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I have encountered the following question for homework, with our lecturer only requiring us to have a basic idea about stopping times. The question is as follows:

Let $X(t)$ be an Ito process and suppose that $a\neq b$ and:

$$\tau_{a} = inf\{ t>0: X(t) = a \}$$


$$\tau_{b} = inf\{ t>0: X(t) = b \}$$

Is $\tau_{a} \wedge \tau_{b}$ a stopping time. Explain your answer?

I understand that $\tau_{a}$ and $\tau_{b}$ are both first passage times, however considering them in conjunction, my reasoning has stalled.

Any feedback would be greatly appreciated!

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Welcome to math.SE. Since you are new, I want to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Arturo Magidin Jun 23 '12 at 21:41
So: begin with definition of "stopping time" for this setting. Are $\tau_a$ and $\tau_b$ stopping times? Then go on to the minimum. – GEdgar Jun 23 '12 at 21:43
Arturo, sorry about that I will take it on board. I judged incorrectly that posts were not waffly thus I just posted the question as is from my homework sheet. I certainly didn't mean to offend and just demand people to "Explain" etc – Lindah Jun 23 '12 at 21:54
@GEdgar, thank you for the hint regarding looking at the minimum of $\tau_{a} \wedge \tau_{b}$, as I understand this is a stopping time also – Lindah Jun 23 '12 at 22:03
$\tau_a \wedge \tau_b$ occurs when the first of these two times happens. Of course without the definition of "stopping time" it is hopeless to prove anything about stopping times. – GEdgar Jun 24 '12 at 0:13

The good news here is that one can (should?) forget completely this context of Itô processes, passage times and the like. The result to prove is that, for every stopping times $T$ and $S$ in any filtration $(\mathcal F_t)_{t\geqslant0}$, $T\wedge S$ is another $(\mathcal F_t)_{t\geqslant0}$ stopping time.

The other good news is that this result is direct from the definition of a stopping time. To wit, the task is to check that $[T\wedge S\leqslant t]$ is in $\mathcal F_t$, for every $t\geqslant0$. Now, $[T\wedge S\leqslant t]=\underline{\qquad}\cup\underline{\qquad}$ and $T$ and $S$ are stopping times for $(\mathcal F_t)_{t\geqslant0}$ hence $\underline{\qquad\qquad\qquad\qquad}$.

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