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

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

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.

Let ${\displaystyle B:[0,+\infty )\times \Omega \to \mathbb {R} }$  be 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 ${\displaystyle t\geq \tau (\omega )}$ .

• Killed process

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