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Suppose $W$ is a Brownian motion, let $H_B$ be the hitting of $B \in \mathbb{R}$ and let $\tau$ be another stopping time that is taken to be minimal, i.e $(W_{t\wedge \tau})_{t \geq 0}$ is uniformly integrable.

Show that $\mathbb{E}[(W_{H_B} - W_\tau)\mathbb{1}_{H_B \leq \tau}]=0$.

The "solution" that I got and needs checking is the following:

We apply the optional stopping theorem which states that for a cadlag adapted integrable process $X$ and any stopping times $T$, $S$, we have that $\mathbb{E}[X_T|\mathcal{F}_S]=X_{T\wedge S}$ iff X is a uniformly integrable martingale. Since $W^\tau$ is UI a mart, we get that $\mathbb{E}[W_\tau|\mathcal{F}_{H_B}]=W_{\tau \wedge H_B}$. In particular $\mathbb{E}[W_\tau 1_A]=\mathbb{E}[W_{\tau \wedge H_B}1_A]$ for any $A$ in $\mathcal{F}_{H_B}$, thus it is sufficient to show that $(H_B\leq \tau )$ is in $\mathcal{F}_{H_B}$.

The definition that I have is $\mathcal{F}_{H_B}:=\{A \in \mathcal{F}_{\infty}:A \cap (H_B \leq t) \in \mathcal{F}_{t}\}$

Fix $t$, then $(H_B \leq \tau)=(H_B < \tau) \cup (H_B=\tau)$. Sufficient to show that both $(H_B < \tau)$ and $(H_B=\tau)$ are in $\mathcal{F}_{H_B}$.

Consider first $(H_B <\tau)=\cup_{q\in \mathbb{Q}}(H_B\leq q) \cap (q<\tau)$. Then $(H_B< \tau) \cap (H_B \leq t)=\big(\cup_{q\in \mathbb{Q}, q\leq t} (H_B \leq q\wedge t) \cap (q < \tau)\big) \cup (t < \tau)$. Clearly $(H_B \leq q \wedge t) \in \mathcal{F}_t$ and $(t < \tau) = (\tau \leq t)^C \in \mathcal{F}_t$. Therefore, $(H_B<\tau) \cap (H_B\leq t) \in \mathcal{F}_t$.

Now let's look at $(H_B=\tau)$. Note that $(H_B\leq t)\cap (\tau \leq t)=\big((H_B=\tau)\cap (H_B \leq t)\big) \cup \big( (H_B < \tau)\cap (\tau \leq t)\big) \cup \big((\tau < H_B) \cap (H_B \leq t)\big).$

Observe that all three sets in $\big(\big)$ are disjoint. It is obvious that $(H_B \leq t) \cap (\tau \leq t) \in \mathcal{F}_t$. From before $(H_B <\tau) \in \mathcal{F}_t$ so clearly $(H_B<\tau)\cap(\tau \leq t) \in \mathcal{F}_t$. Similarly, $(\tau < H_B) \cap (H_B \leq t) \in \mathcal{F}_t$. Hence we get that $(H_B = \tau) \cap (H_B \leq t) \in \mathcal{F}_t$. This concludes our case by case analysis.

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  • $\begingroup$ Do you have any thoughts on the problem? What have you tried? $\endgroup$ – saz Apr 17 at 12:21
  • $\begingroup$ @saz I edited in the solution that I just got. Please check it. $\endgroup$ – tergarg Apr 17 at 13:57
  • $\begingroup$ @saz Pretty sure that is what the Optional Stoping theorem states. If $X$ is a regular martingale (not UI) we need $T$ bounded and $S$ any stopping time. If $X$ is UI then $T$ can also be any martingale. I got it from James Norris' lecture notes: statslab.cam.ac.uk/~james/Lectures/ap.pdf page 25. $\endgroup$ – tergarg Apr 17 at 15:44
  • $\begingroup$ I see, my fault. As far as I can see, the equation for $(H_B \leq \tau)$ is not correct because e.g. the case that $H_B = \tau \in \mathbb{R} \backslash \mathbb{Q}$ is not covered by the right-hand side... but perhaps I'm once more missing something. $\endgroup$ – saz Apr 17 at 16:43
  • $\begingroup$ @saz, okay I edited the solution again, please have a look. If you know a different method to solve the problem, I would be quite happy to see it. $\endgroup$ – tergarg Apr 17 at 19:08

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