What can be said about i.i.d. $X$ and $Y$ such that $XY=(X+Y)/2$ in distribution?












6












$begingroup$


Let $X$ and $Y$ be i.i.d.



If $(X+Y)/2$ is equal in distribution to $XY$, then what do we know about the distributions of $X$ and $Y$?



I feel like I can't say much about these distributions. I can only think of degenerate examples, but I have a feeling that there is something crafty going on here that I am missing.






probability probability-theory probability-distributions independence







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$endgroup$












  • $begingroup$
    Good question. My guess is that $X$ must be degenerate. As you have observed some clever argument may give this.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 6:30


















6












$begingroup$


Let $X$ and $Y$ be i.i.d.



If $(X+Y)/2$ is equal in distribution to $XY$, then what do we know about the distributions of $X$ and $Y$?



I feel like I can't say much about these distributions. I can only think of degenerate examples, but I have a feeling that there is something crafty going on here that I am missing.






probability probability-theory probability-distributions independence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Good question. My guess is that $X$ must be degenerate. As you have observed some clever argument may give this.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 6:30
















6












6








6


3



$begingroup$


Let $X$ and $Y$ be i.i.d.



If $(X+Y)/2$ is equal in distribution to $XY$, then what do we know about the distributions of $X$ and $Y$?



I feel like I can't say much about these distributions. I can only think of degenerate examples, but I have a feeling that there is something crafty going on here that I am missing.






probability probability-theory probability-distributions independence







share|cite|improve this question











$endgroup$




Let $X$ and $Y$ be i.i.d.



If $(X+Y)/2$ is equal in distribution to $XY$, then what do we know about the distributions of $X$ and $Y$?



I feel like I can't say much about these distributions. I can only think of degenerate examples, but I have a feeling that there is something crafty going on here that I am missing.






probability probability-theory probability-distributions independence




probability probability-theory probability-distributions independence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 13:17









Davide Giraudo

126k16150261




126k16150261










asked Dec 10 '18 at 6:12









kpr62kpr62

704




704












  • $begingroup$
    Good question. My guess is that $X$ must be degenerate. As you have observed some clever argument may give this.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 6:30




















  • $begingroup$
    Good question. My guess is that $X$ must be degenerate. As you have observed some clever argument may give this.
    $endgroup$
    – Kavi Rama Murthy
    Dec 10 '18 at 6:30


















$begingroup$
Good question. My guess is that $X$ must be degenerate. As you have observed some clever argument may give this.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 6:30






$begingroup$
Good question. My guess is that $X$ must be degenerate. As you have observed some clever argument may give this.
$endgroup$
– Kavi Rama Murthy
Dec 10 '18 at 6:30












2 Answers
2






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$begingroup$

I assume the first two moments of $X$ exist.



$$E[X] = frac{E[X] + E[X]}{2} = Eleft[frac{X+Y}{2}right] = E[XY] = E[X]E[Y] = E[X]^2$$
so $E[X]$ is either $0$ or $1$.



Similarly
$$frac{E[X^2]+E[X]^2}{2} = E frac{(X+Y)^2}{4} = E[X^2 Y^2] = E[X^2] E[Y^2] = E[X^2]^2.$$




  • In the case $E[X]=0$ we have $E[X^2] = 2 E[X^2]^2$ so $E[X^2] = text{Var}(X)$ is either $0$ or $1/2$.

  • In the case $E[X]=1$ we have $E[X^2] = 2 E[X^2]^2 - 1$, so $E[X^2] = 1$. But this means $text{Var}(X) = 0$.


The only non-degenerate case is $E[X] = 0$ and $text{Var}(X) = 1/2$. I am not sure how to construct an example for this case or show that this case is impossible.






share|cite|improve this answer











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    0












    $begingroup$

    Try something like $X=sin 2pi T$ with $Tsim U(0,1)$






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      3












      $begingroup$

      I assume the first two moments of $X$ exist.



      $$E[X] = frac{E[X] + E[X]}{2} = Eleft[frac{X+Y}{2}right] = E[XY] = E[X]E[Y] = E[X]^2$$
      so $E[X]$ is either $0$ or $1$.



      Similarly
      $$frac{E[X^2]+E[X]^2}{2} = E frac{(X+Y)^2}{4} = E[X^2 Y^2] = E[X^2] E[Y^2] = E[X^2]^2.$$




      • In the case $E[X]=0$ we have $E[X^2] = 2 E[X^2]^2$ so $E[X^2] = text{Var}(X)$ is either $0$ or $1/2$.

      • In the case $E[X]=1$ we have $E[X^2] = 2 E[X^2]^2 - 1$, so $E[X^2] = 1$. But this means $text{Var}(X) = 0$.


      The only non-degenerate case is $E[X] = 0$ and $text{Var}(X) = 1/2$. I am not sure how to construct an example for this case or show that this case is impossible.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        I assume the first two moments of $X$ exist.



        $$E[X] = frac{E[X] + E[X]}{2} = Eleft[frac{X+Y}{2}right] = E[XY] = E[X]E[Y] = E[X]^2$$
        so $E[X]$ is either $0$ or $1$.



        Similarly
        $$frac{E[X^2]+E[X]^2}{2} = E frac{(X+Y)^2}{4} = E[X^2 Y^2] = E[X^2] E[Y^2] = E[X^2]^2.$$




        • In the case $E[X]=0$ we have $E[X^2] = 2 E[X^2]^2$ so $E[X^2] = text{Var}(X)$ is either $0$ or $1/2$.

        • In the case $E[X]=1$ we have $E[X^2] = 2 E[X^2]^2 - 1$, so $E[X^2] = 1$. But this means $text{Var}(X) = 0$.


        The only non-degenerate case is $E[X] = 0$ and $text{Var}(X) = 1/2$. I am not sure how to construct an example for this case or show that this case is impossible.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          I assume the first two moments of $X$ exist.



          $$E[X] = frac{E[X] + E[X]}{2} = Eleft[frac{X+Y}{2}right] = E[XY] = E[X]E[Y] = E[X]^2$$
          so $E[X]$ is either $0$ or $1$.



          Similarly
          $$frac{E[X^2]+E[X]^2}{2} = E frac{(X+Y)^2}{4} = E[X^2 Y^2] = E[X^2] E[Y^2] = E[X^2]^2.$$




          • In the case $E[X]=0$ we have $E[X^2] = 2 E[X^2]^2$ so $E[X^2] = text{Var}(X)$ is either $0$ or $1/2$.

          • In the case $E[X]=1$ we have $E[X^2] = 2 E[X^2]^2 - 1$, so $E[X^2] = 1$. But this means $text{Var}(X) = 0$.


          The only non-degenerate case is $E[X] = 0$ and $text{Var}(X) = 1/2$. I am not sure how to construct an example for this case or show that this case is impossible.






          share|cite|improve this answer











          $endgroup$



          I assume the first two moments of $X$ exist.



          $$E[X] = frac{E[X] + E[X]}{2} = Eleft[frac{X+Y}{2}right] = E[XY] = E[X]E[Y] = E[X]^2$$
          so $E[X]$ is either $0$ or $1$.



          Similarly
          $$frac{E[X^2]+E[X]^2}{2} = E frac{(X+Y)^2}{4} = E[X^2 Y^2] = E[X^2] E[Y^2] = E[X^2]^2.$$




          • In the case $E[X]=0$ we have $E[X^2] = 2 E[X^2]^2$ so $E[X^2] = text{Var}(X)$ is either $0$ or $1/2$.

          • In the case $E[X]=1$ we have $E[X^2] = 2 E[X^2]^2 - 1$, so $E[X^2] = 1$. But this means $text{Var}(X) = 0$.


          The only non-degenerate case is $E[X] = 0$ and $text{Var}(X) = 1/2$. I am not sure how to construct an example for this case or show that this case is impossible.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 20:29

























          answered Dec 10 '18 at 7:08









          angryavianangryavian

          40.4k23280




          40.4k23280























              0












              $begingroup$

              Try something like $X=sin 2pi T$ with $Tsim U(0,1)$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Try something like $X=sin 2pi T$ with $Tsim U(0,1)$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Try something like $X=sin 2pi T$ with $Tsim U(0,1)$






                  share|cite|improve this answer









                  $endgroup$



                  Try something like $X=sin 2pi T$ with $Tsim U(0,1)$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 14:00









                  Empy2Empy2

                  33.5k12261




                  33.5k12261






























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