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Calibrating Noise to Sensitivity in Private Data Analysis

Lecture notes in computer science, 2006 · paper · Computer science

Studies privacy for statistical databases where random noise is added to query answers. Proves privacy can be kept by scaling the noise to the query function's sensitivity, which for several applications needs substantially less noise than previously thought.

Checked: PDF from people.csail.mit.edu, fetched 4 Oct 2026. Page numbers are those of that PDF.

10 findings

Substance p.6: Lap(λ) "has density function h(y) ∝ exp(−|y|/λ), mean 0, and standard deviation λ". The standard deviation of this density is √2·λ. The same slip recurs on p.8: "add Laplace noise with standard deviation S(f)/ϵ".
  • p. 6 Section 2, definition of Laplace distribution
    mean 0, and standard deviation λ
  • p. 8 Section 3.1, sentence before Proposition 1
    add Laplace noise with standard deviation S(f)/ϵ in each coordinate
  • p. 8 Section 3.1, sentence before Proposition 1
    to add Laplace noise with standard deviation S(f)/ϵ in each coordinate
  • p. 8 Section 3.1
    add Laplace noise with standard deviation S(f)/ϵ

Judge's note: I reworked it myself. With h(y)=(1/2λ)e^{−|y|/λ}, the variance is ∫y²h(y)dy = (1/λ)∫_0^∞ y² e^{−y/λ} dy = 2λ², so SD = √2λ. Proposition 1 and the proofs use the scale parameter Lap(S(f)/ϵ), so the privacy results stand. But the stated standard deviation is wrong by a factor √2, and the p.8 instruction taken literally would give noise with scale S(f)/(√2ϵ).

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Substance Lemma 1 (p.11) assumes success "with probability at least β = (1+α)/2", yet the proof asserts the conditional probability "is strictly greater than (β−α)/(1−α) ≥ 1/2". At this threshold the bound equals exactly 1/2, and the lemma as stated fails.
  • p. 11 Lemma 1 and its proof
    is strictly greater than (β −α)/(1−α) ≥ 1/2
  • p. 11 Lemma 1 statement
    with probability at least β = 1+α

Only the authors can say what is right here.

Judge's note: My working: Pr[success | not read] ≥ (β−α)/(1−α) = ((1−α)/2)/(1−α) = 1/2, with no strict inequality available. I checked a counterexample against the stated hypotheses: n=1, D={0,1}, f(x)=10x, σ=1. A reads x_1 with probability α and outputs f(x); otherwise it outputs 0 or 10 with probability 1/2 each, without reading. Then Pr[read] = α and Pr[|A(x)−f(x)| ≤ 1] = α+(1−α)/2 = β. Conditional on not reading, the support is {0,10}, and no point lies within 1 of both. So S(f) = 10 > 2σ. The lemma as printed is false. It holds if β > (1+α)/2.

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Substance p.12: "h_{z,ϵ}(y) ∝ exp(ϵ·d_M(y,z)/(2·S_M(f)))" and Eqn. (4) both have a positive exponent. A minus sign is needed, otherwise the weight grows with distance from f(x).
  • p. 12 Section 3.3, definition of h_{z,ϵ}(y)
    hz,ϵ(y) ∝ exp
  • p. 12 Eqn. (4)
    Pr[T (x) = y] =
  • p. 12 Definition 3 / Eqn. 4
    hz,ϵ(y) ∝ exp(ϵ · dM(y, z) / (2 · SM(f)))

Judge's note: The page image and crops show no minus sign in the definition or in either exponent of Eqn. (4). With a positive exponent the density never integrates on R^d, the output does not approximate f(x), and the Hamming-cube example (flip probability ≈ 1/2 − ϵ/(2S(f))) contradicts it, since the printed form gives a flip probability above 1/2. The ratio argument in the proof works for either sign, so Theorem 2's privacy claim survives once the sign is corrected.

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Substance Hamming-cube example (p.12): "By Theorem 2, one can release f(x) safely by flipping each bit of the output f(x) independently with probability roughly 1/2 − ϵ/(2S(f))". The density of Eqn. 4 gives a flip probability of about 1/2 − ϵ/(8S), or 1/2 − ϵ/(4S) without the factor 2 (Remark 1). The printed rate gives about 2ϵ leakage.
  • p. 12 Last paragraph of Section 3.3
    independently with probability roughly 1/2 − ϵ/(2S(f))
  • p. 12 Remark 1 example
    with probability roughly 1/2 − ε/(2S(f))
  • p. 12 after Remark 1
    probability roughly 1 2 − ϵ 2S(f)

Only the authors can say what is right here.

Judge's note: I checked the page 12 image. Under Eqn. 4 with d_M = Hamming distance (taking the intended negative exponent), the density factorises over bits. Each bit flips with p = e^{-a}/(1+e^{-a}) = 1/(1+e^{a}), where a = ϵ/(2S), so p ≈ 1/2 − a/4 = 1/2 − ϵ/(8S). With Remark 1 (a = ϵ/S), p ≈ 1/2 − ϵ/(4S). Neither matches the printed 1/2 − ϵ/(2S). With the printed p, each differing bit has likelihood ratio (1−p)/p ≈ 1 + 2ϵ/S. Over S differing bits that compounds to about e^{2ϵ}, which is more than the claimed ϵ. 'Roughly' does not cover a factor of 4 (or 2) in the first-order term. Some reviewers' own constants (ϵ/(4S) for a = ϵ/(2S)) were also miscomputed, but the printed value fails under every reading. The authors should confirm the intended rate.

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Reference or label p.12: "one can sample a value according to h_{f(x),ϵ/S}()" — h_{z,ϵ} already contains S_M(f), so the subscript divides by the sensitivity twice. Eqn. (4) and Theorem 2 use h_{f(x),ϵ}.
  • p. 12 Section 3.3, sentence before Eqn. 4
    one can sample a value according to hf(x),ϵ/S()

Judge's note: The crop shows that the definition of h_{z,ϵ} has ϵ/(2·S_M(f)) in the exponent. Substituting ϵ/S gives ϵ/(2S²), which disagrees with Eqn. (4) (ϵ/(2S_M(f))) and with Theorem 2's 'h_{f(x),ϵ}()'. The subscript should be ϵ.

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Reference or label Reference [1] (p.17): "ACM Computing Surveys, 25(4), December 1989". The volume number appears inconsistent with the 1989 date.
  • p. 17 Reference [1]
    ACM Computing Surveys, 25(4), December 1989.
  • p. 17 Reference [1]
    ACM Computing Surveys, 25(4), December 1989

Only the authors can say what is right here.

Judge's note: Rests on outside knowledge: that the December 1989 issue of ACM Computing Surveys is volume 21(4) and that volume 25 is 1993. The page 17 image confirms that the entry prints 25(4), December 1989. Nothing in the files settles the true volume.

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Wording p.8: "we will let f_t : D → R^d be the function" — the domain should be D^n, as for every other query function, including f_i : D^n → R a few lines later and Theorem 1.
  • p. 8 Section 3.1, adaptive transcript notation
    will let ft : D → Rd be the function
  • p. 9 Theorem 1
    let ft(x) : Dn → Rd be its query

Judge's note: The crop shows 'f_t : D → R^d'. The same paragraph and Theorem 1 on p.9 use D^n.

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Wording p.10, Eqn. (2): "µµ^T − (µ+δ)(µ+δ)^T = µδ^T + δµ^T + δδ^T" has the wrong overall sign. The left-hand side equals −(µδ^T + δµ^T + δδ^T).
  • p. 10 Linear Algebraic Functions, Eqn. (2)–(3)
    µµT −(µ + δ)(µ + δ)T = µδT + δµT + δδT

Only the authors can say what is right here.

Judge's note: My own expansion: (µ+δ)(µ+δ)^T = µµ^T + µδ^T + δµ^T + δδ^T, so the printed LHS is the negative of the RHS. Scalar check with µ=δ=1: LHS = −3, RHS = 3. Only the L1 norm is used afterwards, so the 8γ²/n bound still holds. The step is faulty but the stated result stands, hence S3.

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Wording Proof of Lemma 2 (p.16): "Setting δ = ∛(2αϵ̃²/2^d)" puts α in the numerator. Making γ = δ requires δ³ = 2ϵ̃²/(α·2^d).
  • p. 16 Proof of Lemma 2
    Setting δ = ∛(2α˜ϵ²/2^d)
  • p. 16 Proof of Lemma 2, setting of δ
    Setting δ = ∛(2αε̃²/2^d)
  • p. 16 proof of Lemma 2
    Setting δ = 3q 2α˜ϵ2 2d
  • p. 15 Lemma 2 statement
    ϵ2 α · 2d

Only the authors can say what is right here.

Judge's note: I checked the page 16 image and the crop: α is in the numerator under the cube root. My own working: the bad fraction is γ = 2ϵ̃²/(αδ²2^d), and the SD bound is 2(γ+δ), which equals 4δ only if γ = δ, i.e. δ³ = 2ϵ̃²/(α2^d). With the printed δ, γ = δ/α², which is not δ for α < 1. Lemma 2 as stated, O((ϵ²/(α2^d))^{1/3}), follows once α is in the denominator, so the stated result stands. The faulty step is in the proof only.

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Wording Proof of Claim 1 (p.17): "for x ≠ y, the expectation of (2χ0(x)−1)(2χ0(y)−1) is exactly 1/2^d". For r uniform on nonzero strings this expectation is −1/(2^d−1).
  • p. 17 Proof of Claim 1
    is exactly 1/2d (if we chose r with no restriction it would be 0
  • p. 17 Proof of Claim 1, first paragraph
    is exactly 1/2d

Only the authors can say what is right here.

Judge's note: My own working from the page 17 image: the product equals (−1)^{r⊙(x⊕y)}, and b cancels. For w = x⊕y ≠ 0, Σ over all r of (−1)^{r⊙w} = 0. Removing r = 0, which contributes +1, leaves −1, so the mean over the 2^d−1 nonzero r is −1/(2^d−1). Check at d = 1: the only nonzero r is 1, and with x = 0, y = 1 the product is −1, not +1/2. The next display's +1/2^{3d} cross-term coefficient inherits the error. The claim's bound still holds, because the true cross term is non-positive (q_x ≥ 0), giving Var ≤ 2^{−2d}Σq_x² ≤ 2^{−d}max q_x².

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20 minor (typos and the like; most checked by the judge only)
Wording Intro p.3 says "Previous privacy-preserving approximations to these quantities used noise proportional to the dimension", but §3.2 (p.9) gives the earlier histogram noise as "O(√d/ϵ)" per coordinate, an "O(√d) factor larger".
  • p. 3 Section 1.1, Examples of Sensitivity-Based Analysis
    used noise proportional to the dimension; the new analysis permits noise of size O(1)
  • p. 9 Section 3.2, Histograms and Disjoint Analyses
    the noise added in the original framework to each coordinate is O(√d/ϵ)
  • p. 3 Examples of Sensitivity-Based Analysis
    used noise proportional to the dimension
  • p. 9 Histograms
    the noise added in the original framework to each coordinate is O(

Only the authors can say what is right here.

Judge's note: On p.9 the paper says the earlier framework [6] adds O(√d/ϵ) to each histogram coordinate, a factor of √d in the output dimension d. On p.10 the covariance case (d² queries) gives an O(d) factor, where d is the dimension of v, not of the d²-dimensional output. So the blanket 'proportional to the dimension' in the Introduction does not match the histogram analysis it summarises, unless it means variance or the underlying vector dimension. No statement in the paper resolves this, so the authors need to clarify.

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Wording Proof of Theorem 3 (p.15): "with probability at least 1 − 1/(6n) over r_{i+1}, Z(D_{r_i}) is within statistical difference σ of Z(D)". The hybrid step requires Z(D_{r_{i+1}}).
  • p. 15 Proof of Theorem 3
    over ri+1, Z(Dri) is within statistical difference σ of Z(D)
  • p. 15 Proof of Theorem 3
    over ri+1, Z(Dri)
  • p. 15 Proof of Theorem 3
    Z(Dri) is within statistical difference σ of Z(D)

Only the authors can say what is right here.

Judge's note: On the page 15 image, the preceding sentence identifies Z(D_{r_{i+1}}) as the (i+1)-st hybrid step, and the probability is taken over r_{i+1}. Z(D_{r_i}) does not depend on r_{i+1}, so the subscript is an index slip. Applying Lemma 2 to Z with r = r_{i+1} gives the intended statement, so the theorem stands.

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Wording End of proof of Lemma 2 (p.16): "the total distance between p̂(·) and p(·) is at most 4∛(12ϵ²2^{−d}), for at least a 1−α fraction of the pairs (r, b)". The final bound has lost the α of Lemma 2, and the constant 12 does not follow from ϵ̃ < 2ϵ.
  • p. 16 Proof of Lemma 2
    4∛(12ϵ²2^{−d}), for at least a 1−α fraction of the pairs (r, b)
  • p. 15 Lemma 2 statement
    SD (Z(Dr) , Z(D)) ≤ O
  • p. 16 Proof of Lemma 2
    for at least a 1−α fraction of the pairs (r, b)
  • p. 16 end of Lemma 2 proof
    4∛(12ε²2^{-d})

Only the authors can say what is right here.

Judge's note: Page 16 image. With δ³ = 2ϵ̃²/(α2^d) and ϵ̃ < 2ϵ, 4δ < 4∛(8ϵ²/(α2^d)). The printed δ instead gives 4∛(8αϵ²2^{−d}). Neither matches 4∛(12ϵ²2^{−d}), which has no α and constant 12 rather than 8. Lemma 2's statement (p.15) keeps α, and the proof of Theorem 3 uses σ = O(∛(nϵ²2^{−d})) with α = 1/(6n). The stated lemma therefore stands, and only the last line of the proof is inconsistent.

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Wording Proof of Claim 3 (p.20): "Noting that d_H(x1, x2) ≤ k" refers to an undefined x2 where the databases are x0 and x1. Part (2) also introduces an undefined "(k, ϵ′)-indistinguishable" while Eqn. (11) uses ϵ.
  • p. 20 Proof of Claim 3, part (1)
    Noting that dH(x1, x2) ≤ k
  • p. 20 Proof of Claim 3 part (2)
    not (k, ϵ′)-indistinguishable
  • p. 15 Proof of Theorem 3
    over ri+1, Z(Dri)
  • p. 20
    Noting that dH(x1, x2) ≤k we get a contradiction

Only the authors can say what is right here.

Judge's note: Page 20 image: part (1) constructs x0 and x1 and compares them in Eqn. (10); no x2 is defined. Part (2) writes ϵ′, but Eqn. (11) and the claim use ϵ. Both are notation slips with an evident intended reading (d_H(x0,x1), ϵ). The Z(D_{r_i}) pointer in this cluster belongs to C022.

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Typo Introduction, p.2: "beyond the orignal motivation of a purely statistical" — 'orignal' should be 'original'.
  • p. 2 Introduction
    beyond the orignal motivation
  • p. 2 Introduction, paragraph on "holistic" functions
    beyond the orignal motivation of a purely statistical

Judge's note: The page 2 crop shows the misspelling in the typeset text, so it is not an extraction artefact.

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Typo Copy-editing slips: "yields O(1/n)-close distributions O(1/n) on neighboring databases" (p.13, Remark 2, O(1/n) printed twice); "beyond the orignal motivation" (p.2); "non-negligible or leakage is inherent" (p.7, stray 'or').
  • p. 13 Remark 2 after Theorem 3
    yields O(1/n)-close distributions O(1/n) on neighboring databases
  • p. 2 Intro
    beyond the orignal motivation
  • p. 7 Non-negligible Leakage
    non-negligible or leakage is inherent
  • p. 13
    but yields O(1/n)-close distributions O(1/n) on neighboring databases

Judge's note: All three slips are visible on the page images of pp.13, 2 and 7. None is an extraction artefact. The θ(1) point is handled separately under C018.

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Typo p.3: "(e.g minimum cut size in a network)" — the final period of 'e.g.' is missing.
  • p. 3 Section 1.1, Examples of Sensitivity-Based Analysis
    (e.g minimum cut size in a network)

Judge's note: The crop shows 'e.g' with no closing period.

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Typo The author's name is spelt "Evfimievsky et al. [12]" in the text on p.5 but "Alexandre V. Evfimievski" in reference [12] on p.17.
  • p. 5 Definitional Work
    Evfimievsky et al. [12]
  • p. 17 Reference [12]
    Alexandre V. Evfimievski
  • p. 5 Related Work, Definitional Work paragraph
    Evfimievsky et al. [12] noted
  • p. 17 References, entry [12]
    Alexandre V. Evfimievski, Johannes Gehrke, and Ramakrishnan Srikant

Judge's note: Both spellings are visible in the images, so the paper is internally inconsistent. Which spelling is correct (the reference-list form 'Evfimievski' is the usual one) rests on outside knowledge, but the inconsistency itself is in the files.

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Typo p.5: "protection against isolation [7, 8]) have also been" has a closing parenthesis with no matching opening one.
  • p. 5 Definitional Work, k-anonymity paragraph
    isolation [7, 8]) have also been
  • p. 5 Related Work, paragraph on k-anonymity and isolation
    tion against isolation [7, 8]) have also been in the eye of privacy research

Judge's note: The crop shows the stray ')' and no '(' earlier in the sentence.

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Typo p.6, Example 1: "the sums f(x) and f(x′) differs by one" — the verb should be 'differ'.
  • p. 6 Example 1
    the sums f(x) and f(x′) differs by one

Judge's note: The full page 6 image confirms the wording.

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Typo p.7: "However, non-negligible or leakage is inherent for statistical utility" — the word 'or' is stray.
  • p. 7 Non-negligible Leakage paragraph
    non-negligible or leakage is inherent
  • p. 7 Non-negligible Leakage paragraph
    However, non-negligible or leakage is inherent
  • p. 7
    However, non-negligible or leakage is inherent for

Judge's note: The page 7 image shows the text as quoted.

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Typo p.7, Example 3: "(viewed as an real number)" — should read 'a real number'.
  • p. 7 Example 3
    viewed as an real number

Judge's note: The page 7 image confirms the wording.

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Typo Example 3 defines "d disjoint bins B1, ..., Bd" (p.7) but then says the result "is called a histogram for B1, ..., Bm" (p.8); m is undefined here.
  • p. 8 Example 3, continued
    is called a histogram for B1, ..., Bm
  • p. 7 Example 3
    joint bins B1, ..., Bd
  • p. 8 Example 3
    called a histogram for B1, ..., Bm
  • p. 8
    called a histogram for B1, ..., Bm.

Judge's note: Pages 7 and 8 show d bins and f : D^n → Z^d, followed by 'B1,...,Bm'. On p.11, m denotes the number of graph vertices. This is an index slip.

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Typo Remark 2 (p.13) writes "When m = θ(1)" with a lowercase theta where asymptotic Θ(1) is meant.
  • p. 13 Remark 2
    When m = θ(1) this is

Judge's note: The page 13 image shows a lowercase θ. Elsewhere the paper uses capital Ω and O for asymptotic notation. Cosmetic.

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Typo Theorem 3 statement (p.13): "San is an ϵ-indistinguishablenon-interactive mechanism" has a missing space.
  • p. 13 Theorem 3 statement
    ϵ-indistinguishablenon-interactive

Judge's note: The page 13 image shows the two words run together in the typeset text.

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Typo Section 4.1 (p.14) opens with "”Randomized response” refers", using a closing quotation mark in the opening position.
  • p. 14 Section 4.1, first word
    ”Randomized response” refers

Judge's note: The page 14 image shows ” at the start, whereas the paper uses “ ” correctly elsewhere (e.g. “how many people…”).

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Typo Proposition 2 (p.14): "Suppose that San is a ϵ-indistinguishable randomized response mechanism" should read 'an ϵ-indistinguishable'.
  • p. 14 Proposition 2
    Suppose that San is a ϵ-indistinguishable

Judge's note: The page 14 image confirms 'a ϵ-indistinguishable'. Page 13 writes 'an ϵ-indistinguishable', so this is an inconsistent article. Cosmetic.

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Typo Proof of Theorem 3 (p.15): "The ϵ-indistinguishabilityof San" has a missing space.
  • p. 15 Proof of Theorem 3
    The ϵ-indistinguishabilityof San

Judge's note: The run-together words are visible on the page 15 image.

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Typo Proof of Theorem 3 (p.15): "this is close the transcript distribution" is missing 'to'.
  • p. 15 Proof of Theorem 3, first sentence
    this is close the transcript distribution

Judge's note: Visible on the page 15 image.

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Typo Proof of Claim 3, part (2), p. 20: "assume San is not (k, ϵ′)-indistinguishable". ϵ′ is never defined, and Eqn. (11) and the rest of the proof use ϵ, so this should read (k, ϵ)-indistinguishable.
  • p. 20 Proof of Claim 3, part (2)
    assume San is not (k, ϵ′)-indistinguishable
  • p. 20 Proof of Claim 3 (2)
    not (k, ϵ′)-indistinguishable

Judge's note: I checked the page image p020 and crop C031-a. Claim 3 states that a mechanism is (k, ϵ)-indistinguishable iff it is (k, ϵ)-semantically-secure, so both sides use the same ϵ. Part (2) assumes the mechanism is not (k, ϵ′)-indistinguishable, but the condition that follows, Eqn. (11) |ln(...)| > ϵ, uses ϵ. So does the bound in (13), ln(1−α+αe^ϵ). The text layer has only this one occurrence of ϵ′, and it is not defined anywhere. This is a notational typo with an evident fix (ϵ′ → ϵ), and the proof's logic is unaffected.

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