Show that if $X$ and $Y$ are independent then $E_X(g(X,Y))=E(g(X,Y)|Y).$

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Let $X$ and $Y$ be random variables and let $g:mathbb R^2 to mathbb R$ be a non-negative measurable function. If you like assume that $g(X,Y)$ has finite expectation. Let
$$E_X(g(X,Y))=int_mathbb R g(x,Y)dP_X(x),$$
where $P_X=Pcirc X^-1$ is the distribution of $X.$



I am interested in the relationship between $E_X(g(X,Y))$ and $E(g(X,Y)|Y).$ I suspect that the following holds:




If $X$ and $Y$ are independent then
$$E_X(g(X,Y))=E(g(X,Y)|Y).tag*$$




Since I don't have a lot of experience writing proofs with measure-theoretic arguments it would be nice if someone with more experience could have a look at this proof I wrote:



Proof.



In order to show $(*)$ we need to verify



  1. $E_X(g(X,Y))$ is $sigma(Y)$-measurable

  2. $forall Ainsigma(Y)$ we have
    $$E[1_A E_X(g(X,Y))] = E[1_A g(X,Y)].$$

Regarding 1.:



It is
$$E_X(g(X,Y))=int_mathbb R g(x,Y)dP_X(x)=h(Y),$$
where $h(y)=int_mathbb R g(x,y)dP_X(x)$ is a measurable function. Hence $E_X(g(X,Y))$ is $sigma(Y)$-measurable.



Regarding 2.:



Let $Ainsigma(Y).$ Note that $1_A$ can be written as $1_A=f(Y)$ for some measurable function $f.$ Moreover, since $X$ and $Y$ are independent, it holds that
$$P_(X,Y)=P_Xotimes P_Y.$$
Now write
beginalign
E[1_A g(X,Y)] &= int_Omega 1_A g(X,Y) dP\
&= int_Omega f(Y) g(X,Y) dP\
&= int_mathbb R^2 f(y) g(x,y) dP_(X,Y)(x,y)\
&= int_mathbb R^2 f(y) g(x,y) d(P_Xotimes P_Y)(x,y)\
&= int_mathbb Rint_mathbb R f(y) g(x,y) dP_X(x) dP_Y(y) quadtextby Fubini\
&= int_mathbb Rf(y) int_mathbb R g(x,y) dP_X(x) dP_Y(y)\
&= int_mathbb Rf(y) h(y) dP_Y(y)\
&= int_Omega f(Y) h(Y) dP\
&= int_Omega 1_A E_X(g(X,Y)) dP\
&= E[1_A E_X(g(X,Y))].\
endalign




real-analysis probability-theory measure-theory proof-verification conditional-expectation




share|cite|improve this question
















  • 1




    Your proof looks fine. Just want to point out that Fubini/Tonelli is applicable because $f$ is actually non-negative.
    – Kavi Rama Murthy
    Aug 22 at 9:14










  • Great, thank you.
    – Epiousios
    Aug 22 at 9:20










  • Nothing to add. Well done.
    – Did
    Aug 22 at 10:23






  • 2




    Possible duplicate of Define the function $g (y) = E[f(X,y)]$. Show that $g$ is Borel-measurable, and that $E[f (X,Y)|Y=y] = g(y)$
    – Gordon
    Aug 22 at 16:32














up vote
2
down vote

favorite












Let $X$ and $Y$ be random variables and let $g:mathbb R^2 to mathbb R$ be a non-negative measurable function. If you like assume that $g(X,Y)$ has finite expectation. Let
$$E_X(g(X,Y))=int_mathbb R g(x,Y)dP_X(x),$$
where $P_X=Pcirc X^-1$ is the distribution of $X.$



I am interested in the relationship between $E_X(g(X,Y))$ and $E(g(X,Y)|Y).$ I suspect that the following holds:




If $X$ and $Y$ are independent then
$$E_X(g(X,Y))=E(g(X,Y)|Y).tag*$$




Since I don't have a lot of experience writing proofs with measure-theoretic arguments it would be nice if someone with more experience could have a look at this proof I wrote:



Proof.



In order to show $(*)$ we need to verify



  1. $E_X(g(X,Y))$ is $sigma(Y)$-measurable

  2. $forall Ainsigma(Y)$ we have
    $$E[1_A E_X(g(X,Y))] = E[1_A g(X,Y)].$$

Regarding 1.:



It is
$$E_X(g(X,Y))=int_mathbb R g(x,Y)dP_X(x)=h(Y),$$
where $h(y)=int_mathbb R g(x,y)dP_X(x)$ is a measurable function. Hence $E_X(g(X,Y))$ is $sigma(Y)$-measurable.



Regarding 2.:



Let $Ainsigma(Y).$ Note that $1_A$ can be written as $1_A=f(Y)$ for some measurable function $f.$ Moreover, since $X$ and $Y$ are independent, it holds that
$$P_(X,Y)=P_Xotimes P_Y.$$
Now write
beginalign
E[1_A g(X,Y)] &= int_Omega 1_A g(X,Y) dP\
&= int_Omega f(Y) g(X,Y) dP\
&= int_mathbb R^2 f(y) g(x,y) dP_(X,Y)(x,y)\
&= int_mathbb R^2 f(y) g(x,y) d(P_Xotimes P_Y)(x,y)\
&= int_mathbb Rint_mathbb R f(y) g(x,y) dP_X(x) dP_Y(y) quadtextby Fubini\
&= int_mathbb Rf(y) int_mathbb R g(x,y) dP_X(x) dP_Y(y)\
&= int_mathbb Rf(y) h(y) dP_Y(y)\
&= int_Omega f(Y) h(Y) dP\
&= int_Omega 1_A E_X(g(X,Y)) dP\
&= E[1_A E_X(g(X,Y))].\
endalign




real-analysis probability-theory measure-theory proof-verification conditional-expectation




share|cite|improve this question
















  • 1




    Your proof looks fine. Just want to point out that Fubini/Tonelli is applicable because $f$ is actually non-negative.
    – Kavi Rama Murthy
    Aug 22 at 9:14










  • Great, thank you.
    – Epiousios
    Aug 22 at 9:20










  • Nothing to add. Well done.
    – Did
    Aug 22 at 10:23






  • 2




    Possible duplicate of Define the function $g (y) = E[f(X,y)]$. Show that $g$ is Borel-measurable, and that $E[f (X,Y)|Y=y] = g(y)$
    – Gordon
    Aug 22 at 16:32












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $X$ and $Y$ be random variables and let $g:mathbb R^2 to mathbb R$ be a non-negative measurable function. If you like assume that $g(X,Y)$ has finite expectation. Let
$$E_X(g(X,Y))=int_mathbb R g(x,Y)dP_X(x),$$
where $P_X=Pcirc X^-1$ is the distribution of $X.$



I am interested in the relationship between $E_X(g(X,Y))$ and $E(g(X,Y)|Y).$ I suspect that the following holds:




If $X$ and $Y$ are independent then
$$E_X(g(X,Y))=E(g(X,Y)|Y).tag*$$




Since I don't have a lot of experience writing proofs with measure-theoretic arguments it would be nice if someone with more experience could have a look at this proof I wrote:



Proof.



In order to show $(*)$ we need to verify



  1. $E_X(g(X,Y))$ is $sigma(Y)$-measurable

  2. $forall Ainsigma(Y)$ we have
    $$E[1_A E_X(g(X,Y))] = E[1_A g(X,Y)].$$

Regarding 1.:



It is
$$E_X(g(X,Y))=int_mathbb R g(x,Y)dP_X(x)=h(Y),$$
where $h(y)=int_mathbb R g(x,y)dP_X(x)$ is a measurable function. Hence $E_X(g(X,Y))$ is $sigma(Y)$-measurable.



Regarding 2.:



Let $Ainsigma(Y).$ Note that $1_A$ can be written as $1_A=f(Y)$ for some measurable function $f.$ Moreover, since $X$ and $Y$ are independent, it holds that
$$P_(X,Y)=P_Xotimes P_Y.$$
Now write
beginalign
E[1_A g(X,Y)] &= int_Omega 1_A g(X,Y) dP\
&= int_Omega f(Y) g(X,Y) dP\
&= int_mathbb R^2 f(y) g(x,y) dP_(X,Y)(x,y)\
&= int_mathbb R^2 f(y) g(x,y) d(P_Xotimes P_Y)(x,y)\
&= int_mathbb Rint_mathbb R f(y) g(x,y) dP_X(x) dP_Y(y) quadtextby Fubini\
&= int_mathbb Rf(y) int_mathbb R g(x,y) dP_X(x) dP_Y(y)\
&= int_mathbb Rf(y) h(y) dP_Y(y)\
&= int_Omega f(Y) h(Y) dP\
&= int_Omega 1_A E_X(g(X,Y)) dP\
&= E[1_A E_X(g(X,Y))].\
endalign




real-analysis probability-theory measure-theory proof-verification conditional-expectation




share|cite|improve this question












Let $X$ and $Y$ be random variables and let $g:mathbb R^2 to mathbb R$ be a non-negative measurable function. If you like assume that $g(X,Y)$ has finite expectation. Let
$$E_X(g(X,Y))=int_mathbb R g(x,Y)dP_X(x),$$
where $P_X=Pcirc X^-1$ is the distribution of $X.$



I am interested in the relationship between $E_X(g(X,Y))$ and $E(g(X,Y)|Y).$ I suspect that the following holds:




If $X$ and $Y$ are independent then
$$E_X(g(X,Y))=E(g(X,Y)|Y).tag*$$




Since I don't have a lot of experience writing proofs with measure-theoretic arguments it would be nice if someone with more experience could have a look at this proof I wrote:



Proof.



In order to show $(*)$ we need to verify



  1. $E_X(g(X,Y))$ is $sigma(Y)$-measurable

  2. $forall Ainsigma(Y)$ we have
    $$E[1_A E_X(g(X,Y))] = E[1_A g(X,Y)].$$

Regarding 1.:



It is
$$E_X(g(X,Y))=int_mathbb R g(x,Y)dP_X(x)=h(Y),$$
where $h(y)=int_mathbb R g(x,y)dP_X(x)$ is a measurable function. Hence $E_X(g(X,Y))$ is $sigma(Y)$-measurable.



Regarding 2.:



Let $Ainsigma(Y).$ Note that $1_A$ can be written as $1_A=f(Y)$ for some measurable function $f.$ Moreover, since $X$ and $Y$ are independent, it holds that
$$P_(X,Y)=P_Xotimes P_Y.$$
Now write
beginalign
E[1_A g(X,Y)] &= int_Omega 1_A g(X,Y) dP\
&= int_Omega f(Y) g(X,Y) dP\
&= int_mathbb R^2 f(y) g(x,y) dP_(X,Y)(x,y)\
&= int_mathbb R^2 f(y) g(x,y) d(P_Xotimes P_Y)(x,y)\
&= int_mathbb Rint_mathbb R f(y) g(x,y) dP_X(x) dP_Y(y) quadtextby Fubini\
&= int_mathbb Rf(y) int_mathbb R g(x,y) dP_X(x) dP_Y(y)\
&= int_mathbb Rf(y) h(y) dP_Y(y)\
&= int_Omega f(Y) h(Y) dP\
&= int_Omega 1_A E_X(g(X,Y)) dP\
&= E[1_A E_X(g(X,Y))].\
endalign





real-analysis probability-theory measure-theory proof-verification conditional-expectation





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 22 at 8:56









Epiousios

1,501522




1,501522







  • 1




    Your proof looks fine. Just want to point out that Fubini/Tonelli is applicable because $f$ is actually non-negative.
    – Kavi Rama Murthy
    Aug 22 at 9:14










  • Great, thank you.
    – Epiousios
    Aug 22 at 9:20










  • Nothing to add. Well done.
    – Did
    Aug 22 at 10:23






  • 2




    Possible duplicate of Define the function $g (y) = E[f(X,y)]$. Show that $g$ is Borel-measurable, and that $E[f (X,Y)|Y=y] = g(y)$
    – Gordon
    Aug 22 at 16:32












  • 1




    Your proof looks fine. Just want to point out that Fubini/Tonelli is applicable because $f$ is actually non-negative.
    – Kavi Rama Murthy
    Aug 22 at 9:14










  • Great, thank you.
    – Epiousios
    Aug 22 at 9:20










  • Nothing to add. Well done.
    – Did
    Aug 22 at 10:23






  • 2




    Possible duplicate of Define the function $g (y) = E[f(X,y)]$. Show that $g$ is Borel-measurable, and that $E[f (X,Y)|Y=y] = g(y)$
    – Gordon
    Aug 22 at 16:32







1




1




Your proof looks fine. Just want to point out that Fubini/Tonelli is applicable because $f$ is actually non-negative.
– Kavi Rama Murthy
Aug 22 at 9:14




Your proof looks fine. Just want to point out that Fubini/Tonelli is applicable because $f$ is actually non-negative.
– Kavi Rama Murthy
Aug 22 at 9:14












Great, thank you.
– Epiousios
Aug 22 at 9:20




Great, thank you.
– Epiousios
Aug 22 at 9:20












Nothing to add. Well done.
– Did
Aug 22 at 10:23




Nothing to add. Well done.
– Did
Aug 22 at 10:23




2




2




Possible duplicate of Define the function $g (y) = E[f(X,y)]$. Show that $g$ is Borel-measurable, and that $E[f (X,Y)|Y=y] = g(y)$
– Gordon
Aug 22 at 16:32




Possible duplicate of Define the function $g (y) = E[f(X,y)]$. Show that $g$ is Borel-measurable, and that $E[f (X,Y)|Y=y] = g(y)$
– Gordon
Aug 22 at 16:32















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