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:
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How do I calculate $\mathbb{E}(W_\tau)$, and
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Is this a sufficient argument, or could there be other potential reasons why the two expectations are unequal?
Best Answer
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 $\mathbb{E}[W_\sigma] \ne \mathbb{E}[W_0]$.