$E(X^2|X-Y) E(X^3|X-2Y)$ for Gaussians?












0












$begingroup$


For independent gaussians with following the normal distribution with expectation zero and variance one, how do I compute:



$E(X^2|X-2Y), E(X^3|X-2Y)$



I know that $X-2Y$,$X+2Y$ are independent. However, this does not seem to be enough to deduce the result, without a restriction such as:



$E(X^2|X-2Y)=-2E(Y^2|X-2Y)$



(I am not sure if this holds).






conditional-expectation







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    For independent gaussians with following the normal distribution with expectation zero and variance one, how do I compute:



    $E(X^2|X-2Y), E(X^3|X-2Y)$



    I know that $X-2Y$,$X+2Y$ are independent. However, this does not seem to be enough to deduce the result, without a restriction such as:



    $E(X^2|X-2Y)=-2E(Y^2|X-2Y)$



    (I am not sure if this holds).






    conditional-expectation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      For independent gaussians with following the normal distribution with expectation zero and variance one, how do I compute:



      $E(X^2|X-2Y), E(X^3|X-2Y)$



      I know that $X-2Y$,$X+2Y$ are independent. However, this does not seem to be enough to deduce the result, without a restriction such as:



      $E(X^2|X-2Y)=-2E(Y^2|X-2Y)$



      (I am not sure if this holds).






      conditional-expectation







      share|cite|improve this question











      $endgroup$




      For independent gaussians with following the normal distribution with expectation zero and variance one, how do I compute:



      $E(X^2|X-2Y), E(X^3|X-2Y)$



      I know that $X-2Y$,$X+2Y$ are independent. However, this does not seem to be enough to deduce the result, without a restriction such as:



      $E(X^2|X-2Y)=-2E(Y^2|X-2Y)$



      (I am not sure if this holds).






      conditional-expectation




      conditional-expectation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 '18 at 9:38







      Dole

















      asked Nov 27 '18 at 9:32









      DoleDole

      901514




      901514






















          1 Answer
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          $begingroup$

          $X=frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=frac 5 4+frac {(X-2Y)^{2}} 4.$$



          This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
            $endgroup$
            – Dole
            Nov 27 '18 at 11:02








          • 1




            $begingroup$
            Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
            $endgroup$
            – Kavi Rama Murthy
            Nov 27 '18 at 11:49













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          $begingroup$

          $X=frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=frac 5 4+frac {(X-2Y)^{2}} 4.$$



          This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
            $endgroup$
            – Dole
            Nov 27 '18 at 11:02








          • 1




            $begingroup$
            Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
            $endgroup$
            – Kavi Rama Murthy
            Nov 27 '18 at 11:49


















          1












          $begingroup$

          $X=frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=frac 5 4+frac {(X-2Y)^{2}} 4.$$



          This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
            $endgroup$
            – Dole
            Nov 27 '18 at 11:02








          • 1




            $begingroup$
            Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
            $endgroup$
            – Kavi Rama Murthy
            Nov 27 '18 at 11:49
















          1












          1








          1





          $begingroup$

          $X=frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=frac 5 4+frac {(X-2Y)^{2}} 4.$$



          This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.






          share|cite|improve this answer











          $endgroup$



          $X=frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=frac 5 4+frac {(X-2Y)^{2}} 4.$$



          This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 27 '18 at 11:55

























          answered Nov 27 '18 at 9:38









          Kavi Rama MurthyKavi Rama Murthy

          56k42158




          56k42158












          • $begingroup$
            Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
            $endgroup$
            – Dole
            Nov 27 '18 at 11:02








          • 1




            $begingroup$
            Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
            $endgroup$
            – Kavi Rama Murthy
            Nov 27 '18 at 11:49




















          • $begingroup$
            Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
            $endgroup$
            – Dole
            Nov 27 '18 at 11:02








          • 1




            $begingroup$
            Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
            $endgroup$
            – Kavi Rama Murthy
            Nov 27 '18 at 11:49


















          $begingroup$
          Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
          $endgroup$
          – Dole
          Nov 27 '18 at 11:02






          $begingroup$
          Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$.
          $endgroup$
          – Dole
          Nov 27 '18 at 11:02






          1




          1




          $begingroup$
          Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
          $endgroup$
          – Kavi Rama Murthy
          Nov 27 '18 at 11:49






          $begingroup$
          Use $X=frac {(X-2Y)+2(2X+Y)} 5$.
          $endgroup$
          – Kavi Rama Murthy
          Nov 27 '18 at 11:49




















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