Mixing
Moment Estimate for Simple Linear SDE
$begingroup$
In a paper I am reading, it is claimed that the SDE
$$
dX_t = b X_t dt + (sigma X_t + beta_t) dB_t , quad t in [0,T],$$
satisfies
$$
Eleft[ sup_{t in [0,T]} |X_t|^p right] < infty, forall p in [2,infty),
$$
where $b, sigma in mathbb{R}$ are constants and $beta$ is a process satisfying
$$
E left[ int_0^T | beta_t |^2 dt right] < infty .
$$
Is this true?
sde
asked Jan 6 at 12:15
WhiteWhite
939
$endgroup$
add a comment |
$begingroup$
In a paper I am reading, it is claimed that the SDE
$$
dX_t = b X_t dt + (sigma X_t + beta_t) dB_t , quad t in [0,T],$$
satisfies
$$
Eleft[ sup_{t in [0,T]} |X_t|^p right] < infty, forall p in [2,infty),
$$
where $b, sigma in mathbb{R}$ are constants and $beta$ is a process satisfying
$$
E left[ int_0^T | beta_t |^2 dt right] < infty .
$$
Is this true?
sde
asked Jan 6 at 12:15
WhiteWhite
939
$endgroup$
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
1
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53
add a comment |
$begingroup$
In a paper I am reading, it is claimed that the SDE
$$
dX_t = b X_t dt + (sigma X_t + beta_t) dB_t , quad t in [0,T],$$
satisfies
$$
Eleft[ sup_{t in [0,T]} |X_t|^p right] < infty, forall p in [2,infty),
$$
where $b, sigma in mathbb{R}$ are constants and $beta$ is a process satisfying
$$
E left[ int_0^T | beta_t |^2 dt right] < infty .
$$
Is this true?
sde
asked Jan 6 at 12:15
WhiteWhite
939
$endgroup$
In a paper I am reading, it is claimed that the SDE
$$
dX_t = b X_t dt + (sigma X_t + beta_t) dB_t , quad t in [0,T],$$
satisfies
$$
Eleft[ sup_{t in [0,T]} |X_t|^p right] < infty, forall p in [2,infty),
$$
where $b, sigma in mathbb{R}$ are constants and $beta$ is a process satisfying
$$
E left[ int_0^T | beta_t |^2 dt right] < infty .
$$
Is this true?
sde
sde
asked Jan 6 at 12:15
WhiteWhite
939
asked Jan 6 at 12:15
WhiteWhite
939
edited Jan 6 at 12:21
White
asked Jan 6 at 12:15
WhiteWhite
939
asked Jan 6 at 12:15
WhiteWhite
939
asked Jan 6 at 12:15
WhiteWhite
939
939
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
1
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53
add a comment |
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
1
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
1
1
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53
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%2f3063784%2fmoment-estimate-for-simple-linear-sde%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%2f3063784%2fmoment-estimate-for-simple-linear-sde%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
$begingroup$
What is the definition of $X_t$ ?
$endgroup$
– user617446
Jan 6 at 12:16
1
$begingroup$
Without additional assumptions on $beta$ this is not true. Consider for instance $b=0$, $sigma=0$ and $beta_t := Z$ for some random variable $Z in L^2(mathbb{P})$ which is independent of $(B_t)_t$ and which fails to have a finite moment of order $p>2$ (i.e. $mathbb{E}(|Z|^p)=infty$)
$endgroup$
– saz
Jan 6 at 12:48
$begingroup$
that's what I thought, thank you.
$endgroup$
– White
Jan 6 at 12:53