Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 32 (Rough Path Theory and Applications)

Put $V={\mathbb{R}}^d$ and write $a^{(k)}$ for the degree-$k$ component of a tensor. We use the truncated tensor algebraHere $V^{\otimes 0}=\mathbb{R}$, and products of degree above $N$ are discarded. The answers to the three printed parts follow with these conventions; the middle part is missing from the converted TeX.

In the truncated tensor algebra, let ${\mathfrak{g}}^N(V)$ be the linear span of all iterated Lie brackets of elements of $V$ of length at most $N$, with bracket $[a,b]=a\otimes b-b\otimes a$. Brackets of length greater than $N$ vanish. The free step-N nilpotent Lie group isThe exponential and logarithm are the finite seriesIts multiplication is the tensor multiplication above, its identity is $e=(1,0,\dots ,0)$, and its inverse isThe truncated Baker--Campbell--Hausdorff formula shows that this multiplication stays in the exponential of the Lie algebra. This is the connected simply connected Lie group of the free step-$N$ nilpotent Lie algebra, rather than a discrete free nilpotent group.

Give the first layer $V$ its usual Euclidean norm. An absolutely continuous curve $\gamma$ in $G^N(V)$ is horizontal iffor an absolutely continuous control $x$ in $V$. In the truncated tensor algebra this equation is ${\dot{\gamma }}_t={\gamma }_t\otimes {\dot{x}}_t$, truncated at degree $N$. Its horizontal length is $\int |{\dot{x}}_t|dt$. The Carnot-Carathéodory distance isThe first-layer Lie brackets span the full Lie algebra, so horizontal curves reach every endpoint. Thus this infimum is finite; it is a left-invariant distance inducing the usual manifold topology. Dilations ${\delta }_{\lambda }(a{)}^{(k)}={\lambda }^k{a}^{(k)}$ give $d_{CC}({\delta }_{\lambda }g,{\delta }_{\lambda }h)=\lambda d_{CC}(g,h)$ for $\lambda >0$. Equivalently $d_{CC}(e,g)$ is the least length of a bounded variation control whose step-$N$ path signature is $g$.

For $p\ge 1$ set $N=\lfloor p\rfloor$. A weak geometric p-rough path is a continuous path $\mathbf{X}:[0,T]\to G^N(V)$, usually based at $e$, withIts increments ${\mathbf{X}}_{s,t}={\mathbf{X}}_s^{-1}\otimes {\mathbf{X}}_t$ obey ${\mathbf{X}}_{s,u}={\mathbf{X}}_{s,t}\otimes {\mathbf{X}}_{t,u}$. Equivalently their degree-$k$ tensor components have finite $p/k$-variation and satisfy the path signature algebraic constraints. This is a group-valued definition; being a limit of smooth path signatures in the same $p$-variation topology is the stronger definition of a geometric p-rough path.

For $V={\mathbb{R}}^2$ with basis $e_1,e_2$, the step-two Lie algebra has basis $e_1,e_2,[e_1,e_2]$ and its last element is central. WriteIts first level is $v=(x,y)$ and its second level isThe step-two multiplication givesConsequently the mapis an isomorphism onto the real Heisenberg group. The shift by $xy/2$ converts exponential coordinates into the matrix coordinates used for that group. In exponential coordinates the left-invariant horizontal vector fields areTaking $X,Y$ orthonormal gives exactly the Carnot-Carathéodory distance inherited from $G^2({\mathbb{R}}^2)$.

Euclidean Lipschitz continuity does not imply Carnot-Carathéodory Lipschitz continuity. In the exponential coordinates above take the vertical path $\gamma (t)=(0,0,t)$. It is Euclidean Lipschitz continuous, butIndeed left invariance reduces to the central increment, graded dilation gives the square-root factor, and reversal changes the sign without changing the distance.

One can also see the scale directly. A horizontal curve starting at the identity satisfies $\dot{z}=(x\dot{y}-y\dot{x})/2$. If it ends on the vertical axis, its planar projection is a closed loop andFor a projection of length $L$ starting at zero, $\sup |(x,y)|\le L$, so $|z|\le L^2/2$. Thus reaching a central displacement $z$ requires length at least $\sqrt{2|z|}$, while a square loop with the appropriate signed area gives length at most $4\sqrt{|z|}$. In particular $d_{CC}(\gamma (s),\gamma (t))/|t-s|$ is unbounded as $t\to s$. Vertical displacement records area, which scales quadratically with horizontal length.

For a Lipschitz continuous path $x:[0,T]\to V$, define its signature of a bounded variation path in the truncated tensor algebra byThe Riemann-Stieltjes integrals exist since $x$ has bounded variation. Equivalently, since $x$ is absolutely continuous, the integrand is $x^{\prime }(u_1)\otimes \cdots \otimes x^{\prime }(u_k)$ with Lebesgue integration on the ordered simplex. The degree-zero component is $1$, andThe lift solves $dS=S\otimes dx$, starting at the identity. Since its velocities belong to the first layer, its endpoint lies in the free step-N nilpotent Lie group.

For $s\le t\le u$, partition the degree-$k$ integration simplex according to how many integration variables lie before $t$. If exactly $j$ do, the integral factors as the first $j$-fold integral on $[s,t]$ tensor the last $(k-j)$-fold integral on $[t,u]$. The boundaries with a time equal to $t$ have zero measure. Thuswhich is precisely the Chen identity:This proves the identity at every level, including $k=0$, rather than just checking the first two levels.

For a strictly increasing continuously differentiable map $\psi :[a,b]\to [0,T]$, the actual change of variables formula isTo prove it, substitute $u_j=\psi (v_j)$ in each ordered integral. Strict increase preserves the order, and the almost-everywhere chain rule gives $d(x\circ \psi )(v)=x^{\prime }(\psi (v)){\psi }^{\prime }(v)dv$. The resulting integral is over $\psi (a)<u_1<\cdots <u_k<\psi (b)$. A vanishing derivative at some points causes no difficulty: monotone continuously differentiable substitution still applies, and the zero derivative contributes no variation.

Therefore orientation-preserving reparametrisation of the whole segment preserves its path signature. On $[0,1]$ the printed equality needs $\psi (0)=0$ and $\psi (1)=1$, equivalently that $\psi$ be onto. These endpoints are not forced by the stated assumptions. For example $x(t)=t{e}_1$ and $\psi (t)=t/2$ satisfy those assumptions, butwhich already differ at level one for every $N\ge 1$. The endpoint-preserving interpretation proves the intended invariance, while the general formula states exactly what remains true without it.

Fix $T<\infty$ and a standard $d$-dimensional Brownian motion $B$ starting at zero. Let $D_n$ be deterministic nested finite partitions of $[0,T]$ with mesh tending to zero; dyadic partitions are the basic example. Let $B^n$ be the polygonal interpolation through the partition values and put ${\mathbf{B}}^n=S_2(B^n)$. The limit is enhanced Brownian motion, with incrementsHere the integral is a Stratonovich integral. In terms of Itô integrals,Its symmetric part is $\frac{1}{2}{B}_{s,t}^{\otimes 2}$, while its antisymmetric part is Lévy area,These identities and splitting the integral at an intermediate time give the Chen identity, so $\mathbf{B}$ has values in $G^2({\mathbb{R}}^d)$.

A precise nested polygonal approximation of Brownian rough paths statement is: for every $0<\alpha <1/2$,where the rough path metric used here isIn particular, for each $2<p<3$, choosing $1/p<\alpha <1/2$ gives convergence in the inhomogeneous $p$-variation rough path metric. The limiting signed area is the Lévy area, and the limiting full second level is Stratonovich, not Itô. This distinction is necessary: a polygonal lift has the prescribed symmetric second level, whereas an Itô lift would lack half the quadratic-variation correction.

Here is a proof sketch emphasizing the nested martingale structure. SetA Brownian bridge between two prescribed endpoints has the affine interpolation as conditional mean. ConsequentlyFor $i\ne j$ the coordinate processes remain independent when their respective partition values are conditioned on. Approximating the Itô integral by its left sums, conditioning each product, and then taking its $L^2$ limit givesIndeed each conditioned factor is its polygonal interpolation, and their sums converge to the ordinary integral of those piecewise linear functions. For a fixed pair $s,t$, the right side is therefore a martingale in the refinement index $n$. It is not being asserted to be a martingale in physical time; polygonal interpolation uses the right endpoint of its current cell.

The union of the partitions is dense. Continuity of $B$ implies that ${\mathcal{G}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{G}}_n)$ contains the full Brownian path and its stochastic integrals, up to completion. The martingale convergence theorem for conditional expectations therefore yields convergence of $B_t^n$ and all off-diagonal ${\mathbb{B}}_{s,t}^{n,ij}$ at every fixed pair, almost surely and in $L^q$ for finite $q$. Simultaneously it holds for a countable dense collection of pairs. The diagonal terms must be handled separately:They are not conditional expectations of the diagonal Brownian second level: the latter would also contain a conditional variance.

Pointwise convergence alone is insufficient for the rough path metric. The Burkholder-Davis-Gundy inequalities, Brownian scaling and Gaussian increment moments give, for arbitrarily large $q$,Equivalently, using the homogeneous norm on $G^2$, the $q$-th moment of the Carnot-Carathéodory distance of an increment is bounded by $C_q|t-s{|}^{q/2}$. The metric form of the Kolmogorov continuity theorem, with $q$ chosen large, gives for every $\beta <1/2$ a random $K_{\beta }$ having all finite moments such thatConditioning these bounds givesThe diagonal bounds follow by squaring the first bound. The two conditional-expectation processes are nonnegative martingales; the Doob Lp maximal inequality makes their suprema over $n$ finite almost surely, with all needed moments. Thus all the polygonal lifts have a common random bound at exponent $\beta$.

These bounds and the Chen identity give joint equicontinuity of the increments as functions of $(s,t)$. Convergence on the dense collection of pairs consequently gives uniform convergence of both levels on the time triangle. Take $\alpha <\beta <1/2$. For either level $k=1,2$, with $M$ a common bound and ${\epsilon }_n$ its uniform difference, interpolation givesThis proves the stated Hölder rough path metric convergence. Finally, for a partition $D$, the $k$-th level difference satisfiesbecause $\alpha p>1$. Taking the appropriate powers and suprema proves $p$-variation convergence too. Nesting is what permits the conditional-expectation martingale proof; without nesting one instead needs direct approximation estimates.

Work on a finite interval $[0,T]$ with a fixed initial point $y_0\in {\mathbb{R}}^m$. To give a clean global statement, assume $V_0,V_1,\dots ,V_d$ are bounded vector fields with bounded continuous derivatives through order three. Consider the Stratonovich integral form of the stochastic differential equationFor a control in the Cameron-Martin space of Wiener measure,let $y^h$ solve the controlled ordinary differential equationThe Stroock-Varadhan support theorem says, in the uniform metric on continuous paths,Thus a path is in the support of a measure of the diffusion precisely when it can be approximated by these deterministic controlled trajectories. There is no ellipticity assumption: the allowed controls and vector fields already encode any restrictions on reachability.

The central point of the rough-path proof is that ordinary uniform approximation of the driver does not make the solution map continuous. Small planar loops can have small amplitude but accumulated area of fixed size, and vector fields with a nonzero Lie bracket detect that area. The correct driver is enhanced Brownian motion $\mathbf{B}$, including Lévy area. Fix $2<p<3$. The universal limit theorem gives a continuous mapfrom the geometric $p$-variation rough-path space to solution paths. It agrees with the controlled ordinary differential equation when $\mathbf{X}=S_2(h)$. Adding the time coordinate, with its mixed integrals defined by integration against $dt$ and integration by parts, incorporates $V_0$ without changing the continuity argument. This is the role of the universal limit theorem: it turns a support problem for all these nonlinear diffusions into a support problem for one enhanced driving process.

For $\mathbf{X}=\mathbf{B}$, the rough differential equation is the stated Stratonovich equation. Its second-order increment includesReplacing the Stratonovich integral in the second level by its Itô integral expression contributes $\frac{1}{2}{\sum }_{i}D{V}_i{V}_{i}dt$, which is exactly the Stratonovich-to-Itô correction. Polygonal convergence from Question 3 and the universal limit theorem therefore give the Wong-Zakai approximation: the ordinary equations driven by $B^n$ converge to this Stratonovich diffusion.

It remains to establish the support of enhanced Brownian motion:One inclusion follows immediately from Question 3. Every polygonal $B^n$ belongs to $H$, and $S_2(B^n)\to \mathbf{B}$ almost surely in the rough path metric. Since $K$ is closed, $\mathbf{B}\in K$ almost surely, whence its support is contained in $K$. This alone does not prove the reverse inclusion: convergence of random approximations does not automatically give positive mass near every control.

For that inclusion, take dyadic mesh ${\delta }_n=T{2}^{-n}$ and writeOn each grid interval $R^n$ is a pinned Brownian bridge. The bridges are mutually independent and independent of the grid values defining $B^n$, by the Gaussian projection formula. Their canonical Stratonovich enhancement ${\mathbf{R}}^n$ is therefore independent of $B^n$. It is important to enhance the bridge residual itself, including its area, rather than infer area smallness from its uniform size.

The bridge enhancement tends to the zero rough path in probability. Here are the estimates behind this assertion. A bridge on an interval of length ${\delta }_n$ scales as $\sqrt{{\delta }_n}$ times a unit bridge; its full antisymmetric area has mean zero, moments of order ${\delta }_n$, and second moment of order ${\delta }_n^2$. Since its endpoint displacement is zero, at grid points the cumulative second level is just the sum of the independent bridge areas. The Doob L2 maximal inequality givesPartial intervals at the two ends contribute bridge increments and at most two local areas. Bridge scaling, Gaussian moment bounds and the corresponding iterated-integral estimates yield, uniformly in $n$ and for arbitrarily large $q$,For an interval crossing several grid cells, the full-cell second levels are independent centered areas, whose sum has moments bounded by $C_q((t-s){\delta }_n{)}^{q/2}$; the two boundary pieces and their cross product satisfy the displayed bound as well. For an interval inside one cell these are the usual scaled bridge increment and area bounds. At each fixed pair $(s,t)$ both levels converge to zero in probability: the first-level variance is $O({\delta }_n)$, and the second-level second moment is $O((t-s){\delta }_n+{\delta }_n^2)$ for sufficiently fine grids. The Kolmogorov continuity theorem gives uniformly bounded moments of a Hölder rough-path norm at any exponent $\beta <1/2$. Tightness, convergence on a dense set of pairs and interpolation to $\alpha <\beta$ now give ${\rho }_{\alpha }({\mathbf{R}}^n,e)\to 0$ in probability. Taking $1/p<\alpha <1/2$ gives $d_p({\mathbf{R}}^n,e)\to 0$ in probability.

Fix $h\in H$. Its polygonal interpolations $h^n$ converge to $h$ in the total-variation norm: their derivatives are the cell averages of $\dot{h}$, which converge in $L^2$, hence in $L^1$. Moreover $|h^n{|}_{1\text{-var}}\le |h{|}_{1\text{-var}}$. Thus $S_2(h^n)\to S_2(h)$ in the rough path metric.

The rough path translation operation satisfiesFor a bounded variation path $b$, translation adds $b$ at level one and, at level two, the mixed termsThe second mixed term is defined by integration by parts. Their variation bounds show that translation is continuous, uniformly when $|b{|}_{1\text{-var}}$ is bounded, near the zero rough path. More explicitly the mixed second-level norms are bounded by a constant times $\Vert R{\Vert }_{p\text{-var}}|b{|}_{1\text{-var}}$; the other new second-level term is the path signature of $b$.

Given a ball of radius $\epsilon$ around $S_2(h)$, choose a sufficiently large $n$ so that $S_2(h^n)$ is close andfor the small $\eta$ supplied by translation continuity. The relevant variation bounds can be chosen uniformly in $n$, since the variations of the $h^n$ are uniformly bounded. For this fixed $n$, the event $|B^n-h^n{|}_{1\text{-var}}<\eta$ has positive probability: the finite vector of Brownian increments has an everywhere positive Gaussian density, and sufficiently small deviations of its vertices imply this variation bound. These two events are independent. On their intersection, translation continuity gives $d_p(\mathbf{B},S_2(h))<\epsilon$. Every ball around $S_2(h)$ therefore has positive probability. This proves $S_2(h)$ lies in the support, and taking closures gives $K$.

Finally, a continuous map between the separable path spaces in use satisfiesFor one direction the measure is concentrated on its support. For the other, an open neighborhood of $\Phi (x)$ pulls back to an open neighborhood of a support point $x$, which has positive measure. Using continuity of $\Phi$ and the identity $\Phi (S_2(h))=y^h$ proves the boxed Stroock-Varadhan support theorem. This separates the two ingredients: the probabilistic support calculation for Brownian motion and area, and the deterministic continuity supplied by the universal limit theorem.

If the diffusion is instead presented as an Itô integral equation with drift $b$, the controlled equation must useUsing $b$ unchanged in the controlled equation would generally give the wrong support description. The result above states explicit sufficient smoothness and nonexplosion assumptions; extensions to more general coefficients require the usual localization, not an unqualified assertion for arbitrary vector fields.