Expectation of the product of two random variables is the supremum over all identically distributed copies of one of the factors.

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Let $X in L^1$ and $Y in L^infty$ be two random variables (or, more generally, $X in L^p, , Y in L^q$ for $p, , q$ conjugated). Is it true that



$$ mathbbE left[ X Y right] = sup_ hatY sim Y mathbbE left[ X hatY right]$$



where the supremum is taken over all $hatY$ that have the same distribution as $Y$?



Clearly, the right-hand side is well-defined, but could a-priori be $infty$. Also, the right-hand side is clearly larger than the left.



How could one prove the other direction $geq$?




probability probability-theory random-variables




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    down vote

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    Let $X in L^1$ and $Y in L^infty$ be two random variables (or, more generally, $X in L^p, , Y in L^q$ for $p, , q$ conjugated). Is it true that



    $$ mathbbE left[ X Y right] = sup_ hatY sim Y mathbbE left[ X hatY right]$$



    where the supremum is taken over all $hatY$ that have the same distribution as $Y$?



    Clearly, the right-hand side is well-defined, but could a-priori be $infty$. Also, the right-hand side is clearly larger than the left.



    How could one prove the other direction $geq$?




    probability probability-theory random-variables




    share|cite|improve this question





















      up vote
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      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $X in L^1$ and $Y in L^infty$ be two random variables (or, more generally, $X in L^p, , Y in L^q$ for $p, , q$ conjugated). Is it true that



      $$ mathbbE left[ X Y right] = sup_ hatY sim Y mathbbE left[ X hatY right]$$



      where the supremum is taken over all $hatY$ that have the same distribution as $Y$?



      Clearly, the right-hand side is well-defined, but could a-priori be $infty$. Also, the right-hand side is clearly larger than the left.



      How could one prove the other direction $geq$?




      probability probability-theory random-variables




      share|cite|improve this question











      Let $X in L^1$ and $Y in L^infty$ be two random variables (or, more generally, $X in L^p, , Y in L^q$ for $p, , q$ conjugated). Is it true that



      $$ mathbbE left[ X Y right] = sup_ hatY sim Y mathbbE left[ X hatY right]$$



      where the supremum is taken over all $hatY$ that have the same distribution as $Y$?



      Clearly, the right-hand side is well-defined, but could a-priori be $infty$. Also, the right-hand side is clearly larger than the left.



      How could one prove the other direction $geq$?





      probability probability-theory random-variables





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      asked Aug 2 at 7:59









      LittleDoe

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          False. Let $Y$ have standard normal distribution and $X=-Y$. Then $EXY=-1$ but $hat Y =X$ has the same distribution as $Y$ so RHS $geq EX^2=1$. This answers the second part. For the first part you need a bounded random variable, so replace normal distribution by any symmetric bounded distribution.






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            False. Let $Y$ have standard normal distribution and $X=-Y$. Then $EXY=-1$ but $hat Y =X$ has the same distribution as $Y$ so RHS $geq EX^2=1$. This answers the second part. For the first part you need a bounded random variable, so replace normal distribution by any symmetric bounded distribution.






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              up vote
              1
              down vote



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              False. Let $Y$ have standard normal distribution and $X=-Y$. Then $EXY=-1$ but $hat Y =X$ has the same distribution as $Y$ so RHS $geq EX^2=1$. This answers the second part. For the first part you need a bounded random variable, so replace normal distribution by any symmetric bounded distribution.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
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                down vote



                accepted






                False. Let $Y$ have standard normal distribution and $X=-Y$. Then $EXY=-1$ but $hat Y =X$ has the same distribution as $Y$ so RHS $geq EX^2=1$. This answers the second part. For the first part you need a bounded random variable, so replace normal distribution by any symmetric bounded distribution.






                share|cite|improve this answer













                False. Let $Y$ have standard normal distribution and $X=-Y$. Then $EXY=-1$ but $hat Y =X$ has the same distribution as $Y$ so RHS $geq EX^2=1$. This answers the second part. For the first part you need a bounded random variable, so replace normal distribution by any symmetric bounded distribution.







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                answered Aug 2 at 8:33









                Kavi Rama Murthy

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