What is the largest optimal stopping time, and what are the “in-between†times?
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Given an adapted process $X_t$ and it's Snell envelope $S_t$, I know that the smallest optimal stopping time is to stop as soon as the Snell envelope equals $X$. This is very intuitive and makes perfect sense. You could explain it to a kid.
However, I have a hard time grasping any other stopping time, such as the "largest" stopping time.
What is an "informal" description of such other stopping times (take e.g. the largest one). How would you describe it in words?
Could you give an example?
stopping-times
asked Aug 19 at 10:33
nemesis
82
add a comment |Â
up vote
1
down vote
favorite
Given an adapted process $X_t$ and it's Snell envelope $S_t$, I know that the smallest optimal stopping time is to stop as soon as the Snell envelope equals $X$. This is very intuitive and makes perfect sense. You could explain it to a kid.
However, I have a hard time grasping any other stopping time, such as the "largest" stopping time.
What is an "informal" description of such other stopping times (take e.g. the largest one). How would you describe it in words?
Could you give an example?
stopping-times
asked Aug 19 at 10:33
nemesis
82
When you say 'optimal', that implies that there is some objective that you are trying to optimize. What is that objective?
– Ben Derrett
Aug 19 at 10:42
The objective is to maximize the expectation of $X$ relative to stopping times. So e.g. $max_nu E X_nu$ where $nu$ are stopping times. For an optimal stopping time $nu*$, we have $max_nu E X_nu = E X_nu *$.
– nemesis
Aug 19 at 10:44
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given an adapted process $X_t$ and it's Snell envelope $S_t$, I know that the smallest optimal stopping time is to stop as soon as the Snell envelope equals $X$. This is very intuitive and makes perfect sense. You could explain it to a kid.
However, I have a hard time grasping any other stopping time, such as the "largest" stopping time.
What is an "informal" description of such other stopping times (take e.g. the largest one). How would you describe it in words?
Could you give an example?
stopping-times
asked Aug 19 at 10:33
nemesis
82
Given an adapted process $X_t$ and it's Snell envelope $S_t$, I know that the smallest optimal stopping time is to stop as soon as the Snell envelope equals $X$. This is very intuitive and makes perfect sense. You could explain it to a kid.
However, I have a hard time grasping any other stopping time, such as the "largest" stopping time.
What is an "informal" description of such other stopping times (take e.g. the largest one). How would you describe it in words?
Could you give an example?
stopping-times
asked Aug 19 at 10:33
nemesis
82
asked Aug 19 at 10:33
nemesis
82
asked Aug 19 at 10:33
nemesis
82
asked Aug 19 at 10:33
nemesis
82
82
When you say 'optimal', that implies that there is some objective that you are trying to optimize. What is that objective?
– Ben Derrett
Aug 19 at 10:42
The objective is to maximize the expectation of $X$ relative to stopping times. So e.g. $max_nu E X_nu$ where $nu$ are stopping times. For an optimal stopping time $nu*$, we have $max_nu E X_nu = E X_nu *$.
– nemesis
Aug 19 at 10:44
add a comment |Â
When you say 'optimal', that implies that there is some objective that you are trying to optimize. What is that objective?
– Ben Derrett
Aug 19 at 10:42
The objective is to maximize the expectation of $X$ relative to stopping times. So e.g. $max_nu E X_nu$ where $nu$ are stopping times. For an optimal stopping time $nu*$, we have $max_nu E X_nu = E X_nu *$.
– nemesis
Aug 19 at 10:44
When you say 'optimal', that implies that there is some objective that you are trying to optimize. What is that objective?
– Ben Derrett
Aug 19 at 10:42
When you say 'optimal', that implies that there is some objective that you are trying to optimize. What is that objective?
– Ben Derrett
Aug 19 at 10:42
The objective is to maximize the expectation of $X$ relative to stopping times. So e.g. $max_nu E X_nu$ where $nu$ are stopping times. For an optimal stopping time $nu*$, we have $max_nu E X_nu = E X_nu *$.
– nemesis
Aug 19 at 10:44
The objective is to maximize the expectation of $X$ relative to stopping times. So e.g. $max_nu E X_nu$ where $nu$ are stopping times. For an optimal stopping time $nu*$, we have $max_nu E X_nu = E X_nu *$.
– nemesis
Aug 19 at 10:44
add a comment |Â
1 Answer
1
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oldest
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up vote
1
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accepted
Suppose that the problem is formulated in discrete time and that you wish to choose the largest optimal stopping time $nu_*$. Informally, the stopping time $nu_*$ can be described as follows. Start at time 0 and denote the current time by $t$. If $X_t < S_t$, continue to the next timestep. Otherwise, since $S$ dominates $X$, $X_t = S_t$. If $mathbbE_t[S_t+1]=S_t$, continue to the next timestep, otherwise stop.
answered Aug 19 at 11:29
Ben Derrett
2,5521541
Okay, so the Snell envelope $S_t$ equals the maximum of either $X_t$ or $E_t S_t+1$. If they are all equal, i.e. $S_t = X_t = E_t S_t+1$, only then is it still optimal to continue, but if we only have the first equality, then we must stop?
– nemesis
Aug 19 at 11:50
@nemesis, yes, that's right
– Ben Derrett
Aug 19 at 12:00
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Suppose that the problem is formulated in discrete time and that you wish to choose the largest optimal stopping time $nu_*$. Informally, the stopping time $nu_*$ can be described as follows. Start at time 0 and denote the current time by $t$. If $X_t < S_t$, continue to the next timestep. Otherwise, since $S$ dominates $X$, $X_t = S_t$. If $mathbbE_t[S_t+1]=S_t$, continue to the next timestep, otherwise stop.
answered Aug 19 at 11:29
Ben Derrett
2,5521541
Okay, so the Snell envelope $S_t$ equals the maximum of either $X_t$ or $E_t S_t+1$. If they are all equal, i.e. $S_t = X_t = E_t S_t+1$, only then is it still optimal to continue, but if we only have the first equality, then we must stop?
– nemesis
Aug 19 at 11:50
@nemesis, yes, that's right
– Ben Derrett
Aug 19 at 12:00
add a comment |Â
up vote
1
down vote
accepted
Suppose that the problem is formulated in discrete time and that you wish to choose the largest optimal stopping time $nu_*$. Informally, the stopping time $nu_*$ can be described as follows. Start at time 0 and denote the current time by $t$. If $X_t < S_t$, continue to the next timestep. Otherwise, since $S$ dominates $X$, $X_t = S_t$. If $mathbbE_t[S_t+1]=S_t$, continue to the next timestep, otherwise stop.
answered Aug 19 at 11:29
Ben Derrett
2,5521541
Okay, so the Snell envelope $S_t$ equals the maximum of either $X_t$ or $E_t S_t+1$. If they are all equal, i.e. $S_t = X_t = E_t S_t+1$, only then is it still optimal to continue, but if we only have the first equality, then we must stop?
– nemesis
Aug 19 at 11:50
@nemesis, yes, that's right
– Ben Derrett
Aug 19 at 12:00
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Suppose that the problem is formulated in discrete time and that you wish to choose the largest optimal stopping time $nu_*$. Informally, the stopping time $nu_*$ can be described as follows. Start at time 0 and denote the current time by $t$. If $X_t < S_t$, continue to the next timestep. Otherwise, since $S$ dominates $X$, $X_t = S_t$. If $mathbbE_t[S_t+1]=S_t$, continue to the next timestep, otherwise stop.
answered Aug 19 at 11:29
Ben Derrett
2,5521541
Suppose that the problem is formulated in discrete time and that you wish to choose the largest optimal stopping time $nu_*$. Informally, the stopping time $nu_*$ can be described as follows. Start at time 0 and denote the current time by $t$. If $X_t < S_t$, continue to the next timestep. Otherwise, since $S$ dominates $X$, $X_t = S_t$. If $mathbbE_t[S_t+1]=S_t$, continue to the next timestep, otherwise stop.
answered Aug 19 at 11:29
Ben Derrett
2,5521541
answered Aug 19 at 11:29
Ben Derrett
2,5521541
answered Aug 19 at 11:29
Ben Derrett
2,5521541
answered Aug 19 at 11:29
Ben Derrett
2,5521541
2,5521541
Okay, so the Snell envelope $S_t$ equals the maximum of either $X_t$ or $E_t S_t+1$. If they are all equal, i.e. $S_t = X_t = E_t S_t+1$, only then is it still optimal to continue, but if we only have the first equality, then we must stop?
– nemesis
Aug 19 at 11:50
@nemesis, yes, that's right
– Ben Derrett
Aug 19 at 12:00
add a comment |Â
Okay, so the Snell envelope $S_t$ equals the maximum of either $X_t$ or $E_t S_t+1$. If they are all equal, i.e. $S_t = X_t = E_t S_t+1$, only then is it still optimal to continue, but if we only have the first equality, then we must stop?
– nemesis
Aug 19 at 11:50
@nemesis, yes, that's right
– Ben Derrett
Aug 19 at 12:00
Okay, so the Snell envelope $S_t$ equals the maximum of either $X_t$ or $E_t S_t+1$. If they are all equal, i.e. $S_t = X_t = E_t S_t+1$, only then is it still optimal to continue, but if we only have the first equality, then we must stop?
– nemesis
Aug 19 at 11:50
Okay, so the Snell envelope $S_t$ equals the maximum of either $X_t$ or $E_t S_t+1$. If they are all equal, i.e. $S_t = X_t = E_t S_t+1$, only then is it still optimal to continue, but if we only have the first equality, then we must stop?
– nemesis
Aug 19 at 11:50
@nemesis, yes, that's right
– Ben Derrett
Aug 19 at 12:00
@nemesis, yes, that's right
– Ben Derrett
Aug 19 at 12:00
add a comment |Â
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When you say 'optimal', that implies that there is some objective that you are trying to optimize. What is that objective?
– Ben Derrett
Aug 19 at 10:42
The objective is to maximize the expectation of $X$ relative to stopping times. So e.g. $max_nu E X_nu$ where $nu$ are stopping times. For an optimal stopping time $nu*$, we have $max_nu E X_nu = E X_nu *$.
– nemesis
Aug 19 at 10:44