http://www.hairer.org/notes/RoughPaths.pdf here is a textbook, but I am completely lost at the definition. It is defined on page 13, chapter 2. A rough path is defined as an ordered pair, $(X,\mathbb{X})$ where $X$ is a continuous process and $\mathbb{X}$ is a higher ordered area process which defines certain integrals that satisfy "Chen's Relation". What is this area process and what does it have to do with the original continuous function? What are these integrals (which he emphasizes are defined by the area process, not the other way around)? So could the area process be anything satisfying Chen's relations?
Could someone explain rough path theory? More specifically, what is the higher ordered "area process" and what information is it giving us?
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Consider an equation like \begin{equation}\tag{1}dY_t = f(Y_t) dW_t\end{equation} where $Y_t$ is an unknown function and $W_t$ is a continuous, but not differentiable, function.
If $W_t$ is Brownian motion then there is a classical theory for how to understand (1). Brownian motion has finite $p$-variation only for certain values of $p$. As you may know, Brownian motion on the interval $[0,T]$ has quadratic variation equal to $T$.
What if $W_t$ is not Brownian motion, but just some function with finite $p$-variation for some other $p$? Then the approach is to just fix a path of $W_t$ (so it's not a random process anymore, just a single non-differentiable function). Then Lyons shows how to make sense of (1) in terms of $$\tag{2}\int W_s dW_s$$ We don't know what (2) means, so we write down some axioms for what (2) should be like (those are Chen's relations I guess), and use that as a starting point for the theory.
(This is based on my reading of the first page of http://www.maths.ed.ac.uk/~adavie/rpathrev.pdf)
I'm not an expert on this, but I did some research and found this Topics in Gaussian rough paths theory. I suggest you read it from the begining, pay special attention on page 5 and 6.
From all the books and papers I found on Google this paper gives the most comprehensible explanation why you need those iterated integrals. Thy naturaly occur when you construct the solution to SDE by Picard's method. Also they provide necessary extra info which enables you do certain stohastic integrals.
I will cite some things from the above paper to back up this:
This is on page 5.
That abstract setting is what makes it difficult to understand from Lyons work. Here is the book by the way Book
There are other papers which give other interpretations e.g.
Rough path analysis : An Introduction
Rough path theory
Look at page 7. Thhere is also a note on Chen's relation
Differential Equations Driven by Rough Paths Front matter
System Control and Rough Paths
DIFFERENTIAL EQUATIONS DRIVEN BY ROUGH PATHS: AN APPROACH VIA DISCRETE APPROXIMATION
Concerning the Chen's relation, from Lyons books you'll se it arises naturally with introduction of Lie algebra for those iterated integrals.
Here is also something concernig the computational aspect of the theory:
A Distributed Procedure for Computing Stochastic Expansions with Mathematica
And mathematica programs to play with: Christophe Ladroue
I suggest to start studying this from Lyons books. Also refresing the knowledge of the basics from Stohastic analysis would be great. It would also be great if Mr. Lyons answered this question :)
I know this is not worth the bounty but you got me interested.
UPDATE:
Here is an exposition by Lyons, explaining the iterated integral area process (p. 118): The interpretation and solution of ODE driven by rough signals
UPDATE2:
Here is a great book to see how iterated integrals emerge from Taylor expansions of ODE solution Taylor Approximations. Note that most of these authors examine linear ODE with constant coefficients (that operator $L$ is a linear op.) so it is easy to get iterated integrals. Look at Lyons book Book page 13. in PDF. As for linear ODE with variable coeff. look at Christophe Ladroue's paper I posted above. In that case you use integration by parts to compute powers etc. of iterated integrals. page 3. eq. 8.