Mixing
Show that two Markov kernels almost surely agree
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$(E_i,mathcal E_i)$ be a measurable space- $X_1:Omegato E_1$
$X_2:Omegato E_2$ be $(mathcal A,mathcal E_2)$-measurable
$kappa$ be a Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$ with $$operatorname Pleft[X_2in B_2mid X_1right]=kappa(X_1,B_2);;;text{almost surely for all }B_2inmathcal E_2tag1$$
Now, assume $tildekappa$ is another Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$ with $(1)$. Are we able to conclude $$tildekappa(x_1,B_2)=kappa(x_1,B_2);;;text{for }operatorname Pcirc;X_1^{-1}text{-almost all }x_1in E_1text{ and }B_2inmathcal E_2tag2?$$
I guess we need to assume that $mathcal E_2$ is countable generated, i.e. there is a $mathcal G_2subseteqmathcal E_2$ with $|mathcal G_2|le|mathbb N|$ and $sigma(mathcal G_2)=mathcal E_2$. Then, by $(1)$, there is a $Ninmathcal A$ with $operatorname P[N]=0$ and $$tildekappa(X_1(omega),B_2)=kappa(X_1(omega),B_2);;;text{for all }(omega,B_2)in(Omegasetminus N)timesmathcal G_2tag3.$$ Let $$tilde E_1:=bigcap_{B_2inmathcal G_2}left{x_1in E_1:tildekappa(x_1,B_2)=kappa(x_1,B_2)right}.$$ Since $mathcal G_2$ is countable, $tilde E_1inmathcal E_1$ and $$operatorname P[X_1intilde E_1]geoperatorname Pleft[(Omegasetminus N)capleft{X_1intilde E_1right}right]=1tag4.$$ Now, we know that a finite measure is uniquely determined by its values on a $cap$-stable generator. So, the only remaining question is: Do we find a $cap$-stable $mathcal G_2$? I guess we can simply go over to $$tilde{mathcal G}_2:=left{bigcapmathcal H_2:mathcal H_2subseteqmathcal G_2text{ with }|mathcal H_2|inmathbb Nright}.$$
probability-theory measure-theory markov-process
asked Jan 21 at 18:43
0xbadf00d0xbadf00d
1,91441532
$endgroup$
add a comment |
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$(E_i,mathcal E_i)$ be a measurable space- $X_1:Omegato E_1$
$X_2:Omegato E_2$ be $(mathcal A,mathcal E_2)$-measurable
$kappa$ be a Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$ with $$operatorname Pleft[X_2in B_2mid X_1right]=kappa(X_1,B_2);;;text{almost surely for all }B_2inmathcal E_2tag1$$
Now, assume $tildekappa$ is another Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$ with $(1)$. Are we able to conclude $$tildekappa(x_1,B_2)=kappa(x_1,B_2);;;text{for }operatorname Pcirc;X_1^{-1}text{-almost all }x_1in E_1text{ and }B_2inmathcal E_2tag2?$$
I guess we need to assume that $mathcal E_2$ is countable generated, i.e. there is a $mathcal G_2subseteqmathcal E_2$ with $|mathcal G_2|le|mathbb N|$ and $sigma(mathcal G_2)=mathcal E_2$. Then, by $(1)$, there is a $Ninmathcal A$ with $operatorname P[N]=0$ and $$tildekappa(X_1(omega),B_2)=kappa(X_1(omega),B_2);;;text{for all }(omega,B_2)in(Omegasetminus N)timesmathcal G_2tag3.$$ Let $$tilde E_1:=bigcap_{B_2inmathcal G_2}left{x_1in E_1:tildekappa(x_1,B_2)=kappa(x_1,B_2)right}.$$ Since $mathcal G_2$ is countable, $tilde E_1inmathcal E_1$ and $$operatorname P[X_1intilde E_1]geoperatorname Pleft[(Omegasetminus N)capleft{X_1intilde E_1right}right]=1tag4.$$ Now, we know that a finite measure is uniquely determined by its values on a $cap$-stable generator. So, the only remaining question is: Do we find a $cap$-stable $mathcal G_2$? I guess we can simply go over to $$tilde{mathcal G}_2:=left{bigcapmathcal H_2:mathcal H_2subseteqmathcal G_2text{ with }|mathcal H_2|inmathbb Nright}.$$
probability-theory measure-theory markov-process
asked Jan 21 at 18:43
0xbadf00d0xbadf00d
1,91441532
$endgroup$
$begingroup$
If you have a countable numer of sets, the union of sigma algebras over all finite collections of the sets is a pi-system that is also countable.
$endgroup$
– Michael
Jan 25 at 3:13
$begingroup$
@Michael Do you think that anything is wrong with my $tilde{mathcal G_2}$? This should be a countable $cap$-stable system wtih $sigma(tilde{mathcal G_2})=sigma(mathcal G_2)$ too.
$endgroup$
– 0xbadf00d
Jan 29 at 12:59
$begingroup$
No but I only see your last comment in retrospect, I assume what you mean by $cap$-stable is what I mean by $pi$-system. I had written a comment that we seem to be doing similar things but I deleted it since I take a closer look and your $pi$ system seems a bit different from mine. I still wonder then why you are aksing this question if you feel you have already answered it. I also wonder what you mean by equation (1) because it looks like $P[X_2 in B_2 | X_1]$ is just a third kernel, in which case (1) already says that any two kernels differ by a set of measure 0.
$endgroup$
– Michael
Jan 30 at 4:39
$begingroup$
Anyway since I did not quite know what your question was, or what (1) meant, I did not initially follow your discussion after equation (2) and I interpeted your question to mean you have two kernels with properties as in my answer. I suspect I interpreted correctly.
$endgroup$
– Michael
Jan 30 at 4:43
add a comment |
$begingroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$(E_i,mathcal E_i)$ be a measurable space- $X_1:Omegato E_1$
$X_2:Omegato E_2$ be $(mathcal A,mathcal E_2)$-measurable
$kappa$ be a Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$ with $$operatorname Pleft[X_2in B_2mid X_1right]=kappa(X_1,B_2);;;text{almost surely for all }B_2inmathcal E_2tag1$$
Now, assume $tildekappa$ is another Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$ with $(1)$. Are we able to conclude $$tildekappa(x_1,B_2)=kappa(x_1,B_2);;;text{for }operatorname Pcirc;X_1^{-1}text{-almost all }x_1in E_1text{ and }B_2inmathcal E_2tag2?$$
I guess we need to assume that $mathcal E_2$ is countable generated, i.e. there is a $mathcal G_2subseteqmathcal E_2$ with $|mathcal G_2|le|mathbb N|$ and $sigma(mathcal G_2)=mathcal E_2$. Then, by $(1)$, there is a $Ninmathcal A$ with $operatorname P[N]=0$ and $$tildekappa(X_1(omega),B_2)=kappa(X_1(omega),B_2);;;text{for all }(omega,B_2)in(Omegasetminus N)timesmathcal G_2tag3.$$ Let $$tilde E_1:=bigcap_{B_2inmathcal G_2}left{x_1in E_1:tildekappa(x_1,B_2)=kappa(x_1,B_2)right}.$$ Since $mathcal G_2$ is countable, $tilde E_1inmathcal E_1$ and $$operatorname P[X_1intilde E_1]geoperatorname Pleft[(Omegasetminus N)capleft{X_1intilde E_1right}right]=1tag4.$$ Now, we know that a finite measure is uniquely determined by its values on a $cap$-stable generator. So, the only remaining question is: Do we find a $cap$-stable $mathcal G_2$? I guess we can simply go over to $$tilde{mathcal G}_2:=left{bigcapmathcal H_2:mathcal H_2subseteqmathcal G_2text{ with }|mathcal H_2|inmathbb Nright}.$$
probability-theory measure-theory markov-process
asked Jan 21 at 18:43
0xbadf00d0xbadf00d
1,91441532
$endgroup$
Let
$(Omega,mathcal A,operatorname P)$ be a probability space
$(E_i,mathcal E_i)$ be a measurable space- $X_1:Omegato E_1$
$X_2:Omegato E_2$ be $(mathcal A,mathcal E_2)$-measurable
$kappa$ be a Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$ with $$operatorname Pleft[X_2in B_2mid X_1right]=kappa(X_1,B_2);;;text{almost surely for all }B_2inmathcal E_2tag1$$
Now, assume $tildekappa$ is another Markov kernel with source $(E_1,mathcal E_1)$ and target $(E_2,mathcal E_2)$ with $(1)$. Are we able to conclude $$tildekappa(x_1,B_2)=kappa(x_1,B_2);;;text{for }operatorname Pcirc;X_1^{-1}text{-almost all }x_1in E_1text{ and }B_2inmathcal E_2tag2?$$
I guess we need to assume that $mathcal E_2$ is countable generated, i.e. there is a $mathcal G_2subseteqmathcal E_2$ with $|mathcal G_2|le|mathbb N|$ and $sigma(mathcal G_2)=mathcal E_2$. Then, by $(1)$, there is a $Ninmathcal A$ with $operatorname P[N]=0$ and $$tildekappa(X_1(omega),B_2)=kappa(X_1(omega),B_2);;;text{for all }(omega,B_2)in(Omegasetminus N)timesmathcal G_2tag3.$$ Let $$tilde E_1:=bigcap_{B_2inmathcal G_2}left{x_1in E_1:tildekappa(x_1,B_2)=kappa(x_1,B_2)right}.$$ Since $mathcal G_2$ is countable, $tilde E_1inmathcal E_1$ and $$operatorname P[X_1intilde E_1]geoperatorname Pleft[(Omegasetminus N)capleft{X_1intilde E_1right}right]=1tag4.$$ Now, we know that a finite measure is uniquely determined by its values on a $cap$-stable generator. So, the only remaining question is: Do we find a $cap$-stable $mathcal G_2$? I guess we can simply go over to $$tilde{mathcal G}_2:=left{bigcapmathcal H_2:mathcal H_2subseteqmathcal G_2text{ with }|mathcal H_2|inmathbb Nright}.$$
probability-theory measure-theory markov-process
probability-theory measure-theory markov-process
asked Jan 21 at 18:43
0xbadf00d0xbadf00d
1,91441532
asked Jan 21 at 18:43
0xbadf00d0xbadf00d
1,91441532
edited Jan 21 at 20:18
0xbadf00d
asked Jan 21 at 18:43
0xbadf00d0xbadf00d
1,91441532
asked Jan 21 at 18:43
0xbadf00d0xbadf00d
1,91441532
asked Jan 21 at 18:43
0xbadf00d0xbadf00d
1,91441532
1,91441532
$begingroup$
If you have a countable numer of sets, the union of sigma algebras over all finite collections of the sets is a pi-system that is also countable.
$endgroup$
– Michael
Jan 25 at 3:13
$begingroup$
@Michael Do you think that anything is wrong with my $tilde{mathcal G_2}$? This should be a countable $cap$-stable system wtih $sigma(tilde{mathcal G_2})=sigma(mathcal G_2)$ too.
$endgroup$
– 0xbadf00d
Jan 29 at 12:59
$begingroup$
No but I only see your last comment in retrospect, I assume what you mean by $cap$-stable is what I mean by $pi$-system. I had written a comment that we seem to be doing similar things but I deleted it since I take a closer look and your $pi$ system seems a bit different from mine. I still wonder then why you are aksing this question if you feel you have already answered it. I also wonder what you mean by equation (1) because it looks like $P[X_2 in B_2 | X_1]$ is just a third kernel, in which case (1) already says that any two kernels differ by a set of measure 0.
$endgroup$
– Michael
Jan 30 at 4:39
$begingroup$
Anyway since I did not quite know what your question was, or what (1) meant, I did not initially follow your discussion after equation (2) and I interpeted your question to mean you have two kernels with properties as in my answer. I suspect I interpreted correctly.
$endgroup$
– Michael
Jan 30 at 4:43
add a comment |
$begingroup$
If you have a countable numer of sets, the union of sigma algebras over all finite collections of the sets is a pi-system that is also countable.
$endgroup$
– Michael
Jan 25 at 3:13
$begingroup$
@Michael Do you think that anything is wrong with my $tilde{mathcal G_2}$? This should be a countable $cap$-stable system wtih $sigma(tilde{mathcal G_2})=sigma(mathcal G_2)$ too.
$endgroup$
– 0xbadf00d
Jan 29 at 12:59
$begingroup$
No but I only see your last comment in retrospect, I assume what you mean by $cap$-stable is what I mean by $pi$-system. I had written a comment that we seem to be doing similar things but I deleted it since I take a closer look and your $pi$ system seems a bit different from mine. I still wonder then why you are aksing this question if you feel you have already answered it. I also wonder what you mean by equation (1) because it looks like $P[X_2 in B_2 | X_1]$ is just a third kernel, in which case (1) already says that any two kernels differ by a set of measure 0.
$endgroup$
– Michael
Jan 30 at 4:39
$begingroup$
Anyway since I did not quite know what your question was, or what (1) meant, I did not initially follow your discussion after equation (2) and I interpeted your question to mean you have two kernels with properties as in my answer. I suspect I interpreted correctly.
$endgroup$
– Michael
Jan 30 at 4:43
$begingroup$
If you have a countable numer of sets, the union of sigma algebras over all finite collections of the sets is a pi-system that is also countable.
$endgroup$
– Michael
Jan 25 at 3:13
$begingroup$
If you have a countable numer of sets, the union of sigma algebras over all finite collections of the sets is a pi-system that is also countable.
$endgroup$
– Michael
Jan 25 at 3:13
$begingroup$
@Michael Do you think that anything is wrong with my $tilde{mathcal G_2}$? This should be a countable $cap$-stable system wtih $sigma(tilde{mathcal G_2})=sigma(mathcal G_2)$ too.
$endgroup$
– 0xbadf00d
Jan 29 at 12:59
$begingroup$
@Michael Do you think that anything is wrong with my $tilde{mathcal G_2}$? This should be a countable $cap$-stable system wtih $sigma(tilde{mathcal G_2})=sigma(mathcal G_2)$ too.
$endgroup$
– 0xbadf00d
Jan 29 at 12:59
$begingroup$
No but I only see your last comment in retrospect, I assume what you mean by $cap$-stable is what I mean by $pi$-system. I had written a comment that we seem to be doing similar things but I deleted it since I take a closer look and your $pi$ system seems a bit different from mine. I still wonder then why you are aksing this question if you feel you have already answered it. I also wonder what you mean by equation (1) because it looks like $P[X_2 in B_2 | X_1]$ is just a third kernel, in which case (1) already says that any two kernels differ by a set of measure 0.
$endgroup$
– Michael
Jan 30 at 4:39
$begingroup$
No but I only see your last comment in retrospect, I assume what you mean by $cap$-stable is what I mean by $pi$-system. I had written a comment that we seem to be doing similar things but I deleted it since I take a closer look and your $pi$ system seems a bit different from mine. I still wonder then why you are aksing this question if you feel you have already answered it. I also wonder what you mean by equation (1) because it looks like $P[X_2 in B_2 | X_1]$ is just a third kernel, in which case (1) already says that any two kernels differ by a set of measure 0.
$endgroup$
– Michael
Jan 30 at 4:39
$begingroup$
Anyway since I did not quite know what your question was, or what (1) meant, I did not initially follow your discussion after equation (2) and I interpeted your question to mean you have two kernels with properties as in my answer. I suspect I interpreted correctly.
$endgroup$
– Michael
Jan 30 at 4:43
$begingroup$
Anyway since I did not quite know what your question was, or what (1) meant, I did not initially follow your discussion after equation (2) and I interpeted your question to mean you have two kernels with properties as in my answer. I suspect I interpreted correctly.
$endgroup$
– Michael
Jan 30 at 4:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If ${C_i}_{i=1}^{infty}$ is a sequence of subsets then $cup_{n=1}^{infty} sigma(C_1, ..., C_n)$ is a countably infinite $pi$-system, see here: https://en.wikipedia.org/wiki/Pi-system
Thus, if a sigma-algebra is countably generated, it can be countably generated via a $pi$-system. If two measures agree on a $pi$-system then they agree on the entire sigma-algebra.
So, if $mathcal{E}_2$ is countably generated, then there is a countable $pi$-system ${B_2[i]}_{i=1}^{infty}$ that generates it.
Fix $i in {1, 2, 3, ...}$ and consider $B_2[i]$. Any two versions of the conditional expectation $E[ 1_{X_2 in B_2[i]} | X_1]$ agree except for a set of probability measure zero. Define
$$A_i = { omega in Omega : kappa(B_2[i], X_1(omega)) = tilde{kappa}(B_2[i], X_1(omega))}$$
Then $P[A_i]=1$ for all $i in {1, 2, 3, ...}$ and so $P[cap_{i=1}^{infty}A_i]=1$. Now $omega in cap_{i=1}^{infty} A_i$ implies that $kappa(B_2, X_1(omega))$ and $tilde{kappa}(B_2, X_1(omega))$ agree on all sets $B_2$ of the $pi$-system, which means they agree on all sets $B_2$ of the sigma algebra $mathcal{E}_2$. Thus
$$ cap_{i=1}^{infty} A_i = {omega in Omega : kappa(B_2, X_1(omega))=tilde{kappa}(B_2, X_1(omega)) quad forall B_2 in mathcal{E}_2}$$
and so the probability of the latter set is 1.
answered Jan 25 at 5:00
MichaelMichael
13k11429
$endgroup$
add a comment |
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$begingroup$
If ${C_i}_{i=1}^{infty}$ is a sequence of subsets then $cup_{n=1}^{infty} sigma(C_1, ..., C_n)$ is a countably infinite $pi$-system, see here: https://en.wikipedia.org/wiki/Pi-system
Thus, if a sigma-algebra is countably generated, it can be countably generated via a $pi$-system. If two measures agree on a $pi$-system then they agree on the entire sigma-algebra.
So, if $mathcal{E}_2$ is countably generated, then there is a countable $pi$-system ${B_2[i]}_{i=1}^{infty}$ that generates it.
Fix $i in {1, 2, 3, ...}$ and consider $B_2[i]$. Any two versions of the conditional expectation $E[ 1_{X_2 in B_2[i]} | X_1]$ agree except for a set of probability measure zero. Define
$$A_i = { omega in Omega : kappa(B_2[i], X_1(omega)) = tilde{kappa}(B_2[i], X_1(omega))}$$
Then $P[A_i]=1$ for all $i in {1, 2, 3, ...}$ and so $P[cap_{i=1}^{infty}A_i]=1$. Now $omega in cap_{i=1}^{infty} A_i$ implies that $kappa(B_2, X_1(omega))$ and $tilde{kappa}(B_2, X_1(omega))$ agree on all sets $B_2$ of the $pi$-system, which means they agree on all sets $B_2$ of the sigma algebra $mathcal{E}_2$. Thus
$$ cap_{i=1}^{infty} A_i = {omega in Omega : kappa(B_2, X_1(omega))=tilde{kappa}(B_2, X_1(omega)) quad forall B_2 in mathcal{E}_2}$$
and so the probability of the latter set is 1.
answered Jan 25 at 5:00
MichaelMichael
13k11429
$endgroup$
add a comment |
$begingroup$
If ${C_i}_{i=1}^{infty}$ is a sequence of subsets then $cup_{n=1}^{infty} sigma(C_1, ..., C_n)$ is a countably infinite $pi$-system, see here: https://en.wikipedia.org/wiki/Pi-system
Thus, if a sigma-algebra is countably generated, it can be countably generated via a $pi$-system. If two measures agree on a $pi$-system then they agree on the entire sigma-algebra.
So, if $mathcal{E}_2$ is countably generated, then there is a countable $pi$-system ${B_2[i]}_{i=1}^{infty}$ that generates it.
Fix $i in {1, 2, 3, ...}$ and consider $B_2[i]$. Any two versions of the conditional expectation $E[ 1_{X_2 in B_2[i]} | X_1]$ agree except for a set of probability measure zero. Define
$$A_i = { omega in Omega : kappa(B_2[i], X_1(omega)) = tilde{kappa}(B_2[i], X_1(omega))}$$
Then $P[A_i]=1$ for all $i in {1, 2, 3, ...}$ and so $P[cap_{i=1}^{infty}A_i]=1$. Now $omega in cap_{i=1}^{infty} A_i$ implies that $kappa(B_2, X_1(omega))$ and $tilde{kappa}(B_2, X_1(omega))$ agree on all sets $B_2$ of the $pi$-system, which means they agree on all sets $B_2$ of the sigma algebra $mathcal{E}_2$. Thus
$$ cap_{i=1}^{infty} A_i = {omega in Omega : kappa(B_2, X_1(omega))=tilde{kappa}(B_2, X_1(omega)) quad forall B_2 in mathcal{E}_2}$$
and so the probability of the latter set is 1.
answered Jan 25 at 5:00
MichaelMichael
13k11429
$endgroup$
add a comment |
$begingroup$
If ${C_i}_{i=1}^{infty}$ is a sequence of subsets then $cup_{n=1}^{infty} sigma(C_1, ..., C_n)$ is a countably infinite $pi$-system, see here: https://en.wikipedia.org/wiki/Pi-system
Thus, if a sigma-algebra is countably generated, it can be countably generated via a $pi$-system. If two measures agree on a $pi$-system then they agree on the entire sigma-algebra.
So, if $mathcal{E}_2$ is countably generated, then there is a countable $pi$-system ${B_2[i]}_{i=1}^{infty}$ that generates it.
Fix $i in {1, 2, 3, ...}$ and consider $B_2[i]$. Any two versions of the conditional expectation $E[ 1_{X_2 in B_2[i]} | X_1]$ agree except for a set of probability measure zero. Define
$$A_i = { omega in Omega : kappa(B_2[i], X_1(omega)) = tilde{kappa}(B_2[i], X_1(omega))}$$
Then $P[A_i]=1$ for all $i in {1, 2, 3, ...}$ and so $P[cap_{i=1}^{infty}A_i]=1$. Now $omega in cap_{i=1}^{infty} A_i$ implies that $kappa(B_2, X_1(omega))$ and $tilde{kappa}(B_2, X_1(omega))$ agree on all sets $B_2$ of the $pi$-system, which means they agree on all sets $B_2$ of the sigma algebra $mathcal{E}_2$. Thus
$$ cap_{i=1}^{infty} A_i = {omega in Omega : kappa(B_2, X_1(omega))=tilde{kappa}(B_2, X_1(omega)) quad forall B_2 in mathcal{E}_2}$$
and so the probability of the latter set is 1.
answered Jan 25 at 5:00
MichaelMichael
13k11429
$endgroup$
If ${C_i}_{i=1}^{infty}$ is a sequence of subsets then $cup_{n=1}^{infty} sigma(C_1, ..., C_n)$ is a countably infinite $pi$-system, see here: https://en.wikipedia.org/wiki/Pi-system
Thus, if a sigma-algebra is countably generated, it can be countably generated via a $pi$-system. If two measures agree on a $pi$-system then they agree on the entire sigma-algebra.
So, if $mathcal{E}_2$ is countably generated, then there is a countable $pi$-system ${B_2[i]}_{i=1}^{infty}$ that generates it.
Fix $i in {1, 2, 3, ...}$ and consider $B_2[i]$. Any two versions of the conditional expectation $E[ 1_{X_2 in B_2[i]} | X_1]$ agree except for a set of probability measure zero. Define
$$A_i = { omega in Omega : kappa(B_2[i], X_1(omega)) = tilde{kappa}(B_2[i], X_1(omega))}$$
Then $P[A_i]=1$ for all $i in {1, 2, 3, ...}$ and so $P[cap_{i=1}^{infty}A_i]=1$. Now $omega in cap_{i=1}^{infty} A_i$ implies that $kappa(B_2, X_1(omega))$ and $tilde{kappa}(B_2, X_1(omega))$ agree on all sets $B_2$ of the $pi$-system, which means they agree on all sets $B_2$ of the sigma algebra $mathcal{E}_2$. Thus
$$ cap_{i=1}^{infty} A_i = {omega in Omega : kappa(B_2, X_1(omega))=tilde{kappa}(B_2, X_1(omega)) quad forall B_2 in mathcal{E}_2}$$
and so the probability of the latter set is 1.
answered Jan 25 at 5:00
MichaelMichael
13k11429
edited Jan 25 at 5:14
answered Jan 25 at 5:00
MichaelMichael
13k11429
answered Jan 25 at 5:00
MichaelMichael
13k11429
answered Jan 25 at 5:00
MichaelMichael
13k11429
13k11429
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$begingroup$
If you have a countable numer of sets, the union of sigma algebras over all finite collections of the sets is a pi-system that is also countable.
$endgroup$
– Michael
Jan 25 at 3:13
$begingroup$
@Michael Do you think that anything is wrong with my $tilde{mathcal G_2}$? This should be a countable $cap$-stable system wtih $sigma(tilde{mathcal G_2})=sigma(mathcal G_2)$ too.
$endgroup$
– 0xbadf00d
Jan 29 at 12:59
$begingroup$
No but I only see your last comment in retrospect, I assume what you mean by $cap$-stable is what I mean by $pi$-system. I had written a comment that we seem to be doing similar things but I deleted it since I take a closer look and your $pi$ system seems a bit different from mine. I still wonder then why you are aksing this question if you feel you have already answered it. I also wonder what you mean by equation (1) because it looks like $P[X_2 in B_2 | X_1]$ is just a third kernel, in which case (1) already says that any two kernels differ by a set of measure 0.
$endgroup$
– Michael
Jan 30 at 4:39
$begingroup$
Anyway since I did not quite know what your question was, or what (1) meant, I did not initially follow your discussion after equation (2) and I interpeted your question to mean you have two kernels with properties as in my answer. I suspect I interpreted correctly.
$endgroup$
– Michael
Jan 30 at 4:43