The only rough path that I've ever seen discussed are the ones associated with Brownian motion. I could use a "rough path" for any nice function, defeating the point. In particular are there interesting rough paths that are $\alpha$-Holder continuous for $\frac13 \lt \alpha \lt \frac12$?
3 Answers
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An explicit rough path construction for continuous paths with arbitrary Hölder exponent, by Jeremie Unterberger (2009) [published with a different title]:
We construct an explicit geometric rough path over arbitrary $d$-dimensional paths with finite $1/\alpha$-variation for any $\alpha\in (0, 1)$. The method generalizes to arbitrary $\alpha$-Hölder paths the previous constructions that were limited to fractional Brownian motion. Our generalization may be coined as Fourier normal ordering since it consists in a regularization obtained after permuting the order of integration in iterated integrals so that innermost integrals have highest Fourier frequencies. The method is not limited to Brownian motion, because it does not rely on tools belonging exclusively to the Gaussian realm, namely, the equivalence of $L^p$-norms due to the hypercontractivity property of the Ornstein-Uhlenbeck process.
For a broader overview of this topic with many examples, starting from T. Lyon's seminal contributions, see Unterberger's Rough Path Theory.
answered Jun 30, 2015 at 12:58
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1) Fractional Brownian motion (with the lift given by limit of piecewise linear approximations or any reasonable convolution approximation à la Wong-Zakai).
2) Actually as pointed out before any continuous Hölder path can be lifted to a geometric RP (a result originally due to Lions-Victoir).
3) Physical brownian motion in a magnetic field. http://arxiv.org/abs/1302.2531
4) Solutions to SPDEs (see Hairer's "Solving the KPZ equation")
5) Smooth rough paths: https://projecteuclid.org/euclid.rmi/1204128313
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A Levy rough path for a Levy process not a Brownian motion.
Take a look at Friz and Shekhar General rough integration, Levy rough paths and a Levy-Kintchine type formula.
Regarding the Holder condition, don't Brownian motions themselves qualify?
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$\begingroup$ Brownian motion indeed does qualify for Holder condition does qualify, but I am looking for other examples. Thanks for the resource. $\endgroup$– user69208Commented Jun 30, 2015 at 6:32
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$\begingroup$ And other issue: "On the rough-paths approach to non-commutative stochastic calculus". Look her arxiv.org/pdf/1301.6238 $\endgroup$– ZbigniewCommented Nov 14, 2016 at 11:28