Why is $T:=inf{tgeq0:B_tleq at^p-b}$ a stopping time?
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How can I prove that $T:=inf{tgeq0:B_tleq at^p-b}$ is a stopping time w.r.t. a natural filtration of $B$, where $B$ is a $BM$, $p>1/2$ and $a,b>0$?
I can introduce a new random process, $X_t:=e^{B_t-at^p}$, for which $T=inf{tgeq 0:X_tleq e^{-b}}$.
I started:
$${Tleq t}= {exists sleq t:X_sleq e^{-b}}.$$
Is this then equal to $$cup_{sleq t}{X_sleq e^{-b}} = cup_{sleq tcap mathbb{Q}}{X_sleq e^{-b}}?$$
Is it now enough for me to say, as $X_s$ is measurable on the given filtration, that a countable union is also measurable and that gives us the stopping time?
brownian-motion stochastic-analysis stopping-times
edited Dec 9 '18 at 6:44
user1101010
7751730
asked Dec 9 '18 at 0:47
RavonripRavonrip
898
$endgroup$
add a comment |
$begingroup$
How can I prove that $T:=inf{tgeq0:B_tleq at^p-b}$ is a stopping time w.r.t. a natural filtration of $B$, where $B$ is a $BM$, $p>1/2$ and $a,b>0$?
I can introduce a new random process, $X_t:=e^{B_t-at^p}$, for which $T=inf{tgeq 0:X_tleq e^{-b}}$.
I started:
$${Tleq t}= {exists sleq t:X_sleq e^{-b}}.$$
Is this then equal to $$cup_{sleq t}{X_sleq e^{-b}} = cup_{sleq tcap mathbb{Q}}{X_sleq e^{-b}}?$$
Is it now enough for me to say, as $X_s$ is measurable on the given filtration, that a countable union is also measurable and that gives us the stopping time?
brownian-motion stochastic-analysis stopping-times
edited Dec 9 '18 at 6:44
user1101010
7751730
asked Dec 9 '18 at 0:47
RavonripRavonrip
898
$endgroup$
$begingroup$
Seems kosher to me, but I haven't studied this in ~4 years, so don't take my word for it.
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:23
add a comment |
$begingroup$
How can I prove that $T:=inf{tgeq0:B_tleq at^p-b}$ is a stopping time w.r.t. a natural filtration of $B$, where $B$ is a $BM$, $p>1/2$ and $a,b>0$?
I can introduce a new random process, $X_t:=e^{B_t-at^p}$, for which $T=inf{tgeq 0:X_tleq e^{-b}}$.
I started:
$${Tleq t}= {exists sleq t:X_sleq e^{-b}}.$$
Is this then equal to $$cup_{sleq t}{X_sleq e^{-b}} = cup_{sleq tcap mathbb{Q}}{X_sleq e^{-b}}?$$
Is it now enough for me to say, as $X_s$ is measurable on the given filtration, that a countable union is also measurable and that gives us the stopping time?
brownian-motion stochastic-analysis stopping-times
edited Dec 9 '18 at 6:44
user1101010
7751730
asked Dec 9 '18 at 0:47
RavonripRavonrip
898
$endgroup$
How can I prove that $T:=inf{tgeq0:B_tleq at^p-b}$ is a stopping time w.r.t. a natural filtration of $B$, where $B$ is a $BM$, $p>1/2$ and $a,b>0$?
I can introduce a new random process, $X_t:=e^{B_t-at^p}$, for which $T=inf{tgeq 0:X_tleq e^{-b}}$.
I started:
$${Tleq t}= {exists sleq t:X_sleq e^{-b}}.$$
Is this then equal to $$cup_{sleq t}{X_sleq e^{-b}} = cup_{sleq tcap mathbb{Q}}{X_sleq e^{-b}}?$$
Is it now enough for me to say, as $X_s$ is measurable on the given filtration, that a countable union is also measurable and that gives us the stopping time?
brownian-motion stochastic-analysis stopping-times
brownian-motion stochastic-analysis stopping-times
edited Dec 9 '18 at 6:44
user1101010
7751730
asked Dec 9 '18 at 0:47
RavonripRavonrip
898
edited Dec 9 '18 at 6:44
user1101010
7751730
asked Dec 9 '18 at 0:47
RavonripRavonrip
898
edited Dec 9 '18 at 6:44
user1101010
7751730
edited Dec 9 '18 at 6:44
user1101010
7751730
edited Dec 9 '18 at 6:44
user1101010
7751730
7751730
asked Dec 9 '18 at 0:47
RavonripRavonrip
898
asked Dec 9 '18 at 0:47
RavonripRavonrip
898
asked Dec 9 '18 at 0:47
RavonripRavonrip
898
898
$begingroup$
Seems kosher to me, but I haven't studied this in ~4 years, so don't take my word for it.
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:23
add a comment |
$begingroup$
Seems kosher to me, but I haven't studied this in ~4 years, so don't take my word for it.
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:23
$begingroup$
Seems kosher to me, but I haven't studied this in ~4 years, so don't take my word for it.
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:23
$begingroup$
Seems kosher to me, but I haven't studied this in ~4 years, so don't take my word for it.
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your argument is not valid. LHS of the set theoretic identity you have written need not be contained in RHS. For example $X_t leq e^{-b}$ does not imply that there exists $sleq t,s in mathbb Q$ such that $X_t leq e^{-b}$. Instead, consider ${T>t}$. You can write this as union over $s leq t, s in mathbb Q$ of ${X_s>e^{-b}}$ and this proves that ${T>t} in sigma {B_u:u leq t}$.
answered Dec 9 '18 at 5:07
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
$endgroup$
$begingroup$
Thank you for your answer. So if I understand correctly I can do a countable intersection, if I have ${T>t}$? Why is that so, what changes with the change of inequality?
$endgroup$
– Ravonrip
Dec 10 '18 at 20:28
$begingroup$
It is possible that $X_s >e^{-b}$ for all $s <t$ but yet $X_t leq e^{-b}$. But if $X_t >e^{-b}$ then $X_s >e^{-b}$ for all $s$ sufficiently close to $t$ hence for some rational number $s <t$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 23:12
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your argument is not valid. LHS of the set theoretic identity you have written need not be contained in RHS. For example $X_t leq e^{-b}$ does not imply that there exists $sleq t,s in mathbb Q$ such that $X_t leq e^{-b}$. Instead, consider ${T>t}$. You can write this as union over $s leq t, s in mathbb Q$ of ${X_s>e^{-b}}$ and this proves that ${T>t} in sigma {B_u:u leq t}$.
answered Dec 9 '18 at 5:07
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
$endgroup$
$begingroup$
Thank you for your answer. So if I understand correctly I can do a countable intersection, if I have ${T>t}$? Why is that so, what changes with the change of inequality?
$endgroup$
– Ravonrip
Dec 10 '18 at 20:28
$begingroup$
It is possible that $X_s >e^{-b}$ for all $s <t$ but yet $X_t leq e^{-b}$. But if $X_t >e^{-b}$ then $X_s >e^{-b}$ for all $s$ sufficiently close to $t$ hence for some rational number $s <t$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 23:12
add a comment |
$begingroup$
Your argument is not valid. LHS of the set theoretic identity you have written need not be contained in RHS. For example $X_t leq e^{-b}$ does not imply that there exists $sleq t,s in mathbb Q$ such that $X_t leq e^{-b}$. Instead, consider ${T>t}$. You can write this as union over $s leq t, s in mathbb Q$ of ${X_s>e^{-b}}$ and this proves that ${T>t} in sigma {B_u:u leq t}$.
answered Dec 9 '18 at 5:07
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
$endgroup$
$begingroup$
Thank you for your answer. So if I understand correctly I can do a countable intersection, if I have ${T>t}$? Why is that so, what changes with the change of inequality?
$endgroup$
– Ravonrip
Dec 10 '18 at 20:28
$begingroup$
It is possible that $X_s >e^{-b}$ for all $s <t$ but yet $X_t leq e^{-b}$. But if $X_t >e^{-b}$ then $X_s >e^{-b}$ for all $s$ sufficiently close to $t$ hence for some rational number $s <t$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 23:12
add a comment |
$begingroup$
Your argument is not valid. LHS of the set theoretic identity you have written need not be contained in RHS. For example $X_t leq e^{-b}$ does not imply that there exists $sleq t,s in mathbb Q$ such that $X_t leq e^{-b}$. Instead, consider ${T>t}$. You can write this as union over $s leq t, s in mathbb Q$ of ${X_s>e^{-b}}$ and this proves that ${T>t} in sigma {B_u:u leq t}$.
answered Dec 9 '18 at 5:07
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
$endgroup$
Your argument is not valid. LHS of the set theoretic identity you have written need not be contained in RHS. For example $X_t leq e^{-b}$ does not imply that there exists $sleq t,s in mathbb Q$ such that $X_t leq e^{-b}$. Instead, consider ${T>t}$. You can write this as union over $s leq t, s in mathbb Q$ of ${X_s>e^{-b}}$ and this proves that ${T>t} in sigma {B_u:u leq t}$.
answered Dec 9 '18 at 5:07
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
answered Dec 9 '18 at 5:07
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
answered Dec 9 '18 at 5:07
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
answered Dec 9 '18 at 5:07
Kavi Rama MurthyKavi Rama Murthy
54.5k32055
54.5k32055
$begingroup$
Thank you for your answer. So if I understand correctly I can do a countable intersection, if I have ${T>t}$? Why is that so, what changes with the change of inequality?
$endgroup$
– Ravonrip
Dec 10 '18 at 20:28
$begingroup$
It is possible that $X_s >e^{-b}$ for all $s <t$ but yet $X_t leq e^{-b}$. But if $X_t >e^{-b}$ then $X_s >e^{-b}$ for all $s$ sufficiently close to $t$ hence for some rational number $s <t$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 23:12
add a comment |
$begingroup$
Thank you for your answer. So if I understand correctly I can do a countable intersection, if I have ${T>t}$? Why is that so, what changes with the change of inequality?
$endgroup$
– Ravonrip
Dec 10 '18 at 20:28
$begingroup$
It is possible that $X_s >e^{-b}$ for all $s <t$ but yet $X_t leq e^{-b}$. But if $X_t >e^{-b}$ then $X_s >e^{-b}$ for all $s$ sufficiently close to $t$ hence for some rational number $s <t$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 23:12
$begingroup$
Thank you for your answer. So if I understand correctly I can do a countable intersection, if I have ${T>t}$? Why is that so, what changes with the change of inequality?
$endgroup$
– Ravonrip
Dec 10 '18 at 20:28
$begingroup$
Thank you for your answer. So if I understand correctly I can do a countable intersection, if I have ${T>t}$? Why is that so, what changes with the change of inequality?
$endgroup$
– Ravonrip
Dec 10 '18 at 20:28
$begingroup$
It is possible that $X_s >e^{-b}$ for all $s <t$ but yet $X_t leq e^{-b}$. But if $X_t >e^{-b}$ then $X_s >e^{-b}$ for all $s$ sufficiently close to $t$ hence for some rational number $s <t$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 23:12
$begingroup$
It is possible that $X_s >e^{-b}$ for all $s <t$ but yet $X_t leq e^{-b}$. But if $X_t >e^{-b}$ then $X_s >e^{-b}$ for all $s$ sufficiently close to $t$ hence for some rational number $s <t$.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 23:12
add a comment |
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$begingroup$
Seems kosher to me, but I haven't studied this in ~4 years, so don't take my word for it.
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:23