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Discrete masked diffusion and finite-state reverse generation — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

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

Finite model error accumulates in a fixed chain

Example

Let $S = {0, 1}$ , let $N = 2$ , and start the true and learned chains at $δ_{0}$ . At both steps the true kernel is the identity. For a fixed $0 < ε < 1$ , the learned kernel sends $0$ to $1$ with probability $ε$ and leaves $0$ at $0$ otherwise, while it leaves $1$ at $1$ . Each step's largest row total-variation error is $ε$ , but after two learned steps the endpoint law assigns mass $2 ε - ε^{2}$ to $1$ , whereas the true endpoint law is $δ_{0}$ . The comparison is with the fixed two-step identity-kernel chain itself, so it has no discretization-error term, while the learned-kernel error is positive.

Example references

Finite-chain reverse-kernel error telescope: the endpoint bound sums the two uniform row errors and clamps the result at one.

Total variation distance for probability laws: total variation uses the event-supremum convention.

Facts & Assumptions

Given: $S = {0, 1}$ , the initial point mass $δ_{0}$ , two identity true kernels, and the two learned kernels in the Example. Verify each row error and the endpoint law.

[F1]

Total variation distance for probability laws: total variation is the supremum of absolute probability differences over events.

[F2]

Finite-chain reverse-kernel error telescope: the endpoint total variation is at most the initial error plus the sum of uniform row errors, clamped at one.

Verification

1.1givenF1algebra

At state $0$ , the true row is $δ_{0}$ and the learned row is $(1 - ε, ε)$ , so their event difference on ${1}$ is $ε$ ; at state $1$ the rows agree. Since these are the only two states, [F1] gives uniform row error exactly $ε$ at each step.

2.1givenF1step 1.1algebra

After one learned step the law is $(1 - ε, ε)$ . Since $1$ is absorbing and $0$ returns to $0$ with probability $1 - ε$ , after two steps its mass at $0$ is $(1 - ε)^{2}$ and its mass at $1$ is $1 - (1 - ε)^{2} = 2 ε - ε^{2}$ . The true identity chain remains $δ_{0}$ , so [F1] gives endpoint TV $2 ε - ε^{2} > 0$ .

3.1F2step 2.1algebra∎

For $ε = 1 / 4$ , the endpoint error is $7 / 16$ , while [F2] gives the valid bound $2 ε = 1 / 2$ . The example has no time-discretization term because the reference is the fixed two-step chain itself.

CounterexampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

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.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

One-token posterior on a finite mask schedule

Example

Take $V = {a, b}$ , $L = 1$ , $N = 2$ , $α_{0} = 1$ , $α_{1} = 1 / 2$ , $α_{2} = 0$ , and clean token $X = a$ . At the reverse pair $(s, t) = (1, 2)$ , observing $Z_{2} = m$ has probability one and the true posterior assigns probability $1 / 2$ to $Z_{1} = a$ and $1 / 2$ to $Z_{1} = m$ . At the reverse pair $(s, t) = (0, 1)$ , conditional on the positive-probability event $Z_{1} = m$ , the true posterior removes the mask with probability one.

Example references

Finite masking diffusion chain: gives the one-token survival schedule and absorbing mask.

Posterior of a finite masking chain: gives the exact conditional probabilities for each reverse pair.

Facts & Assumptions

Given: The one-token finite masking chain in the Example. Verify both posterior rows.

[F1]

Finite masking diffusion chain: $r_{j} = α_{j} / α_{j - 1}$ and the mask is absorbing.

[F2]

Posterior of a finite masking chain: at a masked later state, the earlier clean-token probability is $(α_{s} - α_{t}) / (1 - α_{t})$ ; an unmasked token is copied.

Verification

1.1givenF1F2algebra

For $(s, t) = (1, 2)$ , $c_{1, 2} = (1 / 2 - 0) / (1 - 0) = 1 / 2$ , and $1 - c_{1, 2} = (1 - 1 / 2) / (1 - 0) = 1 / 2$ . Since $α_{2} = 0$ , [F1] gives $Z_{2} = m$ with probability one, so [F2] yields the displayed posterior row.

2.1givenF1F2step 1.1algebra∎

For $(s, t) = (0, 1)$ , $c_{0, 1} = (1 - 1 / 2) / (1 - 1 / 2) = 1$ . The conditioning event $Z_{1} = m$ has probability $1 - α_{1} = 1 / 2 > 0$ , so [F2] gives probability one to $Z_{0} = a$ and zero to the mask, with no zero-denominator endpoint used.

Discrete masked diffusion and finite-state reverse generation — Examples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

Finite model error accumulates in a fixed chain

Example

Example references

Facts & Assumptions

Verification

A factorized reverse kernel fails for correlated tokens

Statement refuted

Statement references

Facts & Assumptions

Counterexample

One-token posterior on a finite mask schedule

Example

Example references

Facts & Assumptions

Verification

Sources