Stopping times problem: $ \tau_+ = \inf \{t \ge 0 \mid W_t>0\}$

Question

Stopping times problem, $\tau_+ = \inf \{t \ge 0 \mid W_t>0\}$

I can not prove the following :

P/S: When I look at the stopping time, I feel that $\{W_0 > 0\} = \{\tau_+ = 0\}$ , is that correct ?

Answer 1

$\{W_0>0\} = \{\tau_+=0\}$ is not correct.

Note that $\tau_+(\omega)=0$ if, and only if, there exists a sequence of numbers $(t_n)_n \subseteq [0,\infty)$ such that $t_n \to 0$ and $W_{t_n}(\omega)>0$. Show the following equalities:

$$\begin{align*} \{\tau_+=0\} &= \{\omega; \exists (t_n)_n \subseteq [0,\infty), t_n \downarrow 0: W_{t_n}(\omega)>0\} \\ &= \{\omega; \exists (t_n)_n \subseteq [0,\infty) \cap \mathbb{Q}, t_n \downarrow 0: W_{t_n}(\omega)>0\} \\ &= \bigcap_{0<r<t, r \in \mathbb{Q}} \bigcup_{0<q<r, q \in \mathbb{Q}} \{W_q>0\}. \end{align*}$$

For the second equality, use that $W_t$ has continuous sample paths.

Remark. Note that the set $\{W_0>0\}$ is contained in $$A:=\bigcap_{0<r<t, r \in \mathbb{Q}} \bigcup_{0<q<r, q \in \mathbb{Q}} \{W_q>0\}.$$ Indeed, if $W_0(\omega)>0$, then it follows from the continuity of the sample paths that $W_s(\omega)>0$ for $s \leq \delta$ sufficiently small. Hence, $W_s(\omega)>0$ holds in particular for all $s \in \mathbb{Q}$, $s \leq \delta$. This shows $\omega \in A$.