Burkholder's inequality on $[s,t]$
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Let $T>0$, $F$ be an adapted squared-integrable process and $pge 2$, then the Burkholder's inequality implies
$$
Ebigg(bigg|int_0^T F(s)dB(s)bigg|^pbigg)le C(p) Ebigg[bigg(int_0^T F^2(s)dsbigg)^{p/2}bigg],
$$
for some constant $C(p)$.
I was wondering why most textbooks present the inequality from $[0,T]$, instead of any subinterval $[s,t]subset[0,T]$. I follow the proof and I didn't notice any technical difficulty to extend the result to the integral on $[s,t]$:
$$
Ebigg(bigg|int_s^t F(r)dB(r)bigg|^pbigg)le C(p) Ebigg[bigg(int_s^t F^2(r)drbigg)^{p/2}bigg].
$$
Did I overlook anything? Or the result on $[s,t]$ is a simple consequence of the result on $[0,t]$?
Similarly, I was wondering whether the $L^p$ estimate of Levy integral:
$$Ebigg(bigg|int_0^Tint_E F(s)dtilde{N}(ds,de)bigg|^pbigg),$$
the so-called Kunita's inequality, works on arbitrary fixed finite interval $[s,t]$.
stochastic-calculus stochastic-integrals stochastic-analysis quadratic-variation
asked Nov 14 at 15:06
John
9,41411334
add a comment |
up vote
0
down vote
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Let $T>0$, $F$ be an adapted squared-integrable process and $pge 2$, then the Burkholder's inequality implies
$$
Ebigg(bigg|int_0^T F(s)dB(s)bigg|^pbigg)le C(p) Ebigg[bigg(int_0^T F^2(s)dsbigg)^{p/2}bigg],
$$
for some constant $C(p)$.
I was wondering why most textbooks present the inequality from $[0,T]$, instead of any subinterval $[s,t]subset[0,T]$. I follow the proof and I didn't notice any technical difficulty to extend the result to the integral on $[s,t]$:
$$
Ebigg(bigg|int_s^t F(r)dB(r)bigg|^pbigg)le C(p) Ebigg[bigg(int_s^t F^2(r)drbigg)^{p/2}bigg].
$$
Did I overlook anything? Or the result on $[s,t]$ is a simple consequence of the result on $[0,t]$?
Similarly, I was wondering whether the $L^p$ estimate of Levy integral:
$$Ebigg(bigg|int_0^Tint_E F(s)dtilde{N}(ds,de)bigg|^pbigg),$$
the so-called Kunita's inequality, works on arbitrary fixed finite interval $[s,t]$.
stochastic-calculus stochastic-integrals stochastic-analysis quadratic-variation
asked Nov 14 at 15:06
John
9,41411334
1
The inequality for subintervals $[s,t]$ is indeed a direct and simple consequence of the result on $[0,T]$; simply apply the inequality on $[0,T]$ to the truncated function $tilde{F}(r) := F(r) 1_{[s,t]}(r)$.
– saz
Nov 14 at 15:25
@saz Thank you very much. How stupid of I!
– John
Nov 14 at 15:34
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $T>0$, $F$ be an adapted squared-integrable process and $pge 2$, then the Burkholder's inequality implies
$$
Ebigg(bigg|int_0^T F(s)dB(s)bigg|^pbigg)le C(p) Ebigg[bigg(int_0^T F^2(s)dsbigg)^{p/2}bigg],
$$
for some constant $C(p)$.
I was wondering why most textbooks present the inequality from $[0,T]$, instead of any subinterval $[s,t]subset[0,T]$. I follow the proof and I didn't notice any technical difficulty to extend the result to the integral on $[s,t]$:
$$
Ebigg(bigg|int_s^t F(r)dB(r)bigg|^pbigg)le C(p) Ebigg[bigg(int_s^t F^2(r)drbigg)^{p/2}bigg].
$$
Did I overlook anything? Or the result on $[s,t]$ is a simple consequence of the result on $[0,t]$?
Similarly, I was wondering whether the $L^p$ estimate of Levy integral:
$$Ebigg(bigg|int_0^Tint_E F(s)dtilde{N}(ds,de)bigg|^pbigg),$$
the so-called Kunita's inequality, works on arbitrary fixed finite interval $[s,t]$.
stochastic-calculus stochastic-integrals stochastic-analysis quadratic-variation
asked Nov 14 at 15:06
John
9,41411334
Let $T>0$, $F$ be an adapted squared-integrable process and $pge 2$, then the Burkholder's inequality implies
$$
Ebigg(bigg|int_0^T F(s)dB(s)bigg|^pbigg)le C(p) Ebigg[bigg(int_0^T F^2(s)dsbigg)^{p/2}bigg],
$$
for some constant $C(p)$.
I was wondering why most textbooks present the inequality from $[0,T]$, instead of any subinterval $[s,t]subset[0,T]$. I follow the proof and I didn't notice any technical difficulty to extend the result to the integral on $[s,t]$:
$$
Ebigg(bigg|int_s^t F(r)dB(r)bigg|^pbigg)le C(p) Ebigg[bigg(int_s^t F^2(r)drbigg)^{p/2}bigg].
$$
Did I overlook anything? Or the result on $[s,t]$ is a simple consequence of the result on $[0,t]$?
Similarly, I was wondering whether the $L^p$ estimate of Levy integral:
$$Ebigg(bigg|int_0^Tint_E F(s)dtilde{N}(ds,de)bigg|^pbigg),$$
the so-called Kunita's inequality, works on arbitrary fixed finite interval $[s,t]$.
stochastic-calculus stochastic-integrals stochastic-analysis quadratic-variation
stochastic-calculus stochastic-integrals stochastic-analysis quadratic-variation
asked Nov 14 at 15:06
John
9,41411334
asked Nov 14 at 15:06
John
9,41411334
edited Nov 14 at 15:22
asked Nov 14 at 15:06
John
9,41411334
asked Nov 14 at 15:06
John
9,41411334
asked Nov 14 at 15:06
John
9,41411334
9,41411334
1
The inequality for subintervals $[s,t]$ is indeed a direct and simple consequence of the result on $[0,T]$; simply apply the inequality on $[0,T]$ to the truncated function $tilde{F}(r) := F(r) 1_{[s,t]}(r)$.
– saz
Nov 14 at 15:25
@saz Thank you very much. How stupid of I!
– John
Nov 14 at 15:34
add a comment |
1
The inequality for subintervals $[s,t]$ is indeed a direct and simple consequence of the result on $[0,T]$; simply apply the inequality on $[0,T]$ to the truncated function $tilde{F}(r) := F(r) 1_{[s,t]}(r)$.
– saz
Nov 14 at 15:25
@saz Thank you very much. How stupid of I!
– John
Nov 14 at 15:34
1
1
The inequality for subintervals $[s,t]$ is indeed a direct and simple consequence of the result on $[0,T]$; simply apply the inequality on $[0,T]$ to the truncated function $tilde{F}(r) := F(r) 1_{[s,t]}(r)$.
– saz
Nov 14 at 15:25
The inequality for subintervals $[s,t]$ is indeed a direct and simple consequence of the result on $[0,T]$; simply apply the inequality on $[0,T]$ to the truncated function $tilde{F}(r) := F(r) 1_{[s,t]}(r)$.
– saz
Nov 14 at 15:25
@saz Thank you very much. How stupid of I!
– John
Nov 14 at 15:34
@saz Thank you very much. How stupid of I!
– John
Nov 14 at 15:34
add a comment |
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1
The inequality for subintervals $[s,t]$ is indeed a direct and simple consequence of the result on $[0,T]$; simply apply the inequality on $[0,T]$ to the truncated function $tilde{F}(r) := F(r) 1_{[s,t]}(r)$.
– saz
Nov 14 at 15:25
@saz Thank you very much. How stupid of I!
– John
Nov 14 at 15:34