Solution

Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 25 / 3 / d / Solution

{title2=$\mathbb E\sup_{t\le1}|A_t^{(n)}-A_t^{(m)}|^2\to0$}

Let $X$ be a continuous martingale on $[0,1]$ with $X_0=0$ and $|X_t|\le C$. For the dyadic squared-increment sums, including the unfinished increment at time $t$, set $A_t^{(n)}=\sum_k(X_{k2^{-n}\wedge t}-X_{(k-1)2^{-n}\wedge t})^2$. The associated martingale transform $M_t^{(n)}=(X_t^2-A_t^{(n)})/2$ satisfies $\mathbb E(M_1^{(n)})^2\le C^4$ and $\mathbb E(A_1^{(n)})^2\le10C^4$. Discrete martingale orthogonality, the path modulus of continuity and the <Cauchy-Schwarz inequality> show that the terminal transforms are Cauchy in $L^2$. The <Doob L2 maximal inequality> then gives $\mathbb E\sup_{t\le1}|A_t^{(n)}-A_t^{(m)}|^2\to0$. The sums need not be increasing in time before taking the limit. This provides an elementary construction of <quadratic variation> under boundedness.

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 Incoming links: Dyadic quadratic variation of a bounded continuous martingale