Showing that $\tau=\inf\{t\ge0:M_t\le\epsilon\}$ is a stopping time, where $M_t=M_0e^{W_t-t/2}$

Question

Let $(W_t,\mathscr{F_t})$ be a Wiener process and let $$M_t=M_0e^{W_t-t/2}\qquad t\ge0$$ where $M_0$ is deterministic. Show that, for $\epsilon>0$, $\tau=\inf\{t\ge0:M_t\le\epsilon\}$ is a stopping time.

I have been able to show that $M_t$ is a martingale but I'm a bit stuck here. I tried to use the fact that $W_t$ is a normal distribution to show that $P(M_t<\epsilon)>0$ but that got me no where.

Answer 1

Hints:

1. Prove that $$\omega \mapsto \inf_{s \leq t} M_s(\omega)$$ is $\mathcal{F}_t$-measurable. To this end, use the continuity of the sample paths of the Wiener process to show that $$\inf_{s \leq t} M_s(\omega) = \inf_{\substack{s \leq t \\ s \in \mathbb{Q}}} M_s(\omega).$$
2. Show that $$\{\tau \leq t\} = \left\{ \inf_{s \leq t} M_s \leq \epsilon \right\}.$$
3. Conclude.

Answer 2

Recall that a random variable $\tau : \Omega \to [0,\infty]$ is called a stopping time with respect to the filtration $(\mathscr{F}_t)_{t \geq 0}$ if $\{ \tau \leq t \} \in \mathscr{F}_t$ for any $t \geq 0$. To this end, let us observe that \begin{eqnarray} \{ \tau \leq t \} &=& \{ \omega \in \mathbb{R} : M_t(\omega) \leq \varepsilon \} \nonumber \\ &=& \bigcup_{0 \leq s \leq t} \{ \omega \in \mathbb{R} : M_s(\omega) \leq \varepsilon \} \nonumber \\ &=& \bigcup_{s \in [0,t] \cap \mathbb{Q}} \{ \omega \in \mathbb{R} : M_s(\omega) \leq \varepsilon \}, \end{eqnarray} where (1) follows from the fact that for each $s \in [0,t]$, $M_s(\omega)$ is continuous. Now the sets $\{ \omega \in \mathbb{R} : M_s(\omega) \leq \varepsilon \}$ are all $\mathscr{F}_s$-measurable. Since the filtration is increasing, this implies that these sets are $\mathscr{F}_t$-measurable also. Thus, since we have a countable union of $\mathscr{F}_t$-measurable sets, it follows that $\{ \tau \leq t \} \in \mathscr{F}_t$ and so $\tau$ is a stopping time.