Is a continuous function of (integrable) Brownian motion always integrable?
$begingroup$
Let $Z_{t}=e^{-W(t)}$ with ${W(t):tgeq0}$ a Brownian motion. Is $Z_{t}$ integrable, since it is a continuous function of Brownian motion? Furthermore, are all continuous functions of Brownian motion integrable, since Bownian motion itself is integrable?
measure-theory brownian-motion uniform-integrability
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
$endgroup$
add a comment |
$begingroup$
Let $Z_{t}=e^{-W(t)}$ with ${W(t):tgeq0}$ a Brownian motion. Is $Z_{t}$ integrable, since it is a continuous function of Brownian motion? Furthermore, are all continuous functions of Brownian motion integrable, since Bownian motion itself is integrable?
measure-theory brownian-motion uniform-integrability
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
$endgroup$
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27
add a comment |
$begingroup$
Let $Z_{t}=e^{-W(t)}$ with ${W(t):tgeq0}$ a Brownian motion. Is $Z_{t}$ integrable, since it is a continuous function of Brownian motion? Furthermore, are all continuous functions of Brownian motion integrable, since Bownian motion itself is integrable?
measure-theory brownian-motion uniform-integrability
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
$endgroup$
Let $Z_{t}=e^{-W(t)}$ with ${W(t):tgeq0}$ a Brownian motion. Is $Z_{t}$ integrable, since it is a continuous function of Brownian motion? Furthermore, are all continuous functions of Brownian motion integrable, since Bownian motion itself is integrable?
measure-theory brownian-motion uniform-integrability
measure-theory brownian-motion uniform-integrability
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
asked Jan 22 at 9:25
rs4rs35rs4rs35
237
237
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27
add a comment |
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Continuous functions of integrable functions need not be intergable. In this case we can use the fact that if $X$ is normally distributed then $Ee^{cX}$ is integrable for any constant $c$. In fact $Ee^{cX}=e^{cmu} e^{sigma^{2}c^{2}/2}$ where $mu$ is the mean and $sigma^{2}$ the variance.
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
$endgroup$
add a comment |
$begingroup$
$e^{-W_t}$ is integrable: $Ee^{-W_t}=e^{frac{1}{2}t}$.
Other continuous functions of $W_t$ need not be integrable. For example, $e^{W_1^2}$.
answered Jan 22 at 9:36
AddSupAddSup
4731316
$endgroup$
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%2f3082921%2fis-a-continuous-function-of-integrable-brownian-motion-always-integrable%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Continuous functions of integrable functions need not be intergable. In this case we can use the fact that if $X$ is normally distributed then $Ee^{cX}$ is integrable for any constant $c$. In fact $Ee^{cX}=e^{cmu} e^{sigma^{2}c^{2}/2}$ where $mu$ is the mean and $sigma^{2}$ the variance.
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
$endgroup$
add a comment |
$begingroup$
Continuous functions of integrable functions need not be intergable. In this case we can use the fact that if $X$ is normally distributed then $Ee^{cX}$ is integrable for any constant $c$. In fact $Ee^{cX}=e^{cmu} e^{sigma^{2}c^{2}/2}$ where $mu$ is the mean and $sigma^{2}$ the variance.
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
$endgroup$
add a comment |
$begingroup$
Continuous functions of integrable functions need not be intergable. In this case we can use the fact that if $X$ is normally distributed then $Ee^{cX}$ is integrable for any constant $c$. In fact $Ee^{cX}=e^{cmu} e^{sigma^{2}c^{2}/2}$ where $mu$ is the mean and $sigma^{2}$ the variance.
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
$endgroup$
Continuous functions of integrable functions need not be intergable. In this case we can use the fact that if $X$ is normally distributed then $Ee^{cX}$ is integrable for any constant $c$. In fact $Ee^{cX}=e^{cmu} e^{sigma^{2}c^{2}/2}$ where $mu$ is the mean and $sigma^{2}$ the variance.
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
answered Jan 22 at 9:28
Kavi Rama MurthyKavi Rama Murthy
63k42362
63k42362
add a comment |
add a comment |
$begingroup$
$e^{-W_t}$ is integrable: $Ee^{-W_t}=e^{frac{1}{2}t}$.
Other continuous functions of $W_t$ need not be integrable. For example, $e^{W_1^2}$.
answered Jan 22 at 9:36
AddSupAddSup
4731316
$endgroup$
add a comment |
$begingroup$
$e^{-W_t}$ is integrable: $Ee^{-W_t}=e^{frac{1}{2}t}$.
Other continuous functions of $W_t$ need not be integrable. For example, $e^{W_1^2}$.
answered Jan 22 at 9:36
AddSupAddSup
4731316
$endgroup$
add a comment |
$begingroup$
$e^{-W_t}$ is integrable: $Ee^{-W_t}=e^{frac{1}{2}t}$.
Other continuous functions of $W_t$ need not be integrable. For example, $e^{W_1^2}$.
answered Jan 22 at 9:36
AddSupAddSup
4731316
$endgroup$
$e^{-W_t}$ is integrable: $Ee^{-W_t}=e^{frac{1}{2}t}$.
Other continuous functions of $W_t$ need not be integrable. For example, $e^{W_1^2}$.
answered Jan 22 at 9:36
AddSupAddSup
4731316
answered Jan 22 at 9:36
AddSupAddSup
4731316
answered Jan 22 at 9:36
AddSupAddSup
4731316
answered Jan 22 at 9:36
AddSupAddSup
4731316
4731316
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%2f3082921%2fis-a-continuous-function-of-integrable-brownian-motion-always-integrable%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
$begingroup$
Do you meant the integrability in $t$? Then yes, any continuous function is (locally) integrable.
$endgroup$
– zhoraster
Jan 22 at 9:28
$begingroup$
Yes, for a fixed $t$. So, my question: is the expectation of a continuous function of Brownian motion always finite implying that this function is integrable?
$endgroup$
– rs4rs35
Jan 22 at 9:41
$begingroup$
"For a fixed $t$" means integrability in $omega$, not in $t$. See the response of Kavi Rama Murthy.
$endgroup$
– zhoraster
Jan 23 at 11:27