Prove the upper bound of expectations using almost sure convergence
$begingroup$
Let $(X_n)_{nin mathbb{N}}$ be a sequence of iid random variables. Given that $X_n$ converge almost surely to $x_0in [m, M]$, where $0<m<M<1$, I try to find the upper bounds of $mathbb{E}left(dfrac{1}{X_n}right)$ and $mathbb{E}left(dfrac{1}{1-X_n}right)$.
My calculations:
Since $X_n underset{nto +infty}{longrightarrow} x_0$ a.s, there exists $epsilon$ such that $X_n ge x_0 -epsilon ge m - epsilon$. If we choose $epsilon = dfrac{m}{2}$, we obtain $X_n ge dfrac{m}{2}$. We get $dfrac{1}{X_n} le dfrac{2}{m}$ a.s.
Similarly, using again the property $X_n underset{nto +infty}{longrightarrow} x_0$ a.s, there exists $delta$ such that $X_n le x_0 + delta le 1 - m + delta$, this implies $1-X_n ge m - delta$. Therefore if we choose $delta = m/2$, we obtain $dfrac{1}{1-X_n} le dfrac{2}{m}$ a.s.
Then we can conclude that for $n$ large enough
$$
mathbb{E}left(dfrac{1}{X_n}right) le frac{2}{m}quadtext{ and }quad mathbb{E}left(dfrac{1}{1-X_n}right) le frac{2}{m}.
$$
However, I have a doubt about my results. Indeed, the upper bounds of $1/X_n$ and $1/(1-X_n)$ are only true for $n ge N(omega)$ (where $omegainOmega$ of the probability space), so we have also a dependence on $omega$ and taking the expectations over $Omega$ makes a problem!! Furthermore, when we say the upper bounds of the expectations are valid for "$n$ large enough", this seems vague. But I'm not sure. So I would be appreciated for any comment of suggestion that helps to make clear this point. Thank you very much.
probability probability-theory convergence almost-everywhere
asked Dec 3 '18 at 21:39
EmilyEmily
61
$endgroup$
add a comment |
$begingroup$
Let $(X_n)_{nin mathbb{N}}$ be a sequence of iid random variables. Given that $X_n$ converge almost surely to $x_0in [m, M]$, where $0<m<M<1$, I try to find the upper bounds of $mathbb{E}left(dfrac{1}{X_n}right)$ and $mathbb{E}left(dfrac{1}{1-X_n}right)$.
My calculations:
Since $X_n underset{nto +infty}{longrightarrow} x_0$ a.s, there exists $epsilon$ such that $X_n ge x_0 -epsilon ge m - epsilon$. If we choose $epsilon = dfrac{m}{2}$, we obtain $X_n ge dfrac{m}{2}$. We get $dfrac{1}{X_n} le dfrac{2}{m}$ a.s.
Similarly, using again the property $X_n underset{nto +infty}{longrightarrow} x_0$ a.s, there exists $delta$ such that $X_n le x_0 + delta le 1 - m + delta$, this implies $1-X_n ge m - delta$. Therefore if we choose $delta = m/2$, we obtain $dfrac{1}{1-X_n} le dfrac{2}{m}$ a.s.
Then we can conclude that for $n$ large enough
$$
mathbb{E}left(dfrac{1}{X_n}right) le frac{2}{m}quadtext{ and }quad mathbb{E}left(dfrac{1}{1-X_n}right) le frac{2}{m}.
$$
However, I have a doubt about my results. Indeed, the upper bounds of $1/X_n$ and $1/(1-X_n)$ are only true for $n ge N(omega)$ (where $omegainOmega$ of the probability space), so we have also a dependence on $omega$ and taking the expectations over $Omega$ makes a problem!! Furthermore, when we say the upper bounds of the expectations are valid for "$n$ large enough", this seems vague. But I'm not sure. So I would be appreciated for any comment of suggestion that helps to make clear this point. Thank you very much.
probability probability-theory convergence almost-everywhere
asked Dec 3 '18 at 21:39
EmilyEmily
61
$endgroup$
$begingroup$
Try Fatour's lemma better.
$endgroup$
– Will M.
Dec 3 '18 at 22:18
1
$begingroup$
Since the sequence converges to a constant, then the elements of the sequence cannot be identically distributed. Independent, they may.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:40
1
$begingroup$
The fact the the sequence ${X_n}$ converges to the constant $x_0$ says nothing at all about what happens to the elements of the sequence, their domain, support, moments, etc. So really these bounds cannot be determined at all.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:47
add a comment |
$begingroup$
Let $(X_n)_{nin mathbb{N}}$ be a sequence of iid random variables. Given that $X_n$ converge almost surely to $x_0in [m, M]$, where $0<m<M<1$, I try to find the upper bounds of $mathbb{E}left(dfrac{1}{X_n}right)$ and $mathbb{E}left(dfrac{1}{1-X_n}right)$.
My calculations:
Since $X_n underset{nto +infty}{longrightarrow} x_0$ a.s, there exists $epsilon$ such that $X_n ge x_0 -epsilon ge m - epsilon$. If we choose $epsilon = dfrac{m}{2}$, we obtain $X_n ge dfrac{m}{2}$. We get $dfrac{1}{X_n} le dfrac{2}{m}$ a.s.
Similarly, using again the property $X_n underset{nto +infty}{longrightarrow} x_0$ a.s, there exists $delta$ such that $X_n le x_0 + delta le 1 - m + delta$, this implies $1-X_n ge m - delta$. Therefore if we choose $delta = m/2$, we obtain $dfrac{1}{1-X_n} le dfrac{2}{m}$ a.s.
Then we can conclude that for $n$ large enough
$$
mathbb{E}left(dfrac{1}{X_n}right) le frac{2}{m}quadtext{ and }quad mathbb{E}left(dfrac{1}{1-X_n}right) le frac{2}{m}.
$$
However, I have a doubt about my results. Indeed, the upper bounds of $1/X_n$ and $1/(1-X_n)$ are only true for $n ge N(omega)$ (where $omegainOmega$ of the probability space), so we have also a dependence on $omega$ and taking the expectations over $Omega$ makes a problem!! Furthermore, when we say the upper bounds of the expectations are valid for "$n$ large enough", this seems vague. But I'm not sure. So I would be appreciated for any comment of suggestion that helps to make clear this point. Thank you very much.
probability probability-theory convergence almost-everywhere
asked Dec 3 '18 at 21:39
EmilyEmily
61
$endgroup$
Let $(X_n)_{nin mathbb{N}}$ be a sequence of iid random variables. Given that $X_n$ converge almost surely to $x_0in [m, M]$, where $0<m<M<1$, I try to find the upper bounds of $mathbb{E}left(dfrac{1}{X_n}right)$ and $mathbb{E}left(dfrac{1}{1-X_n}right)$.
My calculations:
Since $X_n underset{nto +infty}{longrightarrow} x_0$ a.s, there exists $epsilon$ such that $X_n ge x_0 -epsilon ge m - epsilon$. If we choose $epsilon = dfrac{m}{2}$, we obtain $X_n ge dfrac{m}{2}$. We get $dfrac{1}{X_n} le dfrac{2}{m}$ a.s.
Similarly, using again the property $X_n underset{nto +infty}{longrightarrow} x_0$ a.s, there exists $delta$ such that $X_n le x_0 + delta le 1 - m + delta$, this implies $1-X_n ge m - delta$. Therefore if we choose $delta = m/2$, we obtain $dfrac{1}{1-X_n} le dfrac{2}{m}$ a.s.
Then we can conclude that for $n$ large enough
$$
mathbb{E}left(dfrac{1}{X_n}right) le frac{2}{m}quadtext{ and }quad mathbb{E}left(dfrac{1}{1-X_n}right) le frac{2}{m}.
$$
However, I have a doubt about my results. Indeed, the upper bounds of $1/X_n$ and $1/(1-X_n)$ are only true for $n ge N(omega)$ (where $omegainOmega$ of the probability space), so we have also a dependence on $omega$ and taking the expectations over $Omega$ makes a problem!! Furthermore, when we say the upper bounds of the expectations are valid for "$n$ large enough", this seems vague. But I'm not sure. So I would be appreciated for any comment of suggestion that helps to make clear this point. Thank you very much.
probability probability-theory convergence almost-everywhere
probability probability-theory convergence almost-everywhere
asked Dec 3 '18 at 21:39
EmilyEmily
61
asked Dec 3 '18 at 21:39
EmilyEmily
61
asked Dec 3 '18 at 21:39
EmilyEmily
61
asked Dec 3 '18 at 21:39
EmilyEmily
61
asked Dec 3 '18 at 21:39
EmilyEmily
61
61
$begingroup$
Try Fatour's lemma better.
$endgroup$
– Will M.
Dec 3 '18 at 22:18
1
$begingroup$
Since the sequence converges to a constant, then the elements of the sequence cannot be identically distributed. Independent, they may.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:40
1
$begingroup$
The fact the the sequence ${X_n}$ converges to the constant $x_0$ says nothing at all about what happens to the elements of the sequence, their domain, support, moments, etc. So really these bounds cannot be determined at all.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:47
add a comment |
$begingroup$
Try Fatour's lemma better.
$endgroup$
– Will M.
Dec 3 '18 at 22:18
1
$begingroup$
Since the sequence converges to a constant, then the elements of the sequence cannot be identically distributed. Independent, they may.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:40
1
$begingroup$
The fact the the sequence ${X_n}$ converges to the constant $x_0$ says nothing at all about what happens to the elements of the sequence, their domain, support, moments, etc. So really these bounds cannot be determined at all.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:47
$begingroup$
Try Fatour's lemma better.
$endgroup$
– Will M.
Dec 3 '18 at 22:18
$begingroup$
Try Fatour's lemma better.
$endgroup$
– Will M.
Dec 3 '18 at 22:18
1
1
$begingroup$
Since the sequence converges to a constant, then the elements of the sequence cannot be identically distributed. Independent, they may.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:40
$begingroup$
Since the sequence converges to a constant, then the elements of the sequence cannot be identically distributed. Independent, they may.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:40
1
1
$begingroup$
The fact the the sequence ${X_n}$ converges to the constant $x_0$ says nothing at all about what happens to the elements of the sequence, their domain, support, moments, etc. So really these bounds cannot be determined at all.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:47
$begingroup$
The fact the the sequence ${X_n}$ converges to the constant $x_0$ says nothing at all about what happens to the elements of the sequence, their domain, support, moments, etc. So really these bounds cannot be determined at all.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:47
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%2f3024719%2fprove-the-upper-bound-of-expectations-using-almost-sure-convergence%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%2f3024719%2fprove-the-upper-bound-of-expectations-using-almost-sure-convergence%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
$begingroup$
Try Fatour's lemma better.
$endgroup$
– Will M.
Dec 3 '18 at 22:18
1
$begingroup$
Since the sequence converges to a constant, then the elements of the sequence cannot be identically distributed. Independent, they may.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:40
1
$begingroup$
The fact the the sequence ${X_n}$ converges to the constant $x_0$ says nothing at all about what happens to the elements of the sequence, their domain, support, moments, etc. So really these bounds cannot be determined at all.
$endgroup$
– Alecos Papadopoulos
Dec 4 '18 at 3:47