Convergence in total variation distance of Markov kernel $n$-fold composition to the stationary measure
$begingroup$
Let
$(E,mathcal E)$ be a measurable space
$mu$ be a measure on $(E,mathcal E)$
$p:Eto(0,infty)$ with $$int p:{rm d}mu=1$$ and $pi$ denote the measure with density $p$ with respect to $mu$
$kappa$ be a Markov kernel on $(E,mathcal E)$
Assume $kappa$ is reversible with respect to $pi$, i.e. $$intpi({rm d}x)intkappa(x,{rm d}y)f(x,y)=intpi({rm d}y)intkappa(y,{rm d}x)f(x,y)tag1$$ for all bounded $mathcal Eotimesmathcal E$-measurable $f:Etimes Etomathbb R$. From $(1)$, we're able to conclude that $pi$ is invariant with respect to $kappa$, i.e. $$pikappa=pitag2,$$ where the left-hand side denotes the composition of $pi$ and $kappa$.
I want to show that $$left|kappa^n(x,;cdot;)-piright|xrightarrow{ntoinfty}0;;;text{for }pitext{-almost all }xin Etag3,$$ where the left-hand side denotes the total variation distance of $kappa^n(x,;cdot;)$ and $pi$ and $kappa^n$ denotes the $n$-fold composition of $kappa$.
As usual, $$kappa f:=intkappa(;cdot;,{rm d}y)f(y)$$ for all $mathcal E$-measurable $f:Etomathbb R$ such that the integral is well-defined. In addition, let $$hatkappa f:=intkappa({rm d}x,;cdot;)f(x).$$
Now, suppose that we're able to show $$left|hatkappa^n f-pright|_{L^1(mu)}xrightarrow{ntoinfty}0tag4$$ for all $mathcal E$-measurable $f:Eto[0,infty)$ with $$int f:{rm d}mu=1tag5.$$ Are we able to conclude $(3)$?
Clearly, the left-hand side in $(4)$ is equal to $$2left|(hatkappa^n f)mu-pmuright|,$$ where $(hatkappa^n f)mu$ and $pmu$ denote the measures with density $hatkappa^n f$ and $p$ with respect to $mu$, respectively.
probability-theory measure-theory markov-chains markov-process monte-carlo
asked Dec 31 '18 at 12:19
0xbadf00d0xbadf00d
1,88741533
$endgroup$
add a comment |
$begingroup$
Let
$(E,mathcal E)$ be a measurable space
$mu$ be a measure on $(E,mathcal E)$
$p:Eto(0,infty)$ with $$int p:{rm d}mu=1$$ and $pi$ denote the measure with density $p$ with respect to $mu$
$kappa$ be a Markov kernel on $(E,mathcal E)$
Assume $kappa$ is reversible with respect to $pi$, i.e. $$intpi({rm d}x)intkappa(x,{rm d}y)f(x,y)=intpi({rm d}y)intkappa(y,{rm d}x)f(x,y)tag1$$ for all bounded $mathcal Eotimesmathcal E$-measurable $f:Etimes Etomathbb R$. From $(1)$, we're able to conclude that $pi$ is invariant with respect to $kappa$, i.e. $$pikappa=pitag2,$$ where the left-hand side denotes the composition of $pi$ and $kappa$.
I want to show that $$left|kappa^n(x,;cdot;)-piright|xrightarrow{ntoinfty}0;;;text{for }pitext{-almost all }xin Etag3,$$ where the left-hand side denotes the total variation distance of $kappa^n(x,;cdot;)$ and $pi$ and $kappa^n$ denotes the $n$-fold composition of $kappa$.
As usual, $$kappa f:=intkappa(;cdot;,{rm d}y)f(y)$$ for all $mathcal E$-measurable $f:Etomathbb R$ such that the integral is well-defined. In addition, let $$hatkappa f:=intkappa({rm d}x,;cdot;)f(x).$$
Now, suppose that we're able to show $$left|hatkappa^n f-pright|_{L^1(mu)}xrightarrow{ntoinfty}0tag4$$ for all $mathcal E$-measurable $f:Eto[0,infty)$ with $$int f:{rm d}mu=1tag5.$$ Are we able to conclude $(3)$?
Clearly, the left-hand side in $(4)$ is equal to $$2left|(hatkappa^n f)mu-pmuright|,$$ where $(hatkappa^n f)mu$ and $pmu$ denote the measures with density $hatkappa^n f$ and $p$ with respect to $mu$, respectively.
probability-theory measure-theory markov-chains markov-process monte-carlo
asked Dec 31 '18 at 12:19
0xbadf00d0xbadf00d
1,88741533
$endgroup$
add a comment |
$begingroup$
Let
$(E,mathcal E)$ be a measurable space
$mu$ be a measure on $(E,mathcal E)$
$p:Eto(0,infty)$ with $$int p:{rm d}mu=1$$ and $pi$ denote the measure with density $p$ with respect to $mu$
$kappa$ be a Markov kernel on $(E,mathcal E)$
Assume $kappa$ is reversible with respect to $pi$, i.e. $$intpi({rm d}x)intkappa(x,{rm d}y)f(x,y)=intpi({rm d}y)intkappa(y,{rm d}x)f(x,y)tag1$$ for all bounded $mathcal Eotimesmathcal E$-measurable $f:Etimes Etomathbb R$. From $(1)$, we're able to conclude that $pi$ is invariant with respect to $kappa$, i.e. $$pikappa=pitag2,$$ where the left-hand side denotes the composition of $pi$ and $kappa$.
I want to show that $$left|kappa^n(x,;cdot;)-piright|xrightarrow{ntoinfty}0;;;text{for }pitext{-almost all }xin Etag3,$$ where the left-hand side denotes the total variation distance of $kappa^n(x,;cdot;)$ and $pi$ and $kappa^n$ denotes the $n$-fold composition of $kappa$.
As usual, $$kappa f:=intkappa(;cdot;,{rm d}y)f(y)$$ for all $mathcal E$-measurable $f:Etomathbb R$ such that the integral is well-defined. In addition, let $$hatkappa f:=intkappa({rm d}x,;cdot;)f(x).$$
Now, suppose that we're able to show $$left|hatkappa^n f-pright|_{L^1(mu)}xrightarrow{ntoinfty}0tag4$$ for all $mathcal E$-measurable $f:Eto[0,infty)$ with $$int f:{rm d}mu=1tag5.$$ Are we able to conclude $(3)$?
Clearly, the left-hand side in $(4)$ is equal to $$2left|(hatkappa^n f)mu-pmuright|,$$ where $(hatkappa^n f)mu$ and $pmu$ denote the measures with density $hatkappa^n f$ and $p$ with respect to $mu$, respectively.
probability-theory measure-theory markov-chains markov-process monte-carlo
asked Dec 31 '18 at 12:19
0xbadf00d0xbadf00d
1,88741533
$endgroup$
Let
$(E,mathcal E)$ be a measurable space
$mu$ be a measure on $(E,mathcal E)$
$p:Eto(0,infty)$ with $$int p:{rm d}mu=1$$ and $pi$ denote the measure with density $p$ with respect to $mu$
$kappa$ be a Markov kernel on $(E,mathcal E)$
Assume $kappa$ is reversible with respect to $pi$, i.e. $$intpi({rm d}x)intkappa(x,{rm d}y)f(x,y)=intpi({rm d}y)intkappa(y,{rm d}x)f(x,y)tag1$$ for all bounded $mathcal Eotimesmathcal E$-measurable $f:Etimes Etomathbb R$. From $(1)$, we're able to conclude that $pi$ is invariant with respect to $kappa$, i.e. $$pikappa=pitag2,$$ where the left-hand side denotes the composition of $pi$ and $kappa$.
I want to show that $$left|kappa^n(x,;cdot;)-piright|xrightarrow{ntoinfty}0;;;text{for }pitext{-almost all }xin Etag3,$$ where the left-hand side denotes the total variation distance of $kappa^n(x,;cdot;)$ and $pi$ and $kappa^n$ denotes the $n$-fold composition of $kappa$.
As usual, $$kappa f:=intkappa(;cdot;,{rm d}y)f(y)$$ for all $mathcal E$-measurable $f:Etomathbb R$ such that the integral is well-defined. In addition, let $$hatkappa f:=intkappa({rm d}x,;cdot;)f(x).$$
Now, suppose that we're able to show $$left|hatkappa^n f-pright|_{L^1(mu)}xrightarrow{ntoinfty}0tag4$$ for all $mathcal E$-measurable $f:Eto[0,infty)$ with $$int f:{rm d}mu=1tag5.$$ Are we able to conclude $(3)$?
Clearly, the left-hand side in $(4)$ is equal to $$2left|(hatkappa^n f)mu-pmuright|,$$ where $(hatkappa^n f)mu$ and $pmu$ denote the measures with density $hatkappa^n f$ and $p$ with respect to $mu$, respectively.
probability-theory measure-theory markov-chains markov-process monte-carlo
probability-theory measure-theory markov-chains markov-process monte-carlo
asked Dec 31 '18 at 12:19
0xbadf00d0xbadf00d
1,88741533
asked Dec 31 '18 at 12:19
0xbadf00d0xbadf00d
1,88741533
asked Dec 31 '18 at 12:19
0xbadf00d0xbadf00d
1,88741533
asked Dec 31 '18 at 12:19
0xbadf00d0xbadf00d
1,88741533
asked Dec 31 '18 at 12:19
0xbadf00d0xbadf00d
1,88741533
1,88741533
add a comment |
add a comment |
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