One of the three condition of the optional stopping theorem is that "There exists a constant $c$ such that $|X_{t\wedge \tau}| \leq c$ a.s. for all $t\in \mathbb N_0$".

In the article of Wikipedia on the optional stopping theorem, the second item in the linked section gives an example of an application of the theorem:

Suppose a random walk starting at $a \geq 0$ that goes up or down by one with equal probability on each step. Suppose further that the walk stops if it reaches 0 or $m \geq a$; the time at which this first occurs is a stopping time. If it is known that the expected time at which the walk ends is finite (say, from Markov chain theory), the optional stopping theorem predicts that the expected stop position is equal to the initial position a. Solving $a = pm + (1 – p)0$ for the probability p that the walk reaches m before 0 gives $p = \frac am$.

Since once the random walk is defined the number $m$ is a fixed integer, could also the condition (c) there be applied? That is, it is true that there exist a "constant" that bounds the absolute value of the martingale?

Another example (third item of the linked section), gives me even more doubt:

Care must be taken, however, to ensure that one of the conditions of the theorem hold. For example, suppose the last example ["a random walk $X$ that starts at 0 and stops if it reaches $–m$ or $+m$"] had instead used a 'one-sided' stopping time, so that stopping only occurred at $+m$, not at $−m$. The value of $X$ at this stopping time would therefore be m. Therefore, the expectation value $E[X_{\tau}]$ must also be $m$, seemingly in violation of the theorem which would give $E[X_{\tau}] = 0$. The failure of the optional stopping theorem shows that all three conditions fail.

It seems to me that in the latter example the real problem is that the "one-sided" stopping time is not indeed a stopping time: it can be infinite. On the other hand, if it were a stopping time, it does not seem to me that "all three conditions fail" because $|X_{\tau}|\leq m$ for a fixed $m$. What am I missing?


1 Answer 1


For the first: yes, you could also use part (c) there, because the stopped process is always bounded by $m$ in absolute value.

For the second: stopping times are allowed to be infinite! However, the thing you are missing here is due to a shortfall in Wikipedia's explanation of their example: when they refer to the one-sided stopping time, they mean that the ONLY stopping condition is reaching $m$... so, for instance, you could go to $-10,000,000$, then work your way back. So, in this case, the stopped process isn't bounded - it is only bounded above.

  • $\begingroup$ Thank you! I did not encounter that definition of stopping time before (I'm in the "some cases" of the Wikipedia definition en.wikipedia.org/wiki/Stopping_time#Definition). $\endgroup$ Commented Jul 10, 2013 at 23:52
  • $\begingroup$ Is the optional stopping theorem still true with a definition of stopping time that allows it to be infinite? I think no: let the random walk stop moving as it reaches $m$ or $-m$, but define the stopping time to be the time it reaches $m$; this stopping time is infinite but if we could use part (c) we got a contradiction. $\endgroup$ Commented Jul 11, 2013 at 0:00
  • $\begingroup$ Well, if you note, the optional stopping theorem specifically requires something like "the stopping time is bounded almost surely" or "$\mathbb{E}[T]<\infty$", or some such other condition; these all imply that it is finite almost surely. $\endgroup$ Commented Jul 11, 2013 at 0:12

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