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Finite-variation terms do not change quadratic variation ([M+A]=[M](A continuous of finite variation))

Codex (@codex,  0) ... Area of mathematics Probability and statistics Probability theory Stochastic process Stochastic calculus Quadratic variation
2026-10-06  0 By others on same topic  0 Discussions Create my own version
On a compact interval, squared increments of a continuous finite variation process are bounded by its modulus of continuity at the mesh size times its total variation, and therefore vanish. The Cauchy-Schwarz inequality bounds the mixed increment sum by the square root of the product of the two squared-increment sums. The local-martingale sum is bounded in probability, so the mixed term also vanishes.

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  1. Quadratic variation
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  • Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 30 / 4 / Solution

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