the-probability-measure-and-martingales-course-of-the-university-of-oxford/2023/2/b/iv/solution

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Past exam of the probability, measure and martingales course of the University of Oxford / 2023 / 2 / b / iv / Solution by

Codex  0 2026-10-25

For

1 \leq j \leq n

, repeated use of the floor function identity gives

X_{j} = 2^{- j} ⌊ 2^{j} X_{n} ⌋ .

Thus the filtration satisfies

F_{n} = σ (X_{n})

. Each positive-probability atom of a sigma-algebra here has the form

A_{n, k} = {k 2^{- n} \leq X < (k + 1) 2^{- n}}, 0 \leq k < 2^{n},

with probability

2^{- n}

;

{X = 1}

is a possible additional null set atom. Averaging over each dyadic interval gives a version

E [h (X) ∣ F_{n}] = 2^{n} \int_{X_{n}}^{X_{n} + 2^{- n}} h (u) d u almost surely .

For a globally defined version of the right side, assign any value, for example zero, on

{X = 1}

.

Indeed, the displayed random variable is

F_{n}

-measurable and bounded by

sup ∣ h ∣

. On each

A_{n, k}

its integral is

2^{- n} 2^{n} \int_{k 2^{- n}}^{(k + 1) 2^{- n}} h (u) d u = E [h (X) 1_{A_{n, k}}],

by the uniform distribution. Every member of

F_{n}

is a union of the displayed sets and the possible zero-probability atom. Summing these identities therefore verifies the defining conditional expectation property on all of

F_{n}

.

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