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I've been given a crash course in stochastic processes and martingales for the purposes of a semester project on them. The guy I'm working with has been, I feel, a little vague in the definition of stopping times, and I can't seem to find anything on Wikipedia or Google that clarifies the issue for me.

My problem is that stopping times are defined as random variables. But given the motivation for the concept of stopping times (aren't they basically meant to represent betting strategies?), that doesn't at all seem like how I would personally define stopping times.

I would define a stopping time as a, in some sense, predicate, or a two valued function, that maps sequences of values to either STOP or NOT STOP. So given a sequence of values (representing the values of the stochastic process up to the present), the function tells you whether or not to stop.

But instead, a stopping time is given by Wikipedia as:

A stopping time with respect to a sequence of random variables $X_1, X_2, X_3,\ldots$ is a random variable $\tau$ with the property that for each $t$, the occurrence or non-occurrence of the event $\{\tau = t\}$ depends only on the values of $X_1, X_2, X_3, \ldots, X_t$.

I can't see at all how to relate that to the notion of stopping times as betting strategies. If stopping time is a random variable with respect to a given sequence of values, doesn't that mean it's not a "determined" strategy? That sounds like you look at how much money you've made/lost so far, and then based on both that and the result of a coin flip (or something), decide whether or not to stop playing.

I'm sure I'm wildly misunderstanding either the definition, the motivation, or both. Please avoid dipping too deeply into measure theory or filtrations, if possible.

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Take coin flipping, suppose you decide to stop when you obtain heads, thus $\tau=\min\{t|X_t=H\}$. Now, take the space of outcomes of toincossing experiments. Don't you agree that $\tau$ unambiguously defines a random variable on that space? –  Raskolnikov Mar 15 '13 at 18:00
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The decision of whether to stop may depend on a coin flip. But whether it does or not, it almost always in addition depends on the results so far. So it is a random variable. –  André Nicolas Mar 15 '13 at 18:04
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3 Answers 3

up vote 6 down vote accepted

It sometimes is a useful exercise to separate the random from the non-random pieces of the puzzle.

Let's build up a stopping time, starting without randomness, along the lines of your intuition. Suppose that the observations $X_j$ take values in the space $S$, and let $S^\mathbb{N}$ be the space of $S$-valued sequences.

For any strategy or stopping policy, and any $0\leq n<\infty$ we may define a two-valued map $\phi_n:S^\mathbb{N}\to\{\mbox{GO},\mbox{STOP}\}$ which tells me what to do at time $n$ if I were to observe $s=(s_0,s_1,\dots)$. We require that $\phi_n(s)$ only depends on the first part of the sequence $(s_0,s_1,\dots,s_n)$. That is, the decision to stop at time $n$ must only depend on the observations up to time $n$. No peeking into the future!

Now define $\phi(s)=\inf(n\geq 0: \phi_n(s)=\mbox{STOP})$, where the infimum over the empty set is $\infty$. This gives a map $\phi:S^\mathbb{N}\to \mathbb{N}\cup \{\infty\}$ which expresses our policy, by telling us when to stop.

Finally we can put probability back into the picture by defining $\tau:\Omega\to \mathbb{N}\cup \{\infty\}$ by $$\tau(\omega)=\phi(X_0(\omega), X_1(\omega), X_2(\omega), \dots ).$$ This random variable is the stopping strategy applied to the random sequence $(X_0(\omega), X_1(\omega), X_2(\omega), \dots)$.

Every stopping time $\tau$ can be expressed like this for some such $\phi$.
For $0\leq n< \infty$, by the Doob-Dynkin lemma, there is a measurable map $\varphi_n:(S^\mathbb{N},{\cal G}_n) \to \{0,1\}$ so that $1_{[\tau=n]}=\varphi_n(X_0,X_1,X_2,\dots)$. Here ${\cal G}_n$ is the $\sigma$-field generated by the coordinate maps $s_j$ for $0\leq j\leq n$. Now let $\phi(s)=\inf(n\geq 0: \varphi_n(s)=1)$.

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What is $\omega$ ? An event representing the results of a particular infinite sequence of trials? –  Jack M Mar 15 '13 at 20:04
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@JackM $\omega$ is a point in the probability space that represents the "randomness" in the model. –  Byron Schmuland Mar 15 '13 at 20:17
    
That is a good explanation - one I've never heard before. +1 –  Stefan Hansen Mar 16 '13 at 12:07
    
@StefanHansen Thanks. It's the way I think of stopping times. –  Byron Schmuland Mar 16 '13 at 13:16
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@Sai I've added some explanation. I don't believe you can find this in any book. –  Byron Schmuland May 23 at 11:27
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Whether $\tau\le n$ is determined by $X_1,\ldots,X_n$. It depends on those, and those are random, so $\tau$ is random. One can ask, for example, for the expected value of $\tau$ or the probability that $\tau=6$. One could not do that if $\tau$ were not a random variable, which has a probability distribution.

Notice, though, that the value of $\tau$ is not determined by $X_1,\ldots,X_n$ if they determine (as they sometimes do) that $\tau>n$.

An unpleasant complication is that a different conventional definition of "stopping time" is sometimes also used: A random variable $\tau$ is a stopping time for a sequence $X_1,X_2,X_3,\ldots$ if for each $n$ the even $\tau=n$ is independent of $X_{n+1},X_{n+2},X_{n+3},\ldots$.

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I think I see. So the betting strategy (what I would have modeled with a two-valued function) is represented as the precise sense in which $X_1...X_n$ determine $\tau$, and the stopping time itself represents the value that, for a particular game, that strategy results in a stop? –  Jack M Mar 15 '13 at 18:17
    
Except that $X_1,\ldots,X_n$ don't determine $\tau$ unless they determine that $\tau\le n$ rather than that $\tau>n$. –  Michael Hardy Mar 15 '13 at 18:20
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I think the answer is that if you are unaware of the stopping criteria and observe a the stopping time over several runs, it will appear to behave randomly because the values that it operates on are random. Lets say if you always stop when you accumulate a value in some variable. No matter which is simply an entirely "add to variable x(t) to y until Y = A. Then the time at which you reach Y= A as long as x(t) is grows randomly. Hope this helps. It is the way I understand it.

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