Equivalence of independence and a conditional expectation equality
$begingroup$
Let $E_1$, $E_2$ be Polish, let $D_{E_2}([0,infty])$ be the space of cadlag functions with values in $E_2$ and set $mathcal{F}_t^Y$ to be the $sigma$-algebra generated by $Y_s$, $s leq t$, similarly for $X$ and finally $mathcal{F}_t^{X,Y} := mathcal{F}_t^Y vee mathcal{F}_t^X$. Let $X$ be an $E_1$-valued stochastic process and $Y$ an $E_2$-valued stochastic process.
I am looking for a proof of the following statement:
Assume $Y$ to have independent increments and paths in $D_{E_2}([0,infty])$. Then the following are equivalent.
(i) $(Y_{t+s}-Y_t)_{s geq 0}$ is independent of $mathcal{F}_t^{X,Y}$ for every $t geq 0$.
(ii) $mathbb{E}[h(Y)|mathcal{F}_t^{X,Y}] = mathbb{E}[h(Y)|mathcal{F}_t^{Y}]$ for every $t geq 0$ and every measurable and bounded $h:D_{E_i}([0,infty]) to mathbb{R}$.
Intuively, at least $(i) implies (ii)$ "feels" right and easy to believe to me. However, I could not come up with a rigorous proof. I would appreciate your help very much!
measure-theory stochastic-processes conditional-expectation independence
$endgroup$
add a comment |
$begingroup$
Let $E_1$, $E_2$ be Polish, let $D_{E_2}([0,infty])$ be the space of cadlag functions with values in $E_2$ and set $mathcal{F}_t^Y$ to be the $sigma$-algebra generated by $Y_s$, $s leq t$, similarly for $X$ and finally $mathcal{F}_t^{X,Y} := mathcal{F}_t^Y vee mathcal{F}_t^X$. Let $X$ be an $E_1$-valued stochastic process and $Y$ an $E_2$-valued stochastic process.
I am looking for a proof of the following statement:
Assume $Y$ to have independent increments and paths in $D_{E_2}([0,infty])$. Then the following are equivalent.
(i) $(Y_{t+s}-Y_t)_{s geq 0}$ is independent of $mathcal{F}_t^{X,Y}$ for every $t geq 0$.
(ii) $mathbb{E}[h(Y)|mathcal{F}_t^{X,Y}] = mathbb{E}[h(Y)|mathcal{F}_t^{Y}]$ for every $t geq 0$ and every measurable and bounded $h:D_{E_i}([0,infty]) to mathbb{R}$.
Intuively, at least $(i) implies (ii)$ "feels" right and easy to believe to me. However, I could not come up with a rigorous proof. I would appreciate your help very much!
measure-theory stochastic-processes conditional-expectation independence
$endgroup$
add a comment |
$begingroup$
Let $E_1$, $E_2$ be Polish, let $D_{E_2}([0,infty])$ be the space of cadlag functions with values in $E_2$ and set $mathcal{F}_t^Y$ to be the $sigma$-algebra generated by $Y_s$, $s leq t$, similarly for $X$ and finally $mathcal{F}_t^{X,Y} := mathcal{F}_t^Y vee mathcal{F}_t^X$. Let $X$ be an $E_1$-valued stochastic process and $Y$ an $E_2$-valued stochastic process.
I am looking for a proof of the following statement:
Assume $Y$ to have independent increments and paths in $D_{E_2}([0,infty])$. Then the following are equivalent.
(i) $(Y_{t+s}-Y_t)_{s geq 0}$ is independent of $mathcal{F}_t^{X,Y}$ for every $t geq 0$.
(ii) $mathbb{E}[h(Y)|mathcal{F}_t^{X,Y}] = mathbb{E}[h(Y)|mathcal{F}_t^{Y}]$ for every $t geq 0$ and every measurable and bounded $h:D_{E_i}([0,infty]) to mathbb{R}$.
Intuively, at least $(i) implies (ii)$ "feels" right and easy to believe to me. However, I could not come up with a rigorous proof. I would appreciate your help very much!
measure-theory stochastic-processes conditional-expectation independence
$endgroup$
Let $E_1$, $E_2$ be Polish, let $D_{E_2}([0,infty])$ be the space of cadlag functions with values in $E_2$ and set $mathcal{F}_t^Y$ to be the $sigma$-algebra generated by $Y_s$, $s leq t$, similarly for $X$ and finally $mathcal{F}_t^{X,Y} := mathcal{F}_t^Y vee mathcal{F}_t^X$. Let $X$ be an $E_1$-valued stochastic process and $Y$ an $E_2$-valued stochastic process.
I am looking for a proof of the following statement:
Assume $Y$ to have independent increments and paths in $D_{E_2}([0,infty])$. Then the following are equivalent.
(i) $(Y_{t+s}-Y_t)_{s geq 0}$ is independent of $mathcal{F}_t^{X,Y}$ for every $t geq 0$.
(ii) $mathbb{E}[h(Y)|mathcal{F}_t^{X,Y}] = mathbb{E}[h(Y)|mathcal{F}_t^{Y}]$ for every $t geq 0$ and every measurable and bounded $h:D_{E_i}([0,infty]) to mathbb{R}$.
Intuively, at least $(i) implies (ii)$ "feels" right and easy to believe to me. However, I could not come up with a rigorous proof. I would appreciate your help very much!
measure-theory stochastic-processes conditional-expectation independence
measure-theory stochastic-processes conditional-expectation independence
asked Nov 29 '18 at 16:01
MarcoMarco
485
485
add a comment |
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%2f3018826%2fequivalence-of-independence-and-a-conditional-expectation-equality%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%2f3018826%2fequivalence-of-independence-and-a-conditional-expectation-equality%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