Mixing
Brownian motion maximum inequality
$begingroup$
I am supposed to show this inequality for a 2 or 3 dimensional Brownian motion:
$textbf{P}^0{suplimits_{tleq k}{|B(t)|}geqfrac{1}{2}}leq 2textbf{P}^0{{|B(k)|geqfrac{1}{2}}}$
(where $textbf{P}^0$ means the BM is started in zero).
This looks a lot like Doob's maximum inequality to me, but on the RHS there is a probabilty instead of an expectation and also $|B|$ is not a martingale for d=2,3. I think you can fix the last part by just taking $ln|B|$ and $|B|^{-1}$ instead, but I don't see how to replace the expectation with a probability.
Thanks in advance!
probability brownian-motion martingales
asked Jan 22 at 21:43
John DoeJohn Doe
263
$endgroup$
|
show 1 more comment
$begingroup$
I am supposed to show this inequality for a 2 or 3 dimensional Brownian motion:
$textbf{P}^0{suplimits_{tleq k}{|B(t)|}geqfrac{1}{2}}leq 2textbf{P}^0{{|B(k)|geqfrac{1}{2}}}$
(where $textbf{P}^0$ means the BM is started in zero).
This looks a lot like Doob's maximum inequality to me, but on the RHS there is a probabilty instead of an expectation and also $|B|$ is not a martingale for d=2,3. I think you can fix the last part by just taking $ln|B|$ and $|B|^{-1}$ instead, but I don't see how to replace the expectation with a probability.
Thanks in advance!
probability brownian-motion martingales
asked Jan 22 at 21:43
John DoeJohn Doe
263
$endgroup$
1
$begingroup$
en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– d.k.o.
Jan 23 at 2:14
$begingroup$
Thanks! But I thought the reflection principle was only applicable to linear Brownian motion? Can I also use it for higher dimensions?
$endgroup$
– John Doe
Jan 23 at 16:55
$begingroup$
What is $|B(t)|$ for higher dimensions?
$endgroup$
– d.k.o.
Jan 23 at 23:06
$begingroup$
I am not sure what you mean ... it's of course length of the random vector with independent Brownian motions in it's entries. But I don't think $|B(t)|$ is a Brownian motion again, is it ...?
$endgroup$
– John Doe
Jan 24 at 0:08
$begingroup$
Is it $ell_2$ norm?
$endgroup$
– d.k.o.
Jan 24 at 7:52
|
show 1 more comment
$begingroup$
I am supposed to show this inequality for a 2 or 3 dimensional Brownian motion:
$textbf{P}^0{suplimits_{tleq k}{|B(t)|}geqfrac{1}{2}}leq 2textbf{P}^0{{|B(k)|geqfrac{1}{2}}}$
(where $textbf{P}^0$ means the BM is started in zero).
This looks a lot like Doob's maximum inequality to me, but on the RHS there is a probabilty instead of an expectation and also $|B|$ is not a martingale for d=2,3. I think you can fix the last part by just taking $ln|B|$ and $|B|^{-1}$ instead, but I don't see how to replace the expectation with a probability.
Thanks in advance!
probability brownian-motion martingales
asked Jan 22 at 21:43
John DoeJohn Doe
263
$endgroup$
I am supposed to show this inequality for a 2 or 3 dimensional Brownian motion:
$textbf{P}^0{suplimits_{tleq k}{|B(t)|}geqfrac{1}{2}}leq 2textbf{P}^0{{|B(k)|geqfrac{1}{2}}}$
(where $textbf{P}^0$ means the BM is started in zero).
This looks a lot like Doob's maximum inequality to me, but on the RHS there is a probabilty instead of an expectation and also $|B|$ is not a martingale for d=2,3. I think you can fix the last part by just taking $ln|B|$ and $|B|^{-1}$ instead, but I don't see how to replace the expectation with a probability.
Thanks in advance!
probability brownian-motion martingales
probability brownian-motion martingales
asked Jan 22 at 21:43
John DoeJohn Doe
263
asked Jan 22 at 21:43
John DoeJohn Doe
263
asked Jan 22 at 21:43
John DoeJohn Doe
263
asked Jan 22 at 21:43
John DoeJohn Doe
263
asked Jan 22 at 21:43
John DoeJohn Doe
263
263
1
$begingroup$
en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– d.k.o.
Jan 23 at 2:14
$begingroup$
Thanks! But I thought the reflection principle was only applicable to linear Brownian motion? Can I also use it for higher dimensions?
$endgroup$
– John Doe
Jan 23 at 16:55
$begingroup$
What is $|B(t)|$ for higher dimensions?
$endgroup$
– d.k.o.
Jan 23 at 23:06
$begingroup$
I am not sure what you mean ... it's of course length of the random vector with independent Brownian motions in it's entries. But I don't think $|B(t)|$ is a Brownian motion again, is it ...?
$endgroup$
– John Doe
Jan 24 at 0:08
$begingroup$
Is it $ell_2$ norm?
$endgroup$
– d.k.o.
Jan 24 at 7:52
|
show 1 more comment
1
$begingroup$
en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– d.k.o.
Jan 23 at 2:14
$begingroup$
Thanks! But I thought the reflection principle was only applicable to linear Brownian motion? Can I also use it for higher dimensions?
$endgroup$
– John Doe
Jan 23 at 16:55
$begingroup$
What is $|B(t)|$ for higher dimensions?
$endgroup$
– d.k.o.
Jan 23 at 23:06
$begingroup$
I am not sure what you mean ... it's of course length of the random vector with independent Brownian motions in it's entries. But I don't think $|B(t)|$ is a Brownian motion again, is it ...?
$endgroup$
– John Doe
Jan 24 at 0:08
$begingroup$
Is it $ell_2$ norm?
$endgroup$
– d.k.o.
Jan 24 at 7:52
1
1
$begingroup$
en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– d.k.o.
Jan 23 at 2:14
$begingroup$
en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– d.k.o.
Jan 23 at 2:14
$begingroup$
Thanks! But I thought the reflection principle was only applicable to linear Brownian motion? Can I also use it for higher dimensions?
$endgroup$
– John Doe
Jan 23 at 16:55
$begingroup$
Thanks! But I thought the reflection principle was only applicable to linear Brownian motion? Can I also use it for higher dimensions?
$endgroup$
– John Doe
Jan 23 at 16:55
$begingroup$
What is $|B(t)|$ for higher dimensions?
$endgroup$
– d.k.o.
Jan 23 at 23:06
$begingroup$
What is $|B(t)|$ for higher dimensions?
$endgroup$
– d.k.o.
Jan 23 at 23:06
$begingroup$
I am not sure what you mean ... it's of course length of the random vector with independent Brownian motions in it's entries. But I don't think $|B(t)|$ is a Brownian motion again, is it ...?
$endgroup$
– John Doe
Jan 24 at 0:08
$begingroup$
I am not sure what you mean ... it's of course length of the random vector with independent Brownian motions in it's entries. But I don't think $|B(t)|$ is a Brownian motion again, is it ...?
$endgroup$
– John Doe
Jan 24 at 0:08
$begingroup$
Is it $ell_2$ norm?
$endgroup$
– d.k.o.
Jan 24 at 7:52
$begingroup$
Is it $ell_2$ norm?
$endgroup$
– d.k.o.
Jan 24 at 7:52
|
show 1 more comment
1 Answer
1
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oldest
votes
$begingroup$
For $a>0$ let $tau_a:=inf{tge 0:|B_t|=a}$. Then
$$
mathsf{P}^0(tau_ale t)=mathsf{P}^0(tau_ale t,|B_t|ge a)+mathsf{P}^0(tau_ale t,|B_t|< a).tag{1}
$$
Using the strong Markov property (e.g. Theorem 8.3.7 on page 314 here),
begin{align}
mathsf{P}^0(tau_ale t,|B_t|< a)&=mathsf{E}^0[1{tau_ale t}mathsf{P}^0(|B_{tau_a+(t-tau_a)}|<amid mathcal{F}_{tau_a})] \
&=mathsf{E}^0[1{tau_ale t}mathsf{P}^{B_{tau_a}}(|B_{t-tau_a}|<a)]le frac{1}{2}mathsf{P}^0(tau_ale t).tag{2}
end{align}
Combining $(1)$ and $(2)$, we get
$$
mathsf{P}^0left(sup_{sle t}|B_t|ge aright)=mathsf{P}^0(tau_ale t)le 2mathsf{P}^0(|B_t|ge a).
$$
It remains to verify the last inequality in (2), i.e. for any $x$ with $|x|=a$ and $sge 0$,
$$
mathsf{P}^x(|B_s|<a)le 1/2.
$$
answered Jan 25 at 7:40
d.k.o.d.k.o.
10.2k629
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
For $a>0$ let $tau_a:=inf{tge 0:|B_t|=a}$. Then
$$
mathsf{P}^0(tau_ale t)=mathsf{P}^0(tau_ale t,|B_t|ge a)+mathsf{P}^0(tau_ale t,|B_t|< a).tag{1}
$$
Using the strong Markov property (e.g. Theorem 8.3.7 on page 314 here),
begin{align}
mathsf{P}^0(tau_ale t,|B_t|< a)&=mathsf{E}^0[1{tau_ale t}mathsf{P}^0(|B_{tau_a+(t-tau_a)}|<amid mathcal{F}_{tau_a})] \
&=mathsf{E}^0[1{tau_ale t}mathsf{P}^{B_{tau_a}}(|B_{t-tau_a}|<a)]le frac{1}{2}mathsf{P}^0(tau_ale t).tag{2}
end{align}
Combining $(1)$ and $(2)$, we get
$$
mathsf{P}^0left(sup_{sle t}|B_t|ge aright)=mathsf{P}^0(tau_ale t)le 2mathsf{P}^0(|B_t|ge a).
$$
It remains to verify the last inequality in (2), i.e. for any $x$ with $|x|=a$ and $sge 0$,
$$
mathsf{P}^x(|B_s|<a)le 1/2.
$$
answered Jan 25 at 7:40
d.k.o.d.k.o.
10.2k629
$endgroup$
add a comment |
$begingroup$
For $a>0$ let $tau_a:=inf{tge 0:|B_t|=a}$. Then
$$
mathsf{P}^0(tau_ale t)=mathsf{P}^0(tau_ale t,|B_t|ge a)+mathsf{P}^0(tau_ale t,|B_t|< a).tag{1}
$$
Using the strong Markov property (e.g. Theorem 8.3.7 on page 314 here),
begin{align}
mathsf{P}^0(tau_ale t,|B_t|< a)&=mathsf{E}^0[1{tau_ale t}mathsf{P}^0(|B_{tau_a+(t-tau_a)}|<amid mathcal{F}_{tau_a})] \
&=mathsf{E}^0[1{tau_ale t}mathsf{P}^{B_{tau_a}}(|B_{t-tau_a}|<a)]le frac{1}{2}mathsf{P}^0(tau_ale t).tag{2}
end{align}
Combining $(1)$ and $(2)$, we get
$$
mathsf{P}^0left(sup_{sle t}|B_t|ge aright)=mathsf{P}^0(tau_ale t)le 2mathsf{P}^0(|B_t|ge a).
$$
It remains to verify the last inequality in (2), i.e. for any $x$ with $|x|=a$ and $sge 0$,
$$
mathsf{P}^x(|B_s|<a)le 1/2.
$$
answered Jan 25 at 7:40
d.k.o.d.k.o.
10.2k629
$endgroup$
add a comment |
$begingroup$
For $a>0$ let $tau_a:=inf{tge 0:|B_t|=a}$. Then
$$
mathsf{P}^0(tau_ale t)=mathsf{P}^0(tau_ale t,|B_t|ge a)+mathsf{P}^0(tau_ale t,|B_t|< a).tag{1}
$$
Using the strong Markov property (e.g. Theorem 8.3.7 on page 314 here),
begin{align}
mathsf{P}^0(tau_ale t,|B_t|< a)&=mathsf{E}^0[1{tau_ale t}mathsf{P}^0(|B_{tau_a+(t-tau_a)}|<amid mathcal{F}_{tau_a})] \
&=mathsf{E}^0[1{tau_ale t}mathsf{P}^{B_{tau_a}}(|B_{t-tau_a}|<a)]le frac{1}{2}mathsf{P}^0(tau_ale t).tag{2}
end{align}
Combining $(1)$ and $(2)$, we get
$$
mathsf{P}^0left(sup_{sle t}|B_t|ge aright)=mathsf{P}^0(tau_ale t)le 2mathsf{P}^0(|B_t|ge a).
$$
It remains to verify the last inequality in (2), i.e. for any $x$ with $|x|=a$ and $sge 0$,
$$
mathsf{P}^x(|B_s|<a)le 1/2.
$$
answered Jan 25 at 7:40
d.k.o.d.k.o.
10.2k629
$endgroup$
For $a>0$ let $tau_a:=inf{tge 0:|B_t|=a}$. Then
$$
mathsf{P}^0(tau_ale t)=mathsf{P}^0(tau_ale t,|B_t|ge a)+mathsf{P}^0(tau_ale t,|B_t|< a).tag{1}
$$
Using the strong Markov property (e.g. Theorem 8.3.7 on page 314 here),
begin{align}
mathsf{P}^0(tau_ale t,|B_t|< a)&=mathsf{E}^0[1{tau_ale t}mathsf{P}^0(|B_{tau_a+(t-tau_a)}|<amid mathcal{F}_{tau_a})] \
&=mathsf{E}^0[1{tau_ale t}mathsf{P}^{B_{tau_a}}(|B_{t-tau_a}|<a)]le frac{1}{2}mathsf{P}^0(tau_ale t).tag{2}
end{align}
Combining $(1)$ and $(2)$, we get
$$
mathsf{P}^0left(sup_{sle t}|B_t|ge aright)=mathsf{P}^0(tau_ale t)le 2mathsf{P}^0(|B_t|ge a).
$$
It remains to verify the last inequality in (2), i.e. for any $x$ with $|x|=a$ and $sge 0$,
$$
mathsf{P}^x(|B_s|<a)le 1/2.
$$
answered Jan 25 at 7:40
d.k.o.d.k.o.
10.2k629
answered Jan 25 at 7:40
d.k.o.d.k.o.
10.2k629
answered Jan 25 at 7:40
d.k.o.d.k.o.
10.2k629
answered Jan 25 at 7:40
d.k.o.d.k.o.
10.2k629
10.2k629
add a comment |
add a comment |
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1
$begingroup$
en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– d.k.o.
Jan 23 at 2:14
$begingroup$
Thanks! But I thought the reflection principle was only applicable to linear Brownian motion? Can I also use it for higher dimensions?
$endgroup$
– John Doe
Jan 23 at 16:55
$begingroup$
What is $|B(t)|$ for higher dimensions?
$endgroup$
– d.k.o.
Jan 23 at 23:06
$begingroup$
I am not sure what you mean ... it's of course length of the random vector with independent Brownian motions in it's entries. But I don't think $|B(t)|$ is a Brownian motion again, is it ...?
$endgroup$
– John Doe
Jan 24 at 0:08
$begingroup$
Is it $ell_2$ norm?
$endgroup$
– d.k.o.
Jan 24 at 7:52