Mixing
![How to deny HTTP POST requests from other applications? [closed]](https://lh5.googleusercontent.com/-6eoYCvmkRKs/AAAAAAAAAAI/AAAAAAAAADQ/rzGku2EiCCk/s72-c/photo.jpg?sz=32)
Reversal of Brownian motion from first hitting time
$begingroup$
Let $(B_t)_{t ge 0}$ be a (standard) Brownian motion, and for fixed $C>0$ define the first hitting time
$$T_C := inf {t ge 0: B_t = C}.$$
I am interested in the reversal of the Brownian motion from the first hitting time $T_C$, i.e. a description of the distribution of the path
$$(B_{T_C - s} - C)_{s le T_C}.$$
My question is whether this has the same law as
$$(widetilde{B}_s)_{s le S_{-C}}$$
where $(widetilde{B}_s)_{s ge 0}$ is a Brownian motion conditioned to stay non-positive, and $S_{-C}$ is the last time that the process $(B_s)$ hits $-C$.
My guess above comes from the analogous problem for Brownian motion with negative drift. I was told that the following is true (I would appreciate a reference for that):
Fix $m > 0$ and $C > 0$ as before. Let $T_C$ (resp. $S_{-C}$) be the first hitting time of $C$ (resp. last hittimg time of $-C$) of the Brownian motion with negative drift $(B_s -ms)_{s ge 0}$. Then
$$(B_{T_C - s} + m(T_C - s) - C)_{s le T_C} overset{d}{=} (widetilde{B}_s - ms)_{s le S_{-C}}$$
where $(widetilde{B}_s - ms)_{s ge 0}$ is a Brownian motion with drift $-m$ conditioned to stay non-positive.
As a heuristic I can send $m to 0$ and that would recover my claim above, but this does not constitute a mathematical proof. It would be great if someone could tell me that my claim is correct and point me to references where a proof can be easily found.
probability-theory stochastic-processes brownian-motion stopping-times
asked Jul 24 '18 at 11:28
random_personrandom_person
1437
$endgroup$
add a comment |
$begingroup$
Let $(B_t)_{t ge 0}$ be a (standard) Brownian motion, and for fixed $C>0$ define the first hitting time
$$T_C := inf {t ge 0: B_t = C}.$$
I am interested in the reversal of the Brownian motion from the first hitting time $T_C$, i.e. a description of the distribution of the path
$$(B_{T_C - s} - C)_{s le T_C}.$$
My question is whether this has the same law as
$$(widetilde{B}_s)_{s le S_{-C}}$$
where $(widetilde{B}_s)_{s ge 0}$ is a Brownian motion conditioned to stay non-positive, and $S_{-C}$ is the last time that the process $(B_s)$ hits $-C$.
My guess above comes from the analogous problem for Brownian motion with negative drift. I was told that the following is true (I would appreciate a reference for that):
Fix $m > 0$ and $C > 0$ as before. Let $T_C$ (resp. $S_{-C}$) be the first hitting time of $C$ (resp. last hittimg time of $-C$) of the Brownian motion with negative drift $(B_s -ms)_{s ge 0}$. Then
$$(B_{T_C - s} + m(T_C - s) - C)_{s le T_C} overset{d}{=} (widetilde{B}_s - ms)_{s le S_{-C}}$$
where $(widetilde{B}_s - ms)_{s ge 0}$ is a Brownian motion with drift $-m$ conditioned to stay non-positive.
As a heuristic I can send $m to 0$ and that would recover my claim above, but this does not constitute a mathematical proof. It would be great if someone could tell me that my claim is correct and point me to references where a proof can be easily found.
probability-theory stochastic-processes brownian-motion stopping-times
asked Jul 24 '18 at 11:28
random_personrandom_person
1437
$endgroup$
add a comment |
$begingroup$
Let $(B_t)_{t ge 0}$ be a (standard) Brownian motion, and for fixed $C>0$ define the first hitting time
$$T_C := inf {t ge 0: B_t = C}.$$
I am interested in the reversal of the Brownian motion from the first hitting time $T_C$, i.e. a description of the distribution of the path
$$(B_{T_C - s} - C)_{s le T_C}.$$
My question is whether this has the same law as
$$(widetilde{B}_s)_{s le S_{-C}}$$
where $(widetilde{B}_s)_{s ge 0}$ is a Brownian motion conditioned to stay non-positive, and $S_{-C}$ is the last time that the process $(B_s)$ hits $-C$.
My guess above comes from the analogous problem for Brownian motion with negative drift. I was told that the following is true (I would appreciate a reference for that):
Fix $m > 0$ and $C > 0$ as before. Let $T_C$ (resp. $S_{-C}$) be the first hitting time of $C$ (resp. last hittimg time of $-C$) of the Brownian motion with negative drift $(B_s -ms)_{s ge 0}$. Then
$$(B_{T_C - s} + m(T_C - s) - C)_{s le T_C} overset{d}{=} (widetilde{B}_s - ms)_{s le S_{-C}}$$
where $(widetilde{B}_s - ms)_{s ge 0}$ is a Brownian motion with drift $-m$ conditioned to stay non-positive.
As a heuristic I can send $m to 0$ and that would recover my claim above, but this does not constitute a mathematical proof. It would be great if someone could tell me that my claim is correct and point me to references where a proof can be easily found.
probability-theory stochastic-processes brownian-motion stopping-times
asked Jul 24 '18 at 11:28
random_personrandom_person
1437
$endgroup$
Let $(B_t)_{t ge 0}$ be a (standard) Brownian motion, and for fixed $C>0$ define the first hitting time
$$T_C := inf {t ge 0: B_t = C}.$$
I am interested in the reversal of the Brownian motion from the first hitting time $T_C$, i.e. a description of the distribution of the path
$$(B_{T_C - s} - C)_{s le T_C}.$$
My question is whether this has the same law as
$$(widetilde{B}_s)_{s le S_{-C}}$$
where $(widetilde{B}_s)_{s ge 0}$ is a Brownian motion conditioned to stay non-positive, and $S_{-C}$ is the last time that the process $(B_s)$ hits $-C$.
My guess above comes from the analogous problem for Brownian motion with negative drift. I was told that the following is true (I would appreciate a reference for that):
Fix $m > 0$ and $C > 0$ as before. Let $T_C$ (resp. $S_{-C}$) be the first hitting time of $C$ (resp. last hittimg time of $-C$) of the Brownian motion with negative drift $(B_s -ms)_{s ge 0}$. Then
$$(B_{T_C - s} + m(T_C - s) - C)_{s le T_C} overset{d}{=} (widetilde{B}_s - ms)_{s le S_{-C}}$$
where $(widetilde{B}_s - ms)_{s ge 0}$ is a Brownian motion with drift $-m$ conditioned to stay non-positive.
As a heuristic I can send $m to 0$ and that would recover my claim above, but this does not constitute a mathematical proof. It would be great if someone could tell me that my claim is correct and point me to references where a proof can be easily found.
probability-theory stochastic-processes brownian-motion stopping-times
probability-theory stochastic-processes brownian-motion stopping-times
asked Jul 24 '18 at 11:28
random_personrandom_person
1437
asked Jul 24 '18 at 11:28
random_personrandom_person
1437
edited Jul 24 '18 at 14:01
random_person
asked Jul 24 '18 at 11:28
random_personrandom_person
1437
asked Jul 24 '18 at 11:28
random_personrandom_person
1437
asked Jul 24 '18 at 11:28
random_personrandom_person
1437
1437
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer is positive, in the sense that $(widetilde{B}_s)_{s ge 0}$ is a $BES(3)$-process starting from $0$, and $$S_{-C} := sup {t > 0: widetilde{B}_t = -C }$$ is the last hitting time of $-C$ by the Bessel process. This is a classical result due to Williams, probably proved in his paper on path decomposition on diffusion, and can also be found in the book by Revuz-Yor.
answered Jan 11 at 10:44
random_personrandom_person
1437
$endgroup$
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%2f2861238%2freversal-of-brownian-motion-from-first-hitting-time%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
$begingroup$
The answer is positive, in the sense that $(widetilde{B}_s)_{s ge 0}$ is a $BES(3)$-process starting from $0$, and $$S_{-C} := sup {t > 0: widetilde{B}_t = -C }$$ is the last hitting time of $-C$ by the Bessel process. This is a classical result due to Williams, probably proved in his paper on path decomposition on diffusion, and can also be found in the book by Revuz-Yor.
answered Jan 11 at 10:44
random_personrandom_person
1437
$endgroup$
add a comment |
$begingroup$
The answer is positive, in the sense that $(widetilde{B}_s)_{s ge 0}$ is a $BES(3)$-process starting from $0$, and $$S_{-C} := sup {t > 0: widetilde{B}_t = -C }$$ is the last hitting time of $-C$ by the Bessel process. This is a classical result due to Williams, probably proved in his paper on path decomposition on diffusion, and can also be found in the book by Revuz-Yor.
answered Jan 11 at 10:44
random_personrandom_person
1437
$endgroup$
add a comment |
$begingroup$
The answer is positive, in the sense that $(widetilde{B}_s)_{s ge 0}$ is a $BES(3)$-process starting from $0$, and $$S_{-C} := sup {t > 0: widetilde{B}_t = -C }$$ is the last hitting time of $-C$ by the Bessel process. This is a classical result due to Williams, probably proved in his paper on path decomposition on diffusion, and can also be found in the book by Revuz-Yor.
answered Jan 11 at 10:44
random_personrandom_person
1437
$endgroup$
The answer is positive, in the sense that $(widetilde{B}_s)_{s ge 0}$ is a $BES(3)$-process starting from $0$, and $$S_{-C} := sup {t > 0: widetilde{B}_t = -C }$$ is the last hitting time of $-C$ by the Bessel process. This is a classical result due to Williams, probably proved in his paper on path decomposition on diffusion, and can also be found in the book by Revuz-Yor.
answered Jan 11 at 10:44
random_personrandom_person
1437
answered Jan 11 at 10:44
random_personrandom_person
1437
answered Jan 11 at 10:44
random_personrandom_person
1437
answered Jan 11 at 10:44
random_personrandom_person
1437
1437
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%2f2861238%2freversal-of-brownian-motion-from-first-hitting-time%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
