Equivalent condition for a stochastic process to be independent of a $sigma$-algebra [closed]












0












$begingroup$


Let





  • $(Omega,mathcal A,operatorname P)$ be a probability space


  • $mathcal Fsubseteqmathcal A$ be a $sigma$-algebra on $Omega$


  • $I$ be a set


  • $(E_i,mathcal E_i)$ be a measurable space for $iin I$, $$(E,mathcal E):=left(⨉_{iin I}E_i,bigotimes_{iin I}mathcal E_iright)$$ and $$pi_J:Eto⨉_{jin J};,;;;xmapstoleft.xright|_J$$ for $Jsubseteq I$ (if $J=left{iright}$, we write $pi_i$ instead of $pi_J$)

  • $X:Omegato E$



How can we show that $X$ is independent of $mathcal F$ if and only if $pi_Jcirc X$ is independent of $mathcal F$ for all $Jsubseteq I$ with $|J|inmathbb N$?







probability-theory proof-verification stochastic-processes independence







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$endgroup$



closed as off-topic by Saad, Did, Eevee Trainer, mrtaurho, KReiser Jan 9 at 9:12


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, Eevee Trainer, mrtaurho, KReiser

If this question can be reworded to fit the rules in the help center, please edit the question.





















    0












    $begingroup$


    Let





    • $(Omega,mathcal A,operatorname P)$ be a probability space


    • $mathcal Fsubseteqmathcal A$ be a $sigma$-algebra on $Omega$


    • $I$ be a set


    • $(E_i,mathcal E_i)$ be a measurable space for $iin I$, $$(E,mathcal E):=left(⨉_{iin I}E_i,bigotimes_{iin I}mathcal E_iright)$$ and $$pi_J:Eto⨉_{jin J};,;;;xmapstoleft.xright|_J$$ for $Jsubseteq I$ (if $J=left{iright}$, we write $pi_i$ instead of $pi_J$)

    • $X:Omegato E$



    How can we show that $X$ is independent of $mathcal F$ if and only if $pi_Jcirc X$ is independent of $mathcal F$ for all $Jsubseteq I$ with $|J|inmathbb N$?







    probability-theory proof-verification stochastic-processes independence







    share|cite|improve this question









    $endgroup$



    closed as off-topic by Saad, Did, Eevee Trainer, mrtaurho, KReiser Jan 9 at 9:12


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, Eevee Trainer, mrtaurho, KReiser

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      0












      0








      0





      $begingroup$


      Let





      • $(Omega,mathcal A,operatorname P)$ be a probability space


      • $mathcal Fsubseteqmathcal A$ be a $sigma$-algebra on $Omega$


      • $I$ be a set


      • $(E_i,mathcal E_i)$ be a measurable space for $iin I$, $$(E,mathcal E):=left(⨉_{iin I}E_i,bigotimes_{iin I}mathcal E_iright)$$ and $$pi_J:Eto⨉_{jin J};,;;;xmapstoleft.xright|_J$$ for $Jsubseteq I$ (if $J=left{iright}$, we write $pi_i$ instead of $pi_J$)

      • $X:Omegato E$



      How can we show that $X$ is independent of $mathcal F$ if and only if $pi_Jcirc X$ is independent of $mathcal F$ for all $Jsubseteq I$ with $|J|inmathbb N$?







      probability-theory proof-verification stochastic-processes independence







      share|cite|improve this question









      $endgroup$




      Let





      • $(Omega,mathcal A,operatorname P)$ be a probability space


      • $mathcal Fsubseteqmathcal A$ be a $sigma$-algebra on $Omega$


      • $I$ be a set


      • $(E_i,mathcal E_i)$ be a measurable space for $iin I$, $$(E,mathcal E):=left(⨉_{iin I}E_i,bigotimes_{iin I}mathcal E_iright)$$ and $$pi_J:Eto⨉_{jin J};,;;;xmapstoleft.xright|_J$$ for $Jsubseteq I$ (if $J=left{iright}$, we write $pi_i$ instead of $pi_J$)

      • $X:Omegato E$



      How can we show that $X$ is independent of $mathcal F$ if and only if $pi_Jcirc X$ is independent of $mathcal F$ for all $Jsubseteq I$ with $|J|inmathbb N$?







      probability-theory proof-verification stochastic-processes independence




      probability-theory proof-verification stochastic-processes independence






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 0:47









      0xbadf00d0xbadf00d

      1,84341531




      1,84341531




      closed as off-topic by Saad, Did, Eevee Trainer, mrtaurho, KReiser Jan 9 at 9:12


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, Eevee Trainer, mrtaurho, KReiser

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Saad, Did, Eevee Trainer, mrtaurho, KReiser Jan 9 at 9:12


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Did, Eevee Trainer, mrtaurho, KReiser

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






          active

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          0












          $begingroup$

          Let's start with "$Rightarrow$": Let $Jsubseteq I$. Note that $pi_J$ is $left(mathcal E,bigotimes_{jin J}mathcal E_jright)$-measurable and hence $pi_Jcirc X$ is $left(sigma(X),bigotimes_{jin J}mathcal E_jright)$-measurable. Thus, $$sigmaleft(pi_Jcirc Xright)subseteqsigma(X)tag1$$ yielding the claim.



          For the other direction, we'll need the following fact:




          If $mathcal E_isubseteqmathcal A$ and $mathcal E_icupleft{emptysetright}$ is a $pi$-System, then $mathcal E_1$ and $mathcal E_2$ are independent if and only if $sigma(mathcal E_1)$ and $sigma(mathcal E_2)$ are Independent.




          $Leftarrow$: By Definition, $$mathcal E=sigma(pi_i:iin I)tag2$$ and hence $$sigma(X)=sigmaleft(X^{-1}left(bigcup_{iin I}sigma(pi_i)right)right)tag3.$$ Now, $$X^{-1}left(bigcup_{iin I}sigma(pi_i)right)=bigcup_{iin I}sigma(pi_icirc X).tag4$$ The only problem is that $(4)$ doesn't need to be a $pi$-system (if I'm not mistaken). However, $$mathcal R:=bigcup_{substack{Jsubseteq I\|J|inmathbb N}}sigmaleft(pi_Jcirc Xright)$$ is a $pi$-System. By assumption, $mathcal R$ and $mathcal F$ are independent and hence we obtain the claim by $(3)$ and the mentioned fact.






          share|cite|improve this answer









          $endgroup$




















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            Let's start with "$Rightarrow$": Let $Jsubseteq I$. Note that $pi_J$ is $left(mathcal E,bigotimes_{jin J}mathcal E_jright)$-measurable and hence $pi_Jcirc X$ is $left(sigma(X),bigotimes_{jin J}mathcal E_jright)$-measurable. Thus, $$sigmaleft(pi_Jcirc Xright)subseteqsigma(X)tag1$$ yielding the claim.



            For the other direction, we'll need the following fact:




            If $mathcal E_isubseteqmathcal A$ and $mathcal E_icupleft{emptysetright}$ is a $pi$-System, then $mathcal E_1$ and $mathcal E_2$ are independent if and only if $sigma(mathcal E_1)$ and $sigma(mathcal E_2)$ are Independent.




            $Leftarrow$: By Definition, $$mathcal E=sigma(pi_i:iin I)tag2$$ and hence $$sigma(X)=sigmaleft(X^{-1}left(bigcup_{iin I}sigma(pi_i)right)right)tag3.$$ Now, $$X^{-1}left(bigcup_{iin I}sigma(pi_i)right)=bigcup_{iin I}sigma(pi_icirc X).tag4$$ The only problem is that $(4)$ doesn't need to be a $pi$-system (if I'm not mistaken). However, $$mathcal R:=bigcup_{substack{Jsubseteq I\|J|inmathbb N}}sigmaleft(pi_Jcirc Xright)$$ is a $pi$-System. By assumption, $mathcal R$ and $mathcal F$ are independent and hence we obtain the claim by $(3)$ and the mentioned fact.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let's start with "$Rightarrow$": Let $Jsubseteq I$. Note that $pi_J$ is $left(mathcal E,bigotimes_{jin J}mathcal E_jright)$-measurable and hence $pi_Jcirc X$ is $left(sigma(X),bigotimes_{jin J}mathcal E_jright)$-measurable. Thus, $$sigmaleft(pi_Jcirc Xright)subseteqsigma(X)tag1$$ yielding the claim.



              For the other direction, we'll need the following fact:




              If $mathcal E_isubseteqmathcal A$ and $mathcal E_icupleft{emptysetright}$ is a $pi$-System, then $mathcal E_1$ and $mathcal E_2$ are independent if and only if $sigma(mathcal E_1)$ and $sigma(mathcal E_2)$ are Independent.




              $Leftarrow$: By Definition, $$mathcal E=sigma(pi_i:iin I)tag2$$ and hence $$sigma(X)=sigmaleft(X^{-1}left(bigcup_{iin I}sigma(pi_i)right)right)tag3.$$ Now, $$X^{-1}left(bigcup_{iin I}sigma(pi_i)right)=bigcup_{iin I}sigma(pi_icirc X).tag4$$ The only problem is that $(4)$ doesn't need to be a $pi$-system (if I'm not mistaken). However, $$mathcal R:=bigcup_{substack{Jsubseteq I\|J|inmathbb N}}sigmaleft(pi_Jcirc Xright)$$ is a $pi$-System. By assumption, $mathcal R$ and $mathcal F$ are independent and hence we obtain the claim by $(3)$ and the mentioned fact.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let's start with "$Rightarrow$": Let $Jsubseteq I$. Note that $pi_J$ is $left(mathcal E,bigotimes_{jin J}mathcal E_jright)$-measurable and hence $pi_Jcirc X$ is $left(sigma(X),bigotimes_{jin J}mathcal E_jright)$-measurable. Thus, $$sigmaleft(pi_Jcirc Xright)subseteqsigma(X)tag1$$ yielding the claim.



                For the other direction, we'll need the following fact:




                If $mathcal E_isubseteqmathcal A$ and $mathcal E_icupleft{emptysetright}$ is a $pi$-System, then $mathcal E_1$ and $mathcal E_2$ are independent if and only if $sigma(mathcal E_1)$ and $sigma(mathcal E_2)$ are Independent.




                $Leftarrow$: By Definition, $$mathcal E=sigma(pi_i:iin I)tag2$$ and hence $$sigma(X)=sigmaleft(X^{-1}left(bigcup_{iin I}sigma(pi_i)right)right)tag3.$$ Now, $$X^{-1}left(bigcup_{iin I}sigma(pi_i)right)=bigcup_{iin I}sigma(pi_icirc X).tag4$$ The only problem is that $(4)$ doesn't need to be a $pi$-system (if I'm not mistaken). However, $$mathcal R:=bigcup_{substack{Jsubseteq I\|J|inmathbb N}}sigmaleft(pi_Jcirc Xright)$$ is a $pi$-System. By assumption, $mathcal R$ and $mathcal F$ are independent and hence we obtain the claim by $(3)$ and the mentioned fact.






                share|cite|improve this answer









                $endgroup$



                Let's start with "$Rightarrow$": Let $Jsubseteq I$. Note that $pi_J$ is $left(mathcal E,bigotimes_{jin J}mathcal E_jright)$-measurable and hence $pi_Jcirc X$ is $left(sigma(X),bigotimes_{jin J}mathcal E_jright)$-measurable. Thus, $$sigmaleft(pi_Jcirc Xright)subseteqsigma(X)tag1$$ yielding the claim.



                For the other direction, we'll need the following fact:




                If $mathcal E_isubseteqmathcal A$ and $mathcal E_icupleft{emptysetright}$ is a $pi$-System, then $mathcal E_1$ and $mathcal E_2$ are independent if and only if $sigma(mathcal E_1)$ and $sigma(mathcal E_2)$ are Independent.




                $Leftarrow$: By Definition, $$mathcal E=sigma(pi_i:iin I)tag2$$ and hence $$sigma(X)=sigmaleft(X^{-1}left(bigcup_{iin I}sigma(pi_i)right)right)tag3.$$ Now, $$X^{-1}left(bigcup_{iin I}sigma(pi_i)right)=bigcup_{iin I}sigma(pi_icirc X).tag4$$ The only problem is that $(4)$ doesn't need to be a $pi$-system (if I'm not mistaken). However, $$mathcal R:=bigcup_{substack{Jsubseteq I\|J|inmathbb N}}sigmaleft(pi_Jcirc Xright)$$ is a $pi$-System. By assumption, $mathcal R$ and $mathcal F$ are independent and hence we obtain the claim by $(3)$ and the mentioned fact.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 0:47









                0xbadf00d0xbadf00d

                1,84341531




                1,84341531















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