How to bound increments of Ito integral?

Given a Brownian motion $$\{W_t\}_{t\in[0;T]}$$ and a continuous, adapted and square-integrable (bounded if you want) process $$\{\sigma_t\}_{t\in[0;T]}$$ and $$\varepsilon > 0$$, I want to prove that there is a $$\delta > 0$$ such that:

For all $$s \in [0;T]$$ and all $$M \in \mathcal F_s$$, it is

$$\mathbb E\bigg( 1_M \max_{s \le t \le (s+\delta) \wedge T} \bigg| \int_s^t \sigma_u \mathrm dW_u \bigg| \bigg) \le \varepsilon \cdot \mathbb P(M).$$

For $$\sigma \equiv 1$$, this is easy because we just consider $$\mathbb E\Big(1_M \max_{s \le t \le (s+\delta) \wedge T} |W_t - W_s|\Big)$$ for which we have a bound due to the distribution of the running maximum of $$W$$ and the increments of the Brownian motion are independent from the past.

Is there a similar thing for arbitrary Ito integrals (or those satisfying some assumptions)?

Since $$0 \leq 1_M \leq 1$$, we really want to show that we can pick $$\delta > 0$$ to guarantee that for all $$s \in [0,T]$$ $$\mathbb{E}\bigg[\max_{s \le t \le (s+\delta) \wedge T} \bigg| \int_s^t \sigma_u \mathrm dW_u \bigg| \bigg] \le \varepsilon.$$
We can do this using an appropriate form of Doob's Martingale inequality. We have \begin{align} \mathbb{E}\bigg[\max_{s \le t \le (s+\delta) \wedge T} \bigg| \int_s^t \sigma_u \mathrm dW_u \bigg| \bigg] \leq& \mathbb{E}\bigg[\max_{s \le t \le (s+\delta) \wedge T} \bigg| \int_s^t \sigma_u \mathrm dW_u \bigg|^2 \bigg]^{\frac12} \\ \leq& 2\mathbb{E} \bigg[ \bigg(\int_s^{(s+ \delta) \wedge T} \sigma_u dW_u \bigg)^2 \bigg]^{\frac12} \\ =& 2\mathbb{E} \bigg[ \int_s^{(s+ \delta) \wedge T} \sigma_u^2 du \bigg]^{\frac12} \end{align} where the first inequality is just monotonicity of $$L^p$$-norms on probability spaces, the second is Doob's inequality and the last line then follows by the Ito isometry. It's clear that your assumptions on $$\sigma$$ let you choose $$\delta$$ to make this last quantity as small as you like, giving the desired result.
• Thank you! But why is the martingale increment independent from past? For instance, $\sigma|_{[s;T]}$ could be equal to 0 or equal to 1, depending on an event in $\mathcal F_s$. Then the martingale increment would not be independent from $\mathcal F_s$. Or did I get something wrong? – Kolodez Mar 30 at 20:40
• Whoops, I hadn't read your question carefully enough and had assumed $\sigma$ was deterministic. But this doesn't matter - see my edit. – Rhys Steele Mar 30 at 20:51
• Oh, I am so sorry, I also forgot one tiny, but important thing in my original question. :( I want the expectation to be smaller than $\varepsilon \cdot \mathbb P(M)$ instead of just $\varepsilon$, and $\delta$ should not depend on $M$. If $\sigma$ is deterministic, you are right: We just use the independence. Do you also see a way if $\sigma$ is not deterministic? – Kolodez Mar 30 at 21:16
• Unfortunately, it's not clear to me that this method will yield that bound. Of course, one could initially apply Cauchy-Schwarz instead of the bound $0 \leq 1_M \leq 1$ and that would yield a bound of the form $\varepsilon \cdot \mathbb{P}(M)^{\frac12}$. Using Holders inequality and Burkholder-Davis-Gundy, I think it will be possible to improve this to $\varepsilon \cdot \mathbb{P}(M)^{1- \alpha}$ for any $\alpha >0$ but unfortunately I don't see a way to get the $\alpha = 0$ case without independence. – Rhys Steele Mar 30 at 22:33
• What if the martingale is Markovian, i.e. $X_t = \int_0^t \sigma(u,X_u) dW_u$? We can then consider $\mathbb E(1_M \mathbb E(\max_{s\le t \le s+\delta} \ldots | \mathcal F_s))$ and replace $\mathcal F_s$ by $X_s$. Can we also apply a modified version of the Doob martingale inequality to not the expectation of the maximum value of the martingale, but to the expectation conditioned on the shifted starting value $X_s$? I have no idea how to modify the Doob's inequality into a version with conditional expectations. – Kolodez Apr 5 at 13:19