ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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.

Two-spin pseudo-likelihood is not the exact joint posterior input

Example

Take $Λ = {1, 2}$ , $S = {- 1, + 1}$ , and any fixed exterior boundary; the sole interaction set ${1, 2}$ has no exterior sites, so its energy is independent of that boundary. Use parameter space $Θ = {0, 1}$ with its full power-set sigma-algebra, prior mass $1 / 2$ at each parameter, and the single potential $Φ_{{1, 2}} ((σ_{1}, σ_{2}); J) = - J σ_{1} σ_{2}$ . Use the real exponential $\exp$ , put $e : = \exp (1)$ , and define $\cosh J : = (\exp (J) + \exp (- J)) / 2$ . For the hypothetical observation $(+ 1, + 1)$ , the exact joint likelihood is $\exp (J) / (4 \cosh J)$ , whereas the product of the two one-site Gibbs conditional probabilities is $[\exp (J) / (2 \cosh J)]^{2}$ . By pseudo-posterior I mean the probability vector obtained by multiplying this product by the prior weights and normalizing over $J \in {0, 1}$ . The exact and pseudo-posterior distributions are different.

Example references

Finite-volume Gibbs pseudo-likelihood with fixed boundary

The real exponential function and the number $e$ by a power series

The exponential function is strictly increasing

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$

Facts & Assumptions

Given: The two-spin finite-volume Gibbs model, the equal prior on $J \in {0, 1}$ , and the fixed hypothetical observation $(+ 1, + 1)$ .

[A1]

The observation is a mathematical example and is not reported experimental data.

[F1]

Finite-volume Gibbs pseudo-likelihood with fixed boundary: pseudo-likelihood is the product of the exact one-site conditional probabilities for the fixed interaction neighborhood and boundary.

[F2]

The real exponential function and the number $e$ by a power series: $\exp (x) = \sum_{n = 0}^{\infty} x^{n} / n!$ and $e = \exp (1)$ ; evaluating the series at zero leaves its constant term, so $\exp (0) = 1$ .

[F3]

The exponential function is strictly increasing: $\exp$ is strictly increasing on $R$ , so $e = \exp (1) > \exp (0) = 1$ .

[F4]

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ : $\exp (x) > 0$ and $\exp (- x) = 1 / \exp (x)$ for every real $x$ .

Verification

Given: The data and two-parameter prior in the Example.

1.1A1F1F2algebra

The four unnormalized Gibbs weights are $\exp (J), \exp (- J), \exp (- J), \exp (J)$ , so the partition function is $Z_{J} = 2 \exp (J) + 2 \exp (- J) = 4 \cosh J$ . Therefore the joint likelihood of $(+ 1, + 1)$ is $\exp (J) / (4 \cosh J)$ .

2.1F1F2step 1.1algebra

Given either other spin is $+ 1$ , the conditional probability for the remaining spin to be $+ 1$ is $\exp (J) / (\exp (J) + \exp (- J)) = \exp (J) / (2 \cosh J)$ . Thus the pseudo-likelihood is the product $[\exp (J) / (2 \cosh J)]^{2}$ .

3.1A1F1F2F3F4step 1.1step 2.1algebra

The exact likelihood ratio for $J = 1$ against $J = 0$ is $r = e / \cosh 1$ , while the pseudo-likelihood ratio is $r^{2}$ . By [F3] and [F4], $e > 1$ and $e^{- 1} = 1 / e < 1$ , so $\cosh 1 = (e + e^{- 1}) / 2 < e$ . Hence $r > 1$ and $r \neq r^{2}$ . Under the equal prior, posterior odds equal likelihood ratios, so the exact and pseudo posterior probabilities differ. The pseudo-likelihood product therefore cannot be substituted as the joint likelihood for this model.

4.1F1F2F3F4step 1.1step 2.1algebra∎

Put $c = \cosh 1$ . For each equal-spin configuration, both site conditionals are $e / (2 c)$ ; for each opposite-spin configuration, both are $\exp (- 1) / (2 c) = 1 / (2 e c)$ by [F1], [F2], and [F4]. Summing the four products gives $\sum_{σ \in S^{Λ}} P L_{1} (σ) = \frac{e^{2} + (1 / e)^{2}}{2 c^{2}} = 2 \frac{(e + 1 / e)^{2} - 2}{(e + 1 / e)^{2}} = 2 - \frac{1}{c^{2}} .$ By [F3] and [F4], $e > 1$ and $e^{- 1} = 1 / e$ , so $c - 1 = (e - 1)^{2} / (2 e) > 0$ ; hence the sum is greater than one. Thus the pseudo-likelihood is not a probability mass function on the configuration space, whereas the joint Gibbs likelihood is.

Depends on

Dependency tree · 2 levels

Sources

Julian Besag, Statistical Analysis of Non-Lattice Data, The Statistician 24(3) (1975), 179–195