In Itô calculus, it is easy to construct an associativity rule. Namely, if $B\_t$ is a Brownian motion and $M\_t = \int\_0^t X\_s dB\_s$ for suitable $X\_t$, then we have the following associativity rule: $Z\_t = \int\_0^t Y\_s dM\_s = \int\_0^t Y\_s X\_s dB\_s$. Such a rule can be derived by defining a martingale calculus, or more generally can be extracted from semimartingale calculus.


Is it possible to obtain a similar expression when the integrator $B\_t$ is replaced by $B\_t^{(H)}$, a fractional Brownian motion with Hurst index $H \neq \frac{1}{2}$? In this setting, one can't pass through a semimartingale calculus as $B\_t^{(H)}$ is no longer a semimartingale. My feeling is that the answer may depend on the choice of $H$, as this affects the construction of the integral.



Associativity rule for integration against fractional Brownian motion
For $H > 1/2$ and assuming that both $X$ and $Y$ have trajectories that are almost surely $\alpha$-Hölder continuous for some $\alpha > 1/2$, there is only one sensible definition of the stochastic integral (Riemann-Stieltjes) and associativity holds since it does so for smooth functions and the integral is stable under mollification. For $H < 1/2$ things are more subtle.


If in addition $H > 1/3$, then the three integrals appearing in your statement are well-defined for $X$ and $Y$ that are controlled rough paths, with the underlying reference rough path being any rough path lift of $B$. (For the definition of the integral of $Y$ against $M$, see Remark 4.11 in [my book with Peter](http://www.hairer.org/notes/RoughPaths.pdf).) Associativity then holds again, as can easily be checked by unpacking the definitions. This is true even if you consider "exotic" integrations by choosing a rough path lift of $B$ other than the canonical one.


For $H \le 1/3$ the same is still true, but it gets more cumbersome to verify: Section 4.5 of that book gives hints on how one would go about it. Note also that for $H \le 1/4$ there is no sensible theory of integration against $B$ anyway, so the point becomes moot then.


