Exist $epsilon>0$ such that $P(X>epsilon)>0$












0












$begingroup$


I have the next exercice:



Let $(X_n)_n$ a sequence of random variables i.i.d with law $X$, where $Xge 0$ almost surely et $P(X>0)>0$.




  1. Show that exists $epsilon ge 0$ such that $P(X>epsilon)>0$.

  2. Let $A_n={X_nge epsilon}$. What can be said about the set $limsup A_n$?


My ideas:





  1. Let $B_k={Xgefrac{1}{k}}$



    $bigcup_{k=1}^{infty}B_k={X>0}$



    $sum_{k=1}^{infty}P(B_k)ge P(bigcup_{k=1}^{infty}B_k)=P(X>0)>0$, i.e exist $min mathbb{N}$ such that $P(Xgefrac{1}{m})=P(B_m)>0$.




Where $epsilon =frac{1}{m}$.



My idea is correct? and they could give me suggestions for the second item.






probability-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    The $B_k$ are not disjoint so the equality between the sum of probabilities and the probability of the union does not hold. Replace it with an inequality, and your reasoning is sound.
    $endgroup$
    – Math1000
    Dec 13 '18 at 22:43
















0












$begingroup$


I have the next exercice:



Let $(X_n)_n$ a sequence of random variables i.i.d with law $X$, where $Xge 0$ almost surely et $P(X>0)>0$.




  1. Show that exists $epsilon ge 0$ such that $P(X>epsilon)>0$.

  2. Let $A_n={X_nge epsilon}$. What can be said about the set $limsup A_n$?


My ideas:





  1. Let $B_k={Xgefrac{1}{k}}$



    $bigcup_{k=1}^{infty}B_k={X>0}$



    $sum_{k=1}^{infty}P(B_k)ge P(bigcup_{k=1}^{infty}B_k)=P(X>0)>0$, i.e exist $min mathbb{N}$ such that $P(Xgefrac{1}{m})=P(B_m)>0$.




Where $epsilon =frac{1}{m}$.



My idea is correct? and they could give me suggestions for the second item.






probability-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    The $B_k$ are not disjoint so the equality between the sum of probabilities and the probability of the union does not hold. Replace it with an inequality, and your reasoning is sound.
    $endgroup$
    – Math1000
    Dec 13 '18 at 22:43














0












0








0





$begingroup$


I have the next exercice:



Let $(X_n)_n$ a sequence of random variables i.i.d with law $X$, where $Xge 0$ almost surely et $P(X>0)>0$.




  1. Show that exists $epsilon ge 0$ such that $P(X>epsilon)>0$.

  2. Let $A_n={X_nge epsilon}$. What can be said about the set $limsup A_n$?


My ideas:





  1. Let $B_k={Xgefrac{1}{k}}$



    $bigcup_{k=1}^{infty}B_k={X>0}$



    $sum_{k=1}^{infty}P(B_k)ge P(bigcup_{k=1}^{infty}B_k)=P(X>0)>0$, i.e exist $min mathbb{N}$ such that $P(Xgefrac{1}{m})=P(B_m)>0$.




Where $epsilon =frac{1}{m}$.



My idea is correct? and they could give me suggestions for the second item.






probability-theory







share|cite|improve this question











$endgroup$




I have the next exercice:



Let $(X_n)_n$ a sequence of random variables i.i.d with law $X$, where $Xge 0$ almost surely et $P(X>0)>0$.




  1. Show that exists $epsilon ge 0$ such that $P(X>epsilon)>0$.

  2. Let $A_n={X_nge epsilon}$. What can be said about the set $limsup A_n$?


My ideas:





  1. Let $B_k={Xgefrac{1}{k}}$



    $bigcup_{k=1}^{infty}B_k={X>0}$



    $sum_{k=1}^{infty}P(B_k)ge P(bigcup_{k=1}^{infty}B_k)=P(X>0)>0$, i.e exist $min mathbb{N}$ such that $P(Xgefrac{1}{m})=P(B_m)>0$.




Where $epsilon =frac{1}{m}$.



My idea is correct? and they could give me suggestions for the second item.






probability-theory




probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 22:44







Alex Pozo

















asked Dec 13 '18 at 22:36









Alex PozoAlex Pozo

515214




515214












  • $begingroup$
    The $B_k$ are not disjoint so the equality between the sum of probabilities and the probability of the union does not hold. Replace it with an inequality, and your reasoning is sound.
    $endgroup$
    – Math1000
    Dec 13 '18 at 22:43


















  • $begingroup$
    The $B_k$ are not disjoint so the equality between the sum of probabilities and the probability of the union does not hold. Replace it with an inequality, and your reasoning is sound.
    $endgroup$
    – Math1000
    Dec 13 '18 at 22:43
















$begingroup$
The $B_k$ are not disjoint so the equality between the sum of probabilities and the probability of the union does not hold. Replace it with an inequality, and your reasoning is sound.
$endgroup$
– Math1000
Dec 13 '18 at 22:43




$begingroup$
The $B_k$ are not disjoint so the equality between the sum of probabilities and the probability of the union does not hold. Replace it with an inequality, and your reasoning is sound.
$endgroup$
– Math1000
Dec 13 '18 at 22:43










1 Answer
1






active

oldest

votes


















1












$begingroup$

Hint



For the second question apply the Borel cantelli lemma, i.e. compute $sum P(A_n)$. Is it finite or infinite? What does it tell you about $P(limsup A_n)$?



Your proof for one does not work. It is not true that $sum_{k=1}^{infty}P(B_k)=P(bigcup_{k=1}^{infty}B_k)$. It is true by measure continuity that $sup P(B_k)=lim P(B_k)=P(X>0)>0$. Hence there exists at least one $k$ such that $P(B_k)>0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I found that $sum P(A_n)$ is finite and by Borel Cantelli Lemma $P(limsup A_n) =0$
    $endgroup$
    – Alex Pozo
    Dec 13 '18 at 22:56












  • $begingroup$
    By the proof of a) we know that there exists $epsilon>0$ such that $P(Xgeq epsilon)>0$. In particular $P(A_n)=P(X_ngeq epsilon)=P(Xgeq epsilon)>0$, so the sum diverges (as $sum P(A_n)=sum P(Xgeq epsilon)=infty$) By Borel cantelli, $P(limsup A_n)=1$ since the $X_i$ are independent.
    $endgroup$
    – Foobaz John
    Dec 13 '18 at 23:07













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%2f3038668%2fexist-epsilon0-such-that-px-epsilon0%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









1












$begingroup$

Hint



For the second question apply the Borel cantelli lemma, i.e. compute $sum P(A_n)$. Is it finite or infinite? What does it tell you about $P(limsup A_n)$?



Your proof for one does not work. It is not true that $sum_{k=1}^{infty}P(B_k)=P(bigcup_{k=1}^{infty}B_k)$. It is true by measure continuity that $sup P(B_k)=lim P(B_k)=P(X>0)>0$. Hence there exists at least one $k$ such that $P(B_k)>0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I found that $sum P(A_n)$ is finite and by Borel Cantelli Lemma $P(limsup A_n) =0$
    $endgroup$
    – Alex Pozo
    Dec 13 '18 at 22:56












  • $begingroup$
    By the proof of a) we know that there exists $epsilon>0$ such that $P(Xgeq epsilon)>0$. In particular $P(A_n)=P(X_ngeq epsilon)=P(Xgeq epsilon)>0$, so the sum diverges (as $sum P(A_n)=sum P(Xgeq epsilon)=infty$) By Borel cantelli, $P(limsup A_n)=1$ since the $X_i$ are independent.
    $endgroup$
    – Foobaz John
    Dec 13 '18 at 23:07


















1












$begingroup$

Hint



For the second question apply the Borel cantelli lemma, i.e. compute $sum P(A_n)$. Is it finite or infinite? What does it tell you about $P(limsup A_n)$?



Your proof for one does not work. It is not true that $sum_{k=1}^{infty}P(B_k)=P(bigcup_{k=1}^{infty}B_k)$. It is true by measure continuity that $sup P(B_k)=lim P(B_k)=P(X>0)>0$. Hence there exists at least one $k$ such that $P(B_k)>0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I found that $sum P(A_n)$ is finite and by Borel Cantelli Lemma $P(limsup A_n) =0$
    $endgroup$
    – Alex Pozo
    Dec 13 '18 at 22:56












  • $begingroup$
    By the proof of a) we know that there exists $epsilon>0$ such that $P(Xgeq epsilon)>0$. In particular $P(A_n)=P(X_ngeq epsilon)=P(Xgeq epsilon)>0$, so the sum diverges (as $sum P(A_n)=sum P(Xgeq epsilon)=infty$) By Borel cantelli, $P(limsup A_n)=1$ since the $X_i$ are independent.
    $endgroup$
    – Foobaz John
    Dec 13 '18 at 23:07
















1












1








1





$begingroup$

Hint



For the second question apply the Borel cantelli lemma, i.e. compute $sum P(A_n)$. Is it finite or infinite? What does it tell you about $P(limsup A_n)$?



Your proof for one does not work. It is not true that $sum_{k=1}^{infty}P(B_k)=P(bigcup_{k=1}^{infty}B_k)$. It is true by measure continuity that $sup P(B_k)=lim P(B_k)=P(X>0)>0$. Hence there exists at least one $k$ such that $P(B_k)>0$.






share|cite|improve this answer









$endgroup$



Hint



For the second question apply the Borel cantelli lemma, i.e. compute $sum P(A_n)$. Is it finite or infinite? What does it tell you about $P(limsup A_n)$?



Your proof for one does not work. It is not true that $sum_{k=1}^{infty}P(B_k)=P(bigcup_{k=1}^{infty}B_k)$. It is true by measure continuity that $sup P(B_k)=lim P(B_k)=P(X>0)>0$. Hence there exists at least one $k$ such that $P(B_k)>0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 22:45









Foobaz JohnFoobaz John

22.3k41452




22.3k41452












  • $begingroup$
    I found that $sum P(A_n)$ is finite and by Borel Cantelli Lemma $P(limsup A_n) =0$
    $endgroup$
    – Alex Pozo
    Dec 13 '18 at 22:56












  • $begingroup$
    By the proof of a) we know that there exists $epsilon>0$ such that $P(Xgeq epsilon)>0$. In particular $P(A_n)=P(X_ngeq epsilon)=P(Xgeq epsilon)>0$, so the sum diverges (as $sum P(A_n)=sum P(Xgeq epsilon)=infty$) By Borel cantelli, $P(limsup A_n)=1$ since the $X_i$ are independent.
    $endgroup$
    – Foobaz John
    Dec 13 '18 at 23:07




















  • $begingroup$
    I found that $sum P(A_n)$ is finite and by Borel Cantelli Lemma $P(limsup A_n) =0$
    $endgroup$
    – Alex Pozo
    Dec 13 '18 at 22:56












  • $begingroup$
    By the proof of a) we know that there exists $epsilon>0$ such that $P(Xgeq epsilon)>0$. In particular $P(A_n)=P(X_ngeq epsilon)=P(Xgeq epsilon)>0$, so the sum diverges (as $sum P(A_n)=sum P(Xgeq epsilon)=infty$) By Borel cantelli, $P(limsup A_n)=1$ since the $X_i$ are independent.
    $endgroup$
    – Foobaz John
    Dec 13 '18 at 23:07


















$begingroup$
I found that $sum P(A_n)$ is finite and by Borel Cantelli Lemma $P(limsup A_n) =0$
$endgroup$
– Alex Pozo
Dec 13 '18 at 22:56






$begingroup$
I found that $sum P(A_n)$ is finite and by Borel Cantelli Lemma $P(limsup A_n) =0$
$endgroup$
– Alex Pozo
Dec 13 '18 at 22:56














$begingroup$
By the proof of a) we know that there exists $epsilon>0$ such that $P(Xgeq epsilon)>0$. In particular $P(A_n)=P(X_ngeq epsilon)=P(Xgeq epsilon)>0$, so the sum diverges (as $sum P(A_n)=sum P(Xgeq epsilon)=infty$) By Borel cantelli, $P(limsup A_n)=1$ since the $X_i$ are independent.
$endgroup$
– Foobaz John
Dec 13 '18 at 23:07






$begingroup$
By the proof of a) we know that there exists $epsilon>0$ such that $P(Xgeq epsilon)>0$. In particular $P(A_n)=P(X_ngeq epsilon)=P(Xgeq epsilon)>0$, so the sum diverges (as $sum P(A_n)=sum P(Xgeq epsilon)=infty$) By Borel cantelli, $P(limsup A_n)=1$ since the $X_i$ are independent.
$endgroup$
– Foobaz John
Dec 13 '18 at 23:07




















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%2f3038668%2fexist-epsilon0-such-that-px-epsilon0%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

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Brian Clough

Image
Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
Read more

Cáceres

Image
Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
Read more