0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ If $B_t^{(H)}$ is no longer a semi-martingale, what other nice properties could it have? $\endgroup$ Mar 5, 2021 at 18:14

1 Answer 1

3
$\begingroup$

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.) 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.

$\endgroup$
3
  • $\begingroup$ I am confused by the comment on $H\leq 1/4$. In one dimension, you can define a rough path above any fBm trivially. Also, I have seen this paper projecteuclid.org/journals/annals-of-probability/volume-39/… What was the statement you were trying to make? Thanks for your time. $\endgroup$
    – user168590
    Mar 6, 2021 at 1:37
  • 3
    $\begingroup$ @user168590 What I was trying to say is that for $d>1$ there is no known way of lifting fBm with $H \le 1/4$ such that increments of the RP are stationary, time reversible, and only depend on the increments of the underlying fBm in the same interval. With constructions like Nualart-Tindel, Unterberger, or Lyons-Victoir, integrals like $\int_a^b X_s\, dB_s$ will depend on increments of $B$ outside of $[a,b]$ (even if $X$ doesn't) which doesn't seem sensible. Also, none of these constructions is much more "canonical" than any other of a similar type. $\endgroup$ Mar 6, 2021 at 9:31
  • 2
    $\begingroup$ For example, the construction by Nualart-Tindel depends on the choice of representation of the fBm as a moving average. $\endgroup$ Mar 6, 2021 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.