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Point masses, Jensen–Shannon saturation, and weight clipping

Example

Let X=[0,1] with distance d(x,y)=∣x−y∣, let P=δ0, and for t∈[0,1] let Qt=δt. For each t>0, let Ft={0,t} with the induced finite metric and ordering (0,t), and use probability vectors p=(1,0) and q=(0,1); for t=0, take F0={0} and p=q=(1). For finite-support probabilities R,S, define KL⁡(R∥S)=+∞ if R gives positive mass where S gives zero mass, and otherwise define KL⁡(R∥S)=∑x:R(x)>0R(x)log⁡(R(x)/S(x)), with natural logarithms. Define JS⁡(R,S)=12KL⁡(R∥(R+S)/2)+12KL⁡(S∥(R+S)/2). For every t>0, the finite-space transport cost W1(p,q) on Ft, the compact-space coupling cost W1(P,Qt), and the integral probability metric over all real functions satisfying ∣f(x)−f(y)∣≤∣x−y∣ on X are all t, whereas JS⁡(P,Qt)=log⁡2; at t=0 these three transport/IPM values and the JS value are zero. Thus the compact-space W1(P,Qt)→0 as t↓0 while JS⁡(P,Qt) stays equal to log⁡2 for every t>0.

For the same data space, consider the two-layer scalar affine critic fa,b(x)=a(bx) whose two scalar weights are constrained by ∣a∣,∣b∣≤c for a fixed c>0. When c>1, the clipped family contains f(x)=c2x, which is not 1-Lipschitz; when 0<c<1, it cannot contain f(x)=x; when c=1, it cannot contain the nonlinear 1-Lipschitz function f(x)=x2/2. Hence this fixed clipped affine family is not exactly the class of 1-Lipschitz critics for any c>0.

Facts & Assumptions

Given: X=[0,1] with its usual metric, P=δ0, Qt=δt for 0≤t≤1, the finite-support definitions of KL⁡ and JS⁡ in the Example, and the critic family fa,b(x)=abx with ∣a∣,∣b∣≤c.

[F1]

A Dirac set function assigns mass 1 exactly to measurable sets containing its point, and it is a probability measure. The Dirac set function at a point A Dirac set function is a probability measure

[F2]

For a nonnegative measurable function g, ∫g dδx=g(x). Integration against a Dirac measure evaluates at its point

[F3]

For finite probability vectors p,q, W1(p,q) is the minimum of ∑i,jd(xi,xj)πij over nonnegative coupling matrices with row sums p and column sums q. The finite-metric 1-Wasserstein transport cost

[F4]

W1 on a compact metric space is the infimum of ∫d(x,y) dγ(x,y) over Borel probability couplings with the prescribed marginals. The coupling definition of the 1-Wasserstein cost on a compact metric space

[F5]

The IPM is the supremum of the absolute differences of the test-function integrals over its specified class. Integral probability metrics

[F6]

A measure is countably additive on pairwise disjoint measurable sets; finite unions of null sets are null. Measures on sigma-algebras

Verification

1.1givenF3algebra

For t>0, the second row sum 0 and nonnegative entries force π21=π22=0; the first column sum 0 then forces π11=0, and the first row sum 1 forces π12=1, which also gives the required second column sum 1. This is the unique feasible matrix, and [F3] gives finite-space cost d(0,t)=t. For t=0, the singleton probability vector has unique coupling (1) and cost 0.

1.2givenF1F2F4F6algebra

By [F1], P and Qt are Borel probability measures. If γ is any coupling of P and Qt, then γ({0}×X)=1 and γ(X×{t})=1. The complement of their intersection is contained in the union of their two null complements, so finite subadditivity from [F6] gives γ({(0,t)})=1. A probability measure concentrated on the singleton (0,t) equals its Dirac measure there: sets containing (0,t) have mass 1, and sets omitting it have mass 0. The Dirac measure at (0,t) has the required marginals, so it is the unique coupling. By [F2] and [F4], its cost is d(0,t)=t, so W1(P,Qt)=t, including t=0.

1.3givenF5algebra

Every function in the specified 1-Lipschitz class satisfies ∣f(0)−f(t)∣≤t, and f(x)=x belongs to the class and has ∣f(0)−f(t)∣=t. By [F5], the IPM is exactly t.

1.4givenalgebra

If t>0, the mixture (P+Qt)/2 assigns mass 1/2 to each of 0 and t. The finite sum definition gives KL⁡(P∥(P+Qt)/2)=log⁡2 and KL⁡(Qt∥(P+Qt)/2)=log⁡2, so JS⁡(P,Qt)=log⁡2. If t=0, the two laws agree, each KL term is 0, and the JS value is 0.

2.1givenalgebra∎

Under the weight bounds, the slope of fa,b(x)=abx has absolute value at most c2. If c>1, the allowed choice a=b=c gives slope c2>1. If 0<c<1, every allowed slope has magnitude less than 1, so the function x cannot occur. If c=1, every member is linear, while x2/2 is 1-Lipschitz because ∣x2−y2∣/2=∣x−y∣(x+y)/2≤∣x−y∣ on [0,1]; it is not linear, since its value at 1 is 1/2 and at 1/2 is 1/8. These three exhaustive cases prove the stated failure of exactness for every c>0.

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