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Given an adapted process $X_t$ and it's Snell envelope $S_t$, I know that the smallest optimal stopping time is to stop as soon as the Snell envelope equals $X$. This is very intuitive and makes perfect sense. You could explain it to a kid.

However, I have a hard time grasping any other stopping time, such as the "largest" stopping time.

What is an "informal" description of such other stopping times (take e.g. the largest one). How would you describe it in words?

Could you give an example?

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  • $\begingroup$ When you say 'optimal', that implies that there is some objective that you are trying to optimize. What is that objective? $\endgroup$ – Ben Derrett Aug 19 '18 at 10:42
  • $\begingroup$ The objective is to maximize the expectation of $X$ relative to stopping times. So e.g. $\max_{\nu} E X_\nu$ where $\nu$ are stopping times. For an optimal stopping time $\nu*$, we have $\max_{\nu} E X_\nu = E X_{\nu *}$. $\endgroup$ – nemesis Aug 19 '18 at 10:44
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Suppose that the problem is formulated in discrete time and that you wish to choose the largest optimal stopping time $\nu_*$. Informally, the stopping time $\nu_*$ can be described as follows. Start at time 0 and denote the current time by $t$. If $X_t < S_t$, continue to the next timestep. Otherwise, since $S$ dominates $X$, $X_t = S_t$. If $\mathbb{E}_t[S_{t+1}]=S_t$, continue to the next timestep, otherwise stop.

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  • $\begingroup$ Okay, so the Snell envelope $S_t$ equals the maximum of either $X_t$ or $E_t S_{t+1}$. If they are all equal, i.e. $S_t = X_t = E_t S_{t+1}$, only then is it still optimal to continue, but if we only have the first equality, then we must stop? $\endgroup$ – nemesis Aug 19 '18 at 11:50
  • $\begingroup$ @nemesis, yes, that's right $\endgroup$ – Ben Derrett Aug 19 '18 at 12:00

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