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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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
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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.