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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-201/4/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 201 / 4 / b / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

The

2^{n}

increments are independent normal random variables, each distributed as

2^{- n /2} B_{1}

. Their squared means are

2^{- n}

and, using the permitted variance, their squared variances are

2 \cdot 2^{- 2 n}

. Independence therefore gives

E [B]_{n} = 2^{n} 2^{- n} = 1, var ([B]_{n}) = 2^{n} 2 \cdot 2^{- 2 n} = 2^{1 - n} .

Consequently the dyadic quadratic variation converges to one:

E ([B]_{n} - 1)^{2} = 2^{1 - n} ⟶ 0, [B]_{n} ⟶ 1 in L^{2} .

This is dyadic quadratic variation of Brownian motion on the time interval of length one. The calculation requires independence only within each partition; increments from different resolutions need not be independent.

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