Conditional expectation of a random variable conditioned to another conditionally independent variable












1












$begingroup$


I have a two variables $X$ and $Y$ conditionally independent given a thirs variable $Z$.



Now, assume that $Z$ can take values in ${1,...,k}$.



I will have:
$$
E[XY] = sum_{i=1}^k P[Z = i]E[X|Z = i]E[Y|Z = i]
$$



Now, can I say what follows?
$$
E[X | Y] = E[X]
$$



I have a proof for this, but I am not sure:
$$
E[X | Y] = sum_{i=1}^k P[Z = i]E[X|Z = i, Y] = sum_{i=1}^k P[Z = i]E[X|Z = i] = E[X]
$$



Is this correct? If not, where am I wrong?






probability conditional-expectation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
    $endgroup$
    – Did
    Dec 1 '18 at 16:02










  • $begingroup$
    @Did an infallible way of assessing the soundness of proof passages :)
    $endgroup$
    – Easymode44
    Dec 1 '18 at 17:13










  • $begingroup$
    Related math.stackexchange.com/questions/730702/…
    $endgroup$
    – leonbloy
    Dec 2 '18 at 20:55
















1












$begingroup$


I have a two variables $X$ and $Y$ conditionally independent given a thirs variable $Z$.



Now, assume that $Z$ can take values in ${1,...,k}$.



I will have:
$$
E[XY] = sum_{i=1}^k P[Z = i]E[X|Z = i]E[Y|Z = i]
$$



Now, can I say what follows?
$$
E[X | Y] = E[X]
$$



I have a proof for this, but I am not sure:
$$
E[X | Y] = sum_{i=1}^k P[Z = i]E[X|Z = i, Y] = sum_{i=1}^k P[Z = i]E[X|Z = i] = E[X]
$$



Is this correct? If not, where am I wrong?






probability conditional-expectation







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
    $endgroup$
    – Did
    Dec 1 '18 at 16:02










  • $begingroup$
    @Did an infallible way of assessing the soundness of proof passages :)
    $endgroup$
    – Easymode44
    Dec 1 '18 at 17:13










  • $begingroup$
    Related math.stackexchange.com/questions/730702/…
    $endgroup$
    – leonbloy
    Dec 2 '18 at 20:55














1












1








1


1



$begingroup$


I have a two variables $X$ and $Y$ conditionally independent given a thirs variable $Z$.



Now, assume that $Z$ can take values in ${1,...,k}$.



I will have:
$$
E[XY] = sum_{i=1}^k P[Z = i]E[X|Z = i]E[Y|Z = i]
$$



Now, can I say what follows?
$$
E[X | Y] = E[X]
$$



I have a proof for this, but I am not sure:
$$
E[X | Y] = sum_{i=1}^k P[Z = i]E[X|Z = i, Y] = sum_{i=1}^k P[Z = i]E[X|Z = i] = E[X]
$$



Is this correct? If not, where am I wrong?






probability conditional-expectation







share|cite|improve this question









$endgroup$




I have a two variables $X$ and $Y$ conditionally independent given a thirs variable $Z$.



Now, assume that $Z$ can take values in ${1,...,k}$.



I will have:
$$
E[XY] = sum_{i=1}^k P[Z = i]E[X|Z = i]E[Y|Z = i]
$$



Now, can I say what follows?
$$
E[X | Y] = E[X]
$$



I have a proof for this, but I am not sure:
$$
E[X | Y] = sum_{i=1}^k P[Z = i]E[X|Z = i, Y] = sum_{i=1}^k P[Z = i]E[X|Z = i] = E[X]
$$



Is this correct? If not, where am I wrong?






probability conditional-expectation




probability conditional-expectation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 '18 at 15:56









Ulderique DemoitreUlderique Demoitre

19810




19810








  • 1




    $begingroup$
    Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
    $endgroup$
    – Did
    Dec 1 '18 at 16:02










  • $begingroup$
    @Did an infallible way of assessing the soundness of proof passages :)
    $endgroup$
    – Easymode44
    Dec 1 '18 at 17:13










  • $begingroup$
    Related math.stackexchange.com/questions/730702/…
    $endgroup$
    – leonbloy
    Dec 2 '18 at 20:55














  • 1




    $begingroup$
    Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
    $endgroup$
    – Did
    Dec 1 '18 at 16:02










  • $begingroup$
    @Did an infallible way of assessing the soundness of proof passages :)
    $endgroup$
    – Easymode44
    Dec 1 '18 at 17:13










  • $begingroup$
    Related math.stackexchange.com/questions/730702/…
    $endgroup$
    – leonbloy
    Dec 2 '18 at 20:55








1




1




$begingroup$
Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
$endgroup$
– Did
Dec 1 '18 at 16:02




$begingroup$
Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
$endgroup$
– Did
Dec 1 '18 at 16:02












$begingroup$
@Did an infallible way of assessing the soundness of proof passages :)
$endgroup$
– Easymode44
Dec 1 '18 at 17:13




$begingroup$
@Did an infallible way of assessing the soundness of proof passages :)
$endgroup$
– Easymode44
Dec 1 '18 at 17:13












$begingroup$
Related math.stackexchange.com/questions/730702/…
$endgroup$
– leonbloy
Dec 2 '18 at 20:55




$begingroup$
Related math.stackexchange.com/questions/730702/…
$endgroup$
– leonbloy
Dec 2 '18 at 20:55










1 Answer
1






active

oldest

votes


















0












$begingroup$

This was trickier than it looked ... to me.
It can be put in a more general-compact form thus:




Let $X,Y$ be conditionally independent given $Z$; that is, $P(X ,Y mid Z ) = P(X mid Z) P(Y mid Z)$. Then
$$E[X mid Y] = E[E[X mid Y, Z]] = E[E[X mid Z]] = E[X] tag{1}$$
where in the first and third equality we have applied total expectation (in the third to $X$, in the first to $X mid Y$), and in the second we've applied the conditional independence.




This is wrong, of course (it would imply that for any Markov chain $X to Z to Y$ the extremes $X,Y$ are uncorrelated - false in general).



The wrong equality is the first one. Because $X mid Y$ is not a random variable. Actually it's... nothing.



In your question, the first equality in your "proof" is (like my first equality in $(1)$) wrong, it's a mistaken application of the law of total expectation.






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%2f3021496%2fconditional-expectation-of-a-random-variable-conditioned-to-another-conditionall%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









    0












    $begingroup$

    This was trickier than it looked ... to me.
    It can be put in a more general-compact form thus:




    Let $X,Y$ be conditionally independent given $Z$; that is, $P(X ,Y mid Z ) = P(X mid Z) P(Y mid Z)$. Then
    $$E[X mid Y] = E[E[X mid Y, Z]] = E[E[X mid Z]] = E[X] tag{1}$$
    where in the first and third equality we have applied total expectation (in the third to $X$, in the first to $X mid Y$), and in the second we've applied the conditional independence.




    This is wrong, of course (it would imply that for any Markov chain $X to Z to Y$ the extremes $X,Y$ are uncorrelated - false in general).



    The wrong equality is the first one. Because $X mid Y$ is not a random variable. Actually it's... nothing.



    In your question, the first equality in your "proof" is (like my first equality in $(1)$) wrong, it's a mistaken application of the law of total expectation.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      This was trickier than it looked ... to me.
      It can be put in a more general-compact form thus:




      Let $X,Y$ be conditionally independent given $Z$; that is, $P(X ,Y mid Z ) = P(X mid Z) P(Y mid Z)$. Then
      $$E[X mid Y] = E[E[X mid Y, Z]] = E[E[X mid Z]] = E[X] tag{1}$$
      where in the first and third equality we have applied total expectation (in the third to $X$, in the first to $X mid Y$), and in the second we've applied the conditional independence.




      This is wrong, of course (it would imply that for any Markov chain $X to Z to Y$ the extremes $X,Y$ are uncorrelated - false in general).



      The wrong equality is the first one. Because $X mid Y$ is not a random variable. Actually it's... nothing.



      In your question, the first equality in your "proof" is (like my first equality in $(1)$) wrong, it's a mistaken application of the law of total expectation.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        This was trickier than it looked ... to me.
        It can be put in a more general-compact form thus:




        Let $X,Y$ be conditionally independent given $Z$; that is, $P(X ,Y mid Z ) = P(X mid Z) P(Y mid Z)$. Then
        $$E[X mid Y] = E[E[X mid Y, Z]] = E[E[X mid Z]] = E[X] tag{1}$$
        where in the first and third equality we have applied total expectation (in the third to $X$, in the first to $X mid Y$), and in the second we've applied the conditional independence.




        This is wrong, of course (it would imply that for any Markov chain $X to Z to Y$ the extremes $X,Y$ are uncorrelated - false in general).



        The wrong equality is the first one. Because $X mid Y$ is not a random variable. Actually it's... nothing.



        In your question, the first equality in your "proof" is (like my first equality in $(1)$) wrong, it's a mistaken application of the law of total expectation.






        share|cite|improve this answer











        $endgroup$



        This was trickier than it looked ... to me.
        It can be put in a more general-compact form thus:




        Let $X,Y$ be conditionally independent given $Z$; that is, $P(X ,Y mid Z ) = P(X mid Z) P(Y mid Z)$. Then
        $$E[X mid Y] = E[E[X mid Y, Z]] = E[E[X mid Z]] = E[X] tag{1}$$
        where in the first and third equality we have applied total expectation (in the third to $X$, in the first to $X mid Y$), and in the second we've applied the conditional independence.




        This is wrong, of course (it would imply that for any Markov chain $X to Z to Y$ the extremes $X,Y$ are uncorrelated - false in general).



        The wrong equality is the first one. Because $X mid Y$ is not a random variable. Actually it's... nothing.



        In your question, the first equality in your "proof" is (like my first equality in $(1)$) wrong, it's a mistaken application of the law of total expectation.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 23:15

























        answered Dec 2 '18 at 21:19









        leonbloyleonbloy

        40.5k645107




        40.5k645107






























            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%2f3021496%2fconditional-expectation-of-a-random-variable-conditioned-to-another-conditionall%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

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
            Read more

            Le Mesnil-Réaume

            Image
            Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more