I'm trying to find the rate (or a bound for it) with which an iterated integral of the type

$$\int_{-h}^0 \int_{-h}^{t} A_s d B_s A_t d B_t$$

converges to zero (in probability/distribution) for $h \to 0$. Here, $B$ is a two-sided Brownian motion and $A$ is a bounded process with $A_0 = 0$. In my specific case, I already know that the rate has to be faster than $h^3$.

In the case of a single integral, there's something like the fundamental theorem of calculus (see Isaacson, https://www.jstor.org/stable/2239551) which says (in a more general form than stated here) that under certain conditions $$ \lim_{h \to 0} \frac{1}{B_{t+h} - B_t} \int_t^{t+h} A_s d B_s = A_t$$, thus implicitly giving the rate as $\sqrt{h \log(1/h)}$ by Levy's modulus of continuity for the Brownian motion. Is a result like this known for iterated integrals?

Even if not, I'd be thankful for possible (even vague) ways of bounding a (multiple) stochastic integral in terms of the integration interval.

  • $\begingroup$ What exactly do you know about $A$? Since you say that the rate is faster than $h^3$, that seems to indicate that $A$ is differentiable at $0$ with vanishing derivative. Is that the case? $\endgroup$ – Martin Hairer Apr 19 '14 at 2:34
  • $\begingroup$ The exact form of $A$ is $A_t = t L_u^t$, where $L_u^t$ is the local time in $t$ at time $u$ of another Brownian motion (sorry for interchanging the conventional notation of time and space variables here, I didn't think of the possibility that $A$ might be specified precisely here). $B$ and $L$ are related through the semimartingale decomposition of $L$, i.e. $L_u^t - L_u^0 = 2 \int_0^t \sqrt{L_u^s} d B_s + \mathrm{\, finite \, variation\, part}$ $\endgroup$ – R. Faszanatas Apr 20 '14 at 10:21
  • $\begingroup$ It doesn't seem to fit exactly, but the following paper of Cheridito, Soner, and Touzi may be relevant to you: princeton.edu/~dito/papers/dsti.pdf $\endgroup$ – Dan Apr 22 '14 at 17:52
  • $\begingroup$ Thanks Dan, I will try to get some new ideas from this paper! $\endgroup$ – R. Faszanatas Apr 29 '14 at 7:53

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