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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.

Chill2Macht
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Ben Derrett
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  • A small thing: of X is continuous then $X_{\tau_{s,a,b}}$ is equal a or b. Sounds strange. – Kolmo Jan 17 '12 at 17:24
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    Seems to me that a local supermartingale would still satisfy your property. – Nate Eldredge Jan 17 '12 at 18:16
  • @Kolmo Yes, it will equal $a$ or $b$. – Ben Derrett Jan 17 '12 at 19:02
  • @NateEldredge Thank you! I've edited the question to reflect this (unfortunate) reality. – Ben Derrett Jan 17 '12 at 19:37
  • @Ben Derrett : I don't see why your example is a (local)supermartingale, don't we have $1=E[X_1]>X_0=0$ ? – TheBridge Jan 17 '12 at 21:21
  • @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. – Ben Derrett Jan 17 '12 at 21:34
  • @Ben Derrett : ok I got it, but to be consistent wuith yourproblem you should use $-X_t$ instead of $X_t$, I think. Regards – TheBridge Jan 18 '12 at 07:50
  • @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. – Ben Derrett Jan 18 '12 at 10:06
  • @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 regards – TheBridge Jan 18 '12 at 10:38
  • @Ben derret : Thank's for the preicisions, now I really got what your example was about. – TheBridge Jan 18 '12 at 11:11
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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? – mike Apr 26 '12 at 22:34
  • 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. – jochen Oct 26 '16 at 17:41
  • How is $X_{\tau_{s, a, b}}$ interpreted, if $X$ never leaves the interval $(a, b)$ after time $s$? – jochen Oct 26 '16 at 17:51
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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. – Ben Derrett Oct 29 '16 at 16:26
  • For continuous X, it will be a local supermartingale (for non-cts X, there may be integrability issues). I could write up an an answer but, after more than 11 years, I don't know if there is still any interest? – George Lowther Apr 13 '23 at 16:35
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    @GeorgeLowther I'm still interested. – Ben Derrett Apr 14 '23 at 19:33

1 Answers1

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To show that $ X $ is a local supermartingale, we will construct an increasing sequence of stopping times $ (\sigma_n)_{n \geq 1} $ such that each stopped process $ X^{\sigma_n} = (X_{t \wedge \sigma_n})_{t \geq 0} $ is a supermartingale.

For each $ n \in \mathbb{N} $, define $$ \sigma_n = \inf \{ t \geq 0 : |X_t| \geq n \} \wedge n. $$ Since $ X $ is continuous, $ \sigma_n $ is a stopping time, and $ \sigma_n \uparrow \infty $ as $ n \to \infty $.

Fix $ n $ and let $ 0 \leq s \leq t $. We aim to show that $$ \mathbb{E}[ X_{t \wedge \sigma_n} \mid \mathcal{F}_s ] \leq X_{s \wedge \sigma_n}. $$

Consider $ a = -n $ and $ b = n $. Note that $ \sigma_n = \tau_{0,a,b} \wedge n $. For $ s \leq t $, define $$ \tau = \tau_{s,a,b} = \inf \{ u \geq s : X_u \notin (a, b) \} \wedge n. $$ Since $ \sigma_n \leq n $, we have $ \tau \leq \sigma_n $. Then, $$ X_{t \wedge \sigma_n} = \begin{cases} X_t, & \text{if } t \leq \tau, \\ X_{\tau}, & \text{if } \tau < t \leq \sigma_n, \\ X_{\sigma_n}, & \text{if } t \geq \sigma_n. \end{cases} $$ Because $ X $ is continuous and $ |X_u| \leq n $ for $ u \leq \sigma_n $, $ X_{t \wedge \sigma_n} $ is bounded.

Using the assumption, $$ \mathbb{E}[ X_{\tau} \mid \mathcal{F}_s ] \leq X_s. $$ Since $ X_{t \wedge \sigma_n} \leq X_{\tau} $ for $ t \geq \tau $, we have $$ \mathbb{E}[ X_{t \wedge \sigma_n} \mid \mathcal{F}_s ] \leq \mathbb{E}[ X_{\tau} \mid \mathcal{F}_s ] \leq X_s = X_{s \wedge \sigma_n}. $$ Therefore, $ X^{\sigma_n} $ is a supermartingale.

As $ \sigma_n \uparrow \infty $, it follows that $ X $ is a local supermartingale.

Furdzik Zbignew
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