Stopped 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
* $(Omega,\; mathcal\{F\},\; mathbb\{P\})$ be a probability space;
* $(mathbb\{X\},\; mathcal\{A\})$ be a measurable space;
* $X\; :\; [0,\; +\; infty)\; imes\; Omega\; o\; mathbb\{X\}$ be a stochastic process;
* $au\; :\; Omega\; o\; [0,\; +\; infty]$ be a stopping time with respect to some filtration $\{\; mathcal\{F\}\_\{t\}\; |\; t\; geq\; 0\; \}$ of $\{\}mathcal\{F\}$.

Then the stopped process $X^\{\; au\}$ is defined for $t\; geq\; 0$ and $omega\; in\; Omega$ by

:$X\_\{t\}^\{\; au\}\; (omega)\; :=\; X\_\{min\; \{\; t,\; au\; (omega)\; (omega).$

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

:$au\; (omega)\; :=\; inf\; \{\; t\; geq\; 0\; |\; Y\_\{t\}\; (omega)\; =\; 0\; \}$

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 $B\; :\; [0,\; +\; infty)\; imes\; Omega\; o\; mathbb\{R\}$ be a one-dimensional standard Brownian motion starting at zero.

* Stopping at a deterministic time $T\; >\; 0$: if $au\; (omega)\; equiv\; T$, then the stopped Brownian motion $B^\{\; au\}$ will evolve as per usual up until time $T$, and thereafter will stay constant: i.e., $B\_\{t\}^\{\; au\}\; (omega)\; equiv\; B\_\{T\}\; (omega)$ for all $t\; geq\; T$.

* Stopping at a random time: define a random stopping time $au$ by the first hitting time for the region $\{\; x\; in\; mathbb\{R\}\; |\; x\; geq\; a\; \}$:

::$au\; (omega)\; :=\; inf\; \{\; t\; >\; 0\; |\; B\_\{t\}\; (omega)\; geq\; a\; \}.$

Then the stopped Brownian motion $B^\{\; au\}$ will evolve as per usual up until the random time $au$, and will thereafter be constant with value $a$: i.e., $B\_\{t\}^\{\; au\}\; (omega)\; equiv\; a$ for all $t\; geq\; au\; (omega)$.

ee also

* Killed process