What is the canonical process of a measure space $(Omega, mathscr{F}, mathbb{Q})$?
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I was reading paper which referred to a canonical process within the context of a measure space $(Omega, mathscr{F}, mathbb{Q})$. The surrounding discussion was of functions $a:[t,T] rightarrow Omega$, and the canonical process was denoted $(mathbb{S}_t)^T_{t=0}$.
It may be useful to know that the term was used when defining a "martingale measure". A measure $mathbb{Q}$ was said to be a martingale measure if the canonical process is a local martingale with respect to the measure and $mathbb{S}(0) = 1$ a.s.
What could this canonical process be? Any help is appreciated.
measure-theory stochastic-processes
asked Jan 12 at 12:13
DavenDaven
39519
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add a comment |
$begingroup$
I was reading paper which referred to a canonical process within the context of a measure space $(Omega, mathscr{F}, mathbb{Q})$. The surrounding discussion was of functions $a:[t,T] rightarrow Omega$, and the canonical process was denoted $(mathbb{S}_t)^T_{t=0}$.
It may be useful to know that the term was used when defining a "martingale measure". A measure $mathbb{Q}$ was said to be a martingale measure if the canonical process is a local martingale with respect to the measure and $mathbb{S}(0) = 1$ a.s.
What could this canonical process be? Any help is appreciated.
measure-theory stochastic-processes
asked Jan 12 at 12:13
DavenDaven
39519
$endgroup$
add a comment |
$begingroup$
I was reading paper which referred to a canonical process within the context of a measure space $(Omega, mathscr{F}, mathbb{Q})$. The surrounding discussion was of functions $a:[t,T] rightarrow Omega$, and the canonical process was denoted $(mathbb{S}_t)^T_{t=0}$.
It may be useful to know that the term was used when defining a "martingale measure". A measure $mathbb{Q}$ was said to be a martingale measure if the canonical process is a local martingale with respect to the measure and $mathbb{S}(0) = 1$ a.s.
What could this canonical process be? Any help is appreciated.
measure-theory stochastic-processes
asked Jan 12 at 12:13
DavenDaven
39519
$endgroup$
I was reading paper which referred to a canonical process within the context of a measure space $(Omega, mathscr{F}, mathbb{Q})$. The surrounding discussion was of functions $a:[t,T] rightarrow Omega$, and the canonical process was denoted $(mathbb{S}_t)^T_{t=0}$.
It may be useful to know that the term was used when defining a "martingale measure". A measure $mathbb{Q}$ was said to be a martingale measure if the canonical process is a local martingale with respect to the measure and $mathbb{S}(0) = 1$ a.s.
What could this canonical process be? Any help is appreciated.
measure-theory stochastic-processes
measure-theory stochastic-processes
asked Jan 12 at 12:13
DavenDaven
39519
asked Jan 12 at 12:13
DavenDaven
39519
asked Jan 12 at 12:13
DavenDaven
39519
asked Jan 12 at 12:13
DavenDaven
39519
asked Jan 12 at 12:13
DavenDaven
39519
39519
add a comment |
add a comment |
1 Answer
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If ${X_s:sleq t leq T}$ is a given process on some probability space then, by Kolmpgorov's Theorem we can construct a measure $Q$ on $mathbb R ^{[t,T]}$ (with the sigma algebra generated by the projection maps) such that the random variables $Y_s:tleq sleq T$ defined by $Y_s(omega)=omega (s)$ defined on $(mathbb R ^{[t,T]},Q)$ have the same finite dimensional distributions as the given process. This new process is called the canonical process corresponding to the original process. You can also start with a probability measure $Q$ on $mathbb R ^{[t,T]}$ instead of starting with a process ${X_s:sleq t leq T}$. For example if $Q$ is the Wiener measure on $C[0,1]$ then the canonical process associated with $Q$ is deined by $Y_s(omega)=omega (s)$ for $omega in C[0,1]$ and $s in [0,1]$.
answered Jan 12 at 12:25
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
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You have started by assuming we have process X. The context in which I am working does not have this; we have only a measure space and we are defining what it means to be a martingale measure. I am not saying what you have said is wrong- this objection is just to highlight where my lack of understanding lies
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– Daven
Jan 12 at 12:44
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@Daven I have edited my answer.
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– Kavi Rama Murthy
Jan 12 at 12:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If ${X_s:sleq t leq T}$ is a given process on some probability space then, by Kolmpgorov's Theorem we can construct a measure $Q$ on $mathbb R ^{[t,T]}$ (with the sigma algebra generated by the projection maps) such that the random variables $Y_s:tleq sleq T$ defined by $Y_s(omega)=omega (s)$ defined on $(mathbb R ^{[t,T]},Q)$ have the same finite dimensional distributions as the given process. This new process is called the canonical process corresponding to the original process. You can also start with a probability measure $Q$ on $mathbb R ^{[t,T]}$ instead of starting with a process ${X_s:sleq t leq T}$. For example if $Q$ is the Wiener measure on $C[0,1]$ then the canonical process associated with $Q$ is deined by $Y_s(omega)=omega (s)$ for $omega in C[0,1]$ and $s in [0,1]$.
answered Jan 12 at 12:25
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
$endgroup$
$begingroup$
You have started by assuming we have process X. The context in which I am working does not have this; we have only a measure space and we are defining what it means to be a martingale measure. I am not saying what you have said is wrong- this objection is just to highlight where my lack of understanding lies
$endgroup$
– Daven
Jan 12 at 12:44
$begingroup$
@Daven I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 12 at 12:49
add a comment |
$begingroup$
If ${X_s:sleq t leq T}$ is a given process on some probability space then, by Kolmpgorov's Theorem we can construct a measure $Q$ on $mathbb R ^{[t,T]}$ (with the sigma algebra generated by the projection maps) such that the random variables $Y_s:tleq sleq T$ defined by $Y_s(omega)=omega (s)$ defined on $(mathbb R ^{[t,T]},Q)$ have the same finite dimensional distributions as the given process. This new process is called the canonical process corresponding to the original process. You can also start with a probability measure $Q$ on $mathbb R ^{[t,T]}$ instead of starting with a process ${X_s:sleq t leq T}$. For example if $Q$ is the Wiener measure on $C[0,1]$ then the canonical process associated with $Q$ is deined by $Y_s(omega)=omega (s)$ for $omega in C[0,1]$ and $s in [0,1]$.
answered Jan 12 at 12:25
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
$endgroup$
$begingroup$
You have started by assuming we have process X. The context in which I am working does not have this; we have only a measure space and we are defining what it means to be a martingale measure. I am not saying what you have said is wrong- this objection is just to highlight where my lack of understanding lies
$endgroup$
– Daven
Jan 12 at 12:44
$begingroup$
@Daven I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 12 at 12:49
add a comment |
$begingroup$
If ${X_s:sleq t leq T}$ is a given process on some probability space then, by Kolmpgorov's Theorem we can construct a measure $Q$ on $mathbb R ^{[t,T]}$ (with the sigma algebra generated by the projection maps) such that the random variables $Y_s:tleq sleq T$ defined by $Y_s(omega)=omega (s)$ defined on $(mathbb R ^{[t,T]},Q)$ have the same finite dimensional distributions as the given process. This new process is called the canonical process corresponding to the original process. You can also start with a probability measure $Q$ on $mathbb R ^{[t,T]}$ instead of starting with a process ${X_s:sleq t leq T}$. For example if $Q$ is the Wiener measure on $C[0,1]$ then the canonical process associated with $Q$ is deined by $Y_s(omega)=omega (s)$ for $omega in C[0,1]$ and $s in [0,1]$.
answered Jan 12 at 12:25
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
$endgroup$
If ${X_s:sleq t leq T}$ is a given process on some probability space then, by Kolmpgorov's Theorem we can construct a measure $Q$ on $mathbb R ^{[t,T]}$ (with the sigma algebra generated by the projection maps) such that the random variables $Y_s:tleq sleq T$ defined by $Y_s(omega)=omega (s)$ defined on $(mathbb R ^{[t,T]},Q)$ have the same finite dimensional distributions as the given process. This new process is called the canonical process corresponding to the original process. You can also start with a probability measure $Q$ on $mathbb R ^{[t,T]}$ instead of starting with a process ${X_s:sleq t leq T}$. For example if $Q$ is the Wiener measure on $C[0,1]$ then the canonical process associated with $Q$ is deined by $Y_s(omega)=omega (s)$ for $omega in C[0,1]$ and $s in [0,1]$.
answered Jan 12 at 12:25
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
edited Jan 12 at 12:49
answered Jan 12 at 12:25
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
answered Jan 12 at 12:25
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
answered Jan 12 at 12:25
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
66.9k53067
$begingroup$
You have started by assuming we have process X. The context in which I am working does not have this; we have only a measure space and we are defining what it means to be a martingale measure. I am not saying what you have said is wrong- this objection is just to highlight where my lack of understanding lies
$endgroup$
– Daven
Jan 12 at 12:44
$begingroup$
@Daven I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 12 at 12:49
add a comment |
$begingroup$
You have started by assuming we have process X. The context in which I am working does not have this; we have only a measure space and we are defining what it means to be a martingale measure. I am not saying what you have said is wrong- this objection is just to highlight where my lack of understanding lies
$endgroup$
– Daven
Jan 12 at 12:44
$begingroup$
@Daven I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 12 at 12:49
$begingroup$
You have started by assuming we have process X. The context in which I am working does not have this; we have only a measure space and we are defining what it means to be a martingale measure. I am not saying what you have said is wrong- this objection is just to highlight where my lack of understanding lies
$endgroup$
– Daven
Jan 12 at 12:44
$begingroup$
You have started by assuming we have process X. The context in which I am working does not have this; we have only a measure space and we are defining what it means to be a martingale measure. I am not saying what you have said is wrong- this objection is just to highlight where my lack of understanding lies
$endgroup$
– Daven
Jan 12 at 12:44
$begingroup$
@Daven I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 12 at 12:49
$begingroup$
@Daven I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 12 at 12:49
add a comment |
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