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

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

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

${\displaystyle 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

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

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

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

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