Conditional Expectation from a joint pdf
$begingroup$
Suppose that both $X$ and $Y$ have joint pdf $f(x,y)$ and $Y$ has mariginal density $f(y)=int_{mathbb R}f(x,y)dx$. The conditional density of $X$ given $Y$ is given by $f(x|y)=dfrac{f(x,y)}{f(y)}$ and $E(X|Y=y)=int_{mathbb R}xf(x|y)dx$. Let $sigma(Y)$ be the sigma field generated by $Y$ and show that $E(X|sigma(Y))=theta(Y)$ where $theta(y)=E(X|Y=y)$
My question here is what is $theta(Y)$? Is it $E(X|Y=Y)$ and if so what does this mean?
I know I need to show that $$ int_AE(X|sigma(Y)dP=int_Atheta(Y)dP ;;; forall; Ainsigma(Y) $$
but im unsure how to start since im unclear on what $theta(Y)$ means.
probability-theory conditional-expectation
asked Dec 22 '18 at 16:43
alpastalpast
468314
$endgroup$
|
show 3 more comments
$begingroup$
Suppose that both $X$ and $Y$ have joint pdf $f(x,y)$ and $Y$ has mariginal density $f(y)=int_{mathbb R}f(x,y)dx$. The conditional density of $X$ given $Y$ is given by $f(x|y)=dfrac{f(x,y)}{f(y)}$ and $E(X|Y=y)=int_{mathbb R}xf(x|y)dx$. Let $sigma(Y)$ be the sigma field generated by $Y$ and show that $E(X|sigma(Y))=theta(Y)$ where $theta(y)=E(X|Y=y)$
My question here is what is $theta(Y)$? Is it $E(X|Y=Y)$ and if so what does this mean?
I know I need to show that $$ int_AE(X|sigma(Y)dP=int_Atheta(Y)dP ;;; forall; Ainsigma(Y) $$
but im unsure how to start since im unclear on what $theta(Y)$ means.
probability-theory conditional-expectation
asked Dec 22 '18 at 16:43
alpastalpast
468314
$endgroup$
$begingroup$
$theta (Y)$ is a r.v. s.t. $mathbb E[theta (Y)]=mathbb E[X]$. It's the best approximation of $X$ knowing $Y$ is measurable. Normally, we write $theta (Y)=mathbb E[Xmid Y]$.
$endgroup$
– idm
Dec 22 '18 at 16:47
1
$begingroup$
$theta$ must be looked at as a function $mathbb Rtomathbb R$ prescribed by $ymapstomathbb E[Xmid Y=y]$. Then $theta(Y)$ is just the composition $thetacirc Y:Omegatomathbb R$. The interpretation $theta(Y)=mathbb E[Xmid Y=Y]$ is confusing and must be avoided.
$endgroup$
– drhab
Dec 22 '18 at 16:50
$begingroup$
@drhab so I should think of $theta(Y)$ as $theta(Y)(omega)=E(X|Y=y)$ where $Y(omega)=y$?
$endgroup$
– alpast
Dec 22 '18 at 16:53
$begingroup$
@alpast: No ! as $theta (Y)=mathbb E[Xmid Y]$, where $theta : ymapsto mathbb E[Xmid Y=y]$ is a function. When $f(x)=x^2$, how do you think $f(X)$ ? as $X^2(omega )=x^2$ ? Of course not :)
$endgroup$
– idm
Dec 22 '18 at 17:14
$begingroup$
@idm sorry I dont understand? I get that $theta : ymapsto mathbb E[Xmid Y=y]$. And then $theta(Y)$ is the composition of $Y$ and $theta$, so $theta(Y(omega)) : omegamapsto mathbb E[Xmid Y=y]$. is this right?
$endgroup$
– alpast
Dec 22 '18 at 17:23
|
show 3 more comments
$begingroup$
Suppose that both $X$ and $Y$ have joint pdf $f(x,y)$ and $Y$ has mariginal density $f(y)=int_{mathbb R}f(x,y)dx$. The conditional density of $X$ given $Y$ is given by $f(x|y)=dfrac{f(x,y)}{f(y)}$ and $E(X|Y=y)=int_{mathbb R}xf(x|y)dx$. Let $sigma(Y)$ be the sigma field generated by $Y$ and show that $E(X|sigma(Y))=theta(Y)$ where $theta(y)=E(X|Y=y)$
My question here is what is $theta(Y)$? Is it $E(X|Y=Y)$ and if so what does this mean?
I know I need to show that $$ int_AE(X|sigma(Y)dP=int_Atheta(Y)dP ;;; forall; Ainsigma(Y) $$
but im unsure how to start since im unclear on what $theta(Y)$ means.
probability-theory conditional-expectation
asked Dec 22 '18 at 16:43
alpastalpast
468314
$endgroup$
Suppose that both $X$ and $Y$ have joint pdf $f(x,y)$ and $Y$ has mariginal density $f(y)=int_{mathbb R}f(x,y)dx$. The conditional density of $X$ given $Y$ is given by $f(x|y)=dfrac{f(x,y)}{f(y)}$ and $E(X|Y=y)=int_{mathbb R}xf(x|y)dx$. Let $sigma(Y)$ be the sigma field generated by $Y$ and show that $E(X|sigma(Y))=theta(Y)$ where $theta(y)=E(X|Y=y)$
My question here is what is $theta(Y)$? Is it $E(X|Y=Y)$ and if so what does this mean?
I know I need to show that $$ int_AE(X|sigma(Y)dP=int_Atheta(Y)dP ;;; forall; Ainsigma(Y) $$
but im unsure how to start since im unclear on what $theta(Y)$ means.
probability-theory conditional-expectation
probability-theory conditional-expectation
asked Dec 22 '18 at 16:43
alpastalpast
468314
asked Dec 22 '18 at 16:43
alpastalpast
468314
asked Dec 22 '18 at 16:43
alpastalpast
468314
asked Dec 22 '18 at 16:43
alpastalpast
468314
asked Dec 22 '18 at 16:43
alpastalpast
468314
468314
$begingroup$
$theta (Y)$ is a r.v. s.t. $mathbb E[theta (Y)]=mathbb E[X]$. It's the best approximation of $X$ knowing $Y$ is measurable. Normally, we write $theta (Y)=mathbb E[Xmid Y]$.
$endgroup$
– idm
Dec 22 '18 at 16:47
1
$begingroup$
$theta$ must be looked at as a function $mathbb Rtomathbb R$ prescribed by $ymapstomathbb E[Xmid Y=y]$. Then $theta(Y)$ is just the composition $thetacirc Y:Omegatomathbb R$. The interpretation $theta(Y)=mathbb E[Xmid Y=Y]$ is confusing and must be avoided.
$endgroup$
– drhab
Dec 22 '18 at 16:50
$begingroup$
@drhab so I should think of $theta(Y)$ as $theta(Y)(omega)=E(X|Y=y)$ where $Y(omega)=y$?
$endgroup$
– alpast
Dec 22 '18 at 16:53
$begingroup$
@alpast: No ! as $theta (Y)=mathbb E[Xmid Y]$, where $theta : ymapsto mathbb E[Xmid Y=y]$ is a function. When $f(x)=x^2$, how do you think $f(X)$ ? as $X^2(omega )=x^2$ ? Of course not :)
$endgroup$
– idm
Dec 22 '18 at 17:14
$begingroup$
@idm sorry I dont understand? I get that $theta : ymapsto mathbb E[Xmid Y=y]$. And then $theta(Y)$ is the composition of $Y$ and $theta$, so $theta(Y(omega)) : omegamapsto mathbb E[Xmid Y=y]$. is this right?
$endgroup$
– alpast
Dec 22 '18 at 17:23
|
show 3 more comments
$begingroup$
$theta (Y)$ is a r.v. s.t. $mathbb E[theta (Y)]=mathbb E[X]$. It's the best approximation of $X$ knowing $Y$ is measurable. Normally, we write $theta (Y)=mathbb E[Xmid Y]$.
$endgroup$
– idm
Dec 22 '18 at 16:47
1
$begingroup$
$theta$ must be looked at as a function $mathbb Rtomathbb R$ prescribed by $ymapstomathbb E[Xmid Y=y]$. Then $theta(Y)$ is just the composition $thetacirc Y:Omegatomathbb R$. The interpretation $theta(Y)=mathbb E[Xmid Y=Y]$ is confusing and must be avoided.
$endgroup$
– drhab
Dec 22 '18 at 16:50
$begingroup$
@drhab so I should think of $theta(Y)$ as $theta(Y)(omega)=E(X|Y=y)$ where $Y(omega)=y$?
$endgroup$
– alpast
Dec 22 '18 at 16:53
$begingroup$
@alpast: No ! as $theta (Y)=mathbb E[Xmid Y]$, where $theta : ymapsto mathbb E[Xmid Y=y]$ is a function. When $f(x)=x^2$, how do you think $f(X)$ ? as $X^2(omega )=x^2$ ? Of course not :)
$endgroup$
– idm
Dec 22 '18 at 17:14
$begingroup$
@idm sorry I dont understand? I get that $theta : ymapsto mathbb E[Xmid Y=y]$. And then $theta(Y)$ is the composition of $Y$ and $theta$, so $theta(Y(omega)) : omegamapsto mathbb E[Xmid Y=y]$. is this right?
$endgroup$
– alpast
Dec 22 '18 at 17:23
$begingroup$
$theta (Y)$ is a r.v. s.t. $mathbb E[theta (Y)]=mathbb E[X]$. It's the best approximation of $X$ knowing $Y$ is measurable. Normally, we write $theta (Y)=mathbb E[Xmid Y]$.
$endgroup$
– idm
Dec 22 '18 at 16:47
$begingroup$
$theta (Y)$ is a r.v. s.t. $mathbb E[theta (Y)]=mathbb E[X]$. It's the best approximation of $X$ knowing $Y$ is measurable. Normally, we write $theta (Y)=mathbb E[Xmid Y]$.
$endgroup$
– idm
Dec 22 '18 at 16:47
1
1
$begingroup$
$theta$ must be looked at as a function $mathbb Rtomathbb R$ prescribed by $ymapstomathbb E[Xmid Y=y]$. Then $theta(Y)$ is just the composition $thetacirc Y:Omegatomathbb R$. The interpretation $theta(Y)=mathbb E[Xmid Y=Y]$ is confusing and must be avoided.
$endgroup$
– drhab
Dec 22 '18 at 16:50
$begingroup$
$theta$ must be looked at as a function $mathbb Rtomathbb R$ prescribed by $ymapstomathbb E[Xmid Y=y]$. Then $theta(Y)$ is just the composition $thetacirc Y:Omegatomathbb R$. The interpretation $theta(Y)=mathbb E[Xmid Y=Y]$ is confusing and must be avoided.
$endgroup$
– drhab
Dec 22 '18 at 16:50
$begingroup$
@drhab so I should think of $theta(Y)$ as $theta(Y)(omega)=E(X|Y=y)$ where $Y(omega)=y$?
$endgroup$
– alpast
Dec 22 '18 at 16:53
$begingroup$
@drhab so I should think of $theta(Y)$ as $theta(Y)(omega)=E(X|Y=y)$ where $Y(omega)=y$?
$endgroup$
– alpast
Dec 22 '18 at 16:53
$begingroup$
@alpast: No ! as $theta (Y)=mathbb E[Xmid Y]$, where $theta : ymapsto mathbb E[Xmid Y=y]$ is a function. When $f(x)=x^2$, how do you think $f(X)$ ? as $X^2(omega )=x^2$ ? Of course not :)
$endgroup$
– idm
Dec 22 '18 at 17:14
$begingroup$
@alpast: No ! as $theta (Y)=mathbb E[Xmid Y]$, where $theta : ymapsto mathbb E[Xmid Y=y]$ is a function. When $f(x)=x^2$, how do you think $f(X)$ ? as $X^2(omega )=x^2$ ? Of course not :)
$endgroup$
– idm
Dec 22 '18 at 17:14
$begingroup$
@idm sorry I dont understand? I get that $theta : ymapsto mathbb E[Xmid Y=y]$. And then $theta(Y)$ is the composition of $Y$ and $theta$, so $theta(Y(omega)) : omegamapsto mathbb E[Xmid Y=y]$. is this right?
$endgroup$
– alpast
Dec 22 '18 at 17:23
$begingroup$
@idm sorry I dont understand? I get that $theta : ymapsto mathbb E[Xmid Y=y]$. And then $theta(Y)$ is the composition of $Y$ and $theta$, so $theta(Y(omega)) : omegamapsto mathbb E[Xmid Y=y]$. is this right?
$endgroup$
– alpast
Dec 22 '18 at 17:23
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Here $theta:mathbb Rtomathbb R$ is prescribed by:$$ymapstomathbb E[Xmid X=y]=int xf(xmid y);dx$$
So actually: $$theta(Y)=int xf(xmid Y);dx$$
From here it is our intention to prove that $theta(Y)$ has the characteristics of $mathbb E[Xmid Y]$ which are:
$theta(Y)$ is measurable wrt $sigma(Y)$.
$mathbb Emathbf1_AX=mathbb Emathbf1_Atheta(Y)$ for every $Ainsigma(Y)$
Be aware that $mathbb E[Xmid Y]$ is the same thing as $mathbb E[Xmid sigma(Y)]$.
It is immediate that $theta(Y)$ is measurable wrt $sigma(Y)$.
Further we have: $$Ainsigma(Y)iff A={Yin B}text{ for some Borel set }B$$
With this in mind we find:
$$mathbb Emathbf1_Atheta(Y)=mathbb Emathbf1_B(Y)theta(Y)=int f(y)1_B(y)theta(y);dy=int f(y)1_B(y)left[int xf(xmid y);dxright];dx;dy=$$$$intint 1_B(y)xf(x,y);dx;dy=mathbb Emathbf1_B(Y)X=mathbb Emathbf1_AX$$
Now we are done and this together justifies to state that $$mathbb E[Xmid sigma(Y)]=mathbb E[Xmid Y]=theta(Y)$$
answered Dec 22 '18 at 18:05
drhabdrhab
103k545136
$endgroup$
$begingroup$
Why is $theta(Y)$ measurable w.r.t. $sigma(Y)$?
$endgroup$
– Robert W.
Dec 22 '18 at 18:13
$begingroup$
It is direct that $Y$ is measurable wrt $sigma(Y)$. Then $theta(Y)$ also because the preimage of a Borel set $B$ under $thetacirc Y$ takes the form $Y^{-1}(theta^{-1}(B))$ where $theta^{-1}(B)$ is again a Borel set, so the preimage is element of $sigma(Y)$. More shortly ${theta(Y)in B}={Yintheta^{-1}(B)}insigma(Y)$.
$endgroup$
– drhab
Dec 22 '18 at 18:20
$begingroup$
Sure. But is why $theta$ (Borel) measurable?
$endgroup$
– Robert W.
Dec 22 '18 at 18:25
1
$begingroup$
@RobertW. That is indeed a good question. It goes back to something like: if $g(x,y)$ is a measurable function then also $xmapstoint g(x,y)dy$ is a measurable function. It is treated by proving Fubini. I am not quite willing to go deeper into this here.
$endgroup$
– drhab
Dec 22 '18 at 18:29
add a comment |
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$begingroup$
Here $theta:mathbb Rtomathbb R$ is prescribed by:$$ymapstomathbb E[Xmid X=y]=int xf(xmid y);dx$$
So actually: $$theta(Y)=int xf(xmid Y);dx$$
From here it is our intention to prove that $theta(Y)$ has the characteristics of $mathbb E[Xmid Y]$ which are:
$theta(Y)$ is measurable wrt $sigma(Y)$.
$mathbb Emathbf1_AX=mathbb Emathbf1_Atheta(Y)$ for every $Ainsigma(Y)$
Be aware that $mathbb E[Xmid Y]$ is the same thing as $mathbb E[Xmid sigma(Y)]$.
It is immediate that $theta(Y)$ is measurable wrt $sigma(Y)$.
Further we have: $$Ainsigma(Y)iff A={Yin B}text{ for some Borel set }B$$
With this in mind we find:
$$mathbb Emathbf1_Atheta(Y)=mathbb Emathbf1_B(Y)theta(Y)=int f(y)1_B(y)theta(y);dy=int f(y)1_B(y)left[int xf(xmid y);dxright];dx;dy=$$$$intint 1_B(y)xf(x,y);dx;dy=mathbb Emathbf1_B(Y)X=mathbb Emathbf1_AX$$
Now we are done and this together justifies to state that $$mathbb E[Xmid sigma(Y)]=mathbb E[Xmid Y]=theta(Y)$$
answered Dec 22 '18 at 18:05
drhabdrhab
103k545136
$endgroup$
$begingroup$
Why is $theta(Y)$ measurable w.r.t. $sigma(Y)$?
$endgroup$
– Robert W.
Dec 22 '18 at 18:13
$begingroup$
It is direct that $Y$ is measurable wrt $sigma(Y)$. Then $theta(Y)$ also because the preimage of a Borel set $B$ under $thetacirc Y$ takes the form $Y^{-1}(theta^{-1}(B))$ where $theta^{-1}(B)$ is again a Borel set, so the preimage is element of $sigma(Y)$. More shortly ${theta(Y)in B}={Yintheta^{-1}(B)}insigma(Y)$.
$endgroup$
– drhab
Dec 22 '18 at 18:20
$begingroup$
Sure. But is why $theta$ (Borel) measurable?
$endgroup$
– Robert W.
Dec 22 '18 at 18:25
1
$begingroup$
@RobertW. That is indeed a good question. It goes back to something like: if $g(x,y)$ is a measurable function then also $xmapstoint g(x,y)dy$ is a measurable function. It is treated by proving Fubini. I am not quite willing to go deeper into this here.
$endgroup$
– drhab
Dec 22 '18 at 18:29
add a comment |
$begingroup$
Here $theta:mathbb Rtomathbb R$ is prescribed by:$$ymapstomathbb E[Xmid X=y]=int xf(xmid y);dx$$
So actually: $$theta(Y)=int xf(xmid Y);dx$$
From here it is our intention to prove that $theta(Y)$ has the characteristics of $mathbb E[Xmid Y]$ which are:
$theta(Y)$ is measurable wrt $sigma(Y)$.
$mathbb Emathbf1_AX=mathbb Emathbf1_Atheta(Y)$ for every $Ainsigma(Y)$
Be aware that $mathbb E[Xmid Y]$ is the same thing as $mathbb E[Xmid sigma(Y)]$.
It is immediate that $theta(Y)$ is measurable wrt $sigma(Y)$.
Further we have: $$Ainsigma(Y)iff A={Yin B}text{ for some Borel set }B$$
With this in mind we find:
$$mathbb Emathbf1_Atheta(Y)=mathbb Emathbf1_B(Y)theta(Y)=int f(y)1_B(y)theta(y);dy=int f(y)1_B(y)left[int xf(xmid y);dxright];dx;dy=$$$$intint 1_B(y)xf(x,y);dx;dy=mathbb Emathbf1_B(Y)X=mathbb Emathbf1_AX$$
Now we are done and this together justifies to state that $$mathbb E[Xmid sigma(Y)]=mathbb E[Xmid Y]=theta(Y)$$
answered Dec 22 '18 at 18:05
drhabdrhab
103k545136
$endgroup$
$begingroup$
Why is $theta(Y)$ measurable w.r.t. $sigma(Y)$?
$endgroup$
– Robert W.
Dec 22 '18 at 18:13
$begingroup$
It is direct that $Y$ is measurable wrt $sigma(Y)$. Then $theta(Y)$ also because the preimage of a Borel set $B$ under $thetacirc Y$ takes the form $Y^{-1}(theta^{-1}(B))$ where $theta^{-1}(B)$ is again a Borel set, so the preimage is element of $sigma(Y)$. More shortly ${theta(Y)in B}={Yintheta^{-1}(B)}insigma(Y)$.
$endgroup$
– drhab
Dec 22 '18 at 18:20
$begingroup$
Sure. But is why $theta$ (Borel) measurable?
$endgroup$
– Robert W.
Dec 22 '18 at 18:25
1
$begingroup$
@RobertW. That is indeed a good question. It goes back to something like: if $g(x,y)$ is a measurable function then also $xmapstoint g(x,y)dy$ is a measurable function. It is treated by proving Fubini. I am not quite willing to go deeper into this here.
$endgroup$
– drhab
Dec 22 '18 at 18:29
add a comment |
$begingroup$
Here $theta:mathbb Rtomathbb R$ is prescribed by:$$ymapstomathbb E[Xmid X=y]=int xf(xmid y);dx$$
So actually: $$theta(Y)=int xf(xmid Y);dx$$
From here it is our intention to prove that $theta(Y)$ has the characteristics of $mathbb E[Xmid Y]$ which are:
$theta(Y)$ is measurable wrt $sigma(Y)$.
$mathbb Emathbf1_AX=mathbb Emathbf1_Atheta(Y)$ for every $Ainsigma(Y)$
Be aware that $mathbb E[Xmid Y]$ is the same thing as $mathbb E[Xmid sigma(Y)]$.
It is immediate that $theta(Y)$ is measurable wrt $sigma(Y)$.
Further we have: $$Ainsigma(Y)iff A={Yin B}text{ for some Borel set }B$$
With this in mind we find:
$$mathbb Emathbf1_Atheta(Y)=mathbb Emathbf1_B(Y)theta(Y)=int f(y)1_B(y)theta(y);dy=int f(y)1_B(y)left[int xf(xmid y);dxright];dx;dy=$$$$intint 1_B(y)xf(x,y);dx;dy=mathbb Emathbf1_B(Y)X=mathbb Emathbf1_AX$$
Now we are done and this together justifies to state that $$mathbb E[Xmid sigma(Y)]=mathbb E[Xmid Y]=theta(Y)$$
answered Dec 22 '18 at 18:05
drhabdrhab
103k545136
$endgroup$
Here $theta:mathbb Rtomathbb R$ is prescribed by:$$ymapstomathbb E[Xmid X=y]=int xf(xmid y);dx$$
So actually: $$theta(Y)=int xf(xmid Y);dx$$
From here it is our intention to prove that $theta(Y)$ has the characteristics of $mathbb E[Xmid Y]$ which are:
$theta(Y)$ is measurable wrt $sigma(Y)$.
$mathbb Emathbf1_AX=mathbb Emathbf1_Atheta(Y)$ for every $Ainsigma(Y)$
Be aware that $mathbb E[Xmid Y]$ is the same thing as $mathbb E[Xmid sigma(Y)]$.
It is immediate that $theta(Y)$ is measurable wrt $sigma(Y)$.
Further we have: $$Ainsigma(Y)iff A={Yin B}text{ for some Borel set }B$$
With this in mind we find:
$$mathbb Emathbf1_Atheta(Y)=mathbb Emathbf1_B(Y)theta(Y)=int f(y)1_B(y)theta(y);dy=int f(y)1_B(y)left[int xf(xmid y);dxright];dx;dy=$$$$intint 1_B(y)xf(x,y);dx;dy=mathbb Emathbf1_B(Y)X=mathbb Emathbf1_AX$$
Now we are done and this together justifies to state that $$mathbb E[Xmid sigma(Y)]=mathbb E[Xmid Y]=theta(Y)$$
answered Dec 22 '18 at 18:05
drhabdrhab
103k545136
edited Dec 22 '18 at 18:11
answered Dec 22 '18 at 18:05
drhabdrhab
103k545136
answered Dec 22 '18 at 18:05
drhabdrhab
103k545136
answered Dec 22 '18 at 18:05
drhabdrhab
103k545136
103k545136
$begingroup$
Why is $theta(Y)$ measurable w.r.t. $sigma(Y)$?
$endgroup$
– Robert W.
Dec 22 '18 at 18:13
$begingroup$
It is direct that $Y$ is measurable wrt $sigma(Y)$. Then $theta(Y)$ also because the preimage of a Borel set $B$ under $thetacirc Y$ takes the form $Y^{-1}(theta^{-1}(B))$ where $theta^{-1}(B)$ is again a Borel set, so the preimage is element of $sigma(Y)$. More shortly ${theta(Y)in B}={Yintheta^{-1}(B)}insigma(Y)$.
$endgroup$
– drhab
Dec 22 '18 at 18:20
$begingroup$
Sure. But is why $theta$ (Borel) measurable?
$endgroup$
– Robert W.
Dec 22 '18 at 18:25
1
$begingroup$
@RobertW. That is indeed a good question. It goes back to something like: if $g(x,y)$ is a measurable function then also $xmapstoint g(x,y)dy$ is a measurable function. It is treated by proving Fubini. I am not quite willing to go deeper into this here.
$endgroup$
– drhab
Dec 22 '18 at 18:29
add a comment |
$begingroup$
Why is $theta(Y)$ measurable w.r.t. $sigma(Y)$?
$endgroup$
– Robert W.
Dec 22 '18 at 18:13
$begingroup$
It is direct that $Y$ is measurable wrt $sigma(Y)$. Then $theta(Y)$ also because the preimage of a Borel set $B$ under $thetacirc Y$ takes the form $Y^{-1}(theta^{-1}(B))$ where $theta^{-1}(B)$ is again a Borel set, so the preimage is element of $sigma(Y)$. More shortly ${theta(Y)in B}={Yintheta^{-1}(B)}insigma(Y)$.
$endgroup$
– drhab
Dec 22 '18 at 18:20
$begingroup$
Sure. But is why $theta$ (Borel) measurable?
$endgroup$
– Robert W.
Dec 22 '18 at 18:25
1
$begingroup$
@RobertW. That is indeed a good question. It goes back to something like: if $g(x,y)$ is a measurable function then also $xmapstoint g(x,y)dy$ is a measurable function. It is treated by proving Fubini. I am not quite willing to go deeper into this here.
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– drhab
Dec 22 '18 at 18:29
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Why is $theta(Y)$ measurable w.r.t. $sigma(Y)$?
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– Robert W.
Dec 22 '18 at 18:13
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Why is $theta(Y)$ measurable w.r.t. $sigma(Y)$?
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– Robert W.
Dec 22 '18 at 18:13
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It is direct that $Y$ is measurable wrt $sigma(Y)$. Then $theta(Y)$ also because the preimage of a Borel set $B$ under $thetacirc Y$ takes the form $Y^{-1}(theta^{-1}(B))$ where $theta^{-1}(B)$ is again a Borel set, so the preimage is element of $sigma(Y)$. More shortly ${theta(Y)in B}={Yintheta^{-1}(B)}insigma(Y)$.
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– drhab
Dec 22 '18 at 18:20
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It is direct that $Y$ is measurable wrt $sigma(Y)$. Then $theta(Y)$ also because the preimage of a Borel set $B$ under $thetacirc Y$ takes the form $Y^{-1}(theta^{-1}(B))$ where $theta^{-1}(B)$ is again a Borel set, so the preimage is element of $sigma(Y)$. More shortly ${theta(Y)in B}={Yintheta^{-1}(B)}insigma(Y)$.
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– drhab
Dec 22 '18 at 18:20
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Sure. But is why $theta$ (Borel) measurable?
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– Robert W.
Dec 22 '18 at 18:25
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Sure. But is why $theta$ (Borel) measurable?
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– Robert W.
Dec 22 '18 at 18:25
1
1
$begingroup$
@RobertW. That is indeed a good question. It goes back to something like: if $g(x,y)$ is a measurable function then also $xmapstoint g(x,y)dy$ is a measurable function. It is treated by proving Fubini. I am not quite willing to go deeper into this here.
$endgroup$
– drhab
Dec 22 '18 at 18:29
$begingroup$
@RobertW. That is indeed a good question. It goes back to something like: if $g(x,y)$ is a measurable function then also $xmapstoint g(x,y)dy$ is a measurable function. It is treated by proving Fubini. I am not quite willing to go deeper into this here.
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– drhab
Dec 22 '18 at 18:29
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$theta (Y)$ is a r.v. s.t. $mathbb E[theta (Y)]=mathbb E[X]$. It's the best approximation of $X$ knowing $Y$ is measurable. Normally, we write $theta (Y)=mathbb E[Xmid Y]$.
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– idm
Dec 22 '18 at 16:47
1
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$theta$ must be looked at as a function $mathbb Rtomathbb R$ prescribed by $ymapstomathbb E[Xmid Y=y]$. Then $theta(Y)$ is just the composition $thetacirc Y:Omegatomathbb R$. The interpretation $theta(Y)=mathbb E[Xmid Y=Y]$ is confusing and must be avoided.
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– drhab
Dec 22 '18 at 16:50
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@drhab so I should think of $theta(Y)$ as $theta(Y)(omega)=E(X|Y=y)$ where $Y(omega)=y$?
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– alpast
Dec 22 '18 at 16:53
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@alpast: No ! as $theta (Y)=mathbb E[Xmid Y]$, where $theta : ymapsto mathbb E[Xmid Y=y]$ is a function. When $f(x)=x^2$, how do you think $f(X)$ ? as $X^2(omega )=x^2$ ? Of course not :)
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– idm
Dec 22 '18 at 17:14
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@idm sorry I dont understand? I get that $theta : ymapsto mathbb E[Xmid Y=y]$. And then $theta(Y)$ is the composition of $Y$ and $theta$, so $theta(Y(omega)) : omegamapsto mathbb E[Xmid Y=y]$. is this right?
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– alpast
Dec 22 '18 at 17:23