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Exact-density normalizing flows

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

A normalizing flow and its base density

Definition

Let d≥1, let U,V⊆Rd be nonempty open sets, and let T:U→V be a bijection such that both T and T−1 are continuously differentiable. Here C1 means totally differentiable at every point with continuous derivative matrix. Such a map is a global C1 diffeomorphism. A finite composition T=TK∘⋯∘T1 of global C1 diffeomorphisms between open subsets of Rd is a normalizing flow.

A base density on U is a Lebesgue-measurable function pU:U→[0,∞] with ∫UpU dλd=1. It defines the probability measure μU(A)=∫ApU dλd for Lebesgue-measurable A⊆U.

For the model-measure construction, assume the Axiom of Countable Choice. The model measure on V is μX(B)=μU(T−1(B)) for Borel B⊆V. This value is defined because the inverse C1 diffeomorphism maps Borel sets to Lebesgue-measurable sets under Countable Choice. When this measure has a Lebesgue density, that density is the model density.

Definition references

Continuously differentiable maps, local inverses, and local diffeomorphisms — Total differentiability, continuous derivative matrices, and local diffeomorphisms.

The measure with density f relative to μ — The measure defined by integrating a nonnegative measurable density.

The Axiom of Countable Choice (ACω) — Countable Choice is assumed for the model-measure construction.

Assuming countable choice, every Borel subset of Rn is Lebesgue measurable — Under Countable Choice, Borel subsets of Euclidean space are Lebesgue measurable.

A C^1 diffeomorphism maps Lebesgue measurable sets to Lebesgue measurable sets — A C1 diffeomorphism sends Lebesgue-measurable sets to Lebesgue-measurable sets.

Papamakarios et al., §2.1, PDF pp. 2–3

Sheldon, COMPSCI 688 Lecture 23, PDF pp. 1–5

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

A diffeomorphic flow has the inverse-Jacobian density

Statement

Assume the Axiom of Countable Choice. Let d≥1, let U,V⊆Rd be nonempty open sets, and let T:U→V be a bijection such that T and T−1 are both C1. Let pU:U→[0,∞] be Lebesgue measurable with ∫UpU dλd=1, and define μU(A)=∫ApU dλd for Lebesgue-measurable A⊆U and μX(B)=μU(T−1(B)) for Borel B⊆V. Then

q(x)=pU(T−1(x)) ∣det⁡D(T−1)(x)∣

is Lebesgue measurable and satisfies μX(B)=∫Bq dλd for every Borel B⊆V. Thus q is a version of the density of the transformed measure.

Let K≥1, let U0,…,UK be nonempty open subsets of Rd, and let Tj:Uj−1→Uj be a C1 diffeomorphism for j=1,…,K. For T=TK∘⋯∘T1, u∈U0, and z0=u, zj=Tj(zj−1), the log-Jacobian terms add:

log⁡∣det⁡DT(u)∣=∑j=1Klog⁡∣det⁡DTj(zj−1)∣.

Equivalently, for zK=x∈UK and zj−1=Tj−1(zj),

log⁡∣det⁡DT−1(x)∣=∑j=1Klog⁡∣det⁡DTj−1(zj)∣.

Statement references

Facts & Assumptions

Given: Countable Choice, a flow T:U→V, and a Lebesgue probability density pU as in the Statement.

[A1]

The Axiom of Countable Choice, written ACω, asserts that every sequence of nonempty sets has a choice function. The Axiom of Countable Choice (ACω)

[F1]

A normalizing flow is a finite composition of bijections between open equal-dimensional Euclidean domains whose maps and inverses are C1; a base density pU defines μU(A)=∫ApU dλd, and the transformed measure is μX(B)=μU(T−1(B)). A normalizing flow and its base density

[F2]

For a nonnegative measurable function f, the measure with density f relative to λd assigns A the value ∫Af dλd. The measure with density f relative to μ

[F3]

If S:U0→V0 is a C1 diffeomorphism and f:V0→[0,∞] is Lebesgue measurable, then under Countable Choice ∫V0f(y) dλd(y)=∫U0f(S(x))∣det⁡DS(x)∣ dλd(x). A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions

[F4]

Under Countable Choice, Borel sets are Lebesgue measurable, and a C1 diffeomorphism maps every Lebesgue-measurable subset of its domain to a Lebesgue-measurable subset of its range. Assuming countable choice, every Borel subset of Rn is Lebesgue measurable, A C^1 diffeomorphism maps Lebesgue measurable sets to Lebesgue measurable sets

[F5]

For totally differentiable maps, D(g∘f)(a)=Dg(f(a))∘Df(a). The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)

[F6]

For same-sized square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B). For same-sized finite square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B)

[F7]

The Jacobian matrix is the matrix of the total derivative in standard bases, and its Jacobian determinant is the determinant of that matrix. The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, The Jacobian determinant of a square-dimensional C1 map is the determinant of its Jacobian matrix

[F9]

Products of extended-real measurable functions are measurable whenever defined, including the nonnegative density and determinant factors here. Arithmetic and lattice operations preserve measurability whenever they are defined

Proof

technique · direct
1.1A1F1F4F7F8F9

For every a∈R, the superlevel set Ea={u∈U:pU(u)>a} is Lebesgue measurable, and {x∈V:pU(T−1(x))>a}=T(Ea) is Lebesgue measurable by [F4]. Thus pU∘T−1 is measurable. The entries of D(T−1) are continuous by the C1 hypothesis, and [F8] makes its determinant continuous. Thus [F9] shows that q is nonnegative measurable.

1.2givenF1F5F7

For z0=u and zj=Tj(zj−1), repeated use of [F5] gives DT(u)=DTK(zK−1)⋯DT1(z0). For x=zK, applying the chain rule to T−1=T1−1∘⋯∘TK−1 gives D(T−1)(x)=DT1−1(z1)⋯DTK−1(zK).

2.1A1F1F2F3F4F9step 1.1

Fix a Borel set B⊆V and put E=T−1(B). By [F4], E is Lebesgue measurable. The function h(u)=pU(u)1E(u) is nonnegative measurable by [F9]. Apply [F3] to S=T−1:V→U and h to obtain μX(B)=∫EpU(u) dλd(u)=∫VpU(T−1(x))1B(x)∣det⁡D(T−1)(x)∣ dλd(x)=∫Bq(x) dλd(x). Therefore q is a density version of μX.

3.1F5F6F8step 1.2algebra∎

By [F6], determinants of each derivative composition are the products of the stage determinants. Each factor is nonzero: differentiating Tj−1∘Tj=id⁡ by [F5] gives a product equal to the identity matrix, whose determinant is 1 by [F8]. Taking absolute values and then real logarithms yields both displayed additive formulas.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Positive densities with smooth conditional CDFs admit an exact triangular-flow representation

Statement

Assume the Axiom of Countable Choice. Let d≥1 and let p:Rd→(0,∞) be a Lebesgue probability density. For i=1,…,d, write x<i=(x1,…,xi−1), with x<1 the unique element of R0. For each i, let qi:R×Ri−1→(0,∞) be a positive continuous function, written qi(t∣x<i), and suppose p(x)=∏i=1dqi(xi∣x<i),∫Rqi(t∣x<i) dt=1 for every x∈Rd and every preceding-coordinate vector x<i. Define the conditional cumulative distribution functions Fi(t∣x<i)=∫−∞tqi(s∣x<i) ds. Assume each Fi is jointly C1 in (t,x<i), satisfies ∂tFi(t∣x<i)=qi(t∣x<i), and, for every fixed x<i, the slice t↦Fi(t∣x<i) is a strictly increasing bijection from R onto (0,1). In each inverse below, Fi−1 means the inverse in its first argument with the preceding coordinates held fixed.

Then there is a global C1 diffeomorphism T:(0,1)d→Rd whose ith output coordinate depends only on the first i input coordinates and which maps the uniform density 1 on (0,1)d to the density p. Its inverse is Si(x)=Fi(xi∣x<i), and T=S−1 is determined recursively by x1=F1−1(u1),xi=Fi−1(ui∣x<i)(i=2,…,d). This is an exact representational existence result; it makes no claim about finite-network approximation, trainability, or numerical performance.

Statement references

Facts & Assumptions

Given: Countable Choice, the density p, the factorization by the positive conditional density factors qi, and the stated smoothness and bijectivity of each Fi.

[A1]

The Axiom of Countable Choice, written ACω, asserts that every sequence of nonempty sets has a choice function. The Axiom of Countable Choice (ACω)

[F1]

A normalizing flow is a finite composition of global C1 diffeomorphisms between open equal-dimensional Euclidean domains; its base density defines a measure by integration. A normalizing flow and its base density

[F2]

A nonnegative measurable density f defines the measure A↦∫Af dλd. The measure with density f relative to μ

[F3]

The determinant of a triangular matrix is the product of its diagonal entries. The determinant of a triangular matrix is the product of its diagonal entries

[F4]

If f:U→Rd is C1 and Df(a) is invertible, then f has a C1 local inverse near a. The Euclidean inverse function theorem

[F5]

Under Countable Choice, a C1 diffeomorphism transforms a nonnegative Lebesgue density r by the formula r∘T−1 ∣det⁡D(T−1)∣. A diffeomorphic flow has the inverse-Jacobian density

[F6]

The Jacobian matrix is the matrix of the total derivative in standard bases, and its Jacobian determinant is the determinant of that matrix. The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, The Jacobian determinant of a square-dimensional C1 map is the determinant of its Jacobian matrix

[F7]

The open box (0,1)d is Lebesgue measurable and has measure ∏i=1d(1−0)=1 under Countable Choice. A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included

Proof

technique · direct
1.1givenF3F6

Define S:Rd→(0,1)d by Si(x)=Fi(xi∣x<i). Joint C1 regularity makes S C1. Each Si depends only on x1,…,xi, so DS(x) is lower triangular; its ith diagonal entry is ∂xiFi(xi∣x<i)=qi(xi∣x<i)>0. By [F3] and the factorization hypothesis, det⁡DS(x)=∏i=1dqi(xi∣x<i)=p(x)>0.

1.2givenconstruct

For any u∈(0,1)d, choose x1=F1−1(u1), then recursively choose the unique xi=Fi−1(ui∣x<i). The slice bijections make every coordinate exist and be unique, and construction gives S(x)=u. Conversely, any x with S(x)=u must have these same coordinates by induction. Hence S is bijective, and the recursive formula shows that T=S−1 is triangular.

2.1F4step 1.1step 1.2

By step 1.1, DS(x) is invertible at every x. Applying [F4] at each point gives a C1 local inverse; because S is globally bijective, each local inverse agrees with the same set-theoretic map S−1. Therefore S−1 is globally C1, so T is a global C1 diffeomorphism.

3.1A1F1F2F5F7step 1.1step 2.1∎

By [F7], density 1 on (0,1)d is a probability density. Apply [F5] to T=S−1: the output density at x is 1⋅∣det⁡DS(x)∣=p(x) by step 1.1. Thus the uniform base pushes forward to the target density p, as claimed. Countable Choice is used here through [F5] and [F7]; the recursive inverse construction uses only uniqueness of each conditional quantile.

4 · Examples, counterexamples and false statements

None yet.

Sources