Mixing
If $X$ is strongly Markovian at $tau$ is $X_{tau+;cdot;}$ a Markov process?
0
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$I=[0,infty)$ or $I=mathbb N_0$
$(mathcal F_t)_{tin I}$ be a filtration on $(Omega,mathcal A)$
$(E,mathcal E)$ be a measurable space
$delta_x$ denote the Dirac measure on $(E,mathcal E)$ at $xin E$
$(X_t)_{tin I}$ be an $(E,mathcal E)$-valued $mathcal F$-Markov process on $(Omega,mathcal A,operatorname P)$ with transition semigroup $(kappa_t)_{tin I}$
$kappa$ be a Markov kernel with source $(E,mathcal E)$ and target $(E^I,mathcal E^{otimes I})$ with $$kappa(x,left{yin E^I:(y_{t_0},ldots,y_{t_n})in Bright})=left(delta_xotimesbigotimes_{i=1}^nkappa_{t_i-t_{i-1}}right)(B)$$ for all $ninmathbb N$, $Binmathcal E^{n+1}$ and $t_0,ldots,t_nin I$ with $0=t_0<ldots<t_n$
$tau$ be an $mathcal F$-stopping time
We say that $X$ is strongly $mathcal F$-Markovian at $tau$ if $$1_{left{:tau:<:infty:right}}operatorname Eleft[fleft(left(X_{tau+t}right)_{tin I}right)midmathcal F_tauright]=1_{left{:tau:<:infty:right}}(kappa f)(X_tau);;;text{almost surely}tag1$$ for all bounded and $mathcal E^{otimes I}$-measurable $f:E^Itomathbb R$ (As usual, $kappa f:=intkappa(;cdot;,{rm d}y)f(y)$).
What can we conclude from $(1)$? In particular,
- Does $(1)$ imply that $left(X_{tau+t}right)_{tin I}$ is again a Markov process? If so, with respect to which filtration and how do we prove it?
At Wikipedia, they say that $(1)$ implies that "conditioned on $left{tau<inftyright}$, $X_{tau+t}$ is independent of $$mathcal F_{tau^+}:=left{Ainmathcal A:left{tau=tright}cap Ainmathcal F_{t+}:=bigcap_{substack{sin I\s>t}}mathcal F_stext{ for all }tin Iright}$$ given $X_tau$". What does that mean? Maybe $$operatorname Pleft[Acap Bmidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]=operatorname Pleft[Amidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]operatorname Pleft[Bmidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]tag2$$ almost surely for all $Ainsigma(X_{tau+t})$ and $Binsigma(X_tau)$, where $left.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}$ denotes the trace $sigma$-algebra?
probability-theory stochastic-processes markov-chains markov-process
asked Jan 9 at 16:25
0xbadf00d0xbadf00d
1,89841531
$endgroup$
add a comment |
0
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$I=[0,infty)$ or $I=mathbb N_0$
$(mathcal F_t)_{tin I}$ be a filtration on $(Omega,mathcal A)$
$(E,mathcal E)$ be a measurable space
$delta_x$ denote the Dirac measure on $(E,mathcal E)$ at $xin E$
$(X_t)_{tin I}$ be an $(E,mathcal E)$-valued $mathcal F$-Markov process on $(Omega,mathcal A,operatorname P)$ with transition semigroup $(kappa_t)_{tin I}$
$kappa$ be a Markov kernel with source $(E,mathcal E)$ and target $(E^I,mathcal E^{otimes I})$ with $$kappa(x,left{yin E^I:(y_{t_0},ldots,y_{t_n})in Bright})=left(delta_xotimesbigotimes_{i=1}^nkappa_{t_i-t_{i-1}}right)(B)$$ for all $ninmathbb N$, $Binmathcal E^{n+1}$ and $t_0,ldots,t_nin I$ with $0=t_0<ldots<t_n$
$tau$ be an $mathcal F$-stopping time
We say that $X$ is strongly $mathcal F$-Markovian at $tau$ if $$1_{left{:tau:<:infty:right}}operatorname Eleft[fleft(left(X_{tau+t}right)_{tin I}right)midmathcal F_tauright]=1_{left{:tau:<:infty:right}}(kappa f)(X_tau);;;text{almost surely}tag1$$ for all bounded and $mathcal E^{otimes I}$-measurable $f:E^Itomathbb R$ (As usual, $kappa f:=intkappa(;cdot;,{rm d}y)f(y)$).
What can we conclude from $(1)$? In particular,
- Does $(1)$ imply that $left(X_{tau+t}right)_{tin I}$ is again a Markov process? If so, with respect to which filtration and how do we prove it?
At Wikipedia, they say that $(1)$ implies that "conditioned on $left{tau<inftyright}$, $X_{tau+t}$ is independent of $$mathcal F_{tau^+}:=left{Ainmathcal A:left{tau=tright}cap Ainmathcal F_{t+}:=bigcap_{substack{sin I\s>t}}mathcal F_stext{ for all }tin Iright}$$ given $X_tau$". What does that mean? Maybe $$operatorname Pleft[Acap Bmidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]=operatorname Pleft[Amidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]operatorname Pleft[Bmidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]tag2$$ almost surely for all $Ainsigma(X_{tau+t})$ and $Binsigma(X_tau)$, where $left.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}$ denotes the trace $sigma$-algebra?
probability-theory stochastic-processes markov-chains markov-process
asked Jan 9 at 16:25
0xbadf00d0xbadf00d
1,89841531
$endgroup$
add a comment |
0
0
0
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$I=[0,infty)$ or $I=mathbb N_0$
$(mathcal F_t)_{tin I}$ be a filtration on $(Omega,mathcal A)$
$(E,mathcal E)$ be a measurable space
$delta_x$ denote the Dirac measure on $(E,mathcal E)$ at $xin E$
$(X_t)_{tin I}$ be an $(E,mathcal E)$-valued $mathcal F$-Markov process on $(Omega,mathcal A,operatorname P)$ with transition semigroup $(kappa_t)_{tin I}$
$kappa$ be a Markov kernel with source $(E,mathcal E)$ and target $(E^I,mathcal E^{otimes I})$ with $$kappa(x,left{yin E^I:(y_{t_0},ldots,y_{t_n})in Bright})=left(delta_xotimesbigotimes_{i=1}^nkappa_{t_i-t_{i-1}}right)(B)$$ for all $ninmathbb N$, $Binmathcal E^{n+1}$ and $t_0,ldots,t_nin I$ with $0=t_0<ldots<t_n$
$tau$ be an $mathcal F$-stopping time
We say that $X$ is strongly $mathcal F$-Markovian at $tau$ if $$1_{left{:tau:<:infty:right}}operatorname Eleft[fleft(left(X_{tau+t}right)_{tin I}right)midmathcal F_tauright]=1_{left{:tau:<:infty:right}}(kappa f)(X_tau);;;text{almost surely}tag1$$ for all bounded and $mathcal E^{otimes I}$-measurable $f:E^Itomathbb R$ (As usual, $kappa f:=intkappa(;cdot;,{rm d}y)f(y)$).
What can we conclude from $(1)$? In particular,
- Does $(1)$ imply that $left(X_{tau+t}right)_{tin I}$ is again a Markov process? If so, with respect to which filtration and how do we prove it?
At Wikipedia, they say that $(1)$ implies that "conditioned on $left{tau<inftyright}$, $X_{tau+t}$ is independent of $$mathcal F_{tau^+}:=left{Ainmathcal A:left{tau=tright}cap Ainmathcal F_{t+}:=bigcap_{substack{sin I\s>t}}mathcal F_stext{ for all }tin Iright}$$ given $X_tau$". What does that mean? Maybe $$operatorname Pleft[Acap Bmidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]=operatorname Pleft[Amidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]operatorname Pleft[Bmidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]tag2$$ almost surely for all $Ainsigma(X_{tau+t})$ and $Binsigma(X_tau)$, where $left.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}$ denotes the trace $sigma$-algebra?
probability-theory stochastic-processes markov-chains markov-process
asked Jan 9 at 16:25
0xbadf00d0xbadf00d
1,89841531
$endgroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$I=[0,infty)$ or $I=mathbb N_0$
$(mathcal F_t)_{tin I}$ be a filtration on $(Omega,mathcal A)$
$(E,mathcal E)$ be a measurable space
$delta_x$ denote the Dirac measure on $(E,mathcal E)$ at $xin E$
$(X_t)_{tin I}$ be an $(E,mathcal E)$-valued $mathcal F$-Markov process on $(Omega,mathcal A,operatorname P)$ with transition semigroup $(kappa_t)_{tin I}$
$kappa$ be a Markov kernel with source $(E,mathcal E)$ and target $(E^I,mathcal E^{otimes I})$ with $$kappa(x,left{yin E^I:(y_{t_0},ldots,y_{t_n})in Bright})=left(delta_xotimesbigotimes_{i=1}^nkappa_{t_i-t_{i-1}}right)(B)$$ for all $ninmathbb N$, $Binmathcal E^{n+1}$ and $t_0,ldots,t_nin I$ with $0=t_0<ldots<t_n$
$tau$ be an $mathcal F$-stopping time
We say that $X$ is strongly $mathcal F$-Markovian at $tau$ if $$1_{left{:tau:<:infty:right}}operatorname Eleft[fleft(left(X_{tau+t}right)_{tin I}right)midmathcal F_tauright]=1_{left{:tau:<:infty:right}}(kappa f)(X_tau);;;text{almost surely}tag1$$ for all bounded and $mathcal E^{otimes I}$-measurable $f:E^Itomathbb R$ (As usual, $kappa f:=intkappa(;cdot;,{rm d}y)f(y)$).
What can we conclude from $(1)$? In particular,
- Does $(1)$ imply that $left(X_{tau+t}right)_{tin I}$ is again a Markov process? If so, with respect to which filtration and how do we prove it?
At Wikipedia, they say that $(1)$ implies that "conditioned on $left{tau<inftyright}$, $X_{tau+t}$ is independent of $$mathcal F_{tau^+}:=left{Ainmathcal A:left{tau=tright}cap Ainmathcal F_{t+}:=bigcap_{substack{sin I\s>t}}mathcal F_stext{ for all }tin Iright}$$ given $X_tau$". What does that mean? Maybe $$operatorname Pleft[Acap Bmidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]=operatorname Pleft[Amidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]operatorname Pleft[Bmidleft.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}right]tag2$$ almost surely for all $Ainsigma(X_{tau+t})$ and $Binsigma(X_tau)$, where $left.mathcal F_{tau+}right|_{left{:tau:<:infty:right}}$ denotes the trace $sigma$-algebra?
probability-theory stochastic-processes markov-chains markov-process
probability-theory stochastic-processes markov-chains markov-process
asked Jan 9 at 16:25
0xbadf00d0xbadf00d
1,89841531
asked Jan 9 at 16:25
0xbadf00d0xbadf00d
1,89841531
asked Jan 9 at 16:25
0xbadf00d0xbadf00d
1,89841531
asked Jan 9 at 16:25
0xbadf00d0xbadf00d
1,89841531
asked Jan 9 at 16:25
0xbadf00d0xbadf00d
1,89841531
1,89841531
add a comment |
add a comment |
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