ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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.

Visible RBM units are conditionally independent but dependent marginally

Example

Assume AC. In a binary restricted Boltzmann machine with two visible units, one hidden unit, $a = (0, 0)$ , $b = 0$ and $W_{11} = W_{21} = \log 3$ , the visible units are independent conditional on the hidden unit. Each conditional success probability is $1 / 2$ when $H = 0$ and $3 / 4$ when $H = 1$ . Marginally, $P (V_{1} = 1) = P (V_{2} = 1) = 7 / 10$ and $P (V_{1} = 1, V_{2} = 1) = 1 / 2 \neq 49 / 100$ , so $V_{1}$ and $V_{2}$ are dependent.

Example references

The Elements of Statistical Learning, §17.4.4

Hinton, A Practical Guide to Training Restricted Boltzmann Machines, §2

Facts & Assumptions

Given: The RBM parameters and binary state spaces in the Example.

[F1]

The binary RBM conditionals are products of Bernoulli laws with success probabilities $σ (b_{j} + (W^{T} v)_{j})$ and $σ (a_{i} + (W h)_{i})$ . Binary restricted Boltzmann machine, exact likelihood gradient, and finite contrastive divergence.

[F2]

Conditional independence given a sigma-algebra is the bounded-function conditional-expectation factorization. Conditional independence given a sigma-algebra.

[A1]

AC is the explicit hypothesis of the conditional-independence convention, and here means every family of nonempty sets has a choice function. Conditional independence given a sigma-algebra The Axiom of Choice.

Verification

technique · direct

1.1A1F1F2algebra

With $a_{i} = 0$ , $b = 0$ and $W_{i 1} = \log 3$ , [F1] gives $P (V_{i} = 1 ∣ H = h) = σ (h \log 3)$ , which is $1 / 2$ at $h = 0$ and $3 / 4$ at $h = 1$ . The conditional visible mass factors as the product of these two Bernoulli masses, so [A1] and [F2] give conditional independence.

2.1givenstep 1.1algebra

Summing the unnormalized joint mass over the two visible bits gives weight $(1 + 1)^{2} = 4$ to $h = 0$ and $(1 + 3)^{2} = 16$ to $h = 1$ . Thus $Z = 20$ and $P (H = 0) = 1 / 5$ , $P (H = 1) = 4 / 5$ .

3.1step 1.1step 2.1algebra∎

Averaging the conditional success probabilities from step 1.1 yields $P (V_{i} = 1) = \frac{1}{5} \cdot \frac{1}{2} + \frac{4}{5} \cdot \frac{3}{4} = \frac{7}{10}$ for each i. Likewise, conditional independence gives $P (V_{1} = 1, V_{2} = 1) = \frac{1}{5} \cdot \frac{1}{4} + \frac{4}{5} \cdot \frac{9}{16} = \frac{1}{2}$ , whereas the product of the marginals is $\frac{49}{100}$ . Since these differ, the visible variables are marginally dependent.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

The Elements of Statistical Learning, second edition (standard reference, not scraped)
Geoffrey Hinton, A Practical Guide to Training Restricted Boltzmann Machines (standard reference, not scraped)