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Let $(X_t)_{t\geq 0}$ be a continuous (or càdlàg), real-valued process, and define stopping times $$\tau_{s,a,b}=\inf~ [s,\infty)\cap\{t:X_t\notin (a,b)\}.$$ We can interpret $\tau_{s,a,b}$ as the first time after time $s$ that the process hits $a$ or $b$.

Suppose that for all $s,a,b$ we have:

$$\mathbb{E}[X_{\tau_{s,a,b}}|\mathcal{F}_s]\leq X_s$$

Then is $X$ necessarily a local supermartingale?

Context:

At first I thought that perhaps $X$ was necessarily a supermartingale, but Nate pointed out that there are local supermartingales with this property. For example,

$$ X_t = \begin{cases} W_{\min(\frac{t}{1-t},T)} &\text{for } 0 \le t < 1,\\ 1 &\text{for } 1 \le t < \infty, \end{cases}$$

where $(W_t)_{t\geq 0}$ is a Wiener process and $T=\inf\{t\geq 0:W_t=1\}$, seems to fit the bill.

(The question was edited in response to Nate's comment.)

I can solve the analogous problem in discrete time by induction, but don't know where to go from there, if indeed it's of any use.

Thank you.

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    A small thing: of X is continuous then $X_{\tau_{s,a,b}}$ is equal a or b. Sounds strange.2012-01-17
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    Seems to me that a *local* supermartingale would still satisfy your property.2012-01-17
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    @Kolmo Yes, it will equal $a$ or $b$.2012-01-17
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    @NateEldredge Thank you! I've edited the question to reflect this (unfortunate) reality.2012-01-17
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    @Ben Derrett : I don't see why your example is a (local)supermartingale, don't we have $1=E[X_1]>X_0=0$ ?2012-01-17
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    @TheBridge Yes, but that's OK. It's a trivial modification ($X_t\to -X_t$) of the local martingale given in the [Wikipedia article on local martingales](http://en.wikipedia.org/wiki/Local_martingale).2012-01-17
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    @Ben Derrett : ok I got it, but to be consistent wuith yourproblem you should use $-X_t$ instead of $X_t$, I think. Regards2012-01-18
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    @TheBridge Rereading, the example isn't as clear as it could be. The example is to show that not all processes with the property I defined are supermartingales. $(X)$ of the example has this property, but is not a supermartingale. If I used $-X_t$ instead of $X_t$, it would be a supermartingale.2012-01-18
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    @Ben Derrett: two questions we might ask about this family $\tau_{s,a,b}$, is first how general this family of stopping times is in the space of stopping times, and second if a stopping time doesn't pertains to this family is it possible to approach it by a sequence of element of this family ? But if there is a totally inaccessible stopping time which is not in this family then it might be possible to construct a counterexample of a local supermaringale where your condition is not true for this inaccessible s.t. but true for all elements of your family. Best regards2012-01-18
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    @Ben derret : Thank's for the preicisions, now I really got what your example was about.2012-01-18
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    what do you do when $X_s \notin (a,b)$? $ \tau = s $ I guess that's clear. I'd get rid of this if I knew how?2012-04-26
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    Does the definition of a local supermartingale require that $\mathbb{E}(|X_0|) < \infty$? I don't think integrability can be concluded from the conditions you gave.2016-10-26
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    How is $X_{\tau_{s, a, b}}$ interpreted, if $X$ never leaves the interval $(a, b)$ after time $s$?2016-10-26
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    @jochen Thanks. For the main assumption to make sense, we probably also need to require: (1) that $\tau_{s,a,b} < \infty$ a.s.; and (2) $X_{\tau_{s,a,b}}$ is integrable. Yes, if $X$ is a local supermartingale then $X_0$ is integrable.2016-10-29

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