High-Accuracy Sampling for Diffusion Models and Log-Concave Distributions

• p. 3 Section 1.3 Notation

  Dχ2(µ ∥ν) := Eµ(µ/ν)2 −1

• p. 15 Proof of Lemma B.9

  1 + Dχ2(µ ∥bµ) = Ex∼µ

• p. 3 Section 1.3 Notation

  the chi-squared divergence

• p. 15 Proof of Lemma B.8

  Eµ[eh] = 1 + Dχ2(µ ∥bµ)

Judge's note: The page 3 image clearly shows E_µ with the square outside (µ/ν). E_µ(µ/ν)^2 = ∫µ^3/ν^2, which is not the chi-squared divergence. The proofs use the standard form: on p.15, the proof of Lemma B.8 writes E_µ[e^h] = 1 + D_χ²(µ∥µ̂) with h = log dµ/dµ̂, and the proof of Lemma B.9 writes 1 + D_χ²(µ∥µ̂) = E_{x∼µ}[µ(x)/µ̂(x)]. So the printed definition contradicts how the quantity is used. The fix is evident: E_ν(µ/ν)^2 − 1.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 4 Theorem 3.1

  by 3Be^{2B} log(2/δ)

• p. 18 Appendix C proof

  Note that if any J_i = 0, then I ≤ i

Only the authors can say what is right here.

Judge's note: I checked the counterexample against Algorithm 1 (B > 0, W supported on [−B,B]). With W ≡ −B, every factor (B+W_j)/(2B) is 0, so an iteration accepts only when J = 0 (probability e^{−2B}), and the accepting iteration adds no draws. With B = 0.1 and δ = 0.01, the bound is m = 0.3·e^{0.2}·ln200 ≈ 1.94, so the theorem claims P(N ≤ 1) ≥ 0.99. But P(N ≤ 1) = P(J1=0) + P(J1=1)P(J2=0) = e^{−0.2} + 0.2e^{−0.4} ≈ 0.953. The explicit statement therefore needs a lower bound on B (or a different constant). Downstream uses take B = Θ(1).

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 18 Proof of Theorem 3.1

  c = e−2B suffices to bound both terms by δ/2

• p. 4 Theorem 3.1

  3Be^{2B} log(2/δ)

• p. 18 Proof of Theorem 3.1

  suffices to bound both terms by δ/2

• p. 18 Appendix C, proof of Theorem 3.1, second statement

  Setting i0 = e^{2B} log(2/δ) and c = e^{−2B} suffices to bound both terms by δ/2.

Only the authors can say what is right here.

Judge's note: I read Theorem 3.1 on p.4 and its proof on p.18. With the printed tail and the printed choices, the exponent is c²/(c+2)·i0·B = Be^{−2B}log(2/δ)/(2+e^{−2B}). At B=1 this is about 0.063·log(2/δ), so the second term is about (δ/2)^{0.063}, far above δ/2. Also m = (2+c)i0B equals 3Be^{2B}log(2/δ) only when c=1. Even with c=1 the exponent is (Be^{2B}/3)log(2/δ), which is below log(2/δ) when Be^{2B} < 3. The theorem's query bound itself fails for small B: if W_j ≡ B, every proposal is accepted and N ~ Poisson(2B). For B=0.01 and δ=0.001 the bound is about 0.23 < 1, so it needs N=0, which has probability e^{−0.02} ≈ 0.980 < 0.999. The second statement of Theorem 3.1, as printed, is therefore not established, and it fails for small B > 0.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 34 Theorem F.1 item 2

  d(1−s)/s

• p. 8 Section 5, Poincaré bullet (s=1)

  log(1/ε)) log(χ2/ε2))

• p. 34 Theorem F.1, item 2

  (1 + log^{1/(1+s)}(1/ε)/d^{(1−s)/s})

• p. 8 Section 5, Poincaré bullet (s = 1)

  Õ(κ (d^{1/2} log^{1/2}(1/ε) + log(1/ε)) log(χ²/ε²))

Only the authors can say what is right here.

Judge's note: P.34 image: η⁻¹ = C(β_s²d^s log(1/ε) + β_s²d^{−(1−s)}log²(1/ε))^{1/(1+s)}. Factoring out (β_s²d^s log)^{1/(1+s)} leaves (1 + log/d)^{1/(1+s)} ≍ 1 + (log/d)^{1/(1+s)}, so the denominator is d^{1/(1+s)}. The printed d^{(1−s)/s} equals d^0 at s=1 and is undefined at s=0, although s ∈ [0,1] is allowed. The p.8 crop states the s=1 Poincaré result as κ(d^{1/2}log^{1/2}(1/ε) + log(1/ε)). This matches d^{1/(1+s)}, not the printed theorem: the two differ by a d^{1/2} factor on the second term. The proof is omitted ('exactly as in Altschuler & Chewi'), so the authors would need to confirm the intended form.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 13 Appendix B, Eq. (18)

  Dλ(p ∥q) := Ep

• p. 21 Proof of Proposition E.1

  We consider the λ-divergence

• p. 17 Proof of Lemma B.13

  1 + Dℓ(µf ∥µg) = Eµf

• p. 13 Eq. (18)

  D_λ(p ∥ q) := E_p(dp/dq)^λ − 1

Judge's note: Page 13 image: Eq. (18) shows E_p. The Rényi definition just above it takes the expectation under the second argument. Page 21: E.1 uses E_q. Page 17: B.13's first identity is E_{µf}(µf/µg)^{ℓ−1} = E_{µg}(µf/µg)^ℓ, which is the E_q convention. Under (18), E_p(dp/dq)^λ = E_q(dp/dq)^{λ+1}, so the order is off by one relative to every use. The definition is inconsistent with its uses; the fix is evident (E_q).

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 23 Proposition E.3 statement

  η ≤ σ2 / 4BG

• p. 30 End of proof of Proposition E.3

  we use η ≤ σ2 / 8BG

• p. 31 Lemma E.10 (3)

  P(|WᵀHU| ≥ t + 4η²/σ² (|VᵀHV| + |tr(H)|) | x̄, V)

• p. 30 Lemma E.10 (2) bounds

  |c_U| ≤ η/σ², |c_W| ≤ 2η/σ²

Only the authors can say what is right here.

Judge's note: (1) On p.23 the hypothesis is η ≤ σ²/(4B̄G). Corollary E.6 (p.25) applies Prop. E.3 under the same condition. The proof on p.30 invokes the stricter η ≤ σ²/(8B̄G), which does not follow from the hypothesis. (2) I reworked the proof of Lemma E.10 on p.31 myself. The centring term is 2(c_Uc_W VᵀHV + σ̄_UW trH), with |c_U| ≤ η/σ² and |c_W| ≤ 2η/σ² from Lemma E.10(2) on p.30. That gives |2c_Uc_W VᵀHV| ≤ 4η²/σ⁴|VᵀHV|, not 4η²/σ², while |2σ̄_UW trH| ≤ 4η²/σ²|trH| is correct. Dimensional check: η²/σ²·VᵀHV is not scale-invariant. I checked the reviewer's numerical counterexample: d=1, r=0, y=1, H=1, σ=0.1, η=1e−4, V=1e4. Then σ_V² = 0.0102, c_U ≈ 9.8e−3, c_W ≈ −1.54e−2, conditional centre ≈ −1.5e4 with std ≈ 2.2. The stated shift is ≈ 400, so with t=7500 the left side is ≈ 1 while the right side is ≈ 2exp(−c·5.6e7). The p.30 condition '|VᵀHV|+|tr(H)| ≤ Bσ²/(16η²)' inherits the same slip. Prop. E.3 appears repairable with the σ⁴ version and a constant change, but Lemma E.10(3) is wrong as stated.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 28 Lemma E.9, first moment bound

  ∥s̃⋆(X)∥

Only the authors can say what is right here.

Judge's note: I read p.28 (Lemma E.9) and p.22 (definition of s̃⋆). Checked counterexample: p̄ = δ_m gives p = N(m, σ²I) and s⋆(x) = −(x−m)/σ². With λ = 1/σ², which the lemma permits, s̃⋆ ≡ m/σ², so the left side is ‖m‖/σ, which is unbounded in m. I found no centring assumption in the text (search for center/mean zero/WLOG). The second bound, for ∇s̃⋆, is unaffected. The first bound is false as stated whenever λ > 0.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 28 Lemma E.9

  0 ≤ λ ≤ 1/σ². Then the following holds with an absolute constant C > 0

• p. 28 Proof of Prop. E.8

  X0 ∼ ¯p, X ∼ N X, σ2I

Only the authors can say what is right here.

Judge's note: Page 28 image: Lemma E.9 allows 0 ≤ λ ≤ 1/σ² and claims the first moment bound C√d/σ. On p.22 (Notation, E.2) s̃⋆(x)=s⋆(x)+λx with s⋆=∇log p, and Assumption E.2 places no restriction on where p̄ is located. My own counterexample: take p̄=δ_m in R^d, so p=N(m,σ²I), with η'=0 and λ=1/σ². Then s̃⋆(x) = −(x−m)/σ² + x/σ² = m/σ², so E‖s̃⋆(X)‖ = ‖m‖/σ². This is unbounded in m, so no absolute constant C gives C√d/σ. The nuclear-norm Hessian bound is unaffected, because λI adds at most dλ ≤ d/σ². The lemma is not cited by number elsewhere in the text, so its downstream role is unclear. As a stated bound it is false as printed, hence S1. The image also confirms the secondary point in the proof of Prop. E.8: 'X ∼ N(X, σ²I)' is printed where the mean should be X0.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 31 Lemma E.10 (3) statement and end of proof

  The desired upper bound then follows from bounding

Only the authors can say what is right here.

Judge's note: P.30/31 images: the proof ends 'follows from bounding |c_U| ≤ η/σ², |c_W| ≤ 2η/σ², σ̄_UW ≤ 2η²/σ²'. This gives 2|c_Uc_W| ≤ 4η²/σ⁴, not 4η²/σ². My own check that this is attained: take y=0 and r=1/2, with a_r = sin(πr/2) as defined on p.7. Then σ_VU = −ηa_r, σ_VW = −ηa_r' and σ_V² = σ²+η, so |c_Uc_W| = (π/4)η²/(σ²+η)². With H = h·diag(1,−1) (so trH = 0), V = (v,0) and σ² small (e.g. 0.01, η = 1e-4), the conditional mean (π/4)η²/(σ²+η)²·hv² exceeds (4η²/σ²)hv². The fluctuations grow only linearly in v while the gap grows like v², and the lemma's right-hand side tends to 0 as v→∞. So the printed inequality (3) fails when σ² is small. Downstream use on p.30 may be repairable through the E_{G,K} constraints.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 31 Lemma E.10(3)

  4η²/σ² (|VᵀHV| + |tr(H)|)

Only the authors can say what is right here.

Judge's note: On p.31 the proof gives |c_U| ≤ η/σ² and |c_W| ≤ 2η/σ², so |c_Uc_W| ≤ 2η²/σ⁴, and σ̄_UW ≤ 2η²/σ². I recomputed the reviewer's case: d=1, r=0, y=1, H=1, σ²=0.01, η=1e−4. Then c_Uc_W = −(π/2)η²/(σ²+2η)² ≈ −1.51e−4, against the printed 4η²/σ² = 4e−6. Take t = Av²/2 with A = |c_Uc_W|. As |v| → ∞, the conditional mean ≈ Av² exceeds the threshold by order v², while the fluctuations are O(|v|), so the left probability tends to 1. The right side tends to 0. Nothing in the paper normalises σ, so the statement is false as printed.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 25 Proof of Corollary E.6, last line

  Applying Corollary E.6 completes the proof.

Judge's note: The p.25 image shows the self-citation. The analogous proof of Corollary E.5 (p.24) ends 'applying Proposition E.4 (2) gives the desired result'. The step needed here turns the KL bound against ρ* into the score-error bound, which is Proposition E.4 (1) on p.23.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 16 Proof of Lemma B.11

  It holds that Eµ[h2] ≤2 Eν[h2] + 2B2D2

• p. 16 Proof of Lemma B.11

  E_μ[h²] ≤ 2 E_ν[h²] + 2B²D_H²(ν, μ)

• p. 16 Proof of Lemma B.11

  It holds that E_µ[h²] ≤ 2 E_ν[h²] + 2B²D²_H(ν, µ) because

• p. 16 Proof of Lemma B.11

  Eµ[h2] ≤ 2 Eν[h2] + 2B2D2H(ν, µ) because

Only the authors can say what is right here.

Judge's note: On p.16, the sentence states the coefficient 2B². The display beneath it ends with 4B²D_H², and the next chain uses 3E_ν[h²]+4B²D_H². The paper uses the ½-normalised Hellinger distance: Lemma B.10's proof has ∫(√P−√Q)² = 2D_H². So 2∫h²(√µ−√ν)² ≤ 4B²D_H², and the claimed 2B² is false in general. Counterexample checked: h is the indicator of one point, B=1, µ gives it 0.7, ν gives it 0.1. Then D_H² = 1−√0.07−√0.27 ≈ 0.2158, and 0.7 > 0.2+2(0.2158) = 0.632. The lemma's conclusion is unaffected because 4B² is what the proof carries forward.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 17 Proof of Lemma B.13, first chain

  ≤ (E_{µf} e^{(ℓ−1)|f−g|})² ≤ E_{µf}[e^{2ℓ|f−g|}]

Only the authors can say what is right here.

Judge's note: Page 17. The step needs (E e^X)^{a} ≤ E e^{a|X|} with a = ℓ−1. For a < 1, x^a is concave, so Jensen goes the other way. Counterexample (realisable by choosing g = f − X): X = f−g under µf is 0 w.p. 1−1e−10 and ln(1e16) w.p. 1e−10, with ℓ = 1.5. Then the LHS is ≈ (1+1e6)^{0.5}·1 ≈ 1000, while (E e^{0.5|X|})^2 = (1 + 0.01)^2 ≈ 1.02. The final bound E e^{2ℓ|f−g|} still holds (here it is about e^{110}·1e−10), and Theorem D.1 uses ℓ ≥ 2, so the stated lemma stands; only the intermediate step is wrong.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 19 Claim 1 and its proof

  log Ex∼q, z∼P exp(λ|Wr,z,x|) ≤10dsη1+sλ2β2

• p. 19 Proof of Theorem D.1, Claim 1

  log E exp(λ|W_{r,z,x}|) ≤ 10d^sη^{1+s}λ²β_s²

• p. 19 Claim 1 statement vs last line of its proof

  log E_{x∼q, z∼P} exp(λ|W_{r,z,x}|) ≤ 10d^sη^{1+s}λ²β²_s

Only the authors can say what is right here.

Judge's note: On p.19 the last line of the proof keeps the factor 2, and Claim 2's proof on p.20 uses the factor-2 form. As printed, the claim is false: Jensen gives log E e^{λ|W|} ≥ λE|W|, which is linear in λ, while the right side is quadratic in λ. So the claim fails for small λ whenever E|W| > 0, for example f = x²/2 with x̂ = x₊, where E|W| = η. Adding log 2 repairs it, and Theorem D.1 is unaffected.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 20 Proof of Theorem 4.1

  where we used |a −ClipB(b)| ≤ClipB(|a −b|) + τB(b) for a, b ∈R

• p. 20 Appendix E.1, after the definition of u(x), v(x)

  where we used |a − Clip_B(b)| ≤ Clip_B(|a − b|) + τ_B(b) for a, b ∈ R

Only the authors can say what is right here.

Judge's note: I checked the p.20 image. Counterexample: B=1, a=−1, b=1. The left side is |−1−1| = 2. The right side is Clip_1(2)+τ_1(1) = 1. A second case, a=10B, b=0, gives 10B > B. In the display, the clipped argument is ⟨x−x̄,s_t⟩ (b) and τ_B acts on the ∇log p_t term (a). Putting τ_B(a) in place of τ_B(b) still fails on B=1, a=−1, b=1. My own reworking gives |Clip_B(b)−a| ≤ |Clip_B(b)−Clip_B(a)| + τ_B(a) ≤ min{|a−b|,2B} + τ_B(a). This is a Clip_{2B}-type bound, so the result likely holds with changed constants (e^{2B}). The step as printed is false.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 20 Proof of Theorem 4.1, bound after 'Applying Lemma B.12 gives'

  exp(⟨x −x′, ∇log pt(rx + (1 −r)¯xt)⟩) −eB

• p. 20 First bound after 'Applying Lemma B.12 gives'

  exp(⟨x − x′, ∇log p_t(rx + (1 − r)x̄_t)⟩) − e^B

Only the authors can say what is right here.

Judge's note: On p.20, v(x) is defined through τ_B, which involves |·|, and the next display restores |⟨Z,·⟩|. Without the absolute value the bound is wrong: for p_t Gaussian with the exact score, the printed tail term is identically 0 while the KL divergence is positive. x′ is not defined in this proof; it should be x̄_t. This is a typographical defect in an intermediate step, and the next display carries the correct form.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 20 Opening of the proof of Theorem 4.1

  D_KL(p₁ ∥ p̂₁) = D_KL(p_T ∥ p̂_T) +

• p. 20 Appendix E.1, first display

  D_KL(p1 ∥ p̂1) = D_KL(p_T ∥ p̂_T) + Σ_{t=1}^{T−1} E_{X_{t+1}∼p_{t+1}} D_KL(ρ_t(· | X_{t+1}) ∥ ρ̂_t(· | X_{t+1}))

Only the authors can say what is right here.

Judge's note: On p.20, the chain rule gives equality for the joint path laws. For the marginal of X₁, data processing gives only an upper bound. Checked counterexample: if both backward kernels are the same and ignore the conditioning variable, the left side is 0 while the right side is D_KL(p_T∥p̂_T) > 0. Theorem 4.1 is an upper bound, so replacing '=' by '≤' preserves it.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 20 Proof of Theorem 4.1, before applying Lemma B.12

  |a − Clip_B(b)| ≤ Clip_B(|a − b|) + τ_B(b)

Only the authors can say what is right here.

Judge's note: Clip_B is defined as max{−B, min{B, x}}, so its range is [−B, B]. Counterexample: B=1, a=1, b=−1. The left side is |1−(−1)| = 2; the right side is Clip_1(2)+τ_1(−1) = 1+0 = 1. The intended bound, for |Clip_B(a)−b| as it is used here, is min{|a−b|, 2B} + τ_B(b); the cap has to be 2B. This changes the constant in u(x) when Lemma B.12 is applied, from B to 2B, but not the order of the result.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 26 Proof of Proposition E.7, definition of g_i

  + ⟨V, z_i − X₀⟩/σ²

• p. 27 First displayed inequality

  g_i(V) exp(⟨V, x̄ − z_i⟩/σ²)

• p. 27 Appendix E.4, proof of Prop. E.7, first display

  √M(x̄ + V) ≤ e^{2r²/σ²}/(p(B_j)u(V)) Σ_{i∈I} g_i(V) exp(⟨V, x̄ − z_i⟩/σ²)

• p. 26 definition of g_i(V) and u(V)

  g_i(V) := E_{X0∼p̄} I{X0 ∈ B_i} exp(−∥x̄ − X0∥²/(4σ²) + ⟨V, z_i − X0⟩/σ²)

Only the authors can say what is right here.

Judge's note: Reworked from p.26: by (25), √M ≤ E_q exp(‖X₀−x̄‖²/4σ²). Its numerator is E_p̄ exp(−‖x̄−X₀‖²/4σ² − ⟨V,x̄−X₀⟩/σ²). The printed sum equals E_p̄ exp(−‖x̄−X₀‖²/4σ² + ⟨V,x̄−X₀⟩/σ²), which has the opposite sign. Checked counterexample: prior uniform on {±2}, σ=1, r=0, x̄=2, V=−2. Then √M(0) = e^{1/4} ≈ 1.284, while the printed right side is 1+e^{−12} ≈ 1.000006. Because V is symmetric, the argument can be redone with −V, so the proposition likely stands.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 29 Start of Appendix E.6

  u(γz,r,x0(x)) −u(x0)

Only the authors can say what is right here.

Judge's note: A text search finds sans-serif u only in E.6 (p.29: the W⋆ definition, the ∇W⋆ formula and the sub-exponential norm bounds). The only other u, at p.21 (E.1), is an unrelated scalar function u(x). Later on p.29, and in Eq. (27) on p.31, W⋆ is written as ⟨γ̇, s̃⋆(γ) − s̃⋆(x0)⟩ with no σ⁻². This reconciles only if u = σ²s̃⋆, which is never stated.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 29 Start of Appendix E.6

  = const + Er,z τB(W ⋆

• p. 29 Appendix E.6, opening displays

  log ν_{x0}(x) − log ρ⋆_{x0}(x) = const + E_{r,z} τ_B(W⋆_{z,r,x0}(x))

Only the authors can say what is right here.

Judge's note: Reworked from p.23, where log ν + ‖·‖²/2η = const + E⟨γ̇, s̃⋆(γ) − s̃⋆(x0)⟩, and from Eq. (24), where log ρ⋆ uses E Clip_B(·) with the same quadratic term. The difference is therefore E[W⋆ − Clip_B(W⋆)]. On p.20 (around line 2037 of the text layer) τ_B(y) := (|y|−B)_+ = |y − Clip_B(y)|. For y < −B, y − Clip_B(y) = y + B < 0 while τ_B(y) > 0, so the displayed identity is false as printed. Similarly, the gradient of the true log-ratio is ∇W⋆·1{|W⋆|>B}, not ∇W⋆·h_B(W⋆). The next step uses only E‖∇W⋆‖²1{|W⋆|>B}, so the stated result stands. The step is faulty but repairable.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 29 Opening density identity and its derivative

  log ν_{x₀}(x) − log ρ⋆_{x₀}(x) = const + E_{r,z} τ_B(W⋆)

Only the authors can say what is right here.

Judge's note: From the p.23 identity, log ν + quadratic = const + E⟨γ̇, s̃⋆(γ) − s̃⋆(x₀)⟩. From (24), log ρ⋆ + the same quadratic = const + E Clip_B(·). So the difference is E[W−Clip_B(W)], which is negative when W < −B, while τ_B ≥ 0. The derivative h_B (−1 for y < −B) then has the wrong sign there. The downstream bound uses only h_B² = I{|W|>B}, so the KL estimate stands.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 30 End of proof of Proposition E.3

  (EG,K) + 2 exp(−c2K)

• p. 30 last display of Appendix E.6

  ≤ P_{x̄∼p̄, V∼N(0,σ²_V I)}(E_{G,K}) + 2 exp(−c2K)

Only the authors can say what is right here.

Judge's note: The p.30 image shows plain E_{G,K} with no complement mark. The preceding display gives the conditional bound only for (x̄,V) ∈ E_{G,K}, and Prop. E.3 (p.23) is stated with P((x̄,V) ∉ E_{G,K}). As printed, the bound is trivial. The stated proposition stands, so this is a faulty step rather than a wrong result.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 31 Proof of Lemma E.10(3), final probability display

  2(c_Uc_W VᵀHVᵀ + σ̄_UW tr(H))

• p. 31 proof of Lemma E.10 (3)

  P(|WᵀHU − 2(c_Uc_W VᵀHVᵀ + σ̄_UW tr(H))| > t | x̄, V)

Only the authors can say what is right here.

Judge's note: On p.31, with A = [0 H; H 0] we have xᵀAx = 2WᵀHU, vᵀAv = 2c_Uc_W VᵀHV and tr(ΣA) = 2σ̄_UW tr H. Lemma B.5 therefore controls |2WᵀHU − 2(mean)|. In the printed display only the centring carries the 2, so when the mean dominates the fluctuations the left side is close to 1 while the right side is tiny. The correct form is |WᵀHU − (c_Uc_WVᵀHV + σ̄_UW trH)| > t/2. This changes constants only.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send

• p. 33 Putting Everything Together, last displayed inequality

  min{∥s(X) − s⋆(X)∥², 1}

Only the authors can say what is right here.

Judge's note: On pp.32–33, min{Gx, B²} ≤ B̄² min{x,1} needs G ≲ B̄², and G = η + η²(d+log(1/δ))/σ² is not bounded by the assumptions. Prop. E.4(2) on p.23 keeps G. Gaussian check, reworked: p = N(0,σ²), λ = 1/σ², s = s⋆ + n^{−1/2}, η = n, σ² = n³, δ = 1/n. Then s̃⋆ ≡ 0, so ν_{x₀} = ρ⋆_{x₀} exactly, and the KL term is 0. ρ̂ is shifted by η̄n^{−1/2} ≈ √n with variance η̄ ≈ n, so D_H² → 1 − e^{−1/8} ≈ 0.118, while the printed right side is O(1/n). The step-size assumption of E.4(2) holds for large n. So the final line is false as printed, while E.4(2) as stated, which keeps G, is consistent with this example.

Second check: another model agreed

Report a problem with this finding

This finding is

wrong right unclear or overstated already corrected (erratum, later version)

I am an author of this paper Send