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Finite reversible stochastic rate model and thermal coupling — Examples

1 · Prerequisites

2 · Summary

A three-state reversible rate model shows how hiding microscopic states can destroy a single coarse Markov law valid for every microscopic preparation. The counterexample checks the Gibbs weights, transition rates and unequal exit rates directly.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Reversible finite chain with hidden-state exit-rate obstruction

Statement refuted

Every finite reversible stochastic rate model induces, for every partition of its states, one time-homogeneous Markov law for the block process that is valid for all microscopic initial preparations.

Facts & Assumptions

Given: The Axiom of Choice, a three-state rate model, and the partition below.

[A1]

The Axiom of Choice: every family of nonempty sets has a choice function.

[F1]

Finite reversible stochastic rate model: a valid model has a finite state set, positive Gibbs weights, connected bidirectional positive-rate edges, and rates satisfying detailed balance.

[F2]

Nonexplosive finite continuous-time chain and reversible spectral gap: the transition matrix is $P_{t} = e^{t Q}$ and its finite matrix exponential has the expansion $P_{t} = I + t Q + O (t^{2})$ as $t ↓ 0$ .

[F3]

Exact all-preparations coarse Markov criterion: a single time-homogeneous block generator works for every microscopic initial law exactly when each outgoing block-rate sum is constant within its source block.

Counterexample

Given: [A1], [F1]–[F3]. Take $E = {1, 2, 3}$ , $e_{1} = e_{2} = e_{3} = 0$ , $β = 1$ , and rates $w_{12} = w_{21} = 1, w_{23} = w_{32} = 1, w_{13} = w_{31} = 0.$ Let $C_{1} = {1, 2}$ and $C_{2} = {3}$ .

1.1F1constructalgebra

The positive-rate graph is the connected path $1 \leftrightarrow 2 \leftrightarrow 3$ . Equal state energies give $π = (1 / 3, 1 / 3, 1 / 3)$ , and every positive edge has equal forward and reverse rates, so $π_{a} w_{a b} = π_{b} w_{b a}$ . Thus [F1] holds.

1.2A1F1F2algebra

The exit rate from state $1$ to $C_{2}$ is $q (1, C_{2}) = w_{13} = 0$ , while from state $2$ it is $q (2, C_{2}) = w_{23} = 1$ . Starting with observed block $C_{1}$ , [F2] gives $\Pr_{1} ({\overset{ˉ}{X}}_{h} = C_{2}) = O (h^{2})$ and $\Pr_{2} ({\overset{ˉ}{X}}_{h} = C_{2}) = h + O (h^{2})$ , because a transition from $1$ to $3$ needs at least two jumps, while the direct $2 \to 3$ rate is $1$ . The two preparations have the same initial block but different first-order block-transition probabilities.

2.1A1F2F3step 1.1step 1.2∎

Therefore no single transition law on ${C_{1}, C_{2}}$ can govern the projection for both microscopic preparations. Equivalently, the required block-rate constancy in [F3] fails. This refutes the universal all-preparations claim; it makes no assertion about special initial preparations.

Remarks

The following graph displays the same finite model. Each double arrow denotes two directed transitions, one in each direction, with the rate label shown; the missing $1$ – $3$ edge means both corresponding rates are zero. Node labels are the states, not coordinates of an observed trajectory.

The proof uses the explicitly listed rates and generator expansion, not the drawing.

Sources

David Aldous and James Allen Fill, Reversible Markov Chains and Random Walks on Graphs