This is a technical post researching infinite factor spaces

I define a general measurable factor space that
shows that the history can't be straightforwardly generalized to the infinite case.
The example shows the following:

Unlike in the finite setting, the condition
∀P∈▲∗(F):πJ⫫Pπ¯J|Z in the definition of
generation (see below) can't be characterized by requiring that the atoms of
Z are rectangles. In the example, we show that this would
imply an orthogonality where the implication X⊥Y∣∣Z⇒∀P∈▲∗(F):X⫫PY∣∣Z is not true.

In measurable spaces, conditional histories can only be well-defined
up to a chosen system of nullsets. In the example, we show that a
minimal generating index function J:F→P(I)
does not exist. (We will show in a later post that the history exists if we fix nullsets.)

To talk about orthogonality, we first have to introduce the concepts for
measurable spaces. For now, we focus on the index set I being finite. A
straightforward generalization from the finite case is the following.
We have measurable spaces (Fi,Ai) and
construct F=×i∈IFi. Let Z:F→Z(F) be a feature
(i.e. measurable). Let J:F→P(I) be
Z-measurable. We define
πJ(x)=πJ(x)(x) We note, that we assume w.l.o.g. that we
assume that Fi are pairwise disjoint. Recall that ▲∗(F) is
the set of product probability distributions on F.

J⊢X∣∣Z:⇔X≲(πJ,Z)∧∀P∈▲∗(F):πJ⫫Pπ¯J∣∣Z.

The history should now be the smallest such J. Furthermore, for
orthogonality defined by X⊥Y|Z:⇔H(X|Z)∩H(Y|Z)=∅
we should have X⊥Y|Z⇔∀P∈▲∗(F):X⫫PY|Z

The idea for the example consists of choosing Z, such that
Z−1(z) consists of four points, where the points will
always fall into one unit square each. Call those unit squares Uij.
Choosing Z|Uij to be a diffeomorphism allows us to
calculate the conditional probabilities explicitly. Almost all choices
of Z where Z−1(z) is a rectangle for all
z, will fail to satisfy ∀P∈▲∗(F):X⫫PY|Z.

Example 1

Let I={1,2}. For i∈I, let
Fi=[0,1]∪[2,3] endowed with the
Borel sigma-algebra.

Now
F=F1×F2=([0,1]∪[2,3])^2.

F consists of 4 unit squares.

We will now define a Z, such that Z−1 is a rectangle of four
points.

Let Uij={[0,1] if i=1[2,3]else×{[0,1] if j=1[2,3]else be the four unit squares that F consists of.

Let
φx,φy:[0,1]2→[2,3].
φx will be the function that chooses the new x coordinate.

Let πx:[0,1]2→[2,3],x,y|→x+2πy:[0,1]2→[2,3],x,y|→y+2

Let φ11=(πx,πy)φ21=(φx,πy)φ12=(πx,φy)φ22=(φx,φy)

We assume that all φij are diffeomorphisms.

Define Zij=φ−1ij and Z by
Z|Uij=Zij.

First we assure ourselves, that Z indeed has rectangle
atoms.

Let z=(z1,z2)∈[0,1]2, then
Z−1(z)={φij(z)}={(z1,z2),(z1,φy(z2))(φx(z1),φy(z2))(φx(z1),z2)}={z1,z2}×{φx(z1),φy(z2)}

Now the projections πi:F→Fi would be orthogonal,
if we had ∀P∈▲∗(F):πJ⫫Pπ¯J|Z⇔∀z∈Z(F):Z−1(z) is a rectangle.

We will now calculate the conditional probabilities by Lebesgue
differentiation.

We first note that Z is continuous. Each
Z−1(z) consists of four points, one in each Uij
that form a rectangle.

Because F is compact, it is easy to show that
Z−1(Bδ(z)) concentrates around
Z−1(z) for small delta, i.e.

∀ε>0:∃δ>0:d(Z−1(Bδ(z)),Z−1(z))<ε,
where d(A,B)=supa∈Ainfb∈Bd(a,b) is the
Hausdorff metric for the standard metric d=||⋅−⋅||2.

We will use the uniform distribution
P=P1×P2=λ|F1λ(F1)×λ|F2λ(F2)=λ2|Fλ2(F)

to show π1⫫Pπ2|Z.

Let
A=A1×A2=([0,a1]∩F1)×([0,a2]∩F2)
for ai∈Fi∖{0}.

Take a z such that Z−1 does not lie on the edge of
A, i.e.
Z−1(z)∩F1×{a2}∪{a1}×F1=∅

By Lebesgue differentiation, we know that
1Bδ(z)∫Bδ(z)P(A|Z=⋅)dPZδ→0→P(A|Z=z) for PZ-a.a z
Since the left side equals
P(A|{Z∈Bδ(z)}), we
have
z↦limδ→0P(A|{Z∈Bδ(z)})
is a version of P(A|Z=⋅).

Now we will split up {Z∈Bδ(z)}
into its parts in the unit squares Uij.

Let
Cijδ={Zij∈Bδ(z)}=φij(C11δ).
Let c=1λ2(F).

We have P(Cijδ)=c∫φij(C11δ)1dλ2=c∫C11δ1∘φij∣∣detDφij∣∣dλ2=c∫C11δ∣∣detDφij∣∣dλ2

Since ∣∣detDφij∣∣ is continuous, we have
1P(C11δ)∫C11δ∣∣detDφij∣∣dP=1λ2(C11δ)∫C11δ∣∣detDφij∣∣dλ2→∣∣detDφij(z)∣∣≕dij

Now
|detDφ22|=|detDφ12Dφ21|⇔∀x:p2q1(x)=0∨p2q1(x)=2p1q2(x)

It is clear that you can choose φx and φy without
this property.

Let for example

φx(a,b)=ab+(1−b)a2φy(a,b)=ab+(1−a)b2

We have Dφx=(2a(1−b)+b,a−a2)=(p1p2)Dφy=(2b(1−a)+a,b−b2)=(q1q2) let x=(12,12). Clearly,
p2q1=2ab(1−b)^2+b2(1−b) and
p2q1(x)=14≠0

Furthermore, 2p1q2=4ba(1−a)^2+2a2(1−a) and
2p1q2(x)=12≠14

With these chosen, we have that π1⫫Pπ2|Z.

Example 2

We will now argue, that the history
H(π1|Z) can't exist. Let
z0∈[0,1]2. Let J0(z)={{1}if z=z0Ielse and J=J0∘Z. We will show
J⊢π1|Z. Then, since the history would be a
subset of all such J, we would have
H(π1|Z)={1}. But we have
just seen that π1⫫Pπ2|Z and therefore the history is not a
generating index function.

Now, πJ(x,y)={xif Z(x,y)=z0(x,y)else and π¯J(x,y)={yif Z(x,y)=z0()else

If {z0} is a PZ nullset, clearly
πJ⫫PZπ¯J|Z, since
π¯J is constant a.s. Otherwise, let
{x1,x2}×{y1,y2}=Z−1(z0)
and px1=P1(x1) etc.

Now since Z−1(z0) is a (finite) rectangle with
positive measure, we clearly have for
A⊆{x1,x2},B⊆{y1,y2}
that
P(A×F2|Z=z)P(F1×B|Z=z)=P(A×B|Z=z)
for all z∈Z(F). For
A⊆F∖{x1,x2}×{y1,y2},
we clearly have A⫫P∅|Z.

Since these sets A are closed under intersections and generate
πJ and the sets B generate π¯J, we have
that πJ⫫Pπ¯J|Z.

PDF

This is a technical post researching infinite factor spaces

I define a general measurable factor space that shows that the history can't be straightforwardly generalized to the infinite case. The example shows the following:

Unlike in the finite setting, the condition ∀P∈▲∗(F):πJ⫫Pπ¯J | Z in the definition of generation (see below) can't be characterized by requiring that the atoms of Z are rectangles. In the example, we show that this would imply an orthogonality where the implication X⊥Y∣∣Z⇒∀P∈▲∗(F):X⫫PY∣∣Z is not true.

In measurable spaces, conditional histories can only be well-defined up to a chosen system of nullsets. In the example, we show that a minimal generating index function J:F→P(I) does not exist. (We will show in a later post that the history exists if we fix nullsets.)

To talk about orthogonality, we first have to introduce the concepts for measurable spaces. For now, we focus on the index set I being finite. A straightforward generalization from the finite case is the following. We have measurable spaces (Fi,Ai) and construct F=×i∈IFi. Let Z:F→Z(F) be a feature (i.e. measurable). Let J:F→P(I) be Z-measurable. We define πJ(x)=πJ(x)(x) We note, that we assume w.l.o.g. that we assume that Fi are pairwise disjoint. Recall that ▲∗(F) is the set of product probability distributions on F.

J⊢X∣∣Z:⇔X≲(πJ,Z)∧∀P∈▲∗(F):πJ⫫Pπ¯J∣∣Z.

The history should now be the smallest such J. Furthermore, for orthogonality defined by X⊥Y | Z:⇔H(X|Z)∩H(Y|Z)=∅ we should have X⊥Y | Z⇔∀P∈▲∗(F):X⫫PY|Z

The idea for the example consists of choosing Z, such that Z−1(z) consists of four points, where the points will always fall into one unit square each. Call those unit squares Uij. Choosing Z|Uij to be a diffeomorphism allows us to calculate the conditional probabilities explicitly. Almost all choices of Z where Z−1(z) is a rectangle for all z, will fail to satisfy ∀P∈▲∗(F):X⫫PY | Z.

## Example 1

Let I={1,2}. For i∈I, let Fi=[0,1]∪[2,3] endowed with the Borel sigma-algebra.

Now F=F1×F2=([0,1]∪[2,3])^2.

F consists of 4 unit squares.

We will now define a Z, such that Z−1 is a rectangle of four points.

Let Uij={[0,1] if i=1[2,3]else×{[0,1] if j=1[2,3]else be the four unit squares that F consists of.

Let φx,φy:[0,1]2→[2,3]. φx will be the function that chooses the new x coordinate.

Let πx:[0,1]2→[2,3],x,y|→x+2πy:[0,1]2→[2,3],x,y|→y+2

Let φ11=(πx,πy)φ21=(φx,πy)φ12=(πx,φy)φ22=(φx,φy)

We assume that all φij are diffeomorphisms.

Define Zij=φ−1ij and Z by Z|Uij=Zij.

First we assure ourselves, that Z indeed has rectangle atoms.

Let z=(z1,z2)∈[0,1]2, then Z−1(z)={φij(z)}={(z1,z2),(z1,φy(z2))(φx(z1),φy(z2))(φx(z1),z2)}={z1,z2}×{φx(z1),φy(z2)}

Now the projections πi:F→Fi would be orthogonal, if we had ∀P∈▲∗(F):πJ⫫Pπ¯J | Z⇔∀z∈Z(F):Z−1(z) is a rectangle.

We will now calculate the conditional probabilities by Lebesgue differentiation.

We first note that Z is continuous. Each Z−1(z) consists of four points, one in each Uij that form a rectangle.

Because F is compact, it is easy to show that Z−1(Bδ(z)) concentrates around Z−1(z) for small delta, i.e.

∀ε>0:∃δ>0:d(Z−1(Bδ(z)),Z−1(z))<ε, where d(A,B)=supa∈Ainfb∈Bd(a,b) is the Hausdorff metric for the standard metric d=||⋅−⋅||2.

We will use the uniform distribution P=P1×P2=λ|F1λ(F1)×λ|F2λ(F2)=λ2|Fλ2(F)

to show π1⫫Pπ2|Z.

Let A=A1×A2=([0,a1]∩F1)×([0,a2]∩F2) for ai∈Fi∖{0}.

Take a z such that Z−1 does not lie on the edge of A, i.e. Z−1(z)∩F1×{a2}∪{a1}×F1=∅

By Lebesgue differentiation, we know that 1Bδ(z)∫Bδ(z)P(A|Z=⋅)dPZδ→0→P(A|Z=z) for PZ-a.a z Since the left side equals P(A|{Z∈Bδ(z)}), we have z↦limδ→0P(A | {Z∈Bδ(z)}) is a version of P(A|Z=⋅).

Now we will split up {Z∈Bδ(z)} into its parts in the unit squares Uij.

Let Cijδ={Zij∈Bδ(z)}=φij(C11δ). Let c=1λ2(F).

We have P(Cijδ)=c∫φij(C11δ)1dλ2=c∫C11δ1∘φij∣∣detDφij∣∣dλ2=c∫C11δ∣∣detDφij∣∣dλ2

Since ∣∣detDφij∣∣ is continuous, we have 1P(C11δ)∫C11δ∣∣detDφij∣∣dP=1λ2(C11δ)∫C11δ∣∣detDφij∣∣dλ2→∣∣detDφij(z)∣∣≕dij

Let zij=Z−1ij(z) and let J={(i,j):zi,j∈A}, then

P(A | {Z∈Bδ(z)})=∑i,j∈{1,2}P(A∩Cijδ)∑i,j∈{1,2}P(Cijδ)δ small=∑(i,j)∈JP(Cijδ)∑i,j∈{1,2}P(Cijδ)=∑(i,j)∈J1P(C11δ)P(Cijδ)∑i,j∈{1,2}1P(C11δ)P(Cijδ)→∑(i,j)∈Jdij∑i,j∈{0,1}dij

Now clearly, (P(A1×F2 | Z)P(F1×A2 | Z))(z)=P(A | Z)(z) if and only if d12d21=d22

Since this has to hold for almost all z and the determinants are continuous, it has to hold for all z, we have to have

|detDφ22|=|detDφ12detDφ21|=|detDφ12Dφ21|

Let Dφx=(p1p2) and Dφy=(q1q2). Now

|detDφ12|=∣∣∣det(p1p201)∣∣∣=|p1||detDφ21|=∣∣∣det(10q1q2)∣∣∣=|q2||detDφ22|=∣∣∣det(p1p2q1q2)∣∣∣=|p1q2−p2q1||detDφ12Dφ21|=|p1q2|

Now |detDφ22|=|detDφ12Dφ21|⇔∀x:p2q1(x)=0∨p2q1(x)=2p1q2(x)

It is clear that you can choose φx and φy without this property.

Let for example

φx(a,b)=ab+(1−b)a2φy(a,b)=ab+(1−a)b2

We have Dφx=(2a(1−b)+b,a−a2)=(p1p2)Dφy=(2b(1−a)+a,b−b2)=(q1q2) let x=(12,12). Clearly, p2q1=2ab(1−b)^2+b2(1−b) and p2q1(x)=14≠0

Furthermore, 2p1q2=4ba(1−a)^2+2a2(1−a) and 2p1q2(x)=12≠14

With these chosen, we have that π1⫫Pπ2 | Z.

## Example 2

We will now argue, that the history H(π1|Z) can't exist. Let z0∈[0,1]2. Let J0(z)={{1}if z=z0Ielse and J=J0∘Z. We will show J⊢π1 | Z. Then, since the history would be a subset of all such J, we would have H(π1 | Z)={1}. But we have just seen that π1⫫Pπ2 | Z and therefore the history is not a generating index function.

Now, πJ(x,y)={xif Z(x,y)=z0(x,y)else and π¯J(x,y)={yif Z(x,y)=z0()else

If {z0} is a PZ nullset, clearly πJ⫫PZπ¯J | Z, since π¯J is constant a.s. Otherwise, let {x1,x2}×{y1,y2}=Z−1(z0) and px1=P1(x1) etc.

Now since Z−1(z0) is a (finite) rectangle with positive measure, we clearly have for A⊆{x1,x2},B⊆{y1,y2} that P(A×F2|Z=z)P(F1×B|Z=z)=P(A×B | Z=z) for all z∈Z(F). For A⊆F∖{x1,x2}×{y1,y2}, we clearly have A⫫P∅ | Z.

Since these sets A are closed under intersections and generate πJ and the sets B generate π¯J, we have that πJ⫫Pπ¯J | Z.