The basic martingale equation \( \E(X_t \mid \mathscr{F}_s) = X_s \) for \( s, \, t \in T \) with \( s \le t \) can be generalized by replacing both \( s \) and \( t \) by bounded stopping times. , where each The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. A family of \( \sigma \)-algebras \(\mathfrak{F} = \{\mathscr{F}_t: t \in T\} \) is a filtration on \( (\Omega, \mathscr{F}) \) if \(s, \, t \in T\) and \(s \le t\) imply \(\mathscr{F}_s \subseteq \mathscr{F}_t \subseteq \mathscr{F}\). The random time \( t \wedge \tau \) is a stopping time for each \( t \in T \) by the result above, so \( \mathscr{F}^\tau_t \) is a sub \( \sigma \)-algebra of \( \mathscr{F} \). Total stopping time is made up of: a) reaction time b) braking time c Suppose that \( a_k \gt a_{k-1} \) for some \( k \in \N_+ \). Sometimes we need \( \sigma \)-algebras that are a bit larger than the ones in the last paragraph. ( Consider a gambler betting on games of chance. , Multiplying by $2^k$ is the same as looking at the direct next branch of these numbers in the tree, which means they all have the same stopping time too (well, it depends on what you are looking at when determining the stopping time: Odd only, odd/even,.looking at "odd/even" stopping time is just a bit trickier, the exponent of 2 of the . = Finally, let \[ Y_n = \sum_{k=1}^n (X_k - \mu) \, \quad n \in \N_+ \] Then \( \bs Y = (Y_n: n \in \N_+) \) is a martingale relative to \( \mathfrak F \), with mean 0. Once a driver perceives a need to slow or stop, a small amount of time passes. Note that \( 0 = a_0 \lt a_1 = \frac{1}{2} \). The proof is a simple consequence of the fact that the subset relation defines a partial order. In discrete time, there is a related definition. The basic idea behind the definition is that if the filtration \( \mathfrak{F} \) encodes our information as time goes by, then the process \( \bs{X} \) is observable. Let \( t \in T \). We showed in the Introduction that \( \bs X \) is a martingale if \( p = \frac{1}{2} \) (the fair case), a sub-martingale if \( p \gt \frac{1}{2} \) (the favorable case), and a super-martingale if \( p \lt \frac{1}{2} \) (the unfair case). If \( \tau \) is a stopping time relative to \( \mathfrak{F} \), then the stopped process \( \bs{X}^\tau = \{X^\tau_t: t \in T\} \) is progressively measurable with respect to the stopped filtration \( \mathfrak{F}^\tau \). Let \(\tau = \sup\{\tau_n: n \in \N_+\}\). Then \(\{\tau \ge t\} = \bigcap_{n=1}^\infty\{\tau_n \ge t\} \in \mathscr{F}_t\) for \(t \in T\). If she does not exist, of course, we must select candidate 5. ( The linear recurrence relation can be solved explicitly, but all that we care about is the fact that the solution is finite. Let \( r \in T \). Suppose \( s, \, t \in T \) with \( s \le t \). Both events in the set difference are in \( \mathscr{F}_t \). If \( \bs X \) is a super-martingale relative to \( \mathfrak F \) then \( \E(X_\tau) \le \E(X_0) \). So in particular, \( \mathfrak{F}^\tau \) is coarser than \( \mathfrak{F} \). When two stopping times are ordered, their \( \sigma \)-algebras are also ordered. Find the expected number of trials until each of the following strings is completed. vol. From the result above, \( \tau \) is also a stopping time relative to \( \mathfrak{G} \), so the statement makes sense. In terms of gambling, our gambler plays a sequence of independent and identical games, and on each game, wins 1 with probability \( p \) and loses 1 with probability \( 1 - p \). But \(\{\tau \lt t\} \subseteq \{\tau \le t\}\) and \(\{\tau \lt t\} \in \mathscr{F}_t\). Since this is true for every \( s \gt t \) it follows \(\{\tau \le t\} \in \mathscr{F}_{t+}\). When braking hard, the weight of the vehicle noticeably shifts where? Highest proven Collatz Conjecture stopping time [closed] F If the number is odd, triple it and add one. Suppose that there are \( n \in \N_+ \) candidates for a job, or perhaps potential marriage partners. \( \{\tau \ge t\} \in \mathscr{F}_t\) for every \( t \in T \). Thus the sequence is an infinite string of letters from our alphabet \( S \). Suppose again that \( p = \frac{1}{2} \). T Suppose that \( \mathfrak{F} = \{\mathscr{F}_t: t \in T\} \) and \( \mathfrak{G} = \{\mathscr{G}_t: t \in T\} \) are filtrations on \( (\Omega, \mathscr{F}) \) and that \( \mathfrak{G} \) is finer than \( \mathfrak{F} \). is called a stopping time (with respect to the filtration This is also true: Suppose again that \( \mathfrak{F} = \{\mathscr{F}_t: t \in T\} \) is a filtration on \( (\Omega, \mathscr{F}) \) and that \( \tau \) is a stopping time relative to \( \mathfrak{F} \). By definition, \( \omega \mapsto X_t(\omega) \) is measurable for each \( t \in T \), but it is often necessary for the process to be jointly measurable in \( \omega \) and \( t \). T At any rate, the next theorem gives the solution. Suppose again that \( \bs{X} = \{X_t: t \in T\} \) is a stochastic process with sample space \( (\Omega, \mathscr{F}) \) and state space \( (S, \mathscr{S}) \), and that \( \mathfrak{F} = \{\mathscr{F}_t: t \in T\} \) is a filtration. ( Our general goal in this section is to see if some of the important martingale properties are preserved if the deterministic time \( t \in T \) is replaced by a (random) stopping time. A random variable \( \tau \) taking values in \( T_\infty \) is called a random time. Since all of the bets are fair, \( \bs X = \{X_n: n \in \N_+\} \) is a martingale with mean 0. For \( t \in T \) define \( \mathscr{F}^\tau_t = \mathscr{F}_{t \wedge \tau} \). Hence by the martingale property, \[ \E(X_k ; A \cap \{\tau = j\}) = \E(X_j ; A \cap \{\tau = j\}) = \E(X_\tau ; A \cap \{\tau = j\})\] Since \( k \) is an upper bound on \( \tau \), the events \( A \cap \{\tau = j\} \) for \( j = 0, 1, \ldots, k \) partition \( A \), so summing the displayed equation over \( j \) gives \( \E(X_k ; A) = \E(X_\tau ; A) \). The last definition must seem awfully obscure, but it does have a place. T A stopping time is predictable if it is equal to the limit of an increasing sequence of stopping times n satisfying n < whenever > 0. 1. Let \( S = \{0, 1\} \) and \( \mathscr{S} = \mathscr{P}(S) = \{\emptyset, \{0\}, \{1\}, \{0, 1\}\} \). Follow answered Jan 1, 2012 at 10:10. One handed steering. She continues in this way: as long as she wins, she bets her entire fortune on the next trial on the next letter of the word, until either she loses or completes the word \( \bs a \). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. This variable has the geometric distribution on \( \N_+ \) with success parameter \( f(\bs a) \) and so in particular, \( \E(M_\bs{a}) = 1 / f(\bs a) \). Note that \( X_\tau: \Omega \to S \) is the composition of the function \( \omega \mapsto (\omega, \tau(\omega)) \) from \( \Omega \) to \( \Omega \times T\) with the function \((\omega, t) \mapsto X_t(\omega) \) from \( \Omega \times T \) to \( S \). If a vehicle's speed doubles from 20 mph to 40 mph, the distance needed to stop the vehicle increases by how many times? , We also extend the topology on \( T \) to \( T_\infty \) by the rule that for each \( s \in T \), the set \( \{t \in T_\infty: t \gt s\} \) is an open neighborhood of \( \infty\). We then give \( T_\infty \) the Borel \( \sigma \)-algebra \( \mathscr{T}_\infty \) as before. Description Determine the stopping time, or "total" stopping time, for an initial value. For \( s, \, t \in T \), \[\{\tau \le t\} \cap \{\tau \le s\} = \{\tau \le s \wedge t\} \in \mathscr{F}_{s \wedge t} \subseteq \mathscr{F}_t\] Hence \(\{\tau \le s\} \in \mathscr{F}_\tau\). T However, we usually do have a stochastic process \( \bs{X} = \{X_t: t \in T\} \), and in this case the filtration \( \mathfrak{F}^0 = \{\mathscr{F}^0_t: t \in T\} \) where \( \mathscr{F}^0_t = \sigma\{X_s: s \in T, \, s \le t\} \) is the natural filtration associated with \( \bs{X} \). Parts (a) and (c) hold since \( \bs X \) is a null recurrent Markov chain. The difference between the two words is that the word in (b) has a prefix (a proper string at the beginning of the word) that is also a suffix (a proper string at the end of the word). Set meeting participants as presenters and co-organizers, as necessary. ) To simplify the notation, let \( \N_n = \{0, 1, \ldots, n\} \) and \( \N_n^+ = \{1, 2, \ldots, n\} \). Next recall that the \( \sigma \)-algebra associated with the stopping time \( \tau \) is \[ \mathscr{F}_\tau = \left\{A \in \mathscr{F}: A \cap \{\tau \le t\} \in \mathscr{F}_t \text{ for all } t \in T\right\} \] So \( \mathscr{F}_\tau \) is the collection of events up to the random time \( \tau \) just as \( \mathscr{F}_t \) is the collection of events up to the deterministic time \( t \in T \). Then, X is said to locally satisfy some property P if there exists a sequence of stopping times n, which increases to infinity and for which the processes. Check to ensure the parking brake is set 2. Finally using the tower property we have \[ X_\rho = \E(X_k \mid \mathscr{F}_\rho) = \E[\E(X_k \mid \mathscr{F}_\rho) \mid \mathscr{F}_\tau] = \E[\E(X_k \mid \mathscr{F}_\tau) \mid \mathscr{F}_\rho] = \E(X_\tau \mid \mathscr{F}_\rho)\]. 0 If \( s \gt t \) then there exists \( k \in \N_+ \) such that \( t_n \lt s \) for each \( n \ge k \). That is, the time when the gambler decides to stop playing can only depend on the information that the gambler has up to that point in time. Then. Sci. This result is trivial since \( \{\tau \gt t\} = \{\tau \le t\}^c \) for \( t \in T \). The converse to part (a) (or equivalently (b)) is not true, but in continuous time there is a connection to the right-continuous refinement of the filtration. If \( \bs a \) is compound, then \( \nu(\bs a) = 1 / f(\bs a) + \nu(\bs b) \) where \( \bs b \) is the longest word that is both a prefix and a suffix of \( \bs a \). That is, \( \mathfrak{F} = \mathfrak{F}^0_+ \), so that \( \mathscr{F}_t = \sigma\{X_s: s \in [0, t]\}_+ \) for \( t \in [0, \infty) \). Conversely, suppose that this condition holds. If \( \bs X \) is a martingale relative to \( \mathfrak F \) then \( \E(X_\tau \mid \mathscr{F}_\rho) = X_\rho \). If \( \tau \) is a stopping time relative to a filtration, then it is also a stoping time relative to any finer filtration: Suppose that \( \mathfrak{F} = \{\mathscr{F}_t: t \in T\} \) and \( \mathfrak{G} = \{\mathscr{G}_t: t \in T\} \) are filtrations on \( (\Omega, \mathscr{F}) \), and that \(\mathfrak{G}\) is finer than \( \mathfrak{F} \). The proofs in parts (b) and (c) are as in the discrete time. The concept of stopping time is closely related to that of filtration of a stochastic process. Since a stopped martingale is still a martingale, the the mean property holds. But for \( r \in T \), \( \{t \wedge \tau \le r\} = \Omega \) if \( r \ge t \) and \( \{t \wedge \tau \le r\} = \{\tau \le r\} \) if \( r \lt t \). If \( \bs a \) is simple then \( \nu(\bs a) = 1 / f(\bs a) \). For many random processes, the first time that the process enters or hits a set of states is particularly important. The relation \( \preceq \) is a partial order on the collection of filtrations on \( (\Omega, \mathscr{F}) \). Hence \( \mathscr{F}_t \subseteq \mathscr{H}_t \) for each \( t \in T \) and so \( \mathfrak{F} \preceq \mathfrak{H} \). Let \( \rho_n = \lceil 2^n \rho \rceil / 2^n \) and \( \tau_n = \lceil 2^n \tau \rceil / 2^n \) for \( n \in \N \). {\displaystyle n} A random time \( \tau \) is a stopping time for \( \mathfrak{F} \) if and only if \( \{\tau = n\} \in \mathscr{F}_n \) for every \( n \in \N \). If \(\tau\) is a stopping time then as shown above, \(\{\tau = n\} \in \mathscr{F}_n\) for every \( n \in \N \). stochastic processes - Sum of two stopping times is a stopping time Then \( \tau_a \lt \tau_b \lt \infty \) but \[ b = \E\left(X_{\tau_b} \mid \mathscr{F}_{\tau_a} \right) \ne X_{\tau_a} = a \]. Sometimes, particularly in continuous time, there are technical reasons for somewhat different \( \sigma \)-algebras. The probability measure \( P \) can be extended to \( \mathscr{F}^P \) as described above, and hence is defined on \( \mathscr{F}^P_t \) for each \( t \in T \). Let \( \mathscr{F}^* = \bigcap \{\mathscr{F}^P: P \in \mathscr{P}\} \), and let \( \mathfrak{F}^* = \bigwedge \{\mathfrak{F}^P: P \in \mathscr{P}\} \). The first function is measurable because the two coordinate functions are measurable. No knowledge of the future is required, since such a rule would surely result in an unfair game. The term stopping time comes from gambling. + N Right continuous filtrations have some nice properties, as we will see later. For \(t \in T\), \(\{\tau \le t\} = \bigcap_{n=1}^\infty \{\tau_n \le t\}\). The strategy of waiting until the net winnings reaches a specified goal \( c \) is unsustainable. If a random time \( \tau \) is a stopping time relative to \( \mathfrak{F} \) then \( \tau \) is a stopping time relative to \( \mathfrak{G} \). Examples of totally inaccessible stopping times include the jump times of Poisson processes. Vehicle Braking Time/Distance Human Perception time/distanceis the distance your vehicle travels If the original filtration is not right continuous, the slightly refined filtration is: Suppose again that \( \mathfrak{F} = \{\mathscr{F}_t: t \in [0, \infty)\} \) is a filtration. We need to show that \( \E(X_\tau; A) = \E(X_\rho; A) \) for every \( A \in \mathscr{F}_\rho \). Hence \( \nu(001) = \frac{1}{q^2 p} \), For the word 010, gambler \( N - 2 \) wins \( \frac{1}{q^2 p} \) on her three bets as before. F , First, if X is a process and is a stopping time, then X is used to denote the process X stopped at time . So if the filtration \( \mathfrak{F} \) encodes the information available as time goes by, then the filtration \( \mathfrak{F}_+ \) allows an infinitesimal peak into the future at each \( t \in [0, \infty) \). = In discrete time, it's easy to see that these are stopping times. Threshold braking is the application of brake pressure to a point just short of locking up the brakes. It is shown that there are in nitely many positive integers nhaving a nite total stopping time 1(n) such that Clearly if \( \bs{X} \) is predictable by \( \mathfrak{F} \) then \( \bs{X} \) is adapted to \( \mathfrak{F} \). For a finite word \( \bs a \) from the alphabet \( S \), \( \nu(\bs a) \) is the total winnings by all of the players at time \( N_{\bs a} \). The expected value of this bet is \[ f(a) \frac{c}{f(a)} - c = 0 \] and so the bet is fair. Is that a workable strategy? Then \( \mathfrak{F}^P = \{\mathscr{F}^P_t: t \in T\} \) is a filtration on \( \left(\Omega, \mathscr{F}^P\right) \) that is finer than \( \mathfrak{F} \) and is complete relative to \( P \). If she wins, she bet her entire fortune \( 1 / f(a_1) \) on the next trial on \( a_2 \). F About us. Define \( \bs Y = \{Y_k: k \in \N_n\} \) by \( Y_0 = 0 \) and \( Y_k = X_{\rho \wedge k} \vee a_{n-k} \) for \( k \in \N_n^+ \). possibly 1) which gives a rule for stopping a random process. If \( \bs X \) is a martingale relative to \( \mathfrak F \) then \( \E(X_{t \wedge \tau}) = E(X_0) \), If \( \bs X \) is a sub-martingale relative to \( \mathfrak F \) then \( \E(X_{t \wedge \tau}) \ge E(X_0) \), If \( \bs X \) is a super-martingale relative to \( \mathfrak F \) then \( \E(X_{t \wedge \tau}) \le E(X_0) \). Suppose also that \( \bs{X} = \{X_t: t \in [0, \infty)\} \) is right continuous. ( If \( \bs X \) is a martingale relative to \( \mathfrak F \) then \( \E(X_\tau) = \E(X_\rho) \). Reaction Time. You might think that \(\tau_A\) and \( \rho_A \) should always be a stopping times, since \(\tau_A \le t\) if and only if \(X_s \in A\) for some \( s \in T_+ \) with \(s \le t\), and \( \rho_A \le t \) if and only if \( X_s \in A \) for some \( s \in T \) with \( s \le t \). So again letting \( n \to \infty \) in the displayed equation gives \( \E(X_\tau) \le \E(X_0) \). Then \( \mathfrak{F}_+ = \{\mathscr{F}_{t+}: t \in T\} \) is also a filtration on \( (\Omega, \mathscr{F}) \) and is finer than \( \mathfrak{F} \). , must be based only on the information present at time Let \( \mathfrak F = \{\mathscr{F}_k: k \in \N_n^+\} \) be the natural filtration of \( \bs X \), and suppose that \( \rho \) is a stopping time for \( \mathfrak F \). F If \( A \in \mathscr{F}_\tau \) then for \( t \in T \), \( A \cap \{\tau \le t\} \in \mathscr{F}_t \subseteq \mathscr{G}_t \), so \( A \in \mathscr{G}_\tau \). Suppose instead that \(T = [0, \infty)\) and \(t \in T\). That is, stopping time is accessible if, P( = n for some n) = 1, where n are predictable times. t 3-4 seconds. So a filtration is simply an increasing family of sub-\(\sigma\)-algebras of \( \mathscr{F} \), indexed by \( T \). For \( k \in \N \), \[ \E(X \vee a_k) = \int_0^1 (x \vee a_k) dx = \int_0^{a_k} a_k \, dx + \int_{a_k}^1 x \, dx = \frac{1}{2}(1 + a_k^2) = a_{k+1} \]. a) Set the parking brake b) Put the shift lever in park c) Turn the ignition to off d) Put the shift lever in neutral A The first step to ensure the path behind the vehicle is clear prior to backing is: a) Use the inside rear view mirror b) Use the outside mirrors c) Look over your right shoulder {\displaystyle \mathbb {F} =(({\mathcal {F}}_{n})_{n\in \mathbb {N} }} Note again that we can have a filtration without an underlying stochastic process in the background.
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