How much can language models memorize?

• p. 2 Figure 1

  Memorization plateaus at the capacity of different-sized models

• p. 1 Introduction

  can store be-tween 3.5 and 4 bits of information in each model parameter

Only the authors can say what is right here.

Judge's note: Read the Figure 1 plateau levels from the page 2 image (log y-axis, ticks 10^5 to 10^9). The figure readings are approximate, but the gap is about a factor of 2. Section 3.2 Results presents Figure 1 as the source of the capacity estimates, and the Introduction (page 1) says GPT-style transformers 'can store between 3.5 and 4 bits'. Table 1 gives 2.51e7 bits for the 6.86M bf16 model, and Figure 10 shows ≈2.39e7 bits for a 6.86M model, whereas the ~7M curve in Figure 1 plateaus near 1.3e7. No different training setup is given for Figure 1.

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• p. 4 Section 3.1

  ing all models for 10^6 steps, well past saturation

• p. 19 A.13 Sweep

  10^5 training steps, bfloat16

• p. 4 Section 3.1

  ing all models for 10^6 steps, well past saturation for every configuration we tested (10)

• p. 5 Section 3.2, Experimental details

  We train models for 10^6 steps with a batch size of 2048

Only the authors can say what is right here.

Judge's note: Page 4 (Section 3.1) and page 5 (Section 3.2) give 10^6 steps. Page 6 (Section 4) says 'We repeat the experiments from 3.2'. Page 19 (A.13) gives 10^5 training steps for the main FineWeb sweep. These are two different step counts for the same main-text text experiments. Figure 10 (page 13) is labelled in steps but its axis runs from 10^0 to about 4×10^2, which fits neither count unless the unit is something other than steps. Only the authors can say which count is correct.

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• p. 17 A.11 end of proof

  independent of independent of

• p. 17 A.11

  the shortest description of f independent of independent of

• p. 17 A.11

  Let f be a the density function

• p. 17 A.11, paragraph defining f

  Note that this function is independent of n and θ.

Only the authors can say what is right here.

Judge's note: I checked p.17 (A.11) and the Proposition 4 statement on p.4 ("some constant ϵ independent of θ, ℓ, ℓ′ and n"). By its own definition, f is the joint law of (X_i, Θ̂) given θ. Two θ with different marginals for X_i give different f, and the law of Θ̂ generally changes with n. The proof's justification, that f is described by the training algorithm and "the sampling algorithm that uses θ", builds θ (and n) into the description. Counting argument: there are fewer than 2^(c+1) programs of length ≤ c, so over any infinite family of distinct f_θ (or f_n), H_K(f) is unbounded. Hence ε = 2H_K(f) is not shown to be a constant independent of θ and n, and the uniformity in the stated proposition is unproven (S1 under the derivation rule). Other defects in the merged quotes: the duplicated words "independent of independent of" (p.17, S4) and "worst-cast drift" (p.13, S4). Both are confirmed on the images and covered here. "Let f be a the" is judged separately under C050.

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• p. 5 Section 3.2 Results

  datasets from 16,000 to 4M samples fall within a range of 3.56 − 3.65 × 10^6 bits memorized

• p. 13 Figure 10 legend/caption

  6.86M parameters and a capacity of 23.9 MB

• p. 5 Section 3.2, Results

  All datasets from 16,000 to 4M samples fall within a range of 3.56 −3.65 × 10^6 bits memorized

• p. 13 Figure 10 legend

  65536

Only the authors can say what is right here.

Judge's note: Page 5 text against the Figure 10 image on page 13. The figure's legend sizes (65536 to 8388608) and its plateau (≈2.39e7 bits) match neither the stated dataset range nor the stated bit range. No other figure shows a model plateauing at ~3.6e6 bits across 16K–4M samples. A separate point in the same figure: the line at ≈2.39e7 bits is labelled 23.9 MB, but 2.39e7 bits is about 3 MB.

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• p. 7 Section 4.1 TF-IDF equation

  TF-IDF(d; D) = 1/|d| Σ log |D|/tf(w, D)

• p. 9 Figure 9 y-axis

  6 × 10^4

• p. 18 Table 5 TFIDF column

  78553.72

Only the authors can say what is right here.

Judge's note: Each word w in d occurs in D, so tf(w,D) ≥ 1 and each summand is at most log|D|. The 1/|d| average is therefore at most log|D|. For 2^16 documents, or 2^16×64 tokens, that bound is roughly 11–22 depending on the log base, many orders of magnitude below the plotted and tabulated values. Even dropping the 1/|d| factor (64 terms) gives at most about 1400. The formula as printed cannot have produced the reported values. Only the authors can say what was actually computed.

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• p. 8 Section 5.2.1 equation

  c2 Capacity(θ)/|D| + c3

• p. 9 Fitting

  c1 = 1.34, c2 = −0.034, and c3 = −33.14

• p. 8 Section 5.2.1, equation

  Membership_F1(θ, D) = 1/2(1 + c1σ(c2 Capacity(θ)/|D| + c3))

• p. 8 Section 5.2.1, MembershipF1 equation

  + c3

Only the authors can say what is right here.

Judge's note: The p8 image shows σ(c2·Capacity(θ)/|D| + c3), with c3 outside the product. Reworked: for x = Capacity/|D| > 0, c2 < 0 gives an argument ≤ c3 = −33.14, so σ ≤ about 4e-15 and F1 ≤ 0.5(1 + 1.34·4e-15) ≈ 0.5. Reaching F1 = 0.55 needs σ = 0.1/1.34 ≈ 0.075 (argument ≈ −2.5), and F1 = 0.95 needs σ ≈ 0.67 (argument ≈ 0.72), neither reachable. Also, since c2 < 0, F1 would decrease in Capacity/|D|, i.e. rise with |D|, opposite to the stated decay. The functional form, its argument or the constants are misstated. Only the authors can say which.

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• p. 9 Figure 9 caption

  trained past its capacity on 2^16 sequences

• p. 15 A.7

  just one out of 260, 000 training samples

• p. 18 Table 5 caption

  trained past its capacity on 2^16 sequences of English text

• p. 9 Figure 9 caption

  trained past its capacity on 2^16 sequences of English text

Only the authors can say what is right here.

Judge's note: Captions on p9 and p18 show 2^16 = 65,536. A.7 (p15) discusses the trainset TF-IDF/memorization correlation and the highest-TF-IDF sample, and points to Table 5, so it describes the same analysis. 260,000 ≈ 2^18 = 262,144. I found no statement that a different run is meant. (The 'third-highest' point is not contradicted: Table 5 lists only the top TF-IDF rows, not all memorization values.)

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• p. 15 A.5 text

  and 1.8% when scaling V

• p. 16 Figure 13 caption

  with an average error rate of 1.8%

• p. 15 Table 4 Error column

  0.36

• p. 15 A.5

  1.8% when scaling V

Judge's note: Table 4 on page 15 lists errors of 0.36, 2.78, 2.70, 1.11 and 0.47, which average 1.48. The same check on Table 3 gives 1.70, which matches the stated 1.7% for S. The paper gives no alternative averaging rule.

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• p. 15 A.5

  we substitute our previous estimate of α = 3.642

• p. 15 Table 3, S=64 row

  6.67 × 10^5 2.29 × 10^6 2.36 × 10^6

Only the authors can say what is right here.

Judge's note: Tables 3 and 4, page 15. The data-limited rows match N·S·log2V exactly: S=4 gives 1.80e5, S=32 gives 1.44e6, and V=4096 gives 3.15e6. For the capacity-limited rows, H(X) is larger than α·Params, so Expected should be 3.642·Params. It is consistently about 3% lower, a ratio near 3.54. The rounded Params cannot account for a constant offset. Only the authors can say which α was used.

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• p. 19 A.13 Sensitivity panel

  The slope of the decrease is steeper for cells with more memorization (smaller |X| or larger dmodel)

• p. 20 Figure 18

  n=2^14

Only the authors can say what is right here.

Judge's note: I looked at Figure 18 on p.20 (crop). d=64: n=2^14 goes from ≈2.5e4 to ≈1e4; n=2^24 goes from ≈4e4 to ≈15. d=256: n=2^14 is nearly flat (≈8.5e4 to ≈7e4); n=2^24 goes from ≈1.7e6 to ≈4e3. The ordering by |X| is the reverse of the text on both log and absolute scales. The d=256 small-|X| lines are also flat, which does not support "larger dmodel" either. Only the authors can say which description is intended.

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• p. 2 Figure 2 caption

  with re-spect to a large oracle model (1B params)

• p. 7 Section 4

  We use the largest reference (774M) throughout the main text

• p. 18 A.13

  (774M parameters, FineWeb-finetuned)

• p. 2 Figure 2 caption

  large oracle model (1B params)

Only the authors can say what is right here.

Judge's note: Checked the page 2 caption, page 7 (Section 4, 'On the role of the reference model'), and page 18 (A.13, where the bullet list names gpt2-774M-fineweb-150B as the baseline used in the main paper). Figure 2 is a main-text figure, and Section 4 cites it as the oracle-reference result. Nothing in the paper describes a separate 1B oracle, so either the caption or the text gives the wrong size. Only the authors can say which.

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• p. 13 Figure 10 legend and caption

  a capacity of 23.9 MB

• p. 13 Figure 10 caption and legend

  6.86M parameters and a capacity of 23.9 MB

• p. 5 text referencing Figure 10

  All datasets from 16,000 to 4M samples fall within a range of 3.56 −3.65 × 106 bits memorized

• p. 5 Sec. 3.2

  3.56 −3.65 × 106 bits memorized

Only the authors can say what is right here.

Judge's note: On page 13 the dashed line labelled 'Estimated capacity (23.9 MB)' lies just above the 10^7 gridline, at about 2.4×10^7 bits. Table 1 gives the 6.86M bf16 model a capacity of 2.51×10^7 bits. 23.9 MB would be about 1.9×10^8 bits, roughly 28 bits per parameter, so the unit should read Mbit. Separately, the page 5 sentence refers to Figure 10, but its sample range and its 3.56–3.65×10^6-bit range belong to a different model or set of runs than the one plotted. Only the authors can say which they meant.

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• p. 9 Section 5.2.2

  We train models on the estimated dataset size and measure F1 score (Figure 5).

• p. 6 Figure 5 y-axis

  Model capacity (bits)

Judge's note: Figure 5 (p6) has a model-capacity axis from about 1e6 to about 1e8 bits, with circles on four capacity rows (the small text models). The GPT-2 models would sit at about 4.5e8 and 5.7e9 bits (params × 3.64), off the axis. Their F1 results appear only in Table 2 (p15), which the text never cites (grep finds only the caption).

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• p. 7 Section 4, 'Generalization explains nonzero extraction rates'

  non-overlapping test samples (Figure 16)

• p. 17 Figure 16 caption

  Our sigmoidal scaling law for membership inference fit to experimental data.

• p. 7 Generalization explains nonzero extraction rates

  10,000 non-overlapping test samples (Figure 16)

Judge's note: Figure 16 (p17) plots membership inference F1 against capacity ratio (prediction vs empirical) and contains no extraction rates. Figure 6 (p8) shows 'Extraction rates of 64-token training sequences across prefix lengths' with train/test curves for prefixes 8/16/32, which matches the paragraph.

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• p. 8 Figure 8 caption

  trained on 2^14 sequences of

• p. 8 Figure 8 panel titles

  Compression rate - bitstrings (N = 262144)

• p. 8 Figure 8 caption

  trained on 2^14 sequences of equal-length random bitstrings

• p. 8 Figure 8 caption

  trained on 214 sequences of

Only the authors can say what is right here.

Judge's note: The p8 image confirms the superscript 14 in the caption and N = 262144 in both panel titles. In the bitstring panel, about ten single bins hold close to 3e4 train counts, which is incompatible with 16,384 training sequences, so the caption value appears to be the wrong one. Nothing in the paper defines N as a different quantity.

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• p. 15 Table 2, second model row

  GPT2-Medium

• p. 9 Section 5.2.2

  we select GPT-2 small (125M params)

• p. 9 Section 5.2.2

  For model sizes we select GPT-2 small (125M params) and GPT-2 XL (1.5B params).

• p. 9 5.2.2

  GPT-2 small (125M params)

Judge's note: Page 15, Table 2: the row reads 'GPT2-Medium 768 12 123,702,528'. Page 9 says the models are GPT-2 small and GPT-2 XL, and A.13 (page 18) gives gpt2-medium as 355M. The dimensions, layer count and parameter count in the row are those of GPT-2 small, so the row label is wrong.

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• p. 19 A.14 Limitations

  a new perspective on the phenomenon of grokking (Nakkiran et al., 2019)

• p. 11 References

  Deep double descent: Where bigger models and more data hurt

• p. 11 References

  Deep double descent: Where big-ger models and more data hurt, 2019.

• p. 19 A.14

  grokking (Nakkiran et al., 2019)

Judge's note: I checked p.19 and the reference entry on p.11. The entry's title is about double descent. "Grokking" appears only once in the text. The paper's results are about double descent (e.g., "Dataset-to-capacity ratio predicts double descent", which cites Nakkiran et al. 2019). The term and the citation do not match; the evident fix is "double descent".

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• p. 13 Figure 11 axes

  Total memorization (Bits)

Judge's note: In the crop, the leftmost point sits just left of x=10^5 and below y=10^6 (about 3e5). The top-right point is at about x=6.5e6, y=2.5e7. These match Table 1 (params 8.04e4–6.86e6; bf16 capacity 3.16e5–2.51e7). Read as labelled, the points would give fewer than 1 bit per parameter, which contradicts the 3.64 bits-per-parameter annotation on the plot. The labels should be exchanged.

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• p. 3 Section 2.1, memI definition

  memI(Θ̂, X, Θ) = mem(X, Θ) − memU(X, Θ̂, Θ)

• p. 3 Section 2.1, definition of mem_I

  mem_I(Θ̂, X, Θ) = mem(X, Θ) − mem_U(X, Θ̂, Θ)

• p. 3 Section 2.1, display defining mem_I

  memI(ˆΘ, X, Θ) = mem(X, Θ) −memU(X, ˆΘ, Θ)

• p. 3 Sec. 2.1

  = mem(X, Θ) −memU(X, ˆΘ, Θ)

Judge's note: The page 3 image shows mem(X, Θ) with no hat. The paper defines mem(X, Θ̂) = I(X; Θ̂). The second line of the display, and A.12 (page 17: H(X)−H(X|Θ̂)−H(X|Θ)+H(X|Θ,Θ̂)), both correspond to I(X;Θ̂) − I(X;Θ̂|Θ), so the printed first equality does not match its own second line. The fix is evident (add the hat), and the results are unaffected. A minor related point: the argument order is memI(Θ̂, X, Θ) here but memI(X, Θ̂, Θ) in Proposition 6.

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• p. 4 Definition 3

  memK U (x, θ, ˆθ) = HK(x | θ) −HK(x | (θ, ˆθ))

• p. 4 Proposition 4

  memK U (xi, ˆθ, θ)]

• p. 17 A.11, display after 'Now using Lemma 8 we have'

  xi∼Xi|θ[IK(xi, ˆθ)]

Only the authors can say what is right here.

Judge's note: Read Definition 3 and Proposition 4 on page 4, and Lemma 8 and the proof on pages 16–17. Reworking: Lemma 8 concerns I_K(x,y) = H_K(x) − H_K(x|y). The proof's middle term is E[H_K(x_i) − H_K(x_i|θ̂)]. memK_U is H_K(x_i|θ) − H_K(x_i|θ,θ̂). These differ in general. Counterexample: x_i = (θ, u_i) and θ̂ = θ give memK_U ≈ 0, but I_K(x_i, θ̂) ≈ H_K(θ). So the final inference does not follow as printed. The proof also says H_K(f) is independent of θ, yet it describes f as using θ. The proposition plausibly holds with a version of Lemma 8 conditioned on θ, so I rate this a faulty step rather than a wrong result (S3). The argument-order swap between Definition 3 and Proposition 4 is also confirmed. The brackets in Proposition 4 are balanced, so that part of reviewer note 1 is not a defect.

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• p. 5 Section 3.2 Results

  our models consistently memorize between 3.5 and 3.6 bits per parameter

• p. 6 Table 1, α bf16 column

  2.86

• p. 6 Table 1

  Mean (±0.1):

• p. 13 Figure 11 caption

  We estimate α = 3.64 bits-per-parameter

Only the authors can say what is right here.

Judge's note: Checked the Table 1 values on page 6. The bf16 mean (3.51) and fp32 mean (3.83) reproduce from the 16 values. The values in [3.5, 3.6] are 3.56, 3.53 and 3.51. The Figure 11 caption (page 13) gives 3.64 for the same half-precision models, presumably a fitted slope, but the paper does not say how it differs from the Table 1 mean. 'Consistently' overstates the per-model agreement.

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• p. 6 Section 4, 'On the role of the reference model'

  MEMU defines a consistent ordering across target models and dataset sizes

• p. 19 A.13 Results

  relative comparisons across target runs are preserved

• p. 20 Figure 17/18, d_model=64 panel

  n=2^24

• p. 7 Section 4, On the role of the reference model

  absolute MEMU values shift across references, but the qual-itative shape of the curve is preserved

Only the authors can say what is right here.

Judge's note: Figure 17 (p20), d_model=64 panel: against the 124M reference, MEMU is about 2.6e4 at the smallest n and about 4e4 at the largest, with a U-shape that rises at large n. Against the 774M reference it goes from about 1e4 down to about 2 near 1e6, ending near 15, so the n=2^14 vs n=2^24 ordering reverses. The d=128 and d=256 panels show the same pattern: the 124M/355M curves rise to their maxima at the largest n, while the 774M curve collapses. Figure 18 shows crossing lines: at d=256, n=2^14 is the lowest line at 124M and the highest at 774M. The text claims (p6–7, p19, Fig 17 caption) therefore misdescribe the plotted ablation. Only the authors can say what was meant, e.g. orderings at a fixed dataset size.

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• p. 9 Section 5.2.2

  Our predictions are generally within 1.5 points of the true F1 score.

• p. 15 Table 2

  65.69 ± 0.6

• p. 15 Table 2

  71.08 ± 0.4

• p. 9 5.2.2

  generally within 1.5 points of the true F1 score

Only the authors can say what is right here.

Judge's note: The Table 2 image (p15) has predicted 0.55/0.75/0.95 against observed 54.61/71.08/95.85 (GPT2-XL) and 53.44/65.69/97.98 (GPT2-Medium). After aligning scales the absolute deviations are 0.39, 3.92, 0.85, 1.56, 9.31 and 2.98. Only two are within 1.5 points, so the sentence misdescribes the table.

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• p. 16 A.10, last display

  = H(ˆΘ) −H(ˆΘ | (X | Θ))

Only the authors can say what is right here.

Judge's note: I checked the display on p.16 (crop and full page) and reworked it. Read as conditioning on (X,Θ), the printed right-hand side is H(Θ̂) − H(Θ̂|X,Θ) = I(Θ̂; X,Θ). Read as conditioning on X alone, it is I(X;Θ̂). Neither equals I(X;Θ̂|Θ) in general. Counterexample, consistent with X_i i.i.d. given Θ: Θ ~ Bernoulli(1/2), X_i = Θ, Θ̂ = X_1. Then I(X;Θ̂|Θ) = 0, but H(Θ̂) − H(Θ̂|X,Θ) = 1 − 0 = 1 bit. The stated upper bound still holds through I(X;Θ̂|Θ) = H(Θ̂|Θ) − H(Θ̂|X,Θ) ≤ H(Θ̂|Θ) ≤ H(Θ̂). So the step is faulty but the result stands (S3).

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• p. 17 Figure 15 caption

  never falls below 0.54

• p. 17 Figure 15 caption

  membership inference performance never falls below 0.54

• p. 17 Figure 15 plot

  Membership inference score

Only the authors can say what is right here.

Judge's note: I looked at Figure 15 on p.17 in the crop and on the full page. The last point markers (near 10^7 datapoints) of the 1.7e+05 curve (≈0.52) and the 5.4e+05 curve (≈0.53) sit clearly below 0.54: between the 0.5 and 0.6 gridlines, nearer 0.5. These are the point markers themselves, not the shaded bands. I found nothing in the text (Section 5.1) that explains a different floor. Only the authors know the exact values, so this goes to them as a query.

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• p. 17 A.11, application of Lemma 8

  I(Xi, Θ̂; θ) − HK(f) ≤

Only the authors can say what is right here.

Judge's note: I checked p.17. Lemma 8, as printed on the same page, has the same I(X;Y) on both sides. Applied to the pair (X_i, Θ̂) under f (which is conditional on θ), it gives I(X_i; Θ̂ | θ) − H_K(f) on the left. The printed I(X_i, Θ̂; θ) is a different quantity, and it is 0 when θ is fixed, as in this proof. The step is faulty as printed, but it is repaired by the evident substitution, so the stated bound is unaffected by this particular slip (S3).

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