LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generated

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Finite masking path law factors backward by its posterior rows

Statement

Fix a clean string $x \in V^{L}$ and let $q_{x}$ be the finite forward path law of the absorbing-mask chain. For every path $(z_{0}, \dots, z_{N})$ with $q_{x} (z_{0}, \dots, z_{N}) > 0$ , the reversed path probability factors as $q_{x} (z_{0}, \dots, z_{N}) = q_{x} (Z_{N} = z_{N}) \prod_{j = N}^{1} q_{x} (Z_{j - 1} = z_{j - 1} ∣ Z_{j} = z_{j}) .$ Each factor on the right is the true one-step posterior conditional on the clean string $x$ ; the identity is asserted only on positive-probability paths.

Statement references

Finite masking diffusion chain: the conditional forward path mass is the product of its one-step transition rows.

Posterior of a finite masking chain: gives the finite conditional law of each preceding state given $x$ and the current state.

Facts & Assumptions

Given: A fixed clean string $x$ , the finite Markov path mass, and a positive-probability path $z_{0}, \dots, z_{N}$ . Prove the reversed factorization.

[F1]

Finite masking diffusion chain: conditional on $x$ , the path mass is $1_{{z_{0} = x}} \prod_{j = 1}^{N} a_{j} (z_{j - 1}, z_{j})$ , with each transition depending only on its preceding state.

[F2]

Finite masking-chain transitions define a probability law: these path masses form a probability law on the finite path space.

[F3]

Posterior of a finite masking chain: for a positive-probability current state, the conditional law of $Z_{j - 1}$ given $x, Z_{j}$ is the stated posterior row.

Proof

1.1F1F2givenalgebra

Fix $j$ and a positive-probability suffix $(z_{j}, \dots, z_{N})$ . By [F1], for every prefix ending at $z_{j}$ , the probability of adjoining the fixed future suffix is multiplied by the same factor $\prod_{k = j + 1}^{N} a_{k} (z_{k - 1}, z_{k})$ , which does not depend on $z_{j - 1}$ . Summing over earlier coordinates preserves this common factor, so it cancels in the conditional ratio. Thus $q_{x} (Z_{j - 1} = z_{j - 1} ∣ Z_{j} = z_{j}, \dots, Z_{N} = z_{N}) = q_{x} (Z_{j - 1} = z_{j - 1} ∣ Z_{j} = z_{j})$ on this positive suffix.

2.1F2F3step 1.1algebra∎

Repeated finite conditioning in reverse order gives $q_{x} (z_{0 : N}) = q_{x} (z_{N}) \prod_{j = N}^{1} q_{x} (z_{j - 1} ∣ z_{j}, \dots, z_{N})$ . Replace each suffix conditional by step 1.1 and identify the one-step row using [F3]. The resulting equality is the displayed factorization, including $N = 1$ .

Depends on

Used by

Negative expected finite masked-diffusion ELBO is weighted cross-entropy Corollary

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Subham Sekhar Sahoo et al., Simple and Effective Masked Diffusion Language Models, arXiv v1 (standard reference, not scraped)