If $mathbb E[Xmid Y]$ can be seen as a projection why $mathbb E[Xmid Y]=frac{mathbb E[XY]}{mathbb E[Y^2]}Y$...












1












$begingroup$


We know that $mathbb E[XY]$ is a scalaire product on $L^2(mathbb P)$. In a book (an introduction to stochastic differential equation of Evans) page 30-31, it's written that $mathbb E[Xmid Y]$ can be seen as the orthogonal projection of $X$ on $Y$. So, why $$mathbb E[Xmid Y]=frac{mathbb E[XY]}{mathbb E[Y^2]}Y,$$
not always true ?






measure-theory







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    We know that $mathbb E[XY]$ is a scalaire product on $L^2(mathbb P)$. In a book (an introduction to stochastic differential equation of Evans) page 30-31, it's written that $mathbb E[Xmid Y]$ can be seen as the orthogonal projection of $X$ on $Y$. So, why $$mathbb E[Xmid Y]=frac{mathbb E[XY]}{mathbb E[Y^2]}Y,$$
    not always true ?






    measure-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      We know that $mathbb E[XY]$ is a scalaire product on $L^2(mathbb P)$. In a book (an introduction to stochastic differential equation of Evans) page 30-31, it's written that $mathbb E[Xmid Y]$ can be seen as the orthogonal projection of $X$ on $Y$. So, why $$mathbb E[Xmid Y]=frac{mathbb E[XY]}{mathbb E[Y^2]}Y,$$
      not always true ?






      measure-theory







      share|cite|improve this question









      $endgroup$




      We know that $mathbb E[XY]$ is a scalaire product on $L^2(mathbb P)$. In a book (an introduction to stochastic differential equation of Evans) page 30-31, it's written that $mathbb E[Xmid Y]$ can be seen as the orthogonal projection of $X$ on $Y$. So, why $$mathbb E[Xmid Y]=frac{mathbb E[XY]}{mathbb E[Y^2]}Y,$$
      not always true ?






      measure-theory




      measure-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 5 '18 at 9:15









      NewMathNewMath

      4059




      4059






















          3 Answers
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          active

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          2












          $begingroup$

          You are mixing up the notions of projection on the one-dimensional vector space spanned by $Y$ and projection on the space of all random variables measurable with respect to $sigma (Y)$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Iknow it's false. But the question is why it doesn't work ? Maybe $mathbb E[XY]$ is not really a scalar product ?
            $endgroup$
            – NewMath
            Dec 5 '18 at 9:41










          • $begingroup$
            @NewMath I have given an explanation now.
            $endgroup$
            – Kavi Rama Murthy
            Dec 5 '18 at 9:46










          • $begingroup$
            The second sentence of this post is the correct answer, everything else on this page, so far, is off the mark. (+1)
            $endgroup$
            – Did
            Dec 5 '18 at 10:00






          • 1




            $begingroup$
            Ahhh... user Surb just added the argument to their answer. Cool.
            $endgroup$
            – Did
            Dec 5 '18 at 10:01










          • $begingroup$
            @Did: thank you for your comment :)
            $endgroup$
            – NewMath
            Dec 5 '18 at 15:27





















          2












          $begingroup$

          I'm sure that in your book it's written that "$mathbb E[Xmid Y]$ it the orthogonal projection onto $sigma (Y)$" instead of "on $Y$". If $V$ is a finite vector, $W$ a subspace of $V$ and ${w_1,...,w_n}$ an orthogonal basis, then indeed $$Proj_W(v)=sum_{i=1}^nfrac{left<v,w_iright>}{|w_i|^2}w_i.$$



          This can be prolonged on $V$ and $W$ with infinite dimension with certain conditions (like Hilbert). Now, what would be an orthogonal basis of $L^2(mathbb P,sigma (Y))$ ?






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Well if you take $X$ and $Y$ to be independent, the equation reads $$1 = frac{Ymathbb{E}(Y)}{mathbb{E}(Y^2)},$$ which seems a little bit weird.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              You are mixing up the notions of projection on the one-dimensional vector space spanned by $Y$ and projection on the space of all random variables measurable with respect to $sigma (Y)$.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Iknow it's false. But the question is why it doesn't work ? Maybe $mathbb E[XY]$ is not really a scalar product ?
                $endgroup$
                – NewMath
                Dec 5 '18 at 9:41










              • $begingroup$
                @NewMath I have given an explanation now.
                $endgroup$
                – Kavi Rama Murthy
                Dec 5 '18 at 9:46










              • $begingroup$
                The second sentence of this post is the correct answer, everything else on this page, so far, is off the mark. (+1)
                $endgroup$
                – Did
                Dec 5 '18 at 10:00






              • 1




                $begingroup$
                Ahhh... user Surb just added the argument to their answer. Cool.
                $endgroup$
                – Did
                Dec 5 '18 at 10:01










              • $begingroup$
                @Did: thank you for your comment :)
                $endgroup$
                – NewMath
                Dec 5 '18 at 15:27


















              2












              $begingroup$

              You are mixing up the notions of projection on the one-dimensional vector space spanned by $Y$ and projection on the space of all random variables measurable with respect to $sigma (Y)$.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Iknow it's false. But the question is why it doesn't work ? Maybe $mathbb E[XY]$ is not really a scalar product ?
                $endgroup$
                – NewMath
                Dec 5 '18 at 9:41










              • $begingroup$
                @NewMath I have given an explanation now.
                $endgroup$
                – Kavi Rama Murthy
                Dec 5 '18 at 9:46










              • $begingroup$
                The second sentence of this post is the correct answer, everything else on this page, so far, is off the mark. (+1)
                $endgroup$
                – Did
                Dec 5 '18 at 10:00






              • 1




                $begingroup$
                Ahhh... user Surb just added the argument to their answer. Cool.
                $endgroup$
                – Did
                Dec 5 '18 at 10:01










              • $begingroup$
                @Did: thank you for your comment :)
                $endgroup$
                – NewMath
                Dec 5 '18 at 15:27
















              2












              2








              2





              $begingroup$

              You are mixing up the notions of projection on the one-dimensional vector space spanned by $Y$ and projection on the space of all random variables measurable with respect to $sigma (Y)$.






              share|cite|improve this answer











              $endgroup$



              You are mixing up the notions of projection on the one-dimensional vector space spanned by $Y$ and projection on the space of all random variables measurable with respect to $sigma (Y)$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 5 '18 at 15:52









              Did

              247k23223460




              247k23223460










              answered Dec 5 '18 at 9:26









              Kavi Rama MurthyKavi Rama Murthy

              56.6k42159




              56.6k42159








              • 1




                $begingroup$
                Iknow it's false. But the question is why it doesn't work ? Maybe $mathbb E[XY]$ is not really a scalar product ?
                $endgroup$
                – NewMath
                Dec 5 '18 at 9:41










              • $begingroup$
                @NewMath I have given an explanation now.
                $endgroup$
                – Kavi Rama Murthy
                Dec 5 '18 at 9:46










              • $begingroup$
                The second sentence of this post is the correct answer, everything else on this page, so far, is off the mark. (+1)
                $endgroup$
                – Did
                Dec 5 '18 at 10:00






              • 1




                $begingroup$
                Ahhh... user Surb just added the argument to their answer. Cool.
                $endgroup$
                – Did
                Dec 5 '18 at 10:01










              • $begingroup$
                @Did: thank you for your comment :)
                $endgroup$
                – NewMath
                Dec 5 '18 at 15:27
















              • 1




                $begingroup$
                Iknow it's false. But the question is why it doesn't work ? Maybe $mathbb E[XY]$ is not really a scalar product ?
                $endgroup$
                – NewMath
                Dec 5 '18 at 9:41










              • $begingroup$
                @NewMath I have given an explanation now.
                $endgroup$
                – Kavi Rama Murthy
                Dec 5 '18 at 9:46










              • $begingroup$
                The second sentence of this post is the correct answer, everything else on this page, so far, is off the mark. (+1)
                $endgroup$
                – Did
                Dec 5 '18 at 10:00






              • 1




                $begingroup$
                Ahhh... user Surb just added the argument to their answer. Cool.
                $endgroup$
                – Did
                Dec 5 '18 at 10:01










              • $begingroup$
                @Did: thank you for your comment :)
                $endgroup$
                – NewMath
                Dec 5 '18 at 15:27










              1




              1




              $begingroup$
              Iknow it's false. But the question is why it doesn't work ? Maybe $mathbb E[XY]$ is not really a scalar product ?
              $endgroup$
              – NewMath
              Dec 5 '18 at 9:41




              $begingroup$
              Iknow it's false. But the question is why it doesn't work ? Maybe $mathbb E[XY]$ is not really a scalar product ?
              $endgroup$
              – NewMath
              Dec 5 '18 at 9:41












              $begingroup$
              @NewMath I have given an explanation now.
              $endgroup$
              – Kavi Rama Murthy
              Dec 5 '18 at 9:46




              $begingroup$
              @NewMath I have given an explanation now.
              $endgroup$
              – Kavi Rama Murthy
              Dec 5 '18 at 9:46












              $begingroup$
              The second sentence of this post is the correct answer, everything else on this page, so far, is off the mark. (+1)
              $endgroup$
              – Did
              Dec 5 '18 at 10:00




              $begingroup$
              The second sentence of this post is the correct answer, everything else on this page, so far, is off the mark. (+1)
              $endgroup$
              – Did
              Dec 5 '18 at 10:00




              1




              1




              $begingroup$
              Ahhh... user Surb just added the argument to their answer. Cool.
              $endgroup$
              – Did
              Dec 5 '18 at 10:01




              $begingroup$
              Ahhh... user Surb just added the argument to their answer. Cool.
              $endgroup$
              – Did
              Dec 5 '18 at 10:01












              $begingroup$
              @Did: thank you for your comment :)
              $endgroup$
              – NewMath
              Dec 5 '18 at 15:27






              $begingroup$
              @Did: thank you for your comment :)
              $endgroup$
              – NewMath
              Dec 5 '18 at 15:27













              2












              $begingroup$

              I'm sure that in your book it's written that "$mathbb E[Xmid Y]$ it the orthogonal projection onto $sigma (Y)$" instead of "on $Y$". If $V$ is a finite vector, $W$ a subspace of $V$ and ${w_1,...,w_n}$ an orthogonal basis, then indeed $$Proj_W(v)=sum_{i=1}^nfrac{left<v,w_iright>}{|w_i|^2}w_i.$$



              This can be prolonged on $V$ and $W$ with infinite dimension with certain conditions (like Hilbert). Now, what would be an orthogonal basis of $L^2(mathbb P,sigma (Y))$ ?






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                I'm sure that in your book it's written that "$mathbb E[Xmid Y]$ it the orthogonal projection onto $sigma (Y)$" instead of "on $Y$". If $V$ is a finite vector, $W$ a subspace of $V$ and ${w_1,...,w_n}$ an orthogonal basis, then indeed $$Proj_W(v)=sum_{i=1}^nfrac{left<v,w_iright>}{|w_i|^2}w_i.$$



                This can be prolonged on $V$ and $W$ with infinite dimension with certain conditions (like Hilbert). Now, what would be an orthogonal basis of $L^2(mathbb P,sigma (Y))$ ?






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  I'm sure that in your book it's written that "$mathbb E[Xmid Y]$ it the orthogonal projection onto $sigma (Y)$" instead of "on $Y$". If $V$ is a finite vector, $W$ a subspace of $V$ and ${w_1,...,w_n}$ an orthogonal basis, then indeed $$Proj_W(v)=sum_{i=1}^nfrac{left<v,w_iright>}{|w_i|^2}w_i.$$



                  This can be prolonged on $V$ and $W$ with infinite dimension with certain conditions (like Hilbert). Now, what would be an orthogonal basis of $L^2(mathbb P,sigma (Y))$ ?






                  share|cite|improve this answer











                  $endgroup$



                  I'm sure that in your book it's written that "$mathbb E[Xmid Y]$ it the orthogonal projection onto $sigma (Y)$" instead of "on $Y$". If $V$ is a finite vector, $W$ a subspace of $V$ and ${w_1,...,w_n}$ an orthogonal basis, then indeed $$Proj_W(v)=sum_{i=1}^nfrac{left<v,w_iright>}{|w_i|^2}w_i.$$



                  This can be prolonged on $V$ and $W$ with infinite dimension with certain conditions (like Hilbert). Now, what would be an orthogonal basis of $L^2(mathbb P,sigma (Y))$ ?







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 5 '18 at 9:59

























                  answered Dec 5 '18 at 9:46









                  SurbSurb

                  37.6k94375




                  37.6k94375























                      0












                      $begingroup$

                      Well if you take $X$ and $Y$ to be independent, the equation reads $$1 = frac{Ymathbb{E}(Y)}{mathbb{E}(Y^2)},$$ which seems a little bit weird.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Well if you take $X$ and $Y$ to be independent, the equation reads $$1 = frac{Ymathbb{E}(Y)}{mathbb{E}(Y^2)},$$ which seems a little bit weird.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Well if you take $X$ and $Y$ to be independent, the equation reads $$1 = frac{Ymathbb{E}(Y)}{mathbb{E}(Y^2)},$$ which seems a little bit weird.






                          share|cite|improve this answer









                          $endgroup$



                          Well if you take $X$ and $Y$ to be independent, the equation reads $$1 = frac{Ymathbb{E}(Y)}{mathbb{E}(Y^2)},$$ which seems a little bit weird.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 5 '18 at 9:21









                          StockfishStockfish

                          62726




                          62726






























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