Stopping time wiener process

Question

Suppose that $T=\inf\{t\geq 0: W_{t} \notin [-a,b]\}$ where $a,b>0$ and $W_{t}$ denotes a Wiener process. Now I'm wondering if this is a stopping time but don't know how to work this out.

Answer 1

Define $\mathcal F_t:=\sigma\{W_s:0\le s\le t\}$ for $t\ge 0$. If you define $M_t:=\sup_{0\le s\le t}W_s(=\sup_{0\le s\le t,s\in\Bbb Q} W_s)$ and likewise $m_t:=\inf_{0\le s\le t}W_s$ then both $M$ and $m$ are adapted processes (i.e. $M_t$ and $m_t$ are $\mathcal F_t$-measurable for each $t\ge 0$), and $\{T\ge t\}=\{M_t\le b,m_t\ge -a\}\in\mathcal F_t$. Therefore (by complements) $\{T< t\}\in\mathcal F_t$ for each $t\ge 0$. This is the same as saying that $\{T\le t\}\in \mathcal F^+_t$ for each $t\ge 0$. Here $\mathcal F^+_t:=\cap_{\epsilon>o}\mathcal F_{t+\epsilon}$. Thus $T$ is a stopping time of $(\mathcal F^+_t)_{t\ge 0}$, the natural filtration of $W$, namely $\mathcal (\mathcal F_t)_{t\ge 0}$, made right continuous.

If one "completes" $(\mathcal F_t)_{t\ge 0}$ in the usual way, by passing to $\mathcal F^*_t:=\sigma(\mathcal F_t,\mathcal N)$ (where $\mathcal N$ is the class of null sets in $\mathcal F_\infty:=\sigma(W_s:s\ge 0)$) then $(\mathcal F^*_t)_{t\ge 0}$ is right continuous (i.e, $\cap_{\epsilon>0}\mathcal F^*_{t+\epsilon}=\mathcal F^*_t$ for all $t\ge 0$), from which it follows that $T$ is an $(\mathcal F^*_t)_{t\ge 0}$-stopping time. (The right continuity of $(\mathcal F^*_t)_{t\ge 0}$ is a consequence of the simple Markov property of the Wiener process together with Blumenthal's zero-one law.)