Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Hydrodynamic fluctuation action and nonequilibrium quasipotential — Examples

1 · Prerequisites

None. This page is self-contained.

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Two-site boundary-driven SSEP has a nonproduct stationary law for unequal reservoir rates

Example

Fix $α, β \in (0, 1)$ and set $Ω = {0, 1}^{2}$ . For $η = (η_{1}, η_{2}) \in Ω$ , let $η_{i}$ denote its $i$ th coordinate. Time is measured in units of the bond-swap rate, so that this rate is $1$ . For every real function $g : Ω \to R$ , define the generator $L$ acting on observables by

(L g) (η) : = α (1 - η_{1}) (g (1, η_{2}) - g (η)) + (1 - α) η_{1} (g (0, η_{2}) - g (η)) + β (1 - η_{2}) (g (η_{1}, 1) - g (η)) + (1 - β) η_{2} (g (η_{1}, 0) - g (η)) + 1_{{η_{1} \neq η_{2}}} (g (η_{2}, η_{1}) - g (η)) .

Here $1_{A}$ is $1$ when the condition $A$ holds and $0$ otherwise. This rate formula specifies the two reservoir flips and the bond swap of the finite-state continuous-time chain. Let $P (Ω)$ be the set of probability mass functions $π : Ω \to [0, 1]$ , so $\sum_{η \in Ω} π (η) = 1$ . A law $π \in P (Ω)$ is stationary when $\sum_{η \in Ω} π (η) (L g) (η) = 0$ for every real $g : Ω \to R$ . Define $E_{π} [g] : = \sum_{η \in Ω} π (η) g (η)$ , $m_{i} : = E_{π} [η_{i}]$ , and

${Cov}_{π} (η_{1}, η_{2}) : = E_{π} [η_{1} η_{2}] - m_{1} m_{2} .$

For $i = 1, 2$ , the marginal is $π_{i} (a) : = \sum_{η \in Ω : η_{i} = a} π (η)$ for $a \in {0, 1}$ ; $π$ is a product law when $π (a, b) = π_{1} (a) π_{2} (b)$ for all $a, b \in {0, 1}$ . The unique stationary law of this chain has

${Cov}_{π} (η_{1}, η_{2}) = - \frac{(α - β)^{2}}{18} .$

Consequently, the stationary law is not a product when $α \neq β$ . This is an exact finite-chain correlation example, not a derivation of the continuum quasi-potential.

Transition graph

Each vertex is labelled by $(η_{1}, η_{2})$ . Horizontal arrows change site $1$ by a reservoir flip, vertical arrows change site $2$ , and the two diagonal arrows are the bond swap. An arrow label is its transition rate; the diagram omits events that leave the state unchanged. The generator and proof below are complete without relying on the diagram.

Facts & Assumptions

Given: $α, β \in (0, 1)$ and the four-state generator specified in the Example.

[A1]

$L$ is the generator defined in the Example by the rate-weighted increments, and $π$ is stationary in the stated sense $\sum_{η \in Ω} π (η) (L g) (η) = 0$ for every real function $g$ on $Ω$ .

Verification

1.1A1algebra

For the coordinate observables, the generator gives $L η_{1} = α + η_{2} - 2 η_{1}$ and $L η_{2} = β + η_{1} - 2 η_{2}$ . If $π$ is stationary and $m_{i} : = E_{π} η_{i}$ , then $E_{π} L η_{i} = 0$ , so $2 m_{1} - m_{2} = α$ and $2 m_{2} - m_{1} = β$ . Solving gives $m_{1} = (2 α + β) / 3$ and $m_{2} = (α + 2 β) / 3$ .

2.1A1step 1.1algebra

Let $q : = E_{π} (η_{1} η_{2})$ . The swap does not change $η_{1} η_{2}$ , and the two reservoir flips give $L (η_{1} η_{2}) = α η_{2} + β η_{1} - 2 η_{1} η_{2}$ . Stationarity therefore yields $2 q = α m_{2} + β m_{1}$ , hence $q = (α^{2} + 4 α β + β^{2}) / 6$ .

3.1step 1.1step 2.1algebra

Substituting the moments from steps 1.1 and 2.1 gives $q - m_{1} m_{2} = \frac{α^{2} + 4 α β + β^{2}}{6} - \frac{(2 α + β) (α + 2 β)}{9} = - \frac{(α - β)^{2}}{18}$ . If $α \neq β$ , this covariance is nonzero, whereas two independent Bernoulli occupations have zero covariance; thus $π$ is not a product law.

3.2step 1.1step 2.1algebra

Define probabilities by $π_{11} = q$ , $π_{10} = m_{1} - q$ , $π_{01} = m_{2} - q$ , and $π_{00} = 1 - m_{1} - m_{2} + q$ . They sum to one; $π_{11} > 0$ and $π_{00} = [(1 - α)^{2} + 4 (1 - α) (1 - β) + (1 - β)^{2}] / 6 > 0$ . For fixed $β$ , $6 π_{10} = g (α)$ where $g (a) = a (4 - 4 β - a) + β (2 - β)$ is concave on $[0, 1]$ and has positive endpoint values $g (0) = β (2 - β)$ and $g (1) = (1 - β) (3 + β)$ ; hence $π_{10} > 0$ . By interchanging $α$ and $β$ the same argument gives $π_{01} > 0$ .

4.1A1step 1.1step 2.1step 3.1step 3.2algebra∎

The probabilities in step 3.2 have the moments in steps 1.1 and 2.1, so their expectation of $L$ vanishes on $1, η_{1}, η_{2}, η_{1} η_{2}$ . These four functions form a basis of all real functions on $Ω$ , hence the probability vector is stationary. Conversely, every stationary law must satisfy the same three moment equations, which determine $m_{1}, m_{2}, q$ uniquely and then determine all four atom probabilities; therefore this stationary law is unique. If $α = β = r$ , the same probabilities reduce to $π_{00} = (1 - r)^{2}$ , $π_{10} = π_{01} = r (1 - r)$ , and $π_{11} = r^{2}$ , the product Bernoulli law. Step 3.1 gives the claimed covariance.

5 · Examples, experiments and counterexamples

None yet.