Conditional expectation $mathbb{E}[mathbb{E}[Xvert Y,Z]vert Z]$












1












$begingroup$


Problem



When proving one result in the statistical learning theory course, the instructor uses
$$
mathbb{E}[mathbb{E}[Xvert Y,Z]vert Z]=mathbb{E}[Xvert Z]
$$
but I am not sure why this is true.



What I Have Done



I know I could do the following
$$
mathbb{E}[Xvert Y]=int xf_{Xvert Y}(xvert y)dx
$$
But when $X$ becomes complicated like $mathbb{E}[Xvert Y,Z]$ (sorry for the abuse of variable name), I do not know how to proceed.



Could someone help me, thank you in advance.






probability-theory conditional-expectation expected-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    For me I think it's very related to $$ P(A mid B) = P(A mid B,C) P(B mid C) + P(A mid B,C^c) P(B mid C^c) $$
    $endgroup$
    – HJ_beginner
    Mar 27 at 0:24
















1












$begingroup$


Problem



When proving one result in the statistical learning theory course, the instructor uses
$$
mathbb{E}[mathbb{E}[Xvert Y,Z]vert Z]=mathbb{E}[Xvert Z]
$$
but I am not sure why this is true.



What I Have Done



I know I could do the following
$$
mathbb{E}[Xvert Y]=int xf_{Xvert Y}(xvert y)dx
$$
But when $X$ becomes complicated like $mathbb{E}[Xvert Y,Z]$ (sorry for the abuse of variable name), I do not know how to proceed.



Could someone help me, thank you in advance.






probability-theory conditional-expectation expected-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    For me I think it's very related to $$ P(A mid B) = P(A mid B,C) P(B mid C) + P(A mid B,C^c) P(B mid C^c) $$
    $endgroup$
    – HJ_beginner
    Mar 27 at 0:24














1












1








1





$begingroup$


Problem



When proving one result in the statistical learning theory course, the instructor uses
$$
mathbb{E}[mathbb{E}[Xvert Y,Z]vert Z]=mathbb{E}[Xvert Z]
$$
but I am not sure why this is true.



What I Have Done



I know I could do the following
$$
mathbb{E}[Xvert Y]=int xf_{Xvert Y}(xvert y)dx
$$
But when $X$ becomes complicated like $mathbb{E}[Xvert Y,Z]$ (sorry for the abuse of variable name), I do not know how to proceed.



Could someone help me, thank you in advance.






probability-theory conditional-expectation expected-value







share|cite|improve this question









$endgroup$




Problem



When proving one result in the statistical learning theory course, the instructor uses
$$
mathbb{E}[mathbb{E}[Xvert Y,Z]vert Z]=mathbb{E}[Xvert Z]
$$
but I am not sure why this is true.



What I Have Done



I know I could do the following
$$
mathbb{E}[Xvert Y]=int xf_{Xvert Y}(xvert y)dx
$$
But when $X$ becomes complicated like $mathbb{E}[Xvert Y,Z]$ (sorry for the abuse of variable name), I do not know how to proceed.



Could someone help me, thank you in advance.






probability-theory conditional-expectation expected-value




probability-theory conditional-expectation expected-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 15 at 20:09









Mr.RobotMr.Robot

384110




384110












  • $begingroup$
    For me I think it's very related to $$ P(A mid B) = P(A mid B,C) P(B mid C) + P(A mid B,C^c) P(B mid C^c) $$
    $endgroup$
    – HJ_beginner
    Mar 27 at 0:24


















  • $begingroup$
    For me I think it's very related to $$ P(A mid B) = P(A mid B,C) P(B mid C) + P(A mid B,C^c) P(B mid C^c) $$
    $endgroup$
    – HJ_beginner
    Mar 27 at 0:24
















$begingroup$
For me I think it's very related to $$ P(A mid B) = P(A mid B,C) P(B mid C) + P(A mid B,C^c) P(B mid C^c) $$
$endgroup$
– HJ_beginner
Mar 27 at 0:24




$begingroup$
For me I think it's very related to $$ P(A mid B) = P(A mid B,C) P(B mid C) + P(A mid B,C^c) P(B mid C^c) $$
$endgroup$
– HJ_beginner
Mar 27 at 0:24










2 Answers
2






active

oldest

votes


















4












$begingroup$

This is just a special case of the usual
$$mathbb{E}[X] = mathbb{E}[mathbb{E}[X mid Y]]$$
except all expectations are taken under the conditional distribution given the event $Z=z$. If you are still unsure, take your favorite proof of the above equality and replace all PDFs/PMFs with the conditional distribution given $Z=z$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    by tower property



    if $F_1 subset F_2$ so $E(E(X|F_2)|F_1)=E(X|F_1)$



    now



    $mathbb{E}[mathbb{E}[Xvert Y,Z]vert Z]=mathbb{E}[mathbb{E}[Xvert sigma (Y,Z)]vert sigma(Z)]$



    $=mathbb{E}[mathbb{E}[Xvert F_2]vert F_1]=E(X|F_1)$



    $=mathbb{E}[Xvert sigma(Z)]=mathbb{E}[Xvert Z]$



    since $F_1=sigma(Z) subset F_2=sigma (Y,Z)$



    this proof valid for all type of random variable(continues,discrete and mixture). so you do not need any assumption about type of random variable.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      if you want use straight proof you should have 8 proof (at least). X , Y and Z each can be discrete or continues so you have 8 ways. what about mixtures variable?!!!!
      $endgroup$
      – masoud
      Mar 27 at 0:14












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    This is just a special case of the usual
    $$mathbb{E}[X] = mathbb{E}[mathbb{E}[X mid Y]]$$
    except all expectations are taken under the conditional distribution given the event $Z=z$. If you are still unsure, take your favorite proof of the above equality and replace all PDFs/PMFs with the conditional distribution given $Z=z$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      This is just a special case of the usual
      $$mathbb{E}[X] = mathbb{E}[mathbb{E}[X mid Y]]$$
      except all expectations are taken under the conditional distribution given the event $Z=z$. If you are still unsure, take your favorite proof of the above equality and replace all PDFs/PMFs with the conditional distribution given $Z=z$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        This is just a special case of the usual
        $$mathbb{E}[X] = mathbb{E}[mathbb{E}[X mid Y]]$$
        except all expectations are taken under the conditional distribution given the event $Z=z$. If you are still unsure, take your favorite proof of the above equality and replace all PDFs/PMFs with the conditional distribution given $Z=z$.






        share|cite|improve this answer









        $endgroup$



        This is just a special case of the usual
        $$mathbb{E}[X] = mathbb{E}[mathbb{E}[X mid Y]]$$
        except all expectations are taken under the conditional distribution given the event $Z=z$. If you are still unsure, take your favorite proof of the above equality and replace all PDFs/PMFs with the conditional distribution given $Z=z$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 20:21









        angryavianangryavian

        42.5k23481




        42.5k23481























            0












            $begingroup$

            by tower property



            if $F_1 subset F_2$ so $E(E(X|F_2)|F_1)=E(X|F_1)$



            now



            $mathbb{E}[mathbb{E}[Xvert Y,Z]vert Z]=mathbb{E}[mathbb{E}[Xvert sigma (Y,Z)]vert sigma(Z)]$



            $=mathbb{E}[mathbb{E}[Xvert F_2]vert F_1]=E(X|F_1)$



            $=mathbb{E}[Xvert sigma(Z)]=mathbb{E}[Xvert Z]$



            since $F_1=sigma(Z) subset F_2=sigma (Y,Z)$



            this proof valid for all type of random variable(continues,discrete and mixture). so you do not need any assumption about type of random variable.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              if you want use straight proof you should have 8 proof (at least). X , Y and Z each can be discrete or continues so you have 8 ways. what about mixtures variable?!!!!
              $endgroup$
              – masoud
              Mar 27 at 0:14
















            0












            $begingroup$

            by tower property



            if $F_1 subset F_2$ so $E(E(X|F_2)|F_1)=E(X|F_1)$



            now



            $mathbb{E}[mathbb{E}[Xvert Y,Z]vert Z]=mathbb{E}[mathbb{E}[Xvert sigma (Y,Z)]vert sigma(Z)]$



            $=mathbb{E}[mathbb{E}[Xvert F_2]vert F_1]=E(X|F_1)$



            $=mathbb{E}[Xvert sigma(Z)]=mathbb{E}[Xvert Z]$



            since $F_1=sigma(Z) subset F_2=sigma (Y,Z)$



            this proof valid for all type of random variable(continues,discrete and mixture). so you do not need any assumption about type of random variable.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              if you want use straight proof you should have 8 proof (at least). X , Y and Z each can be discrete or continues so you have 8 ways. what about mixtures variable?!!!!
              $endgroup$
              – masoud
              Mar 27 at 0:14














            0












            0








            0





            $begingroup$

            by tower property



            if $F_1 subset F_2$ so $E(E(X|F_2)|F_1)=E(X|F_1)$



            now



            $mathbb{E}[mathbb{E}[Xvert Y,Z]vert Z]=mathbb{E}[mathbb{E}[Xvert sigma (Y,Z)]vert sigma(Z)]$



            $=mathbb{E}[mathbb{E}[Xvert F_2]vert F_1]=E(X|F_1)$



            $=mathbb{E}[Xvert sigma(Z)]=mathbb{E}[Xvert Z]$



            since $F_1=sigma(Z) subset F_2=sigma (Y,Z)$



            this proof valid for all type of random variable(continues,discrete and mixture). so you do not need any assumption about type of random variable.






            share|cite|improve this answer











            $endgroup$



            by tower property



            if $F_1 subset F_2$ so $E(E(X|F_2)|F_1)=E(X|F_1)$



            now



            $mathbb{E}[mathbb{E}[Xvert Y,Z]vert Z]=mathbb{E}[mathbb{E}[Xvert sigma (Y,Z)]vert sigma(Z)]$



            $=mathbb{E}[mathbb{E}[Xvert F_2]vert F_1]=E(X|F_1)$



            $=mathbb{E}[Xvert sigma(Z)]=mathbb{E}[Xvert Z]$



            since $F_1=sigma(Z) subset F_2=sigma (Y,Z)$



            this proof valid for all type of random variable(continues,discrete and mixture). so you do not need any assumption about type of random variable.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 27 at 0:18

























            answered Mar 27 at 0:05









            masoudmasoud

            1506




            1506












            • $begingroup$
              if you want use straight proof you should have 8 proof (at least). X , Y and Z each can be discrete or continues so you have 8 ways. what about mixtures variable?!!!!
              $endgroup$
              – masoud
              Mar 27 at 0:14


















            • $begingroup$
              if you want use straight proof you should have 8 proof (at least). X , Y and Z each can be discrete or continues so you have 8 ways. what about mixtures variable?!!!!
              $endgroup$
              – masoud
              Mar 27 at 0:14
















            $begingroup$
            if you want use straight proof you should have 8 proof (at least). X , Y and Z each can be discrete or continues so you have 8 ways. what about mixtures variable?!!!!
            $endgroup$
            – masoud
            Mar 27 at 0:14




            $begingroup$
            if you want use straight proof you should have 8 proof (at least). X , Y and Z each can be discrete or continues so you have 8 ways. what about mixtures variable?!!!!
            $endgroup$
            – masoud
            Mar 27 at 0:14


















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