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












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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







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    $begingroup$


    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













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      asked Dec 20 '18 at 2:33









      user521337user521337

      1,2041417




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          $begingroup$

          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





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          1 Answer
          1






          active

          oldest

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          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

          8931217




          8931217








          • 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


















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