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A rough path is defined as an ordered pair $ (X, \mathbb X)$, where $X$ is a path mapping from $[0,T]$ to some Banach space $V$ and $\mathbb X:[0,T]^2 \mapsto V^2$ is another mapping for additional information on the curve $X$.

I am not quite into their motivation, although there are some discussions online. In particular, I find the following remark in the first paragraph of chapter 9 of the book (link) ‘Multidimensional Stochastic Processes as Rough Paths: Theory and Applications’ by Friz and Victoir:

Consider $X$ of finite $p$-variation with $p\ge 2$. … the knowledge of higher indefinite iterated integrals up to order $N = [p]$ must be an apriori information, i.e. assumed to be known.

Intuitively, I thought, as long as the curve $X$ is given, all the attached information (like $\mathbb X$) shall not be apriori, i.e. one can obtain (may be hard) $\mathbb X$ from the given $X$, which is contrary to the above.

There are many discussions on the rough path theory. However, is there any explanation on the above statement in a easier way, which can be understood to a person who has knowledge of Ito stochastic analysis but none of rough path theory?

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2 Answers 2

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Some of the confusion may be caused by the use of the word "information". You are right that in a probabilistic context, one would typically like to build $\mathbb{X}$ as a measurable function of $X$, so in this sense $X$ would contain all the information required to build $\mathbb{X}$. The point they are making is that there is no canonical way to do this, so different constructions may produce different choices of $\mathbb{X}$. In this sense, $\mathbb{X}$ does encode some information not contained in $X$, since it indirectly tells you something about which construction you've used to produce it. For example, if $X$ is a Brownian motion, you can construct $\mathbb{X}$ either by Itô integration or by Stratonovich integration (or in some other way), and inspecting $\mathbb{X}$ would reveal information about your choice of integration.

Going back to the motivation, the aim is to use $X$ to solve differential equations of the type $$ \dot Y = F_0(Y) + \sum_{i=1}^m F_i(Y) \dot X_i\;, $$ or, if you prefer, $$ dY = F_0(Y)\,dt + \sum_{i=1}^m F_i(Y) \,dX_i(t)\;. $$ It turns out that the solution map $S\colon X \mapsto Y$ is simply not continuous in the $p$-variation topology as soon as $p \ge 2$. Similarly, it is not continuous in the $\alpha$-Hölder topology as soon as $\alpha \le 1/2$. What this means is that for any given $X$ (even a smooth one), you can find sequences of smooth functions $X^{(n)}$ and $\smash{\bar X}^{(n)}$ that both converge to $X$ in, say, the ${1\over 2}$-Hölder topology, but such that the solutions $S(X^{(n)})$ and $S(\smash{\bar X}^{(n)})$ either converge to different limits, or fail to converge at all. In this sense, $X$ itself contains "not enough information", because different ways of approximating it may lead to different outcomes.

The point of rough path theory is to figure out what additional information should be added to $X$ so that continuity is restored, and this is precisely what $\mathbb{X}$ encodes. Note also that this is quite standard procedure when weak forms of convergence are involved. Think of Young measures or of varifolds: in both cases they encode some "additional information" required to apply some nonlinear transformation to an object obtained as a weak limit.

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  • $\begingroup$ So nice having your wonderful answer. Now it clears my confusion on the use of "additional information". $\endgroup$
    – kenneth
    Commented May 25, 2015 at 13:23
  • $\begingroup$ Excellent and intuitive explanation. Thanks. $\endgroup$ Commented Jan 4, 2020 at 18:21
  • $\begingroup$ TeX note, while this is on the front page: $X^{(n)}\bar X^{(n)}$ X^{(n)}\bar X^{(n)} puts the latter superscript too high, because it bases its position on the full box $\bar X$, not just on $X$. You can fix it with \smash: $X^{(n)}\smash{\bar X}^{(n)}$ X^{(n)}\smash{\bar X}^{(n)}. I edited accordingly. $\endgroup$
    – LSpice
    Commented Nov 13, 2023 at 17:41
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    $\begingroup$ @LSpice Thanks, good to know. $\endgroup$ Commented Nov 13, 2023 at 21:13
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You have seen all kinds of integration theories before — Itô, Stratonovich, and I'm sure plenty others. Rough paths takes a step back and asks what we want from an integration theory. And so long as we want three basic axioms we are forced to study the iterated integrals.

We want to create an integration theory for

$$\int_0^t F(X(r))dX(r),$$

where $X$ is a signal that is $\alpha$-Hölder continuous with $\alpha\in (1/3,1/2]$ and $F$ is a smooth function. We want our integration theory to satisfy the following three properties.

  1. (Integral of $1$) We want $$\int_s^t 1dX(r)=X(t)-X(s).$$

  2. (Chastles' relation) $$\int_s^u F(X(r))dX(r)+\int_u^t F(X(r)) dX(r)=\int_s^t F(X(r)) dX(r).$$

  3. (Linearity) $$\int_s^t F_1(X(r))+cF_2(X(r))dX(r)=\int_s^t F_1(X(r))dX(r)+c\int_s^t F_2(X(r))dX(r).$$

Then for a partition of $[0,t]$, $\mathcal P=\{0=t_0<\dotsb<t_n=t\}$ we have that the integral \begin{align} \int_0^t F(X(r)) dX(r)&=\sum_{k=0}^n \int_{t_k}^{t_{k+1}} F(X(r))dX(r) \nonumber \\ &=\sum_{k=0}^n \int_{t_k}^{t_{k+1}} F(X(t_k))+F'(X(t_k))(X(r)-X(t_k))+O(|r-t_k|^{2\alpha})dX(r) \nonumber \\ &=\sum_{k=0}^n \biggl(F(X(t_k))(X(t_{k+1})-X(t_k))+F'(X(t_k))\int_{t_k}^{t_{k+1}}(X(r)-X(t_k)) dX(r) \nonumber \\ &~+O(|t_{k+1}-t_k|^{3\alpha})\biggr). \end{align} As $3\alpha>1$, the remainder term should go to $0$ as the mesh size goes to $0$. This reduces the problem of defining the integral $\int_0^t F(X(r))dX(r)$ to just defining $\int_{t_k}^{t_{k+1}}(X(r)-X(t_k)) dX(r)$.

The point is that if we want these three basic properties, the integral $\int_0^t F(X(r))dX(r)$ is defined if $\int_{t_k}^{t_{k+1}}(X(r)-X(t_k)) dX(r)$ is. Rough paths reduces defining $\int_0^t F(X(r))dX(r)$ for every $F$ to just $F$ the identity. Not only this, but it proposes a definition of integral as a standard Riemann–Stieltjes integral with a correction term.

We must construct some object $\mathbb X:\Delta_2^T\to \mathbb R^{d\times d}$ where $d$ is the dimension of the signal $X$ where $\Delta_2^T=\{(s,t)\in [0,T]^2:s\leq t\}$. The $(i,j)$th component of $\mathbb X$ represents the integral of the $i$th component of $X$ against the $j$th component.

What properties would we like the iterated integral to satisfy?

Well first it better be $2\alpha$-Hölder. That is $$\sup_{s\neq t}\frac{|\mathbb X_{s,t}|}{|t-s|^{2\alpha}}<\infty.$$

Second, we want $\mathbb X$ to algebraically relate to $X$. What properties should it satisfy? If we want to solve differential equations, we first consider smooth $X$ and look at

$$Y_s'(t)=Y_s(t) X'(t),$$ with initial condition $Y_s(s)=1$. This has solution $$Y_s(t)=\exp(X(t)-X(s))=1+\int_s^t dX(r)+\int_s^t \int_s^{r_2} dX(r_1)dX(r_2)+\dotsb.$$

We want multiplicativity $Y_s(t)=Y_s(u)Y_u(t)$. On the level of iterated integrals we have the following relations

\begin{align*} 1&=1\\ \int_s^t dX(r)&=\int_s^u dX(r)+\int_u^t dX(r)\\ \int_s^t \int_s^{r_2} dX(r_1)dX(r_2)&=\int_s^u \int_s^{r_2} dX(r_1)dX(r_2)+\int_u^t \int_u^{r_2} dX(r_1)dX(r_2)+\int_s^u dX(r)\otimes\int_u^t dX(r)\\ &\cdots. \end{align*}

The third equation is called Chen's relation.

A rough path is therefore a pair, $(X,\mathbb X):\Delta_2^T\to \mathbb R^d\oplus \mathbb R^{d\times d}$, satisfying the following properties

\begin{align*} &X_{s,t}=X(t)-X(s)\\ &\sup_{s\neq t}\frac{|\mathbb X_{s,t}|}{|t-s|^{2\alpha}}<\infty\\ &\mathbb X_{s,t}=\mathbb X_{s,u}+\mathbb X_{u,t}+X_{s,u}\otimes X_{u,t}. \end{align*}

The second order process can always be defined (this is the Lyons-Victoir extension theorem), although the construction given by Lyons-Victoir is abstract.

However, the second order process $\mathbb X$ is not unique as given any rough path $(X_{s,t},\mathbb X_{s,t})$ and any $f\in C^{2\alpha}$ we have that $(X_{s,t},\mathbb X_{s,t}+f(t)-f(s))$ is also a rough path (just check the properties). Conversely, any two rough paths differ by the addition of the increment of some $f\in C^{2\alpha}$. Therefore rough paths give a way of "parameterizing" integration theories that satisfy those three axioms through this function $f$.

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    $\begingroup$ Great explanation! $\endgroup$
    – Nate River
    Commented Nov 13, 2023 at 15:57
  • $\begingroup$ TeX note: instead of using {align}+\nonumber, you can use {aligned} inside an {equation} to get just one number. I forget if MathJax recognises this construct, but it doesn't matter, because MathJax also doesn't automatically number {align}s (although it does respect \tags)—so I assume that this was authored elsewhere. $\endgroup$
    – LSpice
    Commented Nov 13, 2023 at 17:45
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    $\begingroup$ Very good explanation. $\endgroup$
    – kenneth
    Commented Nov 14, 2023 at 21:33

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