Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$
$begingroup$
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
asked Dec 7 '18 at 8:07
bellcirclebellcircle
1,331411
$endgroup$
add a comment |
$begingroup$
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
asked Dec 7 '18 at 8:07
bellcirclebellcircle
1,331411
$endgroup$
1
$begingroup$
The correct hypothesis is that Z is independent of (X,Y).
$endgroup$
– Did
Dec 7 '18 at 8:16
add a comment |
$begingroup$
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
asked Dec 7 '18 at 8:07
bellcirclebellcircle
1,331411
$endgroup$
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
asked Dec 7 '18 at 8:07
bellcirclebellcircle
1,331411
asked Dec 7 '18 at 8:07
bellcirclebellcircle
1,331411
edited Dec 7 '18 at 8:17
bellcircle
asked Dec 7 '18 at 8:07
bellcirclebellcircle
1,331411
asked Dec 7 '18 at 8:07
bellcirclebellcircle
1,331411
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
add a comment |
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
add a comment |
1 Answer
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oldest
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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 $E(Y|X)$.
answered Dec 7 '18 at 8:25
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
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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 $E(Y|X)$.
answered Dec 7 '18 at 8:25
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
$begingroup$
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 $E(Y|X)$.
answered Dec 7 '18 at 8:25
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
add a comment |
$begingroup$
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 $E(Y|X)$.
answered Dec 7 '18 at 8:25
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
$endgroup$
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 $E(Y|X)$.
answered Dec 7 '18 at 8:25
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
edited Dec 7 '18 at 8:30
answered Dec 7 '18 at 8:25
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Dec 7 '18 at 8:25
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
answered Dec 7 '18 at 8:25
Kavi Rama MurthyKavi Rama Murthy
59.2k42161
59.2k42161
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The correct hypothesis is that Z is independent of (X,Y).
$endgroup$
– Did
Dec 7 '18 at 8:16