Show that the solution of an autonomous SDE is a time-homogeneous Markov process
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Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space and $$mathcal N:=left{Ninmathcal A:operatorname P[N]=0right}$$
$(W_t)_{tge0}$ be a Brownian motion on $(Omega,mathcal A,operatorname P)$ and $$W^{(s)}_t:=W_{s+t}-W_s;;;text{for }tge0$$ for $sge0$
$b,sigma:[0,infty)timesmathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le K(1+|x|^2);;;text{for all }xinmathbb Rtag1$$ and $$|b(x)-b(y)|^2+|sigma(x)-sigma(y)|^2le K|x-y|^2;;;text{for all }x,yinmathbb Rtag2$$ for some $Kge0$
We know that for all $sge0$ there is a real-valued continuous process $left(X^{(s,:x)}_tright)_{(t,:x)in[s,:infty)timesmathbb R}$ on $(Omega,mathcal A,operatorname P)$ with $$X^{(s,:x)}_t=x+int_s^tbleft(X^{(s,:x)}_rright):{rm d}r+int_s^tsigmaleft(X^{(s,:x)}_rright):{rm d}W_rtag3$$ for all $tge s$ almost surely for all $xinmathbb R$.
Now, let $xi$ be a real-valued random variable on $(Omega,mathcal A,operatorname P)$ independent of $W$ and $$mathcal F_t:=sigma(xi)veesigma(W_s:sin[0,t])veemathcal N;;;text{for }tge0.$$ Now, there is a real-valued $mathcal F$-adapted continuous process $(X_t)_{tge0}$ with $$X_t=xi+int_0^tb(X_s):{rm d}s+int_0^tsigma(X_s):{rm d}W_s;;;text{for }tge0.$$ $X$ is a $mathcal F$-Markov process with $$operatorname Pleft[X_tin Bmidmathcal F_sright]=kappa_{s,:t}(X_s,B);;;text{almost surely for all }tge0text{ and }Binmathcal B(mathbb R),tag4$$ where $$kappa_{s,:t}(x,B):=operatorname Pleft[X^{(s,:x)}_tin Bright];;;text{for }tge sge0text{ and }(x,B)inmathbb Rtimesmathcal B(mathbb R).$$
How can we show that the transition semigroup $(kappa_{s,:t}:tge sge0)$ of $X$ is time-homogeneous, i.e. that $kappa_{s,:t}$ only depends on $t-s$ for all $tge sge0$?
Clearly, we may observe that $$X^{(0,:x)}_h=x+int_0^hbleft(X^{(0,:x)}_rright):{rm d}r+int_0^hsigmaleft(X^{(0,:x)}_rright):{rm d}W_rtag5$$ for all $hge0$ almost surely for all $xinmathbb R$, while $$X^{(s,:x)}_{s+h}=x+int_0^hbleft(X^{(s,:x)}_{s+r}right):{rm d}r+int_0^hsigmaleft(X^{(s,:x)}_{s+r}right):{rm d}W_r^{(s)}tag6$$ for all $hge0$ almost surely for all $sge0$ and $xinmathbb R$.
My guess is that $(5)$ and $(6)$ imply that $left(X^{(0,:x)}_hright)_{hge0}$ and $left(X^{(s,:x)}_{s+h}right)_{hge0}$ have the same distribution (since $W$ and $W^{(s)}$ have the same distribution) for all $sge0$ and $xinmathbb R$. How can we show that rigorously and how can we conclude from that?
stochastic-processes markov-process stochastic-integrals stochastic-analysis sde
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Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space and $$mathcal N:=left{Ninmathcal A:operatorname P[N]=0right}$$
$(W_t)_{tge0}$ be a Brownian motion on $(Omega,mathcal A,operatorname P)$ and $$W^{(s)}_t:=W_{s+t}-W_s;;;text{for }tge0$$ for $sge0$
$b,sigma:[0,infty)timesmathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le K(1+|x|^2);;;text{for all }xinmathbb Rtag1$$ and $$|b(x)-b(y)|^2+|sigma(x)-sigma(y)|^2le K|x-y|^2;;;text{for all }x,yinmathbb Rtag2$$ for some $Kge0$
We know that for all $sge0$ there is a real-valued continuous process $left(X^{(s,:x)}_tright)_{(t,:x)in[s,:infty)timesmathbb R}$ on $(Omega,mathcal A,operatorname P)$ with $$X^{(s,:x)}_t=x+int_s^tbleft(X^{(s,:x)}_rright):{rm d}r+int_s^tsigmaleft(X^{(s,:x)}_rright):{rm d}W_rtag3$$ for all $tge s$ almost surely for all $xinmathbb R$.
Now, let $xi$ be a real-valued random variable on $(Omega,mathcal A,operatorname P)$ independent of $W$ and $$mathcal F_t:=sigma(xi)veesigma(W_s:sin[0,t])veemathcal N;;;text{for }tge0.$$ Now, there is a real-valued $mathcal F$-adapted continuous process $(X_t)_{tge0}$ with $$X_t=xi+int_0^tb(X_s):{rm d}s+int_0^tsigma(X_s):{rm d}W_s;;;text{for }tge0.$$ $X$ is a $mathcal F$-Markov process with $$operatorname Pleft[X_tin Bmidmathcal F_sright]=kappa_{s,:t}(X_s,B);;;text{almost surely for all }tge0text{ and }Binmathcal B(mathbb R),tag4$$ where $$kappa_{s,:t}(x,B):=operatorname Pleft[X^{(s,:x)}_tin Bright];;;text{for }tge sge0text{ and }(x,B)inmathbb Rtimesmathcal B(mathbb R).$$
How can we show that the transition semigroup $(kappa_{s,:t}:tge sge0)$ of $X$ is time-homogeneous, i.e. that $kappa_{s,:t}$ only depends on $t-s$ for all $tge sge0$?
Clearly, we may observe that $$X^{(0,:x)}_h=x+int_0^hbleft(X^{(0,:x)}_rright):{rm d}r+int_0^hsigmaleft(X^{(0,:x)}_rright):{rm d}W_rtag5$$ for all $hge0$ almost surely for all $xinmathbb R$, while $$X^{(s,:x)}_{s+h}=x+int_0^hbleft(X^{(s,:x)}_{s+r}right):{rm d}r+int_0^hsigmaleft(X^{(s,:x)}_{s+r}right):{rm d}W_r^{(s)}tag6$$ for all $hge0$ almost surely for all $sge0$ and $xinmathbb R$.
My guess is that $(5)$ and $(6)$ imply that $left(X^{(0,:x)}_hright)_{hge0}$ and $left(X^{(s,:x)}_{s+h}right)_{hge0}$ have the same distribution (since $W$ and $W^{(s)}$ have the same distribution) for all $sge0$ and $xinmathbb R$. How can we show that rigorously and how can we conclude from that?
stochastic-processes markov-process stochastic-integrals stochastic-analysis sde
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up vote
0
down vote
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up vote
0
down vote
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Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space and $$mathcal N:=left{Ninmathcal A:operatorname P[N]=0right}$$
$(W_t)_{tge0}$ be a Brownian motion on $(Omega,mathcal A,operatorname P)$ and $$W^{(s)}_t:=W_{s+t}-W_s;;;text{for }tge0$$ for $sge0$
$b,sigma:[0,infty)timesmathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le K(1+|x|^2);;;text{for all }xinmathbb Rtag1$$ and $$|b(x)-b(y)|^2+|sigma(x)-sigma(y)|^2le K|x-y|^2;;;text{for all }x,yinmathbb Rtag2$$ for some $Kge0$
We know that for all $sge0$ there is a real-valued continuous process $left(X^{(s,:x)}_tright)_{(t,:x)in[s,:infty)timesmathbb R}$ on $(Omega,mathcal A,operatorname P)$ with $$X^{(s,:x)}_t=x+int_s^tbleft(X^{(s,:x)}_rright):{rm d}r+int_s^tsigmaleft(X^{(s,:x)}_rright):{rm d}W_rtag3$$ for all $tge s$ almost surely for all $xinmathbb R$.
Now, let $xi$ be a real-valued random variable on $(Omega,mathcal A,operatorname P)$ independent of $W$ and $$mathcal F_t:=sigma(xi)veesigma(W_s:sin[0,t])veemathcal N;;;text{for }tge0.$$ Now, there is a real-valued $mathcal F$-adapted continuous process $(X_t)_{tge0}$ with $$X_t=xi+int_0^tb(X_s):{rm d}s+int_0^tsigma(X_s):{rm d}W_s;;;text{for }tge0.$$ $X$ is a $mathcal F$-Markov process with $$operatorname Pleft[X_tin Bmidmathcal F_sright]=kappa_{s,:t}(X_s,B);;;text{almost surely for all }tge0text{ and }Binmathcal B(mathbb R),tag4$$ where $$kappa_{s,:t}(x,B):=operatorname Pleft[X^{(s,:x)}_tin Bright];;;text{for }tge sge0text{ and }(x,B)inmathbb Rtimesmathcal B(mathbb R).$$
How can we show that the transition semigroup $(kappa_{s,:t}:tge sge0)$ of $X$ is time-homogeneous, i.e. that $kappa_{s,:t}$ only depends on $t-s$ for all $tge sge0$?
Clearly, we may observe that $$X^{(0,:x)}_h=x+int_0^hbleft(X^{(0,:x)}_rright):{rm d}r+int_0^hsigmaleft(X^{(0,:x)}_rright):{rm d}W_rtag5$$ for all $hge0$ almost surely for all $xinmathbb R$, while $$X^{(s,:x)}_{s+h}=x+int_0^hbleft(X^{(s,:x)}_{s+r}right):{rm d}r+int_0^hsigmaleft(X^{(s,:x)}_{s+r}right):{rm d}W_r^{(s)}tag6$$ for all $hge0$ almost surely for all $sge0$ and $xinmathbb R$.
My guess is that $(5)$ and $(6)$ imply that $left(X^{(0,:x)}_hright)_{hge0}$ and $left(X^{(s,:x)}_{s+h}right)_{hge0}$ have the same distribution (since $W$ and $W^{(s)}$ have the same distribution) for all $sge0$ and $xinmathbb R$. How can we show that rigorously and how can we conclude from that?
stochastic-processes markov-process stochastic-integrals stochastic-analysis sde
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space and $$mathcal N:=left{Ninmathcal A:operatorname P[N]=0right}$$
$(W_t)_{tge0}$ be a Brownian motion on $(Omega,mathcal A,operatorname P)$ and $$W^{(s)}_t:=W_{s+t}-W_s;;;text{for }tge0$$ for $sge0$
$b,sigma:[0,infty)timesmathbb Rtomathbb R$ be Borel measurable with $$|b(t,x)|^2+|sigma(t,x)|^2le K(1+|x|^2);;;text{for all }xinmathbb Rtag1$$ and $$|b(x)-b(y)|^2+|sigma(x)-sigma(y)|^2le K|x-y|^2;;;text{for all }x,yinmathbb Rtag2$$ for some $Kge0$
We know that for all $sge0$ there is a real-valued continuous process $left(X^{(s,:x)}_tright)_{(t,:x)in[s,:infty)timesmathbb R}$ on $(Omega,mathcal A,operatorname P)$ with $$X^{(s,:x)}_t=x+int_s^tbleft(X^{(s,:x)}_rright):{rm d}r+int_s^tsigmaleft(X^{(s,:x)}_rright):{rm d}W_rtag3$$ for all $tge s$ almost surely for all $xinmathbb R$.
Now, let $xi$ be a real-valued random variable on $(Omega,mathcal A,operatorname P)$ independent of $W$ and $$mathcal F_t:=sigma(xi)veesigma(W_s:sin[0,t])veemathcal N;;;text{for }tge0.$$ Now, there is a real-valued $mathcal F$-adapted continuous process $(X_t)_{tge0}$ with $$X_t=xi+int_0^tb(X_s):{rm d}s+int_0^tsigma(X_s):{rm d}W_s;;;text{for }tge0.$$ $X$ is a $mathcal F$-Markov process with $$operatorname Pleft[X_tin Bmidmathcal F_sright]=kappa_{s,:t}(X_s,B);;;text{almost surely for all }tge0text{ and }Binmathcal B(mathbb R),tag4$$ where $$kappa_{s,:t}(x,B):=operatorname Pleft[X^{(s,:x)}_tin Bright];;;text{for }tge sge0text{ and }(x,B)inmathbb Rtimesmathcal B(mathbb R).$$
How can we show that the transition semigroup $(kappa_{s,:t}:tge sge0)$ of $X$ is time-homogeneous, i.e. that $kappa_{s,:t}$ only depends on $t-s$ for all $tge sge0$?
Clearly, we may observe that $$X^{(0,:x)}_h=x+int_0^hbleft(X^{(0,:x)}_rright):{rm d}r+int_0^hsigmaleft(X^{(0,:x)}_rright):{rm d}W_rtag5$$ for all $hge0$ almost surely for all $xinmathbb R$, while $$X^{(s,:x)}_{s+h}=x+int_0^hbleft(X^{(s,:x)}_{s+r}right):{rm d}r+int_0^hsigmaleft(X^{(s,:x)}_{s+r}right):{rm d}W_r^{(s)}tag6$$ for all $hge0$ almost surely for all $sge0$ and $xinmathbb R$.
My guess is that $(5)$ and $(6)$ imply that $left(X^{(0,:x)}_hright)_{hge0}$ and $left(X^{(s,:x)}_{s+h}right)_{hge0}$ have the same distribution (since $W$ and $W^{(s)}$ have the same distribution) for all $sge0$ and $xinmathbb R$. How can we show that rigorously and how can we conclude from that?
stochastic-processes markov-process stochastic-integrals stochastic-analysis sde
stochastic-processes markov-process stochastic-integrals stochastic-analysis sde
asked Nov 18 at 23:53
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