Progressively measurable process.

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

Let $b: [0,T] times Omega times mathbbR rightarrow mathbbR$ with the properties:

1. For all $ x in mathbbR$ the process $(t,omega) mapsto b(t,omega, x)$ is progressively measurable.




2. There exits $ C > 0 $ such that for all $omega, t, x_1, x_2 $ we have
   $$ | b(t,omega, x_1) - b(t,omega, x_2) | leq C | x_1 - x_2 |. $$

Let $X = (X_t)_t in [0,T]$ be progressively measurable. I want to show that the process
$$ (t, omega) mapsto b(t,omega, X_t(omega) ) $$
is progressively measurable. This should follow from 1. and 2. but I was not able to prove it. I hope someone can help me out.

Edit: I actually need a more general result. Let $mathcalP_2(mathbbR)$ be the space of measures on $mathbbR$ with finite second moment. Equip this space with the Wasserstein metric $W^2.$ Then, $mathcalP_2(mathbbR)$ is a Polish space.

Let $b: [0,T] times Omega times mathbbR rightarrow mathbbR$ with the properties:

1. For all $ x in mathbbR, mu in mathcalP_2(mathbbR)$ the process $(t,omega) mapsto b(t,omega, x, mu)$ is progressively measurable.




2. There exits $ C > 0 $ such that for all $omega, t, x_1, x_2 , mu_1, mu_2$ we have
   $$ | b(t,omega, x_1, mu_1) - b(t,omega, x_2, mu_2) | leq C big( | x_1 - x_2 | + W^2(mu_1, mu_2) big). $$

Let $X = (X_t)_t in [0,T]$ be progressively measurable. Let $P(X_t)$ be the push forward measure of $X_t$. I want to show that the process
$$ (t, omega) mapsto b(t,omega, X_t(omega), P(X_t) ) $$
is progressively measurable.

I tried to imitate the proof of the easier case. However, the problem seems to be that I cannot split $mathcalP_2(mathbbR)$ into disjoint sets of "small radius". Do you have an idea how to deal with this?

measure-theory stochastic-calculus sde

share|cite|improve this question

edited Aug 7 at 14:16

asked Aug 5 at 20:50

White

305

add a comment | 

up vote
1
down vote

favorite

Let $b: [0,T] times Omega times mathbbR rightarrow mathbbR$ with the properties:

1. For all $ x in mathbbR$ the process $(t,omega) mapsto b(t,omega, x)$ is progressively measurable.




2. There exits $ C > 0 $ such that for all $omega, t, x_1, x_2 $ we have
   $$ | b(t,omega, x_1) - b(t,omega, x_2) | leq C | x_1 - x_2 |. $$

Let $X = (X_t)_t in [0,T]$ be progressively measurable. I want to show that the process
$$ (t, omega) mapsto b(t,omega, X_t(omega) ) $$
is progressively measurable. This should follow from 1. and 2. but I was not able to prove it. I hope someone can help me out.

Edit: I actually need a more general result. Let $mathcalP_2(mathbbR)$ be the space of measures on $mathbbR$ with finite second moment. Equip this space with the Wasserstein metric $W^2.$ Then, $mathcalP_2(mathbbR)$ is a Polish space.

Let $b: [0,T] times Omega times mathbbR rightarrow mathbbR$ with the properties:

1. For all $ x in mathbbR, mu in mathcalP_2(mathbbR)$ the process $(t,omega) mapsto b(t,omega, x, mu)$ is progressively measurable.




2. There exits $ C > 0 $ such that for all $omega, t, x_1, x_2 , mu_1, mu_2$ we have
   $$ | b(t,omega, x_1, mu_1) - b(t,omega, x_2, mu_2) | leq C big( | x_1 - x_2 | + W^2(mu_1, mu_2) big). $$

Let $X = (X_t)_t in [0,T]$ be progressively measurable. Let $P(X_t)$ be the push forward measure of $X_t$. I want to show that the process
$$ (t, omega) mapsto b(t,omega, X_t(omega), P(X_t) ) $$
is progressively measurable.

I tried to imitate the proof of the easier case. However, the problem seems to be that I cannot split $mathcalP_2(mathbbR)$ into disjoint sets of "small radius". Do you have an idea how to deal with this?

measure-theory stochastic-calculus sde

share|cite|improve this question

edited Aug 7 at 14:16

asked Aug 5 at 20:50

White

305

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $b: [0,T] times Omega times mathbbR rightarrow mathbbR$ with the properties:

1. For all $ x in mathbbR$ the process $(t,omega) mapsto b(t,omega, x)$ is progressively measurable.




2. There exits $ C > 0 $ such that for all $omega, t, x_1, x_2 $ we have
   $$ | b(t,omega, x_1) - b(t,omega, x_2) | leq C | x_1 - x_2 |. $$

Let $X = (X_t)_t in [0,T]$ be progressively measurable. I want to show that the process
$$ (t, omega) mapsto b(t,omega, X_t(omega) ) $$
is progressively measurable. This should follow from 1. and 2. but I was not able to prove it. I hope someone can help me out.

Edit: I actually need a more general result. Let $mathcalP_2(mathbbR)$ be the space of measures on $mathbbR$ with finite second moment. Equip this space with the Wasserstein metric $W^2.$ Then, $mathcalP_2(mathbbR)$ is a Polish space.

Let $b: [0,T] times Omega times mathbbR rightarrow mathbbR$ with the properties:

1. For all $ x in mathbbR, mu in mathcalP_2(mathbbR)$ the process $(t,omega) mapsto b(t,omega, x, mu)$ is progressively measurable.




2. There exits $ C > 0 $ such that for all $omega, t, x_1, x_2 , mu_1, mu_2$ we have
   $$ | b(t,omega, x_1, mu_1) - b(t,omega, x_2, mu_2) | leq C big( | x_1 - x_2 | + W^2(mu_1, mu_2) big). $$

Let $X = (X_t)_t in [0,T]$ be progressively measurable. Let $P(X_t)$ be the push forward measure of $X_t$. I want to show that the process
$$ (t, omega) mapsto b(t,omega, X_t(omega), P(X_t) ) $$
is progressively measurable.

I tried to imitate the proof of the easier case. However, the problem seems to be that I cannot split $mathcalP_2(mathbbR)$ into disjoint sets of "small radius". Do you have an idea how to deal with this?

measure-theory stochastic-calculus sde

share|cite|improve this question

edited Aug 7 at 14:16

asked Aug 5 at 20:50

White

305

Let $b: [0,T] times Omega times mathbbR rightarrow mathbbR$ with the properties:

1. For all $ x in mathbbR$ the process $(t,omega) mapsto b(t,omega, x)$ is progressively measurable.




2. There exits $ C > 0 $ such that for all $omega, t, x_1, x_2 $ we have
   $$ | b(t,omega, x_1) - b(t,omega, x_2) | leq C | x_1 - x_2 |. $$

Let $X = (X_t)_t in [0,T]$ be progressively measurable. I want to show that the process
$$ (t, omega) mapsto b(t,omega, X_t(omega) ) $$
is progressively measurable. This should follow from 1. and 2. but I was not able to prove it. I hope someone can help me out.

Edit: I actually need a more general result. Let $mathcalP_2(mathbbR)$ be the space of measures on $mathbbR$ with finite second moment. Equip this space with the Wasserstein metric $W^2.$ Then, $mathcalP_2(mathbbR)$ is a Polish space.

Let $b: [0,T] times Omega times mathbbR rightarrow mathbbR$ with the properties:

1. For all $ x in mathbbR, mu in mathcalP_2(mathbbR)$ the process $(t,omega) mapsto b(t,omega, x, mu)$ is progressively measurable.




2. There exits $ C > 0 $ such that for all $omega, t, x_1, x_2 , mu_1, mu_2$ we have
   $$ | b(t,omega, x_1, mu_1) - b(t,omega, x_2, mu_2) | leq C big( | x_1 - x_2 | + W^2(mu_1, mu_2) big). $$

Let $X = (X_t)_t in [0,T]$ be progressively measurable. Let $P(X_t)$ be the push forward measure of $X_t$. I want to show that the process
$$ (t, omega) mapsto b(t,omega, X_t(omega), P(X_t) ) $$
is progressively measurable.

I tried to imitate the proof of the easier case. However, the problem seems to be that I cannot split $mathcalP_2(mathbbR)$ into disjoint sets of "small radius". Do you have an idea how to deal with this?

measure-theory stochastic-calculus sde

share|cite|improve this question

edited Aug 7 at 14:16

asked Aug 5 at 20:50

White

305

share|cite|improve this question

share|cite|improve this question

edited Aug 7 at 14:16

edited Aug 7 at 14:16

edited Aug 7 at 14:16

asked Aug 5 at 20:50

White

305

asked Aug 5 at 20:50

White

305

asked Aug 5 at 20:50

White

305

305

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

Consider $b_n(t,omega,x):=b(t,omega,k/n)$ for $xin[k/n,(k+1)/n)$ and integer $k$, where $n=1,2,ldots$. The corresponding process
$$
X^(n)_t(omega):=b_n(t,omega,X_t(omega))=sum_k b(t,omega,k/n)1_[k/n,(k+1)/n)(X_t(omega))
$$
is a countable sum of progressive process, and so is progressive. And because of the uniform Lipschitz condition you have imposed, you have the uniform convergence:
$$
lim_nsup_0le tle T|X^(n)_t(omega)-X_t(omega)|=0.
$$
It follows that $X$ is progressive as well.

share|cite|improve this answer

answered Aug 6 at 23:06

John Dawkins

12.5k1917

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot, this is very helpful. I have one remark: I think uniform convergence is not necessary. Pointwise limits of measurable functions are measurable. Therefore, condition 2. can be relaxed to have the $C$ depending on $(t, omega).$ Is this correct?
  – White
  Aug 7 at 9:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No doubt a weaker condition would suffice but th uniformity makes the pointwise convergence easy to see.
  – John Dawkins
  Aug 7 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot, I posted a more general version. Maybe you could also help me out with this one.
  – White
  Aug 7 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For the more general version, I was missing this piece of information: math.stackexchange.com/questions/2875058/…. Using this fact we can use the same proof.
  – White
  Aug 7 at 15:29
  
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873353%2fprogressively-measurable-process%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Consider $b_n(t,omega,x):=b(t,omega,k/n)$ for $xin[k/n,(k+1)/n)$ and integer $k$, where $n=1,2,ldots$. The corresponding process
$$
X^(n)_t(omega):=b_n(t,omega,X_t(omega))=sum_k b(t,omega,k/n)1_[k/n,(k+1)/n)(X_t(omega))
$$
is a countable sum of progressive process, and so is progressive. And because of the uniform Lipschitz condition you have imposed, you have the uniform convergence:
$$
lim_nsup_0le tle T|X^(n)_t(omega)-X_t(omega)|=0.
$$
It follows that $X$ is progressive as well.

share|cite|improve this answer

answered Aug 6 at 23:06

John Dawkins

12.5k1917

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot, this is very helpful. I have one remark: I think uniform convergence is not necessary. Pointwise limits of measurable functions are measurable. Therefore, condition 2. can be relaxed to have the $C$ depending on $(t, omega).$ Is this correct?
  – White
  Aug 7 at 9:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No doubt a weaker condition would suffice but th uniformity makes the pointwise convergence easy to see.
  – John Dawkins
  Aug 7 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot, I posted a more general version. Maybe you could also help me out with this one.
  – White
  Aug 7 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For the more general version, I was missing this piece of information: math.stackexchange.com/questions/2875058/…. Using this fact we can use the same proof.
  – White
  Aug 7 at 15:29
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Consider $b_n(t,omega,x):=b(t,omega,k/n)$ for $xin[k/n,(k+1)/n)$ and integer $k$, where $n=1,2,ldots$. The corresponding process
$$
X^(n)_t(omega):=b_n(t,omega,X_t(omega))=sum_k b(t,omega,k/n)1_[k/n,(k+1)/n)(X_t(omega))
$$
is a countable sum of progressive process, and so is progressive. And because of the uniform Lipschitz condition you have imposed, you have the uniform convergence:
$$
lim_nsup_0le tle T|X^(n)_t(omega)-X_t(omega)|=0.
$$
It follows that $X$ is progressive as well.

share|cite|improve this answer

answered Aug 6 at 23:06

John Dawkins

12.5k1917

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot, this is very helpful. I have one remark: I think uniform convergence is not necessary. Pointwise limits of measurable functions are measurable. Therefore, condition 2. can be relaxed to have the $C$ depending on $(t, omega).$ Is this correct?
  – White
  Aug 7 at 9:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No doubt a weaker condition would suffice but th uniformity makes the pointwise convergence easy to see.
  – John Dawkins
  Aug 7 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot, I posted a more general version. Maybe you could also help me out with this one.
  – White
  Aug 7 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For the more general version, I was missing this piece of information: math.stackexchange.com/questions/2875058/…. Using this fact we can use the same proof.
  – White
  Aug 7 at 15:29
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Consider $b_n(t,omega,x):=b(t,omega,k/n)$ for $xin[k/n,(k+1)/n)$ and integer $k$, where $n=1,2,ldots$. The corresponding process
$$
X^(n)_t(omega):=b_n(t,omega,X_t(omega))=sum_k b(t,omega,k/n)1_[k/n,(k+1)/n)(X_t(omega))
$$
is a countable sum of progressive process, and so is progressive. And because of the uniform Lipschitz condition you have imposed, you have the uniform convergence:
$$
lim_nsup_0le tle T|X^(n)_t(omega)-X_t(omega)|=0.
$$
It follows that $X$ is progressive as well.

share|cite|improve this answer

answered Aug 6 at 23:06

John Dawkins

12.5k1917

Consider $b_n(t,omega,x):=b(t,omega,k/n)$ for $xin[k/n,(k+1)/n)$ and integer $k$, where $n=1,2,ldots$. The corresponding process
$$
X^(n)_t(omega):=b_n(t,omega,X_t(omega))=sum_k b(t,omega,k/n)1_[k/n,(k+1)/n)(X_t(omega))
$$
is a countable sum of progressive process, and so is progressive. And because of the uniform Lipschitz condition you have imposed, you have the uniform convergence:
$$
lim_nsup_0le tle T|X^(n)_t(omega)-X_t(omega)|=0.
$$
It follows that $X$ is progressive as well.

share|cite|improve this answer

answered Aug 6 at 23:06

John Dawkins

12.5k1917

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 23:06

John Dawkins

12.5k1917

answered Aug 6 at 23:06

John Dawkins

12.5k1917

answered Aug 6 at 23:06

John Dawkins

12.5k1917

12.5k1917

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot, this is very helpful. I have one remark: I think uniform convergence is not necessary. Pointwise limits of measurable functions are measurable. Therefore, condition 2. can be relaxed to have the $C$ depending on $(t, omega).$ Is this correct?
  – White
  Aug 7 at 9:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No doubt a weaker condition would suffice but th uniformity makes the pointwise convergence easy to see.
  – John Dawkins
  Aug 7 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot, I posted a more general version. Maybe you could also help me out with this one.
  – White
  Aug 7 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For the more general version, I was missing this piece of information: math.stackexchange.com/questions/2875058/…. Using this fact we can use the same proof.
  – White
  Aug 7 at 15:29
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot, this is very helpful. I have one remark: I think uniform convergence is not necessary. Pointwise limits of measurable functions are measurable. Therefore, condition 2. can be relaxed to have the $C$ depending on $(t, omega).$ Is this correct?
  – White
  Aug 7 at 9:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No doubt a weaker condition would suffice but th uniformity makes the pointwise convergence easy to see.
  – John Dawkins
  Aug 7 at 13:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot, I posted a more general version. Maybe you could also help me out with this one.
  – White
  Aug 7 at 14:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For the more general version, I was missing this piece of information: math.stackexchange.com/questions/2875058/…. Using this fact we can use the same proof.
  – White
  Aug 7 at 15:29
  
  

  
  
  
  

Thanks a lot, this is very helpful. I have one remark: I think uniform convergence is not necessary. Pointwise limits of measurable functions are measurable. Therefore, condition 2. can be relaxed to have the $C$ depending on $(t, omega).$ Is this correct?
– White
Aug 7 at 9:32

Thanks a lot, this is very helpful. I have one remark: I think uniform convergence is not necessary. Pointwise limits of measurable functions are measurable. Therefore, condition 2. can be relaxed to have the $C$ depending on $(t, omega).$ Is this correct?
– White
Aug 7 at 9:32

No doubt a weaker condition would suffice but th uniformity makes the pointwise convergence easy to see.
– John Dawkins
Aug 7 at 13:20

No doubt a weaker condition would suffice but th uniformity makes the pointwise convergence easy to see.
– John Dawkins
Aug 7 at 13:20

Thanks a lot, I posted a more general version. Maybe you could also help me out with this one.
– White
Aug 7 at 14:16

Thanks a lot, I posted a more general version. Maybe you could also help me out with this one.
– White
Aug 7 at 14:16

For the more general version, I was missing this piece of information: math.stackexchange.com/questions/2875058/…. Using this fact we can use the same proof.
– White
Aug 7 at 15:29

For the more general version, I was missing this piece of information: math.stackexchange.com/questions/2875058/…. Using this fact we can use the same proof.
– White
Aug 7 at 15:29

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873353%2fprogressively-measurable-process%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email