Mixing
$mathbb{E}[f(X,Y)|Y]=mathbb{E}[f(X,Y)]$ for all bounded measurable $f$ using the monotone class theorem
$begingroup$
Let $X,Y$ be independent rv's.
For $f:mathbb{R}^2tomathbb{R}$ measurable we have $f_X:mathbb{R}tomathbb{R}$ where
$$f_X(y)= left{
begin{array}{lr}
mathbb{E}[f(X,y)] & : mathbb{E}|f(X,y)|<infty\
0 & : text{else}
end{array}
right..$$
Let $M_b$ be the set of all bounded measurable functions $mathbb{R}^2tomathbb{R}$ and
$$C={fin M_b: mathbb{E}[f(X,Y)|Y]=f_X(Y)}.$$
How do we prove that $C=M_b$?
If we use the monotone class theorem, then for a $pi-$system $Pi$ that generates $mathcal{B}^2$ we can show that $M_bsubset C$.
Therefore we want
1) $1_{mathbb{R}^2}in C$:
This is clear since for $Ainsigma(Y)$ we have $$int_Amathbb{E}[1_{mathbb{R}^2}(X,Y)|Y]dmathbb{P}=int_A1_{mathbb{R}^2}(X,Y)dmathbb{P}=int_A1dmathbb{P}=int_Amathbb{E}[1_{mathbb{R}^2}(X,Y)]dmathbb{P}.$$
2) for an increasing, non-negative sequence $(f_n)_n$ in $C$, where $f_nto f$ with $f$ bounded, then $fin C$:
So we want to show
$$mathbb{E}[f(X,Y)|Y]=mathbb{E}[f(X,Y)]$$
Can we do this by using conditional monotone convergence and say
$$mathbb{E}[f_n(X,Y)|Y]tomathbb{E}[f(X,Y)|Y]$$
and since
$$mathbb{E}[f_n(X,Y)]tomathbb{E}[f(X,Y)]$$
by regular MCT, it follows?
3) for all $Bin Pi$, $1_Bin C$:
Can we take $Pi={(a,b)times (c,d):a<b,c<dinmathbb{R}}$?
Then for $Ainsigma(Y)$
$begin{align*}
int_Amathbb{E}[1_{(a,b)times (c,d)}(X,Y)|Y]dmathbb{P}&=int_A1_{(a,b)times (c,d)}(X,Y)dmathbb{P}\
&=mathbb{P}[Acap(X,Y)in(a,b)times (c,d)]\
&=mathbb{P}[Acap{Xin(a,b)}]mathbb{P}[Acap{Yin (c,d)}]\
&=int_A1_{Xin(a,b)}dmathbb{P}cdotint_A1_{Yin(c,d)}dmathbb{P}\
&=?\
&=int_Amathbb{E}[1_{(a,b)times (c,d)}(X,Y)]dmathbb{P}.
end{align*}$
Is this correct? Which steps am I missing?
measure-theory random-variables conditional-expectation measurable-functions monotone-class-theorem
asked Jan 29 at 11:27
Joachim DoyleJoachim Doyle
818
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be independent rv's.
For $f:mathbb{R}^2tomathbb{R}$ measurable we have $f_X:mathbb{R}tomathbb{R}$ where
$$f_X(y)= left{
begin{array}{lr}
mathbb{E}[f(X,y)] & : mathbb{E}|f(X,y)|<infty\
0 & : text{else}
end{array}
right..$$
Let $M_b$ be the set of all bounded measurable functions $mathbb{R}^2tomathbb{R}$ and
$$C={fin M_b: mathbb{E}[f(X,Y)|Y]=f_X(Y)}.$$
How do we prove that $C=M_b$?
If we use the monotone class theorem, then for a $pi-$system $Pi$ that generates $mathcal{B}^2$ we can show that $M_bsubset C$.
Therefore we want
1) $1_{mathbb{R}^2}in C$:
This is clear since for $Ainsigma(Y)$ we have $$int_Amathbb{E}[1_{mathbb{R}^2}(X,Y)|Y]dmathbb{P}=int_A1_{mathbb{R}^2}(X,Y)dmathbb{P}=int_A1dmathbb{P}=int_Amathbb{E}[1_{mathbb{R}^2}(X,Y)]dmathbb{P}.$$
2) for an increasing, non-negative sequence $(f_n)_n$ in $C$, where $f_nto f$ with $f$ bounded, then $fin C$:
So we want to show
$$mathbb{E}[f(X,Y)|Y]=mathbb{E}[f(X,Y)]$$
Can we do this by using conditional monotone convergence and say
$$mathbb{E}[f_n(X,Y)|Y]tomathbb{E}[f(X,Y)|Y]$$
and since
$$mathbb{E}[f_n(X,Y)]tomathbb{E}[f(X,Y)]$$
by regular MCT, it follows?
3) for all $Bin Pi$, $1_Bin C$:
Can we take $Pi={(a,b)times (c,d):a<b,c<dinmathbb{R}}$?
Then for $Ainsigma(Y)$
$begin{align*}
int_Amathbb{E}[1_{(a,b)times (c,d)}(X,Y)|Y]dmathbb{P}&=int_A1_{(a,b)times (c,d)}(X,Y)dmathbb{P}\
&=mathbb{P}[Acap(X,Y)in(a,b)times (c,d)]\
&=mathbb{P}[Acap{Xin(a,b)}]mathbb{P}[Acap{Yin (c,d)}]\
&=int_A1_{Xin(a,b)}dmathbb{P}cdotint_A1_{Yin(c,d)}dmathbb{P}\
&=?\
&=int_Amathbb{E}[1_{(a,b)times (c,d)}(X,Y)]dmathbb{P}.
end{align*}$
Is this correct? Which steps am I missing?
measure-theory random-variables conditional-expectation measurable-functions monotone-class-theorem
asked Jan 29 at 11:27
Joachim DoyleJoachim Doyle
818
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be independent rv's.
For $f:mathbb{R}^2tomathbb{R}$ measurable we have $f_X:mathbb{R}tomathbb{R}$ where
$$f_X(y)= left{
begin{array}{lr}
mathbb{E}[f(X,y)] & : mathbb{E}|f(X,y)|<infty\
0 & : text{else}
end{array}
right..$$
Let $M_b$ be the set of all bounded measurable functions $mathbb{R}^2tomathbb{R}$ and
$$C={fin M_b: mathbb{E}[f(X,Y)|Y]=f_X(Y)}.$$
How do we prove that $C=M_b$?
If we use the monotone class theorem, then for a $pi-$system $Pi$ that generates $mathcal{B}^2$ we can show that $M_bsubset C$.
Therefore we want
1) $1_{mathbb{R}^2}in C$:
This is clear since for $Ainsigma(Y)$ we have $$int_Amathbb{E}[1_{mathbb{R}^2}(X,Y)|Y]dmathbb{P}=int_A1_{mathbb{R}^2}(X,Y)dmathbb{P}=int_A1dmathbb{P}=int_Amathbb{E}[1_{mathbb{R}^2}(X,Y)]dmathbb{P}.$$
2) for an increasing, non-negative sequence $(f_n)_n$ in $C$, where $f_nto f$ with $f$ bounded, then $fin C$:
So we want to show
$$mathbb{E}[f(X,Y)|Y]=mathbb{E}[f(X,Y)]$$
Can we do this by using conditional monotone convergence and say
$$mathbb{E}[f_n(X,Y)|Y]tomathbb{E}[f(X,Y)|Y]$$
and since
$$mathbb{E}[f_n(X,Y)]tomathbb{E}[f(X,Y)]$$
by regular MCT, it follows?
3) for all $Bin Pi$, $1_Bin C$:
Can we take $Pi={(a,b)times (c,d):a<b,c<dinmathbb{R}}$?
Then for $Ainsigma(Y)$
$begin{align*}
int_Amathbb{E}[1_{(a,b)times (c,d)}(X,Y)|Y]dmathbb{P}&=int_A1_{(a,b)times (c,d)}(X,Y)dmathbb{P}\
&=mathbb{P}[Acap(X,Y)in(a,b)times (c,d)]\
&=mathbb{P}[Acap{Xin(a,b)}]mathbb{P}[Acap{Yin (c,d)}]\
&=int_A1_{Xin(a,b)}dmathbb{P}cdotint_A1_{Yin(c,d)}dmathbb{P}\
&=?\
&=int_Amathbb{E}[1_{(a,b)times (c,d)}(X,Y)]dmathbb{P}.
end{align*}$
Is this correct? Which steps am I missing?
measure-theory random-variables conditional-expectation measurable-functions monotone-class-theorem
asked Jan 29 at 11:27
Joachim DoyleJoachim Doyle
818
$endgroup$
Let $X,Y$ be independent rv's.
For $f:mathbb{R}^2tomathbb{R}$ measurable we have $f_X:mathbb{R}tomathbb{R}$ where
$$f_X(y)= left{
begin{array}{lr}
mathbb{E}[f(X,y)] & : mathbb{E}|f(X,y)|<infty\
0 & : text{else}
end{array}
right..$$
Let $M_b$ be the set of all bounded measurable functions $mathbb{R}^2tomathbb{R}$ and
$$C={fin M_b: mathbb{E}[f(X,Y)|Y]=f_X(Y)}.$$
How do we prove that $C=M_b$?
If we use the monotone class theorem, then for a $pi-$system $Pi$ that generates $mathcal{B}^2$ we can show that $M_bsubset C$.
Therefore we want
1) $1_{mathbb{R}^2}in C$:
This is clear since for $Ainsigma(Y)$ we have $$int_Amathbb{E}[1_{mathbb{R}^2}(X,Y)|Y]dmathbb{P}=int_A1_{mathbb{R}^2}(X,Y)dmathbb{P}=int_A1dmathbb{P}=int_Amathbb{E}[1_{mathbb{R}^2}(X,Y)]dmathbb{P}.$$
2) for an increasing, non-negative sequence $(f_n)_n$ in $C$, where $f_nto f$ with $f$ bounded, then $fin C$:
So we want to show
$$mathbb{E}[f(X,Y)|Y]=mathbb{E}[f(X,Y)]$$
Can we do this by using conditional monotone convergence and say
$$mathbb{E}[f_n(X,Y)|Y]tomathbb{E}[f(X,Y)|Y]$$
and since
$$mathbb{E}[f_n(X,Y)]tomathbb{E}[f(X,Y)]$$
by regular MCT, it follows?
3) for all $Bin Pi$, $1_Bin C$:
Can we take $Pi={(a,b)times (c,d):a<b,c<dinmathbb{R}}$?
Then for $Ainsigma(Y)$
$begin{align*}
int_Amathbb{E}[1_{(a,b)times (c,d)}(X,Y)|Y]dmathbb{P}&=int_A1_{(a,b)times (c,d)}(X,Y)dmathbb{P}\
&=mathbb{P}[Acap(X,Y)in(a,b)times (c,d)]\
&=mathbb{P}[Acap{Xin(a,b)}]mathbb{P}[Acap{Yin (c,d)}]\
&=int_A1_{Xin(a,b)}dmathbb{P}cdotint_A1_{Yin(c,d)}dmathbb{P}\
&=?\
&=int_Amathbb{E}[1_{(a,b)times (c,d)}(X,Y)]dmathbb{P}.
end{align*}$
Is this correct? Which steps am I missing?
measure-theory random-variables conditional-expectation measurable-functions monotone-class-theorem
measure-theory random-variables conditional-expectation measurable-functions monotone-class-theorem
asked Jan 29 at 11:27
Joachim DoyleJoachim Doyle
818
asked Jan 29 at 11:27
Joachim DoyleJoachim Doyle
818
edited Mar 4 at 9:52
Joachim Doyle
asked Jan 29 at 11:27
Joachim DoyleJoachim Doyle
818
asked Jan 29 at 11:27
Joachim DoyleJoachim Doyle
818
asked Jan 29 at 11:27
Joachim DoyleJoachim Doyle
818
818
add a comment |
add a comment |
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