Mixing
Discrete-time martingale
$begingroup$
I am self-learning SDE from the book, Diffusions, Markov Processes, and Martingales, 2nd ed., by Rogers and Williams. There is one proof that puzzles me. It is the proof of Theorem 59.6.
The setting is as follows:
Let $mathcal{F}_n$ be an increasing sequence of $sigma$-algebras. let $xi$ be $L^1(Omega)$ and $xi_n = E(xi | mathcal{F}_n)$. Let $T$ be a stopping time and define $eta = E(xi | mathcal{F}_T)$.
Then the book says the following holds:
$E(eta|mathcal{F}_n) = xi_{n wedge T}$ a.s. for every $n in mathbb{N}$.
How does one prove this relation?
probability
edited Jan 17 at 17:49
hardmath
29k95299
asked Jan 17 at 17:43
W. S.W. S.
82
$endgroup$
add a comment |
$begingroup$
I am self-learning SDE from the book, Diffusions, Markov Processes, and Martingales, 2nd ed., by Rogers and Williams. There is one proof that puzzles me. It is the proof of Theorem 59.6.
The setting is as follows:
Let $mathcal{F}_n$ be an increasing sequence of $sigma$-algebras. let $xi$ be $L^1(Omega)$ and $xi_n = E(xi | mathcal{F}_n)$. Let $T$ be a stopping time and define $eta = E(xi | mathcal{F}_T)$.
Then the book says the following holds:
$E(eta|mathcal{F}_n) = xi_{n wedge T}$ a.s. for every $n in mathbb{N}$.
How does one prove this relation?
probability
edited Jan 17 at 17:49
hardmath
29k95299
asked Jan 17 at 17:43
W. S.W. S.
82
$endgroup$
add a comment |
$begingroup$
I am self-learning SDE from the book, Diffusions, Markov Processes, and Martingales, 2nd ed., by Rogers and Williams. There is one proof that puzzles me. It is the proof of Theorem 59.6.
The setting is as follows:
Let $mathcal{F}_n$ be an increasing sequence of $sigma$-algebras. let $xi$ be $L^1(Omega)$ and $xi_n = E(xi | mathcal{F}_n)$. Let $T$ be a stopping time and define $eta = E(xi | mathcal{F}_T)$.
Then the book says the following holds:
$E(eta|mathcal{F}_n) = xi_{n wedge T}$ a.s. for every $n in mathbb{N}$.
How does one prove this relation?
probability
edited Jan 17 at 17:49
hardmath
29k95299
asked Jan 17 at 17:43
W. S.W. S.
82
$endgroup$
I am self-learning SDE from the book, Diffusions, Markov Processes, and Martingales, 2nd ed., by Rogers and Williams. There is one proof that puzzles me. It is the proof of Theorem 59.6.
The setting is as follows:
Let $mathcal{F}_n$ be an increasing sequence of $sigma$-algebras. let $xi$ be $L^1(Omega)$ and $xi_n = E(xi | mathcal{F}_n)$. Let $T$ be a stopping time and define $eta = E(xi | mathcal{F}_T)$.
Then the book says the following holds:
$E(eta|mathcal{F}_n) = xi_{n wedge T}$ a.s. for every $n in mathbb{N}$.
How does one prove this relation?
probability
probability
edited Jan 17 at 17:49
hardmath
29k95299
asked Jan 17 at 17:43
W. S.W. S.
82
edited Jan 17 at 17:49
hardmath
29k95299
asked Jan 17 at 17:43
W. S.W. S.
82
edited Jan 17 at 17:49
hardmath
29k95299
edited Jan 17 at 17:49
hardmath
29k95299
edited Jan 17 at 17:49
hardmath
29k95299
29k95299
asked Jan 17 at 17:43
W. S.W. S.
82
asked Jan 17 at 17:43
W. S.W. S.
82
asked Jan 17 at 17:43
W. S.W. S.
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $Z_n:=mathsf{E}[etamid mathcal{F}_n]$. Then since $Z_n$ and $xi_{nwedge T}$ are UI martingales converging a.s. to $xi_T$, by Theorem 50.1,
$$
Z_n=mathsf{E}[xi_Tmid mathcal{F}_n]=xi_{nwedge T} quadtext{a.s.}
$$
answered Jan 17 at 18:22
d.k.o.d.k.o.
9,775628
$endgroup$
$begingroup$
Ah, yes, Theorem 50.1, that is what I missed. Thank you very much!
$endgroup$
– W. S.
Jan 17 at 23:32
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%2f3077279%2fdiscrete-time-martingale%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
$begingroup$
Let $Z_n:=mathsf{E}[etamid mathcal{F}_n]$. Then since $Z_n$ and $xi_{nwedge T}$ are UI martingales converging a.s. to $xi_T$, by Theorem 50.1,
$$
Z_n=mathsf{E}[xi_Tmid mathcal{F}_n]=xi_{nwedge T} quadtext{a.s.}
$$
answered Jan 17 at 18:22
d.k.o.d.k.o.
9,775628
$endgroup$
$begingroup$
Ah, yes, Theorem 50.1, that is what I missed. Thank you very much!
$endgroup$
– W. S.
Jan 17 at 23:32
add a comment |
$begingroup$
Let $Z_n:=mathsf{E}[etamid mathcal{F}_n]$. Then since $Z_n$ and $xi_{nwedge T}$ are UI martingales converging a.s. to $xi_T$, by Theorem 50.1,
$$
Z_n=mathsf{E}[xi_Tmid mathcal{F}_n]=xi_{nwedge T} quadtext{a.s.}
$$
answered Jan 17 at 18:22
d.k.o.d.k.o.
9,775628
$endgroup$
$begingroup$
Ah, yes, Theorem 50.1, that is what I missed. Thank you very much!
$endgroup$
– W. S.
Jan 17 at 23:32
add a comment |
$begingroup$
Let $Z_n:=mathsf{E}[etamid mathcal{F}_n]$. Then since $Z_n$ and $xi_{nwedge T}$ are UI martingales converging a.s. to $xi_T$, by Theorem 50.1,
$$
Z_n=mathsf{E}[xi_Tmid mathcal{F}_n]=xi_{nwedge T} quadtext{a.s.}
$$
answered Jan 17 at 18:22
d.k.o.d.k.o.
9,775628
$endgroup$
Let $Z_n:=mathsf{E}[etamid mathcal{F}_n]$. Then since $Z_n$ and $xi_{nwedge T}$ are UI martingales converging a.s. to $xi_T$, by Theorem 50.1,
$$
Z_n=mathsf{E}[xi_Tmid mathcal{F}_n]=xi_{nwedge T} quadtext{a.s.}
$$
answered Jan 17 at 18:22
d.k.o.d.k.o.
9,775628
answered Jan 17 at 18:22
d.k.o.d.k.o.
9,775628
answered Jan 17 at 18:22
d.k.o.d.k.o.
9,775628
answered Jan 17 at 18:22
d.k.o.d.k.o.
9,775628
9,775628
$begingroup$
Ah, yes, Theorem 50.1, that is what I missed. Thank you very much!
$endgroup$
– W. S.
Jan 17 at 23:32
add a comment |
$begingroup$
Ah, yes, Theorem 50.1, that is what I missed. Thank you very much!
$endgroup$
– W. S.
Jan 17 at 23:32
$begingroup$
Ah, yes, Theorem 50.1, that is what I missed. Thank you very much!
$endgroup$
– W. S.
Jan 17 at 23:32
$begingroup$
Ah, yes, Theorem 50.1, that is what I missed. Thank you very much!
$endgroup$
– W. S.
Jan 17 at 23:32
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%2f3077279%2fdiscrete-time-martingale%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
