$X,Y$ are independent iff conditional regular distribution of $X|Y$ is almost surely the same as the...












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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.










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    0












    $begingroup$


    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.










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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.










      share|cite|improve this question











      $endgroup$




      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






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      edited Nov 25 '18 at 22:57







      Dole

















      asked Nov 25 '18 at 22:30









      DoleDole

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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 $P{Xleq x,Yleq y} =P{Xleq x}P{Yleq y}$.






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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 $P{Xleq x,Yleq y} =P{Xleq x}P{Yleq y}$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              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 $P{Xleq x,Yleq y} =P{Xleq x}P{Yleq y}$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                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 $P{Xleq x,Yleq y} =P{Xleq x}P{Yleq y}$.






                share|cite|improve this answer









                $endgroup$



                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 $P{Xleq x,Yleq y} =P{Xleq x}P{Yleq y}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 '18 at 23:43









                Kavi Rama MurthyKavi Rama Murthy

                54.4k32055




                54.4k32055






























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