Definition of predictable process












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$begingroup$


I am trying to understand the notion of predictable process. Let $(Ω,F_t,P)$ be a filtered measure space, satisfying the usual condition. Things starts with the predictable $sigma$-algebra ${mathcal P}$, which is generated by sets of the form $Atimes (a,b]$ with $Ain{mathcal F}_a$ and $Atimes {0}$ with $Ain{mathcal F}_0$.



My question: is it true that $Sin {mathcal P}$ if and only if $S$ is progressive and ${omega|(omega,t)in S}in{mathcal F}_{t−}$ for all $t$? In another word, is it true that $X$ is predictable if and only if $X$ is progressive and $X$ is adapted to the filtration ${mathcal F}_{t−}$?



The only if part is easy but I am not sure about the if part. I feel that $X$ being ${mathcal F}_{t−}$-measurable seems to be a more "reasonable" definition of "predictable", but maybe I am wrong.






measure-theory stochastic-processes stochastic-calculus







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$endgroup$

















    5












    $begingroup$


    I am trying to understand the notion of predictable process. Let $(Ω,F_t,P)$ be a filtered measure space, satisfying the usual condition. Things starts with the predictable $sigma$-algebra ${mathcal P}$, which is generated by sets of the form $Atimes (a,b]$ with $Ain{mathcal F}_a$ and $Atimes {0}$ with $Ain{mathcal F}_0$.



    My question: is it true that $Sin {mathcal P}$ if and only if $S$ is progressive and ${omega|(omega,t)in S}in{mathcal F}_{t−}$ for all $t$? In another word, is it true that $X$ is predictable if and only if $X$ is progressive and $X$ is adapted to the filtration ${mathcal F}_{t−}$?



    The only if part is easy but I am not sure about the if part. I feel that $X$ being ${mathcal F}_{t−}$-measurable seems to be a more "reasonable" definition of "predictable", but maybe I am wrong.






    measure-theory stochastic-processes stochastic-calculus







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      3



      $begingroup$


      I am trying to understand the notion of predictable process. Let $(Ω,F_t,P)$ be a filtered measure space, satisfying the usual condition. Things starts with the predictable $sigma$-algebra ${mathcal P}$, which is generated by sets of the form $Atimes (a,b]$ with $Ain{mathcal F}_a$ and $Atimes {0}$ with $Ain{mathcal F}_0$.



      My question: is it true that $Sin {mathcal P}$ if and only if $S$ is progressive and ${omega|(omega,t)in S}in{mathcal F}_{t−}$ for all $t$? In another word, is it true that $X$ is predictable if and only if $X$ is progressive and $X$ is adapted to the filtration ${mathcal F}_{t−}$?



      The only if part is easy but I am not sure about the if part. I feel that $X$ being ${mathcal F}_{t−}$-measurable seems to be a more "reasonable" definition of "predictable", but maybe I am wrong.






      measure-theory stochastic-processes stochastic-calculus







      share|cite|improve this question











      $endgroup$




      I am trying to understand the notion of predictable process. Let $(Ω,F_t,P)$ be a filtered measure space, satisfying the usual condition. Things starts with the predictable $sigma$-algebra ${mathcal P}$, which is generated by sets of the form $Atimes (a,b]$ with $Ain{mathcal F}_a$ and $Atimes {0}$ with $Ain{mathcal F}_0$.



      My question: is it true that $Sin {mathcal P}$ if and only if $S$ is progressive and ${omega|(omega,t)in S}in{mathcal F}_{t−}$ for all $t$? In another word, is it true that $X$ is predictable if and only if $X$ is progressive and $X$ is adapted to the filtration ${mathcal F}_{t−}$?



      The only if part is easy but I am not sure about the if part. I feel that $X$ being ${mathcal F}_{t−}$-measurable seems to be a more "reasonable" definition of "predictable", but maybe I am wrong.






      measure-theory stochastic-processes stochastic-calculus




      measure-theory stochastic-processes stochastic-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 7:13









      saz

      81.1k861127




      81.1k861127










      asked Oct 29 '18 at 16:47









      Yu DingYu Ding

      3435




      3435






















          1 Answer
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          No, it's not true.



          First of all, recall that any adapted càdlàg process $(X_t)_{t geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t geq 0}$ your assertion reads




          $(X_t)_{t geq 0}$ is predictable $iff$ $X_{t}$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$.




          Now consider for instance a Poisson process $(X_t)_{t geq 0}$, and let $(mathcal{F}_t)_{t geq 0}$ be its completed canonical filtration. By the very definition of $mathcal{F}_{t-}$, we know that $X_{t-} = lim_{s uparrow t} X_s$ is $mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$. However, $(X_t)_{t geq 0}$ is not predictable. Indeed: If $(X_t)_{t geq 0}$ was predictable, then



          $$M_t := X_t -t mathbb{E}(X_1), qquad t geq 0,$$



          would be a predictable martingale which would imply that $(M_t)_{t geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t geq 0}$ is not predictable.






          share|cite|improve this answer











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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            2





            +100







            $begingroup$

            No, it's not true.



            First of all, recall that any adapted càdlàg process $(X_t)_{t geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t geq 0}$ your assertion reads




            $(X_t)_{t geq 0}$ is predictable $iff$ $X_{t}$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$.




            Now consider for instance a Poisson process $(X_t)_{t geq 0}$, and let $(mathcal{F}_t)_{t geq 0}$ be its completed canonical filtration. By the very definition of $mathcal{F}_{t-}$, we know that $X_{t-} = lim_{s uparrow t} X_s$ is $mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$. However, $(X_t)_{t geq 0}$ is not predictable. Indeed: If $(X_t)_{t geq 0}$ was predictable, then



            $$M_t := X_t -t mathbb{E}(X_1), qquad t geq 0,$$



            would be a predictable martingale which would imply that $(M_t)_{t geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t geq 0}$ is not predictable.






            share|cite|improve this answer











            $endgroup$


















              2





              +100







              $begingroup$

              No, it's not true.



              First of all, recall that any adapted càdlàg process $(X_t)_{t geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t geq 0}$ your assertion reads




              $(X_t)_{t geq 0}$ is predictable $iff$ $X_{t}$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$.




              Now consider for instance a Poisson process $(X_t)_{t geq 0}$, and let $(mathcal{F}_t)_{t geq 0}$ be its completed canonical filtration. By the very definition of $mathcal{F}_{t-}$, we know that $X_{t-} = lim_{s uparrow t} X_s$ is $mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$. However, $(X_t)_{t geq 0}$ is not predictable. Indeed: If $(X_t)_{t geq 0}$ was predictable, then



              $$M_t := X_t -t mathbb{E}(X_1), qquad t geq 0,$$



              would be a predictable martingale which would imply that $(M_t)_{t geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t geq 0}$ is not predictable.






              share|cite|improve this answer











              $endgroup$
















                2





                +100







                2





                +100



                2




                +100



                $begingroup$

                No, it's not true.



                First of all, recall that any adapted càdlàg process $(X_t)_{t geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t geq 0}$ your assertion reads




                $(X_t)_{t geq 0}$ is predictable $iff$ $X_{t}$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$.




                Now consider for instance a Poisson process $(X_t)_{t geq 0}$, and let $(mathcal{F}_t)_{t geq 0}$ be its completed canonical filtration. By the very definition of $mathcal{F}_{t-}$, we know that $X_{t-} = lim_{s uparrow t} X_s$ is $mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$. However, $(X_t)_{t geq 0}$ is not predictable. Indeed: If $(X_t)_{t geq 0}$ was predictable, then



                $$M_t := X_t -t mathbb{E}(X_1), qquad t geq 0,$$



                would be a predictable martingale which would imply that $(M_t)_{t geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t geq 0}$ is not predictable.






                share|cite|improve this answer











                $endgroup$



                No, it's not true.



                First of all, recall that any adapted càdlàg process $(X_t)_{t geq 0}$ is progressively measurable. This means that for any such process $(X_t)_{t geq 0}$ your assertion reads




                $(X_t)_{t geq 0}$ is predictable $iff$ $X_{t}$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$.




                Now consider for instance a Poisson process $(X_t)_{t geq 0}$, and let $(mathcal{F}_t)_{t geq 0}$ be its completed canonical filtration. By the very definition of $mathcal{F}_{t-}$, we know that $X_{t-} = lim_{s uparrow t} X_s$ is $mathcal{F}_{t-}$-measurable. Since $X_t = X_{t-}$ almost surely we find that $X_t$ is $mathcal{F}_{t-}$-measurable for any $t geq 0$. However, $(X_t)_{t geq 0}$ is not predictable. Indeed: If $(X_t)_{t geq 0}$ was predictable, then



                $$M_t := X_t -t mathbb{E}(X_1), qquad t geq 0,$$



                would be a predictable martingale which would imply that $(M_t)_{t geq 0}$ has continuous sample paths (see e.g. here) which is clearly not true; hence $(X_t)_{t geq 0}$ is not predictable.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 9 '18 at 18:10

























                answered Dec 8 '18 at 19:44









                sazsaz

                81.1k861127




                81.1k861127






























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