If the expectation and variance of x are both not affected by y, and vice versa, then must x and y be...

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I know that if $mathbb{E}[X]=mathbb{E}[X|Y] , mathbb{E}[Y]=mathbb{E}[Y|X]$, $X$ and $Y$ can be dependent, for example a ‘uniform’ distribution in a unit circle.
Now we add the variance, if
$$mathbb{E}[X]=mathbb{E}[X|Y], mathbb{E}[Y]=mathbb{E}[Y|X], $$$$Var(X)=Var(X|Y), Var(Y)=Var(Y|X).$$
Say the expectation and variance of $X$ are both not affected by $Y$, and vice versa, then must $X$ and $Y$ be independent?
In this case I can not find a counterexample just like the uniform circle.



If they are independent, how to prove it? If not, is there a counterexample?



Thanks!






probability conditional-expectation independence







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    $begingroup$


    I know that if $mathbb{E}[X]=mathbb{E}[X|Y] , mathbb{E}[Y]=mathbb{E}[Y|X]$, $X$ and $Y$ can be dependent, for example a ‘uniform’ distribution in a unit circle.
    Now we add the variance, if
    $$mathbb{E}[X]=mathbb{E}[X|Y], mathbb{E}[Y]=mathbb{E}[Y|X], $$$$Var(X)=Var(X|Y), Var(Y)=Var(Y|X).$$
    Say the expectation and variance of $X$ are both not affected by $Y$, and vice versa, then must $X$ and $Y$ be independent?
    In this case I can not find a counterexample just like the uniform circle.



    If they are independent, how to prove it? If not, is there a counterexample?



    Thanks!






    probability conditional-expectation independence







    share|cite|improve this question









    New contributor




    Brent Wagner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







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      1








      1





      $begingroup$


      I know that if $mathbb{E}[X]=mathbb{E}[X|Y] , mathbb{E}[Y]=mathbb{E}[Y|X]$, $X$ and $Y$ can be dependent, for example a ‘uniform’ distribution in a unit circle.
      Now we add the variance, if
      $$mathbb{E}[X]=mathbb{E}[X|Y], mathbb{E}[Y]=mathbb{E}[Y|X], $$$$Var(X)=Var(X|Y), Var(Y)=Var(Y|X).$$
      Say the expectation and variance of $X$ are both not affected by $Y$, and vice versa, then must $X$ and $Y$ be independent?
      In this case I can not find a counterexample just like the uniform circle.



      If they are independent, how to prove it? If not, is there a counterexample?



      Thanks!






      probability conditional-expectation independence







      share|cite|improve this question









      New contributor




      Brent Wagner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I know that if $mathbb{E}[X]=mathbb{E}[X|Y] , mathbb{E}[Y]=mathbb{E}[Y|X]$, $X$ and $Y$ can be dependent, for example a ‘uniform’ distribution in a unit circle.
      Now we add the variance, if
      $$mathbb{E}[X]=mathbb{E}[X|Y], mathbb{E}[Y]=mathbb{E}[Y|X], $$$$Var(X)=Var(X|Y), Var(Y)=Var(Y|X).$$
      Say the expectation and variance of $X$ are both not affected by $Y$, and vice versa, then must $X$ and $Y$ be independent?
      In this case I can not find a counterexample just like the uniform circle.



      If they are independent, how to prove it? If not, is there a counterexample?



      Thanks!






      probability conditional-expectation independence




      probability conditional-expectation independence






      share|cite|improve this question









      New contributor




      Brent Wagner is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




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      Check out our Code of Conduct.









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      edited 2 hours ago







      Brent Wagner













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      asked 4 hours ago









      Brent WagnerBrent Wagner

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      Check out our Code of Conduct.






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          2 Answers
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          $begingroup$

          Consider $Xsim N(0,1)$ and $Ysim N(0,2)$ if $X>0$ and $Ysim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).






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            $begingroup$

            Here's a discrete example:
            $$100P(x,y)=begin{array}{c|ccccc}xbackslash y&-2&-1&0&1&2\ hline -2&1&0&6&0&1\-1&0&9&0&9&0\0&6&0&36&0&6\1&0&9&0&9&0\2&1&0&6&0&1end{array}$$
            Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.






            share|cite|improve this answer









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            • $begingroup$
              Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
              $endgroup$
              – Brent Wagner
              2 hours ago











            Your Answer





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            2 Answers
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            2 Answers
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            $begingroup$

            Consider $Xsim N(0,1)$ and $Ysim N(0,2)$ if $X>0$ and $Ysim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Consider $Xsim N(0,1)$ and $Ysim N(0,2)$ if $X>0$ and $Ysim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Consider $Xsim N(0,1)$ and $Ysim N(0,2)$ if $X>0$ and $Ysim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).






                share|cite|improve this answer









                $endgroup$



                Consider $Xsim N(0,1)$ and $Ysim N(0,2)$ if $X>0$ and $Ysim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                DashermanDasherman

                1,065817




                1,065817























                    4












                    $begingroup$

                    Here's a discrete example:
                    $$100P(x,y)=begin{array}{c|ccccc}xbackslash y&-2&-1&0&1&2\ hline -2&1&0&6&0&1\-1&0&9&0&9&0\0&6&0&36&0&6\1&0&9&0&9&0\2&1&0&6&0&1end{array}$$
                    Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
                      $endgroup$
                      – Brent Wagner
                      2 hours ago
















                    4












                    $begingroup$

                    Here's a discrete example:
                    $$100P(x,y)=begin{array}{c|ccccc}xbackslash y&-2&-1&0&1&2\ hline -2&1&0&6&0&1\-1&0&9&0&9&0\0&6&0&36&0&6\1&0&9&0&9&0\2&1&0&6&0&1end{array}$$
                    Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
                      $endgroup$
                      – Brent Wagner
                      2 hours ago














                    4












                    4








                    4





                    $begingroup$

                    Here's a discrete example:
                    $$100P(x,y)=begin{array}{c|ccccc}xbackslash y&-2&-1&0&1&2\ hline -2&1&0&6&0&1\-1&0&9&0&9&0\0&6&0&36&0&6\1&0&9&0&9&0\2&1&0&6&0&1end{array}$$
                    Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.






                    share|cite|improve this answer









                    $endgroup$



                    Here's a discrete example:
                    $$100P(x,y)=begin{array}{c|ccccc}xbackslash y&-2&-1&0&1&2\ hline -2&1&0&6&0&1\-1&0&9&0&9&0\0&6&0&36&0&6\1&0&9&0&9&0\2&1&0&6&0&1end{array}$$
                    Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    jmerryjmerry

                    4,982514




                    4,982514












                    • $begingroup$
                      Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
                      $endgroup$
                      – Brent Wagner
                      2 hours ago


















                    • $begingroup$
                      Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
                      $endgroup$
                      – Brent Wagner
                      2 hours ago
















                    $begingroup$
                    Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
                    $endgroup$
                    – Brent Wagner
                    2 hours ago




                    $begingroup$
                    Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
                    $endgroup$
                    – Brent Wagner
                    2 hours ago










                    Brent Wagner is a new contributor. Be nice, and check out our Code of Conduct.










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