If $(X_t)_{tge0}$ is a Markov process with invariant measure $mu$, does the distribution of $X_t$ weakly...
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$(X_t)_{tge0}$ be a time-homogeneous Markov process on $(Omega,mathcal A,operatorname P)$ and $kappa_t$ denote a regular version of the conditional distribution of $X_t$ given $X_0$ for $tge0$, i.e. $$kappa_t(x,B)=operatorname Pleft[X_tin Bmid X_0=xright];;;text{for all }(x,B)inmathbb Rtimesmathcal B(mathbb R)tag1$$
$C_0(mathbb R)$ denote the space of continuous functions on $mathbb R$ vanishing at infinity
$mu$ be a probability measure on $(mathbb R,mathcal B(mathbb R))$
As usual, let $$kappa_tf:=intkappa_t(x,{rm d}y)f(y)$$ for bounded Borel measurable $f:mathbb Rtomathbb R$ and $tge0$.
Assume $$kappa_tC_0(mathbb R)subseteq C_0(mathbb R);;;text{for all }tge0.tag2$$
Let $$mu_t(B):=operatorname Pleft[X_tin Bright];;;text{for }Binmathcal B(mathbb R)$$ for $tge0$. It's easy to see that weak convergence of $mu_t$ to $mu$ as $ttoinfty$ implies that $mu$ is invariant with respect to $(kappa_t)_{tge0}$, i.e. $$mukappa_t=mu;;;text{for all }tge0,$$ where $mukappa_t$ denotes the composition of $mu$ and $kappa_t$.
Does the converse hold as well, i.e. if $mu$ is invariant with respect to $(kappa_t)_{tge0}$, are we able to conclude that $mu_t$ converges weakly to $mu$ as $ttoinfty$?
probability-theory measure-theory stochastic-processes markov-process stochastic-analysis
asked Dec 28 '18 at 0:45
0xbadf00d
1,75441430
add a comment |
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$(X_t)_{tge0}$ be a time-homogeneous Markov process on $(Omega,mathcal A,operatorname P)$ and $kappa_t$ denote a regular version of the conditional distribution of $X_t$ given $X_0$ for $tge0$, i.e. $$kappa_t(x,B)=operatorname Pleft[X_tin Bmid X_0=xright];;;text{for all }(x,B)inmathbb Rtimesmathcal B(mathbb R)tag1$$
$C_0(mathbb R)$ denote the space of continuous functions on $mathbb R$ vanishing at infinity
$mu$ be a probability measure on $(mathbb R,mathcal B(mathbb R))$
As usual, let $$kappa_tf:=intkappa_t(x,{rm d}y)f(y)$$ for bounded Borel measurable $f:mathbb Rtomathbb R$ and $tge0$.
Assume $$kappa_tC_0(mathbb R)subseteq C_0(mathbb R);;;text{for all }tge0.tag2$$
Let $$mu_t(B):=operatorname Pleft[X_tin Bright];;;text{for }Binmathcal B(mathbb R)$$ for $tge0$. It's easy to see that weak convergence of $mu_t$ to $mu$ as $ttoinfty$ implies that $mu$ is invariant with respect to $(kappa_t)_{tge0}$, i.e. $$mukappa_t=mu;;;text{for all }tge0,$$ where $mukappa_t$ denotes the composition of $mu$ and $kappa_t$.
Does the converse hold as well, i.e. if $mu$ is invariant with respect to $(kappa_t)_{tge0}$, are we able to conclude that $mu_t$ converges weakly to $mu$ as $ttoinfty$?
probability-theory measure-theory stochastic-processes markov-process stochastic-analysis
asked Dec 28 '18 at 0:45
0xbadf00d
1,75441430
What is the relation between $mu_t$, $kappa_t$ and $mu$? Is it something like $$mu_t (B) = int kappa_t(x,B) mu(, d x) = P^{mu}(X_t in B) ?$$ Because otherwise $mu_t to mu$ weakly will not imply $mu kappa_t = mu$.
– Sayantan
Dec 28 '18 at 1:43
@Sayantan We've got $mu_t=mu_0otimes kappa_t$ (product of $mu_0$ and $kappa_t$).
– 0xbadf00d
Dec 28 '18 at 10:35
add a comment |
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$(X_t)_{tge0}$ be a time-homogeneous Markov process on $(Omega,mathcal A,operatorname P)$ and $kappa_t$ denote a regular version of the conditional distribution of $X_t$ given $X_0$ for $tge0$, i.e. $$kappa_t(x,B)=operatorname Pleft[X_tin Bmid X_0=xright];;;text{for all }(x,B)inmathbb Rtimesmathcal B(mathbb R)tag1$$
$C_0(mathbb R)$ denote the space of continuous functions on $mathbb R$ vanishing at infinity
$mu$ be a probability measure on $(mathbb R,mathcal B(mathbb R))$
As usual, let $$kappa_tf:=intkappa_t(x,{rm d}y)f(y)$$ for bounded Borel measurable $f:mathbb Rtomathbb R$ and $tge0$.
Assume $$kappa_tC_0(mathbb R)subseteq C_0(mathbb R);;;text{for all }tge0.tag2$$
Let $$mu_t(B):=operatorname Pleft[X_tin Bright];;;text{for }Binmathcal B(mathbb R)$$ for $tge0$. It's easy to see that weak convergence of $mu_t$ to $mu$ as $ttoinfty$ implies that $mu$ is invariant with respect to $(kappa_t)_{tge0}$, i.e. $$mukappa_t=mu;;;text{for all }tge0,$$ where $mukappa_t$ denotes the composition of $mu$ and $kappa_t$.
Does the converse hold as well, i.e. if $mu$ is invariant with respect to $(kappa_t)_{tge0}$, are we able to conclude that $mu_t$ converges weakly to $mu$ as $ttoinfty$?
probability-theory measure-theory stochastic-processes markov-process stochastic-analysis
asked Dec 28 '18 at 0:45
0xbadf00d
1,75441430
Let
$(Omega,mathcal A,operatorname P)$ be a complete probability space
$(X_t)_{tge0}$ be a time-homogeneous Markov process on $(Omega,mathcal A,operatorname P)$ and $kappa_t$ denote a regular version of the conditional distribution of $X_t$ given $X_0$ for $tge0$, i.e. $$kappa_t(x,B)=operatorname Pleft[X_tin Bmid X_0=xright];;;text{for all }(x,B)inmathbb Rtimesmathcal B(mathbb R)tag1$$
$C_0(mathbb R)$ denote the space of continuous functions on $mathbb R$ vanishing at infinity
$mu$ be a probability measure on $(mathbb R,mathcal B(mathbb R))$
As usual, let $$kappa_tf:=intkappa_t(x,{rm d}y)f(y)$$ for bounded Borel measurable $f:mathbb Rtomathbb R$ and $tge0$.
Assume $$kappa_tC_0(mathbb R)subseteq C_0(mathbb R);;;text{for all }tge0.tag2$$
Let $$mu_t(B):=operatorname Pleft[X_tin Bright];;;text{for }Binmathcal B(mathbb R)$$ for $tge0$. It's easy to see that weak convergence of $mu_t$ to $mu$ as $ttoinfty$ implies that $mu$ is invariant with respect to $(kappa_t)_{tge0}$, i.e. $$mukappa_t=mu;;;text{for all }tge0,$$ where $mukappa_t$ denotes the composition of $mu$ and $kappa_t$.
Does the converse hold as well, i.e. if $mu$ is invariant with respect to $(kappa_t)_{tge0}$, are we able to conclude that $mu_t$ converges weakly to $mu$ as $ttoinfty$?
probability-theory measure-theory stochastic-processes markov-process stochastic-analysis
probability-theory measure-theory stochastic-processes markov-process stochastic-analysis
asked Dec 28 '18 at 0:45
0xbadf00d
1,75441430
asked Dec 28 '18 at 0:45
0xbadf00d
1,75441430
asked Dec 28 '18 at 0:45
0xbadf00d
1,75441430
asked Dec 28 '18 at 0:45
0xbadf00d
1,75441430
asked Dec 28 '18 at 0:45
0xbadf00d
1,75441430
1,75441430
What is the relation between $mu_t$, $kappa_t$ and $mu$? Is it something like $$mu_t (B) = int kappa_t(x,B) mu(, d x) = P^{mu}(X_t in B) ?$$ Because otherwise $mu_t to mu$ weakly will not imply $mu kappa_t = mu$.
– Sayantan
Dec 28 '18 at 1:43
@Sayantan We've got $mu_t=mu_0otimes kappa_t$ (product of $mu_0$ and $kappa_t$).
– 0xbadf00d
Dec 28 '18 at 10:35
add a comment |
What is the relation between $mu_t$, $kappa_t$ and $mu$? Is it something like $$mu_t (B) = int kappa_t(x,B) mu(, d x) = P^{mu}(X_t in B) ?$$ Because otherwise $mu_t to mu$ weakly will not imply $mu kappa_t = mu$.
– Sayantan
Dec 28 '18 at 1:43
@Sayantan We've got $mu_t=mu_0otimes kappa_t$ (product of $mu_0$ and $kappa_t$).
– 0xbadf00d
Dec 28 '18 at 10:35
What is the relation between $mu_t$, $kappa_t$ and $mu$? Is it something like $$mu_t (B) = int kappa_t(x,B) mu(, d x) = P^{mu}(X_t in B) ?$$ Because otherwise $mu_t to mu$ weakly will not imply $mu kappa_t = mu$.
– Sayantan
Dec 28 '18 at 1:43
What is the relation between $mu_t$, $kappa_t$ and $mu$? Is it something like $$mu_t (B) = int kappa_t(x,B) mu(, d x) = P^{mu}(X_t in B) ?$$ Because otherwise $mu_t to mu$ weakly will not imply $mu kappa_t = mu$.
– Sayantan
Dec 28 '18 at 1:43
@Sayantan We've got $mu_t=mu_0otimes kappa_t$ (product of $mu_0$ and $kappa_t$).
– 0xbadf00d
Dec 28 '18 at 10:35
@Sayantan We've got $mu_t=mu_0otimes kappa_t$ (product of $mu_0$ and $kappa_t$).
– 0xbadf00d
Dec 28 '18 at 10:35
add a comment |
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What is the relation between $mu_t$, $kappa_t$ and $mu$? Is it something like $$mu_t (B) = int kappa_t(x,B) mu(, d x) = P^{mu}(X_t in B) ?$$ Because otherwise $mu_t to mu$ weakly will not imply $mu kappa_t = mu$.
– Sayantan
Dec 28 '18 at 1:43
@Sayantan We've got $mu_t=mu_0otimes kappa_t$ (product of $mu_0$ and $kappa_t$).
– 0xbadf00d
Dec 28 '18 at 10:35