Stopped process

Short description: Stochastic process

In mathematics, a stopped process is a stochastic process that is forced to assume the same value after a prescribed (possibly random) time.

Definition

Let

• [math]\displaystyle{ (\Omega, \mathcal{F}, \mathbb{P}) }[/math] be a probability space;
• [math]\displaystyle{ (\mathbb{X}, \mathcal{A}) }[/math] be a measurable space;
• [math]\displaystyle{ X : [0, + \infty) \times \Omega \to \mathbb{X} }[/math] be a stochastic process;
• [math]\displaystyle{ \tau : \Omega \to [0, + \infty] }[/math] be a stopping time with respect to some filtration [math]\displaystyle{ \{ \mathcal{F}_{t} | t \geq 0 \} }[/math] of [math]\displaystyle{ {}\mathcal{F} }[/math].

Then the stopped process [math]\displaystyle{ X^{\tau} }[/math] is defined for [math]\displaystyle{ t \geq 0 }[/math] and [math]\displaystyle{ \omega \in \Omega }[/math] by

[math]\displaystyle{ X_{t}^{\tau} (\omega) := X_{\min \{ t, \tau (\omega) \}} (\omega). }[/math]

Examples

Gambling

Consider a gambler playing roulette. X_t denotes the gambler's total holdings in the casino at time t ≥ 0, which may or may not be allowed to be negative, depending on whether or not the casino offers credit. Let Y_t denote what the gambler's holdings would be if he/she could obtain unlimited credit (so Y can attain negative values).

• Stopping at a deterministic time: suppose that the casino is prepared to lend the gambler unlimited credit, and that the gambler resolves to leave the game at a predetermined time T, regardless of the state of play. Then X is really the stopped process Y^T, since the gambler's account remains in the same state after leaving the game as it was in at the moment that the gambler left the game.
• Stopping at a random time: suppose that the gambler has no other sources of revenue, and that the casino will not extend its customers credit. The gambler resolves to play until and unless he/she goes broke. Then the random time

[math]\displaystyle{ \tau (\omega) := \inf \{ t \geq 0 | Y_{t} (\omega) = 0 \} }[/math]

is a stopping time for Y, and, since the gambler cannot continue to play after he/she has exhausted his/her resources, X is the stopped process Y^τ.

Brownian motion

Let [math]\displaystyle{ B : [0, + \infty) \times \Omega \to \mathbb{R} }[/math] be a one-dimensional standard Brownian motion starting at zero.

• Stopping at a deterministic time [math]\displaystyle{ T \gt 0 }[/math]: if [math]\displaystyle{ \tau (\omega) \equiv T }[/math], then the stopped Brownian motion [math]\displaystyle{ B^{\tau} }[/math] will evolve as per usual up until time [math]\displaystyle{ T }[/math], and thereafter will stay constant: i.e., [math]\displaystyle{ B_{t}^{\tau} (\omega) \equiv B_{T} (\omega) }[/math] for all [math]\displaystyle{ t \geq T }[/math].
• Stopping at a random time: define a random stopping time [math]\displaystyle{ \tau }[/math] by the first hitting time for the region [math]\displaystyle{ \{ x \in \mathbb{R} | x \geq a \} }[/math]:

[math]\displaystyle{ \tau (\omega) := \inf \{ t \gt 0 | B_{t} (\omega) \geq a \}. }[/math]

Then the stopped Brownian motion [math]\displaystyle{ B^{\tau} }[/math] will evolve as per usual up until the random time [math]\displaystyle{ \tau }[/math], and will thereafter be constant with value [math]\displaystyle{ a }[/math]: i.e., [math]\displaystyle{ B_{t}^{\tau} (\omega) \equiv a }[/math] for all [math]\displaystyle{ t \geq \tau (\omega) }[/math].

See also

• Killed process

References

• Robert G. Gallager. Stochastic Processes: Theory for Applications. Cambridge University Press, Dec 12, 2013 pg. 450