Mixing
Autocorrelation in a Brownian model
$begingroup$
I have the following brownian model:
$$
dot{x}=v_0cos(theta(t))+sqrt{2D_t}xi_x(t) \
dot{y}=v_0sin(theta(t))+sqrt{2D_t}xi_y(t) \
dot{theta}=sqrt{2D_r}xi_theta(t)\
$$
with $v_0$ constant and $xi_theta$ is a white noise, this is: $<xi_theta>=0 and <xi_i(t)xi_j(s)>=delta(t-s)delta_{ij}$
Now, I want to solve the following integral:
$$
int_0^t dt int_0^t ds <cos(theta(t)+theta(s))>
$$
Unfortunately, I do not know how to solve it. If instead of a sum there is a subtraction inside the cos, I know I can do the following:
$$
eta=theta(t)-theta(s) \
<cos(theta(t)-theta(s))>=<Re(e^eta)>=Re<e^eta>=e^{<eta^2>}\
<eta^2>=<(theta(t)-theta(s))^2>=2D_rt+2D_rt-2int dtint ds<2D_rxi_theta(t)xi_theta(s)>=2D_rt+2D_rt-2Dr(s or t)
$$
where the last "s or t" means that if you are integrating with t>s then it is an s and if it is s>t then it is a t. If I try to solve the same here, I can see how the solution is not compatible with my simulations results. Any idea on how to solve this? What is wrong if I try to replicate the sum scenario with the subtraction one?
Basically, instead of calculating the MSD ($<x(t)^2+y(t)^2>$), I am calculating $<x(t)^2-y(t)^2>$.
integration normal-distribution brownian-motion correlation noise
asked Jan 31 at 14:28
Learning from mastersLearning from masters
1375
$endgroup$
add a comment |
$begingroup$
I have the following brownian model:
$$
dot{x}=v_0cos(theta(t))+sqrt{2D_t}xi_x(t) \
dot{y}=v_0sin(theta(t))+sqrt{2D_t}xi_y(t) \
dot{theta}=sqrt{2D_r}xi_theta(t)\
$$
with $v_0$ constant and $xi_theta$ is a white noise, this is: $<xi_theta>=0 and <xi_i(t)xi_j(s)>=delta(t-s)delta_{ij}$
Now, I want to solve the following integral:
$$
int_0^t dt int_0^t ds <cos(theta(t)+theta(s))>
$$
Unfortunately, I do not know how to solve it. If instead of a sum there is a subtraction inside the cos, I know I can do the following:
$$
eta=theta(t)-theta(s) \
<cos(theta(t)-theta(s))>=<Re(e^eta)>=Re<e^eta>=e^{<eta^2>}\
<eta^2>=<(theta(t)-theta(s))^2>=2D_rt+2D_rt-2int dtint ds<2D_rxi_theta(t)xi_theta(s)>=2D_rt+2D_rt-2Dr(s or t)
$$
where the last "s or t" means that if you are integrating with t>s then it is an s and if it is s>t then it is a t. If I try to solve the same here, I can see how the solution is not compatible with my simulations results. Any idea on how to solve this? What is wrong if I try to replicate the sum scenario with the subtraction one?
Basically, instead of calculating the MSD ($<x(t)^2+y(t)^2>$), I am calculating $<x(t)^2-y(t)^2>$.
integration normal-distribution brownian-motion correlation noise
asked Jan 31 at 14:28
Learning from mastersLearning from masters
1375
$endgroup$
$begingroup$
Aren't you also using that eta is gaussian distributed in your computation of the expectation of the exponential? Why is this true?
$endgroup$
– Thomas
Jan 31 at 14:48
$begingroup$
Yes! well in fact, since the noise is gaussian distributed, theta should be gaussian. And since a subtraction or summ of two gaussians it is also a gaussian, eta is gaussian distributed.
$endgroup$
– Learning from masters
Jan 31 at 14:50
$begingroup$
Mmm... In a lagevin equation under an external potential V the noise is gaussian but X is not gaussian distributed (the distribution depends on the potential). I think you need a bit more to say this?
$endgroup$
– Thomas
Jan 31 at 17:08
$begingroup$
One more question: do I understand well or in your sytem the variable theta is completely decoupled from $x$ and $y$? I do not see $x$ and $y$ in the third differential equation. theta seems to respect the differential equation of pure brownian motion ?
$endgroup$
– Thomas
Feb 1 at 10:40
$begingroup$
theta does not depend from x and y, it evolves by itself, but x and y depend on theta value. Yes, theta respects the pure brownian motion. About the Gaussian distribution, since "the speed of " theta is not depending in any potential, shouldn't theta be gaussian?
$endgroup$
– Learning from masters
Feb 1 at 12:50
add a comment |
$begingroup$
I have the following brownian model:
$$
dot{x}=v_0cos(theta(t))+sqrt{2D_t}xi_x(t) \
dot{y}=v_0sin(theta(t))+sqrt{2D_t}xi_y(t) \
dot{theta}=sqrt{2D_r}xi_theta(t)\
$$
with $v_0$ constant and $xi_theta$ is a white noise, this is: $<xi_theta>=0 and <xi_i(t)xi_j(s)>=delta(t-s)delta_{ij}$
Now, I want to solve the following integral:
$$
int_0^t dt int_0^t ds <cos(theta(t)+theta(s))>
$$
Unfortunately, I do not know how to solve it. If instead of a sum there is a subtraction inside the cos, I know I can do the following:
$$
eta=theta(t)-theta(s) \
<cos(theta(t)-theta(s))>=<Re(e^eta)>=Re<e^eta>=e^{<eta^2>}\
<eta^2>=<(theta(t)-theta(s))^2>=2D_rt+2D_rt-2int dtint ds<2D_rxi_theta(t)xi_theta(s)>=2D_rt+2D_rt-2Dr(s or t)
$$
where the last "s or t" means that if you are integrating with t>s then it is an s and if it is s>t then it is a t. If I try to solve the same here, I can see how the solution is not compatible with my simulations results. Any idea on how to solve this? What is wrong if I try to replicate the sum scenario with the subtraction one?
Basically, instead of calculating the MSD ($<x(t)^2+y(t)^2>$), I am calculating $<x(t)^2-y(t)^2>$.
integration normal-distribution brownian-motion correlation noise
asked Jan 31 at 14:28
Learning from mastersLearning from masters
1375
$endgroup$
I have the following brownian model:
$$
dot{x}=v_0cos(theta(t))+sqrt{2D_t}xi_x(t) \
dot{y}=v_0sin(theta(t))+sqrt{2D_t}xi_y(t) \
dot{theta}=sqrt{2D_r}xi_theta(t)\
$$
with $v_0$ constant and $xi_theta$ is a white noise, this is: $<xi_theta>=0 and <xi_i(t)xi_j(s)>=delta(t-s)delta_{ij}$
Now, I want to solve the following integral:
$$
int_0^t dt int_0^t ds <cos(theta(t)+theta(s))>
$$
Unfortunately, I do not know how to solve it. If instead of a sum there is a subtraction inside the cos, I know I can do the following:
$$
eta=theta(t)-theta(s) \
<cos(theta(t)-theta(s))>=<Re(e^eta)>=Re<e^eta>=e^{<eta^2>}\
<eta^2>=<(theta(t)-theta(s))^2>=2D_rt+2D_rt-2int dtint ds<2D_rxi_theta(t)xi_theta(s)>=2D_rt+2D_rt-2Dr(s or t)
$$
where the last "s or t" means that if you are integrating with t>s then it is an s and if it is s>t then it is a t. If I try to solve the same here, I can see how the solution is not compatible with my simulations results. Any idea on how to solve this? What is wrong if I try to replicate the sum scenario with the subtraction one?
Basically, instead of calculating the MSD ($<x(t)^2+y(t)^2>$), I am calculating $<x(t)^2-y(t)^2>$.
integration normal-distribution brownian-motion correlation noise
integration normal-distribution brownian-motion correlation noise
asked Jan 31 at 14:28
Learning from mastersLearning from masters
1375
asked Jan 31 at 14:28
Learning from mastersLearning from masters
1375
asked Jan 31 at 14:28
Learning from mastersLearning from masters
1375
asked Jan 31 at 14:28
Learning from mastersLearning from masters
1375
asked Jan 31 at 14:28
Learning from mastersLearning from masters
1375
1375
$begingroup$
Aren't you also using that eta is gaussian distributed in your computation of the expectation of the exponential? Why is this true?
$endgroup$
– Thomas
Jan 31 at 14:48
$begingroup$
Yes! well in fact, since the noise is gaussian distributed, theta should be gaussian. And since a subtraction or summ of two gaussians it is also a gaussian, eta is gaussian distributed.
$endgroup$
– Learning from masters
Jan 31 at 14:50
$begingroup$
Mmm... In a lagevin equation under an external potential V the noise is gaussian but X is not gaussian distributed (the distribution depends on the potential). I think you need a bit more to say this?
$endgroup$
– Thomas
Jan 31 at 17:08
$begingroup$
One more question: do I understand well or in your sytem the variable theta is completely decoupled from $x$ and $y$? I do not see $x$ and $y$ in the third differential equation. theta seems to respect the differential equation of pure brownian motion ?
$endgroup$
– Thomas
Feb 1 at 10:40
$begingroup$
theta does not depend from x and y, it evolves by itself, but x and y depend on theta value. Yes, theta respects the pure brownian motion. About the Gaussian distribution, since "the speed of " theta is not depending in any potential, shouldn't theta be gaussian?
$endgroup$
– Learning from masters
Feb 1 at 12:50
add a comment |
$begingroup$
Aren't you also using that eta is gaussian distributed in your computation of the expectation of the exponential? Why is this true?
$endgroup$
– Thomas
Jan 31 at 14:48
$begingroup$
Yes! well in fact, since the noise is gaussian distributed, theta should be gaussian. And since a subtraction or summ of two gaussians it is also a gaussian, eta is gaussian distributed.
$endgroup$
– Learning from masters
Jan 31 at 14:50
$begingroup$
Mmm... In a lagevin equation under an external potential V the noise is gaussian but X is not gaussian distributed (the distribution depends on the potential). I think you need a bit more to say this?
$endgroup$
– Thomas
Jan 31 at 17:08
$begingroup$
One more question: do I understand well or in your sytem the variable theta is completely decoupled from $x$ and $y$? I do not see $x$ and $y$ in the third differential equation. theta seems to respect the differential equation of pure brownian motion ?
$endgroup$
– Thomas
Feb 1 at 10:40
$begingroup$
theta does not depend from x and y, it evolves by itself, but x and y depend on theta value. Yes, theta respects the pure brownian motion. About the Gaussian distribution, since "the speed of " theta is not depending in any potential, shouldn't theta be gaussian?
$endgroup$
– Learning from masters
Feb 1 at 12:50
$begingroup$
Aren't you also using that eta is gaussian distributed in your computation of the expectation of the exponential? Why is this true?
$endgroup$
– Thomas
Jan 31 at 14:48
$begingroup$
Aren't you also using that eta is gaussian distributed in your computation of the expectation of the exponential? Why is this true?
$endgroup$
– Thomas
Jan 31 at 14:48
$begingroup$
Yes! well in fact, since the noise is gaussian distributed, theta should be gaussian. And since a subtraction or summ of two gaussians it is also a gaussian, eta is gaussian distributed.
$endgroup$
– Learning from masters
Jan 31 at 14:50
$begingroup$
Yes! well in fact, since the noise is gaussian distributed, theta should be gaussian. And since a subtraction or summ of two gaussians it is also a gaussian, eta is gaussian distributed.
$endgroup$
– Learning from masters
Jan 31 at 14:50
$begingroup$
Mmm... In a lagevin equation under an external potential V the noise is gaussian but X is not gaussian distributed (the distribution depends on the potential). I think you need a bit more to say this?
$endgroup$
– Thomas
Jan 31 at 17:08
$begingroup$
Mmm... In a lagevin equation under an external potential V the noise is gaussian but X is not gaussian distributed (the distribution depends on the potential). I think you need a bit more to say this?
$endgroup$
– Thomas
Jan 31 at 17:08
$begingroup$
One more question: do I understand well or in your sytem the variable theta is completely decoupled from $x$ and $y$? I do not see $x$ and $y$ in the third differential equation. theta seems to respect the differential equation of pure brownian motion ?
$endgroup$
– Thomas
Feb 1 at 10:40
$begingroup$
One more question: do I understand well or in your sytem the variable theta is completely decoupled from $x$ and $y$? I do not see $x$ and $y$ in the third differential equation. theta seems to respect the differential equation of pure brownian motion ?
$endgroup$
– Thomas
Feb 1 at 10:40
$begingroup$
theta does not depend from x and y, it evolves by itself, but x and y depend on theta value. Yes, theta respects the pure brownian motion. About the Gaussian distribution, since "the speed of " theta is not depending in any potential, shouldn't theta be gaussian?
$endgroup$
– Learning from masters
Feb 1 at 12:50
$begingroup$
theta does not depend from x and y, it evolves by itself, but x and y depend on theta value. Yes, theta respects the pure brownian motion. About the Gaussian distribution, since "the speed of " theta is not depending in any potential, shouldn't theta be gaussian?
$endgroup$
– Learning from masters
Feb 1 at 12:50
add a comment |
0
active
oldest
votes
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094941%2fautocorrelation-in-a-brownian-model%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094941%2fautocorrelation-in-a-brownian-model%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Aren't you also using that eta is gaussian distributed in your computation of the expectation of the exponential? Why is this true?
$endgroup$
– Thomas
Jan 31 at 14:48
$begingroup$
Yes! well in fact, since the noise is gaussian distributed, theta should be gaussian. And since a subtraction or summ of two gaussians it is also a gaussian, eta is gaussian distributed.
$endgroup$
– Learning from masters
Jan 31 at 14:50
$begingroup$
Mmm... In a lagevin equation under an external potential V the noise is gaussian but X is not gaussian distributed (the distribution depends on the potential). I think you need a bit more to say this?
$endgroup$
– Thomas
Jan 31 at 17:08
$begingroup$
One more question: do I understand well or in your sytem the variable theta is completely decoupled from $x$ and $y$? I do not see $x$ and $y$ in the third differential equation. theta seems to respect the differential equation of pure brownian motion ?
$endgroup$
– Thomas
Feb 1 at 10:40
$begingroup$
theta does not depend from x and y, it evolves by itself, but x and y depend on theta value. Yes, theta respects the pure brownian motion. About the Gaussian distribution, since "the speed of " theta is not depending in any potential, shouldn't theta be gaussian?
$endgroup$
– Learning from masters
Feb 1 at 12:50