Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 31 / 3

For Brownian motion $B$ starting at zero, define its enhanced Brownian motion by the Stratonovich second level:The associated group-valued path is ${\mathbf{B}}_t={\mathbf{B}}_{0,t}$. The Stratonovich chain rule givesConsequently its symmetric second level is $\frac{1}{2}{B}_{s,t}\otimes B_{s,t}$ and the remaining antisymmetric part $A_{s,t}=\frac{1}{2}({\mathbb{B}}_{s,t}-{\mathbb{B}}_{s,t}^{\mathsf{T}})$ is Lévy area. Splitting the integral at $u$ proves the Chen identity, so these increments lie in $G^2({\mathbb{R}}^d)$ and are the relative increments of ${\mathbf{B}}_t$.

The group norm for the Carnot-Carathéodory distance is comparable to $|B_{s,t}|+|A_{s,t}{|}^{1/2}$. The planar construction in Question 2 extends to dimension $d$: prescribe every coordinate-plane area with a closed rectangle, concatenate these central-area loops, and finish with the displacement segment. The converse bound follows from length bounds on displacement and on each area integral. Thus the comparison constants depend only on $d$.

Brownian scaling and stationarity of increments imply that $(B_{s,t},A_{s,t})$ has the distribution of $(\sqrt{t-s}{B}_{0,1},(t-s)A_{0,1})$. The permitted moment estimates for Lévy area, together with Gaussian moments of Brownian displacement, therefore give, for every finite $q\ge 1$,This also verifies the hypothesis of the metric-valued Kolmogorov continuity theorem. We give the dyadic argument to obtain a simultaneous version explicitly.

Fix $0<\beta <1/2$ and choose $q$ so large that $q(1/2-\beta )>1$. The Markov inequality and a union bound giveThe right-hand side is summable. By the Borel-Cantelli first lemma, almost surely there is a finite random constant $K_{\beta }$ bounding every adjacent dyadic increment by $K_{\beta }{2}^{-n\beta }$, including the finitely many exceptional levels.

Decompose any interval with dyadic endpoints into successive dyadic intervals, with at most two intervals of each size and no size exceeding its length. The metric triangle inequality and the sum of the geometric series givefor all dyadic $s,t$. The finite-dimensional Carnot group is complete, so the path extends uniquely from dyadic times to a continuous $\beta$-Hölder path. At each fixed time the original stochastic enhancement and its values at approaching dyadic times converge in probability by the moment estimate. The extension has the same limit almost surely, hence is a modification of the original enhancement.

Perform this construction on one countable sequence of exponents increasing to $1/2$. The extensions agree on the dense set of dyadic times, so they give one modification that is Hölder at every exponent below $1/2$. On $[0,1]$, a $\beta$-Hölder estimate implies the same estimate at every $0\le \alpha \le \beta$. Therefore

Put ${\mathcal{F}}_n=\sigma (B_{k{2}^{-n}}:0\le k\le 2^n)$. These sigma-fields increase, and their union generates the continuous Brownian path. Let ${\mathbf{B}}^{(n)}=S_2(B^{(n)})$ be the polygonal enhancement. Its coordinates are determined by its displacement $b_{s,t}^{(n)}$ and antisymmetric area $a_{s,t}^{(n)}$; the symmetric second level is $\frac{1}{2}{b}_{s,t}^{(n)}\otimes b_{s,t}^{(n)}$.

The given conditional-expectation identities implyFor the second assertion it is important to use area coordinates rather than asserting that the whole tensor signature is a conditional expectation. Write $I_t={\int }_0^t\beta d\tilde{\beta }$. The local off-diagonal second integral isConditional on their separate dyadic values, the two independent Brownian components remain independent. Thus the conditional expectation of the product in the last term is ${\beta }_s^{(n)}({\tilde{\beta }}_t^{(n)}-{\tilde{\beta }}_s^{(n)})$. The assumed identity for $I$ gives the polygonal $12$ entry. The reverse entry follows by integration by parts, since the product of the two component increments also has the product of their conditional means. Taking the antisymmetric part proves the area identity. Diagonal second levels are recovered from the geometric identity, rather than from a conditional-expectation claim about squares.

For each fixed dyadic time $t$, the first-level and area conditional expectations are uniformly integrable martingales in $n$. Their targets are measurable with respect to ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$: Brownian area is a limit of stochastic integral sums determined by the path. The martingale convergence theorem gives convergence to $B_t$ and $A_{0,t}$, respectively. The continuous group coordinate formulas then giveat all dyadic times on a common event of probability one.

For the uniform Hölder estimate, fix $\alpha <1/2$ and put $K=\Vert \mathbf{B}{\Vert }_{\alpha \text{-Ho¨lder};[0,1]}$. The given all-order integrability and the group norm comparison implyUse the conditional-expectation formulas, the conditional Jensen inequality, and $\mathbb{E}[K\mid {\mathcal{F}}_n]\le \sqrt{\mathbb{E}[K^2\mid {\mathcal{F}}_n]}$. The group norm comparison yieldsFirst take all dyadic pairs, a countable collection, and then extend the inequality to all $s,t$ by continuity. Hence the Brownian rough path martingale Hölder bound isFor any $q>1$, the Doob Lp maximal inequality givesTherefore

To upgrade convergence from dyadic times to every time simultaneously, apply this bound at one positive exponent $\beta <1/2$. The polygonal enhancements are then equicontinuous in the group metric, with a common finite random constant. For a finite dyadic time net, convergence holds at every net point. The triangle inequality bounds the errors at the remaining times by the two Hölder errors to the nearest net point and the error at that point. First let $n\to \infty$, and then let the net mesh tend to zero. This proves uniform convergence, and in particular convergence for every $t\in [0,1]$ on one event of probability one.

There is also convergence in each weaker Hölder rough path metric. Uniform convergence of the group coordinates gives uniform convergence of their levelwise increments through the Chen formulas. If $\alpha <\beta <1/2$, the uniform $\beta$ bounds bound the level-$k$ increment differences by a random constant times $|t-s{|}^{k\beta }$, while their unweighted suprema tend to zero. Interpolating these two bounds proves convergence with denominators $|t-s{|}^{k\alpha }$, for $k=1,2$. This establishes the nested polygonal approximation of Brownian rough paths in the topology needed for the support argument.