// Workers AI · dad joke modeWhy was the stopped process sad? It was terminated.

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;
• ${\displaystyle \tau$ :\Omega \to [0,+\infty ]} be a stopping time with respect to some filtration $\{{\mathcal {F}}_{t}|t\geq 0\}$ of ${}{\mathcal {F}}$.

Then the stopped process $X^{\tau }$ 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 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 $t\geq \tau (\omega )$.

See also

• Killed process

References

• Gallager, Robert G. (2013). Stochastic Processes: Theory for Applications. Cambridge University Press. p. 450. ISBN 978-1-107-03975-9.