Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$












1












$begingroup$



Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$




My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?



Thanks in advance!






probability-theory conditional-expectation independence







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    $begingroup$
    The correct hypothesis is that Z is independent of (X,Y).
    $endgroup$
    – Did
    Dec 7 '18 at 8:16
















1












$begingroup$



Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$




My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?



Thanks in advance!






probability-theory conditional-expectation independence







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The correct hypothesis is that Z is independent of (X,Y).
    $endgroup$
    – Did
    Dec 7 '18 at 8:16














1












1








1





$begingroup$



Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$




My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?



Thanks in advance!






probability-theory conditional-expectation independence







share|cite|improve this question











$endgroup$





Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$




My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?



Thanks in advance!






probability-theory conditional-expectation independence




probability-theory conditional-expectation independence






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share|cite|improve this question













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share|cite|improve this question








edited Dec 7 '18 at 8:17







bellcircle

















asked Dec 7 '18 at 8:07









bellcirclebellcircle

1,331411




1,331411








  • 1




    $begingroup$
    The correct hypothesis is that Z is independent of (X,Y).
    $endgroup$
    – Did
    Dec 7 '18 at 8:16














  • 1




    $begingroup$
    The correct hypothesis is that Z is independent of (X,Y).
    $endgroup$
    – Did
    Dec 7 '18 at 8:16








1




1




$begingroup$
The correct hypothesis is that Z is independent of (X,Y).
$endgroup$
– Did
Dec 7 '18 at 8:16




$begingroup$
The correct hypothesis is that Z is independent of (X,Y).
$endgroup$
– Did
Dec 7 '18 at 8:16










1 Answer
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$begingroup$

Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.






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    $begingroup$

    Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.






        share|cite|improve this answer











        $endgroup$



        Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 8:30

























        answered Dec 7 '18 at 8:25









        Kavi Rama MurthyKavi Rama Murthy

        58.2k42160




        58.2k42160






























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