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Let $W$ be a standard Brownian motion, and $\mathcal F_t$ it’s natural filtration. Let $H$ be a continuous process, adapted to $\mathcal F_t$ and integrable with respect to $W$.

Question: Is it true that for all a.s. finite $\mathcal F_t$-stopping times $\sigma$ we have

$$\lim_{h \to 0+} \frac{1}{W_{\sigma + h} - W_{(\sigma - h) \vee 0}} \int_{(\sigma - h)\vee 0}^{\sigma + h} H_s \, dW_s = H_\sigma$$

in probability?

Note: By convention we set $\frac{1}{0} = \infty.$

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    $\begingroup$ What does integrable with respect to $W$ mean here? For example, for all $T>0$, $E \int_{0 }^{T} H_s^2 ds < \infty$ or $P( \int_{0}^{T} H_s^2 ds < \infty) = 1$? $\endgroup$ Sep 17 at 13:07
  • $\begingroup$ Ah, the latter, sorry. $\endgroup$
    – Nate River
    Sep 17 at 13:27
  • $\begingroup$ This does not seem to be true even for deterministic $\sigma$: $W_{\sigma+h}-W_{\sigma-h} = 0$ along some sequence $h = h_n$ which converges to zero, and there is no reason to expect the integral to be zero for $h = h_n$. Or am I missing something? $\endgroup$ Sep 17 at 21:02
  • $\begingroup$ Hmm I think it’s because the limit is taken in probability that it has a chance of existing. The sequence $h_n$ converges to $0$ but is itself random. @MateuszKwaśnicki $\endgroup$
    – Nate River
    Sep 18 at 0:33
  • $\begingroup$ Ah, okay, sorry. Somehow I did not notice that the limit is in probability. $\endgroup$ Sep 18 at 8:49

1 Answer 1

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Yes. Let $h \in (0,1)$, $Q_h = \frac{1}{W_{\sigma + h} - W_{(\sigma - h) \vee 0}} \int_{(\sigma - h)\vee 0}^{\sigma + h} (H_s - H_{(\sigma - h) \vee 0}) \, dW_s$ and set $M_h = \sqrt{-\log(h)}$. Because $H$ is continuous, and hence, $\lim_{h \to 0} H_{(\sigma - h) \vee 0} = H_{\sigma} $ almost surely, the claim holds if we can prove that $Q_h$ converges to zero in probability.

For arbitrary $\epsilon>0$ \begin{align*} & P( |Q_h| > \epsilon) = A_h + B_h \quad \text{where} \\ & A_h := P( |Q_h| > \epsilon , |W_{\sigma + h} - W_{(\sigma - h) \vee 0} | > M_h \sqrt{\sigma + h - (\sigma - h) \vee 0 } )\\ & B_h := P( |Q_h| > \epsilon , |W_{\sigma + h} - W_{(\sigma - h) \vee 0} | < M_h \sqrt{\sigma + h - (\sigma - h) \vee 0} ) \end{align*} By the strong Markov property of BM (given $\mathcal{F}_{(\sigma-h) \vee 0}$), $$ B_h \le P(|W_{\sigma + h} - W_{(\sigma - h) \vee 0} | > M_h \sqrt{\sigma + h - (\sigma - h) \vee 0} ) = \mathcal{N}(0,1)[M_h, \infty] \le e^{-M_h^2 } \;. $$ By Markov's inequality, the strong Markov property of BM (given $\mathcal{F}_{(\sigma-h) \vee 0}$), and Itô isometry, \begin{align*} & A_h \le P\left(\left| \frac{1}{\sqrt{\sigma + h - (\sigma - h) \vee 0}} \int_{(\sigma - h)\vee 0}^{\sigma + h} (H_s - H_{(\sigma - h) \vee 0}) \, dW_s \right| > M_h \epsilon \right) \\ & \le E\left(\left| \frac{1}{\sqrt{\sigma + h - (\sigma - h) \vee 0}} \int_{(\sigma - h)\vee 0}^{\sigma + h} (H_s - H_{(\sigma - h) \vee 0}) \, dW_s \right|^2 \right) M_h^{-2} \epsilon^{-2} \\ & \le E\left( \frac{1}{\sigma + h - (\sigma - h) \vee 0} \int_{(\sigma - h)\vee 0}^{\sigma + h} (H_s - H_{(\sigma - h) \vee 0})^2 ds \right) M_h^{-2} \epsilon^{-2} \end{align*} By continuity of $H$, we see that $A_h\searrow 0$ and $B_h \searrow 0$, and in turn, $P(|Q_h| > \epsilon) \searrow 0$ as $h \searrow 0$.

Thus, $Q_h$ converges to zero in probability, as required.

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    $\begingroup$ Nice! One comment: I think one has to use the adapted process $(H_s - H_{(\sigma - h)\vee 0})$ rather than the non-adapted one $(H_s - H_\sigma)$ as the integrand. $\endgroup$ Sep 18 at 8:54
  • $\begingroup$ @MateuszKwaśnicki sorry for overlooking that, and thanks for pointing it out! $\endgroup$ Sep 18 at 11:33
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    $\begingroup$ Very nice! Thanks for the answer. $\endgroup$
    – Nate River
    Sep 18 at 19:11

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