Stochastic version of the Radon-Nikodým theorem

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Let



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

  • $T>0$

  • $I:=(0,T]$

  • $(mathcal F_t)_tinoverline I$ be a filtration on $(Omega,mathcal A)$

  • $mathcal P$ denote the predictable $sigma$-algebra on $Omegatimesoverline I$

  • $A:Omegatimesoverline Itomathbb R$ be $mathcal Aotimesmathcal B(overline I)$-measurable such that $A_0(omega)=0$ and $A(omega)$ is continuous and of bounded variation for all $omegainOmega$ and $$operatorname Eleft[|A|_Tright]<inftytag1,$$ where $|A(omega)|$ denotes the total variation function of $A(omega)$ for $omegainOmega$

  • $B:Omegatimesoverline Itomathbb R$ satisfy all assumptions imposed on $A$ and assume that $B(omega)$ is nondecreasing for all $omegainOmega$

Assume $$int_(0,:t]xi:rm dA=0;;;textalmost surely for all tin Itag1$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$ with $$int_(0,:t]xi:rm dB=0;;;textalmost surely for all tin I.tag2$$



Now, let $$mu(P):=operatorname Eleft[int1_P:rm dAright]$$ and $$nu(P):=operatorname Eleft[int1_P:rm dBright]$$ for $Pinmathcal P$. By assumption, $mu$ is absolutely continuous with respect to $nu$ and hence the Radon-Nikodým derivative $$alpha:=fracrm dmurm dnu$$ is well-defined and $mathcal P$-measurable with $$operatorname Eleft[int|alpha|:rm dBright]<inftytag3$$ and $$operatorname Eleft[intxi:rm dBright]=intxi:rm dmu=intalphaxi:rm dnutag4$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$.




Let $$M:=A-intalpha:rm dB$$ and note that $$operatorname Eleft[intxi:rm dMright]=0tag5$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$. Why does $(5)$ yield that $M$ is an $mathcal F$-martingale?





probability-theory stochastic-processes martingales stochastic-analysis radon-nikodym




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    up vote
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    Let



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

    • $T>0$

    • $I:=(0,T]$

    • $(mathcal F_t)_tinoverline I$ be a filtration on $(Omega,mathcal A)$

    • $mathcal P$ denote the predictable $sigma$-algebra on $Omegatimesoverline I$

    • $A:Omegatimesoverline Itomathbb R$ be $mathcal Aotimesmathcal B(overline I)$-measurable such that $A_0(omega)=0$ and $A(omega)$ is continuous and of bounded variation for all $omegainOmega$ and $$operatorname Eleft[|A|_Tright]<inftytag1,$$ where $|A(omega)|$ denotes the total variation function of $A(omega)$ for $omegainOmega$

    • $B:Omegatimesoverline Itomathbb R$ satisfy all assumptions imposed on $A$ and assume that $B(omega)$ is nondecreasing for all $omegainOmega$

    Assume $$int_(0,:t]xi:rm dA=0;;;textalmost surely for all tin Itag1$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$ with $$int_(0,:t]xi:rm dB=0;;;textalmost surely for all tin I.tag2$$



    Now, let $$mu(P):=operatorname Eleft[int1_P:rm dAright]$$ and $$nu(P):=operatorname Eleft[int1_P:rm dBright]$$ for $Pinmathcal P$. By assumption, $mu$ is absolutely continuous with respect to $nu$ and hence the Radon-Nikodým derivative $$alpha:=fracrm dmurm dnu$$ is well-defined and $mathcal P$-measurable with $$operatorname Eleft[int|alpha|:rm dBright]<inftytag3$$ and $$operatorname Eleft[intxi:rm dBright]=intxi:rm dmu=intalphaxi:rm dnutag4$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$.




    Let $$M:=A-intalpha:rm dB$$ and note that $$operatorname Eleft[intxi:rm dMright]=0tag5$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$. Why does $(5)$ yield that $M$ is an $mathcal F$-martingale?





    probability-theory stochastic-processes martingales stochastic-analysis radon-nikodym




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let



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

      • $T>0$

      • $I:=(0,T]$

      • $(mathcal F_t)_tinoverline I$ be a filtration on $(Omega,mathcal A)$

      • $mathcal P$ denote the predictable $sigma$-algebra on $Omegatimesoverline I$

      • $A:Omegatimesoverline Itomathbb R$ be $mathcal Aotimesmathcal B(overline I)$-measurable such that $A_0(omega)=0$ and $A(omega)$ is continuous and of bounded variation for all $omegainOmega$ and $$operatorname Eleft[|A|_Tright]<inftytag1,$$ where $|A(omega)|$ denotes the total variation function of $A(omega)$ for $omegainOmega$

      • $B:Omegatimesoverline Itomathbb R$ satisfy all assumptions imposed on $A$ and assume that $B(omega)$ is nondecreasing for all $omegainOmega$

      Assume $$int_(0,:t]xi:rm dA=0;;;textalmost surely for all tin Itag1$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$ with $$int_(0,:t]xi:rm dB=0;;;textalmost surely for all tin I.tag2$$



      Now, let $$mu(P):=operatorname Eleft[int1_P:rm dAright]$$ and $$nu(P):=operatorname Eleft[int1_P:rm dBright]$$ for $Pinmathcal P$. By assumption, $mu$ is absolutely continuous with respect to $nu$ and hence the Radon-Nikodým derivative $$alpha:=fracrm dmurm dnu$$ is well-defined and $mathcal P$-measurable with $$operatorname Eleft[int|alpha|:rm dBright]<inftytag3$$ and $$operatorname Eleft[intxi:rm dBright]=intxi:rm dmu=intalphaxi:rm dnutag4$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$.




      Let $$M:=A-intalpha:rm dB$$ and note that $$operatorname Eleft[intxi:rm dMright]=0tag5$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$. Why does $(5)$ yield that $M$ is an $mathcal F$-martingale?





      probability-theory stochastic-processes martingales stochastic-analysis radon-nikodym




      share|cite|improve this question











      Let



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

      • $T>0$

      • $I:=(0,T]$

      • $(mathcal F_t)_tinoverline I$ be a filtration on $(Omega,mathcal A)$

      • $mathcal P$ denote the predictable $sigma$-algebra on $Omegatimesoverline I$

      • $A:Omegatimesoverline Itomathbb R$ be $mathcal Aotimesmathcal B(overline I)$-measurable such that $A_0(omega)=0$ and $A(omega)$ is continuous and of bounded variation for all $omegainOmega$ and $$operatorname Eleft[|A|_Tright]<inftytag1,$$ where $|A(omega)|$ denotes the total variation function of $A(omega)$ for $omegainOmega$

      • $B:Omegatimesoverline Itomathbb R$ satisfy all assumptions imposed on $A$ and assume that $B(omega)$ is nondecreasing for all $omegainOmega$

      Assume $$int_(0,:t]xi:rm dA=0;;;textalmost surely for all tin Itag1$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$ with $$int_(0,:t]xi:rm dB=0;;;textalmost surely for all tin I.tag2$$



      Now, let $$mu(P):=operatorname Eleft[int1_P:rm dAright]$$ and $$nu(P):=operatorname Eleft[int1_P:rm dBright]$$ for $Pinmathcal P$. By assumption, $mu$ is absolutely continuous with respect to $nu$ and hence the Radon-Nikodým derivative $$alpha:=fracrm dmurm dnu$$ is well-defined and $mathcal P$-measurable with $$operatorname Eleft[int|alpha|:rm dBright]<inftytag3$$ and $$operatorname Eleft[intxi:rm dBright]=intxi:rm dmu=intalphaxi:rm dnutag4$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$.




      Let $$M:=A-intalpha:rm dB$$ and note that $$operatorname Eleft[intxi:rm dMright]=0tag5$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$. Why does $(5)$ yield that $M$ is an $mathcal F$-martingale?






      probability-theory stochastic-processes martingales stochastic-analysis radon-nikodym





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      asked Aug 5 at 22:46









      0xbadf00d

      2,04141028




      2,04141028




















          1 Answer
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          active

          oldest

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          up vote
          1
          down vote



          accepted










          Fix $s leq t$. For fixed $F in mathcalF_s$ the process



          $$xi(u,omega) := 1_F(omega) 1_(s,t](u)$$



          is bounded, non-negative and progressively measurable. By assumption, this implies that



          $$mathbbE left( int xi , dM right)=0$$



          which means that



          $$mathbbE(1_F (M_t-M_s)) =0.$$



          Since $F in mathcalF_s$ is arbitrary, this is equivalent to saying that



          $$mathbbE(M_t-M_s mid mathcalF_s)=0,$$



          and so $(M_t,mathcalF_t)_t geq 0$ is a martingale.






          share|cite|improve this answer





















          • Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
            – 0xbadf00d
            Aug 8 at 20:27










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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Fix $s leq t$. For fixed $F in mathcalF_s$ the process



          $$xi(u,omega) := 1_F(omega) 1_(s,t](u)$$



          is bounded, non-negative and progressively measurable. By assumption, this implies that



          $$mathbbE left( int xi , dM right)=0$$



          which means that



          $$mathbbE(1_F (M_t-M_s)) =0.$$



          Since $F in mathcalF_s$ is arbitrary, this is equivalent to saying that



          $$mathbbE(M_t-M_s mid mathcalF_s)=0,$$



          and so $(M_t,mathcalF_t)_t geq 0$ is a martingale.






          share|cite|improve this answer





















          • Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
            – 0xbadf00d
            Aug 8 at 20:27














          up vote
          1
          down vote



          accepted










          Fix $s leq t$. For fixed $F in mathcalF_s$ the process



          $$xi(u,omega) := 1_F(omega) 1_(s,t](u)$$



          is bounded, non-negative and progressively measurable. By assumption, this implies that



          $$mathbbE left( int xi , dM right)=0$$



          which means that



          $$mathbbE(1_F (M_t-M_s)) =0.$$



          Since $F in mathcalF_s$ is arbitrary, this is equivalent to saying that



          $$mathbbE(M_t-M_s mid mathcalF_s)=0,$$



          and so $(M_t,mathcalF_t)_t geq 0$ is a martingale.






          share|cite|improve this answer





















          • Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
            – 0xbadf00d
            Aug 8 at 20:27












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Fix $s leq t$. For fixed $F in mathcalF_s$ the process



          $$xi(u,omega) := 1_F(omega) 1_(s,t](u)$$



          is bounded, non-negative and progressively measurable. By assumption, this implies that



          $$mathbbE left( int xi , dM right)=0$$



          which means that



          $$mathbbE(1_F (M_t-M_s)) =0.$$



          Since $F in mathcalF_s$ is arbitrary, this is equivalent to saying that



          $$mathbbE(M_t-M_s mid mathcalF_s)=0,$$



          and so $(M_t,mathcalF_t)_t geq 0$ is a martingale.






          share|cite|improve this answer













          Fix $s leq t$. For fixed $F in mathcalF_s$ the process



          $$xi(u,omega) := 1_F(omega) 1_(s,t](u)$$



          is bounded, non-negative and progressively measurable. By assumption, this implies that



          $$mathbbE left( int xi , dM right)=0$$



          which means that



          $$mathbbE(1_F (M_t-M_s)) =0.$$



          Since $F in mathcalF_s$ is arbitrary, this is equivalent to saying that



          $$mathbbE(M_t-M_s mid mathcalF_s)=0,$$



          and so $(M_t,mathcalF_t)_t geq 0$ is a martingale.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 7 at 17:56









          saz

          73k552112




          73k552112











          • Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
            – 0xbadf00d
            Aug 8 at 20:27
















          • Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
            – 0xbadf00d
            Aug 8 at 20:27















          Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
          – 0xbadf00d
          Aug 8 at 20:27




          Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
          – 0xbadf00d
          Aug 8 at 20:27












           

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