$X,Y$ are independent iff conditional regular distribution of $X|Y$ is almost surely the same as the...
$begingroup$
I want to prove that $X,Y$ scalar random variables are independent iff conditional regular distribution of $X|Y$ is almost surely the same as the distribution of $X$. A simple result to prove for the... "regular" conditional distribution.
The definition employed for the regular distribution is the following.
Let $(Omega,mathcal F,P)$ be a probability space, and $(Omega',mathcal F')$ a measurable space. Now $mu:Omegatimesmathcal F'rightarrow mathbb R$ is the regular conditional distribution if:
1) For every $Ain mathcal F'$, $mu(.,A)=E(I_{Xin A}|mathcal G)$ almost surely. $mathcal Gsubseteq mathcal F$, and in this case $mathcal G=sigma (Y)$, which is the pre-image of the borel sigma algebra.
2) For almost every $omegain Omega$, $mu(omega,.)$ is a probability measure.
This definition I find very hard to use in order to prove almost anything.
measure-theory
asked Nov 25 '18 at 22:30
DoleDole
899514
$endgroup$
add a comment |
$begingroup$
I want to prove that $X,Y$ scalar random variables are independent iff conditional regular distribution of $X|Y$ is almost surely the same as the distribution of $X$. A simple result to prove for the... "regular" conditional distribution.
The definition employed for the regular distribution is the following.
Let $(Omega,mathcal F,P)$ be a probability space, and $(Omega',mathcal F')$ a measurable space. Now $mu:Omegatimesmathcal F'rightarrow mathbb R$ is the regular conditional distribution if:
1) For every $Ain mathcal F'$, $mu(.,A)=E(I_{Xin A}|mathcal G)$ almost surely. $mathcal Gsubseteq mathcal F$, and in this case $mathcal G=sigma (Y)$, which is the pre-image of the borel sigma algebra.
2) For almost every $omegain Omega$, $mu(omega,.)$ is a probability measure.
This definition I find very hard to use in order to prove almost anything.
measure-theory
asked Nov 25 '18 at 22:30
DoleDole
899514
$endgroup$
add a comment |
$begingroup$
I want to prove that $X,Y$ scalar random variables are independent iff conditional regular distribution of $X|Y$ is almost surely the same as the distribution of $X$. A simple result to prove for the... "regular" conditional distribution.
The definition employed for the regular distribution is the following.
Let $(Omega,mathcal F,P)$ be a probability space, and $(Omega',mathcal F')$ a measurable space. Now $mu:Omegatimesmathcal F'rightarrow mathbb R$ is the regular conditional distribution if:
1) For every $Ain mathcal F'$, $mu(.,A)=E(I_{Xin A}|mathcal G)$ almost surely. $mathcal Gsubseteq mathcal F$, and in this case $mathcal G=sigma (Y)$, which is the pre-image of the borel sigma algebra.
2) For almost every $omegain Omega$, $mu(omega,.)$ is a probability measure.
This definition I find very hard to use in order to prove almost anything.
measure-theory
asked Nov 25 '18 at 22:30
DoleDole
899514
$endgroup$
I want to prove that $X,Y$ scalar random variables are independent iff conditional regular distribution of $X|Y$ is almost surely the same as the distribution of $X$. A simple result to prove for the... "regular" conditional distribution.
The definition employed for the regular distribution is the following.
Let $(Omega,mathcal F,P)$ be a probability space, and $(Omega',mathcal F')$ a measurable space. Now $mu:Omegatimesmathcal F'rightarrow mathbb R$ is the regular conditional distribution if:
1) For every $Ain mathcal F'$, $mu(.,A)=E(I_{Xin A}|mathcal G)$ almost surely. $mathcal Gsubseteq mathcal F$, and in this case $mathcal G=sigma (Y)$, which is the pre-image of the borel sigma algebra.
2) For almost every $omegain Omega$, $mu(omega,.)$ is a probability measure.
This definition I find very hard to use in order to prove almost anything.
measure-theory
measure-theory
asked Nov 25 '18 at 22:30
DoleDole
899514
asked Nov 25 '18 at 22:30
DoleDole
899514
edited Nov 25 '18 at 22:57
Dole
asked Nov 25 '18 at 22:30
DoleDole
899514
asked Nov 25 '18 at 22:30
DoleDole
899514
asked Nov 25 '18 at 22:30
DoleDole
899514
899514
add a comment |
add a comment |
1 Answer
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Take $A=(-infty , x]$. If $X$ and $Y$ are independent then $mu(omega, A)=E(I_{(X in A)} |Y)=P(X^{-1}(A))=P{Xleq x}$ so the conditional distribution coincides with teh distribution of $X$. Conversely if this condition holds integrate both sides over ${Yleq y}$ to get $P{Xleq x,Yleq y} =P{Xleq x}P{Yleq y}$.
answered Nov 25 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
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add a comment |
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$begingroup$
Take $A=(-infty , x]$. If $X$ and $Y$ are independent then $mu(omega, A)=E(I_{(X in A)} |Y)=P(X^{-1}(A))=P{Xleq x}$ so the conditional distribution coincides with teh distribution of $X$. Conversely if this condition holds integrate both sides over ${Yleq y}$ to get $P{Xleq x,Yleq y} =P{Xleq x}P{Yleq y}$.
answered Nov 25 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
add a comment |
$begingroup$
Take $A=(-infty , x]$. If $X$ and $Y$ are independent then $mu(omega, A)=E(I_{(X in A)} |Y)=P(X^{-1}(A))=P{Xleq x}$ so the conditional distribution coincides with teh distribution of $X$. Conversely if this condition holds integrate both sides over ${Yleq y}$ to get $P{Xleq x,Yleq y} =P{Xleq x}P{Yleq y}$.
answered Nov 25 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
add a comment |
$begingroup$
Take $A=(-infty , x]$. If $X$ and $Y$ are independent then $mu(omega, A)=E(I_{(X in A)} |Y)=P(X^{-1}(A))=P{Xleq x}$ so the conditional distribution coincides with teh distribution of $X$. Conversely if this condition holds integrate both sides over ${Yleq y}$ to get $P{Xleq x,Yleq y} =P{Xleq x}P{Yleq y}$.
answered Nov 25 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
$endgroup$
Take $A=(-infty , x]$. If $X$ and $Y$ are independent then $mu(omega, A)=E(I_{(X in A)} |Y)=P(X^{-1}(A))=P{Xleq x}$ so the conditional distribution coincides with teh distribution of $X$. Conversely if this condition holds integrate both sides over ${Yleq y}$ to get $P{Xleq x,Yleq y} =P{Xleq x}P{Yleq y}$.
answered Nov 25 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Nov 25 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Nov 25 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
answered Nov 25 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
54.4k32055
54.4k32055
add a comment |
add a comment |
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