# Minimal stopping time of brownian motion

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.

• Do you have any thoughts on the problem? What have you tried? – saz Apr 17 at 12:21
• @saz I edited in the solution that I just got. Please check it. – tergarg Apr 17 at 13:57
• @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. – tergarg Apr 17 at 15:44
• 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. – saz Apr 17 at 16:43
• @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. – tergarg Apr 17 at 19:08