Reverse kernel of a Markov kernel with density
$begingroup$
Let
$(E_i,mathcal E_i)$ be a measurable space
$kappa$ be a Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$
Assume $kappa$ has a positive density with respect to a measure $mu$ on $(E_2,mathcal E_2)$, i.e. there is a $mathcal E_1otimesmathcal E_2$-measurable $f:E_1times E_2to(0,infty)$ with $$kappa(x,;cdot;)=f(x,;cdot;)mu;;;text{for all }xin E_1.$$ Now, let $nu$ be a probability measure on $(E_1,mathcal E_1)$ and $$overleftarrowkappa_nu(y,;cdot;):=frac1{c(y)}f(;cdot;,y)nu;;;text{for }yin E_2,$$ where $c(y):=intnu({rm d}x)f(x,y)$ (and we assume that $c(y)<infty$) for $yin E_2$.
How can we show that $overleftarrowkappa_nu$ is the reverse kernel of $kappa$ with respect to $nu$ (see Definition 2.1.2), i.e.$^1$ $$intnu({rm d}x)intkappa(x,{rm d}y)g(x,y)=intnukappa({rm d}y)intoverleftarrowkappa_nu(y,{rm d}x)g(x,y)tag1$$ for all bounded and $mathcal E_1otimesmathcal E_2$-measurable $g:E_1times E_2tomathbb R$?
Let $$pi:E_1times E_2to E_2times E_1;,;;;(x,y)mapsto(y,x).$$ It's easy to observe that the left-hand side of $(1)$ is equal to$^2$ $$int g:{rm d}(nuotimeskappa)tag2$$ and the right-hand side is equal to $$int gcircpi^{-1}:{rm d}(nukappaotimesoverleftarrowkappa_nu).tag3$$
Now, it's easy to see that$^3$ $$pi_ast(nuotimeskappa)=muotimesoverleftarrowkappa_nutag4$$ and $$nuotimeskappa=left(pi^{-1}right)_ast(muotimesoverleftarrowkappa_nu)tag5.$$
$(1)$ is claimed in the linked document below the Definition. Could it be the case that their definition of "reverse kernel" is broken? From a terminological point of view it would make more sense to me if $nukappa$ on the right-hand side of $(1)$ would be replaced by $nu$.
$^1$ $nukappa$ denotes the composition of $nu$ and $kappa$.
$^2$ $nuotimeskappa$ denotes the product of $nu$ and $kappa$.
$^3$ $pi_ast(nuotimeskappa)$ denotes the pushforward measure of $pi$ with respect to $nuotimeskappa$.
real-analysis probability-theory measure-theory markov-chains markov-process
asked Jan 11 at 22:55
0xbadf00d0xbadf00d
1,90441530
$endgroup$
add a comment |
$begingroup$
Let
$(E_i,mathcal E_i)$ be a measurable space
$kappa$ be a Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$
Assume $kappa$ has a positive density with respect to a measure $mu$ on $(E_2,mathcal E_2)$, i.e. there is a $mathcal E_1otimesmathcal E_2$-measurable $f:E_1times E_2to(0,infty)$ with $$kappa(x,;cdot;)=f(x,;cdot;)mu;;;text{for all }xin E_1.$$ Now, let $nu$ be a probability measure on $(E_1,mathcal E_1)$ and $$overleftarrowkappa_nu(y,;cdot;):=frac1{c(y)}f(;cdot;,y)nu;;;text{for }yin E_2,$$ where $c(y):=intnu({rm d}x)f(x,y)$ (and we assume that $c(y)<infty$) for $yin E_2$.
How can we show that $overleftarrowkappa_nu$ is the reverse kernel of $kappa$ with respect to $nu$ (see Definition 2.1.2), i.e.$^1$ $$intnu({rm d}x)intkappa(x,{rm d}y)g(x,y)=intnukappa({rm d}y)intoverleftarrowkappa_nu(y,{rm d}x)g(x,y)tag1$$ for all bounded and $mathcal E_1otimesmathcal E_2$-measurable $g:E_1times E_2tomathbb R$?
Let $$pi:E_1times E_2to E_2times E_1;,;;;(x,y)mapsto(y,x).$$ It's easy to observe that the left-hand side of $(1)$ is equal to$^2$ $$int g:{rm d}(nuotimeskappa)tag2$$ and the right-hand side is equal to $$int gcircpi^{-1}:{rm d}(nukappaotimesoverleftarrowkappa_nu).tag3$$
Now, it's easy to see that$^3$ $$pi_ast(nuotimeskappa)=muotimesoverleftarrowkappa_nutag4$$ and $$nuotimeskappa=left(pi^{-1}right)_ast(muotimesoverleftarrowkappa_nu)tag5.$$
$(1)$ is claimed in the linked document below the Definition. Could it be the case that their definition of "reverse kernel" is broken? From a terminological point of view it would make more sense to me if $nukappa$ on the right-hand side of $(1)$ would be replaced by $nu$.
$^1$ $nukappa$ denotes the composition of $nu$ and $kappa$.
$^2$ $nuotimeskappa$ denotes the product of $nu$ and $kappa$.
$^3$ $pi_ast(nuotimeskappa)$ denotes the pushforward measure of $pi$ with respect to $nuotimeskappa$.
real-analysis probability-theory measure-theory markov-chains markov-process
asked Jan 11 at 22:55
0xbadf00d0xbadf00d
1,90441530
$endgroup$
add a comment |
$begingroup$
Let
$(E_i,mathcal E_i)$ be a measurable space
$kappa$ be a Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$
Assume $kappa$ has a positive density with respect to a measure $mu$ on $(E_2,mathcal E_2)$, i.e. there is a $mathcal E_1otimesmathcal E_2$-measurable $f:E_1times E_2to(0,infty)$ with $$kappa(x,;cdot;)=f(x,;cdot;)mu;;;text{for all }xin E_1.$$ Now, let $nu$ be a probability measure on $(E_1,mathcal E_1)$ and $$overleftarrowkappa_nu(y,;cdot;):=frac1{c(y)}f(;cdot;,y)nu;;;text{for }yin E_2,$$ where $c(y):=intnu({rm d}x)f(x,y)$ (and we assume that $c(y)<infty$) for $yin E_2$.
How can we show that $overleftarrowkappa_nu$ is the reverse kernel of $kappa$ with respect to $nu$ (see Definition 2.1.2), i.e.$^1$ $$intnu({rm d}x)intkappa(x,{rm d}y)g(x,y)=intnukappa({rm d}y)intoverleftarrowkappa_nu(y,{rm d}x)g(x,y)tag1$$ for all bounded and $mathcal E_1otimesmathcal E_2$-measurable $g:E_1times E_2tomathbb R$?
Let $$pi:E_1times E_2to E_2times E_1;,;;;(x,y)mapsto(y,x).$$ It's easy to observe that the left-hand side of $(1)$ is equal to$^2$ $$int g:{rm d}(nuotimeskappa)tag2$$ and the right-hand side is equal to $$int gcircpi^{-1}:{rm d}(nukappaotimesoverleftarrowkappa_nu).tag3$$
Now, it's easy to see that$^3$ $$pi_ast(nuotimeskappa)=muotimesoverleftarrowkappa_nutag4$$ and $$nuotimeskappa=left(pi^{-1}right)_ast(muotimesoverleftarrowkappa_nu)tag5.$$
$(1)$ is claimed in the linked document below the Definition. Could it be the case that their definition of "reverse kernel" is broken? From a terminological point of view it would make more sense to me if $nukappa$ on the right-hand side of $(1)$ would be replaced by $nu$.
$^1$ $nukappa$ denotes the composition of $nu$ and $kappa$.
$^2$ $nuotimeskappa$ denotes the product of $nu$ and $kappa$.
$^3$ $pi_ast(nuotimeskappa)$ denotes the pushforward measure of $pi$ with respect to $nuotimeskappa$.
real-analysis probability-theory measure-theory markov-chains markov-process
asked Jan 11 at 22:55
0xbadf00d0xbadf00d
1,90441530
$endgroup$
Let
$(E_i,mathcal E_i)$ be a measurable space
$kappa$ be a Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$
Assume $kappa$ has a positive density with respect to a measure $mu$ on $(E_2,mathcal E_2)$, i.e. there is a $mathcal E_1otimesmathcal E_2$-measurable $f:E_1times E_2to(0,infty)$ with $$kappa(x,;cdot;)=f(x,;cdot;)mu;;;text{for all }xin E_1.$$ Now, let $nu$ be a probability measure on $(E_1,mathcal E_1)$ and $$overleftarrowkappa_nu(y,;cdot;):=frac1{c(y)}f(;cdot;,y)nu;;;text{for }yin E_2,$$ where $c(y):=intnu({rm d}x)f(x,y)$ (and we assume that $c(y)<infty$) for $yin E_2$.
How can we show that $overleftarrowkappa_nu$ is the reverse kernel of $kappa$ with respect to $nu$ (see Definition 2.1.2), i.e.$^1$ $$intnu({rm d}x)intkappa(x,{rm d}y)g(x,y)=intnukappa({rm d}y)intoverleftarrowkappa_nu(y,{rm d}x)g(x,y)tag1$$ for all bounded and $mathcal E_1otimesmathcal E_2$-measurable $g:E_1times E_2tomathbb R$?
Let $$pi:E_1times E_2to E_2times E_1;,;;;(x,y)mapsto(y,x).$$ It's easy to observe that the left-hand side of $(1)$ is equal to$^2$ $$int g:{rm d}(nuotimeskappa)tag2$$ and the right-hand side is equal to $$int gcircpi^{-1}:{rm d}(nukappaotimesoverleftarrowkappa_nu).tag3$$
Now, it's easy to see that$^3$ $$pi_ast(nuotimeskappa)=muotimesoverleftarrowkappa_nutag4$$ and $$nuotimeskappa=left(pi^{-1}right)_ast(muotimesoverleftarrowkappa_nu)tag5.$$
$(1)$ is claimed in the linked document below the Definition. Could it be the case that their definition of "reverse kernel" is broken? From a terminological point of view it would make more sense to me if $nukappa$ on the right-hand side of $(1)$ would be replaced by $nu$.
$^1$ $nukappa$ denotes the composition of $nu$ and $kappa$.
$^2$ $nuotimeskappa$ denotes the product of $nu$ and $kappa$.
$^3$ $pi_ast(nuotimeskappa)$ denotes the pushforward measure of $pi$ with respect to $nuotimeskappa$.
real-analysis probability-theory measure-theory markov-chains markov-process
real-analysis probability-theory measure-theory markov-chains markov-process
asked Jan 11 at 22:55
0xbadf00d0xbadf00d
1,90441530
asked Jan 11 at 22:55
0xbadf00d0xbadf00d
1,90441530
edited Jan 12 at 22:09
0xbadf00d
asked Jan 11 at 22:55
0xbadf00d0xbadf00d
1,90441530
asked Jan 11 at 22:55
0xbadf00d0xbadf00d
1,90441530
asked Jan 11 at 22:55
0xbadf00d0xbadf00d
1,90441530
1,90441530
add a comment |
add a comment |
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