Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$












1












$begingroup$



Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$




My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?



Thanks in advance!






probability-theory conditional-expectation independence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The correct hypothesis is that Z is independent of (X,Y).
    $endgroup$
    – Did
    Dec 7 '18 at 8:16
















1












$begingroup$



Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$




My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?



Thanks in advance!






probability-theory conditional-expectation independence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The correct hypothesis is that Z is independent of (X,Y).
    $endgroup$
    – Did
    Dec 7 '18 at 8:16














1












1








1





$begingroup$



Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$




My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?



Thanks in advance!






probability-theory conditional-expectation independence







share|cite|improve this question











$endgroup$





Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$




My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?



Thanks in advance!






probability-theory conditional-expectation independence




probability-theory conditional-expectation independence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 8:17







bellcircle

















asked Dec 7 '18 at 8:07









bellcirclebellcircle

1,331411




1,331411








  • 1




    $begingroup$
    The correct hypothesis is that Z is independent of (X,Y).
    $endgroup$
    – Did
    Dec 7 '18 at 8:16














  • 1




    $begingroup$
    The correct hypothesis is that Z is independent of (X,Y).
    $endgroup$
    – Did
    Dec 7 '18 at 8:16








1




1




$begingroup$
The correct hypothesis is that Z is independent of (X,Y).
$endgroup$
– Did
Dec 7 '18 at 8:16




$begingroup$
The correct hypothesis is that Z is independent of (X,Y).
$endgroup$
– Did
Dec 7 '18 at 8:16










1 Answer
1






active

oldest

votes


















1












$begingroup$

Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.






share|cite|improve this answer











$endgroup$













    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%2f3029631%2fprove-that-mathbfey-sigmax-mathbfey-sigmax-z%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












    $begingroup$

    Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.






        share|cite|improve this answer











        $endgroup$



        Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 8:30

























        answered Dec 7 '18 at 8:25









        Kavi Rama MurthyKavi Rama Murthy

        53.5k32055




        53.5k32055






























            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%2f3029631%2fprove-that-mathbfey-sigmax-mathbfey-sigmax-z%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

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more