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A unit-scale affine coupling has a triangular Jacobian

Example

Assume the Axiom of Countable Choice and let $p_{U}$ be a Lebesgue probability density on $R^{2}$ . Define the unit-scale affine coupling $T : R^{2} \to R^{2}$ by $T (u, v) = (u, u + v) .$ It is a global $C^{1}$ diffeomorphism with inverse $T^{- 1} (x, y) = (x, y - x)$ . Its Jacobian matrix is lower triangular, $D T (u, v) = (\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}), \det D T (u, v) = 1.$ The transformed model has density $p_{X} (x, y) = p_{U} (x, y - x)$ . The first coordinate is unchanged and the second is shifted by the first.

Example references

Papamakarios et al., §3.1.1, equations (32)–(33), PDF pp. 12–13

Sheldon, COMPSCI 688 Lecture 23, Real-NVP, PDF pp. 5–6

Facts & Assumptions

Given: Countable Choice and a Lebesgue probability density $p_{U}$ on $R^{2}$ .

[A1]

The Axiom of Countable Choice, written ${A C}_{ω}$ , asserts that every sequence of nonempty sets has a choice function. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

A $C^{1}$ map is totally differentiable at every point with continuous derivative matrix; a global $C^{1}$ diffeomorphism has a $C^{1}$ inverse, and a flow transports its base measure by $μ_{X} (B) = μ_{U} (T^{- 1} (B))$ . A normalizing flow and its base density

[F2]

Under Countable Choice, a $C^{1}$ diffeomorphism transforms density $p_{U}$ by $p_{U} \circ T^{- 1} ∣ \det D (T^{- 1}) ∣$ . 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 $C^{1}$ 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 $\hat{T} (x, y) = (x, y - x)$ . Direct substitution gives $\hat{T} (T (u, v)) = (u, v)$ and $T (\hat{T} (x, y)) = (x, y)$ . Both maps are linear, so their first-order remainders vanish and their derivative matrices are constant; by [F1] they are $C^{1}$ . Thus they are inverse bijections and $T$ is a global $C^{1}$ diffeomorphism.

1.2givenF3F4algebra

The linear map $T (u, v) = (u, u + v)$ has matrix $(\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array})$ , which is its Jacobian matrix by [F3]. It is lower triangular, so [F4] gives determinant $1 \cdot 1 = 1$ . Its inverse has Jacobian matrix $(\begin{array}{cc} 1 & 0 \\ - 1 & 1 \end{array})$ , also lower triangular with determinant $1$ .

2.1A1F1F2step 1.1step 1.2∎

Apply [F2] to the base density $p_{U}$ and the inverse from step 1.1. Step 1.2 gives $∣ \det D (T^{- 1}) (x, y) ∣ = 1$ , so the transformed density version is $p_{X} (x, y) = p_{U} (x, y - x)$ .

Depends on

Used by

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Sources

George Papamakarios, Eric Nalisnick, Danilo Jimenez Rezende, Shakir Mohamed, and Balaji Lakshminarayanan, Normalizing Flows for Probabilistic Modeling and Inference, arXiv:1912.02762v1 (standard reference, not scraped)
Dan Sheldon, COMPSCI 688: Probabilistic Graphical Models, Lecture 23: Normalizing Flows (annotated course notes) (standard reference, not scraped)