Showing a stopping time is finite












1














Let $T = inf{ n : S_n = a text{ or } S_n = -b}$ be a stopping time, where $S_n = X_1 + dots +X_n$ and each $X_n$ is a martingale. I am looking at a proof which shows that $T < infty$ almost surely. They state:



$$P(T = infty) leq P(T > n) leq P(|S_n| leq text{max}{a,b})$$



Could someone explain these inequalities for me? The first one holds for all $n$ which I can somewhat see, but I have no idea about the second one. Surely $|S_n| leq text{max}{a,b}$ doesn't make sense, as if $S_n = a$ or $-b$ then $ Tnot > n$?






probability probability-theory measure-theory martingales stopping-times







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  • But if $S_n=a$ or $S_n=-b$, then $Tnot>n$
    – Jimmy R.
    Mar 30 '16 at 19:30








  • 1




    @JimmyR. Yes...but you want the implication to go the other way, so it doesn't matter.
    – Ian
    Mar 30 '16 at 19:31










  • everything else is copied up correctly
    – lampj20la
    Mar 30 '16 at 19:32
















1














Let $T = inf{ n : S_n = a text{ or } S_n = -b}$ be a stopping time, where $S_n = X_1 + dots +X_n$ and each $X_n$ is a martingale. I am looking at a proof which shows that $T < infty$ almost surely. They state:



$$P(T = infty) leq P(T > n) leq P(|S_n| leq text{max}{a,b})$$



Could someone explain these inequalities for me? The first one holds for all $n$ which I can somewhat see, but I have no idea about the second one. Surely $|S_n| leq text{max}{a,b}$ doesn't make sense, as if $S_n = a$ or $-b$ then $ Tnot > n$?






probability probability-theory measure-theory martingales stopping-times







share|cite|improve this question
























  • But if $S_n=a$ or $S_n=-b$, then $Tnot>n$
    – Jimmy R.
    Mar 30 '16 at 19:30








  • 1




    @JimmyR. Yes...but you want the implication to go the other way, so it doesn't matter.
    – Ian
    Mar 30 '16 at 19:31










  • everything else is copied up correctly
    – lampj20la
    Mar 30 '16 at 19:32














1












1








1


2





Let $T = inf{ n : S_n = a text{ or } S_n = -b}$ be a stopping time, where $S_n = X_1 + dots +X_n$ and each $X_n$ is a martingale. I am looking at a proof which shows that $T < infty$ almost surely. They state:



$$P(T = infty) leq P(T > n) leq P(|S_n| leq text{max}{a,b})$$



Could someone explain these inequalities for me? The first one holds for all $n$ which I can somewhat see, but I have no idea about the second one. Surely $|S_n| leq text{max}{a,b}$ doesn't make sense, as if $S_n = a$ or $-b$ then $ Tnot > n$?






probability probability-theory measure-theory martingales stopping-times







share|cite|improve this question















Let $T = inf{ n : S_n = a text{ or } S_n = -b}$ be a stopping time, where $S_n = X_1 + dots +X_n$ and each $X_n$ is a martingale. I am looking at a proof which shows that $T < infty$ almost surely. They state:



$$P(T = infty) leq P(T > n) leq P(|S_n| leq text{max}{a,b})$$



Could someone explain these inequalities for me? The first one holds for all $n$ which I can somewhat see, but I have no idea about the second one. Surely $|S_n| leq text{max}{a,b}$ doesn't make sense, as if $S_n = a$ or $-b$ then $ Tnot > n$?






probability probability-theory measure-theory martingales stopping-times




probability probability-theory measure-theory martingales stopping-times






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share|cite|improve this question













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edited Mar 30 '16 at 19:33









Jimmy R.

33k42157




33k42157










asked Mar 30 '16 at 19:15









lampj20la

16613




16613












  • But if $S_n=a$ or $S_n=-b$, then $Tnot>n$
    – Jimmy R.
    Mar 30 '16 at 19:30








  • 1




    @JimmyR. Yes...but you want the implication to go the other way, so it doesn't matter.
    – Ian
    Mar 30 '16 at 19:31










  • everything else is copied up correctly
    – lampj20la
    Mar 30 '16 at 19:32


















  • But if $S_n=a$ or $S_n=-b$, then $Tnot>n$
    – Jimmy R.
    Mar 30 '16 at 19:30








  • 1




    @JimmyR. Yes...but you want the implication to go the other way, so it doesn't matter.
    – Ian
    Mar 30 '16 at 19:31










  • everything else is copied up correctly
    – lampj20la
    Mar 30 '16 at 19:32
















But if $S_n=a$ or $S_n=-b$, then $Tnot>n$
– Jimmy R.
Mar 30 '16 at 19:30






But if $S_n=a$ or $S_n=-b$, then $Tnot>n$
– Jimmy R.
Mar 30 '16 at 19:30






1




1




@JimmyR. Yes...but you want the implication to go the other way, so it doesn't matter.
– Ian
Mar 30 '16 at 19:31




@JimmyR. Yes...but you want the implication to go the other way, so it doesn't matter.
– Ian
Mar 30 '16 at 19:31












everything else is copied up correctly
– lampj20la
Mar 30 '16 at 19:32




everything else is copied up correctly
– lampj20la
Mar 30 '16 at 19:32










1 Answer
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oldest

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To have $T=infty$ you must have $T>n$ for all $n$, so in particular you must have $T>n$ for a fixed $n$. To have $T>n$ for a fixed $n$, $S_k$ can't be $a$ or $-b$ for any $k leq n$. But this would have to happen at some point if $|S_k|$ ever became larger than both $a$ and $b$. (The process doesn't let you skip over $a$ or $-b$ on the way to the outside of the interval $(-b,a)$.) So $max_{k leq n} |S_n|$ has to be less than $max { a,b }$.



So ${ T=infty } subseteq { T>n } subseteq { max_{k leq n} |S_n| leq max { a,b } }$. Then use monotonicity of $P$ to get the result.



By the way, in the terminology that I learned, a "Markov time" is what you are referring to and a "stopping time" is an a.s. finite Markov time. I'm not sure how universal this terminology is, but I find it a useful distinction.






share|cite|improve this answer





















  • Thanks for your response - I don't follow why $|S_n| leq text{max}{a,b}$. Surely if it equaled either $a$ or $b$ we would not have $T>n$?
    – lampj20la
    Mar 30 '16 at 19:35






  • 1




    @lampj20la That's true, but we're saying $T>n Rightarrow |S_n| leq max { a,b }$ not the other way around. You could say $<$ if you wanted, but there is no real gain from doing that.
    – Ian
    Mar 30 '16 at 19:36








  • 1




    I see - thanks for the quick answer!
    – lampj20la
    Mar 30 '16 at 19:37










  • @Ian May I ask, how does this show the martingale has almost surely finite stopping time? It seems to show only that the probability of the stopping time being infinite is less than that of the sum being a or b.
    – Dole
    Dec 2 at 4:04











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To have $T=infty$ you must have $T>n$ for all $n$, so in particular you must have $T>n$ for a fixed $n$. To have $T>n$ for a fixed $n$, $S_k$ can't be $a$ or $-b$ for any $k leq n$. But this would have to happen at some point if $|S_k|$ ever became larger than both $a$ and $b$. (The process doesn't let you skip over $a$ or $-b$ on the way to the outside of the interval $(-b,a)$.) So $max_{k leq n} |S_n|$ has to be less than $max { a,b }$.



So ${ T=infty } subseteq { T>n } subseteq { max_{k leq n} |S_n| leq max { a,b } }$. Then use monotonicity of $P$ to get the result.



By the way, in the terminology that I learned, a "Markov time" is what you are referring to and a "stopping time" is an a.s. finite Markov time. I'm not sure how universal this terminology is, but I find it a useful distinction.






share|cite|improve this answer





















  • Thanks for your response - I don't follow why $|S_n| leq text{max}{a,b}$. Surely if it equaled either $a$ or $b$ we would not have $T>n$?
    – lampj20la
    Mar 30 '16 at 19:35






  • 1




    @lampj20la That's true, but we're saying $T>n Rightarrow |S_n| leq max { a,b }$ not the other way around. You could say $<$ if you wanted, but there is no real gain from doing that.
    – Ian
    Mar 30 '16 at 19:36








  • 1




    I see - thanks for the quick answer!
    – lampj20la
    Mar 30 '16 at 19:37










  • @Ian May I ask, how does this show the martingale has almost surely finite stopping time? It seems to show only that the probability of the stopping time being infinite is less than that of the sum being a or b.
    – Dole
    Dec 2 at 4:04
















3














To have $T=infty$ you must have $T>n$ for all $n$, so in particular you must have $T>n$ for a fixed $n$. To have $T>n$ for a fixed $n$, $S_k$ can't be $a$ or $-b$ for any $k leq n$. But this would have to happen at some point if $|S_k|$ ever became larger than both $a$ and $b$. (The process doesn't let you skip over $a$ or $-b$ on the way to the outside of the interval $(-b,a)$.) So $max_{k leq n} |S_n|$ has to be less than $max { a,b }$.



So ${ T=infty } subseteq { T>n } subseteq { max_{k leq n} |S_n| leq max { a,b } }$. Then use monotonicity of $P$ to get the result.



By the way, in the terminology that I learned, a "Markov time" is what you are referring to and a "stopping time" is an a.s. finite Markov time. I'm not sure how universal this terminology is, but I find it a useful distinction.






share|cite|improve this answer





















  • Thanks for your response - I don't follow why $|S_n| leq text{max}{a,b}$. Surely if it equaled either $a$ or $b$ we would not have $T>n$?
    – lampj20la
    Mar 30 '16 at 19:35






  • 1




    @lampj20la That's true, but we're saying $T>n Rightarrow |S_n| leq max { a,b }$ not the other way around. You could say $<$ if you wanted, but there is no real gain from doing that.
    – Ian
    Mar 30 '16 at 19:36








  • 1




    I see - thanks for the quick answer!
    – lampj20la
    Mar 30 '16 at 19:37










  • @Ian May I ask, how does this show the martingale has almost surely finite stopping time? It seems to show only that the probability of the stopping time being infinite is less than that of the sum being a or b.
    – Dole
    Dec 2 at 4:04














3












3








3






To have $T=infty$ you must have $T>n$ for all $n$, so in particular you must have $T>n$ for a fixed $n$. To have $T>n$ for a fixed $n$, $S_k$ can't be $a$ or $-b$ for any $k leq n$. But this would have to happen at some point if $|S_k|$ ever became larger than both $a$ and $b$. (The process doesn't let you skip over $a$ or $-b$ on the way to the outside of the interval $(-b,a)$.) So $max_{k leq n} |S_n|$ has to be less than $max { a,b }$.



So ${ T=infty } subseteq { T>n } subseteq { max_{k leq n} |S_n| leq max { a,b } }$. Then use monotonicity of $P$ to get the result.



By the way, in the terminology that I learned, a "Markov time" is what you are referring to and a "stopping time" is an a.s. finite Markov time. I'm not sure how universal this terminology is, but I find it a useful distinction.






share|cite|improve this answer












To have $T=infty$ you must have $T>n$ for all $n$, so in particular you must have $T>n$ for a fixed $n$. To have $T>n$ for a fixed $n$, $S_k$ can't be $a$ or $-b$ for any $k leq n$. But this would have to happen at some point if $|S_k|$ ever became larger than both $a$ and $b$. (The process doesn't let you skip over $a$ or $-b$ on the way to the outside of the interval $(-b,a)$.) So $max_{k leq n} |S_n|$ has to be less than $max { a,b }$.



So ${ T=infty } subseteq { T>n } subseteq { max_{k leq n} |S_n| leq max { a,b } }$. Then use monotonicity of $P$ to get the result.



By the way, in the terminology that I learned, a "Markov time" is what you are referring to and a "stopping time" is an a.s. finite Markov time. I'm not sure how universal this terminology is, but I find it a useful distinction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 30 '16 at 19:28









Ian

67.3k25386




67.3k25386












  • Thanks for your response - I don't follow why $|S_n| leq text{max}{a,b}$. Surely if it equaled either $a$ or $b$ we would not have $T>n$?
    – lampj20la
    Mar 30 '16 at 19:35






  • 1




    @lampj20la That's true, but we're saying $T>n Rightarrow |S_n| leq max { a,b }$ not the other way around. You could say $<$ if you wanted, but there is no real gain from doing that.
    – Ian
    Mar 30 '16 at 19:36








  • 1




    I see - thanks for the quick answer!
    – lampj20la
    Mar 30 '16 at 19:37










  • @Ian May I ask, how does this show the martingale has almost surely finite stopping time? It seems to show only that the probability of the stopping time being infinite is less than that of the sum being a or b.
    – Dole
    Dec 2 at 4:04


















  • Thanks for your response - I don't follow why $|S_n| leq text{max}{a,b}$. Surely if it equaled either $a$ or $b$ we would not have $T>n$?
    – lampj20la
    Mar 30 '16 at 19:35






  • 1




    @lampj20la That's true, but we're saying $T>n Rightarrow |S_n| leq max { a,b }$ not the other way around. You could say $<$ if you wanted, but there is no real gain from doing that.
    – Ian
    Mar 30 '16 at 19:36








  • 1




    I see - thanks for the quick answer!
    – lampj20la
    Mar 30 '16 at 19:37










  • @Ian May I ask, how does this show the martingale has almost surely finite stopping time? It seems to show only that the probability of the stopping time being infinite is less than that of the sum being a or b.
    – Dole
    Dec 2 at 4:04
















Thanks for your response - I don't follow why $|S_n| leq text{max}{a,b}$. Surely if it equaled either $a$ or $b$ we would not have $T>n$?
– lampj20la
Mar 30 '16 at 19:35




Thanks for your response - I don't follow why $|S_n| leq text{max}{a,b}$. Surely if it equaled either $a$ or $b$ we would not have $T>n$?
– lampj20la
Mar 30 '16 at 19:35




1




1




@lampj20la That's true, but we're saying $T>n Rightarrow |S_n| leq max { a,b }$ not the other way around. You could say $<$ if you wanted, but there is no real gain from doing that.
– Ian
Mar 30 '16 at 19:36






@lampj20la That's true, but we're saying $T>n Rightarrow |S_n| leq max { a,b }$ not the other way around. You could say $<$ if you wanted, but there is no real gain from doing that.
– Ian
Mar 30 '16 at 19:36






1




1




I see - thanks for the quick answer!
– lampj20la
Mar 30 '16 at 19:37




I see - thanks for the quick answer!
– lampj20la
Mar 30 '16 at 19:37












@Ian May I ask, how does this show the martingale has almost surely finite stopping time? It seems to show only that the probability of the stopping time being infinite is less than that of the sum being a or b.
– Dole
Dec 2 at 4:04




@Ian May I ask, how does this show the martingale has almost surely finite stopping time? It seems to show only that the probability of the stopping time being infinite is less than that of the sum being a or b.
– Dole
Dec 2 at 4:04


















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