If $h(x_{t\wedge \tau})$ is a submartingale such that $h(0)=1$ and $\tau$ is a stopping time. Why is it that $h(x_\tau)=1_{{\tau<\infty}}$?. Also $h(x)$ is decreasing.
stopping time and martingale
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stochastic-processes
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12You are a very serious contender for the Least Informative Question Title Of The Year Award! – 2010-09-10
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0thank you very much for all your help. – 2010-09-10
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2@Vaolter, You should try to make more explicit what you want to know. As Robin says, it is very hard to see what you want to know—I do not think I understand your notation, even—and as I jokingly observed earlier, your title does not help :) If you elaborated a bit more, maybe we could work together to flesh out a good question we can answer. – 2010-09-10
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1I would like to add that $h(x_{t\wedge \tau})$ is a submartingale. – 2010-09-10
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2@Vaolter, I would have *never* guessed you had martingales in mind! (Add that to the question, so people reading it do not have to go through comments) – 2010-09-10
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1@Vaolter: If you do not edit the question to 1) Make the title more meaningful 2) Make the question make some sense, the question might get closed soon. Also, with that sarcastic attitude you will only cause people to ignore you. – 2010-09-10
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0This question uses standard terminology and notation. The question and its title are less opaque than some on this site, such as a recent one about the kernel of the tangent map (which is perfectly clear to many of us but will be incomprehensible to people from other mathematical fields who know those concepts but are not trained in pure mathematics). The sarcasm began with the very first comment, not with the question. Why do you guys jump all over someone who is interested in something you haven't been exposed to? That's wolfpack behavior... – 2010-09-10
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3@whuber: See the [edit history](http://math.stackexchange.com/posts/4370/revisions) for this question. As you can see, when @Mariano posted his comment, the title was "general question," which I agree doesn't help people figure out what is being asked. – 2010-09-10
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3@whuber: The current form of the question occured _after_ my comment. So please do check the timeline/context (as mentioned by Kaestur) before jumping to conclusions. – 2010-09-11
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0I have voted to reopen, based on whuber's comments. The close votes were for the previous version of the question. – 2010-09-11
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2@Kaestur and @Moron: my apologies. I was unaware substantial changes had occurred (and will learn to look at the edit record in the future). My conclusions were based on the material as it is presented to anyone linking to this page. Even so, the very first comment on record--and the fact that it has been substantially voted up--still gives me pause: is this any way to respond to a newcomer? More graciousness and gentler handling would seem to be in order, both here and generally. – 2010-09-11
1 Answers
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This can't be true as stated, without any relation between $\tau$ and $x_t$ turning into $0$. Under the following assumptions the conclusion is true:
- the stopping time $\tau$ is $\inf\{t: x_t=0\}$
- the decreasing function $h$ decreases to $0$ at $+\infty$.
- the process $x_t$ either hits $0$ or tends to infinity almost surely.
The third property can be probably proved easily (if true) on the basis of some information about $x_t$. But as the question is stated, we could have $x_t\equiv 10$ for all we know.