Why is $E[f(X,Y) mid X]=g(X)$?
Let $X,Y$ be independent and let $f$ be Borel such that $f(X,Y) in L^1(Omega,mathcal{A},P)$. Moreover let
$$
g(x)=E[f(x,Y)] text{ if } vert E[f(x,Y)]vert <infty text{ and } 0 text{ otherwise}
$$
Then show that $g$ is Borel on $mathbb{R}$ and that
$$
E[f(X,Y)mid X]=g(X)
$$
My attempt
1. I am not sure how to go about showing the Borel measurability of $g$ so a hint here would be highly appreciated
- In order to show that $E[f(X,Y)mid X]=g(X)$ we need to show that g(X) is $sigma(X)$ measurable which is the case once part 1)(Borel measurability of $g$) is proved.
And we must also show that for all $A in sigma(X)$ we have
begin{equation} label{1}
E[f(X,Y) 1_A]=E[g(X) 1_A]
end{equation}
Now I can show that if $A$ is of the form $A={X=x}$ then the equation above holds since
$$
E[f(X,Y) 1_A]=E[f(X,Y) 1_{{X=x}}]=E[f(x,Y) 1_{{X=x}}]=E[f(x,Y)] E[1_{{X=x}}]=g(x)E[1_{{X=x}}]=E[g(x)1_{{X=x}}]=E[g(X)1_{{X=x}}]
$$
where I used the independence of $X$ and $Y$ and the properties of expectation. How can I extend this to all sets in $sigma(X)$ if $X$ we discrete taking countably many values this would work but how can I make this argument work in the general case. Even hints would be highly appreciated or a nudge in the right direction
probability-theory conditional-expectation
asked Dec 5 '18 at 12:32
user3503589user3503589
1,2181721
add a comment |
Let $X,Y$ be independent and let $f$ be Borel such that $f(X,Y) in L^1(Omega,mathcal{A},P)$. Moreover let
$$
g(x)=E[f(x,Y)] text{ if } vert E[f(x,Y)]vert <infty text{ and } 0 text{ otherwise}
$$
Then show that $g$ is Borel on $mathbb{R}$ and that
$$
E[f(X,Y)mid X]=g(X)
$$
My attempt
1. I am not sure how to go about showing the Borel measurability of $g$ so a hint here would be highly appreciated
- In order to show that $E[f(X,Y)mid X]=g(X)$ we need to show that g(X) is $sigma(X)$ measurable which is the case once part 1)(Borel measurability of $g$) is proved.
And we must also show that for all $A in sigma(X)$ we have
begin{equation} label{1}
E[f(X,Y) 1_A]=E[g(X) 1_A]
end{equation}
Now I can show that if $A$ is of the form $A={X=x}$ then the equation above holds since
$$
E[f(X,Y) 1_A]=E[f(X,Y) 1_{{X=x}}]=E[f(x,Y) 1_{{X=x}}]=E[f(x,Y)] E[1_{{X=x}}]=g(x)E[1_{{X=x}}]=E[g(x)1_{{X=x}}]=E[g(X)1_{{X=x}}]
$$
where I used the independence of $X$ and $Y$ and the properties of expectation. How can I extend this to all sets in $sigma(X)$ if $X$ we discrete taking countably many values this would work but how can I make this argument work in the general case. Even hints would be highly appreciated or a nudge in the right direction
probability-theory conditional-expectation
asked Dec 5 '18 at 12:32
user3503589user3503589
1,2181721
add a comment |
Let $X,Y$ be independent and let $f$ be Borel such that $f(X,Y) in L^1(Omega,mathcal{A},P)$. Moreover let
$$
g(x)=E[f(x,Y)] text{ if } vert E[f(x,Y)]vert <infty text{ and } 0 text{ otherwise}
$$
Then show that $g$ is Borel on $mathbb{R}$ and that
$$
E[f(X,Y)mid X]=g(X)
$$
My attempt
1. I am not sure how to go about showing the Borel measurability of $g$ so a hint here would be highly appreciated
- In order to show that $E[f(X,Y)mid X]=g(X)$ we need to show that g(X) is $sigma(X)$ measurable which is the case once part 1)(Borel measurability of $g$) is proved.
And we must also show that for all $A in sigma(X)$ we have
begin{equation} label{1}
E[f(X,Y) 1_A]=E[g(X) 1_A]
end{equation}
Now I can show that if $A$ is of the form $A={X=x}$ then the equation above holds since
$$
E[f(X,Y) 1_A]=E[f(X,Y) 1_{{X=x}}]=E[f(x,Y) 1_{{X=x}}]=E[f(x,Y)] E[1_{{X=x}}]=g(x)E[1_{{X=x}}]=E[g(x)1_{{X=x}}]=E[g(X)1_{{X=x}}]
$$
where I used the independence of $X$ and $Y$ and the properties of expectation. How can I extend this to all sets in $sigma(X)$ if $X$ we discrete taking countably many values this would work but how can I make this argument work in the general case. Even hints would be highly appreciated or a nudge in the right direction
probability-theory conditional-expectation
asked Dec 5 '18 at 12:32
user3503589user3503589
1,2181721
Let $X,Y$ be independent and let $f$ be Borel such that $f(X,Y) in L^1(Omega,mathcal{A},P)$. Moreover let
$$
g(x)=E[f(x,Y)] text{ if } vert E[f(x,Y)]vert <infty text{ and } 0 text{ otherwise}
$$
Then show that $g$ is Borel on $mathbb{R}$ and that
$$
E[f(X,Y)mid X]=g(X)
$$
My attempt
1. I am not sure how to go about showing the Borel measurability of $g$ so a hint here would be highly appreciated
- In order to show that $E[f(X,Y)mid X]=g(X)$ we need to show that g(X) is $sigma(X)$ measurable which is the case once part 1)(Borel measurability of $g$) is proved.
And we must also show that for all $A in sigma(X)$ we have
begin{equation} label{1}
E[f(X,Y) 1_A]=E[g(X) 1_A]
end{equation}
Now I can show that if $A$ is of the form $A={X=x}$ then the equation above holds since
$$
E[f(X,Y) 1_A]=E[f(X,Y) 1_{{X=x}}]=E[f(x,Y) 1_{{X=x}}]=E[f(x,Y)] E[1_{{X=x}}]=g(x)E[1_{{X=x}}]=E[g(x)1_{{X=x}}]=E[g(X)1_{{X=x}}]
$$
where I used the independence of $X$ and $Y$ and the properties of expectation. How can I extend this to all sets in $sigma(X)$ if $X$ we discrete taking countably many values this would work but how can I make this argument work in the general case. Even hints would be highly appreciated or a nudge in the right direction
probability-theory conditional-expectation
probability-theory conditional-expectation
asked Dec 5 '18 at 12:32
user3503589user3503589
1,2181721
asked Dec 5 '18 at 12:32
user3503589user3503589
1,2181721
asked Dec 5 '18 at 12:32
user3503589user3503589
1,2181721
asked Dec 5 '18 at 12:32
user3503589user3503589
1,2181721
asked Dec 5 '18 at 12:32
user3503589user3503589
1,2181721
1,2181721
add a comment |
add a comment |
1 Answer
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Second part something like:$$mathbb Ef(X,Y)mathbf1_{{Xin B}}=intint f(x,y)mathbf1_B(x);dF_Y(y)dF_X(x)=intmathbf1_B(x)int f(x,y);dF_Y(y)dF_X(x)=$$$$intmathbf1_B(x) g(x)dF_X(x)=mathbb Eg(X)mathbf1_{{Xin B}}$$where the first equality is based on independence.
What we use here too is that $sigma(X)={{Xin B}mid Binmathcal B(mathbb R)}$
answered Dec 5 '18 at 13:36
drhabdrhab
98.5k544129
add a comment |
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1 Answer
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1 Answer
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Second part something like:$$mathbb Ef(X,Y)mathbf1_{{Xin B}}=intint f(x,y)mathbf1_B(x);dF_Y(y)dF_X(x)=intmathbf1_B(x)int f(x,y);dF_Y(y)dF_X(x)=$$$$intmathbf1_B(x) g(x)dF_X(x)=mathbb Eg(X)mathbf1_{{Xin B}}$$where the first equality is based on independence.
What we use here too is that $sigma(X)={{Xin B}mid Binmathcal B(mathbb R)}$
answered Dec 5 '18 at 13:36
drhabdrhab
98.5k544129
add a comment |
Second part something like:$$mathbb Ef(X,Y)mathbf1_{{Xin B}}=intint f(x,y)mathbf1_B(x);dF_Y(y)dF_X(x)=intmathbf1_B(x)int f(x,y);dF_Y(y)dF_X(x)=$$$$intmathbf1_B(x) g(x)dF_X(x)=mathbb Eg(X)mathbf1_{{Xin B}}$$where the first equality is based on independence.
What we use here too is that $sigma(X)={{Xin B}mid Binmathcal B(mathbb R)}$
answered Dec 5 '18 at 13:36
drhabdrhab
98.5k544129
add a comment |
Second part something like:$$mathbb Ef(X,Y)mathbf1_{{Xin B}}=intint f(x,y)mathbf1_B(x);dF_Y(y)dF_X(x)=intmathbf1_B(x)int f(x,y);dF_Y(y)dF_X(x)=$$$$intmathbf1_B(x) g(x)dF_X(x)=mathbb Eg(X)mathbf1_{{Xin B}}$$where the first equality is based on independence.
What we use here too is that $sigma(X)={{Xin B}mid Binmathcal B(mathbb R)}$
answered Dec 5 '18 at 13:36
drhabdrhab
98.5k544129
Second part something like:$$mathbb Ef(X,Y)mathbf1_{{Xin B}}=intint f(x,y)mathbf1_B(x);dF_Y(y)dF_X(x)=intmathbf1_B(x)int f(x,y);dF_Y(y)dF_X(x)=$$$$intmathbf1_B(x) g(x)dF_X(x)=mathbb Eg(X)mathbf1_{{Xin B}}$$where the first equality is based on independence.
What we use here too is that $sigma(X)={{Xin B}mid Binmathcal B(mathbb R)}$
answered Dec 5 '18 at 13:36
drhabdrhab
98.5k544129
answered Dec 5 '18 at 13:36
drhabdrhab
98.5k544129
answered Dec 5 '18 at 13:36
drhabdrhab
98.5k544129
answered Dec 5 '18 at 13:36
drhabdrhab
98.5k544129
98.5k544129
add a comment |
add a comment |
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