Stopped process

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In mathematics, a stopped process is a stochastic process that is forced to assume the same value after a prescribed (possibly random) time.

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[edit] Definition

Let

Then the stopped process Xτ is defined for t \geq 0 and \omega \in \Omega by

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

[edit] Examples

[edit] Gambling

Consider a gambler playing roulette. Xt denotes the gambler's total holdings in the casino at time t \geq 0, which may or may not be allowed to be negative, depending on whether or not the casino offers credit. Let Yt 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 time T, regardless of the state of play. Then X is really the stopped process YT, 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_{0} (\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^{\tau_{0}}.

[edit] 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 Bτ 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 τ 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 Bτ will evolve as per usual up until the random time τ, and will thereafter be constant with value a: i.e., B_{t}^{\tau} (\omega) \equiv a for all t \geq \tau (\omega).