Bound for probability with almost sure convergence
up vote
0
down vote
favorite
Let $(X_n)_n$ be random variables which converge almost surely to a constant $x in mathbb R$, i.e. $X_n xrightarrow{n to infty} x$ a.s. Let $Y$ be another random variable.
Question: Can I say something like $$mathrm{Pr}(Y geq X_n) leq mathrm{Pr} left(Y geq frac{x}{2} right)$$ for $n$ large enough?
probability probability-theory convergence stochastic-processes random-variables
asked Nov 27 at 8:39
Kariani
51
add a comment |
up vote
0
down vote
favorite
Let $(X_n)_n$ be random variables which converge almost surely to a constant $x in mathbb R$, i.e. $X_n xrightarrow{n to infty} x$ a.s. Let $Y$ be another random variable.
Question: Can I say something like $$mathrm{Pr}(Y geq X_n) leq mathrm{Pr} left(Y geq frac{x}{2} right)$$ for $n$ large enough?
probability probability-theory convergence stochastic-processes random-variables
asked Nov 27 at 8:39
Kariani
51
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(X_n)_n$ be random variables which converge almost surely to a constant $x in mathbb R$, i.e. $X_n xrightarrow{n to infty} x$ a.s. Let $Y$ be another random variable.
Question: Can I say something like $$mathrm{Pr}(Y geq X_n) leq mathrm{Pr} left(Y geq frac{x}{2} right)$$ for $n$ large enough?
probability probability-theory convergence stochastic-processes random-variables
asked Nov 27 at 8:39
Kariani
51
Let $(X_n)_n$ be random variables which converge almost surely to a constant $x in mathbb R$, i.e. $X_n xrightarrow{n to infty} x$ a.s. Let $Y$ be another random variable.
Question: Can I say something like $$mathrm{Pr}(Y geq X_n) leq mathrm{Pr} left(Y geq frac{x}{2} right)$$ for $n$ large enough?
probability probability-theory convergence stochastic-processes random-variables
probability probability-theory convergence stochastic-processes random-variables
asked Nov 27 at 8:39
Kariani
51
asked Nov 27 at 8:39
Kariani
51
asked Nov 27 at 8:39
Kariani
51
asked Nov 27 at 8:39
Kariani
51
asked Nov 27 at 8:39
Kariani
51
51
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44
add a comment |
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $frac 1 {2^{n}}$ and $1-frac 1 {2^{n}}$ respectively. Since $sum P(X_n=0)<infty$ Borel cantalli Lemma shows that $X_n to 1$ almost surely. Now take $Y=frac 1 4$. Then $P(Ygeq X_n) =frac 1 {2^{n}}$ and $P(Y geq frac 1 2)=0$.
answered Nov 27 at 8:46
Kavi Rama Murthy
47.4k31854
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',
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%2f3015505%2fbound-for-probability-with-almost-sure-convergence%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
No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $frac 1 {2^{n}}$ and $1-frac 1 {2^{n}}$ respectively. Since $sum P(X_n=0)<infty$ Borel cantalli Lemma shows that $X_n to 1$ almost surely. Now take $Y=frac 1 4$. Then $P(Ygeq X_n) =frac 1 {2^{n}}$ and $P(Y geq frac 1 2)=0$.
answered Nov 27 at 8:46
Kavi Rama Murthy
47.4k31854
add a comment |
up vote
1
down vote
accepted
No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $frac 1 {2^{n}}$ and $1-frac 1 {2^{n}}$ respectively. Since $sum P(X_n=0)<infty$ Borel cantalli Lemma shows that $X_n to 1$ almost surely. Now take $Y=frac 1 4$. Then $P(Ygeq X_n) =frac 1 {2^{n}}$ and $P(Y geq frac 1 2)=0$.
answered Nov 27 at 8:46
Kavi Rama Murthy
47.4k31854
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $frac 1 {2^{n}}$ and $1-frac 1 {2^{n}}$ respectively. Since $sum P(X_n=0)<infty$ Borel cantalli Lemma shows that $X_n to 1$ almost surely. Now take $Y=frac 1 4$. Then $P(Ygeq X_n) =frac 1 {2^{n}}$ and $P(Y geq frac 1 2)=0$.
answered Nov 27 at 8:46
Kavi Rama Murthy
47.4k31854
No, you cannot have such an inequality. Let $X_n$ take the values $0$ and $1$ with probabilities $frac 1 {2^{n}}$ and $1-frac 1 {2^{n}}$ respectively. Since $sum P(X_n=0)<infty$ Borel cantalli Lemma shows that $X_n to 1$ almost surely. Now take $Y=frac 1 4$. Then $P(Ygeq X_n) =frac 1 {2^{n}}$ and $P(Y geq frac 1 2)=0$.
answered Nov 27 at 8:46
Kavi Rama Murthy
47.4k31854
answered Nov 27 at 8:46
Kavi Rama Murthy
47.4k31854
answered Nov 27 at 8:46
Kavi Rama Murthy
47.4k31854
answered Nov 27 at 8:46
Kavi Rama Murthy
47.4k31854
47.4k31854
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%2f3015505%2fbound-for-probability-with-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
This seems unlikely if $x$ is zero or negative
– Henry
Nov 27 at 8:44