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I'm considering the last hitting time $\tau=\sup\{t\leq1:W_t=1\}$ (taking the supremum of the empty set to be zero), and want to show that it is not a stopping time.

My strategy is to show that $\mathbb{E}(W_\tau)\neq\mathbb{E}(W_0)=0$ and conclude that by the optional stopping theorem and that $\{W_t\}$ is a martingale, $\tau$ fails to be a stopping time, but:

  1. How do I calculate $\mathbb{E}(W_\tau)$, and

  2. Is this a sufficient argument, or could there be other potential reasons why the two expectations are unequal?

Ice Tea
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    Loosely speaking, $\tau$ is a stopping time if we know when it has happened. That is, if $\mathcal F_t$ is the set of all events known at time $t$, then ${\tau \le t} \in \mathcal F_t$. But for this $\tau$, show that the event ${\tau \le 1/2}$ does not belong to $\mathcal F_{1/2}$. Even if we know $W_t$ for $t \in [0,1/2]$, we do not know whether $W_t$ will return to $1$ at some time in the future $t \in (1/2,1]$, so we do not know whether $\tau \le 1/2$. – GEdgar Oct 18 '21 at 00:43

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To elaborate on @GEdgar's comment, here is a specific example of a time $t$ such that $\{\tau \le t\} \not \in \mathcal F_t$.

Since we're taking $\sup (\emptyset)= 0$, the event $\{\tau \le 0\} = \{W_t < 1 \text{ for all }t \in [0,1]\}$. Clearly we wouldn't expect $\{W_t < 1 \text{ for all }t \in [0,1]\}$ to be $\mathcal F_0$ measurable, and we can prove it because $\mathcal F_0$ is a trivial $\sigma$-algebra (in the sense that all events have either probability $0$ or $1$), but $\mathbb{P}(\{W_t < 1 \text{ for all }t \in [0,1]\}) \not \in \{0,1\}$. You can find $\mathbb{P}(\{W_t < 1 \text{ for all }t \in [0,1]\})$ explicitly in terms of the normal CDF if you want, or just take my word for it that it's strictly between $0$ and $1$.

To answer your question, yes, showing $\mathbb{E}[W_\tau] \ne \mathbb{E}[W_0]$ would show $\tau$ is not a stopping time. It's worth mentioning that only works because $\tau$ is bounded: you can have unbounded stopping times $\sigma$ such that $\mathbb{E}[W_\sigma] \ne \mathbb{E}[W_0]$.

user6247850
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    If I understand correctly, $W$ is a Brownisn motion. Then indeed we have $$\mathsf P(W_t < 1\ \forall t\in[0,1])=\mathsf P(|W_1|<1)$$ and this is explicitly computable as $W_1$ has a standard normal distribution. – Maximilian Janisch Oct 18 '21 at 12:44
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    @MaximilianJanisch Thanks for the reference! I knew the formula, I just didn't want to go through the reflection principle argument in this post. – user6247850 Oct 18 '21 at 13:22
  • Sorry, maybe I'm missing something but the reflection principle arguments make sense to me when looking at first hitting time -- how do they appear again in the last hitting time case? – Ice Tea Oct 18 '21 at 20:46
  • @IceTea It won't give you the distribution of $\tau$, it was just to compute the $\mathbb{P}(\tau \le 0)$ that I mentioned is strictly between $0$ and $1$ in my answer. – user6247850 Oct 18 '21 at 20:56
  • @user6247850 ah I see; do you mind expounding a bit more on how to calculate $\mathbb{E}(W_\tau)$? – Ice Tea Oct 19 '21 at 08:42
  • @IceTea There probably isn't an easy way to calculate it, but it isn't needed for this question. If you are interested in how to compute $\mathbb{E}[W_\tau]$, I would suggest asking it as a separate question. – user6247850 Oct 19 '21 at 12:52