Random variables $X,Y$ such that $E(X|Y)=E(Y|X)$ a.s.












1












$begingroup$


Let $X$ and $Y$ be random variables such that $E(|X|), E(|Y|)<infty$ and $E(X|Y)=E(Y|X)$ a.s. Then is it true that $X=Y$ a.s. ? If this is not true in general, what happens if we also assume $X,Y$ are identically distributed ?






probability-theory measure-theory random-variables conditional-expectation expected-value







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $X$ and $Y$ be random variables such that $E(|X|), E(|Y|)<infty$ and $E(X|Y)=E(Y|X)$ a.s. Then is it true that $X=Y$ a.s. ? If this is not true in general, what happens if we also assume $X,Y$ are identically distributed ?






    probability-theory measure-theory random-variables conditional-expectation expected-value







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $X$ and $Y$ be random variables such that $E(|X|), E(|Y|)<infty$ and $E(X|Y)=E(Y|X)$ a.s. Then is it true that $X=Y$ a.s. ? If this is not true in general, what happens if we also assume $X,Y$ are identically distributed ?






      probability-theory measure-theory random-variables conditional-expectation expected-value







      share|cite|improve this question









      $endgroup$




      Let $X$ and $Y$ be random variables such that $E(|X|), E(|Y|)<infty$ and $E(X|Y)=E(Y|X)$ a.s. Then is it true that $X=Y$ a.s. ? If this is not true in general, what happens if we also assume $X,Y$ are identically distributed ?






      probability-theory measure-theory random-variables conditional-expectation expected-value




      probability-theory measure-theory random-variables conditional-expectation expected-value






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 20 '18 at 2:33









      user521337user521337

      1,0801415




      1,0801415






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Not true at all.



          Let $X, Y$ be indepedently, identically distributed.



          $$
          mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
          $$



          but this does not mean $P(X = Y) = 1$.




          Suppose $X, Y overset{iid}{sim} Normal(mu, sigma^2)$.



          Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z sim N(0, 2sigma^2)$ is a continuous R.V.



          So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).



          But since $X, Y$ are independent,



          $$
          mu = mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
          $$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:40












          • $begingroup$
            @user521337 Yes, since $X$ is independent of $Y$.
            $endgroup$
            – Moreblue
            Dec 20 '18 at 2:42












          • $begingroup$
            Also, $Z$ is $N(0 , sigma^2)$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:49












          • $begingroup$
            @user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
            $endgroup$
            – Moreblue
            Dec 20 '18 at 2:50








          • 1




            $begingroup$
            again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:52











          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%2f3047087%2frandom-variables-x-y-such-that-exy-eyx-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









          2












          $begingroup$

          Not true at all.



          Let $X, Y$ be indepedently, identically distributed.



          $$
          mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
          $$



          but this does not mean $P(X = Y) = 1$.




          Suppose $X, Y overset{iid}{sim} Normal(mu, sigma^2)$.



          Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z sim N(0, 2sigma^2)$ is a continuous R.V.



          So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).



          But since $X, Y$ are independent,



          $$
          mu = mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
          $$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:40












          • $begingroup$
            @user521337 Yes, since $X$ is independent of $Y$.
            $endgroup$
            – Moreblue
            Dec 20 '18 at 2:42












          • $begingroup$
            Also, $Z$ is $N(0 , sigma^2)$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:49












          • $begingroup$
            @user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
            $endgroup$
            – Moreblue
            Dec 20 '18 at 2:50








          • 1




            $begingroup$
            again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:52
















          2












          $begingroup$

          Not true at all.



          Let $X, Y$ be indepedently, identically distributed.



          $$
          mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
          $$



          but this does not mean $P(X = Y) = 1$.




          Suppose $X, Y overset{iid}{sim} Normal(mu, sigma^2)$.



          Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z sim N(0, 2sigma^2)$ is a continuous R.V.



          So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).



          But since $X, Y$ are independent,



          $$
          mu = mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
          $$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:40












          • $begingroup$
            @user521337 Yes, since $X$ is independent of $Y$.
            $endgroup$
            – Moreblue
            Dec 20 '18 at 2:42












          • $begingroup$
            Also, $Z$ is $N(0 , sigma^2)$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:49












          • $begingroup$
            @user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
            $endgroup$
            – Moreblue
            Dec 20 '18 at 2:50








          • 1




            $begingroup$
            again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:52














          2












          2








          2





          $begingroup$

          Not true at all.



          Let $X, Y$ be indepedently, identically distributed.



          $$
          mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
          $$



          but this does not mean $P(X = Y) = 1$.




          Suppose $X, Y overset{iid}{sim} Normal(mu, sigma^2)$.



          Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z sim N(0, 2sigma^2)$ is a continuous R.V.



          So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).



          But since $X, Y$ are independent,



          $$
          mu = mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
          $$






          share|cite|improve this answer











          $endgroup$



          Not true at all.



          Let $X, Y$ be indepedently, identically distributed.



          $$
          mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
          $$



          but this does not mean $P(X = Y) = 1$.




          Suppose $X, Y overset{iid}{sim} Normal(mu, sigma^2)$.



          Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z sim N(0, 2sigma^2)$ is a continuous R.V.



          So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).



          But since $X, Y$ are independent,



          $$
          mu = mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 20 '18 at 3:07

























          answered Dec 20 '18 at 2:38









          MoreblueMoreblue

          8791216




          8791216








          • 1




            $begingroup$
            You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:40












          • $begingroup$
            @user521337 Yes, since $X$ is independent of $Y$.
            $endgroup$
            – Moreblue
            Dec 20 '18 at 2:42












          • $begingroup$
            Also, $Z$ is $N(0 , sigma^2)$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:49












          • $begingroup$
            @user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
            $endgroup$
            – Moreblue
            Dec 20 '18 at 2:50








          • 1




            $begingroup$
            again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:52














          • 1




            $begingroup$
            You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:40












          • $begingroup$
            @user521337 Yes, since $X$ is independent of $Y$.
            $endgroup$
            – Moreblue
            Dec 20 '18 at 2:42












          • $begingroup$
            Also, $Z$ is $N(0 , sigma^2)$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:49












          • $begingroup$
            @user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
            $endgroup$
            – Moreblue
            Dec 20 '18 at 2:50








          • 1




            $begingroup$
            again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
            $endgroup$
            – user521337
            Dec 20 '18 at 2:52








          1




          1




          $begingroup$
          You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
          $endgroup$
          – user521337
          Dec 20 '18 at 2:40






          $begingroup$
          You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
          $endgroup$
          – user521337
          Dec 20 '18 at 2:40














          $begingroup$
          @user521337 Yes, since $X$ is independent of $Y$.
          $endgroup$
          – Moreblue
          Dec 20 '18 at 2:42






          $begingroup$
          @user521337 Yes, since $X$ is independent of $Y$.
          $endgroup$
          – Moreblue
          Dec 20 '18 at 2:42














          $begingroup$
          Also, $Z$ is $N(0 , sigma^2)$ ...
          $endgroup$
          – user521337
          Dec 20 '18 at 2:49






          $begingroup$
          Also, $Z$ is $N(0 , sigma^2)$ ...
          $endgroup$
          – user521337
          Dec 20 '18 at 2:49














          $begingroup$
          @user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
          $endgroup$
          – Moreblue
          Dec 20 '18 at 2:50






          $begingroup$
          @user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
          $endgroup$
          – Moreblue
          Dec 20 '18 at 2:50






          1




          1




          $begingroup$
          again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
          $endgroup$
          – user521337
          Dec 20 '18 at 2:52




          $begingroup$
          again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
          $endgroup$
          – user521337
          Dec 20 '18 at 2:52


















          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%2f3047087%2frandom-variables-x-y-such-that-exy-eyx-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

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more