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A factorized reverse kernel fails for correlated tokens

Statement refuted

For every finite clean-sequence law and every finite absorbing-mask schedule with independent forward masking conditional on the clean sequence, the exact reverse conditional law of the earlier sequence given a later masked state factorizes over positions.

Statement references

Posterior of a finite masking chain: its product posterior is conditional on the clean sequence; marginalizing a dependent clean-sequence law need not preserve the product form.

Facts & Assumptions

Given: $V = {0, 1}$ , $L = 2$ , $N = 1$ , $α_{0} = 1$ , $α_{1} = 0$ , and $\Pr (X = (0, 0)) = \Pr (X = (1, 1)) = 1 / 2$ . The forward corruption is the finite absorbing-mask chain, and the proposed learned reverse sampler uses independent position draws with $h_{1, 1} (m^{2}) = h_{2, 1} (m^{2}) = (1 / 2, 1 / 2)$ .

[F1]

Finite masking diffusion chain: at terminal survival $α_{1} = 0$ , every clean token is masked, while the clean sequence itself retains its original law.

[F2]

Posterior of a finite masking chain: for the endpoint pair $s = 0, t = 1$ , the true reverse conditional given $X$ and the all-mask state returns the clean string with probability one.

[F3]

Finite-grid masked reverse sampler: at $c_{1} = 1$ , the learned sampler draws each masked coordinate independently from its position predictor.

Counterexample

Given: The finite model and probability law in the Statement.

1.1givenF1F2algebra

Since $α_{1} = 0$ , [F1] gives $Z_{1} = (m, m)$ with probability one, independent of $X$ . By [F2], conditioning on this state and reversing to $s = 0$ returns $X$ exactly. Hence the true reverse law puts mass $1 / 2$ on $(0, 0)$ and $1 / 2$ on $(1, 1)$ , and zero on the two unequal strings.

2.1givenstep 1.1algebra

Each marginal coordinate of that true law is uniform on ${0, 1}$ , but its joint law is not their product: the product would assign mass $1 / 4$ to $(0, 1)$ , while the true law assigns zero. Therefore the exact reverse conditional is not independent across positions.

3.1F3step 1.1step 2.1algebra∎

The learned sampler has $c_{1} = 1$ and draws the two coordinates independently from their uniform predictors by [F3]. Its output assigns probability $1 / 4$ to each of the four strings, so it cannot equal the true reverse law in step 1.1. This refutes the universal factorization statement while leaving the forward conditional independence assumption intact.

Depends on

Used by

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Sources

Subham Sekhar Sahoo et al., Simple and Effective Masked Diffusion Language Models, arXiv v1 (standard reference, not scraped)
Jacob Austin et al., Structured Denoising Diffusion Models in Discrete State-Spaces (standard reference, not scraped)