Question about stopping time : what would be $M_{\tau}$ if $\{\tau\leq t\}\notin \mathcal F_t$?

I really have difficulties by really understand in why stopping times are interesting.

Let $$(\Omega ,\mathcal F,\mathbb P)$$ a probability space, $$(M_t)_{t\geq 0}$$ a stochastic process and let $$(\mathcal F_t)_{t\geq 0}$$ an adapted filtration. Let $$\tau:\Omega \to [0,\infty )$$. What would be the problem if $$\{\tau\leq t\}\notin \mathcal F_t$$ ?

So we have that $$M_{\tau}:\Omega \to \mathbb R$$ defined by $$(M_{\tau})(\omega )=M_{\tau(\omega )}(\omega )$$. I can imagine that if $$\{\tau(\omega )\leq t\}\notin \mathcal F_t$$ then for $$\omega \in \Omega$$ s.t. $$\tau(\omega )=t$$, then $$M_{\tau(\omega )}=M_t$$ could not be in $$\mathcal F_t$$ (but I'm not even sure). I'm quite confuse with that because $$\mathcal M_t\in \mathcal F_t$$... Any clarification is welcome.

• I think you are correct that the stopped process then may not be measurable, since $M$ is also defined on $\Omega$.
– user88923
Commented May 8, 2019 at 8:18

I think that I can give a motivating example that shows what kinds of problems arise for random times that are not stopping times, i.e. $$\tau:\Omega\to[0,\infty)$$ such that there exists $$s$$ when $$\{\tau\leq s\}\notin\mathcal{F}_s$$. Maybe you've seen this example, so I apologize if so.

So, here's the example. Say that I am gambling, and that $$M_t$$ is my amount of money at time $$t$$. It would be nice if I could know when I should quit, or in other words the time when I have the most money: $\tau=\inf\ \{ t : M_t\geq M_s \forall s \}.$

I claim that $$\tau$$ is not a stopping time. To see this, note that the event $$\{\tau\leq s\}$$ is the same as the event $$\{\exists t\leq s: M_t\geq M_q\forall q>s\}$$. This depends on $$M_q$$ for $$q>s$$ so this event is clearly not contained in $$\mathcal{F}_s$$. The stopped process $$M_\tau$$ represents my maximum cash if I keep playing, and I would really like to know it, but it's not something I could ever actually expect to know.

(This also addresses something in your question -- in this example, the event $$\{M_t=M_\tau\}$$ is definitely not contained in $$\mathcal{F}_t$$. For this example that is the same as the event $$\{\tau=t\}$$, which is the event that I should stop playing right now.)

So, to me this gives an intuition for why people care about stopping times. The idea behind the definition "$$\tau$$ is a stopping time $$\iff$$ $\{ \tau=t \} \in\mathcal{F}_t$ for all $$t$$" is that $$\tau$$ should not make use of information from the future. So, it helps us model quantities that correspond to measurable events in our filtration. If the filtration is meaningful as part of a model of the world, then that's helpful.

The other thing is that the definition of stopping times allows for optional stopping theorems.

I think that maybe looking at an application could be helpful too, since all of these gambling examples can get a little boring. If you've ever seen Erdos-Renyi random graphs, then maybe this paper studying the phase transition to the existence of a giant connected component using martingales and stopping times could be of interest. This paper makes use of optional stopping type theorems.