Equivalence of the condition that the supremum of i.i.d. RVs are finite a.s.












1












$begingroup$


I am proving the following :



Suppose ${X_n : ninmathbb{N}}$ are i.i.d. random variables. Then $P(sup_{ninmathbb{N}}X_n < infty) = 1$ if and only if $ sum_{ninmathbb{N}}{P(X_n > M)} < infty$ for some $M>0.$



I guess by the form of the problem, that Borel-Canteli lemma will be used at some point, but I could not thought of a good way to connect it to the problem.



Any idea or hint will be really helpful. Thanks.



(edit)



Assume $sum_{ninmathbb{N}}{P(X_n > M)} < infty.$ In fact, as they are i.i.d., $P(X_n > M) = P(X_1>M)$ for all $n.$ Thus $P(X_1>M) = 0$ and $P(sup_{ninmathbb{N}}X_n > M) le sum_{ninmathbb{N}}{P(X_n > M)} = 0.$



Conversely, I've realized that ${sup_{ninmathbb{N}}{X_n} < infty} = bigcup_{m=1}^{infty}bigcap_{n=1}^{infty}{X_n le m}.$






probability-theory measure-theory random-variables independence borel-cantelli-lemmas







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
    $endgroup$
    – Did
    Dec 10 '18 at 15:20










  • $begingroup$
    I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
    $endgroup$
    – Euduardo
    Dec 10 '18 at 15:22










  • $begingroup$
    Well then the exam is slightly absurd, but why not.
    $endgroup$
    – Did
    Dec 10 '18 at 15:23










  • $begingroup$
    You're right Did. Now I understand independence is enough, thanks!
    $endgroup$
    – Euduardo
    Dec 11 '18 at 5:09
















1












$begingroup$


I am proving the following :



Suppose ${X_n : ninmathbb{N}}$ are i.i.d. random variables. Then $P(sup_{ninmathbb{N}}X_n < infty) = 1$ if and only if $ sum_{ninmathbb{N}}{P(X_n > M)} < infty$ for some $M>0.$



I guess by the form of the problem, that Borel-Canteli lemma will be used at some point, but I could not thought of a good way to connect it to the problem.



Any idea or hint will be really helpful. Thanks.



(edit)



Assume $sum_{ninmathbb{N}}{P(X_n > M)} < infty.$ In fact, as they are i.i.d., $P(X_n > M) = P(X_1>M)$ for all $n.$ Thus $P(X_1>M) = 0$ and $P(sup_{ninmathbb{N}}X_n > M) le sum_{ninmathbb{N}}{P(X_n > M)} = 0.$



Conversely, I've realized that ${sup_{ninmathbb{N}}{X_n} < infty} = bigcup_{m=1}^{infty}bigcap_{n=1}^{infty}{X_n le m}.$






probability-theory measure-theory random-variables independence borel-cantelli-lemmas







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
    $endgroup$
    – Did
    Dec 10 '18 at 15:20










  • $begingroup$
    I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
    $endgroup$
    – Euduardo
    Dec 10 '18 at 15:22










  • $begingroup$
    Well then the exam is slightly absurd, but why not.
    $endgroup$
    – Did
    Dec 10 '18 at 15:23










  • $begingroup$
    You're right Did. Now I understand independence is enough, thanks!
    $endgroup$
    – Euduardo
    Dec 11 '18 at 5:09














1












1








1





$begingroup$


I am proving the following :



Suppose ${X_n : ninmathbb{N}}$ are i.i.d. random variables. Then $P(sup_{ninmathbb{N}}X_n < infty) = 1$ if and only if $ sum_{ninmathbb{N}}{P(X_n > M)} < infty$ for some $M>0.$



I guess by the form of the problem, that Borel-Canteli lemma will be used at some point, but I could not thought of a good way to connect it to the problem.



Any idea or hint will be really helpful. Thanks.



(edit)



Assume $sum_{ninmathbb{N}}{P(X_n > M)} < infty.$ In fact, as they are i.i.d., $P(X_n > M) = P(X_1>M)$ for all $n.$ Thus $P(X_1>M) = 0$ and $P(sup_{ninmathbb{N}}X_n > M) le sum_{ninmathbb{N}}{P(X_n > M)} = 0.$



Conversely, I've realized that ${sup_{ninmathbb{N}}{X_n} < infty} = bigcup_{m=1}^{infty}bigcap_{n=1}^{infty}{X_n le m}.$






probability-theory measure-theory random-variables independence borel-cantelli-lemmas







share|cite|improve this question











$endgroup$




I am proving the following :



Suppose ${X_n : ninmathbb{N}}$ are i.i.d. random variables. Then $P(sup_{ninmathbb{N}}X_n < infty) = 1$ if and only if $ sum_{ninmathbb{N}}{P(X_n > M)} < infty$ for some $M>0.$



I guess by the form of the problem, that Borel-Canteli lemma will be used at some point, but I could not thought of a good way to connect it to the problem.



Any idea or hint will be really helpful. Thanks.



(edit)



Assume $sum_{ninmathbb{N}}{P(X_n > M)} < infty.$ In fact, as they are i.i.d., $P(X_n > M) = P(X_1>M)$ for all $n.$ Thus $P(X_1>M) = 0$ and $P(sup_{ninmathbb{N}}X_n > M) le sum_{ninmathbb{N}}{P(X_n > M)} = 0.$



Conversely, I've realized that ${sup_{ninmathbb{N}}{X_n} < infty} = bigcup_{m=1}^{infty}bigcap_{n=1}^{infty}{X_n le m}.$






probability-theory measure-theory random-variables independence borel-cantelli-lemmas




probability-theory measure-theory random-variables independence borel-cantelli-lemmas






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 15:10







Euduardo

















asked Dec 10 '18 at 14:15









EuduardoEuduardo

1288




1288








  • 1




    $begingroup$
    Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
    $endgroup$
    – Did
    Dec 10 '18 at 15:20










  • $begingroup$
    I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
    $endgroup$
    – Euduardo
    Dec 10 '18 at 15:22










  • $begingroup$
    Well then the exam is slightly absurd, but why not.
    $endgroup$
    – Did
    Dec 10 '18 at 15:23










  • $begingroup$
    You're right Did. Now I understand independence is enough, thanks!
    $endgroup$
    – Euduardo
    Dec 11 '18 at 5:09














  • 1




    $begingroup$
    Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
    $endgroup$
    – Did
    Dec 10 '18 at 15:20










  • $begingroup$
    I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
    $endgroup$
    – Euduardo
    Dec 10 '18 at 15:22










  • $begingroup$
    Well then the exam is slightly absurd, but why not.
    $endgroup$
    – Did
    Dec 10 '18 at 15:23










  • $begingroup$
    You're right Did. Now I understand independence is enough, thanks!
    $endgroup$
    – Euduardo
    Dec 11 '18 at 5:09








1




1




$begingroup$
Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
$endgroup$
– Did
Dec 10 '18 at 15:20




$begingroup$
Are you sure the random variables $(X_n)$ are supposed to be i.i.d., or only independent?
$endgroup$
– Did
Dec 10 '18 at 15:20












$begingroup$
I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
$endgroup$
– Euduardo
Dec 10 '18 at 15:22




$begingroup$
I am 100 % sure that they are supposed to be i.i.d., as it was exam problem.
$endgroup$
– Euduardo
Dec 10 '18 at 15:22












$begingroup$
Well then the exam is slightly absurd, but why not.
$endgroup$
– Did
Dec 10 '18 at 15:23




$begingroup$
Well then the exam is slightly absurd, but why not.
$endgroup$
– Did
Dec 10 '18 at 15:23












$begingroup$
You're right Did. Now I understand independence is enough, thanks!
$endgroup$
– Euduardo
Dec 11 '18 at 5:09




$begingroup$
You're right Did. Now I understand independence is enough, thanks!
$endgroup$
– Euduardo
Dec 11 '18 at 5:09










1 Answer
1






active

oldest

votes


















3












$begingroup$

Suppose $P(sup_n X_n < infty) = 1$. Since ${sup_n X_n < infty} = bigcup_{M=1}^infty {sup_n X_n le M}$, we have $lim_{M to infty} P(sup_n X_n le M) = 1$.
But if $P(X_i > M) > 0$, $P(sup_n X_n > M) = 1$. So for some $M$ we must have $P(X_i > M) = 0$.



Conversely, if $P(sup_n X_n < infty) < 1$, then for all $M$ we have $P(sup_n X_n > M) > 0$,
and $P(X_i > M) > 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
    $endgroup$
    – copper.hat
    Dec 10 '18 at 17:07










  • $begingroup$
    Second Borel-Cantelli lemma.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 18:21










  • $begingroup$
    Using the fact that $X_n$ are iid, I see. Thanks.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 18:23











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033962%2fequivalence-of-the-condition-that-the-supremum-of-i-i-d-rvs-are-finite-a-s%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









3












$begingroup$

Suppose $P(sup_n X_n < infty) = 1$. Since ${sup_n X_n < infty} = bigcup_{M=1}^infty {sup_n X_n le M}$, we have $lim_{M to infty} P(sup_n X_n le M) = 1$.
But if $P(X_i > M) > 0$, $P(sup_n X_n > M) = 1$. So for some $M$ we must have $P(X_i > M) = 0$.



Conversely, if $P(sup_n X_n < infty) < 1$, then for all $M$ we have $P(sup_n X_n > M) > 0$,
and $P(X_i > M) > 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
    $endgroup$
    – copper.hat
    Dec 10 '18 at 17:07










  • $begingroup$
    Second Borel-Cantelli lemma.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 18:21










  • $begingroup$
    Using the fact that $X_n$ are iid, I see. Thanks.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 18:23
















3












$begingroup$

Suppose $P(sup_n X_n < infty) = 1$. Since ${sup_n X_n < infty} = bigcup_{M=1}^infty {sup_n X_n le M}$, we have $lim_{M to infty} P(sup_n X_n le M) = 1$.
But if $P(X_i > M) > 0$, $P(sup_n X_n > M) = 1$. So for some $M$ we must have $P(X_i > M) = 0$.



Conversely, if $P(sup_n X_n < infty) < 1$, then for all $M$ we have $P(sup_n X_n > M) > 0$,
and $P(X_i > M) > 0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
    $endgroup$
    – copper.hat
    Dec 10 '18 at 17:07










  • $begingroup$
    Second Borel-Cantelli lemma.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 18:21










  • $begingroup$
    Using the fact that $X_n$ are iid, I see. Thanks.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 18:23














3












3








3





$begingroup$

Suppose $P(sup_n X_n < infty) = 1$. Since ${sup_n X_n < infty} = bigcup_{M=1}^infty {sup_n X_n le M}$, we have $lim_{M to infty} P(sup_n X_n le M) = 1$.
But if $P(X_i > M) > 0$, $P(sup_n X_n > M) = 1$. So for some $M$ we must have $P(X_i > M) = 0$.



Conversely, if $P(sup_n X_n < infty) < 1$, then for all $M$ we have $P(sup_n X_n > M) > 0$,
and $P(X_i > M) > 0$.






share|cite|improve this answer









$endgroup$



Suppose $P(sup_n X_n < infty) = 1$. Since ${sup_n X_n < infty} = bigcup_{M=1}^infty {sup_n X_n le M}$, we have $lim_{M to infty} P(sup_n X_n le M) = 1$.
But if $P(X_i > M) > 0$, $P(sup_n X_n > M) = 1$. So for some $M$ we must have $P(X_i > M) = 0$.



Conversely, if $P(sup_n X_n < infty) < 1$, then for all $M$ we have $P(sup_n X_n > M) > 0$,
and $P(X_i > M) > 0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 15:11









Robert IsraelRobert Israel

322k23212465




322k23212465












  • $begingroup$
    How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
    $endgroup$
    – copper.hat
    Dec 10 '18 at 17:07










  • $begingroup$
    Second Borel-Cantelli lemma.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 18:21










  • $begingroup$
    Using the fact that $X_n$ are iid, I see. Thanks.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 18:23


















  • $begingroup$
    How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
    $endgroup$
    – copper.hat
    Dec 10 '18 at 17:07










  • $begingroup$
    Second Borel-Cantelli lemma.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 18:21










  • $begingroup$
    Using the fact that $X_n$ are iid, I see. Thanks.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 18:23
















$begingroup$
How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
$endgroup$
– copper.hat
Dec 10 '18 at 17:07




$begingroup$
How do you go from $P[X_i >M] >0$ to $P[sup_n X_n >M] = 1$?
$endgroup$
– copper.hat
Dec 10 '18 at 17:07












$begingroup$
Second Borel-Cantelli lemma.
$endgroup$
– Robert Israel
Dec 10 '18 at 18:21




$begingroup$
Second Borel-Cantelli lemma.
$endgroup$
– Robert Israel
Dec 10 '18 at 18:21












$begingroup$
Using the fact that $X_n$ are iid, I see. Thanks.
$endgroup$
– copper.hat
Dec 10 '18 at 18:23




$begingroup$
Using the fact that $X_n$ are iid, I see. Thanks.
$endgroup$
– copper.hat
Dec 10 '18 at 18:23


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033962%2fequivalence-of-the-condition-that-the-supremum-of-i-i-d-rvs-are-finite-a-s%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Image
0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
Read more

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
Read more