I have a question that asks to show that $\tau= \inf \{t\geq0:B_t\leq e^t-2\}$ is a stopping time with respect to $(\mathcal{F}^B_t)_{t\geq0}$, where $B=(B_t)_{t\geq0}$ is a standard Brownian motion started at $0$. I'm not quite sure where to start on this sort of question should I first start by rearranging the inequality?


For a continuous function $f: [0,T] \to \mathbb{R}$ the following statements are equivalent:

  1. There exists $t \in [0,T]$ such that $f(t)=0$.
  2. There exists a sequence $(r_j)_{j \in \mathbb{N}} \subseteq \mathbb{Q} \cap [0,T]$ such that $f(r_j) \to 0$ as $j \to \infty$, i.e. for all $\epsilon>0$ there exists $r \in \mathbb{Q} \cap [0,T]$ such that $|f(r)| \leq \epsilon$.

Applying this for $f(t) := B_t(\omega)-e^t+2$ with $\omega \in \Omega$ fixed, we find

$$\{\tau \leq T\} = \{\omega; \forall \epsilon>0 \exists r \in \mathbb{Q} \cap [0,T]: |B_r(\omega)-e^r+2| \leq \epsilon\}.$$


$$\{\tau \leq T\} = \bigcap_{k \in \mathbb{N}} \bigcup_{r \in \mathbb{Q} \cap [0,T]} \underbrace{\{\omega; |B_r(\omega)-e^r+2| \leq 1/k\}}_{\in \mathcal{F}_r^B \subseteq \mathcal{F}_T^B} \in \mathcal{F}_T^B.$$

  • $\begingroup$ Ah, thank you, that's cleared it up for me $\endgroup$ – df12 Jan 19 '17 at 20:30

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