Mixing
How to prove that the exit time of a Brownian motion is a stopping time?
$begingroup$
Given the following setting:
Let ${W_t:tgeq 0}$ be a Brownian motion. for arbitrary $a>0$, define the exit time of the interval $[-a,a]$ as $$tau=inf{tgeq 0:|W_t|>a}$$
The question is to show that this is a stopping time, i.e. that for a filtration $mathscr{F}_t=sigma(W_s,0leq sleq t)$, we have that $(tauleq t)inmathscr{F}_t$.
This is an exercise that came after proving that a hitting time, i.e. $tau=inf{tgeq 0:W_t=a}$, is a stopping time. I know how to prove that this is a stopping time, but the method I used there cannot be applied here because we do not have that $(tauleq t)inmathscr{F}_t$, but instead we have that $(tauleq t)inmathscr{F}_{t^+}$.
I'm not even sure where to start with this exercise, but the hint given was that we can ''make'' $mathscr{F}_{t^+}=mathscr{F}_{t}$ by adding things of measure $0$ to the filtration. However, this hint does not bring me any closer to knowing where to start. What is meant by adding things of measure $0$ to the filtration? Why does this help? How do I proceed with solving this exercise?
Any help is appreciated.
probability probability-theory stochastic-processes brownian-motion stopping-times
asked Jan 25 at 17:08
S. CrimS. Crim
404212
$endgroup$
add a comment |
$begingroup$
Given the following setting:
Let ${W_t:tgeq 0}$ be a Brownian motion. for arbitrary $a>0$, define the exit time of the interval $[-a,a]$ as $$tau=inf{tgeq 0:|W_t|>a}$$
The question is to show that this is a stopping time, i.e. that for a filtration $mathscr{F}_t=sigma(W_s,0leq sleq t)$, we have that $(tauleq t)inmathscr{F}_t$.
This is an exercise that came after proving that a hitting time, i.e. $tau=inf{tgeq 0:W_t=a}$, is a stopping time. I know how to prove that this is a stopping time, but the method I used there cannot be applied here because we do not have that $(tauleq t)inmathscr{F}_t$, but instead we have that $(tauleq t)inmathscr{F}_{t^+}$.
I'm not even sure where to start with this exercise, but the hint given was that we can ''make'' $mathscr{F}_{t^+}=mathscr{F}_{t}$ by adding things of measure $0$ to the filtration. However, this hint does not bring me any closer to knowing where to start. What is meant by adding things of measure $0$ to the filtration? Why does this help? How do I proceed with solving this exercise?
Any help is appreciated.
probability probability-theory stochastic-processes brownian-motion stopping-times
asked Jan 25 at 17:08
S. CrimS. Crim
404212
$endgroup$
add a comment |
$begingroup$
Given the following setting:
Let ${W_t:tgeq 0}$ be a Brownian motion. for arbitrary $a>0$, define the exit time of the interval $[-a,a]$ as $$tau=inf{tgeq 0:|W_t|>a}$$
The question is to show that this is a stopping time, i.e. that for a filtration $mathscr{F}_t=sigma(W_s,0leq sleq t)$, we have that $(tauleq t)inmathscr{F}_t$.
This is an exercise that came after proving that a hitting time, i.e. $tau=inf{tgeq 0:W_t=a}$, is a stopping time. I know how to prove that this is a stopping time, but the method I used there cannot be applied here because we do not have that $(tauleq t)inmathscr{F}_t$, but instead we have that $(tauleq t)inmathscr{F}_{t^+}$.
I'm not even sure where to start with this exercise, but the hint given was that we can ''make'' $mathscr{F}_{t^+}=mathscr{F}_{t}$ by adding things of measure $0$ to the filtration. However, this hint does not bring me any closer to knowing where to start. What is meant by adding things of measure $0$ to the filtration? Why does this help? How do I proceed with solving this exercise?
Any help is appreciated.
probability probability-theory stochastic-processes brownian-motion stopping-times
asked Jan 25 at 17:08
S. CrimS. Crim
404212
$endgroup$
Given the following setting:
Let ${W_t:tgeq 0}$ be a Brownian motion. for arbitrary $a>0$, define the exit time of the interval $[-a,a]$ as $$tau=inf{tgeq 0:|W_t|>a}$$
The question is to show that this is a stopping time, i.e. that for a filtration $mathscr{F}_t=sigma(W_s,0leq sleq t)$, we have that $(tauleq t)inmathscr{F}_t$.
This is an exercise that came after proving that a hitting time, i.e. $tau=inf{tgeq 0:W_t=a}$, is a stopping time. I know how to prove that this is a stopping time, but the method I used there cannot be applied here because we do not have that $(tauleq t)inmathscr{F}_t$, but instead we have that $(tauleq t)inmathscr{F}_{t^+}$.
I'm not even sure where to start with this exercise, but the hint given was that we can ''make'' $mathscr{F}_{t^+}=mathscr{F}_{t}$ by adding things of measure $0$ to the filtration. However, this hint does not bring me any closer to knowing where to start. What is meant by adding things of measure $0$ to the filtration? Why does this help? How do I proceed with solving this exercise?
Any help is appreciated.
probability probability-theory stochastic-processes brownian-motion stopping-times
probability probability-theory stochastic-processes brownian-motion stopping-times
asked Jan 25 at 17:08
S. CrimS. Crim
404212
asked Jan 25 at 17:08
S. CrimS. Crim
404212
asked Jan 25 at 17:08
S. CrimS. Crim
404212
asked Jan 25 at 17:08
S. CrimS. Crim
404212
asked Jan 25 at 17:08
S. CrimS. Crim
404212
404212
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
By considering right-continuous $(mathcal{F}_t)$ it suffices to show that ${tau<t}inmathcal{F}_t$. Then since $tmapsto W_t$ is continuous,
$$
{tau<t}=bigcup_{q<t,qinmathbb{Q}}{|W_q|>a}.
$$
The same applies to any open set $A$ and $tau:=inf{tge 0:W_tin A}$.
The completed natural filtration of a Brownian motion, $(mathcal{F}_tbigvee mathcal{N})$ ($mathcal{N}$ are the $mathsf{P}$-null sets of $mathcal{F}$), is right-continuous (e.g. Theorem 8.2.2 on page 309 here).
answered Jan 25 at 17:51
d.k.o.d.k.o.
10.3k629
$endgroup$
$begingroup$
Thanks for the answer. Any idea on what could be meant with the hint ''adding things of measure zero to the filtration''?
$endgroup$
– S. Crim
Jan 25 at 18:57
$begingroup$
This shows only that $tau$ is a.s. equal to an $(mathscr F_t)$ stopping time.
$endgroup$
– John Dawkins
Jan 26 at 21:07
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
By considering right-continuous $(mathcal{F}_t)$ it suffices to show that ${tau<t}inmathcal{F}_t$. Then since $tmapsto W_t$ is continuous,
$$
{tau<t}=bigcup_{q<t,qinmathbb{Q}}{|W_q|>a}.
$$
The same applies to any open set $A$ and $tau:=inf{tge 0:W_tin A}$.
The completed natural filtration of a Brownian motion, $(mathcal{F}_tbigvee mathcal{N})$ ($mathcal{N}$ are the $mathsf{P}$-null sets of $mathcal{F}$), is right-continuous (e.g. Theorem 8.2.2 on page 309 here).
answered Jan 25 at 17:51
d.k.o.d.k.o.
10.3k629
$endgroup$
$begingroup$
Thanks for the answer. Any idea on what could be meant with the hint ''adding things of measure zero to the filtration''?
$endgroup$
– S. Crim
Jan 25 at 18:57
$begingroup$
This shows only that $tau$ is a.s. equal to an $(mathscr F_t)$ stopping time.
$endgroup$
– John Dawkins
Jan 26 at 21:07
add a comment |
$begingroup$
By considering right-continuous $(mathcal{F}_t)$ it suffices to show that ${tau<t}inmathcal{F}_t$. Then since $tmapsto W_t$ is continuous,
$$
{tau<t}=bigcup_{q<t,qinmathbb{Q}}{|W_q|>a}.
$$
The same applies to any open set $A$ and $tau:=inf{tge 0:W_tin A}$.
The completed natural filtration of a Brownian motion, $(mathcal{F}_tbigvee mathcal{N})$ ($mathcal{N}$ are the $mathsf{P}$-null sets of $mathcal{F}$), is right-continuous (e.g. Theorem 8.2.2 on page 309 here).
answered Jan 25 at 17:51
d.k.o.d.k.o.
10.3k629
$endgroup$
$begingroup$
Thanks for the answer. Any idea on what could be meant with the hint ''adding things of measure zero to the filtration''?
$endgroup$
– S. Crim
Jan 25 at 18:57
$begingroup$
This shows only that $tau$ is a.s. equal to an $(mathscr F_t)$ stopping time.
$endgroup$
– John Dawkins
Jan 26 at 21:07
add a comment |
$begingroup$
By considering right-continuous $(mathcal{F}_t)$ it suffices to show that ${tau<t}inmathcal{F}_t$. Then since $tmapsto W_t$ is continuous,
$$
{tau<t}=bigcup_{q<t,qinmathbb{Q}}{|W_q|>a}.
$$
The same applies to any open set $A$ and $tau:=inf{tge 0:W_tin A}$.
The completed natural filtration of a Brownian motion, $(mathcal{F}_tbigvee mathcal{N})$ ($mathcal{N}$ are the $mathsf{P}$-null sets of $mathcal{F}$), is right-continuous (e.g. Theorem 8.2.2 on page 309 here).
answered Jan 25 at 17:51
d.k.o.d.k.o.
10.3k629
$endgroup$
By considering right-continuous $(mathcal{F}_t)$ it suffices to show that ${tau<t}inmathcal{F}_t$. Then since $tmapsto W_t$ is continuous,
$$
{tau<t}=bigcup_{q<t,qinmathbb{Q}}{|W_q|>a}.
$$
The same applies to any open set $A$ and $tau:=inf{tge 0:W_tin A}$.
The completed natural filtration of a Brownian motion, $(mathcal{F}_tbigvee mathcal{N})$ ($mathcal{N}$ are the $mathsf{P}$-null sets of $mathcal{F}$), is right-continuous (e.g. Theorem 8.2.2 on page 309 here).
answered Jan 25 at 17:51
d.k.o.d.k.o.
10.3k629
edited Jan 25 at 19:47
answered Jan 25 at 17:51
d.k.o.d.k.o.
10.3k629
answered Jan 25 at 17:51
d.k.o.d.k.o.
10.3k629
answered Jan 25 at 17:51
d.k.o.d.k.o.
10.3k629
10.3k629
$begingroup$
Thanks for the answer. Any idea on what could be meant with the hint ''adding things of measure zero to the filtration''?
$endgroup$
– S. Crim
Jan 25 at 18:57
$begingroup$
This shows only that $tau$ is a.s. equal to an $(mathscr F_t)$ stopping time.
$endgroup$
– John Dawkins
Jan 26 at 21:07
add a comment |
$begingroup$
Thanks for the answer. Any idea on what could be meant with the hint ''adding things of measure zero to the filtration''?
$endgroup$
– S. Crim
Jan 25 at 18:57
$begingroup$
This shows only that $tau$ is a.s. equal to an $(mathscr F_t)$ stopping time.
$endgroup$
– John Dawkins
Jan 26 at 21:07
$begingroup$
Thanks for the answer. Any idea on what could be meant with the hint ''adding things of measure zero to the filtration''?
$endgroup$
– S. Crim
Jan 25 at 18:57
$begingroup$
Thanks for the answer. Any idea on what could be meant with the hint ''adding things of measure zero to the filtration''?
$endgroup$
– S. Crim
Jan 25 at 18:57
$begingroup$
This shows only that $tau$ is a.s. equal to an $(mathscr F_t)$ stopping time.
$endgroup$
– John Dawkins
Jan 26 at 21:07
$begingroup$
This shows only that $tau$ is a.s. equal to an $(mathscr F_t)$ stopping time.
$endgroup$
– John Dawkins
Jan 26 at 21:07
add a comment |
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