Mixing
Show that for each Markov kernel admiting a density there is a corresponding reverse kernel
$begingroup$
Let
$(E_i,mathcal E_i)$ be a measurable space
$kappa$ be a Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$
$mu$ be a probability measure on $(E_1,mathcal E_1)$
Assume $kappa$ has a density $f:E_1times E_2to[0,infty)$ with respect to a measure $lambda$ on $(E_2,mathcal E_2)$, i.e. $$kappa(x,B)=int_Bf(x,y):lambda({rm d}y)tag1;;;text{for all }(x,B)in E_1timesmathcal E_2.$$ Note that $$c(y):=intmu({rm d}x)f(x,y);;;text{for }yinmathcal E_2.$$ is $mathcal E_2$-measurable and hence $N_1:=left{c=0right}inmathcal E_2$. It's easy to see that$^1$ $$(mukappa)(N)=0.tag2$$
Now, let $N_2:=left{c=inftyright}$. Again, $N_2inmathcal E_2$, but are we able to show $(mukappa)(N_2)=0$?
My idea was to show that $$0leint c:{rm d}(mukappa)<inftytag3,$$ but I wasn't able to do so. Using Fubini's theorem we obtain $$int c:{rm d}(mukappa)=intmu({rm d}tilde x)intmu({rm d}x)intlambda({rm d}y)f(tilde x,y)f(x,y)tag4$$ and we know that $$intlambda({rm d}y)f(x,y)=1;;;text{for all }xin E_1tag5.$$ How can we conclude?
$^1$ $mukappa$ denotes the composition of $mu$ and $kappa$.
probability-theory measure-theory markov-chains markov-process
asked Jan 12 at 19:17
0xbadf00d0xbadf00d
1,90941531
$endgroup$
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 $(E_2,mathcal E_2)$
$mu$ be a probability measure on $(E_1,mathcal E_1)$
Assume $kappa$ has a density $f:E_1times E_2to[0,infty)$ with respect to a measure $lambda$ on $(E_2,mathcal E_2)$, i.e. $$kappa(x,B)=int_Bf(x,y):lambda({rm d}y)tag1;;;text{for all }(x,B)in E_1timesmathcal E_2.$$ Note that $$c(y):=intmu({rm d}x)f(x,y);;;text{for }yinmathcal E_2.$$ is $mathcal E_2$-measurable and hence $N_1:=left{c=0right}inmathcal E_2$. It's easy to see that$^1$ $$(mukappa)(N)=0.tag2$$
Now, let $N_2:=left{c=inftyright}$. Again, $N_2inmathcal E_2$, but are we able to show $(mukappa)(N_2)=0$?
My idea was to show that $$0leint c:{rm d}(mukappa)<inftytag3,$$ but I wasn't able to do so. Using Fubini's theorem we obtain $$int c:{rm d}(mukappa)=intmu({rm d}tilde x)intmu({rm d}x)intlambda({rm d}y)f(tilde x,y)f(x,y)tag4$$ and we know that $$intlambda({rm d}y)f(x,y)=1;;;text{for all }xin E_1tag5.$$ How can we conclude?
$^1$ $mukappa$ denotes the composition of $mu$ and $kappa$.
probability-theory measure-theory markov-chains markov-process
asked Jan 12 at 19:17
0xbadf00d0xbadf00d
1,90941531
$endgroup$
$begingroup$
The claim would allow us to conclude that $$overleftarrowkappa_mu(y,;cdot;):=frac{f(;cdot;,y)mu}{mu f(;cdot;,y)};;;text{for }yin E_2$$ is a reverse kernel with respect to $(mu,kappa)$.
$endgroup$
– 0xbadf00d
Jan 12 at 19:19
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 $(E_2,mathcal E_2)$
$mu$ be a probability measure on $(E_1,mathcal E_1)$
Assume $kappa$ has a density $f:E_1times E_2to[0,infty)$ with respect to a measure $lambda$ on $(E_2,mathcal E_2)$, i.e. $$kappa(x,B)=int_Bf(x,y):lambda({rm d}y)tag1;;;text{for all }(x,B)in E_1timesmathcal E_2.$$ Note that $$c(y):=intmu({rm d}x)f(x,y);;;text{for }yinmathcal E_2.$$ is $mathcal E_2$-measurable and hence $N_1:=left{c=0right}inmathcal E_2$. It's easy to see that$^1$ $$(mukappa)(N)=0.tag2$$
Now, let $N_2:=left{c=inftyright}$. Again, $N_2inmathcal E_2$, but are we able to show $(mukappa)(N_2)=0$?
My idea was to show that $$0leint c:{rm d}(mukappa)<inftytag3,$$ but I wasn't able to do so. Using Fubini's theorem we obtain $$int c:{rm d}(mukappa)=intmu({rm d}tilde x)intmu({rm d}x)intlambda({rm d}y)f(tilde x,y)f(x,y)tag4$$ and we know that $$intlambda({rm d}y)f(x,y)=1;;;text{for all }xin E_1tag5.$$ How can we conclude?
$^1$ $mukappa$ denotes the composition of $mu$ and $kappa$.
probability-theory measure-theory markov-chains markov-process
asked Jan 12 at 19:17
0xbadf00d0xbadf00d
1,90941531
$endgroup$
Let
$(E_i,mathcal E_i)$ be a measurable space
$kappa$ be a Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$
$mu$ be a probability measure on $(E_1,mathcal E_1)$
Assume $kappa$ has a density $f:E_1times E_2to[0,infty)$ with respect to a measure $lambda$ on $(E_2,mathcal E_2)$, i.e. $$kappa(x,B)=int_Bf(x,y):lambda({rm d}y)tag1;;;text{for all }(x,B)in E_1timesmathcal E_2.$$ Note that $$c(y):=intmu({rm d}x)f(x,y);;;text{for }yinmathcal E_2.$$ is $mathcal E_2$-measurable and hence $N_1:=left{c=0right}inmathcal E_2$. It's easy to see that$^1$ $$(mukappa)(N)=0.tag2$$
Now, let $N_2:=left{c=inftyright}$. Again, $N_2inmathcal E_2$, but are we able to show $(mukappa)(N_2)=0$?
My idea was to show that $$0leint c:{rm d}(mukappa)<inftytag3,$$ but I wasn't able to do so. Using Fubini's theorem we obtain $$int c:{rm d}(mukappa)=intmu({rm d}tilde x)intmu({rm d}x)intlambda({rm d}y)f(tilde x,y)f(x,y)tag4$$ and we know that $$intlambda({rm d}y)f(x,y)=1;;;text{for all }xin E_1tag5.$$ How can we conclude?
$^1$ $mukappa$ denotes the composition of $mu$ and $kappa$.
probability-theory measure-theory markov-chains markov-process
probability-theory measure-theory markov-chains markov-process
asked Jan 12 at 19:17
0xbadf00d0xbadf00d
1,90941531
asked Jan 12 at 19:17
0xbadf00d0xbadf00d
1,90941531
edited Jan 12 at 22:50
0xbadf00d
asked Jan 12 at 19:17
0xbadf00d0xbadf00d
1,90941531
asked Jan 12 at 19:17
0xbadf00d0xbadf00d
1,90941531
asked Jan 12 at 19:17
0xbadf00d0xbadf00d
1,90941531
1,90941531
$begingroup$
The claim would allow us to conclude that $$overleftarrowkappa_mu(y,;cdot;):=frac{f(;cdot;,y)mu}{mu f(;cdot;,y)};;;text{for }yin E_2$$ is a reverse kernel with respect to $(mu,kappa)$.
$endgroup$
– 0xbadf00d
Jan 12 at 19:19
add a comment |
$begingroup$
The claim would allow us to conclude that $$overleftarrowkappa_mu(y,;cdot;):=frac{f(;cdot;,y)mu}{mu f(;cdot;,y)};;;text{for }yin E_2$$ is a reverse kernel with respect to $(mu,kappa)$.
$endgroup$
– 0xbadf00d
Jan 12 at 19:19
$begingroup$
The claim would allow us to conclude that $$overleftarrowkappa_mu(y,;cdot;):=frac{f(;cdot;,y)mu}{mu f(;cdot;,y)};;;text{for }yin E_2$$ is a reverse kernel with respect to $(mu,kappa)$.
$endgroup$
– 0xbadf00d
Jan 12 at 19:19
$begingroup$
The claim would allow us to conclude that $$overleftarrowkappa_mu(y,;cdot;):=frac{f(;cdot;,y)mu}{mu f(;cdot;,y)};;;text{for }yin E_2$$ is a reverse kernel with respect to $(mu,kappa)$.
$endgroup$
– 0xbadf00d
Jan 12 at 19:19
add a comment |
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$begingroup$
The claim would allow us to conclude that $$overleftarrowkappa_mu(y,;cdot;):=frac{f(;cdot;,y)mu}{mu f(;cdot;,y)};;;text{for }yin E_2$$ is a reverse kernel with respect to $(mu,kappa)$.
$endgroup$
– 0xbadf00d
Jan 12 at 19:19