# stopping time and martingale

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.

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You are a very serious contender for the Least Informative Question Title Of The Year Award! – Mariano Suárez-Alvarez Sep 10 '10 at 11:15
@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. – Mariano Suárez-Alvarez Sep 10 '10 at 12:59
@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) – Mariano Suárez-Alvarez Sep 10 '10 at 13:36
@whuber: See the edit history 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. – Larry Wang Sep 10 '10 at 22:40
@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. – Aryabhata Sep 11 '10 at 0:52

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.