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;
$X:[0,+\infty )\times \Omega \to \mathbb {X}$ be a stochastic process;
${\displaystyle \tau$ be a stopping time with respect to some filtration $\{{\mathcal {F}}_{t}|t\geq 0\}$ of ${}{\mathcal {F}}$ .

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

X_{t}^{\tau }(\omega ):=X_{\min\{t,\tau (\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

\tau (\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 $B:[0,+\infty )\times \Omega \to \mathbb {R}$ be a one-dimensional standard Brownian motion starting at zero.

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

\tau (\omega ):=\inf\{t>0|B_{t}(\omega )\geq a\}.

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

References

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

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Definition

Examples

Gambling

Brownian motion

See also

References