Is the strong solution of a SDE adapted to the filtration generated by the driving Brownian motion?












1












$begingroup$


Let





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


  • $(mathcal F_t)_{tge0}$ be a complete and right-continuous filtration on $(Omega,mathcal A,operatorname P)$


  • $xi$ be an $mathcal F_0$-measurable square-integrable random variable on $(Omega,mathcal A,operatorname P)$


  • $W$ be an $mathcal F$-Brownian motion on $(Omega,mathcal A,operatorname P)$


  • $b,sigma:[0,infty)timesmathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le C_1(1+|x|^2);;;text{for all }tge0text{ and }xinmathbb Rtag1$$ for some $C_1ge0$ and $$|b(t,x)-b(t,y)|^2+|sigma(t,x)-sigma(t,y)|^2le C_2|x-y|^2;;;text{for all }tge0text{ and }x,yinmathbb Rtag2$$ for some $C_2ge0$


Note that $$mathcal C^T_b:=left{X:Xtext{ is a continuous }mathcal Ftext{-adapted process on }(Omega,mathcal A,operatorname P)text{ with }left|sup_{tin[0,:T]}|X_t|right|_{L^2(operatorname P)}<inftyright}$$ equipped with $$left|Xright|_{mathcal C_b^T}:=left|sup_{tin[0,:T]}|X_t|right|_{L^2(operatorname P)};;;text{for }Xinmathcal C_b^T$$ is a complete semi-normed space for all $T>0$. Now, let $$Xi_T(X):=xi+left(int_0^tb(s,X_s):{rm d}sright)_{tin[0,:T]}+left(int_0^tsigma(s,X_s):{rm d}W_sright)_{tin[0,:T]};;;text{for }Xin C^T_b$$ for $T>0$. We can show that, for all $T>0$, there is a $ninmathbb N$ such that the $n$-fold composition of $Xi_T$ is a contraction on $C_b^T$ and hence there is a unique $X^{(T)}inmathcal C_b^T$ with $$Xi_Tleft(X^{(T)}right)=X^{(T)}tag3.$$ Clearly, $$X_t:=X^{(N)}_t;;;text{for }tin[0,N]text{ and }Ninmathbb N$$ is well-defined.




By definition, $X$ is $mathcal F$-adapted. Assume $xi$ is independent of $W$. Are we able to show that $X$ is even $(sigma(xi)veemathcal F^W_t)_{tge0}$-adapted, where $mathcal F^W$ denotes the filtration generated by $W$?




Fix $T>0$. The idea is to show that $(X_t)_{tin[0,:T]}$ is $(sigma(xi)veemathcal F^W_t)_{tin[0,:T]}$-adapted. Let $Y^0:=xi$ and $$Y^n:=Xi_T(Y^{n-1})=Xi_T^n(xi);;;text{for }ninmathbb N.$$ Now, $W$ is a $(sigma(xi)veemathcal F^W_t)_{tge0}$-Brownian motion and we can show $$left|(X_t)_{tin[0,:T]}-Y^nright|_Txrightarrow{ntoinfty}0tag5.$$ The desired claim follows.




However, what I don't get is the following: Why didn't we work out the construction of $X$ with $mathcal F$ replaced by $(sigma(xi)veemathcal F^W_t)_{tge0}$ in the first place? Since the latter filtration is smaller, we would still be able to conclude $mathcal F$-adaptedness of $X$. There must be something crucial I'm missing here ...




EDIT: As the result seems to be wrong, in general: How can we show it, if $xi$ is non-random?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let





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


    • $(mathcal F_t)_{tge0}$ be a complete and right-continuous filtration on $(Omega,mathcal A,operatorname P)$


    • $xi$ be an $mathcal F_0$-measurable square-integrable random variable on $(Omega,mathcal A,operatorname P)$


    • $W$ be an $mathcal F$-Brownian motion on $(Omega,mathcal A,operatorname P)$


    • $b,sigma:[0,infty)timesmathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le C_1(1+|x|^2);;;text{for all }tge0text{ and }xinmathbb Rtag1$$ for some $C_1ge0$ and $$|b(t,x)-b(t,y)|^2+|sigma(t,x)-sigma(t,y)|^2le C_2|x-y|^2;;;text{for all }tge0text{ and }x,yinmathbb Rtag2$$ for some $C_2ge0$


    Note that $$mathcal C^T_b:=left{X:Xtext{ is a continuous }mathcal Ftext{-adapted process on }(Omega,mathcal A,operatorname P)text{ with }left|sup_{tin[0,:T]}|X_t|right|_{L^2(operatorname P)}<inftyright}$$ equipped with $$left|Xright|_{mathcal C_b^T}:=left|sup_{tin[0,:T]}|X_t|right|_{L^2(operatorname P)};;;text{for }Xinmathcal C_b^T$$ is a complete semi-normed space for all $T>0$. Now, let $$Xi_T(X):=xi+left(int_0^tb(s,X_s):{rm d}sright)_{tin[0,:T]}+left(int_0^tsigma(s,X_s):{rm d}W_sright)_{tin[0,:T]};;;text{for }Xin C^T_b$$ for $T>0$. We can show that, for all $T>0$, there is a $ninmathbb N$ such that the $n$-fold composition of $Xi_T$ is a contraction on $C_b^T$ and hence there is a unique $X^{(T)}inmathcal C_b^T$ with $$Xi_Tleft(X^{(T)}right)=X^{(T)}tag3.$$ Clearly, $$X_t:=X^{(N)}_t;;;text{for }tin[0,N]text{ and }Ninmathbb N$$ is well-defined.




    By definition, $X$ is $mathcal F$-adapted. Assume $xi$ is independent of $W$. Are we able to show that $X$ is even $(sigma(xi)veemathcal F^W_t)_{tge0}$-adapted, where $mathcal F^W$ denotes the filtration generated by $W$?




    Fix $T>0$. The idea is to show that $(X_t)_{tin[0,:T]}$ is $(sigma(xi)veemathcal F^W_t)_{tin[0,:T]}$-adapted. Let $Y^0:=xi$ and $$Y^n:=Xi_T(Y^{n-1})=Xi_T^n(xi);;;text{for }ninmathbb N.$$ Now, $W$ is a $(sigma(xi)veemathcal F^W_t)_{tge0}$-Brownian motion and we can show $$left|(X_t)_{tin[0,:T]}-Y^nright|_Txrightarrow{ntoinfty}0tag5.$$ The desired claim follows.




    However, what I don't get is the following: Why didn't we work out the construction of $X$ with $mathcal F$ replaced by $(sigma(xi)veemathcal F^W_t)_{tge0}$ in the first place? Since the latter filtration is smaller, we would still be able to conclude $mathcal F$-adaptedness of $X$. There must be something crucial I'm missing here ...




    EDIT: As the result seems to be wrong, in general: How can we show it, if $xi$ is non-random?










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let





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


      • $(mathcal F_t)_{tge0}$ be a complete and right-continuous filtration on $(Omega,mathcal A,operatorname P)$


      • $xi$ be an $mathcal F_0$-measurable square-integrable random variable on $(Omega,mathcal A,operatorname P)$


      • $W$ be an $mathcal F$-Brownian motion on $(Omega,mathcal A,operatorname P)$


      • $b,sigma:[0,infty)timesmathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le C_1(1+|x|^2);;;text{for all }tge0text{ and }xinmathbb Rtag1$$ for some $C_1ge0$ and $$|b(t,x)-b(t,y)|^2+|sigma(t,x)-sigma(t,y)|^2le C_2|x-y|^2;;;text{for all }tge0text{ and }x,yinmathbb Rtag2$$ for some $C_2ge0$


      Note that $$mathcal C^T_b:=left{X:Xtext{ is a continuous }mathcal Ftext{-adapted process on }(Omega,mathcal A,operatorname P)text{ with }left|sup_{tin[0,:T]}|X_t|right|_{L^2(operatorname P)}<inftyright}$$ equipped with $$left|Xright|_{mathcal C_b^T}:=left|sup_{tin[0,:T]}|X_t|right|_{L^2(operatorname P)};;;text{for }Xinmathcal C_b^T$$ is a complete semi-normed space for all $T>0$. Now, let $$Xi_T(X):=xi+left(int_0^tb(s,X_s):{rm d}sright)_{tin[0,:T]}+left(int_0^tsigma(s,X_s):{rm d}W_sright)_{tin[0,:T]};;;text{for }Xin C^T_b$$ for $T>0$. We can show that, for all $T>0$, there is a $ninmathbb N$ such that the $n$-fold composition of $Xi_T$ is a contraction on $C_b^T$ and hence there is a unique $X^{(T)}inmathcal C_b^T$ with $$Xi_Tleft(X^{(T)}right)=X^{(T)}tag3.$$ Clearly, $$X_t:=X^{(N)}_t;;;text{for }tin[0,N]text{ and }Ninmathbb N$$ is well-defined.




      By definition, $X$ is $mathcal F$-adapted. Assume $xi$ is independent of $W$. Are we able to show that $X$ is even $(sigma(xi)veemathcal F^W_t)_{tge0}$-adapted, where $mathcal F^W$ denotes the filtration generated by $W$?




      Fix $T>0$. The idea is to show that $(X_t)_{tin[0,:T]}$ is $(sigma(xi)veemathcal F^W_t)_{tin[0,:T]}$-adapted. Let $Y^0:=xi$ and $$Y^n:=Xi_T(Y^{n-1})=Xi_T^n(xi);;;text{for }ninmathbb N.$$ Now, $W$ is a $(sigma(xi)veemathcal F^W_t)_{tge0}$-Brownian motion and we can show $$left|(X_t)_{tin[0,:T]}-Y^nright|_Txrightarrow{ntoinfty}0tag5.$$ The desired claim follows.




      However, what I don't get is the following: Why didn't we work out the construction of $X$ with $mathcal F$ replaced by $(sigma(xi)veemathcal F^W_t)_{tge0}$ in the first place? Since the latter filtration is smaller, we would still be able to conclude $mathcal F$-adaptedness of $X$. There must be something crucial I'm missing here ...




      EDIT: As the result seems to be wrong, in general: How can we show it, if $xi$ is non-random?










      share|cite|improve this question











      $endgroup$




      Let





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


      • $(mathcal F_t)_{tge0}$ be a complete and right-continuous filtration on $(Omega,mathcal A,operatorname P)$


      • $xi$ be an $mathcal F_0$-measurable square-integrable random variable on $(Omega,mathcal A,operatorname P)$


      • $W$ be an $mathcal F$-Brownian motion on $(Omega,mathcal A,operatorname P)$


      • $b,sigma:[0,infty)timesmathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le C_1(1+|x|^2);;;text{for all }tge0text{ and }xinmathbb Rtag1$$ for some $C_1ge0$ and $$|b(t,x)-b(t,y)|^2+|sigma(t,x)-sigma(t,y)|^2le C_2|x-y|^2;;;text{for all }tge0text{ and }x,yinmathbb Rtag2$$ for some $C_2ge0$


      Note that $$mathcal C^T_b:=left{X:Xtext{ is a continuous }mathcal Ftext{-adapted process on }(Omega,mathcal A,operatorname P)text{ with }left|sup_{tin[0,:T]}|X_t|right|_{L^2(operatorname P)}<inftyright}$$ equipped with $$left|Xright|_{mathcal C_b^T}:=left|sup_{tin[0,:T]}|X_t|right|_{L^2(operatorname P)};;;text{for }Xinmathcal C_b^T$$ is a complete semi-normed space for all $T>0$. Now, let $$Xi_T(X):=xi+left(int_0^tb(s,X_s):{rm d}sright)_{tin[0,:T]}+left(int_0^tsigma(s,X_s):{rm d}W_sright)_{tin[0,:T]};;;text{for }Xin C^T_b$$ for $T>0$. We can show that, for all $T>0$, there is a $ninmathbb N$ such that the $n$-fold composition of $Xi_T$ is a contraction on $C_b^T$ and hence there is a unique $X^{(T)}inmathcal C_b^T$ with $$Xi_Tleft(X^{(T)}right)=X^{(T)}tag3.$$ Clearly, $$X_t:=X^{(N)}_t;;;text{for }tin[0,N]text{ and }Ninmathbb N$$ is well-defined.




      By definition, $X$ is $mathcal F$-adapted. Assume $xi$ is independent of $W$. Are we able to show that $X$ is even $(sigma(xi)veemathcal F^W_t)_{tge0}$-adapted, where $mathcal F^W$ denotes the filtration generated by $W$?




      Fix $T>0$. The idea is to show that $(X_t)_{tin[0,:T]}$ is $(sigma(xi)veemathcal F^W_t)_{tin[0,:T]}$-adapted. Let $Y^0:=xi$ and $$Y^n:=Xi_T(Y^{n-1})=Xi_T^n(xi);;;text{for }ninmathbb N.$$ Now, $W$ is a $(sigma(xi)veemathcal F^W_t)_{tge0}$-Brownian motion and we can show $$left|(X_t)_{tin[0,:T]}-Y^nright|_Txrightarrow{ntoinfty}0tag5.$$ The desired claim follows.




      However, what I don't get is the following: Why didn't we work out the construction of $X$ with $mathcal F$ replaced by $(sigma(xi)veemathcal F^W_t)_{tge0}$ in the first place? Since the latter filtration is smaller, we would still be able to conclude $mathcal F$-adaptedness of $X$. There must be something crucial I'm missing here ...




      EDIT: As the result seems to be wrong, in general: How can we show it, if $xi$ is non-random?







      probability-theory stochastic-processes stochastic-integrals stochastic-analysis sde






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      edited Feb 22 at 16:51







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      asked Dec 21 '18 at 12:21









      0xbadf00d0xbadf00d

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