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Exact-density normalizing flows — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A unit-scale affine coupling has a triangular Jacobian

Example

Assume the Axiom of Countable Choice and let pU be a Lebesgue probability density on R2. Define the unit-scale affine coupling T:R2→R2 by T(u,v)=(u,u+v). It is a global C1 diffeomorphism with inverse T−1(x,y)=(x,y−x). Its Jacobian matrix is lower triangular, DT(u,v)=(1011),det⁡DT(u,v)=1. The transformed model has density pX(x,y)=pU(x,y−x). The first coordinate is unchanged and the second is shifted by the first.

Facts & Assumptions

Given: Countable Choice and a Lebesgue probability density pU on R2.

[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 C1 map is totally differentiable at every point with continuous derivative matrix; a global C1 diffeomorphism has a C1 inverse, and a flow transports its base measure by μX(B)=μU(T−1(B)). A normalizing flow and its base density

[F2]

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

[F3]

The Jacobian matrix records the total derivative in standard coordinates, and the Jacobian determinant is its determinant. 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

[F4]

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

Verification

Given: The objects and hypotheses in the Example.

1.1givenF1constructalgebra

Define T^(x,y)=(x,y−x). Direct substitution gives T^(T(u,v))=(u,v) and T(T^(x,y))=(x,y). Both maps are linear, so their first-order remainders vanish and their derivative matrices are constant; by [F1] they are C1. Thus they are inverse bijections and T is a global C1 diffeomorphism.

1.2givenF3F4algebra

The linear map T(u,v)=(u,u+v) has matrix (1011), which is its Jacobian matrix by [F3]. It is lower triangular, so [F4] gives determinant 1⋅1=1. Its inverse has Jacobian matrix (10−11), also lower triangular with determinant 1.

2.1A1F1F2step 1.1step 1.2∎

Apply [F2] to the base density pU and the inverse from step 1.1. Step 1.2 gives ∣det⁡D(T−1)(x,y)∣=1, so the transformed density version is pX(x,y)=pU(x,y−x).

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A full-dimensional flow cannot represent a line-supported law exactly

Example

Assume the Axiom of Countable Choice. Let ν be the probability measure on [0,1] with density 1 relative to one-dimensional Lebesgue measure, let i:[0,1]→R2 be i(t)=(t,0), and define a measure on Borel sets B⊆R2 by μ(B)=ν(i−1(B)). This probability measure is concentrated on the line segment M=[0,1]×{0} and satisfies μ(M)=1. It is not the output of any normalizing flow T:R2→R2 whose base measure has a Lebesgue density.

Facts & Assumptions

Given: Countable Choice, the interval probability measure ν, the continuous embedding i(t)=(t,0), and its image measure μ.

[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 nonnegative measurable density f defines a measure by A↦∫Af dλ. The measure with density f relative to μ

[F2]

A flow from a Lebesgue density through a global C1 diffeomorphism has density pU∘T−1∣det⁡D(T−1)∣. A diffeomorphic flow has the inverse-Jacobian density

[F3]

A continuous map has Borel preimages of Borel sets. A continuous map has Borel preimages of Borel sets

[F4]

Every closed box [a,b] in Rn is Borel and Lebesgue measurable; its measure is ∏i<n(bi−ai), so [0,1] has one-dimensional Lebesgue measure 1. A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included

[F5]

A rectangle with a degenerate side is Lebesgue null under Countable Choice. A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in Rn

[F6]

A nonnegative measurable density integrates to zero over a measurable null set. A nonnegative integral over a null set vanishes

[F7]

A normalizing flow is a finite composition of global C1 diffeomorphisms between open equal-dimensional Euclidean domains; its output measure is obtained by taking preimages under the flow. A normalizing flow and its base density

Verification

Given: Countable Choice, the image measure μ of ν under i, and an arbitrary normalizing flow T:R2→R2 with Lebesgue base density.

1.1givenA1F1F3F4

The function i is continuous, so [F3] makes i−1(B) Borel for every Borel B⊆R2. For pairwise disjoint Borel sets Bn, their preimages are pairwise disjoint and i−1(⋃nBn)=⋃ni−1(Bn); countable additivity of ν therefore makes μ a measure. By [F1] and [F4], ν([0,1])=∫[0,1]1 dλ1=λ1([0,1])=1, whence μ(R2)=1 and μ(M)=ν([0,1])=1.

1.2givenA1F4F5

The set M=[0,1]×{0}=[0,1]×[0,0] is a closed box, hence Borel by [F4]. Its second side is degenerate, so [F5] gives λ2(M)=0.

1.3givenA1F1F2F6F7

By [F7] and [F2], the flow output has a nonnegative Lebesgue density q on R2. Applying [F1] and [F6] to the null set M gives μX(M)=∫Mq dλ2=0.

2.1givenstep 1.1step 1.3∎

The target measure satisfies μ(M)=1 by step 1.1, whereas every such flow output satisfies μX(M)=0 by step 1.3. Therefore μX≠μ, so no full-dimensional flow in the stated class represents this target law exactly.

Sources