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Can we bound the following expectation by $C\delta^{\frac{1}{2}}$ for some constant $C$:

$$E\sup_{|s-t|\leq \delta,s\in[0,T],t\in[0,T]}|\int_s^t\sigma(r)dB_r|,$$ where $B$ is a standard Brownian motion and $|\sigma|\leq M$.

When $\sigma$ is a constant, I know this can be done by BDG inequality. But when $\sigma$ is a time dependent function or even random, can we have similar estimate?

Guanguan
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  • BDG works fine for processes $\sigma$ which are time-dependent and/or random. – saz Mar 06 '17 at 16:13
  • How do you want to use BDG here, where you have a variable lower integration limit? No way. In fact, the correct asymptotics (for bounded $\sigma$) is $(\delta |\log \delta|)^{1/2}$ (the same for non-random or even constant $\sigma$). – zhoraster Mar 06 '17 at 17:06
  • @zhoraster Could you please show me how to get $(\delta|\log\delta|)^{\frac{1}{2}}$? I can only get $\delta^{1/2}$ when $\sigma$ is a constant by BDG. For example, when $\sigma=1$, $B_t-B_s$ has the same distribution as $B_{t-s}$, so BDG is applicable. – Guanguan Mar 06 '17 at 17:29
  • @saz I do not think BDG works for random $\sigma$ since it is not in the form of BDG inequality. If I overlook something, please remind me. – Guanguan Mar 06 '17 at 17:38
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    (For Ito processes), BDG says, $\mathbb{E}\left|\sup_{t\in[0,T]}\int_0^t\sigma_s dB_s\right|^{2p}\le \mathbb{E}\left(\int_0^T\sigma^2_s ds\right)^{p}$. $\sigma$, of course, can be random. – zhoraster Mar 06 '17 at 19:19
  • Let $T=1$, $\sigma = 1$, $\delta = 1/n$. Let us take points $t,s$ only of particular form $k\delta$, $k=0,\dots,n$. Then the $\sup$ in question is nothing else but the maximum of $n$ iid $N(0,1/n)$. It is well known (you can find it here) that the expectation of the latter is asymptotically $\sqrt{\frac2n\log n}$. From this result it is possible to get $(\delta|\log \delta|)^{1/2}$ in the general case in rather routine matter. – zhoraster Mar 06 '17 at 19:26
  • @zhoraster Did you forget to add a link/reference? ("you can find it here" but as far as I can see there is nothing to find. :)) – saz Mar 06 '17 at 21:00
  • @zhoraster Thanks a lot for your comments. But when $\sigma=1$, $E\sup_{|s-t|\leq\delta}|\int_s^t \sigma(r)dB_r|=E\sup_{|s-t|\leq\delta}|B_t-B_s|=E\sup_{|s-t|\leq\delta}|B_{t-s}=E\sup_{r\leq\delta}|B_r|\leq C\delta^{1/2}$ by BDG. Thus, $\frac{E\sup_{|s-t|\leq\delta}|\int_s^t \sigma(r)dB_r|}{\delta^{1/2}}\leq C$, which contradicts your result $(\delta|\log\delta|)^{\frac{1}{2}}$. So could you please present some more detail about your method? especially how to get the result when $sigma$ is random from the result when $\sigma$ is deterministic. Thanks a lot! – Guanguan Mar 07 '17 at 10:53
  • @zhoraster I am sorry. In my last reply, my result does not contradict yours. But can we go in the following way, by applying monotone class theorem? Assume we are working on $(\Omega,\mathcal{F},P)$, it is easy to check the following four items: 1. define $\mathcal{H}={\sigma: E\sup_{|s-t|\leq\delta}\int_s^t\sigma_rdB_r\leq C\delta^{1/2}, for some constant C}$; 2. $1\in\mathcal{H}$; 3. $1_{A\times[r_1,r_2]}\in\mathcal{H}$, for any $A\in\mathcal{F}$. 4. for any $\sigma_n\in\mathcal{H}$ with $\sigma_n\uparrow\sigma$, we have $\sigma\in\mathcal{H}$. Thus, monotone class thm is applicable. – Guanguan Mar 07 '17 at 12:28
  • @saz I think you are right. BDG works here. Suppose $T=N\delta$ and $|\sigma|\leq M$. We have $E\sup_{|s-t|\leq\delta}|\int_s^t\sigma_r dB_r|\leq 3\max_{1\leq k\leq N}{E\sup_{s\in[k\delta,(k+1)\delta]}|\int_{k\delta}^s\sigma_r dB_r|}\leq E(\int_{k\delta}^{(k+1)\delta}\sigma^2 dr)^{1/2}\leq CM\delta^{1/2}$, by BDG. Note that we can get this $C$ uniformly for each subinterval because in BDG inequality, the coefficient is universal and does not depend on the integral and time, see P166 in the Book by Karatzas and Shreve. If I overlook something, please let me know. – Guanguan Mar 07 '17 at 17:33
  • Again, $\delta^{1/2}$ is impossible, as the asymptotics I wrote is precise. See e.g. here: http://math.stackexchange.com/questions/1456567/expected-maximum-absolute-value-of-n-iid-standard-gaussians – zhoraster Mar 07 '17 at 20:39
  • @Guanguan I didn't claim that BDG is useful for this particular problem; I way just saying that BDG works for time-dependent and random processes. – saz Mar 07 '17 at 21:04
  • @zhoraster thank you very much. But as stated above, when $\sigma=1$, we have $E\sup_{|s-t|\leq\delta}|B_{t}-B_s|= E\sup_{|t-s|\leq\delta}|B_{t-s}|=E\sup_{0\leq r\leq \delta}|B_r|$. So BDG can be applied to have $E\sup_{0\leq r\leq \delta}|B_r|\leq C\delta^{1/2}$. Where am I wrong? – Guanguan Mar 08 '17 at 10:21
  • The first equality isn't true. – zhoraster Mar 08 '17 at 12:02
  • @zhoraster got it. I can get your point now. $T=1$ is fixed. We proceed as follows: Step 1. When $\sigma=1$, $E\sup_{|s-t|\leq\delta}|B_t-B_s|\leq 3 E\max_{1\leq i\leq n}\sup_{s,t\in[(i-1)\delta,i\delta]}|B_t-B_s|=3E\max_{1\leq i\leq n}\sup_{s,t\in[(i-1)\delta,i\delta]}|N^i(0,t-s)|$. $E\max_{1\leq i\leq n}\sup_{s,t\in[(i-1)\delta,i\delta]}|N^i(0,t-s)|=E\max_{1\leq i\leq n}\sup_{s,t\in[(i-1)\delta,i\delta]}\sqrt{t-s}|N^i(0,1)|\leq \sqrt{\delta}E\max_i |N^i(0,1)|\leq \sqrt{2\delta|\log\delta|}$. step 2 is in the next comment due to the character limit. – Guanguan Mar 08 '17 at 13:37
  • @zhoraster Step 2. when $\sigma$ is deterministic and $|\sigma|\leq M$, we have $\int_s^t\sigma_r dB_r$ follows $N(0,\int_s^t\sigma^2_rdr)$. So we have $E\sup_{|s-t|\leq\delta}|\int_s^t\sigma_rdB_r|\leq 3E\max_i \sup_{s,t\in[(i-1)\delta,i\delta]}|N^i(0,\int_s^t\sigma^2_rdr)|$, where $N^i$ are iid normal variables. $E\max_i \sup_{s,t\in[(i-1)\delta,i\delta]}|N^i(0,\int_s^t\sigma^2_rdr)|=E\sup_{s,t}\sqrt{\int_s^t\sigma^2dr}\max_i|N^i(0,1)|\leq M\sqrt{2\delta|\log\delta|}$ – Guanguan Mar 08 '17 at 13:50
  • @zhoraster Step 3. when $\sigma$ is random and bounded. This case is what I am not very sure. Do we apply monotone class theorem here? We define $\mathcal{H}={\sigma:E\sup_{|s-t|\leq \delta}|\int_s^t\sigma_rdB_r|\leq 3M\sqrt{2\delta|\log\delta|}, |\sigma|\leq M}$. then we can check the three items in the monotone class thm are true. So the inequality is true for all bounded and measurable $\sigma$. Am I right? – Guanguan Mar 08 '17 at 13:55
  • It's hard to read the comments. You can answer your own question, no problem. I'll check then. – zhoraster Mar 08 '17 at 19:02
  • @zhoraster I have already post my answer. Please help me have a look when you are free. Thank you very much for all your comments. – Guanguan Mar 09 '17 at 15:03

1 Answers1

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Suppose we are working on the probability space $(\Omega,\mathcal{F},P)$. It is sufficient to consider $T=1$ and $\delta=1/n$. First of all, let us recall a well known result: Expected maximum absolute value of $n$ iid standard Gaussians?

We claim the result: $E\sup_{|s-t|\leq \delta,s,t\in[0,1]}|\int_s^t\sigma_rdB_r|\leq 3M\sqrt{2\delta|\log\delta|}$, where $M$ is the essential supremum of $\sigma$ since it is assumed to be bounded. We proceed by the following two steps.

Step 1: when $\sigma=1$. We have $$E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|B_t-B_s|\leq 3E\max_{1\leq i\leq n}\sup_{s,t\in[(i-1)\delta,i\delta]}|B_t-B_s|=3E\max_{1\leq i\leq n}\sup_{s,t\in[(i-1)\delta,i\delta]}|N^i(0,t-s)|,$$ where $\{N^i(0,t-s)\}_{i=1}^n$ are iid normal variables. So it holds that \begin{equation}\begin{split}&~E\max_{1\leq i\leq n}\sup_{s,t\in[(i-1)\delta,i\delta]}|N^i(0,t-s)|=E\max_i\sup_{s,t\in[(i-1)\delta,i\delta]}\sqrt{t-s}|N^i(0,1)|\\&~\leq\sqrt{\delta}E\max_i|N^i(0,1)|\leq \sqrt{2\delta|\log\delta|},\end{split}\end{equation} where the last inequality is due to the recalled result at the very beginning. So in this case, we have $E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|B_t-B_s|\leq 3\sqrt{2\delta|\log\delta|}$.

Step 2. Define $\mathcal{H}=\{\sigma\text{ measurable and bounded}:E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^t\sigma_rdB_r|\leq 3M\sqrt{2\delta|\log\delta|},\text{M is the essential supremum of }\sigma\}$.

By step 1, we have $1\in\mathcal{H}$. Similarly, for each $r_1,r_2\in[0,1]$ and each $A\in\mathcal{F}$, we have $1_{[r_1,r_2]\times A}\in\mathcal{H}$. Moreover, for any $f,g\in\mathcal{H}$, we have $af+bg\in\mathcal{H}$ for any scalar $a,b$. Finally, for any sequence $f^n\in\mathcal{H}$ and $0\leq f^n\uparrow f$ with $f$ being bounded, we have $E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^tf^n_rdB_r|\leq 3M_n\sqrt{2\delta|\log\delta|}\leq 3M\sqrt{2\delta|\log\delta|}$, where $M_n$ and $M$ are the essential supremum of $f^n$ and $f$ respectively. Note that \begin{equation} \begin{split} &~\left|E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^tf^n_rdB_r|-E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^tf_rdB_r|\right|\\ \leq&~ E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^t(f^n_r-f_r)dB_r|\\ \leq &~2E\sup_{0\leq t\leq 1}|\int_0^t(f^n_r-f_r)dB_r|\leq CE\int_0^1(f^n_r-f_r)^2dr\rightarrow 0, \end{split} \end{equation} by dominated convergence. So we have $$E\sup_{|s-t|\leq\delta,s,t\in[0,1]}|\int_s^tf_rdB_r|\leq 3M\sqrt{2\delta|\log\delta|},$$ which implies $f\in\mathcal{H}$. So by monotone class theorem, $\mathcal{H}$ contains all $\mathcal{B}([0,T]\times\Omega)$ measurable and bounded $\sigma$.

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