If $E(X_n^2)<infty$, then for a Martingale $E(X_n^2)<M$ iff $sum_{n=1}^infty...
$begingroup$
Let ${X_n}_{ngeq0}$ be a martingale with $E(X_n^2)<infty$ for all $n$. How to prove that:
$E(X_n^2)<M$ for all $n$, if and only if $sum_{n=1}^infty E[(X_n-X_{n-1})^2]<infty$.
The hunch is to use the optional stopping time theorem, but it's hard to see how to apply it to the case. A tip is very much appreciated.
measure-theory martingales stopping-times
asked Dec 10 '18 at 23:07
DoleDole
907514
$endgroup$
add a comment |
$begingroup$
Let ${X_n}_{ngeq0}$ be a martingale with $E(X_n^2)<infty$ for all $n$. How to prove that:
$E(X_n^2)<M$ for all $n$, if and only if $sum_{n=1}^infty E[(X_n-X_{n-1})^2]<infty$.
The hunch is to use the optional stopping time theorem, but it's hard to see how to apply it to the case. A tip is very much appreciated.
measure-theory martingales stopping-times
asked Dec 10 '18 at 23:07
DoleDole
907514
$endgroup$
add a comment |
$begingroup$
Let ${X_n}_{ngeq0}$ be a martingale with $E(X_n^2)<infty$ for all $n$. How to prove that:
$E(X_n^2)<M$ for all $n$, if and only if $sum_{n=1}^infty E[(X_n-X_{n-1})^2]<infty$.
The hunch is to use the optional stopping time theorem, but it's hard to see how to apply it to the case. A tip is very much appreciated.
measure-theory martingales stopping-times
asked Dec 10 '18 at 23:07
DoleDole
907514
$endgroup$
Let ${X_n}_{ngeq0}$ be a martingale with $E(X_n^2)<infty$ for all $n$. How to prove that:
$E(X_n^2)<M$ for all $n$, if and only if $sum_{n=1}^infty E[(X_n-X_{n-1})^2]<infty$.
The hunch is to use the optional stopping time theorem, but it's hard to see how to apply it to the case. A tip is very much appreciated.
measure-theory martingales stopping-times
measure-theory martingales stopping-times
asked Dec 10 '18 at 23:07
DoleDole
907514
asked Dec 10 '18 at 23:07
DoleDole
907514
asked Dec 10 '18 at 23:07
DoleDole
907514
asked Dec 10 '18 at 23:07
DoleDole
907514
asked Dec 10 '18 at 23:07
DoleDole
907514
907514
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is in fact related to the quadratic variation process of $X_n$. Note that
$$
X_n^2 - sum_{k=0}^n E[(X_k-X_{k-1})^2;|mathcal{F}_{k-1}],quad ngeq0,
$$ is a ${mathcal{F}_n}$-martingale.
answered Dec 10 '18 at 23:15
SongSong
11.1k628
$endgroup$
$begingroup$
Very nice hints, certainly became manageable!
$endgroup$
– Dole
Dec 11 '18 at 1:50
add a comment |
$begingroup$
By the martingale property you can show
$$E[(X_n - X_{n-1})^2] = E[X_n^2] - E[X_{n-1}^2].$$
answered Dec 10 '18 at 23:11
angryavianangryavian
40.6k23380
$endgroup$
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%2f3034618%2fif-ex-n2-infty-then-for-a-martingale-ex-n2m-iff-sum-n-1-infty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is in fact related to the quadratic variation process of $X_n$. Note that
$$
X_n^2 - sum_{k=0}^n E[(X_k-X_{k-1})^2;|mathcal{F}_{k-1}],quad ngeq0,
$$ is a ${mathcal{F}_n}$-martingale.
answered Dec 10 '18 at 23:15
SongSong
11.1k628
$endgroup$
$begingroup$
Very nice hints, certainly became manageable!
$endgroup$
– Dole
Dec 11 '18 at 1:50
add a comment |
$begingroup$
It is in fact related to the quadratic variation process of $X_n$. Note that
$$
X_n^2 - sum_{k=0}^n E[(X_k-X_{k-1})^2;|mathcal{F}_{k-1}],quad ngeq0,
$$ is a ${mathcal{F}_n}$-martingale.
answered Dec 10 '18 at 23:15
SongSong
11.1k628
$endgroup$
$begingroup$
Very nice hints, certainly became manageable!
$endgroup$
– Dole
Dec 11 '18 at 1:50
add a comment |
$begingroup$
It is in fact related to the quadratic variation process of $X_n$. Note that
$$
X_n^2 - sum_{k=0}^n E[(X_k-X_{k-1})^2;|mathcal{F}_{k-1}],quad ngeq0,
$$ is a ${mathcal{F}_n}$-martingale.
answered Dec 10 '18 at 23:15
SongSong
11.1k628
$endgroup$
It is in fact related to the quadratic variation process of $X_n$. Note that
$$
X_n^2 - sum_{k=0}^n E[(X_k-X_{k-1})^2;|mathcal{F}_{k-1}],quad ngeq0,
$$ is a ${mathcal{F}_n}$-martingale.
answered Dec 10 '18 at 23:15
SongSong
11.1k628
answered Dec 10 '18 at 23:15
SongSong
11.1k628
answered Dec 10 '18 at 23:15
SongSong
11.1k628
answered Dec 10 '18 at 23:15
SongSong
11.1k628
11.1k628
$begingroup$
Very nice hints, certainly became manageable!
$endgroup$
– Dole
Dec 11 '18 at 1:50
add a comment |
$begingroup$
Very nice hints, certainly became manageable!
$endgroup$
– Dole
Dec 11 '18 at 1:50
$begingroup$
Very nice hints, certainly became manageable!
$endgroup$
– Dole
Dec 11 '18 at 1:50
$begingroup$
Very nice hints, certainly became manageable!
$endgroup$
– Dole
Dec 11 '18 at 1:50
add a comment |
$begingroup$
By the martingale property you can show
$$E[(X_n - X_{n-1})^2] = E[X_n^2] - E[X_{n-1}^2].$$
answered Dec 10 '18 at 23:11
angryavianangryavian
40.6k23380
$endgroup$
add a comment |
$begingroup$
By the martingale property you can show
$$E[(X_n - X_{n-1})^2] = E[X_n^2] - E[X_{n-1}^2].$$
answered Dec 10 '18 at 23:11
angryavianangryavian
40.6k23380
$endgroup$
add a comment |
$begingroup$
By the martingale property you can show
$$E[(X_n - X_{n-1})^2] = E[X_n^2] - E[X_{n-1}^2].$$
answered Dec 10 '18 at 23:11
angryavianangryavian
40.6k23380
$endgroup$
By the martingale property you can show
$$E[(X_n - X_{n-1})^2] = E[X_n^2] - E[X_{n-1}^2].$$
answered Dec 10 '18 at 23:11
angryavianangryavian
40.6k23380
answered Dec 10 '18 at 23:11
angryavianangryavian
40.6k23380
answered Dec 10 '18 at 23:11
angryavianangryavian
40.6k23380
answered Dec 10 '18 at 23:11
angryavianangryavian
40.6k23380
40.6k23380
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%2f3034618%2fif-ex-n2-infty-then-for-a-martingale-ex-n2m-iff-sum-n-1-infty%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
