Infinitesimal generator : what is it exactly?
$begingroup$
Let $(X_t)$ an diffusion Itô process, i.e. a solution of $$dX_t=b(X_t)dt+sigma (X_t)dB_t.$$ The infinitesimal generator of $(X_t)$ is $$Af(x)=lim_{tto 0^+}frac{mathbb E^x[f(X_t)]-f(x)}{t},$$
where $mathbb E^x$ is the expectation wrt $mathbb P^x$.
Q1) What represent exactly $Af(x)$ for $X_t$ ? For example, for a Brownian motion, if $f$ is $C^2$ then $$A f(x)=frac{1}{2}Delta f(x).$$
But I don't really understand which information does $A$ give is. Is it a sort of derivative of $X_t$ ?
Q2) What is exactely the measure $mathbb P^x$ ? I know it is $mathbb P^x{X_tin A}=mathbb P({X_tin A}mid {X_0=x}),$ But does it mean that on $(Omega ,mathcal F,mathbb P^x)$ we have that $mathbb P{X_0=x}=1$ ? (i.e. is deterministic).
probability measure-theory
$endgroup$
add a comment |
$begingroup$
Let $(X_t)$ an diffusion Itô process, i.e. a solution of $$dX_t=b(X_t)dt+sigma (X_t)dB_t.$$ The infinitesimal generator of $(X_t)$ is $$Af(x)=lim_{tto 0^+}frac{mathbb E^x[f(X_t)]-f(x)}{t},$$
where $mathbb E^x$ is the expectation wrt $mathbb P^x$.
Q1) What represent exactly $Af(x)$ for $X_t$ ? For example, for a Brownian motion, if $f$ is $C^2$ then $$A f(x)=frac{1}{2}Delta f(x).$$
But I don't really understand which information does $A$ give is. Is it a sort of derivative of $X_t$ ?
Q2) What is exactely the measure $mathbb P^x$ ? I know it is $mathbb P^x{X_tin A}=mathbb P({X_tin A}mid {X_0=x}),$ But does it mean that on $(Omega ,mathcal F,mathbb P^x)$ we have that $mathbb P{X_0=x}=1$ ? (i.e. is deterministic).
probability measure-theory
$endgroup$
1
$begingroup$
(1) Intuitively, if $X_t$ is a deterministic function of $t$, $Af(x)$ is exactly $frac{rm d}{{rm d}t}|_{t=0}f(X_t)$. When $X_t$ is an Ito process, $Af(x)$ measures how "fast" $f(X_t)$ changes with respect to $t$ in the sense of expectation. (2) $mathbb{P}^x$ is a conditional probability, which conditions on $X_0=x$. That is, $X_t$ has a fixed, deterministic starting point $x$. In this sense, yes, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
$endgroup$
– hypernova
Jan 3 at 13:49
1
$begingroup$
You might want to take a look at this question and this question
$endgroup$
– saz
Jan 3 at 13:59
$begingroup$
@saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
$endgroup$
– NewMath
Jan 3 at 23:02
add a comment |
$begingroup$
Let $(X_t)$ an diffusion Itô process, i.e. a solution of $$dX_t=b(X_t)dt+sigma (X_t)dB_t.$$ The infinitesimal generator of $(X_t)$ is $$Af(x)=lim_{tto 0^+}frac{mathbb E^x[f(X_t)]-f(x)}{t},$$
where $mathbb E^x$ is the expectation wrt $mathbb P^x$.
Q1) What represent exactly $Af(x)$ for $X_t$ ? For example, for a Brownian motion, if $f$ is $C^2$ then $$A f(x)=frac{1}{2}Delta f(x).$$
But I don't really understand which information does $A$ give is. Is it a sort of derivative of $X_t$ ?
Q2) What is exactely the measure $mathbb P^x$ ? I know it is $mathbb P^x{X_tin A}=mathbb P({X_tin A}mid {X_0=x}),$ But does it mean that on $(Omega ,mathcal F,mathbb P^x)$ we have that $mathbb P{X_0=x}=1$ ? (i.e. is deterministic).
probability measure-theory
$endgroup$
Let $(X_t)$ an diffusion Itô process, i.e. a solution of $$dX_t=b(X_t)dt+sigma (X_t)dB_t.$$ The infinitesimal generator of $(X_t)$ is $$Af(x)=lim_{tto 0^+}frac{mathbb E^x[f(X_t)]-f(x)}{t},$$
where $mathbb E^x$ is the expectation wrt $mathbb P^x$.
Q1) What represent exactly $Af(x)$ for $X_t$ ? For example, for a Brownian motion, if $f$ is $C^2$ then $$A f(x)=frac{1}{2}Delta f(x).$$
But I don't really understand which information does $A$ give is. Is it a sort of derivative of $X_t$ ?
Q2) What is exactely the measure $mathbb P^x$ ? I know it is $mathbb P^x{X_tin A}=mathbb P({X_tin A}mid {X_0=x}),$ But does it mean that on $(Omega ,mathcal F,mathbb P^x)$ we have that $mathbb P{X_0=x}=1$ ? (i.e. is deterministic).
probability measure-theory
probability measure-theory
asked Jan 3 at 13:37
NewMathNewMath
4029
4029
1
$begingroup$
(1) Intuitively, if $X_t$ is a deterministic function of $t$, $Af(x)$ is exactly $frac{rm d}{{rm d}t}|_{t=0}f(X_t)$. When $X_t$ is an Ito process, $Af(x)$ measures how "fast" $f(X_t)$ changes with respect to $t$ in the sense of expectation. (2) $mathbb{P}^x$ is a conditional probability, which conditions on $X_0=x$. That is, $X_t$ has a fixed, deterministic starting point $x$. In this sense, yes, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
$endgroup$
– hypernova
Jan 3 at 13:49
1
$begingroup$
You might want to take a look at this question and this question
$endgroup$
– saz
Jan 3 at 13:59
$begingroup$
@saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
$endgroup$
– NewMath
Jan 3 at 23:02
add a comment |
1
$begingroup$
(1) Intuitively, if $X_t$ is a deterministic function of $t$, $Af(x)$ is exactly $frac{rm d}{{rm d}t}|_{t=0}f(X_t)$. When $X_t$ is an Ito process, $Af(x)$ measures how "fast" $f(X_t)$ changes with respect to $t$ in the sense of expectation. (2) $mathbb{P}^x$ is a conditional probability, which conditions on $X_0=x$. That is, $X_t$ has a fixed, deterministic starting point $x$. In this sense, yes, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
$endgroup$
– hypernova
Jan 3 at 13:49
1
$begingroup$
You might want to take a look at this question and this question
$endgroup$
– saz
Jan 3 at 13:59
$begingroup$
@saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
$endgroup$
– NewMath
Jan 3 at 23:02
1
1
$begingroup$
(1) Intuitively, if $X_t$ is a deterministic function of $t$, $Af(x)$ is exactly $frac{rm d}{{rm d}t}|_{t=0}f(X_t)$. When $X_t$ is an Ito process, $Af(x)$ measures how "fast" $f(X_t)$ changes with respect to $t$ in the sense of expectation. (2) $mathbb{P}^x$ is a conditional probability, which conditions on $X_0=x$. That is, $X_t$ has a fixed, deterministic starting point $x$. In this sense, yes, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
$endgroup$
– hypernova
Jan 3 at 13:49
$begingroup$
(1) Intuitively, if $X_t$ is a deterministic function of $t$, $Af(x)$ is exactly $frac{rm d}{{rm d}t}|_{t=0}f(X_t)$. When $X_t$ is an Ito process, $Af(x)$ measures how "fast" $f(X_t)$ changes with respect to $t$ in the sense of expectation. (2) $mathbb{P}^x$ is a conditional probability, which conditions on $X_0=x$. That is, $X_t$ has a fixed, deterministic starting point $x$. In this sense, yes, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
$endgroup$
– hypernova
Jan 3 at 13:49
1
1
$begingroup$
You might want to take a look at this question and this question
$endgroup$
– saz
Jan 3 at 13:59
$begingroup$
You might want to take a look at this question and this question
$endgroup$
– saz
Jan 3 at 13:59
$begingroup$
@saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
$endgroup$
– NewMath
Jan 3 at 23:02
$begingroup$
@saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
$endgroup$
– NewMath
Jan 3 at 23:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Q1) It seems that this link could help.
Q2) Also see this link for a formal definition of conditional probability. Even, the existence of regular conditional probabilities is a non-trivial fact even for simple cases like $mathbb{R}^2$ equipped with Borel sigma algebra.
$endgroup$
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',
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%2f3060561%2finfinitesimal-generator-what-is-it-exactly%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Q1) It seems that this link could help.
Q2) Also see this link for a formal definition of conditional probability. Even, the existence of regular conditional probabilities is a non-trivial fact even for simple cases like $mathbb{R}^2$ equipped with Borel sigma algebra.
$endgroup$
add a comment |
$begingroup$
Q1) It seems that this link could help.
Q2) Also see this link for a formal definition of conditional probability. Even, the existence of regular conditional probabilities is a non-trivial fact even for simple cases like $mathbb{R}^2$ equipped with Borel sigma algebra.
$endgroup$
add a comment |
$begingroup$
Q1) It seems that this link could help.
Q2) Also see this link for a formal definition of conditional probability. Even, the existence of regular conditional probabilities is a non-trivial fact even for simple cases like $mathbb{R}^2$ equipped with Borel sigma algebra.
$endgroup$
Q1) It seems that this link could help.
Q2) Also see this link for a formal definition of conditional probability. Even, the existence of regular conditional probabilities is a non-trivial fact even for simple cases like $mathbb{R}^2$ equipped with Borel sigma algebra.
answered Jan 3 at 18:21
AragonAragon
213
213
add a comment |
add a comment |
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%2f3060561%2finfinitesimal-generator-what-is-it-exactly%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
1
$begingroup$
(1) Intuitively, if $X_t$ is a deterministic function of $t$, $Af(x)$ is exactly $frac{rm d}{{rm d}t}|_{t=0}f(X_t)$. When $X_t$ is an Ito process, $Af(x)$ measures how "fast" $f(X_t)$ changes with respect to $t$ in the sense of expectation. (2) $mathbb{P}^x$ is a conditional probability, which conditions on $X_0=x$. That is, $X_t$ has a fixed, deterministic starting point $x$. In this sense, yes, $mathbb{P}left{X_0=xright}=1$, which, rigorously, should be $mathbb{P}left(left{X_0=xright}|left{X_0=xright}right)=1$. This is trivially true.
$endgroup$
– hypernova
Jan 3 at 13:49
1
$begingroup$
You might want to take a look at this question and this question
$endgroup$
– saz
Jan 3 at 13:59
$begingroup$
@saz : Very nice links. Thank you for this contribution, it's very helpful. I'll probably comment them in the future.
$endgroup$
– NewMath
Jan 3 at 23:02