# Does this Condition on Exit Times imply $X_t$ is a Local Supermartingale?

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.

• A small thing: of X is continuous then $X_{\tau_{s,a,b}}$ is equal a or b. Sounds strange. Jan 17, 2012 at 17:24
• Seems to me that a local supermartingale would still satisfy your property. Jan 17, 2012 at 18:16
• 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?
– mike
Apr 26, 2012 at 22:34
• @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. Oct 29, 2016 at 16:26
• @GeorgeLowther I'm still interested. Apr 14, 2023 at 19:33