Deterministic Integral of a Predictable Process is Predictable
$begingroup$
I was reviewing a proof of existence of solutions to stochastic evolution equations which takes the form of a fixed point argument on the space of predictable processes such that
$$
sup_{tleq T}mathbb{E}[|X(t)|]<infty.
$$
This space is certainly Banach under the above norm. I noticed that in the contraction mapping argument, no one ever seems to comment on the deterministic integral that could appear, of the form
$$
int_0^t f(s,X(s))ds
$$
for $f$, say, a globally Lipschitz continuous mapping. While this is certainly well defined in the above norm, to fully address the fixed point argument, one would need to know that this process, denoted $Y(t)$, is also predictable. My question is, is this obvious? The argument that springs to mind is to approximate it by a left Riemann sum which will certainly be predictable and in this Banach space, and then use the completeness of the space - is that correct? Is there another argument that one would make?
stochastic-processes stochastic-calculus stochastic-integrals stochastic-pde
asked Jan 16 at 14:00
user2379888user2379888
24019
$endgroup$
add a comment |
$begingroup$
I was reviewing a proof of existence of solutions to stochastic evolution equations which takes the form of a fixed point argument on the space of predictable processes such that
$$
sup_{tleq T}mathbb{E}[|X(t)|]<infty.
$$
This space is certainly Banach under the above norm. I noticed that in the contraction mapping argument, no one ever seems to comment on the deterministic integral that could appear, of the form
$$
int_0^t f(s,X(s))ds
$$
for $f$, say, a globally Lipschitz continuous mapping. While this is certainly well defined in the above norm, to fully address the fixed point argument, one would need to know that this process, denoted $Y(t)$, is also predictable. My question is, is this obvious? The argument that springs to mind is to approximate it by a left Riemann sum which will certainly be predictable and in this Banach space, and then use the completeness of the space - is that correct? Is there another argument that one would make?
stochastic-processes stochastic-calculus stochastic-integrals stochastic-pde
asked Jan 16 at 14:00
user2379888user2379888
24019
$endgroup$
1
$begingroup$
Hi left continuous (or càg) adapted processes are predictable, see theorem 1 point 6 here : almostsure.wordpress.com/2016/11/22/predictable-processes
$endgroup$
– TheBridge
Jan 21 at 12:25
add a comment |
$begingroup$
I was reviewing a proof of existence of solutions to stochastic evolution equations which takes the form of a fixed point argument on the space of predictable processes such that
$$
sup_{tleq T}mathbb{E}[|X(t)|]<infty.
$$
This space is certainly Banach under the above norm. I noticed that in the contraction mapping argument, no one ever seems to comment on the deterministic integral that could appear, of the form
$$
int_0^t f(s,X(s))ds
$$
for $f$, say, a globally Lipschitz continuous mapping. While this is certainly well defined in the above norm, to fully address the fixed point argument, one would need to know that this process, denoted $Y(t)$, is also predictable. My question is, is this obvious? The argument that springs to mind is to approximate it by a left Riemann sum which will certainly be predictable and in this Banach space, and then use the completeness of the space - is that correct? Is there another argument that one would make?
stochastic-processes stochastic-calculus stochastic-integrals stochastic-pde
asked Jan 16 at 14:00
user2379888user2379888
24019
$endgroup$
I was reviewing a proof of existence of solutions to stochastic evolution equations which takes the form of a fixed point argument on the space of predictable processes such that
$$
sup_{tleq T}mathbb{E}[|X(t)|]<infty.
$$
This space is certainly Banach under the above norm. I noticed that in the contraction mapping argument, no one ever seems to comment on the deterministic integral that could appear, of the form
$$
int_0^t f(s,X(s))ds
$$
for $f$, say, a globally Lipschitz continuous mapping. While this is certainly well defined in the above norm, to fully address the fixed point argument, one would need to know that this process, denoted $Y(t)$, is also predictable. My question is, is this obvious? The argument that springs to mind is to approximate it by a left Riemann sum which will certainly be predictable and in this Banach space, and then use the completeness of the space - is that correct? Is there another argument that one would make?
stochastic-processes stochastic-calculus stochastic-integrals stochastic-pde
stochastic-processes stochastic-calculus stochastic-integrals stochastic-pde
asked Jan 16 at 14:00
user2379888user2379888
24019
asked Jan 16 at 14:00
user2379888user2379888
24019
asked Jan 16 at 14:00
user2379888user2379888
24019
asked Jan 16 at 14:00
user2379888user2379888
24019
asked Jan 16 at 14:00
user2379888user2379888
24019
24019
1
$begingroup$
Hi left continuous (or càg) adapted processes are predictable, see theorem 1 point 6 here : almostsure.wordpress.com/2016/11/22/predictable-processes
$endgroup$
– TheBridge
Jan 21 at 12:25
add a comment |
1
$begingroup$
Hi left continuous (or càg) adapted processes are predictable, see theorem 1 point 6 here : almostsure.wordpress.com/2016/11/22/predictable-processes
$endgroup$
– TheBridge
Jan 21 at 12:25
1
1
$begingroup$
Hi left continuous (or càg) adapted processes are predictable, see theorem 1 point 6 here : almostsure.wordpress.com/2016/11/22/predictable-processes
$endgroup$
– TheBridge
Jan 21 at 12:25
$begingroup$
Hi left continuous (or càg) adapted processes are predictable, see theorem 1 point 6 here : almostsure.wordpress.com/2016/11/22/predictable-processes
$endgroup$
– TheBridge
Jan 21 at 12:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think you only need the Borel measurability of $f$ for the predictability of $Y_t=f(t,X_t)$.
$Y$ can be decomposed into
$$
[0,infty)timesOmegaoverset{phi_X}{rightarrow}[0,infty)timesmathbb R
overset{f}{rightarrow}mathbb R,
$$
so predictability follows once the $mathcal P/mathcal B([0,infty)timesmathbb R)$-measurability of $phi_X$ is established, where $phi_X(t,omega):=(t,X_t(omega))$. And to establish the last measurability it suffices to consider measurable sets of the form $[0,tau]times B$, $tauin[0,infty)$, $Binmathcal B(mathbb R)$, which generate $mathcal B([0,infty)timesmathbb R)equiv mathcal B([0,infty))otimesmathcal B(mathbb R)$. So look at
$$
{(t,omega)in[0,infty)timesOmega:
(t,X_t(omega))in[0,tau]times B
}
=([0,tau]timesOmega)cap X^{-1}(B).
$$
$X^{-1}(B)$ is in $mathcal P$ by the assumption $X$ is predictable; and $([0,tau]timesOmega)$, too, is in $mathcal P$ as $1_{tin[0,tau]}(t,omega)$ is an adapted caglad process.
answered Jan 18 at 10:42
AddSupAddSup
5101318
$endgroup$
1
$begingroup$
The $Y_t$ process I am concerned with is $Y_t = int_0^t f(s,X_s)ds$
$endgroup$
– user2379888
Jan 21 at 13:29
$begingroup$
@user2379888 Sorry for the mistake... Here's an addendum that will complete the answer: Given the predictability of $f(t,X_t)$ as observed above, $f(t,X_t)$ is then a fortiori progressively measurable. Thus the integral $int_0^tf(s,X_s)ds$ is adapted. (Provided of course that the integral is well-defined. This requires more than the Borel measurability of $f$.) And adaptedness and continuity are more than enough for predictability (as TheBridge noted).
$endgroup$
– AddSup
Jan 21 at 15:00
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%2f3075760%2fdeterministic-integral-of-a-predictable-process-is-predictable%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
$begingroup$
I think you only need the Borel measurability of $f$ for the predictability of $Y_t=f(t,X_t)$.
$Y$ can be decomposed into
$$
[0,infty)timesOmegaoverset{phi_X}{rightarrow}[0,infty)timesmathbb R
overset{f}{rightarrow}mathbb R,
$$
so predictability follows once the $mathcal P/mathcal B([0,infty)timesmathbb R)$-measurability of $phi_X$ is established, where $phi_X(t,omega):=(t,X_t(omega))$. And to establish the last measurability it suffices to consider measurable sets of the form $[0,tau]times B$, $tauin[0,infty)$, $Binmathcal B(mathbb R)$, which generate $mathcal B([0,infty)timesmathbb R)equiv mathcal B([0,infty))otimesmathcal B(mathbb R)$. So look at
$$
{(t,omega)in[0,infty)timesOmega:
(t,X_t(omega))in[0,tau]times B
}
=([0,tau]timesOmega)cap X^{-1}(B).
$$
$X^{-1}(B)$ is in $mathcal P$ by the assumption $X$ is predictable; and $([0,tau]timesOmega)$, too, is in $mathcal P$ as $1_{tin[0,tau]}(t,omega)$ is an adapted caglad process.
answered Jan 18 at 10:42
AddSupAddSup
5101318
$endgroup$
1
$begingroup$
The $Y_t$ process I am concerned with is $Y_t = int_0^t f(s,X_s)ds$
$endgroup$
– user2379888
Jan 21 at 13:29
$begingroup$
@user2379888 Sorry for the mistake... Here's an addendum that will complete the answer: Given the predictability of $f(t,X_t)$ as observed above, $f(t,X_t)$ is then a fortiori progressively measurable. Thus the integral $int_0^tf(s,X_s)ds$ is adapted. (Provided of course that the integral is well-defined. This requires more than the Borel measurability of $f$.) And adaptedness and continuity are more than enough for predictability (as TheBridge noted).
$endgroup$
– AddSup
Jan 21 at 15:00
add a comment |
$begingroup$
I think you only need the Borel measurability of $f$ for the predictability of $Y_t=f(t,X_t)$.
$Y$ can be decomposed into
$$
[0,infty)timesOmegaoverset{phi_X}{rightarrow}[0,infty)timesmathbb R
overset{f}{rightarrow}mathbb R,
$$
so predictability follows once the $mathcal P/mathcal B([0,infty)timesmathbb R)$-measurability of $phi_X$ is established, where $phi_X(t,omega):=(t,X_t(omega))$. And to establish the last measurability it suffices to consider measurable sets of the form $[0,tau]times B$, $tauin[0,infty)$, $Binmathcal B(mathbb R)$, which generate $mathcal B([0,infty)timesmathbb R)equiv mathcal B([0,infty))otimesmathcal B(mathbb R)$. So look at
$$
{(t,omega)in[0,infty)timesOmega:
(t,X_t(omega))in[0,tau]times B
}
=([0,tau]timesOmega)cap X^{-1}(B).
$$
$X^{-1}(B)$ is in $mathcal P$ by the assumption $X$ is predictable; and $([0,tau]timesOmega)$, too, is in $mathcal P$ as $1_{tin[0,tau]}(t,omega)$ is an adapted caglad process.
answered Jan 18 at 10:42
AddSupAddSup
5101318
$endgroup$
1
$begingroup$
The $Y_t$ process I am concerned with is $Y_t = int_0^t f(s,X_s)ds$
$endgroup$
– user2379888
Jan 21 at 13:29
$begingroup$
@user2379888 Sorry for the mistake... Here's an addendum that will complete the answer: Given the predictability of $f(t,X_t)$ as observed above, $f(t,X_t)$ is then a fortiori progressively measurable. Thus the integral $int_0^tf(s,X_s)ds$ is adapted. (Provided of course that the integral is well-defined. This requires more than the Borel measurability of $f$.) And adaptedness and continuity are more than enough for predictability (as TheBridge noted).
$endgroup$
– AddSup
Jan 21 at 15:00
add a comment |
$begingroup$
I think you only need the Borel measurability of $f$ for the predictability of $Y_t=f(t,X_t)$.
$Y$ can be decomposed into
$$
[0,infty)timesOmegaoverset{phi_X}{rightarrow}[0,infty)timesmathbb R
overset{f}{rightarrow}mathbb R,
$$
so predictability follows once the $mathcal P/mathcal B([0,infty)timesmathbb R)$-measurability of $phi_X$ is established, where $phi_X(t,omega):=(t,X_t(omega))$. And to establish the last measurability it suffices to consider measurable sets of the form $[0,tau]times B$, $tauin[0,infty)$, $Binmathcal B(mathbb R)$, which generate $mathcal B([0,infty)timesmathbb R)equiv mathcal B([0,infty))otimesmathcal B(mathbb R)$. So look at
$$
{(t,omega)in[0,infty)timesOmega:
(t,X_t(omega))in[0,tau]times B
}
=([0,tau]timesOmega)cap X^{-1}(B).
$$
$X^{-1}(B)$ is in $mathcal P$ by the assumption $X$ is predictable; and $([0,tau]timesOmega)$, too, is in $mathcal P$ as $1_{tin[0,tau]}(t,omega)$ is an adapted caglad process.
answered Jan 18 at 10:42
AddSupAddSup
5101318
$endgroup$
I think you only need the Borel measurability of $f$ for the predictability of $Y_t=f(t,X_t)$.
$Y$ can be decomposed into
$$
[0,infty)timesOmegaoverset{phi_X}{rightarrow}[0,infty)timesmathbb R
overset{f}{rightarrow}mathbb R,
$$
so predictability follows once the $mathcal P/mathcal B([0,infty)timesmathbb R)$-measurability of $phi_X$ is established, where $phi_X(t,omega):=(t,X_t(omega))$. And to establish the last measurability it suffices to consider measurable sets of the form $[0,tau]times B$, $tauin[0,infty)$, $Binmathcal B(mathbb R)$, which generate $mathcal B([0,infty)timesmathbb R)equiv mathcal B([0,infty))otimesmathcal B(mathbb R)$. So look at
$$
{(t,omega)in[0,infty)timesOmega:
(t,X_t(omega))in[0,tau]times B
}
=([0,tau]timesOmega)cap X^{-1}(B).
$$
$X^{-1}(B)$ is in $mathcal P$ by the assumption $X$ is predictable; and $([0,tau]timesOmega)$, too, is in $mathcal P$ as $1_{tin[0,tau]}(t,omega)$ is an adapted caglad process.
answered Jan 18 at 10:42
AddSupAddSup
5101318
answered Jan 18 at 10:42
AddSupAddSup
5101318
answered Jan 18 at 10:42
AddSupAddSup
5101318
answered Jan 18 at 10:42
AddSupAddSup
5101318
5101318
1
$begingroup$
The $Y_t$ process I am concerned with is $Y_t = int_0^t f(s,X_s)ds$
$endgroup$
– user2379888
Jan 21 at 13:29
$begingroup$
@user2379888 Sorry for the mistake... Here's an addendum that will complete the answer: Given the predictability of $f(t,X_t)$ as observed above, $f(t,X_t)$ is then a fortiori progressively measurable. Thus the integral $int_0^tf(s,X_s)ds$ is adapted. (Provided of course that the integral is well-defined. This requires more than the Borel measurability of $f$.) And adaptedness and continuity are more than enough for predictability (as TheBridge noted).
$endgroup$
– AddSup
Jan 21 at 15:00
add a comment |
1
$begingroup$
The $Y_t$ process I am concerned with is $Y_t = int_0^t f(s,X_s)ds$
$endgroup$
– user2379888
Jan 21 at 13:29
$begingroup$
@user2379888 Sorry for the mistake... Here's an addendum that will complete the answer: Given the predictability of $f(t,X_t)$ as observed above, $f(t,X_t)$ is then a fortiori progressively measurable. Thus the integral $int_0^tf(s,X_s)ds$ is adapted. (Provided of course that the integral is well-defined. This requires more than the Borel measurability of $f$.) And adaptedness and continuity are more than enough for predictability (as TheBridge noted).
$endgroup$
– AddSup
Jan 21 at 15:00
1
1
$begingroup$
The $Y_t$ process I am concerned with is $Y_t = int_0^t f(s,X_s)ds$
$endgroup$
– user2379888
Jan 21 at 13:29
$begingroup$
The $Y_t$ process I am concerned with is $Y_t = int_0^t f(s,X_s)ds$
$endgroup$
– user2379888
Jan 21 at 13:29
$begingroup$
@user2379888 Sorry for the mistake... Here's an addendum that will complete the answer: Given the predictability of $f(t,X_t)$ as observed above, $f(t,X_t)$ is then a fortiori progressively measurable. Thus the integral $int_0^tf(s,X_s)ds$ is adapted. (Provided of course that the integral is well-defined. This requires more than the Borel measurability of $f$.) And adaptedness and continuity are more than enough for predictability (as TheBridge noted).
$endgroup$
– AddSup
Jan 21 at 15:00
$begingroup$
@user2379888 Sorry for the mistake... Here's an addendum that will complete the answer: Given the predictability of $f(t,X_t)$ as observed above, $f(t,X_t)$ is then a fortiori progressively measurable. Thus the integral $int_0^tf(s,X_s)ds$ is adapted. (Provided of course that the integral is well-defined. This requires more than the Borel measurability of $f$.) And adaptedness and continuity are more than enough for predictability (as TheBridge noted).
$endgroup$
– AddSup
Jan 21 at 15:00
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%2f3075760%2fdeterministic-integral-of-a-predictable-process-is-predictable%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
$begingroup$
Hi left continuous (or càg) adapted processes are predictable, see theorem 1 point 6 here : almostsure.wordpress.com/2016/11/22/predictable-processes
$endgroup$
– TheBridge
Jan 21 at 12:25