Stopping time with finite expectation
up vote
1
down vote
favorite
Given a filtered space $left(S, mathcal{Sigma}, {mathcal{Sigma_n: nge 0}}, mathbb{P}right)$, and a stopping time $tau: S to {0, 1,2, ldots}$ with respect to ${mathcal{Sigma_n: nge 0}}$, suppose that there exist $minmathbb{N}$ and $epsilon$ such that, for every $nge 0$:
$$mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] > epsilon quad text{a.s.}.$$
Prove that $mathbb{E}[tau]<infty$.
I think one should use $E[tau] < infty Longleftrightarrow sumlimits_{n=0}^infty mathbb{P}(tau ge n)< infty.$
However, I cannot find a tight enough bound for $mathbb{P}(tau ge n)$. The best I can do is
$$mathbb{P}(tau > n+m) = 1- mathbb{P}(tau le n+m) = 1 - mathbb{E}left[mathbb{I}_{tau le n+m}right] = 1-mathbb{E}mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] le 1- mathbb{E}[epsilon] =1 - epsilon.$$
A hint is provided:
For every $k = 1, 2, 3, ldots,$ $$mathbb{P}(tau > (k+1)m) = mathbb{P}(tau > (k+1)m quad text{and} quad tau > km).$$
Use induction to show that $mathbb{P}(tau > km) le (1-epsilon)^k.$
probability-theory measure-theory stopping-times
add a comment |
up vote
1
down vote
favorite
Given a filtered space $left(S, mathcal{Sigma}, {mathcal{Sigma_n: nge 0}}, mathbb{P}right)$, and a stopping time $tau: S to {0, 1,2, ldots}$ with respect to ${mathcal{Sigma_n: nge 0}}$, suppose that there exist $minmathbb{N}$ and $epsilon$ such that, for every $nge 0$:
$$mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] > epsilon quad text{a.s.}.$$
Prove that $mathbb{E}[tau]<infty$.
I think one should use $E[tau] < infty Longleftrightarrow sumlimits_{n=0}^infty mathbb{P}(tau ge n)< infty.$
However, I cannot find a tight enough bound for $mathbb{P}(tau ge n)$. The best I can do is
$$mathbb{P}(tau > n+m) = 1- mathbb{P}(tau le n+m) = 1 - mathbb{E}left[mathbb{I}_{tau le n+m}right] = 1-mathbb{E}mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] le 1- mathbb{E}[epsilon] =1 - epsilon.$$
A hint is provided:
For every $k = 1, 2, 3, ldots,$ $$mathbb{P}(tau > (k+1)m) = mathbb{P}(tau > (k+1)m quad text{and} quad tau > km).$$
Use induction to show that $mathbb{P}(tau > km) le (1-epsilon)^k.$
probability-theory measure-theory stopping-times
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a filtered space $left(S, mathcal{Sigma}, {mathcal{Sigma_n: nge 0}}, mathbb{P}right)$, and a stopping time $tau: S to {0, 1,2, ldots}$ with respect to ${mathcal{Sigma_n: nge 0}}$, suppose that there exist $minmathbb{N}$ and $epsilon$ such that, for every $nge 0$:
$$mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] > epsilon quad text{a.s.}.$$
Prove that $mathbb{E}[tau]<infty$.
I think one should use $E[tau] < infty Longleftrightarrow sumlimits_{n=0}^infty mathbb{P}(tau ge n)< infty.$
However, I cannot find a tight enough bound for $mathbb{P}(tau ge n)$. The best I can do is
$$mathbb{P}(tau > n+m) = 1- mathbb{P}(tau le n+m) = 1 - mathbb{E}left[mathbb{I}_{tau le n+m}right] = 1-mathbb{E}mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] le 1- mathbb{E}[epsilon] =1 - epsilon.$$
A hint is provided:
For every $k = 1, 2, 3, ldots,$ $$mathbb{P}(tau > (k+1)m) = mathbb{P}(tau > (k+1)m quad text{and} quad tau > km).$$
Use induction to show that $mathbb{P}(tau > km) le (1-epsilon)^k.$
probability-theory measure-theory stopping-times
Given a filtered space $left(S, mathcal{Sigma}, {mathcal{Sigma_n: nge 0}}, mathbb{P}right)$, and a stopping time $tau: S to {0, 1,2, ldots}$ with respect to ${mathcal{Sigma_n: nge 0}}$, suppose that there exist $minmathbb{N}$ and $epsilon$ such that, for every $nge 0$:
$$mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] > epsilon quad text{a.s.}.$$
Prove that $mathbb{E}[tau]<infty$.
I think one should use $E[tau] < infty Longleftrightarrow sumlimits_{n=0}^infty mathbb{P}(tau ge n)< infty.$
However, I cannot find a tight enough bound for $mathbb{P}(tau ge n)$. The best I can do is
$$mathbb{P}(tau > n+m) = 1- mathbb{P}(tau le n+m) = 1 - mathbb{E}left[mathbb{I}_{tau le n+m}right] = 1-mathbb{E}mathbb{E}left[mathbb{I}_{tau le n+m} | mathcal{Sigma_n}right] le 1- mathbb{E}[epsilon] =1 - epsilon.$$
A hint is provided:
For every $k = 1, 2, 3, ldots,$ $$mathbb{P}(tau > (k+1)m) = mathbb{P}(tau > (k+1)m quad text{and} quad tau > km).$$
Use induction to show that $mathbb{P}(tau > km) le (1-epsilon)^k.$
probability-theory measure-theory stopping-times
probability-theory measure-theory stopping-times
edited Nov 29 at 15:05
asked Nov 28 at 22:00
math_enthuthiast
16910
16910
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Since ${tau>km} in mathcal{Sigma_{km}}:$
$$mathbb{P}(tau le (k+1)m, tau >km) =mathbb{E}[I_{tau le (k+1)m, tau >km}]= mathbb{E}mathbb{E}[I_{tau le (k+1)m, tau >km}|mathcal{Sigma_{km}}] = mathbb{E}mathbb{E}[I_{tau le (k+1)m} I_{tau >km}|mathcal{Sigma_{km}}] = mathbb{E}left[I_{tau >km}mathbb{E}[I_{tau le (k+1)m} |mathcal{Sigma_{km}}]right] > mathbb{E}left[I_{tau >km}epsilonright] = mathbb{P}(tau>km)epsilon.$$
On the other hand
$$
mathbb{P}(tau le (k+1)m, tau >km) = mathbb{P}(km < tau le (k+1)m ) = mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m),
$$
hence
$$
mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m) > mathbb{P}(tau>km)epsilon,
$$
or
$$
mathbb{P}(tau>(k+1)m) le (1-epsilon)mathbb{P}(tau>km).
$$
Induction gives the desired result.
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%2f3017784%2fstopping-time-with-finite-expectation%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
up vote
1
down vote
accepted
Since ${tau>km} in mathcal{Sigma_{km}}:$
$$mathbb{P}(tau le (k+1)m, tau >km) =mathbb{E}[I_{tau le (k+1)m, tau >km}]= mathbb{E}mathbb{E}[I_{tau le (k+1)m, tau >km}|mathcal{Sigma_{km}}] = mathbb{E}mathbb{E}[I_{tau le (k+1)m} I_{tau >km}|mathcal{Sigma_{km}}] = mathbb{E}left[I_{tau >km}mathbb{E}[I_{tau le (k+1)m} |mathcal{Sigma_{km}}]right] > mathbb{E}left[I_{tau >km}epsilonright] = mathbb{P}(tau>km)epsilon.$$
On the other hand
$$
mathbb{P}(tau le (k+1)m, tau >km) = mathbb{P}(km < tau le (k+1)m ) = mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m),
$$
hence
$$
mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m) > mathbb{P}(tau>km)epsilon,
$$
or
$$
mathbb{P}(tau>(k+1)m) le (1-epsilon)mathbb{P}(tau>km).
$$
Induction gives the desired result.
add a comment |
up vote
1
down vote
accepted
Since ${tau>km} in mathcal{Sigma_{km}}:$
$$mathbb{P}(tau le (k+1)m, tau >km) =mathbb{E}[I_{tau le (k+1)m, tau >km}]= mathbb{E}mathbb{E}[I_{tau le (k+1)m, tau >km}|mathcal{Sigma_{km}}] = mathbb{E}mathbb{E}[I_{tau le (k+1)m} I_{tau >km}|mathcal{Sigma_{km}}] = mathbb{E}left[I_{tau >km}mathbb{E}[I_{tau le (k+1)m} |mathcal{Sigma_{km}}]right] > mathbb{E}left[I_{tau >km}epsilonright] = mathbb{P}(tau>km)epsilon.$$
On the other hand
$$
mathbb{P}(tau le (k+1)m, tau >km) = mathbb{P}(km < tau le (k+1)m ) = mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m),
$$
hence
$$
mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m) > mathbb{P}(tau>km)epsilon,
$$
or
$$
mathbb{P}(tau>(k+1)m) le (1-epsilon)mathbb{P}(tau>km).
$$
Induction gives the desired result.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since ${tau>km} in mathcal{Sigma_{km}}:$
$$mathbb{P}(tau le (k+1)m, tau >km) =mathbb{E}[I_{tau le (k+1)m, tau >km}]= mathbb{E}mathbb{E}[I_{tau le (k+1)m, tau >km}|mathcal{Sigma_{km}}] = mathbb{E}mathbb{E}[I_{tau le (k+1)m} I_{tau >km}|mathcal{Sigma_{km}}] = mathbb{E}left[I_{tau >km}mathbb{E}[I_{tau le (k+1)m} |mathcal{Sigma_{km}}]right] > mathbb{E}left[I_{tau >km}epsilonright] = mathbb{P}(tau>km)epsilon.$$
On the other hand
$$
mathbb{P}(tau le (k+1)m, tau >km) = mathbb{P}(km < tau le (k+1)m ) = mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m),
$$
hence
$$
mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m) > mathbb{P}(tau>km)epsilon,
$$
or
$$
mathbb{P}(tau>(k+1)m) le (1-epsilon)mathbb{P}(tau>km).
$$
Induction gives the desired result.
Since ${tau>km} in mathcal{Sigma_{km}}:$
$$mathbb{P}(tau le (k+1)m, tau >km) =mathbb{E}[I_{tau le (k+1)m, tau >km}]= mathbb{E}mathbb{E}[I_{tau le (k+1)m, tau >km}|mathcal{Sigma_{km}}] = mathbb{E}mathbb{E}[I_{tau le (k+1)m} I_{tau >km}|mathcal{Sigma_{km}}] = mathbb{E}left[I_{tau >km}mathbb{E}[I_{tau le (k+1)m} |mathcal{Sigma_{km}}]right] > mathbb{E}left[I_{tau >km}epsilonright] = mathbb{P}(tau>km)epsilon.$$
On the other hand
$$
mathbb{P}(tau le (k+1)m, tau >km) = mathbb{P}(km < tau le (k+1)m ) = mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m),
$$
hence
$$
mathbb{P}(tau>km) - mathbb{P}(tau>(k+1)m) > mathbb{P}(tau>km)epsilon,
$$
or
$$
mathbb{P}(tau>(k+1)m) le (1-epsilon)mathbb{P}(tau>km).
$$
Induction gives the desired result.
edited Nov 29 at 15:05
answered Nov 29 at 0:56
math_enthuthiast
16910
16910
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3017784%2fstopping-time-with-finite-expectation%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