Mixing
Does solution exit for the stochastic differential equation (SDE) $d X_t = -frac{1}{2} e^{-2 X_t} dt +...
$begingroup$
Let $W_t$ be a standard Brownian Motion. Solve the stochastic differential
equation:
$$d X_t = -frac{1}{2} e^{-2 X_t} dt + e^{-X_t} dW_t$$
Does the solution exist for all times t?
sde
asked Jan 25 at 21:39
GOAT MessiGOAT Messi
304
$endgroup$
add a comment |
$begingroup$
Let $W_t$ be a standard Brownian Motion. Solve the stochastic differential
equation:
$$d X_t = -frac{1}{2} e^{-2 X_t} dt + e^{-X_t} dW_t$$
Does the solution exist for all times t?
sde
asked Jan 25 at 21:39
GOAT MessiGOAT Messi
304
$endgroup$
$begingroup$
What are your ideas? Did you test the assumptions of the existence theorem? You could also try to apply the Ito theorem to $Y_t=e^{2X_t}$ to see if the transformed equation has a more simple structure.
$endgroup$
– LutzL
Jan 25 at 22:16
add a comment |
$begingroup$
Let $W_t$ be a standard Brownian Motion. Solve the stochastic differential
equation:
$$d X_t = -frac{1}{2} e^{-2 X_t} dt + e^{-X_t} dW_t$$
Does the solution exist for all times t?
sde
asked Jan 25 at 21:39
GOAT MessiGOAT Messi
304
$endgroup$
Let $W_t$ be a standard Brownian Motion. Solve the stochastic differential
equation:
$$d X_t = -frac{1}{2} e^{-2 X_t} dt + e^{-X_t} dW_t$$
Does the solution exist for all times t?
sde
sde
asked Jan 25 at 21:39
GOAT MessiGOAT Messi
304
asked Jan 25 at 21:39
GOAT MessiGOAT Messi
304
asked Jan 25 at 21:39
GOAT MessiGOAT Messi
304
asked Jan 25 at 21:39
GOAT MessiGOAT Messi
304
asked Jan 25 at 21:39
GOAT MessiGOAT Messi
304
304
$begingroup$
What are your ideas? Did you test the assumptions of the existence theorem? You could also try to apply the Ito theorem to $Y_t=e^{2X_t}$ to see if the transformed equation has a more simple structure.
$endgroup$
– LutzL
Jan 25 at 22:16
add a comment |
$begingroup$
What are your ideas? Did you test the assumptions of the existence theorem? You could also try to apply the Ito theorem to $Y_t=e^{2X_t}$ to see if the transformed equation has a more simple structure.
$endgroup$
– LutzL
Jan 25 at 22:16
$begingroup$
What are your ideas? Did you test the assumptions of the existence theorem? You could also try to apply the Ito theorem to $Y_t=e^{2X_t}$ to see if the transformed equation has a more simple structure.
$endgroup$
– LutzL
Jan 25 at 22:16
$begingroup$
What are your ideas? Did you test the assumptions of the existence theorem? You could also try to apply the Ito theorem to $Y_t=e^{2X_t}$ to see if the transformed equation has a more simple structure.
$endgroup$
– LutzL
Jan 25 at 22:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Set $Y=e^X-W$, then by the Ito theorem
$$
dY=e^XdX+frac12e^Xd[X]-dW=0implies Y=e^{X_0}
$$
and the equation $e^{X_t}=e^{X_0}+W_t$ can only be true as long as the right side is positive. As there is a non-zero probability that it becomes non-negative, the solution will not exist for all times.
How to get to this idea with a little hand-waving
On an atomic level one has $(dW)^2=dt$. Then the expression $e^{-X}dW-frac12e^{-2X}(dW)^2$ is the start of the Taylor series of $ln(1+e^{-X}dW)$. Taking the exponential gives thus
$$
e^{X_{t+dt}-X_t}=1+e^{-X_t}dW_timplies e^{X_{t+dt}}=e^{X_t}+dW_timplies e^{X_t}=e^{X_0}+W_t
$$
To confirm this heuristic computation, apply as above the Ito theorem to $Y=e^X-W$ to confirm that it is a constant process.
answered Jan 25 at 22:35
LutzLLutzL
59.7k42057
$endgroup$
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%2f3087648%2fdoes-solution-exit-for-the-stochastic-differential-equation-sde-d-x-t-fra%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
$begingroup$
Set $Y=e^X-W$, then by the Ito theorem
$$
dY=e^XdX+frac12e^Xd[X]-dW=0implies Y=e^{X_0}
$$
and the equation $e^{X_t}=e^{X_0}+W_t$ can only be true as long as the right side is positive. As there is a non-zero probability that it becomes non-negative, the solution will not exist for all times.
How to get to this idea with a little hand-waving
On an atomic level one has $(dW)^2=dt$. Then the expression $e^{-X}dW-frac12e^{-2X}(dW)^2$ is the start of the Taylor series of $ln(1+e^{-X}dW)$. Taking the exponential gives thus
$$
e^{X_{t+dt}-X_t}=1+e^{-X_t}dW_timplies e^{X_{t+dt}}=e^{X_t}+dW_timplies e^{X_t}=e^{X_0}+W_t
$$
To confirm this heuristic computation, apply as above the Ito theorem to $Y=e^X-W$ to confirm that it is a constant process.
answered Jan 25 at 22:35
LutzLLutzL
59.7k42057
$endgroup$
add a comment |
$begingroup$
Set $Y=e^X-W$, then by the Ito theorem
$$
dY=e^XdX+frac12e^Xd[X]-dW=0implies Y=e^{X_0}
$$
and the equation $e^{X_t}=e^{X_0}+W_t$ can only be true as long as the right side is positive. As there is a non-zero probability that it becomes non-negative, the solution will not exist for all times.
How to get to this idea with a little hand-waving
On an atomic level one has $(dW)^2=dt$. Then the expression $e^{-X}dW-frac12e^{-2X}(dW)^2$ is the start of the Taylor series of $ln(1+e^{-X}dW)$. Taking the exponential gives thus
$$
e^{X_{t+dt}-X_t}=1+e^{-X_t}dW_timplies e^{X_{t+dt}}=e^{X_t}+dW_timplies e^{X_t}=e^{X_0}+W_t
$$
To confirm this heuristic computation, apply as above the Ito theorem to $Y=e^X-W$ to confirm that it is a constant process.
answered Jan 25 at 22:35
LutzLLutzL
59.7k42057
$endgroup$
add a comment |
$begingroup$
Set $Y=e^X-W$, then by the Ito theorem
$$
dY=e^XdX+frac12e^Xd[X]-dW=0implies Y=e^{X_0}
$$
and the equation $e^{X_t}=e^{X_0}+W_t$ can only be true as long as the right side is positive. As there is a non-zero probability that it becomes non-negative, the solution will not exist for all times.
How to get to this idea with a little hand-waving
On an atomic level one has $(dW)^2=dt$. Then the expression $e^{-X}dW-frac12e^{-2X}(dW)^2$ is the start of the Taylor series of $ln(1+e^{-X}dW)$. Taking the exponential gives thus
$$
e^{X_{t+dt}-X_t}=1+e^{-X_t}dW_timplies e^{X_{t+dt}}=e^{X_t}+dW_timplies e^{X_t}=e^{X_0}+W_t
$$
To confirm this heuristic computation, apply as above the Ito theorem to $Y=e^X-W$ to confirm that it is a constant process.
answered Jan 25 at 22:35
LutzLLutzL
59.7k42057
$endgroup$
Set $Y=e^X-W$, then by the Ito theorem
$$
dY=e^XdX+frac12e^Xd[X]-dW=0implies Y=e^{X_0}
$$
and the equation $e^{X_t}=e^{X_0}+W_t$ can only be true as long as the right side is positive. As there is a non-zero probability that it becomes non-negative, the solution will not exist for all times.
How to get to this idea with a little hand-waving
On an atomic level one has $(dW)^2=dt$. Then the expression $e^{-X}dW-frac12e^{-2X}(dW)^2$ is the start of the Taylor series of $ln(1+e^{-X}dW)$. Taking the exponential gives thus
$$
e^{X_{t+dt}-X_t}=1+e^{-X_t}dW_timplies e^{X_{t+dt}}=e^{X_t}+dW_timplies e^{X_t}=e^{X_0}+W_t
$$
To confirm this heuristic computation, apply as above the Ito theorem to $Y=e^X-W$ to confirm that it is a constant process.
answered Jan 25 at 22:35
LutzLLutzL
59.7k42057
answered Jan 25 at 22:35
LutzLLutzL
59.7k42057
answered Jan 25 at 22:35
LutzLLutzL
59.7k42057
answered Jan 25 at 22:35
LutzLLutzL
59.7k42057
59.7k42057
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%2f3087648%2fdoes-solution-exit-for-the-stochastic-differential-equation-sde-d-x-t-fra%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 are your ideas? Did you test the assumptions of the existence theorem? You could also try to apply the Ito theorem to $Y_t=e^{2X_t}$ to see if the transformed equation has a more simple structure.
$endgroup$
– LutzL
Jan 25 at 22:16