LLM-generated corrections to the published research of P. M. Aronow

Journal article P007

Corrections to: Listwise Deletion in High Dimensions

J. Sophia Wang and P. M. Aronow (2023). Listwise Deletion in High Dimensions. Political Analysis 31(1):149–155.

DOI
10.1017/pan.2022.5
arXiv
2101.11470
Audited edition
Political Analysis, 2023, 31(1):149–155; 7 pages in the audited file
SHA-256 of the audited file
995b10fbd583f2f43964658e09c3d03a8059a504ed6df543b8038edded5ee3cb
Permanent address
https://pmaronow.github.io/llm-audit/works/P007/

Summary

The paper studies listwise deletion when the number of variables kk grows with the number of observations nn. Its main result, Proposition 5, states that under Assumptions 1 and 2, with q∗<1q_*<1 fixed as nn grows, the probability that listwise deletion removes all rows tends to 1 when kk grows faster than log⁡n\log n. Simulations on two datasets illustrate the loss. 9 entries are classified as errors (9 Minor), and 2 are clarity notes. Two entries overlap: P007-E010 and P007-E014 both correct the State Failure average in Table 1; the latter also explains the script ordering and the treatment of empty text entries. The Quality of Government results come from the Basic dataset, not the Standard dataset named and cited (P007-E009). No finding changes the statement of Lemma 3 or of Proposition 5.

What was and was not checked

I checked the following, subject to the corrections cited.

  • Lemma 3 and Proposition 5. I followed both proofs step by step and re-derived the statements with the bound q∗q_* fixed along the sequence. The bound of Lemma 3 is attained when all entries are independent and each is observed with probability q∗q_*. Lemma 8 and Proposition 10 of the supplement follow by the same argument with groups in place of variables. The proofs require the corrections in P007-E003.
  • Section 2.1, p. 151. The values ⌊(log⁡n)1.1⌋=2\lfloor(\log n)^{1.1}\rfloor=2, 8 and 17 at n=10n=10, 1,000 and 1,000,000 are correct, and the lower bound of Lemma 3 is approximately 1.34×10−111.34\times10^{-11} at 1% idiosyncratic missingness, k=150k=150 and n=100n=100.
  • Supporting Information A. All values plotted in Figures 3, 4 and 5 were recomputed from the formulas and from the replication code, and they agree. The text around Figures 4 and 5 is corrected in P007-E005 and P007-E006.
  • Quality of Government. The Table 1 column and saved surviving-row counts reproduce from the Basic file (P007-E009) with the original simulation seed. Estimated row loss first exceeds 50% at 2 variables and 99% at 17; the estimated probability of losing every row first exceeds 0.5 at 14 variables and 0.99 at 52.
  • State Failure. The data file sort11v3.dta is among the replication files. I ran the simulation code on this file with the seed of the replication files (213) and 25,000 draws at each of the 100 values of kk. The run reproduces the saved numbers of surviving rows. The simulation estimates an expected proportion of rows removed above 50% with one variable and above 99% first with 6 variables. The estimated probability that all rows are removed first exceeds 0.5 with 6 variables and 0.99 with 17; the printed condition of more than 17 variables is sufficient. The other cells of the State Failure column of Table 1 match the output of the code. The average missingness and the 99% threshold for the expected proportion of rows removed are corrected in P007-E010. The evidence given in P007-E010 and P007-E014 does not use the data file, and the counts of missing cells in those findings are inferred from the saved log of the simulation code. With the code’s definition of a missing cell, I find from the data file that the average missingness over the 1,205 variables that remain is 0.66316…0.66316\ldots, which rounds to the corrected entry of 66.3%.
  • The simulation code matches the description in Section 3.1: 25,000 replications, a fresh draw of kk variables without replacement in each, and the share of complete rows recorded.
  • The numbering of assumptions, results and figures, and the cross-references to them, are consistent. The six figure files of the replication files match the published figures by eye.

I did not check all the cited books and articles. The quotation from Lehmann (1999) and the page citations to Allison, Berk, Cameron and Trivedi, Little and Rubin, and Schafer are unchecked. The source checks cover glmnet and the package version named in its cited article (P007-E001) and the editors of the volume that contains Berk (1983) (P007-E013). I could not read Lai, Robbins, and Wei (1978) beyond its abstract, and whether that paper supports the remark on p. 151 that cites it is not settled. I did not establish whether the State Failure file contains empty text fields that the code counts as observed (P007-E014).

The Standard dataset, which the article names, is not among the replication files, and no analysis of it was made. These checks cover the items listed, and they are not a verification of the paper as a whole.

Headline claims and related corrections

  1. For nn independent rows and kk variables, if each conditional observation probability is at most q∗<1q_*<1 whenever the preceding variables are all observed with positive probability, the probability that listwise deletion removes every row is at least (1−q∗k)n(1-q_*^k)^n.

    Lemma 3; Assumptions 1 and 2 (p. 150); proof in the Appendix (p. 154)

    See P007-E003.

  2. Under independent rows and a fixed bound q∗<1q_*<1 on conditional observation probabilities, the probability that listwise deletion removes all rows tends to 1 when the number of variables grows faster than log⁡n\log n.

    Proposition 5; Assumptions 1, 2 and 4; Lemma 3; abstract and Introduction (p. 149); proof p. 154

    See P007-E003, P007-N001.

  3. The finite-sample lower bound is substantial when idiosyncratic missingness is 10% or 25%, and can be extremely small when it is 1%, even with k=150k=150 and n=100n=100.

    Section 2.1, last paragraph (p. 151); Supporting Information A, Figures 3-5; Numerical_Illustration.R

    See P007-E005, P007-E006.

  4. For the January 2021 Quality of Government Standard dataset, the paper reports 194 countries, 351 variables and 35.9% missing cells. With 2 randomly chosen variables, expected row loss exceeds 50%; with 17 it exceeds 99%. The probability of losing all rows exceeds 0.50.5 at 14 variables and 0.990.99 at 52.

    Table 1 (Quality of Government column); Figure 1; Section 3.2 (p. 153); Simcode_Qog.R

    See P007-E009.

  5. For the State Failure data, the paper reports 8,580 country-years, 1,205 variables and 66.8% missing cells. Expected row loss exceeds 50% with one randomly chosen variable and 99% with more than three; the probability of losing all rows exceeds 0.50.5 at 6 variables and 0.990.99 with more than 17.

    Table 1 (State Failure column); Figure 2; Section 3.2 (p. 153); Simcode_Statefailure.R

    See P007-E010, P007-E014.

  6. The results extend to groups of variables that share one missingness pattern, with the number of groups gg in place of the number of variables kk.

    Section 2 (p. 150); Supporting Information B: Assumptions 6, 7, 9; Lemma 8; Proposition 10

    See P007-E003.

  7. Listwise deletion is not viable in many data-analytic settings; researchers using it may have to forgo variables they need even when nn is large, and the method for handling missing data should be chosen for the problem at hand.

    Abstract; end of Introduction (p. 150); Section 4 (p. 154); Proposition 5; Figures 1 and 2; Table 1

    See P007-E010, P007-E009.

Corrections

Minor (9)

Real but inconsequential, for example a misprint that does not change a result.

The proofs of Lemmas 3 and 8 wrongly read q∗=0q_* = 0 as every variable being missing and omit the case of a conditioning event of probability zero, and several symbols are misprinted; the two lemmas and Propositions 5 and 10 hold as stated.

Appendix, proofs of Lemma 3 and Proposition 5, p. 154 (PDF p. 6); Supplementary Material B, p. 3 (definition of Mij\mathbf{M}_{ij}) and p. 4 (proofs of Lemma 8 and Proposition 10)

See also under Formal statements (Lean): P007-T3, P007-T4, P007-T5, P007-T6

As printed
  • Proof of Lemma 3, p. 154: “First suppose q∗=0q_* = 0, which is equivalent to say that all variables are fully missing. Then pall=1=(1−q∗k)np_{all} = 1 = (1 - q_*^k)^n. Now suppose q∗∈(0,1)q_* \in (0,1). By the independence assumption, pall=Πi=1n(1−Pr⁡(Mik=0))p_{all} = \Pi_{i=1}^{n}(1 - \Pr(\mathbf{M}_{ik} = 0)). Denote qij=Pr⁡(Mij=0∣Mi(j−1)=0)q_{ij} = \Pr(M_{ij} = 0 | \mathbf{M}_{i(j-1)} = 0) if j>1j > 1, else qij=Pr⁡(Mij=0)q_{ij} = \Pr(M_{ij} = 0). By Assumption 2, qij≤q∗\colorbox{#ffe58a}{$q_{ij} \leq q_*$}, for all j∈{1,2,…,k}j \in \{1, 2, \ldots, k\}. By the chain rule of conditional probability, Pr⁡(Mik=0)=qi1,qi2,…,qik\Pr(\mathbf{M}_{ik} = 0) = \colorbox{#ffe58a}{$q_{i1}, q_{i2}, \ldots, q_{ik}$}. This means that the probability of a single observation containing at least one missing entry is (1−qi1,qi2,…,qik)(1 - \colorbox{#ffe58a}{$q_{i1}, q_{i2}, \ldots, q_{ik}$}). Since q∗≥qijq_* \geq q_{ij} for all j∈{1,2,…,k}j \in \{1, 2, \ldots, k\}, q∗k≥qi1,qi2,…,qikq_*^k \geq \colorbox{#ffe58a}{$q_{i1}, q_{i2}, \ldots, q_{ik}$}. Thus (1−q∗k)≤(1−qi1,qi2,…,qik)(1 - q_*^k) \leq (1 - \colorbox{#ffe58a}{$q_{i1}, q_{i2}, \ldots, q_{ik}$}).”
  • Proof of Proposition 5, p. 154: “Since q∗∈(0,1)q_* \in (0,1) and k=f(n)≥∗1k = f(n) \colorbox{#ffe58a}{$\geq_*$} 1, −q∗f(n)>−1-q_*^{f(n)} > -1 and 1−q∗f(n)≤11 - q_*^{f(n)} \leq 1. By Bernoulli’s Inequality, since n∈Nn \in \mathbb{N}, (1−q∗f(n))n≥∗1+n(−q∗f(n))=1−nq∗f(n)(1 - q_*^{f(n)})^n \colorbox{#ffe58a}{$\geq_*$} 1 + n(-q_*^{f(n)}) = 1 - nq_*^{f(n)}. Thus 1−nq∗f(n)≤(1−q∗f(n)n≤11 - nq_*^{f(n)} \leq \colorbox{#ffe58a}{$(1 - q_*^{f(n})^n$} \leq 1 in the common domain n∈Nn \in \mathbb{N}.”
  • Supplement, p. 3: “we use Mij\mathbf{M}_{ij} to represent the random vector collecting the missingness indicators up to variable jj, (Mi1,Mi2,…,Mij)(M_{i1}, M_{i2}, \ldots, M_{ij}).”
  • Supplement, proof of Lemma 8, p. 4: “Suppose q∗=0q_* = 0. Since this entails that all groups of variables are completely missing, pall=1=(1−q∗g)np_{all} = 1 = (1 - q_*^g)^n. For the second case suppose q∗∈(0,1)q_* \in (0,1). By the group independence assumption, pall=Πi=1n(1−Pr⁡(Mig=0))p_{all} = \Pi_{i=1}^{n}(1 - \Pr(\mathbf{M}_{ig} = 0)). Denote qij=Pr⁡(Mij=0∣Mi(j−1)=0)q_{ij} = \Pr(M_{ij} = 0 | \mathbf{M}_{i(j-1)} = 0) if j>1j > 1, else qij=Pr⁡(Mij=0)q_{ij} = \Pr(M_{ij} = 0). By Assumption 7, qij≤q∗\colorbox{#ffe58a}{$q_{ij} \leq q_*$}, for all j∈{1,2,…,g}j \in \{1, 2, \ldots, g\}.”
  • Supplement, proof of Proposition 10, p. 4: “Since q∗∈(0,1)q_* \in (0,1) and k=l(n)≥∗1\colorbox{#ffe58a}{$k$} = l(n) \colorbox{#ffe58a}{$\geq_*$} 1, −q∗l(n)>−1-q_*^{l(n)} > -1 and 1−q∗l(n)≤11 - q_*^{l(n)} \leq 1. By Bernoulli’s Inequality, since n∈Nn \in \mathbb{N}, (1−q∗l(n))n≥∗1+n(−q∗l(n))=1−nq∗l(n)(1 - q_*^{l(n)})^n \colorbox{#ffe58a}{$\geq_*$} 1 + n(-q_*^{l(n)}) = 1 - nq_*^{l(n)}. Thus 1−nq∗l(n)≤(1−q∗l(n)n≤11 - nq_*^{l(n)} \leq \colorbox{#ffe58a}{$(1 - q_*^{l(n})^n$} \leq 1 in the common” (the sentence runs on to p. 5).
The problem

The printed proofs have two defects of argument and several misprints. None of them makes a stated result false.

The case q∗=0q_* = 0. Both proofs read q∗=0q_* = 0 as saying that all variables (in the supplement, all groups) are fully missing. Assumption 2 with q∗=0q_* = 0 says only that the first variable is missing with probability one in every row. The conditions on later variables are then vacuous, because each is imposed only when the earlier variables are all observed with positive probability, and later variables may be observed. The conclusion pall=1p_{all} = 1 is still correct, since every row has a missing first entry. The printed equivalence also fails in the other direction. Since q∗q_* is an upper bound and need not be the smallest one, data in which every variable is always missing satisfy Assumption 2 with q∗=1/2q_* = 1/2 as well.

Conditioning on an event of probability zero. In the second case the proofs define qijq_{ij}, for j>1j > 1, as the probability that entry jj is observed given that the first j−1j - 1 entries are observed, and assert that qij≤q∗q_{ij} \leq q_* for all jj. Assumption 2 (in the supplement, Assumption 7) bounds this conditional probability only when the conditioning event has positive probability. When that event has probability zero, qijq_{ij} is a ratio 0/00/0 and the proofs state no convention for it. Neither the bound nor the product formula is then available as written. The split into q∗=0q_* = 0 and q∗∈(0,1)q_* \in (0,1) does not cover this, since it can happen in either case.

Misprints. In the proof of Lemma 3 the product qi1qi2⋯qikq_{i1} q_{i2} \cdots q_{ik} is printed four times as a list separated by commas. In the proofs of Propositions 5 and 10 the sign ≥\geq twice carries a subscript star, and in the sandwich inequality the exponent f(n)f(n) (in the supplement, l(n)l(n)) lacks its closing parenthesis. The proof of Proposition 10 writes k=l(n)k = l(n) where the number of groups, g=l(n)g = l(n), is meant; kk is the number of variables, and k≥gk \geq g. On p. 3 of the supplement, “up to variable jj” should be “up to group jj”. The proofs of the two propositions also begin the logarithm argument with q∗∈(0,1)q_* \in (0,1) without saying that the case q∗=0q_* = 0 follows from the lemma.

Evidence an exact counterexample

The evidence consists of exact counterexamples to the two statements in the printed proofs, and a derivation of the lemmas and propositions under the stated assumptions, given under “Corrected”.

Let n=1n = 1 and k=2k = 2, and let the row satisfy (M11,M12)=(1,0)(M_{11}, M_{12}) = (1, 0) with probability one. Assumption 1 holds, since there is only one row. With q∗=0q_* = 0, Assumption 2 holds: I have Pr⁡(M11=0)=0≤q∗\Pr(M_{11} = 0) = 0 \leq q_*, and no condition is imposed for j=2j = 2 because Pr⁡(M11=0)=0\Pr(M_{11} = 0) = 0. Since the second variable is always observed, the variables are not all fully missing, while pall=1=(1−q∗2)1p_{all} = 1 = (1 - q_*^2)^1. With two groups of one variable each, the same example applies to the proof of Lemma 8.

For the same row and q∗=1/2q_* = 1/2, Assumption 2 holds again, but q12=Pr⁡(M12=0∣M11=0)q_{12} = \Pr(M_{12} = 0 | M_{11} = 0) is a ratio 0/00/0. The lemma’s conclusion holds, since pall=1≥3/4p_{all} = 1 \geq 3/4. The event of probability zero need not involve the first variable alone. Let k=3k = 3 and q∗=1/2q_* = 1/2, and let a row equal (1,0,0)(1,0,0) or (0,1,0)(0,1,0) with probability 1/21/2 each. Then Pr⁡(Mi1=0)=1/2\Pr(M_{i1} = 0) = 1/2 and Pr⁡(Mi1=Mi2=0)=0\Pr(M_{i1} = M_{i2} = 0) = 0, and qi3q_{i3} is undefined. For two independent rows with this distribution, pall=1≥49/64p_{all} = 1 \geq 49/64.

A row equal to (1,1)(1,1) with probability one satisfies Assumption 2 with q∗=1/2q_* = 1/2 as well as with q∗=0q_* = 0. Complete missingness does not force q∗=0q_* = 0. For the index in the proof of Proposition 10, two groups of two variables each give g=2g = 2 and k=4k = 4. I read the misprints from the printed pages. The supplement prints the product in the proof of Lemma 8 correctly, as qi1qi2⋯qigq_{i1} q_{i2} \cdots q_{ig}.

What survives

Lemma 3, Lemma 8, Proposition 5 and Proposition 10 hold as stated, under the stated assumptions; no assumption is added and no rate changes. No number changes: the numerical illustrations of Supplementary Material A use the bound of Lemma 3 directly, and the simulations of Section 3 count complete rows directly. What changes is the wording of the case q∗=0q_* = 0, one step in the proofs of the two lemmas, and the misprinted symbols.

Corrected

The statements of Lemma 3, Lemma 8, Proposition 5 and Proposition 10 stand as printed. The proofs are corrected as follows. For the supplement, read groups for variables, gg for kk, l(n)l(n) for f(n)f(n), and Assumptions 6, 7 and 9 for Assumptions 1, 2 and 4.

Proof of Lemma 3. Assume that the rows are mutually independent (Assumption 1) and that there exists q∗∈[0,1)q_* \in [0,1) such that, for all ii, Pr⁡(Mi1=0)≤q∗\Pr(M_{i1} = 0) \leq q_* and Pr⁡(Mij=0∣Mi(j−1)=0)≤q∗\Pr(M_{ij} = 0 | \mathbf{M}_{i(j-1)} = 0) \leq q_* for all j∈{2,…,k}j \in \{2, \ldots, k\} such that Pr⁡(Mi(j−1)=0)>0\Pr(\mathbf{M}_{i(j-1)} = 0) > 0 (Assumption 2). Fix a row ii. Let si0=1s_{i0} = 1 and, for j∈{1,…,k}j \in \{1, \ldots, k\}, let sij=Pr⁡(Mij=0)s_{ij} = \Pr(\mathbf{M}_{ij} = 0) be the probability that the first jj entries of row ii are observed. Then

sij≤q∗ si(j−1)for all j∈{1,…,k}.s_{ij} \leq q_* \, s_{i(j-1)} \quad \text{for all } j \in \{1, \ldots, k\}.

For j=1j = 1 this is the first part of Assumption 2. For j≥2j \geq 2 with si(j−1)=0\colorbox{#ffe58a}{$s_{i(j-1)} = 0$}, the event {Mij=0}\{\mathbf{M}_{ij} = 0\} is contained in {Mi(j−1)=0}\{\mathbf{M}_{i(j-1)} = 0\}, so sij=0s_{ij} = 0. For j≥2j \geq 2 with si(j−1)>0s_{i(j-1)} > 0, the second part of Assumption 2 gives sij/si(j−1)=Pr⁡(Mij=0∣Mi(j−1)=0)≤q∗s_{ij} / s_{i(j-1)} = \Pr(M_{ij} = 0 | \mathbf{M}_{i(j-1)} = 0) \leq q_*. By induction on jj, this implies that sik≤q∗ks_{ik} \leq q_*^k. Since the rows are independent,

pall=∏i=1n(1−sik)≥(1−q∗k)n.p_{all} = \prod_{i=1}^{n} (1 - s_{ik}) \geq (1 - q_*^k)^n.

If q∗=0q_* = 0, then si1=0s_{i1} = 0 for every ii: the first variable is missing with probability one in every row, and pall=1p_{all} = 1. Later variables need not be missing.

Proof of Proposition 5. If q∗=0\colorbox{#ffe58a}{$q_* = 0$}, Lemma 3 gives pall=1p_{all} = 1 for every nn with f(n)≥1f(n) \geq 1. Suppose that q∗∈(0,1)q_* \in (0,1). The limit lim⁡n→∞nq∗f(n)=0\lim_{n \to \infty} n q_*^{f(n)} = 0 is obtained as printed. The paragraph that follows should read: “Since q∗∈(0,1)q_* \in (0,1) and k=f(n)≥1k = f(n) \colorbox{#ffe58a}{$\geq$} 1, −q∗f(n)>−1-q_*^{f(n)} > -1 and 1−q∗f(n)≤11 - q_*^{f(n)} \leq 1. By Bernoulli’s Inequality, since n∈Nn \in \mathbb{N}, (1−q∗f(n))n≥1+n(−q∗f(n))=1−nq∗f(n)(1 - q_*^{f(n)})^n \colorbox{#ffe58a}{$\geq$} 1 + n(-q_*^{f(n)}) = 1 - nq_*^{f(n)}. Thus 1−nq∗f(n)≤(1−q∗f(n))n≤11 - nq_*^{f(n)} \leq \colorbox{#ffe58a}{$(1 - q_*^{f(n)})^n$} \leq 1 in the common domain n∈Nn \in \mathbb{N}.” The rest of the proof is as printed.

Other changes. Each of the four comma lists in the printed proof of Lemma 3 is the product qi1qi2⋯qik\colorbox{#ffe58a}{$q_{i1} q_{i2} \cdots q_{ik}$}. The proof of Proposition 10 should read g=l(n)≥1\colorbox{#ffe58a}{$g$} = l(n) \geq 1, with the same three repairs as in the proof of Proposition 5. On p. 3 of the supplement, read “up to group jj”. In the proof of Lemma 8, “the group independence assumption” is Assumption 6, which is independence across rows.

Crop of the proof of Lemma 3 under the heading 'Appendix: Proofs'. Seven red outlines mark: the words 'which is equivalent to say that all variables are fully missing' (two outlines, over a line break); the expression 'q i j less than or equal to q star, for all j in 1, 2, up to k'; and four occurrences of the comma-separated list 'q i 1, q i 2, up to q i k'.
Proof of Lemma 3, p. 154 (PDF p. 6). Marked: the sentence on the case q∗=0q_* = 0, the bound qij≤q∗q_{ij} \leq q_* asserted for all jj, and the four comma lists printed for the product qi1qi2⋯qikq_{i1} q_{i2} \cdots q_{ik}.
Crop of three lines of the proof of Proposition 5, beginning 'Since q star in (0,1) and k = f(n)'. Three red outlines mark a greater-than-or-equal sign with a subscript star after 'k = f(n)', a second such sign after the n-th power of one minus q star to the f(n), and the expression one minus q star to the power 'f(n' with no closing parenthesis, raised to the power n.
Proof of Proposition 5, p. 154 (PDF p. 6). Marked: the two signs ≥\geq printed with a subscript star, and the expression (1−q∗f(n)n(1 - q_*^{f(n})^n, in which the exponent f(n)f(n) lacks its closing parenthesis.
Crop of Lemma 8 and its proof in the supplement. Three red outlines mark the words 'Since this entails that all groups of variables are completely missing,' and, over a line break, the expression 'q i j less than or equal to q star, for all j in 1, 2, up to g'.
Supplementary Material B, p. 4: Lemma 8 and its proof. Marked: the sentence on the case q∗=0q_* = 0 and the bound qij≤q∗q_{ij} \leq q_* asserted for all jj.
Crop of two lines of the proof of Proposition 10, beginning 'Since q star in (0,1) and k = l(n)'. Four red outlines mark the letter k, two greater-than-or-equal signs that carry a subscript star, and the expression one minus q star to the power 'l(n' with no closing parenthesis, raised to the power n.
Supplementary Material B, foot of p. 4: proof of Proposition 10. Marked: kk printed for gg, the two signs ≥\geq with a subscript star, and the expression (1−q∗l(n)n(1 - q_*^{l(n})^n with the unclosed exponent.
Crop of two paragraphs of Supplementary Material B that define g, the number of variable groups, and the indicator for the j-th group in the i-th row. In the sentence 'we use M i j to represent the random vector collecting the missingness indicators up to variable j', the word 'variable' is outlined in red.
Supplementary Material B, p. 3: definitions for groups of variables. The marked word “variable” should be “group”: in this section jj indexes groups.

Assumption 1 requires mutually independent rows, not i.i.d. rows, and “out theoretical assumptions” should read “our theoretical assumptions”.

Section 3.1, third paragraph, p. 152 (PDF p. 4), first, second and last sentences

See also under Formal statements (Lean): P007-T11

As printed

“We briefly discuss how out theoretical assumptions align with this setting. Assumption 1 asserts i.i.d. missingness across rows, which is unlikely to be met in this setting. […] Our theoretical assumption of i.i.d. missingness across rows can therefore be seen as optimistic when faced with real-world data and, all else equal, the probability that all data will be lost may be higher than theory might dictate.”

The problem

Assumption 1 (p. 150) states that the rows are mutually independent. It does not ask that they be identically distributed, and the proof of Lemma 3 works with row-specific probabilities. The paragraph twice describes the assumption as “i.i.d.”, which adds a condition that the paper neither assumes nor uses. The first sentence also prints “out” for “our”.

Evidence an exact counterexample

The evidence is a check of the page against Assumption 1 as printed, and an example. Two independent rows with one variable each, observed with probabilities 1/41/4 and 1/21/2, satisfy Assumptions 1 and 2 with q∗=1/2q_* = 1/2 and are not identically distributed. Lemma 3 holds for them, since pall=(3/4)(1/2)=3/8≥1/4p_{all} = (3/4)(1/2) = 3/8 \geq 1/4.

What survives

Nothing depends on the label. Assumption 1 as stated, Lemma 3, Proposition 5 and the paragraph’s point about clustered rows are unaffected.

Corrected

“We briefly discuss how our theoretical assumptions align with this setting. Assumption 1 asserts mutually independent missingness across rows, which is unlikely to be met in this setting. […] Our theoretical assumption of mutually independent missingness across rows can therefore be seen as optimistic when faced with real-world data and, all else equal, the probability that all data will be lost may be higher than theory might dictate.”

Crop of the third paragraph of Section 3.1. Three terms are outlined in red: 'out' in 'how out theoretical assumptions', 'i.i.d.' in 'Assumption 1 asserts i.i.d. missingness across rows', and 'i.i.d.' in 'Our theoretical assumption of i.i.d. missingness across rows'.
Section 3.1, third paragraph, p. 152 (PDF p. 4). The misprint “out” and the two places where Assumption 1 is called “i.i.d.” are marked.

The formula plotted in Figure 4 gives the largest kk at which the lower bound (1−q∗k)n(1-q_*^k)^n does not exceed the target, not the smallest kk that reaches it; at k=33k=33 and k=952k=952 the bound is 47.1% and 49.7%, and it first reaches 50% at k=34k=34 and k=953k=953.

Supporting Information A, supplement p. 2: the two paragraphs before Figure 4, and the caption and subcaptions of Figure 4

See also under Formal statements (Lean): P007-T7, P007-T8

As printed

“Figure 4 illustrates, for a given n,q∗n,q_*, how large kk can be while still ensuring that pall≤0.5,0.99\colorbox{#ffe58a}{$p_{all}$}\leq 0.5,0.99. We compute this using the result from Lemma 3, pall‾=(1−q∗k)n\underline{p_{all}}=(1-q_*^k)^n. Since (1−q∗k)n(1-q_*^k)^n is strictly increasing in kk, solving for equality pall=(1−q∗k)n\colorbox{#ffe58a}{$p_{all}$}=(1-q_*^k)^n we will get the smallest possible kk for each pall‾\underline{p_{all}} that

k=⌊log⁡(1−pall‾1/n)log⁡(q∗)⌋.k=\left\lfloor\frac{\log(1-\underline{p_{all}}^{1/n})}{\log(q_*)}\right\rfloor.

…

With more missingness, q∗=0.75,0.90q_*=0.75,0.90, even relatively small kk can yield missingness of pall‾\underline{p_{all}}. For example, even with n=10,000n=10,000 and q∗=0.75q_*=0.75 we need only k=33\colorbox{#ffe58a}{$k=33$} to have a 50% probability that all rows will be missing. However, when missingness is very low, kk needs to be very large to cause all data to be missing. For example, with n=10,000n=10,000 and q∗=0.99q_*=0.99, we need k=952\colorbox{#ffe58a}{$k=952$} to have a 50% probability that all rows will be missing.”

Subcaptions and caption of Figure 4: “(a) pall≤0.5\colorbox{#ffe58a}{$p_{all}$}\leq 0.5”, “(b) pall≤0.99\colorbox{#ffe58a}{$p_{all}$}\leq 0.99”, “Figure 4. Largest kk such that pall≤0.5,0.99\colorbox{#ffe58a}{$p_{all}$}\leq 0.5,0.99 plotted against nn and q∗q_*.”

The problem

Let B(k)=(1−q∗k)nB(k)=(1-q_*^k)^n denote the lower bound of Lemma 3 on the probability pallp_{all} that listwise deletion removes all rows; the supplement writes it pall‾\underline{p_{all}}. For a target aa, the floor in the display is the largest integer kk such that B(k)≤aB(k)\le a. The titles of the two panels (the first reads “Largest k such that probability all data missing is ≤\leq 0.50”) and the caption say so, and the code computes this value. The text calls it the smallest possible kk, and the two examples present k=33k=33 and k=952k=952 as the numbers of variables needed for a 50% probability that all rows are missing. At these values the lower bound is below 50%: for n=10,000n=10{,}000, I find B(33)=0.4708B(33)=0.4708 at q∗=0.75q_*=0.75 and B(952)=0.4969B(952)=0.4969 at q∗=0.99q_*=0.99. The smallest integers at which the lower bound is at least 50% are 34 and 953.

The first sentence, the subcaptions and the caption also print pallp_{all} where the lower bound pall‾\underline{p_{all}} is meant. A lower bound on pallp_{all} cannot ensure that pall≤ap_{all}\le a, and Figure 4 shows the largest kk such that pall‾≤a\underline{p_{all}}\le a.

Evidence an exact counterexample

The evidence is a derivation and a calculation in exact rational arithmetic. Let n≥1n\ge1, 0<q∗<10<q_*<1 and 0<a<10<a<1, and set r=log⁡(1−a1/n)/log⁡q∗r=\log(1-a^{1/n})/\log q_*. For real k≥0k\ge0, I have

B(k)≤a  ⟺  1−q∗k≤a1/n  ⟺  q∗k≥1−a1/n  ⟺  klog⁡q∗≥log⁡(1−a1/n)  ⟺  k≤r,\begin{aligned} B(k)\le a&\iff 1-q_*^k\le a^{1/n}\iff q_*^k\ge 1-a^{1/n}\\ &\iff k\log q_*\ge\log(1-a^{1/n})\iff k\le r, \end{aligned}

where the last step divides by log⁡q∗<0\log q_*<0. Reversing the inequalities, I obtain that B(k)≥aB(k)\ge a if and only if k≥rk\ge r. The largest integer kk such that B(k)≤aB(k)\le a is ⌊r⌋\lfloor r\rfloor, and the smallest integer kk such that B(k)≥aB(k)\ge a is ⌈r⌉\lceil r\rceil. For n=10,000n=10{,}000 and a=1/2a=1/2, the values of BB on both sides of rr are as follows.

The lower bound B(k)=(1−q∗k)nB(k)=(1-q_*^k)^n at n=10,000n=10{,}000, computed in exact rational arithmetic and rounded to five places.
q∗q_*kkB(k)B(k)
0.75330.47075
0.75340.56833
0.999520.49689
0.999530.50037

The bound B(k)B(k) is attained when all entries are independent and each is observed with probability q∗q_*. In that case the probability that listwise deletion removes all rows is 47.1% at k=33k=33 and 49.7% at k=952k=952, below the 50% of the text. The statements printed without the underline fail as well. Let n=10,000n=10{,}000, q∗=0.75q_*=0.75 and k=33k=33, and let all entries be independent, the first variable observed with probability 3/43/4 and each of the other 32 variables with probability 1/41/4. Assumptions 1 and 2 hold, B(33)<0.5B(33)<0.5, and pall=(1−3⋅4−33)10000>0.99p_{all}=(1-3\cdot4^{-33})^{10000}>0.99. All 59,406 values plotted in Figure 4 satisfy B(k)<a<B(k+1)B(k)<a<B(k+1).

What survives

The plotted values in Figure 4 and the code that produces them are correct when read as the largest kk at which the lower bound does not exceed the target, and no plotted value changes. Lemma 3, Proposition 5 and the two applications do not use this inversion. The corrected text is a statement about the lower bound of Lemma 3, and it contains no upper bound on pallp_{all}. The contrast between high and low rates of missingness stands.

Corrected

“Figure 4 shows, for each nn and q∗q_*, the largest nonnegative integer kk for which the sharp lower bound pall‾=(1−q∗k)n\underline{p_{all}}=(1-q_*^k)^n does not exceed aa, with a=0.5a=0.5 or 0.990.99. Solving (1−q∗r)n=a(1-q_*^r)^n=a gives

r=log⁡(1−a1/n)log⁡(q∗),r=\frac{\log(1-a^{1/n})}{\log(q_*)},

and the plotted value is k=⌊r⌋k=\lfloor r\rfloor. This calculation concerns the lowest possible probability that listwise deletion removes all rows under Assumptions 1 and 2, not an upper bound on the actual probability. …

… For n=10,000n=10,000, the lower bound is approximately 47.08% at q∗=0.75q_*=0.75, k=33k=33, and approximately 49.69% at q∗=0.99q_*=0.99, k=952k=952. The smallest integers at which this lower bound is at least 50% are 34 and 953, respectively.”

Subcaptions and caption of Figure 4: “(a) pall‾≤0.5\colorbox{#ffe58a}{$\underline{p_{all}}$}\leq 0.5”, “(b) pall‾≤0.99\colorbox{#ffe58a}{$\underline{p_{all}}$}\leq 0.99”, “Figure 4. Largest kk such that pall‾≤0.5,0.99\colorbox{#ffe58a}{$\underline{p_{all}}$}\leq 0.5,0.99 plotted against nn and q∗q_*.” In the titles of both panels, “probability all data missing” is replaced by “lower bound on probability all data missing”.

The statement behind the figure is the following. Let n≥1n\ge1, 0<q∗<10<q_*<1 and 0<a<10<a<1, and let rr be as in the display. The largest integer k≥0k\ge0 such that (1−q∗k)n≤a(1-q_*^k)^n\le a is ⌊r⌋\lfloor r\rfloor, and the smallest integer k≥0k\ge0 such that (1−q∗k)n≥a(1-q_*^k)^n\ge a is ⌈r⌉\lceil r\rceil.

Crop of two paragraphs. In the first, a red box marks the words 'smallest possible' in 'we will get the smallest possible k for each p_all that k =', followed by the floor of log(1 - p_all^(1/n)) over log(q_*). In the second, red boxes mark 'k = 33' in the example with n = 10,000 and q_* = 0.75 and 'k = 952' in the example with n = 10,000 and q_* = 0.99.
Supporting Information A, supplement p. 2: the two paragraphs that describe Figure 4. The marked words call the plotted value the smallest possible kk, and the marked values are given as yielding a 50% probability.
Crop of the subcaptions '(a) p_all at most 0.5' and '(b) p_all at most 0.99' and of the caption 'Figure 4. Largest k such that p_all is at most 0.5, 0.99 plotted against n and q_*'. Red boxes mark the three occurrences of p_all, none of which is underlined.
Supplement p. 2: the subcaptions and the caption of Figure 4. The caption says “Largest kk”; the three marked symbols are printed without the underline that denotes the lower bound.

The quantity 1−q∗k1-q_*^k of Figure 5 is a lower bound, not an upper bound, on the expected proportion of rows removed, and the caption of Figure 5 labels it with the symbol pall‾\underline{p_{all}} and the dependence on nn of a different quantity.

Supporting Information A, supplement p. 2, last paragraph, first sentence; supplement p. 3, caption of Figure 5

See also under Formal statements (Lean): P007-T9, P007-T10

As printed

Supplement p. 2: “Our final numerical illustration considers an upper bound on the expected proportion of observations that are missing, 1−q∗k1-q_*^k, which does not depend on nn. Figure 5 plots the expected proportion of data missing versus the number of variables kk.”

Supplement p. 3: “Figure 5. Lower bound of expected proportion of all rows missing (pall‾\colorbox{#ffe58a}{$\underline{p_{all}}$}) plotted against values of n,k\colorbox{#ffe58a}{$n$},k and q∗q_*.”

The problem

Under Assumption 2, each row is complete with probability at most q∗kq_*^k, and the expected proportion of rows removed by listwise deletion is at least 1−q∗k1-q_*^k. The quantity is a lower bound, as the title of Figure 5 says, and the first sentence of the paragraph calls it an upper bound. The caption attaches to the curve the symbol pall‾\underline{p_{all}}, which the supplement defines as (1−q∗k)n(1-q_*^k)^n, the lower bound on the probability that all rows are removed, and it lists nn among the quantities that vary. The plotted quantity is 1−q∗k1-q_*^k, which does not depend on nn, and the figure varies kk and q∗q_* only. The two quantities agree when n=1n=1, as the text after the paragraph notes.

Evidence an exact counterexample

The evidence is a derivation and an exact counterexample. Let FnF_n denote the proportion of the nn rows that are removed. By Assumption 2 and the chain rule, each row is complete with probability at most q∗kq_*^k; a row for which some conditioning event of the assumption has probability zero is complete with probability zero. By linearity of expectation, I obtain E[Fn]≥1−q∗k\mathbb{E}[F_n]\ge1-q_*^k, and Assumption 1 is not used. For the counterexample to the upper bound, let n=k=2n=k=2 and q∗=3/4q_*=3/4, and let the four entries be independent, the first variable observed with probability 3/43/4 and the second with probability 1/21/2 in each row. Assumptions 1 and 2 hold, and each row is complete with probability 3/83/8. I find E[F2]=5/8\mathbb{E}[F_2]=5/8, which exceeds 1−q∗2=7/161-q_*^2=7/16. In the same example the lower bound on the probability that both rows are removed is (7/16)2=49/256(7/16)^2=49/256, which differs from 7/167/16. The code computes the plotted curve as 1−q∗k1-q_*^k for k=1,…,150k=1,\dots,150 and the three values of q∗q_*, and all 1,350 values of its grid agree with this formula. At q∗=0.99q_*=0.99 and k=150k=150 the curve equals 0.77850.7785, while (1−q∗k)n=1.3×10−11(1-q_*^k)^n=1.3\times10^{-11} for n=100n=100.

What survives

The curves of Figure 5 are correct and do not change. The correction replaces one word and the caption, and no assumption, proof or calculation in the applications depends on either.

Corrected

Supplement p. 2: “Our final numerical illustration considers a lower bound on the expected proportion of rows removed by listwise deletion, bk=1−q∗kb_k=1-q_*^k, which does not depend on nn. Figure 5 plots this lower bound versus the number of variables kk.”

Supplement p. 3: “Figure 5. Lower bound bk=1−q∗k\colorbox{#ffe58a}{$b_k=1-q_*^k$} on the expected proportion of rows removed by listwise deletion, plotted against k\colorbox{#ffe58a}{$k$} for q∗∈{0.75,0.90,0.99}q_*\in\{0.75,0.90,0.99\}.”

The statement behind the figure is the following. Under Assumption 2, for all n≥1n\ge1 and k≥1k\ge1, the expected proportion of rows removed by listwise deletion is at least 1−q∗k1-q_*^k. Equality holds when all entries are independent and each is observed with probability q∗q_*. The lower bound (1−q∗k)n(1-q_*^k)^n of Lemma 3 on the probability that all rows are removed is strictly smaller than 1−q∗k1-q_*^k when n>1n>1 and 0<q∗<1\colorbox{#ffe58a}{$0<q_*<1$}.

Crop of three lines beginning 'Our final numerical illustration considers an upper bound on the expected proportion of observations that are missing, 1 - q_*^k, which does not depend on n'. A red box marks the words 'upper bound'.
Supporting Information A, supplement p. 2, last paragraph. The marked words should read “lower bound”.
Crop of Figure 5. The title reads 'Lower bound on expected proportion of data missing'. Three curves, labelled q_* = 0.75, 0.90 and 0.99, show the lower bound on the expected proportion missing, from 0 to 1, against the number of variables, from 0 to 150; the curve for 0.99 ends near 0.78. In the caption below, 'Figure 5. Lower bound of expected proportion of all rows missing (p_all underlined) plotted against values of n,k and q_*', red boxes mark the underlined symbol p_all in parentheses and the letters 'n,k'.
Supplement p. 3: Figure 5 with its title and caption. The title says “Lower bound on expected proportion of data missing”; the marked symbol denotes the bound of Figure 3, and the marked parameter list incorrectly includes nn.

The Quality of Government data analysed in Section 3 are the January 2021 Basic cross-sectional dataset, not the Standard dataset that the article names and cites; the reported numbers stand for the Basic file.

Section 3, p. 151 (PDF p. 3), first paragraph; reference list, p. 155 (PDF p. 7), entry for Teorell et al.; replication files, readme.txt and Simcode_Qog.R

As printed

Section 3, p. 151: “… we considered two prominent datasets in use in the fields of comparative politics and international relations: the January 2021 QoG (Teorell et al. 2021) standard cross-sectional dataset, and the State Failure dataset …”

Reference list, p. 155: Teorell, J., A. Sundström, S. Holmberg, B. Rothstein, N. A. Pachon, and C. M. Dalli, 2021. “The Quality of Government Standard Dataset, Version Jan21.” University of Gothenburg, The Quality of Government Institute.

Replication files, readme.txt: “qog_bas_cs_jan21.dta available at: https://www.qogdata.pol.gu.se/data/qog_std_cs_jan21.dta (last accessed Jun 25, 2021)”, followed by the citation of “The Quality of Government Standard Dataset, version Jan21” with doi:10.18157/qogstdjan21.

The problem

Section 3 describes the first dataset as the January 2021 QoG standard cross-sectional dataset, and the entry in the reference list is for the Standard dataset. The file that the replication code reads is qog_bas_cs_jan21.dta. The data label stored in this file is “Quality of Government Basic dataset 2021 - Cross-Section”, and its field version equals QoGBasCSJan21 in all 194 rows. The Quality of Government column of Table 1, Figure 1 and the thresholds 2, 17, 14 and 52 of Section 3.2 are computed from this Basic file. The article attributes these results to a dataset that the replication files do not analyse. The readme and a comment in the script pair the name of the Basic file with the download address and the citation of the Standard file. The citation and download address identify a different edition. The simulation draws variables at random from the 351 variables of the Basic file, and its expected proportions and probabilities refer to that set of variables. The replication files contain no result for the Standard dataset.

How each source identifies the Quality of Government data.
SourceIdentification
Section 3, p. 151“standard cross-sectional dataset”
Reference list, p. 155“The Quality of Government Standard Dataset, Version Jan21”
Readme, download addressqog_std_cs_jan21.dta
Readme, citationStandard Dataset, doi:10.18157/qogstdjan21
File read by the codeqog_bas_cs_jan21.dta
Data label stored in the file“Quality of Government Basic dataset 2021 - Cross-Section”
Field version, all 194 rowsQoGBasCSJan21
Evidence a recalculation from the source

The evidence is a recalculation from the replication files. The code drops six identifier columns (ccode, cname, ccodealp, ccodecow, ccodewb, version), which leaves 194 rows and 351 columns with 24,424 of 68,094 cells missing. The column with the most missing values has 176 of them, and six columns have none. These counts give the five numerical entries of the Quality of Government column of Table 1. Let rir_i denote the number of observed columns in row ii. When kk of the 351 columns are drawn uniformly without replacement, the expected proportion of complete rows is

1194∑i=1194(rik)/(351k),\frac{1}{194}\sum_{i=1}^{194}\binom{r_i}{k}\Big/\binom{351}{k},

which is approximately 0.43640.4364 at k=2k=2 and 0.009130.00913 at k=17k=17. I ran the simulation code on the Basic file with the seed of the replication files (213) and 25,000 draws at each of the 100 values of kk. The run reproduces all 100 estimates of expected proportions and all 100 probability estimates of the saved log at its printed precision. I did not have the Standard file, and I did not compute its dimensions, its missingness or its simulation curves.

The Quality of Government numbers of the article, as printed and as recomputed from the Basic file. The first five rows are the numerical entries of the column of Table 1 (p. 152), and the last four are the thresholds of Section 3.2 (p. 153). The expected proportions are calculated exactly and rounded here, and the two probabilities are simulation estimates from 25,000 draws.
QuantityAs printedFrom the Basic file
Number of observations194194
Number of variables351351
Proportion of missing values, average35.9%35.87% (24,424 of 68,094)
Proportion of missing values, maximum90.7%90.72% (176 of 194)
Number of variables fully observed66
Expected proportion of rows removed, 2 variablesmore than 50%56.36%
Expected proportion of rows removed, 17 variablesmore than 99%99.09%
Probability that all rows are removed, 14 variablesgreater than 0.50.5076
Probability that all rows are removed, 52 variablesmore than 0.990.9902
What survives

The Quality of Government column of Table 1 and the simulation estimates in Figure 1 and Section 3.2 stand for the Basic file. Section 2, the Appendix, the supplement and the State Failure application do not use this file. The correction changes the dataset, and with it the set of variables, that the Quality of Government application describes.

Corrected

Section 3, p. 151: “… the January 2021 QoG Basic cross-sectional dataset, and the State Failure dataset …”

The citation “Teorell et al. 2021” in Section 3 and in the Introduction (p. 149), and the entry in the reference list, are replaced by the reference for the Quality of Government Basic Dataset, Version Jan21. The file is qog_bas_cs_jan21.dta, with version field QoGBasCSJan21. I have not confirmed the author list and the DOI of that reference against the documentation of the Quality of Government Institute, and I do not give them here.

In the replication files, the download address and the citation in readme.txt and in the comment of Simcode_Qog.R are replaced by those of the Basic file. The Quality of Government column of Table 1 and Figure 1 keep their values, and their source is the Basic dataset.

Crop of the heading '3 Application' and the six lines below it. A red box marks the words '(Teorell et al. 2021) standard' in the phrase 'the January 2021 QoG (Teorell et al. 2021) standard cross-sectional dataset'.
Section 3, p. 151 (PDF p. 3), first paragraph. The marked words name and cite the Standard dataset.
Crop of three reference entries, the last partly shown. In the entry for Teorell, Sundström, Holmberg, Rothstein, Pachon and Dalli, 2021, a red box marks the word 'Standard' in the title 'The Quality of Government Standard Dataset, Version Jan21'.
Reference list, p. 155 (PDF p. 7), entry for Teorell et al. The marked word is in the title of the dataset.

For the State Failure data, the average proportion of missing values is 66.3%, not 66.8%, and in the simulation the estimated loss of rows exceeds 99% with six or more variables, not with more than three.

Table 1, p. 152 (PDF p. 4), State Failure column, row “Proportion of missing values (avg.)”; Section 3.2, p. 153 (PDF p. 5), third paragraph, last sentence

See also under Formal statements (Lean): P007-T1, P007-T2, P007-T12, P007-T13

As printed

Table 1, State Failure column: Number of observations 8,580; Number of variables 1,205; Proportion of missing values (avg.) 66.8%; Proportion of missing values (max) 99.99%; Number of variables fully observed 79.

Section 3.2: “The researcher can expect to lose more than 50% of the data if the only variable included is randomly selected. With more than three variables, the loss is more than 99%.”

The problem

Table 1 describes the State Failure data after the 19 variables with 100% missingness were removed: 8,580 country-years and 1,205 variables. The replication script computes the average proportion of missing values before it removes those 19 variables and does not compute it again. The printed 66.8% is an average over 1,224 variables, while the number of variables and the maximum in the same column refer to the 1,205 that remain. For those 1,205 variables the average is 66.3%.

The sentence in Section 3.2 places the 99% threshold too early. In the simulation that the article reports, the estimated loss of rows is 97.2% with four variables and 98.6% with five, and it first exceeds 99% with six. The gap at four variables is far larger than the simulation error. An exact bound shows that, for data of these dimensions with this many observed cells, the expected loss with four variables cannot reach 99%.

Evidence a historical Lean attribution (Generated status: clause support not certified in this release)

I checked the replication script and its saved log, recalculated the average, and derived an exact bound.

The average. The script computes the average at line 29, removes the wholly missing variables at lines 39–40, and builds the table column at line 55 from the reduced data and the earlier average. The log prints the average as 0.668390827582004, taken over 8,580×1,224=10,501,9208{,}580 \times 1{,}224 = 10{,}501{,}920 cells. Allowing for rounding, the only whole number of missing cells consistent with that value is 7,019,387. The 19 removed variables hold 19×8,580=163,02019 \times 8{,}580 = 163{,}020 missing cells and no observed ones. The average over the remaining variables is

7,019,387−163,0208,580×1,205=6,856,36710,338,900=0.66316…,\frac{7{,}019{,}387 - 163{,}020}{8{,}580 \times 1{,}205} = \frac{6{,}856{,}367}{10{,}338{,}900} = 0.66316\ldots,

which rounds to 66.3%. I infer these counts from the log. My recalculation from the data file gives the same average, 0.66316…0.66316\ldots, under the code’s definition of a missing cell, as described under “What was checked”.

The threshold. Let rir_i be the number of observed cells in row ii among the 1,205 variables, so that ∑iri=10,338,900−6,856,367=3,482,533\sum_i r_i = 10{,}338{,}900 - 6{,}856{,}367 = 3{,}482{,}533. When four variables are drawn uniformly without replacement, row ii survives for (ri4)\binom{r_i}{4} of the (1,2054)\binom{1{,}205}{4} equally likely draws. Since the function r↦(r4)r \mapsto \binom{r}{4} is convex on the integers, the sum ∑i(ri4)\sum_i \binom{r_i}{4} is smallest when the rir_i are as equal as possible, which means 405 in 947 rows and 406 in 7,633 rows. I obtain

E[proportion of rows retained]=∑i(ri4)8,580(1,2054)≥947(4054)+7,633(4064)8,580(1,2054)=24,513,560,2171,923,069,889,510=0.01274….\begin{aligned} \mathbb{E}[\text{proportion of rows retained}] &= \frac{\sum_i \binom{r_i}{4}}{8{,}580 \binom{1{,}205}{4}} \geq \frac{947 \binom{405}{4} + 7{,}633 \binom{406}{4}}{8{,}580 \binom{1{,}205}{4}} \\ &= \frac{24{,}513{,}560{,}217}{1{,}923{,}069{,}889{,}510} = 0.01274\ldots. \end{aligned}

This implies that the expected loss with four variables is at most 98.73%, however the missing cells are arranged. The bound uses only the dimensions of the data and the number of observed cells.

The estimates of the archived simulation, with 25,000 draws for each number of variables, are in the table below. A Chernoff bound for the mean of 25,000 independent draws in [0,1][0,1] applies to the two estimates nearest the threshold. If the expected proportion retained with five variables were at most 1%, an estimate as large as the one in the log would have probability at most 2.8×10−72.8 \times 10^{-7}. If it were at least 1% with six variables, an estimate as small as the one in the log would have probability at most 9.6×10−89.6 \times 10^{-8}.

State Failure data: estimates in the archived simulation log, by number of randomly selected variables.
VariablesProportion of rows retainedLoss of rows
10.336160466.38%
20.134565486.54%
30.0605838293.94%
40.0282075397.18%
50.0136500398.63%
60.00664025299.34%
70.00318795399.68%
What survives

Figure 2 and the simulation estimates are unchanged: the simulation uses the 1,205 variables that remain and never reads the average. The other State Failure entries of Table 1 do not change as a consequence. The QoG results, Lemma 3, Proposition 5 and the numerical illustrations do not depend on either number. The simulation still supports the conclusion that a few variables are enough for rapid loss of rows. One table entry and one sentence change. The script’s part in the table entry is the subject of P007-E014.

Corrected

Table 1, State Failure column: Proportion of missing values (avg.) 66.3%.

Section 3.2: “The researcher can expect to lose more than 50% of the data if the only variable included is randomly selected. In the simulation, the estimated loss exceeds 99% with six or more variables.”

In the replication script, recompute the average after the wholly missing columns are removed, with prop.missing <- mean(colMeans(isnamatrix)) after line 44.

Crop of Table 1, summary statistics for the QoG and State Failure datasets. In the State Failure column, the entry 66.8 per cent in the row 'Proportion of missing values (avg.)' is outlined in red. The other entries in that column are Country-years, 8,580, 1,205, 99.99 per cent and 79.
Table 1, p. 152 (PDF p. 4). The marked entry, 66.8%, is an average over 1,224 variables; the rest of the column describes the 1,205 variables that remain after 19 were removed.
Crop of the third paragraph of Section 3.2 and the start of the fourth. In the sentence 'With more than three variables, the loss is more than 99 per cent.' the words 'more than three' are outlined in red.
Section 3.2, p. 153 (PDF p. 5), third paragraph. The marked words place the 99% threshold at four variables; in the simulation it is first passed at six.
Line chart with a logarithmic vertical axis. Seven blue points give the estimated share of rows retained with one to seven variables: 33.6, 13.5, 6.06, 2.82, 1.37, 0.66 and 0.32 per cent. A dashed horizontal line marks 1 per cent of rows retained, that is 99 per cent lost. A shaded orange region below that line, beginning between three and four variables, marks the printed statement. The points for four and five variables lie above the line, and the points for six and seven lie below it. A black diamond at four variables marks the exact lower bound, 1.27 per cent.
State Failure data: estimated proportion of rows retained after listwise deletion, by number of randomly selected variables, from the archived simulation log (logarithmic scale). The shaded region is the printed statement, a loss above 99% with more than three variables. The estimates for four and five variables lie outside it, and the exact lower bound at four variables (diamond) is 1.27%.

The script behind the State Failure column of Table 1 averages missingness before it drops the 19 wholly missing variables, and it counts an empty text entry imported from Stata as observed, a convention that the article does not state.

Table 1, p. 152 (PDF p. 4), State Failure column; replication script Simcode_Statefailure.R, lines 13, 28–29, 39–47 and 55

See also under Formal statements (Lean): P007-T12, P007-T13

As printed

Table 1, State Failure column: Number of observations 8,580; Number of variables 1,205; Proportion of missing values (avg.) 66.8%; Proportion of missing values (max) 99.99%; Number of variables fully observed 79. Below the table: “and in the case of the State Failure data, to apply the principle of charity, we removed 19 variables that exhibited 100% missingness.”

Replication script Simcode_Statefailure.R:

  • line 13: statefailure <- read_dta("sort11v3.dta")
  • line 28: isnamatrix <- is.na(statefailurepure)
  • line 29: prop.missing <- mean(colMeans(isnamatrix))
  • line 44: isnamatrix <- is.na(statefailure_final)
The problem

Two things are at issue. The first is the average in Table 1, corrected in P007-E010. The script computes it at line 29, over 1,224 variables, and line 55 reports it for the 1,205 variables that remain after the 19 wholly missing ones are dropped.

The second is how the script decides that a cell is missing. It applies is.na to data read from a Stata file with read_dta. An empty text entry in a Stata file arrives in R as an empty string, which is.na does not count as missing. An empty text entry is then counted as an observed value. The article does not state this convention. I have not established whether the State Failure file contains empty text entries or how counting them as missing would change the results. For a fixed set of variables, counting them as missing adds missing cells and removes none. It could only show more rows lost.

Evidence a recalculation from the source

The evidence is a derivation for the average and a test of the import on a small constructed file.

Let qbeforeq_{\text{before}} and qafterq_{\text{after}} be the average proportions of missing cells over the 1,224 and the 1,205 variables. Since the 19 removed variables are wholly missing and all variables have the same 8,580 rows, I have

1,224 qbefore=19+1,205 qafter.1{,}224 \, q_{\text{before}} = 19 + 1{,}205 \, q_{\text{after}}.

A value that rounds to 66.8% satisfies qbefore<0.6685q_{\text{before}} < 0.6685. Then the identity implies that

qafter<1,224×0.6685−191,205=199,811301,250<0.6675,q_{\text{after}} < \frac{1{,}224 \times 0.6685 - 19}{1{,}205} = \frac{199{,}811}{301{,}250} < 0.6675,

and the average over the 1,205 variables cannot round to 66.8%. Substituting the value in the saved log, qbefore=0.668390827582004q_{\text{before}} = 0.668390827582004, into the identity, I obtain qafter=0.66316…q_{\text{after}} = 0.66316\ldots, which rounds to 66.3%. The log implies 6,856,367 missing cells out of 10,338,900 retained cells. The pre-removal average exceeds the retained average by 0.52 percentage points.

For the import, a Stata file with four rows was written and read back with read_dta (haven 2.5.5; the authors report version 2.4.0). Its text variable held an empty string, “0”, “1” and a missing value, and its numeric variable had no missing value. The text values came back as an empty string, “0”, “1” and an empty string. Under is.na, none of the 8 cells is missing and all 4 rows are complete. When empty strings are counted as missing, 2 cells are missing and 2 rows are complete. The test shows how the convention works. It does not show that the State Failure file has empty text entries.

What survives

Recomputing the average changes no simulation result, since the simulation uses the 1,205 variables that remain and never reads the average. Figure 2 is unchanged. Under the script’s own definition of a missing cell, the count of 79 fully observed variables is unaffected. The results of Section 2, the numerical illustrations and the QoG calculations do not depend on this script. If the State Failure file does hold empty text entries, the count of fully observed variables and the curves of Figure 2 could change under a definition that counts them as missing; no such change is established here.

Corrected

Table 1, State Failure column: Proportion of missing values (avg.) 66.3%.

In the script, add prop.missing <- mean(colMeans(isnamatrix)) after line 44.

Add to the description of the data: “A cell is counted as missing when is.na is true after import with read_dta; an empty text entry is counted as observed.” An analysis that counts empty text entries as missing should be reported separately, with its own definition of a missing cell and its own set of variables, after the data and their documentation have been inspected.

Crop of Table 1 and the sentence that follows it. In the State Failure column, the entry 66.8 per cent in the row 'Proportion of missing values (avg.)' is outlined in red. The sentence below the table ends 'we removed 19 variables that exhibited 100 per cent missingness.'
Table 1 and the sentence below it, p. 152 (PDF p. 4). The script computes the marked average before the removal of the 19 variables that the sentence describes.

The glmnet function did not omit rows with missing data by default when the article appeared, and it should be removed from the list of software examples.

Introduction, p. 149 (PDF p. 1), first paragraph, second sentence

As printed

“It constitutes the default behavior for standard data analyses in popular software packages: for example, rows with any missing data are by default omitted by the lm function in R (R Core Team 2020), the regress command in Stata (Stata.com 2020), and the glmnet function in the R package of the same name (Friedman, Hastie, and Tibshirani 2010).”

The problem

The sentence says that three functions omit incomplete rows by default, and the third does not. In glmnet 4.1-3, the release current when the article was published online on 2 March 2022, the function glmnet contains no step that removes rows. For a Gaussian response, it keeps the number of rows of its input and passes the matrix of predictors, with any missing values, to the fitting routine. A missing value in the response stops the function with an error. Rows are omitted only if the user removes them, from the predictors and the response together, before the call.

Evidence an exact counterexample

The evidence is a check of the package source as released. I read the files R/glmnet.R and R/elnet.R of glmnet 4.1-3, released on CRAN on 2 November 2021 (https://raw.githubusercontent.com/cran/glmnet/4.1-3/R/glmnet.R), and the source of releases 4.1, 4.1-1, 4.1-2 and 1.1-4, the version named in Friedman, Hastie, and Tibshirani (2010). None of them removes incomplete rows by default. I also ran the R code of 4.1-3 on eight rows, with the compiled fitting routine replaced by a function that records its arguments. With one missing predictor value, all eight rows reach the fitting routine. With one missing response value, the call stops with the error “missing value where TRUE/FALSE needed”. Seven rows reach the fitting routine only when the incomplete row is removed beforehand. I did not run the compiled package of 2021, and I have not established what its fitting routine did with a missing predictor value. I did not check the lm and regress examples against the releases of that date.

What survives

No result, table or figure depends on the example, and the replication code does not use glmnet. The other two examples and the point that the sentence makes about default behavior are not affected.

Corrected

“It constitutes the default behavior for standard data analyses in popular software packages: for example, rows with any missing data are by default omitted by the lm function in R (R Core Team 2020) and the regress command in Stata (Stata.com 2020).”

The citation of Friedman, Hastie, and Tibshirani (2010) is removed with the example. It is the only citation of that reference in the article, and the entry in the reference list is removed as well.

Crop of the first paragraph of the Introduction and the first two lines of the next paragraph. Red boxes mark the words 'the glmnet function in the R package of the same name' at the end of one line and '(Friedman, Hastie, and Tibshirani 2010)' at the start of the next line.
Introduction, p. 149 (PDF p. 1), first paragraph. The marked words are the third software example and its citation.

The editors of the Handbook of Survey Research are P. H. Rossi, J. D. Wright and A. B. Anderson.

References, p. 155 (PDF p. 7), entry for Berk (1983)

As printed

Berk, R. 1983. “Applications of the General Linear Model to Survey Data.” In Handbook of Survey Research, edited by A. B. A. Peter, H. Rossi, and J. D. Wright, pp. 495–546. Quantitative Studies in Social Relations. New York: Academic Press.

The problem

The entry names the editors of the volume as “A. B. A. Peter, H. Rossi, and J. D. Wright”. The editors of the 1983 edition are Peter H. Rossi, James D. Wright and Andy B. Anderson.

Evidence a check of the printed page

The evidence is a check of a public source. The publisher’s catalogue entry for the 1983 Academic Press edition (ISBN 9780125982269) lists the editors as Peter H. Rossi, James D. Wright and Andy B. Anderson, in that order, and the title page reproduced in a public preview of the book has the same names in the same order. The catalogue entry lists Berk’s chapter as chapter 13, pp. 495–546.

What survives

The chapter title, pages, series and publisher in the entry are as printed, and no result depends on it. I did not read the chapter itself.

Corrected

Berk, R. 1983. “Applications of the General Linear Model to Survey Data.” In Handbook of Survey Research, edited by P. H. Rossi, J. D. Wright, and A. B. Anderson, pp. 495–546. Quantitative Studies in Social Relations. New York: Academic Press.

Crop of three entries in the reference list. In the entry for Berk (1983), the words 'A. B. A. Peter, H. Rossi, and J. D. Wright,' after 'edited by' are outlined in red.
References, p. 155 (PDF p. 7), entry for Berk (1983). The marked editor names are garbled.

Clarity notes

Imprecise or ambiguous wording whose intended meaning is correct; not an error.

Assumption 2 rules out a variable that is always observed whenever all the variables before it in the ordering are observed; it does not rule out the reverse, as the sentence before the assumption can suggest.

Section 2, p. 150 (PDF p. 2), paragraph before Assumption 2, last sentence

As printed

“Namely, we will assume that there is a factorization of the data such that all conditional probabilities that an observation is missing are bounded away from zero, given whenever prior variables are observed. Substantively, this assumption rules out the case where, for example, one variable is always observed whenever another variable is observed.”

Intended meaning

Assumption 2 bounds by q∗<1q_*<1 the probability that variable jj is observed given that variables 1,…,j−11,\dots,j-1 are all observed, whenever that event has positive probability. The sentence describes this bound, and its example is to be read in the direction of the ordering: a variable cannot be always observed whenever all the variables before it are observed. Read in this way, the sentence is correct.

The assumption allows the reverse. For example, let k=2k=2, and let (Mi1,Mi2)(M_{i1},M_{i2}) take the values (1,1)(1,1), (0,1)(0,1) and (0,0)(0,0) with probabilities 1/21/2, 1/41/4 and 1/41/4, where Mij=0M_{ij}=0 means that the entry is observed. Then Pr⁡(Mi1=0)=1/2\Pr(M_{i1}=0)=1/2 and Pr⁡(Mi2=0∣Mi1=0)=1/2\Pr(M_{i2}=0\mid M_{i1}=0)=1/2, and Assumption 2 holds with q∗=1/2q_*=1/2. However, Pr⁡(Mi1=0∣Mi2=0)=1\Pr(M_{i1}=0\mid M_{i2}=0)=1: the first variable is always observed whenever the second variable is observed. No result uses the sentence.

Corrected

“Substantively, this assumption rules out the case where, for example, a variable is always observed whenever all preceding variables in the ordering are observed, provided that this event has positive probability.”

Crop of the paragraph beginning 'We also assume that there is some idiosyncratic missingness in each variable', followed by Assumption 2. Red boxes mark the words 'one variable is always observed whenever another variable is observed', which run over two lines. Assumption 2 below bounds Pr(M_i1 = 0) and Pr(M_ij = 0 given that the bold vector M_i(j-1) equals 0) by q_*.
Section 2, p. 150 (PDF p. 2): the paragraph before Assumption 2 and the assumption. The marked words describe what the assumption rules out.

In the abstract, “for large nn, listwise deletion will drop all rows with probability 1” means that the probability of dropping all rows tends to 1 as nn grows, which is the statement of Proposition 5.

Abstract, p. 149 (PDF p. 1), second sentence

As printed

“We show that when (i) all data have some idiosyncratic missingness and (ii) the number of variables grows superlogarithmically in nn, then, for large nn, listwise deletion will drop all rows with probability 1.”

Intended meaning

The clause summarizes Proposition 5 (p. 151), which states that lim⁡n→∞pall=1\lim_{n\to\infty}p_{all}=1 under Assumptions 1, 2 and 4, where pallp_{all} is the probability that listwise deletion removes all rows. This statement is correct. The clause is not a guarantee at a finite nn: if all entries are independent and each is observed with probability q∈(0,1)q\in(0,1), then pall=(1−qk)n<1p_{all}=(1-q^k)^n<1 for every nn and kk. For instance, pall=9/16p_{all}=9/16 when q=1/2q=1/2 and n=k=2n=k=2.

A stronger reading also holds, although the article does not state it. Suppose that the datasets for all nn are defined on one probability space and that the bound q∗q_* of Assumption 2 does not depend on nn. By Assumption 2, each row is complete with probability at most q∗f(n)q_*^{f(n)}, and the probability that some row of the dataset with nn rows is complete is at most nq∗f(n)nq_*^{f(n)}. Since f(n)/log⁡n→∞f(n)/\log n\to\infty, I have nq∗f(n)≤n−2nq_*^{f(n)}\le n^{-2} for all large nn. These bounds are summable, and the first Borel–Cantelli lemma implies that, with probability 1, listwise deletion removes all rows for all sufficiently large nn. The corresponding sentence of the Introduction says “asymptotically” and needs no change.

Corrected

“We show that when (i) all data have some idiosyncratic missingness and (ii) the number of variables grows superlogarithmically in nn, then the probability that listwise deletion drops all rows tends to 1 as nn grows.”

Crop of the abstract. A red box marks the clause 'for large n, listwise deletion will drop all rows with probability 1' in the second sentence.
Abstract, p. 149 (PDF p. 1). The marked clause states the main result.

Formal statements (Lean)

Formal results: 15 whole; 1 restricted; 0 refuted; 1 omitted.

Coverage concerns the identified claim under its stated assumptions. A checked proof establishes the formal statement; correspondence to the published text was reviewed by AI. Repository access is currently required to open the source links.

Each result links to its current formal statement and proof. “Whole” covers the identified claim under its assumptions; “restricted” covers the stated narrower scope; “refuted” gives a counterexample to the identified assertion; “omitted” has no current theorem. Any named published theorem used as a premise remains an assumption. Browse the formal results for this work.

Section 3.2, State Failure paragraph, last sentence (“With more than three variables, the loss is more than 99%”), read with Table 1 and Section 3.1. The refutation is the validator's exact bound and is not in the paper

p. 153 (PDF p. 5), Section 3.2, third paragraph; Table 1 and Section 3.1, p. 152 (PDF p. 4)

Section 3.2, p. 153 (PDF p. 5), with Table 1 and Section 3.1, p. 152 (PDF p. 4). Every fixed 8580 by 1205 array with the specified observed total has four-variable expected loss below 99%, with the complete exact certificate.

Historical component credits
  • Proved AronowAudit.P007.uniform_retention_counting_identity Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The additional parameter values constitute a harmless generalization; both explicit premises make the sampling interpretation well defined.
  • Proved AronowAudit.P007.supportingLine Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Fidelity sweep: ok
  • Proved AronowAudit.P007.supportingLine_equality Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Fidelity sweep: ok
  • Proved AronowAudit.P007.numeratorLowerBound Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Fidelity sweep: ok
  • Proved AronowAudit.P007.denominatorValue Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Fidelity sweep: ok
  • Counterexample proved AronowAudit.P007.fourVariablesCannotLoseMoreThan99Percent Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; This is the universal conditional refutation of the four-variable threshold, not an assertion that the empirical count was independently recounted. The observed-total premise is explicitly part of T1(iii)-(v).
  • Counterexample proved AronowAudit.P007.concrete_four_variable_counterexample Withdrawn support; baseline fidelity: wrong_credit; manifest: not used; not used in Solution; Final status wrong_credit: a synthetic feasibility witness does not identify the named empirical array; the universal conditional bound supplies the refutation.

Table 1, State Failure column, “Proportion of missing values (avg.)”, and Section 3.2, the sentence “With more than three variables, the loss is more than 99%”, as repaired by Sol

Table 1, p. 152 (PDF p. 4); Section 3.2, third paragraph, p. 153 (PDF p. 5)

Table 1, p. 152 (PDF p. 4), and Section 3.2, p. 153 (PDF p. 5). The logged-average hypothesis gives corrected counts and 66.3% missingness, the four-variable population bound, and the comparisons of simulation estimates at four, five and six variables. Restricted: drops the corrected sentence's assertion about every variable count above six; only the three displayed estimates are compared.

Historical component credits
  • Proved AronowAudit.P007.logged_missing_count_unique Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Fidelity sweep: ok
  • Proved AronowAudit.P007.stateFailure_corrected_average Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; This is the arithmetic component T2(i)(b)-(d); R047-10 and R047-11 supply the separate connection to masks and the logged tolerance. The natural subtractions do not truncate in these concrete calculations.
  • Counterexample proved AronowAudit.P007.fourVariablesCannotLoseMoreThan99Percent Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; counterexample_proved describes the refutation component (ii) of this corrected-statement target, not proof of the entire correction by this row alone.
  • Proved AronowAudit.P007.logged_mask_correction Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The premises match the target mathematical text: it designates 19 fully missing removed columns but does not require that all retained columns contain an observation. Its universal conclusion concerns the same retained mask, rather than a newly chosen example.
  • Proved AronowAudit.P007.archived_simulation_threshold Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; This row covers exactly T2(iii), the comparison of three supplied estimates. It does not establish the stronger correction phrase concerning all six-or-more variable counts.

Proof of Lemma 3, the case q∗=0q_* = 0 (“which is equivalent to say that all variables are fully missing”), and the same step in the Proof of Lemma 8. The witness is the validator's and is not in the paper

p. 154 (PDF p. 6), Appendix: Proofs, Proof of Lemma 3, first three sentences; Supplement p. 4, Proof of Lemma 8, first three sentences

proof of Lemma 3, p. 154 (PDF p. 6), and proof of Lemma 8, Supplement p. 4. The two explicit laws refute both directions of the asserted equivalence between a zero observation bound and full missingness, and also apply to two singleton groups.

Historical component credits
  • Counterexample proved AronowAudit.P007.zero_bound_counterexample Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; This refutes the printed assertion about every variable or group being fully missing; it confirms rather than refutes Lemma 3 itself. Two singleton groups give the stated grouped witness.
  • Counterexample proved AronowAudit.P007.fully_missing_all_bounds Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; q is a selected upper bound, not the least bound. Taking q=1/2 gives the counterexample to the converse implication, even though q=0 is also admissible.

Proof of Lemma 3, the definition of qijq_{ij} and the step “By Assumption 2, qij≤q∗q_{ij} \leq q_*, for all j∈{1,2,…,k}j \in \{1,2,\ldots,k\}”, and the same step in the Proof of Lemma 8. The witnesses are the validator's and are not in the paper

p. 154 (PDF p. 6), Appendix: Proofs, Proof of Lemma 3; Supplement p. 4, Proof of Lemma 8

proof of Lemma 3, p. 154 (PDF p. 6), and proof of Lemma 8, Supplement p. 4. Both explicit laws meet the assumptions but have a null conditioning prefix, while retaining the deletion bound; singleton groups give the same counterexamples.

Historical component credits
  • Counterexample proved AronowAudit.P007.null_prefix_counterexample Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The all-v relation is intentionally unrestricted at a null prefix: v=1 exceeds q=1/2 and v=0 also works. This is a witness to the proof gap, not a counterexample to the lemma conclusion.
  • Counterexample proved AronowAudit.P007.later_null_counterexample Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The witness exposes an undefined ordinary conditional at the third variable. No null-event conditional is silently defined to be Lean real 0/0; singleton groups yield the grouped example.

Proofs of Lemma 3 and Lemma 8, as repaired (Sol's smallest repair and the validator's derivation). The statements of the two lemmas are unchanged

p. 154 (PDF p. 6), Appendix: Proofs, Proof of Lemma 3; Supplement p. 3 (definition of Mij\mathbf{M}_{ij}) and p. 4 (Proof of Lemma 8)

repaired proofs of Lemma 3, p. 154 (PDF p. 6), and Lemma 8, Supplement p. 4. The prefix recursion, zero-prefix coefficient convention, product formula and zero endpoint establish the finite bound for variables or groups.

Historical component credits
  • Proved AronowAudit.P007.zero_bound_first_missing Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the first-entry assertion in T5(a). Its almost-sure conclusion is not an everywhere claim about zero-weight arrays.
  • Proved AronowAudit.P007.zero_bound_all_deleted Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the all-deletion consequence in T5(a); this does not claim all later variables are missing.
  • Proved AronowAudit.P007.prefix_step Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: T5(b). Null-prefix survival is forced to remain zero by event inclusion; the input does not assume the recursion.
  • Proved AronowAudit.P007.repairedConditional_bounds Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the bounds in T5(c). The zero convention is part of the corrected target, and the bounds remain a theorem rather than fields of a contract.
  • Proved AronowAudit.P007.repaired_chain_product Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the product in T5(c). At j=0 the product is one; at null prefixes zero factors and zero survival agree, and no division by zero is interpreted as an ordinary conditional probability.
  • Proved AronowAudit.P007.prefix_survival_bound Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: T5(d), with a useful extension to every prefix. The bound applies also at q=0 and after a null prefix.
  • Proved AronowAudit.P007.independent_rows_product Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the equality in T5(e), which uses row independence alone; the lower inequality is supplied by R048-02. No conditional bound is hidden in RowsIndependent.
  • Proved AronowAudit.P007.lemma3 Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the unchanged lemma conclusion in T5(e). ConditionalObservationBound includes the first-entry bound and constrains later ratios only on positive prefixes.

Proofs of Proposition 5 and Proposition 10, third paragraph (the Bernoulli step), as repaired. The statements of the two propositions are unchanged

p. 154 (PDF p. 6), Appendix: Proofs, Proof of Proposition 5, paragraph beginning “Since q∗∈(0,1)q_* \in (0,1) and k=f(n)k = f(n)”; Supplement p. 4 (foot) and p. 5, Proof of Proposition 10, paragraph beginning “Since q∗∈(0,1)q_* \in (0,1) and k=l(n)k = l(n)”

repaired proof of Proposition 5, p. 154 (PDF p. 6), and Proposition 10, Supplement pp. 4-5. The corrected finite Bernoulli sandwich, zero endpoint, positive-domain exponential identity and two-variable group counts hold.

Historical component credits
  • Proved AronowAudit.P007.bernoulli_sandwich Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: T6(a). This is a harmless generalization of the corrected finite inequality, not an asymptotic result.
  • Proved AronowAudit.P007.deletion_probability_sandwich Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: T6(b). All displayed finite conclusions are exported; no finite inequality is credited as a limit theorem.
  • Proved AronowAudit.P007.zero_bound_all_deleted Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the zero endpoint in T6(c). It is the same theorem as R048-07, legitimately credited for two different components.
  • Proved AronowAudit.P007.logarithm_decay_identity Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the finite identity in T6(c), not the decay limit. The positive-domain hypotheses exclude log(0), and q<1 is needed only for the separate limit argument.
  • Proved AronowAudit.P007.groups_are_not_variables Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; {'source': 'adjudication', 'decision': 'rejected', 'what': 'At AronowAudit/P007/P007_Probability.lean:348, groups_are_not_variables states that every natural n > 0 satisfies n ≤ 2*n and 2*n ≠ n, matching the count witness in T6(d): “g_n = l(n) = n” and “k_n = 2n ≠ l(n)”. For every such n, groups Fin n and variables Fin n × Fin 2, assigned to groups by the first projection, realize exactly these counts and two variables per group. The absence of explicit group objects therefore loses no cases of this component and adds no hypothesis beyond the claimed positivity; the other parts of T6 are outside this row’s stated component.'}

Supporting Information A, the two paragraphs on Figure 4 (“the smallest possible kk”; “we need only k=33k = 33”; “we need k=952k = 952”; “ensuring that pall≤0.5,0.99p_{all} \leq 0.5, 0.99”) and the caption of Figure 4. The witnesses are the validator's and are not in the paper

Supplement p. 2 (publisher-supplement.pdf p. 2), Supporting Information A, paragraphs beginning “Figure 4 illustrates” and “With more missingness”, and Figure 4; not in the article PDF

Figure 4 and its discussion, Supplement p. 2. All specified equality, rounding, worked-value and actual-probability upper-guarantee counterexamples hold for the displayed normalized independent-cell laws.

Historical component credits
  • Counterexample proved AronowAudit.P007.floor_is_not_first_crossing Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: T7(a). The witness addresses the equality/first-crossing prose; the floor remains correct as the largest integer with best-case lower bound at most the target.
  • Counterexample proved AronowAudit.P007.figure4_printed_worked_counterexamples Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: T7(b). These are counterexamples to a guaranteed 50% probability or the stated first crossings; they do not say that no other admissible law can have actual probability 50% at those integers.
  • Counterexample proved AronowAudit.P007.lower_bound_is_not_upper_guarantee Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Components: T7(c1) and (c2), for the literal actual-probability reading. The first printed floor is separately exported by R048-16; the second follows from the adjacent bound bracket and the explicit increasing formula.
  • Counterexample proved AronowAudit.P007.printed_parameters_actual_deletion_high Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the high actual probability and Bernoulli estimate in T7(c3). The lower-bound comparison B(33)<1/2 is exported by R048-14, so it is not silently attributed to this statement alone.
  • Counterexample proved AronowAudit.P007.figure4_printed_lower_bound_not_upper Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: T7(c3), read literally as an upper bound on actual p_all. This does not refute using the plotted expression as a best-case lower bound.

Supporting Information A, the two paragraphs on Figure 4 and the caption, subcaptions and plot titles of Figure 4, as repaired by Sol

Supplement p. 2 (publisher-supplement.pdf p. 2), Supporting Information A; not in the article PDF

corrected Figure 4 and its discussion, Supplement p. 2. The real inverse, integer floor and ceiling, sharp probability guarantees and both worked values describe the lower bound, with explicit attaining laws and an upper-guarantee counterexample.

Historical component credits
  • Proved AronowAudit.P007.bestCase_increasing Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The declaration exactly matches the assigned component of the corrected target; there are no conditional external-theorem premises.
  • Proved AronowAudit.P007.threshold_unique_solution Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The declaration exactly matches the assigned component of the corrected target; there are no conditional external-theorem premises.
  • Proved AronowAudit.P007.threshold_integer_inversion Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The declaration exactly matches the assigned component of the corrected target; there are no conditional external-theorem premises.
  • Proved AronowAudit.P007.threshold_floor_ceiling Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The declaration exactly matches the assigned component of the corrected target; there are no conditional external-theorem premises.
  • Proved AronowAudit.P007.threshold_rounding_gap Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The declaration exactly matches the assigned component of the corrected target; there are no conditional external-theorem premises.
  • Proved AronowAudit.P007.threshold_probability_guarantee Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Finite joint laws capture every distribution of the target binary indicators. The target itself specifies the independent equal-observation witness. The separate middle inequality pAll>=B(k) in the target can be recovered from the credited inversion and guarantee rows by taking a=B(k).
  • Proved AronowAudit.P007.bestCase_attained Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The finite joint law is lossless for finitely many binary indicators on an arbitrary probability space. The independent-cell restriction constructs an attaining model; it is not imposed on all admissible laws.
  • Proved AronowAudit.P007.figure4_corrected_worked_values Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The declaration exactly matches the assigned component of the corrected target; there are no conditional external-theorem premises.

Supporting Information A, last paragraph (“an upper bound on the expected proportion of observations that are missing, 1−q∗k1 - q_*^k”), and the caption of Figure 5. The witness is the validator's and is not in the paper

Supplement p. 2 (last paragraph) and p. 3 (its continuation, Figure 5 and the caption); not in the article PDF

Figure 5 and its discussion, Supplement pp. 2-3. The specified two-row law has expected deleted fraction 5/8, exceeding 7/16, and distinguishes that fraction bound from both all-deletion probabilities.

Historical component credits
  • Counterexample proved AronowAudit.P007.figure5_upper_bound_counterexample Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The witness targets the literal word upper and the caption quantity. It does not refute the correct lower-bound reading of the figure.

Supporting Information A, last paragraph, and the caption of Figure 5, as repaired by Sol

Supplement p. 2 (last paragraph) and p. 3 (Figure 5 and its caption); not in the article PDF

corrected Figure 5 and its discussion, Supplement pp. 2-3. The usual measure-theoretic conditions are assumed. For positive integers n,k and any measurable missingness array on a probability space, Assumption 2 gives E[D_i]=1-s_ik >= b_k and E[F_n]=(sum_i (1-s_ik))/n >= b_k, where b_k=1-q^k is independent of n. These conclusions require no row independence. Every law with mutually independent cells of observation probability q satisfies Assumptions 1 and 2 and attains E[F_n]=b_k; a normalized independent-cell law exists for every q in [0,1). The all-deletion lower bound is b_k^n, equals b_k at n=1, and is strictly smaller for n>=2 and 0<q<1, when 0<b_k<1. On the entire domain q in [0,1), the bounds coincide if and only if n=1 or q=0; at q=0 both bounds are one.

Historical component credits
  • Proved AronowAudit.P007.expected_removed_row Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; This is the equality component of T10(i); R049-10 supplies its lower-bound component.
  • Proved AronowAudit.P007.expected_removed_fraction Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; This is the equality component of T10(ii), not a standalone claim to prove its lower-bound inequality; R049-10 supplies that inequality.
  • Proved AronowAudit.P007.expected_deletion_lower_bound Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; All conditional ratios are required only on positive prefixes, exactly as in Assumption 2; q=0 and null prefixes remain allowed.
  • Proved AronowAudit.P007.expected_deletion_bound_attained Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; This row covers the expectation equality in T10(iii); R049-01 explicitly exports both paper assumptions for the same law, so no unexported assumption certificate is being credited here.
  • Proved AronowAudit.P007.fraction_bound_vs_all_bound Limited support; baseline fidelity: narrower; manifest: used; used in Solution; {'source': 'adjudication+second review', 'adjudicator': "P007_Probability.lean:847 imposes 0 < q on every conjunct of fraction_bound_vs_all_bound, including the identities at P007_Probability.lean:848 and P007_Probability.lean:849, whereas T10(iv) gives these identities under the standing domain 0 <= q < 1. The allowed endpoint q = 0, h >= 1 is therefore omitted, although both bounds then equal 1 for every positive n; this is a mathematically trivial omission and warrants minor severity. The outer n >= 2 premise does not independently lose the n = 1 equality for interior q, because its conclusion explicitly mentions bestCaseBound 1 and one can instantiate the outer n with 2. The strict comparison has exactly the target's interior hypotheses, so there is no major narrowing. The target's closing 'only in the case n = 1' must retain the preceding interior restriction, as is appropriate for Figure 5's positive plotted q values; it is false if extended to q = 0.", 'suggested_fix': 'Export bestCaseBound n q h = (deletedFractionBound q h)^n and bestCaseBound 1 q h = deletedFractionBound q h without hypotheses, since both follow immediately from the definitions at P007_Probability.lean:735 and P007_Probability.lean:738. Keep 2 <= n, 0 < h, 0 < q, q < 1 only on the positivity and strict-comparison conclusions; retain the interior qualification when describing equality as occurring only at n = 1.', 'second_review': "T10(iv) states the identities 'underline(p_all) = b_k^n' and 'For n = 1, underline(p_all) = b_k' on the standing domain 0 ≤ q_* < 1, but P007_Probability.lean:847 puts 0 < q before the entire conjunction, including the identities at P007_Probability.lean:848 and P007_Probability.lean:849. The strict comparison has the target's interior hypotheses, and the n = 1 identity for interior q remains available by instantiating the unused outer n with 2. The definitions at P007_Probability.lean:735 and P007_Probability.lean:738 make both identities immediate without hypotheses, including q = 0, where both bounds equal 1 for positive k; exporting the identities separately is therefore a minor statement repair. I read the target's closing 'only in the case n = 1' within its preceding interior-q qualification, so it does not introduce a separate claim-text error.", 'severity': 'minor'}

Section 3.1, third paragraph: the two descriptions of Assumption 1 as “i.i.d. missingness across rows”, as repaired by Sol

p. 152 (PDF p. 4), Section 3.1, third paragraph, second and last sentences; Assumption 1 is on p. 150 (PDF p. 2)

Section 3.1, p. 152 (PDF p. 4), referring to Assumption 1, p. 150 (PDF p. 2). Independent rows need not have identical distributions, as the explicit witness and general product of row-specific deletion probabilities show.

Historical component credits
  • Counterexample proved AronowAudit.P007.mutually_independent_need_not_be_iid Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The corrected-statement target expressly includes this witness to the printed i.i.d. misdescription; its counterexample_proved role is consistent with that component.
  • Proved AronowAudit.P007.independent_rows_product Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The conclusion uses row-specific factors, exactly as required; neither identical distribution nor within-row independence is assumed.

Table 1, State Failure column, “Proportion of missing values (avg.)”, corrected for the 1,205 retained variables (the part of P007-E014 that is established)

Table 1, p. 152 (PDF p. 4); preprocessing sentence, pp. 151-152 (PDF pp. 3-4); replication script Simcode_Statefailure.R, lines 28-55

preprocessing, pp. 151-152 (PDF pp. 3-4), and Table 1, p. 152 (PDF p. 4). Removing wholly missing columns gives the exact average identities, rules out the printed retained average, and yields the corrected average and overstatement under the specified missing-cell count.

Historical component credits
  • Proved AronowAudit.P007.columnAverage_count_identity Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The removal theorem already exports the needed count-to-average identities.
  • Proved AronowAudit.P007.removing_fully_missing_columns Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Reindexing the designated removed and retained columns by a disjoint sum loses no arrays. The target and theorem both allow d=0 and additional fully missing retained columns.
  • Proved AronowAudit.P007.printed_average_cannot_be_retained Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; The partition hypothesis is the identity exported from masks by R049-17, so it is an established input to this arithmetic component rather than a circular assumption of the retained-average bound.
  • Proved AronowAudit.P007.stateFailure_corrected_average Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; This row certifies arithmetic and rounding of the logged counts; it does not independently infer those counts from an actual mask. R049-15 supplies that component of the target.
  • Proved AronowAudit.P007.logged_mask_correction Historical support retained; baseline fidelity: ok; manifest: not used; not used in Solution; Part (iii) uses the canonical missing-count premise A=7019387 directly; the stronger log-tolerance correction is packaged in T2.
  • Proved AronowAudit.P007.stateFailure_average_overstatement Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; This row certifies the exact arithmetic after the logged counts are known; R049-15 provides their link to the same data mask.

Table 1, State Failure column, and the State Failure sentences on expected loss in Section 3.2, as printed

Table 1, p. 152 (PDF p. 4); Section 3.2, third paragraph, p. 153 (PDF p. 5); Section 3.1, p. 152 (PDF p. 4)

No current theorem is supplied for this target.

Historical component credits
  • Not applicable Excluded association; baseline fidelity: None; manifest: not applicable; Solution use unverified; Historical excluded association.

Lemma 3

p. 150 (PDF p. 2), Section 2.1; proof in the Appendix, p. 154 (PDF p. 6)

Lemma 3, p. 150 (PDF p. 2). Under mutual row independence and the printed conditional observation bound, the probability of deleting all rows is at least (1-q^k)^n.

Historical component credits
  • Proved AronowAudit.P007.lemma3 Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Finite normalized weights are an exact representation of the distribution of finitely many Boolean indicators on any underlying probability space. All such joint laws are included, including q=0 and null prefixes; there is no added finite-support restriction on the underlying data.

Lemma 8 (Supporting Information B)

Supplement p. 4 (publisher-supplement.pdf p. 4), Supporting Information B; proof on the same page

Lemma 8, Supporting Information B, Supplement p. 4. Under mutual row independence and the printed group observation bound, the probability of deleting all rows is at least (1-q^g)^n.

Historical component credits
  • Proved AronowAudit.P007.lemma8 Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the whole printed Lemma 8. Independent groups are not assumed; the input is mutual independence of row vectors.

Proposition 5

p. 151 (PDF p. 3), Section 2.1; proof in the Appendix, p. 154 (PDF p. 6)

Proposition 5, p. 151 (PDF p. 3). Arbitrary changing joint laws with independent rows, one fixed conditional observation bound q<1 and superlogarithmic variable counts have all-deletion probability tending to one, including q=0.

Historical component credits
  • Proved AronowAudit.P007.proposition5 Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; This row proves the printed proposition itself. ConditionalObservationBound preserves the positive-prefix proviso exactly: the ratio is tested only when its denominator is positive, and the null-prefix cases remain allowed. The one q is fixed over every positive n, precisely as specified in the standing assumptions; no identical row law or cross-n coupling is required. All custom definitions were expanded, and none assumes the limit or an intermediate result. No metadata inconsistency was found.
  • Proved AronowAudit.P007.uniform_decay Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Auxiliary analytic decay is already used internally by the headline proposition; the Challenge states its probability limit.

Proposition 10 (Supporting Information B)

Supplement p. 4 (publisher-supplement.pdf p. 4), Supporting Information B; proof on pp. 4-5

Proposition 10, Supporting Information B, Supplement p. 4. Arbitrary changing group-indicator laws with independent rows, one fixed conditional observation bound q<1 and superlogarithmic group counts have all-deletion probability tending to one, including q=0.

Historical component credits
  • Proved AronowAudit.P007.proposition10 Historical support retained; baseline fidelity: ok; manifest: used; used in Solution; Component: the whole printed Proposition 10. The single q is outside all n-dependent assumptions, and convergence is along the full natural sequence, not a subsequence.

Reproduction

What I recalculated or reran from public materials. "Agrees", "Agrees in part" or "Disagrees" compares my result with the printed one. "Not compared" means the comparison could not be made.

Status
Full reproduction of the selected headline targets
Agreement
Agrees in part
Reported results that were targeted
TargetWhereMethodStatusAgreement
Table 1p. 152 (main PDF p. 4)I reran the supplied code with changes. It reproduces the printed Table 1 cells to rounding, including the erroneous 66.8% average computed before variables were removed. For the 1,205 retained State Failure variables, the corrected average is 66.3% (P007-E010; P007-E014).ReproducedTo rounding
Figure 1 and QoG thresholdsp. 153 (main PDF p. 5), Figure 1 and results proseI reran the supplied code with changes on the QoG Basic file. All 2.5 million surviving-row counts match the saved output exactly; the plotted curves agree at displayed precision and all four reported thresholds agree. The dataset name is corrected in P007-E009.ReproducedTo rounding
Figure 2 and State Failure thresholdsp. 153 (main PDF p. 5), Figure 2 and results proseI reran the supplied code with changes. All 2.5 million surviving-row counts match the saved output exactly, and the Figure 2 curves agree at displayed precision. The prose threshold for expected loss above 99% disagrees: the first crossing is at six variables, rather than four (P007-E010). The reported probabilities of losing all rows agree.ReproducedAgrees in part
Figure 3Supplement A pp. 1-2 (supplement PDF pp. 1-2); main p. 151 (main PDF p. 3)I reran the supplied numerical code with changes and recalculated all 1,350 grid points from the formula. The plotted curves and approximate endpoints agree at displayed precision.ReproducedTo rounding
Figure 4 and integer examplesSupplement A p. 2 (supplement PDF p. 2)I reran the supplied numerical code with changes and recalculated the threshold grids. The values 33 and 952 reproduce exactly as the largest kk with lower bound at most 0.50.5; the smallest integers reaching 0.50.5 are 34 and 953. The interpretation in the text and caption is corrected in P007-E005.ReproducedExact
Figure 5Supplement A pp. 2-3 (supplement PDF pp. 2-3)I reran the supplied numerical code with changes and recalculated the curves of 1−q∗k1-q_*^k. The plotted curves and approximate endpoints agree at displayed precision; the bound is a lower bound on the expected proportion of rows removed (P007-E006).ReproducedTo rounding
Superlogarithmic growth examplep. 151 (main PDF p. 3), paragraph before Assumption 4I recalculated ⌊(log⁡n)1.1⌋\lfloor(\log n)^{1.1}\rfloor at the three stated values n=10n=10, n=1,000n=1{,}000 and n=1,000,000n=1{,}000{,}000, obtaining the printed integers 22, 88 and 1717.RecalculatedExact