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)?


Edit: The question has been edited to ask for something stronger which this answer doesn't give. I'm leaving this here in case it inspires someone else/I work out a way to improve it to give the stronger claim.

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.

  • $\begingroup$ 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? $\endgroup$ – Kolodez Mar 30 at 20:40
  • $\begingroup$ Whoops, I hadn't read your question carefully enough and had assumed $\sigma$ was deterministic. But this doesn't matter - see my edit. $\endgroup$ – Rhys Steele Mar 30 at 20:51
  • $\begingroup$ 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? $\endgroup$ – Kolodez Mar 30 at 21:16
  • 1
    $\begingroup$ 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. $\endgroup$ – Rhys Steele Mar 30 at 22:33
  • $\begingroup$ 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. $\endgroup$ – Kolodez Apr 5 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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