Suppose $X_n rightarrow X$ a.s. and for each $n$, $X_n perp textit F$. Then is it true that $X perp textit...












2














For random variables $X_n$ and $X$, suppose $X_n rightarrow X$ a.s. and for each $n$, $X_n perp mathcal F$ i.e. independent with $mathcal F$.



My question is:




Is it true that $X perp mathcal F$ ?




I know that I can assume $X_n rightarrow X$ pointwise since measure zero sets can be ignored when considering independence. Then, I know $X$ is $sigma (X_n : n geq1)$ measurable. By assumption, $sigma (X_n) perp mathcal F$ for all $n$. I tried to make use of Dynkin's lemma by constructing a pi system which generates $sigma (X_n : n geq1)$ and is independent with $mathcal F$ , but it gets me nowhere.



Any hint is appreciated.






probability random-variables independence probability-limit-theorems







share|cite|improve this question
























  • Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
    – mathworker21
    Nov 23 at 13:32












  • @mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
    – izimath
    Nov 23 at 13:46


















2














For random variables $X_n$ and $X$, suppose $X_n rightarrow X$ a.s. and for each $n$, $X_n perp mathcal F$ i.e. independent with $mathcal F$.



My question is:




Is it true that $X perp mathcal F$ ?




I know that I can assume $X_n rightarrow X$ pointwise since measure zero sets can be ignored when considering independence. Then, I know $X$ is $sigma (X_n : n geq1)$ measurable. By assumption, $sigma (X_n) perp mathcal F$ for all $n$. I tried to make use of Dynkin's lemma by constructing a pi system which generates $sigma (X_n : n geq1)$ and is independent with $mathcal F$ , but it gets me nowhere.



Any hint is appreciated.






probability random-variables independence probability-limit-theorems







share|cite|improve this question
























  • Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
    – mathworker21
    Nov 23 at 13:32












  • @mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
    – izimath
    Nov 23 at 13:46
















2












2








2







For random variables $X_n$ and $X$, suppose $X_n rightarrow X$ a.s. and for each $n$, $X_n perp mathcal F$ i.e. independent with $mathcal F$.



My question is:




Is it true that $X perp mathcal F$ ?




I know that I can assume $X_n rightarrow X$ pointwise since measure zero sets can be ignored when considering independence. Then, I know $X$ is $sigma (X_n : n geq1)$ measurable. By assumption, $sigma (X_n) perp mathcal F$ for all $n$. I tried to make use of Dynkin's lemma by constructing a pi system which generates $sigma (X_n : n geq1)$ and is independent with $mathcal F$ , but it gets me nowhere.



Any hint is appreciated.






probability random-variables independence probability-limit-theorems







share|cite|improve this question















For random variables $X_n$ and $X$, suppose $X_n rightarrow X$ a.s. and for each $n$, $X_n perp mathcal F$ i.e. independent with $mathcal F$.



My question is:




Is it true that $X perp mathcal F$ ?




I know that I can assume $X_n rightarrow X$ pointwise since measure zero sets can be ignored when considering independence. Then, I know $X$ is $sigma (X_n : n geq1)$ measurable. By assumption, $sigma (X_n) perp mathcal F$ for all $n$. I tried to make use of Dynkin's lemma by constructing a pi system which generates $sigma (X_n : n geq1)$ and is independent with $mathcal F$ , but it gets me nowhere.



Any hint is appreciated.






probability random-variables independence probability-limit-theorems




probability random-variables independence probability-limit-theorems






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 14:13

























asked Nov 23 at 13:28









izimath

366110




366110












  • Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
    – mathworker21
    Nov 23 at 13:32












  • @mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
    – izimath
    Nov 23 at 13:46




















  • Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
    – mathworker21
    Nov 23 at 13:32












  • @mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
    – izimath
    Nov 23 at 13:46


















Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
– mathworker21
Nov 23 at 13:32






Why can't you do something like this: $E[XF] = E[lim_n X_n F] = lim_n E[X_n F] = lim_n E[X_n] E[F] = E[X] E[F]$?
– mathworker21
Nov 23 at 13:32














@mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
– izimath
Nov 23 at 13:46






@mathworker21 I think you are tring to use bounded convergence theorem to $1_B (X_n)$ for some borel set $B$, but $1_B (X_n)$ does not converge to $1_B (X)$
– izimath
Nov 23 at 13:46












1 Answer
1






active

oldest

votes


















2














We can follow these steps:




  1. Using the dominated convergence theorem, show that for all continuous and bounded function $varphicolonmathbb Rtomathbb R$ and each $Finmathcal F$, the following equality holds:
    $$tag{*} mathbb Eleft[varphi(X)mathbf 1_Fright]= mathbb Eleft[varphi(X) right]mathbb P(F).$$


  2. Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $varphi$ is the indicator function of $S$.


  3. This shows that $mathbb P({Xin S}cap F)=mathbb P(Xin S)mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.






share|cite|improve this answer





















  • In the second step, can we use a closed interval $S$, just for simplicity?
    – izimath
    Nov 23 at 14:23












  • Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
    – Davide Giraudo
    Nov 23 at 14:42












  • Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
    – izimath
    Nov 23 at 14:47










  • Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
    – izimath
    Nov 23 at 14:54











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%2f3010363%2fsuppose-x-n-rightarrow-x-a-s-and-for-each-n-x-n-perp-textit-f-then-i%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









2














We can follow these steps:




  1. Using the dominated convergence theorem, show that for all continuous and bounded function $varphicolonmathbb Rtomathbb R$ and each $Finmathcal F$, the following equality holds:
    $$tag{*} mathbb Eleft[varphi(X)mathbf 1_Fright]= mathbb Eleft[varphi(X) right]mathbb P(F).$$


  2. Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $varphi$ is the indicator function of $S$.


  3. This shows that $mathbb P({Xin S}cap F)=mathbb P(Xin S)mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.






share|cite|improve this answer





















  • In the second step, can we use a closed interval $S$, just for simplicity?
    – izimath
    Nov 23 at 14:23












  • Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
    – Davide Giraudo
    Nov 23 at 14:42












  • Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
    – izimath
    Nov 23 at 14:47










  • Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
    – izimath
    Nov 23 at 14:54
















2














We can follow these steps:




  1. Using the dominated convergence theorem, show that for all continuous and bounded function $varphicolonmathbb Rtomathbb R$ and each $Finmathcal F$, the following equality holds:
    $$tag{*} mathbb Eleft[varphi(X)mathbf 1_Fright]= mathbb Eleft[varphi(X) right]mathbb P(F).$$


  2. Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $varphi$ is the indicator function of $S$.


  3. This shows that $mathbb P({Xin S}cap F)=mathbb P(Xin S)mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.






share|cite|improve this answer





















  • In the second step, can we use a closed interval $S$, just for simplicity?
    – izimath
    Nov 23 at 14:23












  • Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
    – Davide Giraudo
    Nov 23 at 14:42












  • Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
    – izimath
    Nov 23 at 14:47










  • Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
    – izimath
    Nov 23 at 14:54














2












2








2






We can follow these steps:




  1. Using the dominated convergence theorem, show that for all continuous and bounded function $varphicolonmathbb Rtomathbb R$ and each $Finmathcal F$, the following equality holds:
    $$tag{*} mathbb Eleft[varphi(X)mathbf 1_Fright]= mathbb Eleft[varphi(X) right]mathbb P(F).$$


  2. Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $varphi$ is the indicator function of $S$.


  3. This shows that $mathbb P({Xin S}cap F)=mathbb P(Xin S)mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.






share|cite|improve this answer












We can follow these steps:




  1. Using the dominated convergence theorem, show that for all continuous and bounded function $varphicolonmathbb Rtomathbb R$ and each $Finmathcal F$, the following equality holds:
    $$tag{*} mathbb Eleft[varphi(X)mathbf 1_Fright]= mathbb Eleft[varphi(X) right]mathbb P(F).$$


  2. Approximation the indicator function of a closed set $S$ by continuous bounded functions, using $d(cdot,S)$ (distance to a set). The monotone convergence theorem allows to extend (*) when $varphi$ is the indicator function of $S$.


  3. This shows that $mathbb P({Xin S}cap F)=mathbb P(Xin S)mathbb P(F)$ for all closed sets, and using complements, this holds also when $S$ is open. Then we can conclude by the Dynkin's lemma.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 23 at 14:10









Davide Giraudo

125k16150259




125k16150259












  • In the second step, can we use a closed interval $S$, just for simplicity?
    – izimath
    Nov 23 at 14:23












  • Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
    – Davide Giraudo
    Nov 23 at 14:42












  • Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
    – izimath
    Nov 23 at 14:47










  • Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
    – izimath
    Nov 23 at 14:54


















  • In the second step, can we use a closed interval $S$, just for simplicity?
    – izimath
    Nov 23 at 14:23












  • Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
    – Davide Giraudo
    Nov 23 at 14:42












  • Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
    – izimath
    Nov 23 at 14:47










  • Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
    – izimath
    Nov 23 at 14:54
















In the second step, can we use a closed interval $S$, just for simplicity?
– izimath
Nov 23 at 14:23






In the second step, can we use a closed interval $S$, just for simplicity?
– izimath
Nov 23 at 14:23














Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
– Davide Giraudo
Nov 23 at 14:42






Yes, and after we have to work the collection of complements of closed intervals instead of the class of open set. So I think the argument can also work.
– Davide Giraudo
Nov 23 at 14:42














Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
– izimath
Nov 23 at 14:47




Thanks very much. What do you think about the approach that I was trying to do? I don't see a way through.
– izimath
Nov 23 at 14:47












Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
– izimath
Nov 23 at 14:54




Well, I guess $sigma (X_n : n geq1)$ is too big to be independent with $mathcal F$.
– izimath
Nov 23 at 14:54


















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.





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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3010363%2fsuppose-x-n-rightarrow-x-a-s-and-for-each-n-x-n-perp-textit-f-then-i%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

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more