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How statement and proof provenance work

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  • 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.

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Discrete masked diffusion and finite-state reverse generation

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Finite masking diffusion chain

Definition

Fix a finite nonempty clean vocabulary V, a mask symbol m∉V, integers L,N≥1, a grid 0=t0<t1<⋯<tN=1, a survival schedule 1=α0>α1>⋯>αN=0, and a probability vector μ on VL (nonnegative entries summing to one). Put S=(V∪{m})L and rj=αj/αj−1 for 1≤j≤N. On one coordinate, the step-j transition from v∈V stays at v with probability rj and moves to m with probability 1−rj; from m it stays at m with probability one. All other one-coordinate transitions have probability zero. The sequence-state transition from z to z′ in S is the product of these one-coordinate probabilities over the L coordinates. Starting with Z0=X∼μ, the conditional path mass is Pr⁡(Z0=z0,…,ZN=zN∣X=x)=1{z0=x}∏j=1N∏i=1Laj(zj−1,i,zj,i), where aj is the stated one-coordinate transition rule. Thus, conditional on the clean sequence, the coordinate trajectories are independent. At each step, the next-coordinate states are independent conditional on the preceding sequence state, with the transition probabilities above; successive states of a coordinate need not be independent. The mask is absorbing.

The following one-coordinate diagram records the three transition formulas at step j; each arrow label is its transition probability. The clean-state loop has probability rN=0 at the terminal step j=N, so it is absent from the support there.

v2Vm1¡rjrj1

Definition references

Measure kernel and probability kernel: a probability kernel assigns a probability measure to each state and has measurable evaluation functions; on the finite power-set space above this is exactly a row-stochastic transition matrix.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

Uniform row error bounds mixture-kernel error

Statement

Let S be a finite nonempty set and let Q and P be probability kernels on S, meaning that each row is a probability vector. For probability vectors u,v on S, use TV⁡(u,v)=sup⁡A⊆S∣u(A)−v(A)∣, where u(A)=∑x∈Au(x). If sup⁡x∈STV⁡(Q(x,⋅),P(x,⋅))≤ε for 0≤ε≤1, then every probability vector μ on S satisfies TV⁡(μQ,μP)≤ε, where (μQ)(A)=∑x∈Sμ(x)Q(x,A) and similarly for μP.

Statement references

Total variation distance for probability laws: total variation is the supremum of absolute event-probability differences.

Measure kernel and probability kernel: each kernel row is a probability measure; on finite power-set spaces this is a row-stochastic matrix.

Facts & Assumptions

Given: A finite nonempty set S, probability kernels Q,P on S, a probability vector μ on S, and 0≤ε≤1 with the stated uniform row bound. Prove TV⁡(μQ,μP)≤ε.

[F1]

Total variation distance for probability laws: for probability laws u,v, TV⁡(u,v)=sup⁡A∣u(A)−v(A)∣.

[F2]

Measure kernel and probability kernel: every row Q(x,⋅) and P(x,⋅) is a probability law.

Proof

1.1givenF1F2algebra

Fix A⊆S. By [F2] and the finite mixture formulas, ∣(μQ)(A)−(μP)(A)∣=∣∑x∈Sμ(x)(Q(x,A)−P(x,A))∣≤∑x∈Sμ(x)∣Q(x,A)−P(x,A)∣. Each summand's event difference is bounded by its row total variation, so the right side is at most ∑xμ(x)ε=ε.

2.1F1step 1.1∎

Taking the supremum over all A⊆S in step 1.1 and applying [F1] gives TV⁡(μQ,μP)≤ε.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

Finite product KL is the sum of coordinate KLs

Statement

Let r≥1, let S1,…,Sr be finite nonempty sets, and let uk,vk be probability vectors on Sk. Put U=⨂k=1ruk and V=⨂k=1rvk on S1×⋯×Sr. With D the finite directional KL divergence, zero source masses contributing zero and positive source mass against zero reference mass contributing +∞, one has D(U∥V)=∑k=1rD(uk∥vk), with equality also when one or both sides are +∞.

Statement references

Finite Kullback–Leibler divergence: D(P∥Q)=∑x:px>0pxlog⁡(px/qx), with +∞ for positive source mass where the reference mass is zero.

Facts & Assumptions

Given: A finite list of finite nonempty sets and probability vectors uk,vk. Prove additivity of the finite KL of their product laws, including singular support.

[F1]

Finite Kullback–Leibler divergence: the finite KL convention is directional; zero source terms vanish and positive source mass over zero reference mass gives +∞.

Proof

1.1givenF1construct

Suppose some coordinate k has uk(ak)>0=vk(ak). For every other coordinate i, the finite probability vector ui has a nonempty support set Ai={a:ui(a)>0}. Choose one point from each of the finitely many Ai by recursion over the finite index list; this uses only finite choice. Then U(a1,…,ar)>0 while V(a1,…,ar)=0. By [F1], D(U∥V)=+∞ and D(uk∥vk)=+∞, so both sides of the claimed identity are +∞.

1.2givenF1algebra

Otherwise, for every i and every ai with ui(ai)>0, vi(ai)>0. For each tuple with U(a)>0, log⁡U(a)V(a)=∑i=1rlog⁡ui(ai)vi(ai). Substitute this finite sum into the definition [F1], interchange the finite sums, and sum out every coordinate except i; each such marginal sum is ∑akuk(ak)=1. The result is D(U∥V)=∑i∑ai:ui(ai)>0ui(ai)log⁡(ui(ai)/vi(ai))=∑iD(ui∥vi).

2.1F1step 1.1step 1.2∎

Steps 1.1 and 1.2 cover all possibilities, including r=1 and coordinates with zero source mass. This proves the extended-valued identity.

TheoremStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

Finite-chain reverse-kernel error telescope

Statement

Let S be a finite nonempty state space and let N≥1. For each j=1,…,N, let Qj and Pj be probability kernels on S, used in reverse order from level j to level j−1. Starting from probability vectors μN,νN, define μj−1=μjQj and νj−1=νjPj. If TV⁡(μN,νN)≤εinit and sup⁡z∈STV⁡(Qj(z,⋅),Pj(z,⋅))≤εj(1≤j≤N), where εinit,ε1,…,εN∈[0,1], then TV⁡(μ0,ν0)≤min⁡{1,εinit+∑j=1Nεj}. Here TV⁡(u,v)=sup⁡A⊆S∣u(A)−v(A)∣.

Statement references

Total variation distance for probability laws: for probability laws, total variation is the supremum of absolute event-probability differences.

Measure kernel and probability kernel: a probability kernel has probability-law rows; on S its action is matrix multiplication.

Facts & Assumptions

Given: The finite state space, the N pairs of probability kernels, the initial probability vectors, and all stated total-variation bounds. Prove the endpoint bound.

[F1]

Total variation distance for probability laws: TV⁡(u,v)=sup⁡A⊆S∣u(A)−v(A)∣ and lies in [0,1] for probability laws.

[F2]

Total variation duality, kernel contraction, and coupling bound: applying the same probability kernel to two input laws cannot increase their total-variation distance.

[F3]

Uniform row error bounds mixture-kernel error: a uniform row-wise total-variation error εj gives TV⁡(ρQj,ρPj)≤εj for every input law ρ.

[F4]

Measure kernel and probability kernel: each Qj and Pj row is a probability law.

Proof

1.1givenF1F2F3F4algebra

Fix j and A⊆S. Insert the intermediate measure νjQj and use the ordinary triangle inequality for real numbers on the two event differences. The first difference is at most TV⁡(μj,νj) by [F2], and the second is at most εj by [F3]. Taking the supremum over A by [F1] gives TV⁡(μj−1,νj−1)≤TV⁡(μj,νj)+εj.

2.1givenstep 1.1algebra

Apply step 1.1 successively for j=N,N−1,…,1. The intermediate distances telescope, so TV⁡(μ0,ν0)≤TV⁡(μN,νN)+∑j=1Nεj≤εinit+∑j=1Nεj.

3.1F1step 2.1algebra∎

Since both endpoint vectors are probability laws, [F1] gives TV⁡(μ0,ν0)≤1. Combining this with step 2.1 yields the stated minimum, including the cases N=1, zero kernel errors, and a sum at least one.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Finite masking-chain transitions define a probability law

Statement

For a finite absorbing-mask chain with finite clean vocabulary, mask symbol, sequence length, strict grid, and survival schedule as specified above, each sequence-state transition row is a probability vector, and the displayed joint path mass μ(x)1{z0=x}∏j=1N∏i=1Laj(zj−1,i,zj,i) sums to one over x∈VL and z0,…,zN∈S. Thus it defines a probability law on the finite clean-sequence and path space.

Statement references

Finite masking diffusion chain: fixes V,m,L,N,α,μ, the one-coordinate rows, product sequence kernels, and the path-mass formula.

Facts & Assumptions

Given: A finite nonempty V, m∉V, L,N≥1, the strict survival schedule, and a probability vector μ on VL as in the defining item. Prove each row and the joint path mass are normalized.

[F1]

Finite masking diffusion chain: rj=αj/αj−1 and aj(v,v)=rj, aj(v,m)=1−rj, aj(m,m)=1, with all other entries zero; Qj(z,z′)=∏iaj(zi,zi′).

[F2]

Measure kernel and probability kernel: a probability kernel has probability-measure rows and measurable evaluation functions.

Proof

1.1givenF1F2algebra

The strict schedule gives 0≤rj≤1. For a clean token v, its one-coordinate row has total mass rj+(1−rj)=1; for m, the row mass is 1. Therefore every product row satisfies ∑z′∈SQj(z,z′)=∏i=1L∑w∈V∪{m}aj(zi,w)=1. Its entries are nonnegative. Since S is finite with its power-set sigma-algebra, [F2] makes Qj a probability kernel.

2.1F1step 1.1algebra

For fixed x, sum the conditional path mass backward: summing first over zN uses the row sum of QN from step 1.1 and removes that factor; repeating for zN−1,…,z1 leaves 1{z0=x}. Summing over z0 then gives 1, so each conditional path mass sums to one.

3.1givenF1step 2.1algebra∎

Multiplying these conditional path laws by μ(x) and summing over the finite set VL gives total mass ∑xμ(x)=1. All masses are nonnegative and the path space is finite, so the joint mass is a probability vector.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Finite-grid masked reverse sampler

Definition

For the finite masking chain with state space S=(V∪{m})L and survival schedule α0,…,αN, take probability vectors hi,j(z)∈Δ(V) for every position i∈{1,…,L}, reverse step j∈{1,…,N}, and current state z∈S, where Δ(V)={p:V→[0,1]:∑v∈Vp(v)=1}. Initialize ZN=mL. For j=N,N−1,…,1, set cj=(αj−1−αj)/(1−αj) and update all positions from the same current state z=Zj as follows, independently across positions conditional on z: if zi≠m, set Zj−1,i=zi; if zi=m and cj=1, draw Zj−1,i from hi,j(z); if zi=m and cj<1, set Zj−1,i=m when Uj,imask≥cj, and otherwise draw the clean token from hi,j(z) using Uj,itoken. The resulting one-coordinate row is pi,j(w∣z)=1{w=zi} when zi≠m, and, when zi=m, has mass 1−cj at m and mass cjhi,j(z)(v) at each v∈V; the sequence-state kernel is Pj(z,z′)=∏i=1Lpi,j(zi′∣z). Return Z0 after the finite loop. The mathematical random source is a finite product of mutually independent uniform [0,1] variables Uj,imask,Uj,itoken, one pair for each step and position; an implementation seed is valid only insofar as its supplied random kernel has this law. For inverse-CDF token draws, fix any order V=(v1,…,vK), choose the first vk whose cumulative mass strictly exceeds Uj,itoken, and return vK if that draw equals 1. At j=1, c1=1, so every remaining mask is unmasked.

Definition references

Finite masking diffusion chain: supplies the finite vocabulary, mask, length, grid, and survival schedule.

Measure kernel and probability kernel: a probability kernel is a family of probability-law rows with measurable evaluation functions; the finite Pj above is a product of such rows.

LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

Finite-grid masked sampler defines a normalized law

Statement

Let V be finite and nonempty, m∉V, L,N≥1, 0=t0<⋯<tN=1, and 1=α0>⋯>αN=0. For each i,j,z, let hi,j(z) be a probability vector on V, put cj=(αj−1−αj)/(1−αj), and define the reverse coordinate rows to copy an unmasked token and, from m, assign mass 1−cj to m and mass cjhi,j(z)(v) to each v∈V. Their product rows are probability vectors. Composing these N kernels from the point mass at mL therefore gives a normalized finite law, and its final state lies in VL because c1=1 removes every remaining mask.

Statement references

Finite-grid masked reverse sampler: defines cj, the position-wise reverse rows, their product kernel, and the start state mL.

Measure kernel and probability kernel: a probability kernel has probability-law rows.

Facts & Assumptions

Given: A finite strict survival schedule and the probability-vector family hi,j(z) from the sampler definition. Prove the reverse kernels are normalized and end in VL.

[F1]

Finite-grid masked reverse sampler: cj=(αj−1−αj)/(1−αj); a masked coordinate stays masked with mass 1−cj and moves to v∈V with mass cjhi,j(z)(v); unmasked coordinates are copied.

[F2]

Measure kernel and probability kernel: on a finite power-set state space, a nonnegative row summing to one is a probability kernel.

Proof

1.1givenF1algebra

For j≥1, strict decrease gives αj<1 and αj−1>αj≥0, so cj>0. Since αj−1≤1, αj−1−αj≤1−αj, hence cj≤1.

2.1F1F2step 1.1algebra

If zi≠m, the coordinate row in [F1] is a point mass and sums to one. If zi=m, its mass is (1−cj)+∑v∈Vcjhi,j(z)(v)=1 because hi,j(z) is a probability vector. The product over positions therefore has nonnegative entries summing to the product of L ones, namely one; by [F2] it is a probability-kernel row.

3.1F1givenstep 2.1algebra

Fix a current state z and position i. If zi=m and cj<1, the uniform threshold satisfies Pr⁡(Uj,imask≥cj)=1−cj and Pr⁡(Uj,imask<cj)=cj. For an ordered vocabulary v1,…,vK, put H0=0 and Hk=∑a=1khi,j(z)(va). The inverse-CDF rule selects vk on the interval [Hk−1,Hk), whose uniform probability is Hk−Hk−1=hi,j(z)(vk). The fallback at Uj,itoken=1 has probability zero and so does not change this law, even if the fallback token has zero mass. Since the mask and token uniforms are independent, the resulting row assigns mass 1−cj to m and cjhi,j(z)(v) to each v. If cj=1, the direct token draw has law hi,j(z) and no mask is retained. Unmasked coordinates are copied, and independence of the random pairs across positions makes the sequence row the product in [F1]. Thus the stated random instructions realize the coordinate and product rows from step 2.1.

4.1F1step 1.1step 2.1step 3.1algebra∎

The loop has exactly N reverse-step iterations, each updates the finite set of L positions once, and each inverse-CDF search checks at most the finite number K=∣V∣ of tokens; the fallback defines the draw also at Uj,itoken=1. Hence every instruction terminates and the algorithm terminates after finitely many operations. Starting from the point mass at mL, composition of the normalized rows from step 2.1 gives a normalized finite path and endpoint law. At j=1, c1=(α0−α1)/(1−α1)=1, so the algorithm's direct token-draw branch removes every remaining mask and copies every already clean coordinate. Thus Z0∈VL, completing the support claim.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

Posterior of a finite masking chain

Statement

Let V be a finite nonempty vocabulary, let m∉V, let L,N≥1, and let 1=α0>α1>⋯>αN=0 be the survival schedule of the finite masking chain. Fix a clean sequence x∈VL and grid indices 0≤s<t≤N. For every state zt∈(V∪{m})L with Pr⁡(Zt=zt∣X=x)>0, the conditional law of Zs given X=x,Zt=zt is a product over coordinates. At coordinate i, if zt,i∈V, then Zs,i=zt,i with probability one. If zt,i=m, then, with cs,t=(αs−αt)/(1−αt), the only possible values are xi and m, with probabilities Pr⁡(Zs,i=xi∣X=x,Zt,i=m)=cs,t,Pr⁡(Zs,i=m∣X=x,Zt,i=m)=1−αs1−αt. The denominator is positive because t≥1 implies αt<1; at s=0 the mask posterior has zero mass.

Statement references

Finite masking diffusion chain: conditional on X=x, coordinates and masking draws are independent, clean tokens never change, and m is absorbing.

Facts & Assumptions

Given: The finite masking chain, a fixed clean string x, grid indices 0≤s<t≤N, and a state zt with positive conditional probability given X=x. Prove the displayed product posterior.

[F1]

Finite masking diffusion chain: at step j, a clean token survives with probability rj=αj/αj−1, the mask is absorbing, and corruption factors over positions conditional on the clean string.

[F2]

Finite masking-chain transitions define a probability law: the finite path mass is a normalized probability law.

Proof

1.1givenF1F2algebra

For one coordinate, the probability of surviving all steps through s is ∏j=1srj=αs, with the empty product equal to one when s=0. Survival through t has probability αt. Since the mask is absorbing, the event “clean at s and masked at t” has probability αs−αt, while “masked at t” has probability 1−αt. The latter is positive for t≥1. Dividing these finite event probabilities gives the two displayed conditional masses, which sum to one.

2.1F1F2step 1.1algebra

If zt,i∈V, positive conditional probability forces zt,i=xi; the token could not have changed or been unmasked after masking, so its earlier value is the same. If zt,i=m, step 1.1 gives the two possible earlier values and their probabilities. By [F1], conditional on x the coordinate paths are independent; conditioning on the coordinatewise event Zt=zt therefore leaves the coordinate posterior laws independent, and their product is the asserted full conditional law.

3.1givenF1step 1.1algebra∎

The strict schedule gives 0<cs,t≤1 and 1−cs,t=(1−αs)/(1−αt). At s=0, cs,t=1; at t=N, αt=0 and the denominator is one. Thus the formula covers both endpoints without conditioning on the zero-probability event Z0=m.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Finite masking path law factors backward by its posterior rows

Statement

Fix a clean string x∈VL and let qx be the finite forward path law of the absorbing-mask chain. For every path (z0,…,zN) with qx(z0,…,zN)>0, the reversed path probability factors as qx(z0,…,zN)=qx(ZN=zN)∏j=N1qx(Zj−1=zj−1∣Zj=zj). 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 z0,…,zN. Prove the reversed factorization.

[F1]

Finite masking diffusion chain: conditional on x, the path mass is 1{z0=x}∏j=1Naj(zj−1,zj), 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 Zj−1 given x,Zj is the stated posterior row.

Proof

1.1F1F2givenalgebra

Fix j and a positive-probability suffix (zj,…,zN). By [F1], for every prefix ending at zj, the probability of adjoining the fixed future suffix is multiplied by the same factor ∏k=j+1Nak(zk−1,zk), which does not depend on zj−1. Summing over earlier coordinates preserves this common factor, so it cancels in the conditional ratio. Thus qx(Zj−1=zj−1∣Zj=zj,…,ZN=zN)=qx(Zj−1=zj−1∣Zj=zj) on this positive suffix.

2.1F2F3step 1.1algebra∎

Repeated finite conditioning in reverse order gives qx(z0:N)=qx(zN)∏j=N1qx(zj−1∣zj,…,zN). 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.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

Masked-diffusion KL reduces to weighted cross-entropy

Statement

Let V be a finite nonempty vocabulary, let m∉V, let L,N≥1, and let 1=α0>⋯>αN=0. Draw X=(X1,…,XL) from any probability law on VL. Conditional on X, at step j each clean coordinate survives with probability rj=αj/αj−1 using fresh Bernoulli innovations independent across coordinates and steps, and otherwise becomes m; an already masked coordinate remains masked. Thus Zj is the state after step j. For j=1,…,N, put cj=(αj−1−αj)/(1−αj),dj=(1−αj−1)/(1−αj). Let hi,j:S→Δ(V) be any predictors, where S=(V∪{m})L and Δ(V) is the set of probability vectors on V. Define the learned reverse kernel Pj to have product rows: an unmasked coordinate is copied, while a masked coordinate assigns mass dj to m and mass cjhi,j(z)(v) to each v∈V. For every positive-probability current state zj, the true conditional law q(Zj−1∣Zj=zj,X) also has product rows: it copies every unmasked coordinate and, at a masked coordinate i, assigns mass cj to Xi and dj to m. For 1≤i≤L, let Mj,i={Zj,i=m}, which has probability 1−αj>0, and let Pj,imask be the conditional law of (Zj,Xi) given Mj,i. View hi,j as a predictor in the finite supervised problem with feature Zj, label Xi, action space Δ(V), its singleton predictor class, and loss ℓ(p,v)=−log⁡p(v), with −log⁡0=+∞. Its population risk under Pj,imask is Rj,i=E[ℓ(hi,j(Zj),Xi)∣Mj,i]. For finite-alphabet probability vectors P=(pa) and Q=(qa), write DKL(P∥Q)=∑a:pa>0palog⁡(pa/qa), with zero source terms set to zero and positive source mass against zero reference mass set to +∞. Define the extended-nonnegative conditional reverse-KL contribution Lj=EX,Zj[DKL(q(Zj−1∣Zj,X)∥Pj(Zj,⋅))]. Then ∑j=1NLj=∑j=1N∑i=1L(αj−1−αj)Rj,i as an equality of extended nonnegative quantities. If a predictor assigns zero mass to the true token on a positive-probability masked event, both sides are +∞; finite cross-entropy is not assumed.

Statement references

Finite Kullback–Leibler divergence: for finite probability vectors, D(P∥Q)=∑x:px>0pxlog⁡(px/qx), with +∞ when px>0=qx.

Finite product KL is the sum of coordinate KLs: the finite KL of product probability vectors is the sum of their coordinate KLs, including the singular-support case.

Posterior of a finite masking chain: conditional on the clean string, the true reverse posterior is a product of copy rows and two-point mask/unmask rows.

Finite-grid masked reverse sampler: the learned reverse rows copy unmasked tokens and use the same mask/unmask probability with clean-token vector hi,j.

Supervised prediction problems: a supervised problem specifies measurable feature, label, action spaces, a joint law, a predictor class and a nonnegative measurable loss.

Population and empirical risk: population risk is the extended nonnegative integral of the loss under the data law.

Population and empirical risks are measurable: the finite-state loss and conditional risk are measurable and defined.

Facts & Assumptions

Given: A finite masking chain, its true posterior, the learned product reverse kernel, finite-alphabet KL, and the finite conditional prediction problems in the Statement. Prove the summed reverse-KL/cross-entropy identity.

[F1]

Finite masking diffusion chain: forward corruption is independent across positions conditional on X; a clean token survives to step j with probability αj, and the mask is absorbing.

[F2]

Posterior of a finite masking chain: for s=j−1<t=j, a masked coordinate has true clean probability cj=(αj−1−αj)/(1−αj) and mask probability dj=(1−αj−1)/(1−αj); an unmasked coordinate is copied; the posterior factors across positions given X,Zj.

[F3]

Finite-grid masked reverse sampler: from a masked coordinate the learned row assigns mass dj=1−cj to m and mass cjhi,j(z)(v) to each clean token v; learned position rows are conditionally independent.

[F4]

Finite Kullback–Leibler divergence: finite relative entropy is directional, zero source masses contribute zero, and positive source mass against zero target mass gives +∞.

[F5]

Supervised prediction problems: the finite feature and label spaces, Borel action simplex Δ(V), singleton predictor, conditional joint law, and cross-entropy loss form a supervised prediction problem.

[F6]

Population and empirical risk: its population risk is the extended conditional expectation Rj,i.

[F7]

Population and empirical risks are measurable: finite-state measurable predictors and losses have a defined measurable risk.

[F8]

Finite product KL is the sum of coordinate KLs: finite product KL is the sum of its coordinate KL terms, with +∞ preserved when a factor has singular support.

Proof

1.1F2F3F4F5F7F8given

Equip the finite feature space S and label space V with their power-set sigma-algebras, and equip Δ(V) with the Borel sigma-algebra inherited from RV. Every predictor hi,j:S→Δ(V) is measurable because its domain is finite and discrete. For each v∈V, the coordinate map p↦p(v) is continuous, and g(t)=−log⁡t for t>0, with g(0)=+∞, is Borel measurable on [0,1]. Therefore p↦−log⁡p(v) is Borel measurable and nonnegative. Since V is finite and discrete, the loss (p,v)↦−log⁡p(v) is measurable on Δ(V)×V. This verifies the loss-measurability requirement in [F5] and the finite-state measurability used in [F7]. Now fix a step j, clean string x, and observed positive-probability state zj. The true and learned conditional laws both factor over positions by [F2] and [F3]. Applying [F8] to these finite product rows makes their joint conditional KL the sum of the coordinate KLs, preserving the singular +∞ case described by [F4].

2.1F2F3F4step 1.1algebra

If zj,i≠m, the true and learned coordinate rows are the same point mass, so their KL is zero. If zj,i=m, put cj=(αj−1−αj)/(1−αj)>0 and dj=(1−αj−1)/(1−αj). The true row has mass cj at xi and dj at m; the learned row has mass cjhi,j(zj)(v) at v∈V and dj at m. If hi,j(zj)(xi)>0, the only nonzero KL terms are cjlog⁡(cj/(cjhi,j(zj)(xi)))=−cjlog⁡hi,j(zj)(xi) and, when dj>0, djlog⁡(dj/dj)=0. If hi,j(zj)(xi)=0, [F4] gives +∞, which is the same extended value as cj[−log⁡hi,j(zj)(xi)]. Thus the full conditional KL is ∑i1Mj,icj[−log⁡hi,j(zj)(xi)].

3.1F1F5F6F7step 2.1algebra

For each i, [F1] gives Pr⁡(Mj,i∣X=x)=1−αj, independent of x. Averaging the ith term of step 2.1 over the forward corruption therefore gives cj(1−αj)E[−log⁡hi,j(Zj)(Xi)∣Mj,i]=(αj−1−αj)Rj,i by [F5]–[F7]. This argument conditions on a positive event and does not assume the clean coordinates of X are independent.

4.1F1F2step 2.1step 3.1algebra∎

Sum the step 3.1 identity over the finite positions and all j=1,…,N. The coefficient identity cj(1−αj)=αj−1−αj holds because αj<1. At j=1, α0=1 gives d1=0, which is the zero mask-retention term already omitted in step 2.1; at j=N, αN=0 gives Pr⁡(MN,i)=1, so the conditional risk in step 3.1 is defined on the whole terminal event. Extended nonnegative sums preserve the +∞ cases from step 2.1, giving the stated equality.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Finite masked-diffusion generative model

Definition

For the finite masking state space S=(V∪{m})L and reverse kernels Pj of the finite-grid sampler, put a point-mass prior at ZN=mL. Define a decoder D from S to VL by D(x∣z0)=1{x=z0} when z0∈VL, and D(x∣z0)=∣V∣−L for every x∈VL when z0∉VL. The finite joint mass of a clean output x∈VL and latent path (z0,…,zN)∈SN+1 is p(x,z0,…,zN)=1{zN=mL}(∏j=1NPj(zj,zj−1))D(x∣z0). Thus reverse step j maps the state at level j to the state at level j−1; the decoder is deterministic on clean endpoint states and is explicitly normalized on all other states.

Definition references

Finite masking diffusion chain: fixes the finite vocabulary, mask, sequence length, grid, survival schedule, and sequence state space.

Finite-grid masked reverse sampler: supplies each reverse kernel Pj and starts at mL.

Measure kernel and probability kernel: a probability kernel supplies normalized transition rows.

LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

Finite masked-diffusion joint model is normalized

Statement

Let V be finite and nonempty, m∉V, L,N≥1, and S=(V∪{m})L. Let Pj be probability kernels on S, put prior mass one at zN=mL, and let D(x∣z0)=1{x=z0} when z0∈VL and D(x∣z0)=∣V∣−L otherwise. Then the finite joint mass p(x,z0,…,zN)=1{zN=mL}(∏j=1NPj(zj,zj−1))D(x∣z0) is nonnegative and sums to one over x∈VL and (z0,…,zN)∈SN+1.

Statement references

Finite masked-diffusion generative model: defines the terminal point mass, reverse-kernel product, normalized decoder, and joint mass.

Facts & Assumptions

Given: The finite generative joint mass and the sampler's probability kernels. Prove the total joint mass is one.

[F1]

Finite masked-diffusion generative model: p(x,z0,…,zN)=1{zN=mL}(∏j=1NPj(zj,zj−1))D(x∣z0).

[F2]

Finite-grid masked sampler defines a normalized law: each reverse-kernel row Pj(z,⋅) is a probability vector.

Proof

1.1F1givenalgebra

For fixed z0∈S, if z0∈VL then D(⋅∣z0) is a point mass; otherwise it assigns ∣V∣−L to each of the ∣V∣L clean outputs. In both cases ∑x∈VLD(x∣z0)=1.

2.1F1F2step 1.1algebra

Sum [F1] first over x, using step 1.1. Then sum over z0 in P1(z1,z0), over z1 in P2(z2,z1), and successively over zN−1 in PN(zN,zN−1). Each inner sum is one by [F2], leaving ∑zN1{zN=mL}=1. All sums are finite, so this iterated summation equals the total joint mass.

3.1F1step 2.1algebra∎

Every factor in [F1] is nonnegative, and step 2.1 shows the finite total mass is one. Therefore the displayed function is a probability law.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Forward path law is admissible for the finite masked model

Statement

Let V be finite and nonempty, m∉V, L,N≥1, and 1=α0>⋯>αN=0. For each position, reverse step, state with a masked position, and clean token, assume hi,j(z)(v)>0. Fix x∈VL, let qx be the forward path probability vector on SN+1 conditioned on X=x, let γx(z0,…,zN)=p(x,z0,…,zN) be the finite joint model mass, and put Zx=∑zγx(z) and πx=γx/Zx. Then 0<Zx≤1, qx=0 wherever πx=0, and ∑z:qx(z)>0qx(z)(∣log⁡γx(z)∣+∣log⁡qx(z)∣)<∞. With weighted log terms set to zero where qx=0, as in the variational definition, qx is an admissible variational density for the finite joint target γx.

Statement references

Finite masked-diffusion generative model: the joint mass is the terminal point mass times the learned reverse kernels and the decoder.

Finite masking diffusion chain: the forward path starts at x, masks are absorbing, and the terminal state is mL.

Variational family and evidence lower bound: admissibility requires support domination and absolute log-integrability for a finite positive normalizer.

Facts & Assumptions

Given: The finite forward path law, the finite reverse generative model, the strict endpoints, and full-support clean-token vectors at every masked reverse state. Prove the forward path density is admissible for the model's conditional joint target.

[F1]

Finite masking diffusion chain: a forward path starts at x, each clean token either survives or becomes m, and m remains masked thereafter.

[F2]

Finite-grid masked reverse sampler: reverse rows copy clean tokens and assign mass cjhi,j(z)(v) to a clean token from a masked position and dj=(1−αj−1)/(1−αj) to retaining the mask.

[F3]

Finite masking-chain transitions define a probability law: qx is a normalized probability vector on the finite path space.

[F4]

Finite masked-diffusion joint model is normalized: the joint model mass is a probability vector, so Zx≤1.

[F5]

Variational family and evidence lower bound: admissibility is qx≪πx and finite ∑qx(∣log⁡γx∣+∣log⁡qx∣), with zero-density terms set to zero.

Proof

1.1givenF1F2constructalgebra

The forward survival ratios telescope to ∏j=1Nrj=αN=0, so for each coordinate the event of remaining clean at N has probability zero; hence every qx-positive path has zN=mL, while the defining path mass forces z0=x. Fix such a path. A clean-to-clean forward transition copies its token, and the learned reverse row assigns that transition probability one. A clean-to-mask transition has learned reverse probability cjhi,j(zj)(xi)>0. A mask-to-mask transition can occur only for j≥2, because z0=x is clean; then αj−1<1 and its learned reverse probability dj=(1−αj−1)/(1−αj)>0. These are all positive-probability forward transitions by [F1]. Thus every factor in the learned reverse path mass is positive, and the terminal prior and clean-state decoder are also positive; hence γx(z0,…,zN)>0.

2.1F3F4step 1.1algebra

By [F3], at least one path has positive qx mass. Step 1.1 makes its joint model mass positive, so Zx>0; [F4] gives Zx≤1. On every qx-positive path, step 1.1 gives γx>0, hence πx>0 and qx≪πx.

3.1F5step 1.1step 2.1algebra∎

The path space is finite. On each of its finitely many qx-positive states, both qx(z) and γx(z) are positive finite real numbers, so both logarithms are finite. Their finite weighted sum is finite; on qx=0 states the variational convention contributes zero. This verifies [F5] and proves admissibility.

CorollaryStatement: Literature-sourcedProof: AI-generatedOpen item page →

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

Statement

Assume the Axiom of Choice. Let V be finite and nonempty, m∉V, L,N≥1, 1=α0>⋯>αN=0, and let X∼μ for a probability vector μ on VL. Put S=(V∪{m})L. Conditional on X=x, let qx be the finite forward masking path law started at Z0=x: at step j, each clean coordinate survives with probability αj/αj−1 and otherwise becomes the absorbing mask m, independently across coordinates. For every position i, reverse step j, state z∈S with zi=m, and token v∈V, assume hi,j(z)(v)>0, where each hi,j(z) is a probability vector on V. Put cj=(αj−1−αj)/(1−αj) and dj=(1−αj−1)/(1−αj). Define the reverse kernel Pj by copying every unmasked coordinate and, independently across masked positions conditional on zj, assigning mass dj to m and mass cjhi,j(zj)(v) to each v∈V. Use the point mass at mL as terminal prior and the decoder Dec⁡(x∣z0)=1{x=z0} when z0∈VL, while Dec⁡(x∣z0)=∣V∣−L otherwise. Define p(x,z0,…,zN)=1{zN=mL}(∏j=1NPj(zj,zj−1))Dec⁡(x∣z0),p(x)=∑z0,…,zNp(x,z0,…,zN). For each x∈VL, let πx(z)=p(x,z)/p(x) whenever p(x)>0 and define the finite ELBO E(x)=∑z:qx(z)>0qx(z)log⁡(p(x,z)/qx(z)). Here admissibility means qx is supported where p(x,⋅)>0 and the finite weighted logarithm sums required for the ELBO are finite. Let Pj,imask be the conditional law of (Zj,Xi) given Zj,i=m, and let Rj,i=E[−log⁡hi,j(Zj)(Xi)∣Zj,i=m], with −log⁡0=+∞. For probability vectors u,v on a finite set, write DKL(u∥v)=∑a:u(a)>0u(a)log⁡(u(a)/v(a)), with zero source terms set to zero and positive source mass against zero reference mass set to +∞. Then p(x)>0, qx is admissible, and for every x, log⁡p(x)=E(x)+DKL(qx∥πx),E(x)≤log⁡p(x). Moreover, averaging over X∼μ gives E[−E(X)]=∑j=1N∑i=1L(αj−1−αj)Rj,i. The reconstruction term and terminal-prior KL vanish here because the endpoints are exactly clean and all-masked; with approximate endpoints, the ordinary ELBO retains those terms and this corollary does not discard them.

Statement references

Finite masked-diffusion generative model: defines Pj, the terminal point mass, decoder D, and finite joint path mass.

Forward path law is admissible for the finite masked model: full support of h makes p(x)>0 and the forward path law admissible.

Masked-diffusion KL reduces to weighted cross-entropy: each averaged finite reverse-KL contribution equals the weighted conditional masked-token population risk.

Finite masking path law factors backward by its posterior rows: the positive forward path law factors in reverse order by its one-step posterior rows.

Evidence lower bound is log evidence minus reverse KL: under AC and admissibility, log evidence is ELBO plus the reverse KL and therefore bounds the ELBO above.

Facts & Assumptions

Given: Choice, the exact finite masked chain and learned reverse model, the full-support hypothesis, the stated decoder/prior, and the finite ELBO definition. Prove admissibility, the ELBO identity and bound, and the averaged cross-entropy formula.

[A1]

The Axiom of Choice: AC is the axiom that every family of nonempty sets has a choice function; it is used only through the library ELBO/KL theorem.

[F1]

Finite masked-diffusion generative model: the reverse kernels, terminal prior, decoder, and joint mass define the model.

[F2]

Finite masked-diffusion joint model is normalized: the finite joint model is a probability vector.

[F3]

Forward path law is admissible for the finite masked model: for each clean x, the forward path law is supported where p(x,z)>0 and satisfies the variational absolute-log-integrability condition.

[F4]

Variational family and evidence lower bound: admissible q has finite evidence and log integrability, and its ELBO is ∫q(log⁡γ−log⁡q).

[F5]

Evidence lower bound is log evidence minus reverse KL: under AC and admissibility, log⁡Z=ELBO⁡(q)+∫qlog⁡(q/π) dμ, the log-ratio integral is the library divergence D(Q∥Π), and D(Q∥Π)≥0.

[F6]

Finite masking path law factors backward by its posterior rows: on positive paths, qx(z0:N)=qx(zN)∏j=N1qx(zj−1∣zj,x).

[F7]

Masked-diffusion KL reduces to weighted cross-entropy: the data-averaged sum of conditional reverse KL terms is the stated weighted cross-entropy risk.

[F8]

Finite Kullback–Leibler divergence: for finite probability vectors, the directional KL is the finite expectation of the log likelihood ratio, with singular positive-mass terms equal to +∞.

Proof

1.1A1F2F3F4F5F8given

Fix x∈VL and take counting measure on the finite path space SN+1. By [F2], γx(z)=p(x,z) has finite normalizer Zx=p(x)≤1; by [F3], 0<Zx and qx is admissible for γx. Under this counting measure, the induced probability laws assign singleton masses qx(z) and πx(z). Admissibility gives qx(z)>0⇒πx(z)>0, so the ratio integral in [F5] is the finite sum ∑z:qx(z)>0qx(z)log⁡(qx(z)/πx(z)), which is exactly DKL(qx∥πx) by [F8]. Applying [F5] with [A1] therefore gives log⁡p(x)=E(x)+DKL(qx∥πx) and E(x)≤log⁡p(x).

1.2F1F3F6F8givenalgebra

The forward path formula forces Z0=x. For each coordinate, the product of forward survival ratios is αN=0, so the finite forward law assigns probability one to ZN=mL; thus on every qx-positive path the decoder and prior factors in [F1] are both one. By [F6], qx(z0:N)=qx(zN)∏j=N1qx(zj−1∣zj,x) on this path, and qx(zN)=1. Therefore log⁡(p(x,z)/qx(z)) is the sum of the reverse-step log ratios. By [F8], the conditional expectation of each log ratio is the negative finite KL term, so E(x)=−∑j=1NLj(x), where Lj(x)=EZj∣X=xDKL(q(Zj−1∣Zj,x)∥Pj(Zj,⋅)).

2.1F7step 1.2algebra

Average the identity from step 1.2 over X∼μ. The resulting expected sum of Lj(X) is exactly the left side of the finite KL-to-risk identity in [F7]; substituting [F7] yields E[−E(X)]=∑j,i(αj−1−αj)Rj,i.

3.1F1F3step 1.2algebra∎

The endpoint cancellation in step 1.2 uses Z0=x from α0=1 and ZN=mL from αN=0. If either endpoint is changed, its decoder reconstruction or terminal-prior KL factor remains in the finite log-ratio expansion unless a separate hypothesis makes it vanish.

RemarkRemark: Literature-sourcedProof: Not applicableOpen item page →

Factorization and finite-endpoint warnings

Remark

The learned sampler assumes conditional independence across positions at each reverse update. Forward masking is independent across positions only after conditioning on the clean sequence; marginalizing a dependent clean-sequence law can make the true reverse coordinates dependent, so a normalized learned sampler need not be the exact reverse law. The counterexample on the companion page gives a finite witness. The identities here concern a fixed finite grid only; they assert no continuous-time or zero-mesh limit, which would require separate convergence hypotheses. The weighted-cross-entropy ELBO identity uses the exact clean and all-mask endpoints and their matching decoder and prior; approximate endpoints retain their reconstruction and terminal-prior terms. Sahoo et al. label their zero-masking and carry-over substitutions as “RB techniques” and call the simplified objective “Rao–Blackwellized”; this finite identity makes no variance-reduction claim, because no estimator and conditioning comparison are specified here.

Remark references

A factorized reverse kernel fails for correlated tokens ↗: a two-token finite witness where the true reverse posterior is dependent but the learned reverse draw is a product.

Negative expected finite masked-diffusion ELBO is weighted cross-entropy: the exact-endpoint finite ELBO identity and the support condition for its finite proof.

Sahoo et al., §3.3 and Appendix B.1: the paper's objective simplifications using zero-masking and carry-over substitutions, termed “RB techniques.”

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