{
 "claims": [
  {
   "claim": "Given only $EY_1 = \\mathrm{E}[Y \\mid Z = 1]$, $EY_0 = \\mathrm{E}[Y \\mid Z = 0]$ and $P_c$ (the difference in take-up), with outcomes in $[0,1]$, the complier average potential outcomes lie in the two closed intervals of Proposition 1, and these bounds are sharp.",
   "claim_id": "P028-H1",
   "related_corrections": [
    "P028-E001",
    "P028-E002",
    "P028-N002"
   ],
   "tied_to": "Abstract; Section 1, last paragraph; Proposition 1 and its proof, Eqs. (1)-(3)"
  },
  {
   "claim": "Under random assignment, exclusion, a nonzero first stage and monotonicity, the two assignment-arm outcome means and the take-up difference suffice to estimate the CACE, $(EY_1 - EY_0)/P_c$, but do not in general identify the two complier average potential outcomes.",
   "claim_id": "P028-H2",
   "related_corrections": [
    "P028-N002"
   ],
   "tied_to": "Section 1, first paragraph; Section 2 LATE display; Proposition 1"
  },
  {
   "claim": "If $\\mathrm{E}[Y \\mid D = d, Z = z]$ and $\\Pr[D = d \\mid Z = z]$ are known for $d,z \\in \\{0,1\\}$, the complier average potential outcomes are point identified by the formula of Corollary 1.",
   "claim_id": "P028-H3",
   "related_corrections": [
    "P028-E002",
    "P028-E003"
   ],
   "tied_to": "Section 1, last sentence; Corollary 1 and its proof; Eqs. (1)-(2); Section 5"
  },
  {
   "claim": "Published reports often omit the information needed to recover complier average potential outcomes: only 6 of the 20 published randomized get-out-the-vote canvassing studies collected for the paper report enough.",
   "claim_id": "P028-H4",
   "related_corrections": [],
   "tied_to": "Abstract ('often fail to report'); Section 1, second paragraph; footnote 1 (meta.csv)"
  },
  {
   "claim": "In the Mullainathan et al. debate experiment the published figures ($EY_0 = 0.45$, $EY_1 = 0.51$, $P_c = 0.21$) give a CACE of 29 percentage points but only wide bounds: $\\mathrm{E}[Y(0) \\mid D(1) > D(0)] \\in [0,0.71]$ and $\\mathrm{E}[Y(1) \\mid D(1) > D(0)] \\in [0.29,1]$.",
   "claim_id": "P028-H5",
   "related_corrections": [
    "P028-N005",
    "P028-N007"
   ],
   "tied_to": "Section 1, last paragraph; Section 4, first paragraph; Proposition 1 (no table or figure)"
  },
  {
   "claim": "With four further quantities taken from the replication data, Corollary 1 gives point estimates $\\mathrm{E}[Y(0) \\mid D(1) > D(0)] = 0.43$ and $\\mathrm{E}[Y(1) \\mid D(1) > D(0)] = 0.72$, a considerable tightening relative to the bounds.",
   "claim_id": "P028-H6",
   "related_corrections": [
    "P028-N005",
    "P028-N007"
   ],
   "tied_to": "Section 4, second paragraph; Corollary 1 (no table or figure)"
  },
  {
   "claim": "Researchers should report estimates of $\\mathrm{E}[Y \\mid D = d, Z = z]$ and $\\Pr[D = d \\mid Z = z]$ for $d,z \\in \\{0,1\\}$, so that complier average potential outcomes can be estimated from published results; for binary outcomes such reporting also facilitates alternative parametric models and robust variance estimators.",
   "claim_id": "P028-H7",
   "related_corrections": [],
   "tied_to": "Section 1, last sentence; Section 5; Corollary 1"
  }
 ],
 "date_published": "2026-10-25",
 "findings": [
  {
   "before_review": null,
   "class": "error",
   "date_modified": "2026-10-25",
   "date_published": "2026-10-25",
   "evidence_type": "complete_derivation",
   "exhibits": [
    {
     "alt": "Crop of the proof of Proposition 1 from equation (2) through equation (3) to the three displayed inequalities. Red outlines surround two items in the sentence after equation (3). The first is the final less-than-or-equal-to 1 in the constraint 0 is at most EY sub 1 minus P sub c EY sub c (1), which is at most 1. The second is the symbol EY sub c (1) in the words: Then substituting the implied value of EY sub c (1) from Eq. (3). The third displayed inequality below begins (EY sub 1 + P sub c minus 1) / P sub c.",
     "caption": "Proof of Proposition 1, p. 678 (PDF p. 2), from equation (2) to the three displayed inequalities. The outlines mark the upper limit $1$ of the third constraint and the symbol $EY_c(1)$ in “the implied value of $EY_c(1)$”.",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/exhibits/P028-E001.png"
    }
   ],
   "formal_release": null,
   "formal_statements": [
    {
     "id": "P028-T4",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T4"
    },
    {
     "id": "P028-T5",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T5"
    },
    {
     "id": "P028-T8",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T8"
    }
   ],
   "headline": "In the proof of Proposition 1, the third constraint after equation (3) has upper limit $1 - P_c$ and not $1$, and the mean substituted out of equation (3) is $EY_c(0)$ and not $EY_c(1)$; the three displayed inequalities and the proposition are unchanged.",
   "id": "P028-E001",
   "location": "Proof of Proposition 1, the sentence between equation (3) and the three displayed inequalities, p. 678 (PDF p. 2)",
   "pdf_page": 2,
   "printed_page": "678",
   "reaches": [],
   "related_entries": [],
   "revisions": [],
   "same_defect_as": [],
   "severity": "Minor",
   "subject": "paper",
   "target_ids": [
    "P028-T4",
    "P028-T5",
    "P028-T8"
   ],
   "text": {
    "as_printed": "Since $EY_n = [EY_1 - P_c EY_c(1)]/(1 - P_c)$, we may reframe the problem as\n\\[\nP_c EY_c(0) + EY_1 - P_c EY_c(1) = EY_0, \\qquad (3)\n\\]\nsuch that $0 \\le EY_c(0) \\le 1$, $0 \\le EY_c(1) \\le 1$, and $0 \\le EY_1 - P_c EY_c(1) \\le \\hl{1}$. Then substituting the implied value of $\\hl{EY_c(1)}$ from Eq.~(3) into the system of inequalities and rearranging terms, we obtain\n\\[\n\\begin{aligned}\n&0 \\le EY_c(1) \\le 1,\\\\\n&(EY_1 - EY_0)/P_c \\le EY_c(1) \\le 1 + (EY_1 - EY_0)/P_c,\\\\\n&(EY_1 + P_c - 1)/P_c \\le EY_c(1) \\le EY_1/P_c,\n\\end{aligned}\n\\]\nwhich yields the interval for $EY_c(1)$ asserted by the proposition. The interval for $EY_c(0)$ follows via substitution from the equality in Eq.~(3).",
    "corrected": "Since $EY_n = [EY_1 - P_c EY_c(1)]/(1 - P_c)$, we may reframe the problem as\n\\[\nP_c EY_c(0) + EY_1 - P_c EY_c(1) = EY_0, \\qquad (3)\n\\]\nsuch that $0 \\le EY_c(0) \\le 1$, $0 \\le EY_c(1) \\le 1$, and $0 \\le EY_1 - P_c EY_c(1) \\le \\hl{1 - P_c}$. Then substituting the implied value of $\\hl{EY_c(0)}$ from Eq.~(3) into the system of inequalities and rearranging terms, we obtain\n\\[\n\\begin{aligned}\n&0 \\le EY_c(1) \\le 1,\\\\\n&(EY_1 - EY_0)/P_c \\le EY_c(1) \\le 1 + (EY_1 - EY_0)/P_c,\\\\\n&(EY_1 + P_c - 1)/P_c \\le EY_c(1) \\le EY_1/P_c,\n\\end{aligned}\n\\]\nwhich yields the interval for $EY_c(1)$ asserted by the proposition. The interval for $EY_c(0)$ follows via substitution from the equality in Eq.~(3).",
    "evidence": "I give a derivation, with an exact example which shows that the printed system is not equivalent to the original one. Suppose that $0 < P_c < 1$. By (2), I have $EY_n = [EY_1 - P_c EY_c(1)]/(1 - P_c)$, and therefore\n\\[\n0 \\le EY_n \\le 1 \\iff 0 \\le EY_1 - P_c EY_c(1) \\le 1 - P_c \\iff \\frac{EY_1 + P_c - 1}{P_c} \\le EY_c(1) \\le \\frac{EY_1}{P_c}.\n\\]\nBy (3), I have $EY_c(0) = EY_c(1) - (EY_1 - EY_0)/P_c$, which implies that\n\\[\n0 \\le EY_c(0) \\le 1 \\iff \\frac{EY_1 - EY_0}{P_c} \\le EY_c(1) \\le 1 + \\frac{EY_1 - EY_0}{P_c}.\n\\]\nTogether with $0 \\le EY_c(1) \\le 1$, these are the three displayed inequalities. Since $(EY_1 + P_c - 1)/P_c = 1 - (1 - EY_1)/P_c$, their intersection is the interval for $EY_c(1)$ in Proposition 1.\n\nFor the example, let $P_c = 1/2$, $EY_0 = EY_1 = 3/4$ and $EY_c(0) = EY_c(1) = 0$. Equation (3) and the three printed constraints hold, since $EY_1 - P_c EY_c(1) = 3/4 \\le 1$. But the implied noncomplier mean is $EY_n = (3/4)/(1/2) = 3/2 > 1$. For these values of $P_c$, $EY_0$ and $EY_1$, I obtain the interval $[0, 1]$ for $EY_c(1)$ from the printed system, and the interval $[1/2, 1]$ of Proposition 1 from the corrected system.",
    "problem": "The third constraint restates $0 \\le EY_n \\le 1$ after the substitution $EY_n = [EY_1 - P_c EY_c(1)]/(1 - P_c)$. Multiplying by $1 - P_c$, I obtain the upper limit $1 - P_c$, and the sentence prints $1$. For $0 < P_c < 1$ the printed constraint is weaker than the one it replaces. It implies only the lower bound $(EY_1 - 1)/P_c$ for $EY_c(1)$, and not the lower bound $(EY_1 + P_c - 1)/P_c$ in the third line of the display. The next sentence names the wrong mean. Since the system that follows is written in $EY_c(1)$, the mean eliminated through equation (3) is $EY_c(0)$. The display itself uses the intended constraint and the intended substitution.",
    "what_survives": "Proposition 1, its sharpness, the three displayed inequalities and the LATE identity are unaffected. The bounds of the application are also unaffected. From the reported $0.45$, $0.51$ and $0.21$ the intervals are $[0, 5/7]$ and $[2/7, 1]$, printed as $[0, 0.71]$ and $[0.29, 1]$. The correction changes one limit and one symbol in the proof and has no cost. The case $P_c = 1$ is treated in \\fid{P028-E002}."
   },
   "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-E001",
   "work": {
    "arxiv_id": null,
    "doi": "10.1016/j.spl.2012.11.012",
    "title": "Sharp Bounds for Complier Average Potential Outcomes in Experiments with Noncompliance and Incomplete Reporting",
    "url": "https://pmaronow.github.io/llm-audit/works/P028/"
   },
   "work_id": "P028"
  },
  {
   "before_review": null,
   "class": "error",
   "date_modified": "2026-10-25",
   "date_published": "2026-10-25",
   "evidence_type": "exact_counterexample",
   "exhibits": [
    {
     "alt": "Crop of the first six lines of the proof of Proposition 1, with equations (1) and (2). Red outlines surround two items. The first is the definition EY sub n is identically E of Y given D(1) = D(0), in the first line of the proof. The second is the divisor, slash open parenthesis 1 minus P sub c close parenthesis, in the sentence: Since EY sub n = [EY sub 1 minus P sub c EY sub c (1)] / (1 minus P sub c), we may reframe the problem as.",
     "caption": "Proof of Proposition 1, p. 678 (PDF p. 2). The outlines mark the definition of the noncomplier mean $EY_n$ and the division by $1 - P_c$.",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/exhibits/P028-E002.png"
    }
   ],
   "formal_release": null,
   "formal_statements": [
    {
     "id": "P028-T6",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T6"
    },
    {
     "id": "P028-T7",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T7"
    },
    {
     "id": "P028-T8",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T8"
    },
    {
     "id": "P028-T9",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T9"
    }
   ],
   "headline": "Since the proof of Proposition 1 defines a mean among noncompliers and divides by $1 - P_c$, it needs a separate case for full compliance, $P_c = 1$, which the assumptions allow; the printed intervals are correct in that case.",
   "id": "P028-E002",
   "location": "Proof of Proposition 1, first sentence and the sentence after equation (2), p. 678 (PDF p. 2); assumptions (iii) and (iv), Section 2, p. 678",
   "pdf_page": 2,
   "printed_page": "678",
   "reaches": [],
   "related_entries": [],
   "revisions": [],
   "same_defect_as": [],
   "severity": "Minor",
   "subject": "paper",
   "target_ids": [
    "P028-T6",
    "P028-T7",
    "P028-T8",
    "P028-T9"
   ],
   "text": {
    "as_printed": "\\emph{Section 2, p.~678.} (iii) non-zero first-stage effect: $E[D_i(1) - D_i(0)] \\neq 0$; (iv) monotonicity: $\\forall i$, $D_i(1) \\geq D_i(0)$.\n\n\\emph{Proof of Proposition 1, p.~678.} Define $\\hl{EY_n \\equiv \\mathrm{E}[Y \\mid D(1) = D(0)]}$, $EY_c(0) \\equiv \\mathrm{E}[Y(0) \\mid D(1) > D(0)]$ and $EY_c(1) \\equiv \\mathrm{E}[Y(1) \\mid D(1) > D(0)]$. By the law of total probability,\n\\[\n\\begin{aligned}\nP_c EY_c(0) + (1 - P_c) EY_n &= EY_0, \\qquad (1)\\\\\nP_c EY_c(1) + (1 - P_c) EY_n &= EY_1, \\qquad (2)\n\\end{aligned}\n\\]\ngiven the constraints that $0 \\le EY_c(0) \\le 1$, $0 \\le EY_c(1) \\le 1$, and $0 \\le EY_n \\le 1$. Since $EY_n = [EY_1 - P_c EY_c(1)]\\hl{/(1 - P_c)}$, we may reframe the problem as",
    "corrected": "Insert at the start of the proof of Proposition 1:\n\n\\hl{If $P_c = 1$, a unit is a complier with probability one, so $EY_c(0) = EY_0$ and $EY_c(1) = EY_1$; the displayed intervals reduce to these singletons. Henceforth suppose $0 < P_c < 1$.}\n\nThe proof can also be written with no mean over the noncompliers, and then no separate case is needed. Assume (i)--(iv) of Section 2 and $Y \\in [0,1]$, and assume that both assignment arms have positive probability. Let $C$ denote the complier event $D(1) > D(0)$, and set $p = P_c = \\Pr(C) > 0$, $e_z = EY_z = \\mathrm{E}[Y \\mid Z = z]$ and $c_d = \\mathrm{E}[Y(d) \\mid C]$. Define the noncomplier contribution\n\\[\nm = \\mathrm{E}[Y \\cdot D \\mid Z = 0] + \\mathrm{E}[Y \\cdot (1 - D) \\mid Z = 1],\n\\]\nwhere $Y$ and $D$ are the observed outcome and treatment and the dot denotes multiplication. Then\n\\[\n0 \\le m \\le 1 - p, \\qquad e_z = p\\,c_z + m \\quad \\text{for } z \\in \\{0, 1\\}.\n\\]\nFor $p < 1$, $m = (1 - p) EY_n$. For $p = 1$, $m = 0$ and $c_z = e_z$, with no mean over an empty group and no division by $1 - p$.\n\nCorollary 1 is correct as printed when the four cell means exist. To state it in a form that allows empty treatment-by-assignment cells, replace its two weighted cell means by the two product moments in $m$, and use $c_0 = (e_0 - m)/p$ and $c_1 = (e_1 - m)/p$. Each product moment equals the printed probability times the cell mean whenever that cell mean exists, and it is zero when the cell is empty.",
    "evidence": "I give an exact counterexample to the step of the proof, together with a derivation that covers $P_c = 1$. For the counterexample, let $Z$ be $0$ or $1$ with probability $1/2$ each, and suppose that $D(0) = 0$, $D(1) = 1$, $Y(0) = 0$, $Y(1) = 1$ and $Y(d, z) = Y(d)$ for every unit. Assumptions (i)--(iv) hold and $Y \\in [0,1]$. I have $P_c = 1$, $EY_0 = 0$, $EY_1 = 1$, $EY_c(0) = 0$ and $EY_c(1) = 1$. However, $\\Pr[D(1) = D(0)] = 0$, and $[EY_1 - P_c EY_c(1)]/(1 - P_c) = 0/0$.\n\nFor the derivation, write $p = P_c$, $e_z = EY_z$ and $c_d = EY_c(d)$, and let $A$ and $N$ denote the events $D(0) = D(1) = 1$ (always-takers) and $D(0) = D(1) = 0$ (never-takers). By monotonicity, $\\Pr(A) + \\Pr(N) = 1 - p$. Define $m = \\mathrm{E}[\\mathbf{1}_A Y(1)] + \\mathrm{E}[\\mathbf{1}_N Y(0)]$. Since $Y \\in [0,1]$, I have $0 \\le m \\le 1 - p$. In the arm $Z = 0$ the observed outcome is $Y(0)$ for compliers and never-takers and $Y(1)$ for always-takers, and in the arm $Z = 1$ it is $Y(1)$ for compliers and always-takers and $Y(0)$ for never-takers. By random assignment and exclusion, it follows that\n\\[\ne_0 = p\\,c_0 + m, \\qquad e_1 = p\\,c_1 + m .\n\\]\nThese equations use no mean over the noncompliers. If $p < 1$, then $m = (1 - p) EY_n$ and they are (1) and (2). If $p = 1$, then $m = 0$, and I have $c_0 = e_0$ and $c_1 = e_1$. Setting $P_c = 1$ in the printed intervals, I obtain\n\\[\n\\begin{aligned}\n\\max\\{0,\\, e_0 - e_1,\\, e_0\\} &= e_0 = \\min\\{1,\\, 1 - e_1 + e_0,\\, e_0\\},\\\\\n\\max\\{0,\\, e_1 - e_0,\\, e_1\\} &= e_1 = \\min\\{1,\\, 1 + e_1 - e_0,\\, e_1\\},\n\\end{aligned}\n\\]\nbecause $0 \\le e_0 \\le 1$ and $0 \\le e_1 \\le 1$. The two intervals are the points $e_0$ and $e_1$. In the arm $Z = 0$ the product $Y \\cdot D$ equals $Y(1)$ on $A$ and $0$ elsewhere, and in the arm $Z = 1$ the product $Y \\cdot (1 - D)$ equals $Y(0)$ on $N$ and $0$ elsewhere. This implies the expression for $m$ in terms of observed variables that is used below. A check in exact arithmetic agrees. The printed intervals are single points in 81 models with full compliance, and the formulas with $m$ hold in 810 finite models, some with empty cells.",
    "problem": "Assumptions (iii) and (iv) imply that $0 < P_c \\le 1$, and no assumption of Section 2 excludes $P_c = 1$. When $P_c = 1$ the event $D(1) = D(0)$ has probability zero. The mean $EY_n$ is then not defined, and the quotient $[EY_1 - P_c EY_c(1)]/(1 - P_c)$ is $0/0$. The proof treats no separate case and does not cover full compliance. The omission is in the proof and not in the result. At $P_c = 1$ the two printed intervals reduce to the points $EY_0$ and $EY_1$, which are the complier average potential outcomes in that case. The ratio in the proof of Corollary 1 has the same denominator, $\\Pr[D = 1 \\mid Z = 0] + \\Pr[D = 0 \\mid Z = 1] = 1 - P_c$. Corollary 1 assumes that the four cell means $\\mathrm{E}[Y \\mid D = d, Z = z]$ are known. This assumption requires that $P_c < 1$, and the corollary is correct as printed.",
    "what_survives": "The intervals of Proposition 1 are correct as printed, for $P_c = 1$ as well as for $P_c < 1$, and they remain sharp. The LATE identity divides by $P_c > 0$ and is unaffected, and Corollary 1 holds under its premise that the four cell means are known. The application has two-sided noncompliance, with $P_c = 0.21$, and no printed number changes. The repair adds a case to the proof and no assumption."
   },
   "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-E002",
   "work": {
    "arxiv_id": null,
    "doi": "10.1016/j.spl.2012.11.012",
    "title": "Sharp Bounds for Complier Average Potential Outcomes in Experiments with Noncompliance and Incomplete Reporting",
    "url": "https://pmaronow.github.io/llm-audit/works/P028/"
   },
   "work_id": "P028"
  },
  {
   "before_review": null,
   "class": "error",
   "date_modified": "2026-10-25",
   "date_published": "2026-10-25",
   "evidence_type": "page_check",
   "exhibits": [
    {
     "alt": "Crop of the four entries of the reference list. A red outline surrounds the numeral 4 in the third entry, which ends: Review of Economic Studies 4, 555 to 574. The first entry ends: Journal of the American Statistical Association 91, 444 to 455.",
     "caption": "References, p. 679 (PDF p. 3). The outline marks the numeral 4 in the Imbens and Rubin (1997) entry, in the position where the first entry gives the volume 91.",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/exhibits/P028-E003.png"
    }
   ],
   "formal_release": null,
   "formal_statements": [],
   "headline": "The Imbens and Rubin (1997) entry prints the issue number, 4, where the volume belongs; the article is in Review of Economic Studies, volume 64, issue 4, pp. 555–574.",
   "id": "P028-E003",
   "location": "References, third entry (Imbens and Rubin, 1997), p. 679 (PDF p. 3)",
   "pdf_page": 3,
   "printed_page": "679",
   "reaches": [],
   "related_entries": [],
   "revisions": [],
   "same_defect_as": [],
   "severity": "Minor",
   "subject": "paper",
   "target_ids": [],
   "text": {
    "as_printed": "Imbens, G.W., Rubin, D.B., 1997. Estimating outcome distributions for compliers in instrumental variables models. Review of Economic Studies \\hl{4}, 555--574.",
    "corrected": "Imbens, G.W., Rubin, D.B., 1997. Estimating outcome distributions for compliers in instrumental variables models. Review of Economic Studies \\hl{64}, 555--574.",
    "evidence": "I checked the journal's table of contents. Oxford Academic lists the article under ``The Review of Economic Studies, Volume 64, Issue 4, October 1997, Pages 555--574''.",
    "problem": "The numeral 4 stands where the other journal entry of the list has its volume (``Journal of the American Statistical Association 91, 444--455''). The article is in volume 64, and 4 is the number of the issue. The entry omits the volume.",
    "what_survives": "The cited work is identified by its authors, title, year and pages, and no result of the paper depends on the entry."
   },
   "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-E003",
   "work": {
    "arxiv_id": null,
    "doi": "10.1016/j.spl.2012.11.012",
    "title": "Sharp Bounds for Complier Average Potential Outcomes in Experiments with Noncompliance and Incomplete Reporting",
    "url": "https://pmaronow.github.io/llm-audit/works/P028/"
   },
   "work_id": "P028"
  },
  {
   "before_review": null,
   "class": "clarity_note",
   "date_modified": "2026-10-25",
   "date_published": "2026-10-25",
   "evidence_type": "exact_counterexample",
   "exhibits": [
    {
     "alt": "Crop of seven lines of Section 2, from the sentence that defines the treatment assigned to unit i through the four assumptions (i) to (iv) to the first two lines of the next paragraph. A red outline surrounds the words: given (i), labels are exchangeable. The sentence reads: First, given (i), labels are exchangeable and so we drop notational dependence on i.",
     "caption": "Section 2, first paragraph, p. 678 (PDF p. 2). The outline marks the words “given (i), labels are exchangeable”, the reason given for dropping the unit index $i$.",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/exhibits/P028-N002.png"
    }
   ],
   "formal_release": null,
   "formal_statements": [],
   "headline": "The unit index is dropped as a change to generic-unit population notation, in which $D$ and $Y$ are the realized treatment and outcome; random assignment does not make the unit labels exchangeable, and no further assumption is needed.",
   "id": "P028-N002",
   "location": "Section 2, first paragraph, p. 678 (PDF p. 2)",
   "pdf_page": 2,
   "printed_page": "678",
   "reaches": [],
   "related_entries": [],
   "revisions": [],
   "same_defect_as": [],
   "severity": null,
   "subject": "paper",
   "target_ids": [],
   "text": {
    "as_printed": "We assume (i) random assignment: $[Y_i(d, z), D_i(z) : d, z \\in \\{0, 1\\}] \\perp\\!\\!\\!\\perp Z_i$; (ii) exclusion restriction: $Y_i(d, z)$ does not depend on $z$, so we may write $Y_i(d, z) = Y_i(d)$; (iii) non-zero first-stage effect: $E[D_i(1) - D_i(0)] \\neq 0$; (iv) monotonicity: $\\forall i$, $D_i(1) \\geq D_i(0)$. We make two notational simplifications. First, \\hl{given (i), labels are exchangeable and so} we drop notational dependence on $i$. Second, we define $P_c \\equiv \\Pr[D(1) > D(0)]$, $EY_1 \\equiv \\mathrm{E}[Y \\mid Z = 1]$, and $EY_0 \\equiv \\mathrm{E}[Y \\mid Z = 0]$.",
    "corrected": "We make two notational simplifications. First, we drop notational dependence on $i$ \\hl{and use generic-unit population notation, writing $D = D(Z)$ and $Y = Y(D)$ for the realized treatment and outcome}. Second, we define $P_c \\equiv \\Pr[D(1) > D(0)]$, $EY_1 \\equiv \\mathrm{E}[Y \\mid Z = 1]$, and $EY_0 \\equiv \\mathrm{E}[Y \\mid Z = 0]$.",
    "intended_meaning": "The sentence changes notation. From this point the paper writes $Y(d)$, $D(z)$ and $Z$ for a generic unit of the population, and $D = D(Z)$ and $Y = Y(D)$ denote the realized treatment and the realized outcome. The paper does not display these two identities, but their usual meaning is clear from the definitions before the sentence, from the discussion of population quantities after it, and from the application. On this reading the LATE identity, Proposition 1 and Corollary 1 hold as stated.\n\nRead literally, the reason given for dropping the index does not hold. Assumption (i) says that the assignment of each unit is independent of the potential outcomes and potential treatments of that unit, and this does not imply that the unit labels are exchangeable. For example, consider two units such that $D_i(0) = 0$ and $D_i(1) = 1$, with $Y_1(0) = Y_1(1) = 0$ and $Y_2(0) = Y_2(1) = 1$, each assigned by an independent fair coin. Since the potential outcomes of each unit are constant, (i) holds. However, the outcome of the first unit is always $0$, and the outcome of the second is always $1$.\n\nThe same example shows that the index can be dropped without assuming that the units are identically distributed. Let the generic unit be one of the two units, chosen with probability $1/2$ each and independently of the assignments. By a direct computation, I find that $\\mathrm{E}[Y \\mid Z = 0] = \\mathrm{E}[Y \\mid Z = 1] = 1/2$. The population quantities of Section 2 are defined in this example, although the two units have different outcome distributions."
   },
   "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-N002",
   "work": {
    "arxiv_id": null,
    "doi": "10.1016/j.spl.2012.11.012",
    "title": "Sharp Bounds for Complier Average Potential Outcomes in Experiments with Noncompliance and Incomplete Reporting",
    "url": "https://pmaronow.github.io/llm-audit/works/P028/"
   },
   "work_id": "P028"
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  {
   "before_review": null,
   "class": "error",
   "date_modified": "2026-10-25",
   "date_published": "2026-10-25",
   "evidence_type": "exact_counterexample",
   "exhibits": [
    {
     "alt": "Crop of the end of the first paragraph of Section 4, which gives the bounds [0, 0.71] and [0.29, 1], and of the whole second paragraph. A red outline surrounds the words: we are able to fill in the necessary missing information. The paragraph continues with the four quantities 0.163, 0.634, 0.506 and 0.442 and with the sentence: Application of Corollary 1 yields E[Y(0) | D(1) > D(0)] = 0.43 and E[Y(1) | D(1) > D(0)] = 0.72.",
     "caption": "Section 4, second paragraph, p. 679 (PDF p. 3), with the last three lines of the first paragraph. The outline marks the words “we are able to fill in the necessary missing information”.",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/exhibits/P028-N005.png"
    },
    {
     "alt": "Chart with a horizontal axis from 0 to 1, labelled complier average potential outcome, the proportion whose view of at least one candidate changed. Title: 71 missing outcomes leave the complier means in a range, not at a point. There are two rows. In the row for the control complier mean, a grey line runs from 0 to 0.71, the bounds printed in Section 4, and below it a blue bar runs from 0.14 to 0.75, the range over all completions of the 71 missing outcomes. An orange diamond on the bar marks the printed estimate 0.43, and black ticks just to its right mark the estimates under three conventional treatments, 0.436 to 0.439. In the row for the treated complier mean, the grey line runs from 0.29 to 1 and the blue bar from 0.66 to 0.75, with the diamond at the printed estimate 0.72 and the ticks at 0.712 to 0.721.",
     "caption": "The two complier means in the example of Section 4 when the 71 missing outcomes are unrestricted. For each mean the blue bar is the range over all completions of the missing binary outcomes, with viewing as coded in the replication data; the diamond is the point estimate printed in the paper; the black ticks are the estimates under three conventional treatments of the missing outcomes; and the grey line is the interval printed in the first paragraph of Section 4, shown for comparison. The ranges are not confidence intervals.",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/exhibits/P028-N005-missing-outcomes.png"
    }
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   "formal_release": null,
   "formal_statements": [
    {
     "id": "P028-T1",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T1"
    },
    {
     "id": "P028-T2",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T2"
    },
    {
     "id": "P028-T3",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T3"
    }
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   "headline": "The point estimates $0.43$ and $0.72$ require an assumption about the 71 respondents whose outcome is missing, and the paper states none; under the assumptions of Section 2 alone, the replication data do not point identify the complier average potential outcomes.",
   "id": "P028-N005",
   "location": "Section 4, second paragraph, p. 679 (PDF p. 3)",
   "pdf_page": 3,
   "printed_page": "679",
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   "severity": "Minor",
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   "text": {
    "as_printed": "Point identification is possible when additional information about the joint distribution of $Y, D, Z$ is known. Since the replication data set is publicly available for the study$^{2}$, \\hl{we are able to fill in the necessary missing information}: $\\Pr[D = 1 \\mid Z = 0] = 0.163$, $\\Pr[D = 0 \\mid Z = 1] = 0.634$, $\\mathrm{E}[Y \\mid D = 1, Z = 0] = 0.506$ and $\\mathrm{E}[Y \\mid D = 0, Z = 1] = 0.442$. Application of Corollary 1 yields $\\mathrm{E}[Y(0) \\mid D(1) > D(0)] = 0.43$ and $\\mathrm{E}[Y(1) \\mid D(1) > D(0)] = 0.72$.",
    "corrected": "Point identification is possible when additional information about the joint distribution of $Y, D, Z$ is known. Since the replication data set is publicly available for the study$^{2}$, we are able to \\hl{estimate the necessary missing quantities}: $\\Pr[D = 1 \\mid Z = 0] = 0.163$, $\\Pr[D = 0 \\mid Z = 1] = 0.634$, $\\mathrm{E}[Y \\mid D = 1, Z = 0] = 0.506$ and $\\mathrm{E}[Y \\mid D = 0, Z = 1] = 0.442$. \\hl{The outcome is missing for 71 of the 1000 respondents, 37 of the 495 assigned to control and 34 of the 505 assigned to encouragement. Assume in addition that, within each assignment group, whether the outcome is recorded is independent of $(Y, D)$, so that means among the respondents whose outcome is recorded estimate the corresponding population means. Under this assumption, the reported values can be interpreted as point estimates for the study population:} $\\mathrm{E}[Y(0) \\mid D(1) > D(0)] = 0.43$ and $\\mathrm{E}[Y(1) \\mid D(1) > D(0)] = 0.72$. \\hl{Without an assumption about the missing outcomes, these data do not point identify the complier average potential outcomes.}\n\nThe added assumption is sufficient for a calculation that takes the outcome means from the respondents whose outcome is recorded and the take-up rates from all respondents. It does not follow from random assignment and does not specify the exact calculation of the printed values; I have not established it for this study. A complete statement would also give the coding of the outcome and of viewing and the respondents who enter each quantity; the paper's own analysis script and original denominators are unavailable. If no assumption about nonresponse is made, the two point estimates should be replaced by the ranges over the missing outcomes: with viewing as coded in the replication data, $0.14$ to $0.75$ for the control complier mean and $0.66$ to $0.75$ for the treated complier mean.",
    "evidence": "The evidence is an exact counterexample. There exist two models that satisfy every assumption of Section 2, in which the observed data have the same distribution and the complier means differ. The observed data are the assignment $Z$, the viewing variable $D$ as coded in the replication data, whether the outcome is recorded, and the outcome when it is recorded. The table lists the counts.\n\\begin{audittable}{llrrrrr}{Respondents in the replication data, by assignment $Z$ and viewing $D$. The outcome $Y$ is coded as in the source study.}\n$Z$ & $D$ & Respondents & Outcome recorded & $Y = 1$ & $Y = 0$ & Outcome missing \\\\\n0 & 0 & 415 & 379 & 167 & 212 & 36 \\\\\n0 & 1 & 80 & 79 & 40 & 39 & 1 \\\\\n1 & 0 & 320 & 294 & 130 & 164 & 26 \\\\\n1 & 1 & 185 & 177 & 110 & 67 & 8 \\\\\n\\end{audittable}\nLet $x_{zd}$ denote the number of missing outcomes in the cell $(Z, D) = (z, d)$ that equal $1$, so that $0 \\le x_{00} \\le 36$, $0 \\le x_{01} \\le 1$, $0 \\le x_{10} \\le 26$ and $0 \\le x_{11} \\le 8$. For the counterexample I treat the proportions within each arm, completed in this way, as a population distribution. Under monotonicity the cell $(Z, D) = (0, 1)$ consists of always-takers and the cell $(1, 0)$ of never-takers. Their probabilities are $p_a = 80/495$ and $p_n = 320/505$, and compliers have probability $p_c = 1 - p_a - p_n = 2047/9999$. Let $\\mu_d = \\mathrm{E}[Y(d) \\mid D(1) > D(0)]$. By random assignment and exclusion, I have\n\\[\n\\frac{167 + x_{00}}{495} = \\frac{130 + x_{10}}{505} + p_c\\,\\mu_0, \\qquad \\frac{110 + x_{11}}{505} = \\frac{40 + x_{01}}{495} + p_c\\,\\mu_1 .\n\\]\nThese are the equations of Corollary 1, written with joint probabilities. Solving them, I obtain\n\\[\n\\mu_0 = \\frac{(167 + x_{00})/495 - (130 + x_{10})/505}{2047/9999}, \\qquad \\mu_1 = \\frac{(110 + x_{11})/505 - (40 + x_{01})/495}{2047/9999}.\n\\]\nChoosing $(x_{00}, x_{01}, x_{10}, x_{11}) = (0, 1, 26, 0)$, I obtain\n\\[\n(\\mu_0, \\mu_1) = (1423/10235,\\ 6749/10235) \\approx (0.139,\\ 0.659),\n\\]\nand the choice $(36, 0, 0, 8)$ gives\n\\[\n(\\mu_0, \\mu_1) = (7633/10235,\\ 7642/10235) \\approx (0.746,\\ 0.747).\n\\]\nSince $\\mu_0$ increases in $x_{00}$ and decreases in $x_{10}$, and $\\mu_1$ increases in $x_{11}$ and decreases in $x_{01}$, these are the extreme values over all completions of the missing outcomes.\n\nFor each of the two choices there exists an explicit population model. It has always-takers, never-takers and compliers with probabilities $p_a$, $p_n$ and $p_c$, binary potential outcomes, and, for each assignment, an indicator that the outcome is recorded whose probability depends on the type of the unit and on its outcome. In each model the assignment is independent of all potential variables, exclusion and monotonicity hold, the first stage is $2047/9999$, and the observed data have the distribution in the table. A check in exact arithmetic verifies these properties for the two choices and for the other 14 choices such that each $x_{zd}$ is $0$ or its maximum.\n\nThe ranges are extrema over the missing binary outcomes, with viewing held as coded. They are not confidence intervals, and they do not show that $0.43$ or $0.72$ is biased. Under three conventional treatments of the missing outcomes, the estimates are close to the printed ones. With outcome means among respondents whose outcome is recorded and take-up rates among all respondents, I obtain $(0.439, 0.721)$ from Corollary 1. With every quantity computed among the 929 respondents whose outcome is recorded, I find $(0.436, 0.719)$. With the cell means among recorded outcomes weighted by the cell probabilities among all respondents, the estimates are $(0.436, 0.712)$.",
    "problem": "The source study codes the outcome as $1$ if the respondent's opinion of either mayoral candidate changed, as $0$ if both opinions were unchanged, and as missing otherwise. With this coding the outcome is missing for 71 of the 1000 respondents in the replication data: 37 of the 495 assigned to control and 34 of the 505 assigned to encouragement, at different rates in the four cells defined by assignment and viewing. The paper does not mention these respondents. It does not say which respondents enter each quantity, and it states no assumption about nonresponse. Section 2 assumes random assignment, exclusion, a non-zero first stage, monotonicity and bounded outcomes, and none of these assumptions concerns which outcomes are recorded. Corollary 1 identifies the two complier average potential outcomes, which I call the complier means, from the joint distribution of $(Y, D, Z)$ in the study population. The replication data determine the distribution of $Y$ only among respondents whose outcome is recorded, and a mean among those respondents equals the corresponding population mean only under a further assumption. The words ``we are able to fill in the necessary missing information'' claim more than the data supply. The values $0.43$ and $0.72$ are presented as point estimates under the assumptions of Section 2, and they also depend on an assumption about nonresponse that the paper does not state and that random assignment does not imply.\n\nThe correction does not say that either number is wrong. Under conventional treatments of the missing outcomes the estimates are about $0.44$ and $0.72$, and no bias is shown. The repair has a cost. To keep $0.43$ and $0.72$ as point estimates for the study population, one must add an assumption about nonresponse that the application may not satisfy. Without such an assumption, the point estimates must be replaced by a range over the missing outcomes, which is a weaker conclusion.",
    "what_survives": "Corollary 1 is unaffected, since it identifies the complier means when the population moments in its statement are known. Proposition 1 and the reporting recommendation of Section 5 do not depend on the example, and no printed number is shown to be wrong. What changes is the status of $0.43$ and $0.72$, and of the statement that the replication data tighten inference to point estimates. Both hold only under an added assumption about nonresponse. The CACE of 29 percentage points and the bounds of the first paragraph use outcome means among respondents whose outcome is recorded. They need the same care if they are read as statements about the whole study population."
   },
   "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-N005",
   "work": {
    "arxiv_id": null,
    "doi": "10.1016/j.spl.2012.11.012",
    "title": "Sharp Bounds for Complier Average Potential Outcomes in Experiments with Noncompliance and Incomplete Reporting",
    "url": "https://pmaronow.github.io/llm-audit/works/P028/"
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   "work_id": "P028"
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   "date_modified": "2026-10-25",
   "date_published": "2026-10-25",
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   "exhibits": [
    {
     "alt": "Crop of Section 4 from its second line to the end of its second paragraph. Red outlines surround five items. In the first paragraph: P sub c = 0.21, in the sentence The authors report estimates of EY sub 0 = 0.45, EY sub 1 = 0.51 and P sub c = 0.21. In the second paragraph: the values 0.163 and 0.634 of the two take-up rates, and the values 0.43 and 0.72 in the sentence Application of Corollary 1 yields E[Y(0) | D(1) > D(0)] = 0.43 and E[Y(1) | D(1) > D(0)] = 0.72.",
     "caption": "Section 4, p. 679 (PDF p. 3). The outlines mark the first stage $P_c = 0.21$ in the first paragraph, and the take-up rates $0.163$ and $0.634$ and the point estimates $0.43$ and $0.72$ in the second.",
     "url": "https://pmaronow.github.io/llm-audit/works/P028/exhibits/P028-N007.png"
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   "formal_statements": [],
   "headline": "The seven figures printed in Section 4 are not one consistently rounded set, and the paper does not state the calculation behind $0.43$ and $0.72$; both values are reproduced when the reported outcome means are combined with unrounded take-up rates and cell means from the replication data.",
   "id": "P028-N007",
   "location": "Section 4, first and second paragraphs, p. 679 (PDF p. 3)",
   "pdf_page": 3,
   "printed_page": "679",
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   "text": {
    "as_printed": "The authors report estimates of $EY_0 = 0.45$, $EY_1 = 0.51$ and $\\hl{P_c = 0.21}$. While the CACE can be estimated as 29 percentage points, the authors do not provide sufficient information for point estimation of complier average potential outcomes. [\\ldots] Since the replication data set is publicly available for the study$^{2}$, we are able to fill in the necessary missing information: $\\Pr[D = 1 \\mid Z = 0] = \\hl{0.163}$, $\\Pr[D = 0 \\mid Z = 1] = \\hl{0.634}$, $\\mathrm{E}[Y \\mid D = 1, Z = 0] = 0.506$ and $\\mathrm{E}[Y \\mid D = 0, Z = 1] = 0.442$. Application of Corollary 1 yields $\\mathrm{E}[Y(0) \\mid D(1) > D(0)] = \\hl{0.43}$ and $\\mathrm{E}[Y(1) \\mid D(1) > D(0)] = \\hl{0.72}$. Our inferences are considerably tightened by knowledge of additional features of the distribution that were not reported.",
    "corrected": "Application of Corollary 1 yields $\\mathrm{E}[Y(0) \\mid D(1) > D(0)] = 0.43$ and $\\mathrm{E}[Y(1) \\mid D(1) > D(0)] = 0.72$. \\hl{These point estimates can be reproduced by retaining the reported outcome means $0.45$ and $0.51$ and using the unrounded take-up rates and cell outcome means from the replication data. With uncertain viewing responses coded as missing, this calculation uses the first stage $P_c = 1 - 80/492 - 320/505 \\approx 0.204$, rather than the $0.21$ reported by the source study.}\n\nThe added sentences describe one calculation that reproduces the printed values. If the original calculation was different, they should describe that calculation.",
    "intended_meaning": "The first paragraph of Section 4 uses the three figures reported by the source study, and the second paragraph reports point estimates that were computed after consulting the replication data. All of these figures are rounded estimates, and on this reading the two paragraphs need not use the same first stage. The paper does not say which first stage and which precision enter the point estimates, and the printed figures alone do not reproduce them.\n\nUnder monotonicity, $P_c = 1 - \\Pr[D = 1 \\mid Z = 0] - \\Pr[D = 0 \\mid Z = 1]$. Rates that round to $0.163$ and $0.634$ imply that $0.202 < P_c \\le 0.204$, while a first stage that rounds to $0.21$ satisfies $0.205 \\le P_c < 0.215$. Hence the seven printed figures are not one consistently rounded set. Substituting them into Corollary 1, I obtain\n\\[\n\\frac{0.45 - 0.163 \\cdot 0.506 - 0.634 \\cdot 0.442}{0.21} = \\frac{0.087294}{0.21} \\approx 0.4157, \\qquad 0.4157 + \\frac{0.51 - 0.45}{0.21} \\approx 0.7014,\n\\]\nwhich round to $0.42$ and $0.70$. With $P_c = 0.203$, the value implied by the two printed rates, the same substitution gives $0.4300$ and $0.7256$, which round to $0.43$ and $0.73$.\n\nOne calculation returns both printed values. I keep the reported outcome means $0.45$ and $0.51$, and I take from the replication data the unrounded take-up rates $80/492$ and $320/505$, the unrounded cell means $40/79$ and $130/294$, and the first stage $P_c = 1 - 80/492 - 320/505 = 2531/12423 \\approx 0.2037$. The rate $80/492$ leaves out three control respondents whose answer to the viewing question was ``not sure'' or a refusal. Using Corollary 1, I find\n\\[\n\\begin{aligned}\n\\frac{0.45 - \\frac{80}{492} \\cdot \\frac{40}{79} - \\frac{320}{505} \\cdot \\frac{130}{294}}{2531/12423} &\\approx 0.4294,\\\\\n\\frac{0.51 - \\frac{80}{492} \\cdot \\frac{40}{79} - \\frac{320}{505} \\cdot \\frac{130}{294}}{2531/12423} &\\approx 0.7239,\n\\end{aligned}\n\\]\nwhich round to $0.43$ and $0.72$. No analysis script for the application is available. This calculation is a reconstruction and may differ from the one originally made.\n\nWith every input computed from the replication data at full precision, the estimates round to $0.44$ and $0.72$ under each of three conventions for the sample and for the coding of viewing: $(0.4393, 0.7207)$, $(0.4390, 0.7217)$ and $(0.4359, 0.7192)$. No printed estimate is shown to be wrong. The printed estimates lie inside the printed bounds $[0, 0.71]$ and $[0.29, 1]$, and this note does not affect the statement that the additional information tightens the inferences. The assumption on which the point estimates rest is the subject of \\fid{P028-N005}."
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    "doi": "10.1016/j.spl.2012.11.012",
    "title": "Sharp Bounds for Complier Average Potential Outcomes in Experiments with Noncompliance and Incomplete Reporting",
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     "body": "<p>Section 4, p. 679, repaired using Corollary 1, p. 678. Joint moments identify complier means without positive treatment cells. The stated mean equalities justify mixed respondent formulas; conditional response independence suffices. Cellwise mean ignorability justifies reweighting but need not justify the mixed formulas. With unrestricted response the stated bounds depend only on the recorded-data law. All source-table values and approximations hold; both endpoints of each source interval are attained, so no valid interval can be narrower. The usual measure-theoretic conditions are assumed.</p>",
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     "location": "p. 679 (PDF p. 3), Section 4, second paragraph; Corollary 1 is on p. 678 (PDF p. 2)"
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     "body": "Section 4, p. 679, repaired using Corollary 1, p. 678. Joint moments identify complier means without positive treatment cells. The stated mean equalities justify mixed respondent formulas; conditional response independence suffices. Cellwise mean ignorability justifies reweighting but need not justify the mixed formulas. With unrestricted response the stated bounds depend only on the recorded-data law. All source-table values and approximations hold; both endpoints of each source interval are attained, so no valid interval can be narrower. The usual measure-theoretic conditions are assumed.",
     "headline": "Section 4, second paragraph, the inference from the replication data to the complier means 0.43 and 0.72, as repaired by Sol: identification of the complier average potential outcomes from incompletely recorded outcomes under a stated response condition",
     "location": "p. 679 (PDF p. 3), Section 4, second paragraph; Corollary 1 is on p. 678 (PDF p. 2)"
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     "body": "<p>No current theorem is supplied for this target.</p>",
     "headline": "Section 4 (Application), second paragraph, as printed, with the sentence of Section 2 on estimation by sample analogues",
     "location": "p. 679 (PDF p. 3), Section 4, second paragraph, with footnote 2; p. 678 (PDF p. 2), Section 2, second paragraph"
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     "body": "No current theorem is supplied for this target.",
     "headline": "Section 4 (Application), second paragraph, as printed, with the sentence of Section 2 on estimation by sample analogues",
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     "body": "<p>proof of Proposition 1, p. 678 (PDF p. 2), corrected intermediate system. For 0 &lt; p &lt; 1 the corrected system is equivalent to the mixture equations and the printed intervals, with the stated substitutions, endpoint translations, unique noncomplier mean, and strict weakening caused by the printed residual constraint.</p>",
     "headline": "Proof of Proposition 1, the sentence between Eq. (3) and the three-line display, corrected: upper limit <span class=\"katex\"><span class=\"katex-mathml\"><math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mn>1</mn><mo>−</mo><msub><mi>P</mi><mi>c</mi></msub></mrow><annotation encoding=\"application/x-tex\">1 - P_c</annotation></semantics></math></span><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.7278em;vertical-align:-0.0833em;\"></span><span class=\"mord\">1</span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span><span class=\"mbin\">−</span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.8333em;vertical-align:-0.15em;\"></span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.1389em;\">P</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.1389em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span></span></span></span> (printed 1) and ‘the implied value of <span class=\"katex\"><span class=\"katex-mathml\"><math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mi>E</mi><msub><mi>Y</mi><mi>c</mi></msub><mo stretchy=\"false\">(</mo><mn>0</mn><mo stretchy=\"false\">)</mo></mrow><annotation encoding=\"application/x-tex\">EY_c(0)</annotation></semantics></math></span><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mopen\">(</span><span class=\"mord\">0</span><span class=\"mclose\">)</span></span></span></span>’ (printed <span class=\"katex\"><span class=\"katex-mathml\"><math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mi>E</mi><msub><mi>Y</mi><mi>c</mi></msub><mo stretchy=\"false\">(</mo><mn>1</mn><mo stretchy=\"false\">)</mo></mrow><annotation encoding=\"application/x-tex\">EY_c(1)</annotation></semantics></math></span><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mopen\">(</span><span class=\"mord\">1</span><span class=\"mclose\">)</span></span></span></span>)",
     "location": "p. 678 (PDF p. 2), proof of Proposition 1"
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    "text": {
     "body": "proof of Proposition 1, p. 678 (PDF p. 2), corrected intermediate system. For 0 < p < 1 the corrected system is equivalent to the mixture equations and the printed intervals, with the stated substitutions, endpoint translations, unique noncomplier mean, and strict weakening caused by the printed residual constraint.",
     "headline": "Proof of Proposition 1, the sentence between Eq. (3) and the three-line display, corrected: upper limit $1 - P_c$ (printed 1) and ‘the implied value of $EY_c(0)$’ (printed $EY_c(1)$)",
     "location": "p. 678 (PDF p. 2), proof of Proposition 1"
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     "body": "<p>proof of Proposition 1 after Eq. (3), p. 678. The exact candidate (1/2,3/4,3/4,0,0) satisfies the printed system and fails the constrained mixture equations: its forced noncomplier mean is 3/2. The two printed feasible coordinate sets are [0,1], while the corrected mixture sets and Proposition 1 intervals are [1/2,1]. The input triple has the specified realization with half compliers of outcome one, half never-takers of outcome one half, and independent fair assignment. The usual measure-theoretic conditions are assumed.</p>",
     "headline": "Proof of Proposition 1, the system printed after Eq. (3) (‘such that ... and <span class=\"katex\"><span class=\"katex-mathml\"><math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mn>0</mn><mo>≤</mo><mi>E</mi><msub><mi>Y</mi><mn>1</mn></msub><mo>−</mo><msub><mi>P</mi><mi>c</mi></msub><mi>E</mi><msub><mi>Y</mi><mi>c</mi></msub><mo stretchy=\"false\">(</mo><mn>1</mn><mo stretchy=\"false\">)</mo><mo>≤</mo><mn>1</mn></mrow><annotation encoding=\"application/x-tex\">0 \\le EY_1 - P_c EY_c(1) \\le 1</annotation></semantics></math></span><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.7804em;vertical-align:-0.136em;\"></span><span class=\"mord\">0</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span><span class=\"mrel\">≤</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.8333em;vertical-align:-0.15em;\"></span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.3011em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span><span class=\"mbin\">−</span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.1389em;\">P</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.1389em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mopen\">(</span><span class=\"mord\">1</span><span class=\"mclose\">)</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span><span class=\"mrel\">≤</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.6444em;\"></span><span class=\"mord\">1</span></span></span></span>’), offered as a reframing of Eqs. (1)–(2) with their constraints; the witness is the validator&#x27;s and is not in the paper",
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     "body": "proof of Proposition 1 after Eq. (3), p. 678. The exact candidate (1/2,3/4,3/4,0,0) satisfies the printed system and fails the constrained mixture equations: its forced noncomplier mean is 3/2. The two printed feasible coordinate sets are [0,1], while the corrected mixture sets and Proposition 1 intervals are [1/2,1]. The input triple has the specified realization with half compliers of outcome one, half never-takers of outcome one half, and independent fair assignment. The usual measure-theoretic conditions are assumed.",
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     "body": "<p>Proposition 1 proof, p. 678, and Corollary 1 proof, p. 679. A bounded causal model has positive complier probability at most one, and full compliance is allowed. The noncomplier contribution gives the mixture identities, the partial- and full-compliance branches, cell-compatible observable formulas, both bounds, and singleton boundary intervals. Positive pure cells give the printed probability-times-mean expression; empty cells have zero product moment. The usual measure-theoretic conditions are assumed.</p>",
     "headline": "Proof of Proposition 1 with the full-compliance case <span class=\"katex\"><span class=\"katex-mathml\"><math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><msub><mi>P</mi><mi>c</mi></msub><mo>=</mo><mn>1</mn></mrow><annotation encoding=\"application/x-tex\">P_c = 1</annotation></semantics></math></span><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.8333em;vertical-align:-0.15em;\"></span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.1389em;\">P</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.1389em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span><span class=\"mrel\">=</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.6444em;\"></span><span class=\"mord\">1</span></span></span></span> supplied, and the substitution in the proof of Corollary 1 written without a non-complier mean, as repaired by Sol",
     "location": "p. 678 (PDF p. 2), proof of Proposition 1 (‘Define <span class=\"katex\"><span class=\"katex-mathml\"><math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mi>E</mi><msub><mi>Y</mi><mi>n</mi></msub><mo>≡</mo><mi mathvariant=\"normal\">E</mi><mo stretchy=\"false\">[</mo><mi>Y</mi><mi mathvariant=\"normal\">∣</mi><mi>D</mi><mo stretchy=\"false\">(</mo><mn>1</mn><mo stretchy=\"false\">)</mo><mo>=</mo><mi>D</mi><mo stretchy=\"false\">(</mo><mn>0</mn><mo stretchy=\"false\">)</mo><mo stretchy=\"false\">]</mo></mrow><annotation encoding=\"application/x-tex\">EY_n \\equiv \\mathrm{E}[Y|D(1) = D(0)]</annotation></semantics></math></span><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.8333em;vertical-align:-0.15em;\"></span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">n</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span><span class=\"mrel\">≡</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord mathrm\">E</span><span class=\"mopen\">[</span><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"mord\">∣</span><span class=\"mord mathnormal\" style=\"margin-right:0.0278em;\">D</span><span class=\"mopen\">(</span><span class=\"mord\">1</span><span class=\"mclose\">)</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span><span class=\"mrel\">=</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord mathnormal\" style=\"margin-right:0.0278em;\">D</span><span class=\"mopen\">(</span><span class=\"mord\">0</span><span class=\"mclose\">)]</span></span></span></span>’ and ‘Since <span class=\"katex\"><span class=\"katex-mathml\"><math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mi>E</mi><msub><mi>Y</mi><mi>n</mi></msub><mo>=</mo><mo stretchy=\"false\">[</mo><mi>E</mi><msub><mi>Y</mi><mn>1</mn></msub><mo>−</mo><msub><mi>P</mi><mi>c</mi></msub><mi>E</mi><msub><mi>Y</mi><mi>c</mi></msub><mo stretchy=\"false\">(</mo><mn>1</mn><mo stretchy=\"false\">)</mo><mo stretchy=\"false\">]</mo><mi mathvariant=\"normal\">/</mi><mo stretchy=\"false\">(</mo><mn>1</mn><mo>−</mo><msub><mi>P</mi><mi>c</mi></msub><mo stretchy=\"false\">)</mo></mrow><annotation encoding=\"application/x-tex\">EY_n = [EY_1 - P_c EY_c(1)]/(1 - P_c)</annotation></semantics></math></span><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.8333em;vertical-align:-0.15em;\"></span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">n</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span><span class=\"mrel\">=</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mopen\">[</span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.3011em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span><span class=\"mbin\">−</span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.1389em;\">P</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.1389em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mopen\">(</span><span class=\"mord\">1</span><span class=\"mclose\">)]</span><span class=\"mord\">/</span><span class=\"mopen\">(</span><span class=\"mord\">1</span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span><span class=\"mbin\">−</span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.1389em;\">P</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.1389em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mclose\">)</span></span></span></span>’); p. 679 (PDF p. 3), proof of Corollary 1"
    },
    "id": "P028-T6",
    "text": {
     "body": "Proposition 1 proof, p. 678, and Corollary 1 proof, p. 679. A bounded causal model has positive complier probability at most one, and full compliance is allowed. The noncomplier contribution gives the mixture identities, the partial- and full-compliance branches, cell-compatible observable formulas, both bounds, and singleton boundary intervals. Positive pure cells give the printed probability-times-mean expression; empty cells have zero product moment. The usual measure-theoretic conditions are assumed.",
     "headline": "Proof of Proposition 1 with the full-compliance case $P_c = 1$ supplied, and the substitution in the proof of Corollary 1 written without a non-complier mean, as repaired by Sol",
     "location": "p. 678 (PDF p. 2), proof of Proposition 1 (‘Define $EY_n \\equiv \\mathrm{E}[Y|D(1) = D(0)]$’ and ‘Since $EY_n = [EY_1 - P_c EY_c(1)]/(1 - P_c)$’); p. 679 (PDF p. 3), proof of Corollary 1"
    },
    "url": "https://pmaronow.github.io/llm-audit/works/P028/#P028-T6"
   },
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      "credit_reason": "Zero joint pure-cell masses are equivalent to zero assignment-conditional probabilities because both arms have probability 1/2. Null-event and zero-denominator assertions faithfully express failed side conditions without asserting that Lean division is undefined.",
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      "says": null,
      "source": "https://github.com/pmaronow/llm-audit-lean/blob/b9e9df44831afed4f2b8f63be5c0cc9169a5fda4/AronowAudit/P028/P028_MissingOutcomes.lean#L359",
      "status": "counterexample_proved",
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      "claim_revision_id": "catalog:6849a029f904e5ef2efbc17ceec99dad4b5150a5d11a7f5f033ad6271c84b19b",
      "conditional_on": [],
      "credit_reason": "This is exactly the concrete F6 evaluation; it is not credited as the general full-compliance repair.",
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      "source": "https://github.com/pmaronow/llm-audit-lean/blob/b9e9df44831afed4f2b8f63be5c0cc9169a5fda4/AronowAudit/P028/P028_Results.lean#L890",
      "status": "proved",
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      "status": "proved",
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    "html": {
     "body": "<p>proof of Proposition 1, p. 678 (PDF p. 2), and Corollary 1 proof, p. 679. An explicit two-atom randomized full-compliance law has the stated outcome means, null noncomplier and pure-cell events, zero proof denominators and numerators, undetermined noncomplier mean, and valid singleton Proposition 1 intervals.</p>",
     "headline": "Proof of Proposition 1, ‘Define <span class=\"katex\"><span class=\"katex-mathml\"><math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mi>E</mi><msub><mi>Y</mi><mi>n</mi></msub><mo>≡</mo><mi mathvariant=\"normal\">E</mi><mo stretchy=\"false\">[</mo><mi>Y</mi><mi mathvariant=\"normal\">∣</mi><mi>D</mi><mo stretchy=\"false\">(</mo><mn>1</mn><mo stretchy=\"false\">)</mo><mo>=</mo><mi>D</mi><mo stretchy=\"false\">(</mo><mn>0</mn><mo stretchy=\"false\">)</mo><mo stretchy=\"false\">]</mo></mrow><annotation encoding=\"application/x-tex\">EY_n \\equiv \\mathrm{E}[Y|D(1) = D(0)]</annotation></semantics></math></span><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.8333em;vertical-align:-0.15em;\"></span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">n</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span><span class=\"mrel\">≡</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord mathrm\">E</span><span class=\"mopen\">[</span><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"mord\">∣</span><span class=\"mord mathnormal\" style=\"margin-right:0.0278em;\">D</span><span class=\"mopen\">(</span><span class=\"mord\">1</span><span class=\"mclose\">)</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span><span class=\"mrel\">=</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord mathnormal\" style=\"margin-right:0.0278em;\">D</span><span class=\"mopen\">(</span><span class=\"mord\">0</span><span class=\"mclose\">)]</span></span></span></span>’ and ‘Since <span class=\"katex\"><span class=\"katex-mathml\"><math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mi>E</mi><msub><mi>Y</mi><mi>n</mi></msub><mo>=</mo><mo stretchy=\"false\">[</mo><mi>E</mi><msub><mi>Y</mi><mn>1</mn></msub><mo>−</mo><msub><mi>P</mi><mi>c</mi></msub><mi>E</mi><msub><mi>Y</mi><mi>c</mi></msub><mo stretchy=\"false\">(</mo><mn>1</mn><mo stretchy=\"false\">)</mo><mo stretchy=\"false\">]</mo><mi mathvariant=\"normal\">/</mi><mo stretchy=\"false\">(</mo><mn>1</mn><mo>−</mo><msub><mi>P</mi><mi>c</mi></msub><mo stretchy=\"false\">)</mo></mrow><annotation encoding=\"application/x-tex\">EY_n = [EY_1 - P_c EY_c(1)]/(1 - P_c)</annotation></semantics></math></span><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.8333em;vertical-align:-0.15em;\"></span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">n</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span><span class=\"mrel\">=</span><span class=\"mspace\" style=\"margin-right:0.2778em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mopen\">[</span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.3011em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span><span class=\"mbin\">−</span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.1389em;\">P</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.1389em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mord mathnormal\" style=\"margin-right:0.0576em;\">E</span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.2222em;\">Y</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.2222em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mopen\">(</span><span class=\"mord\">1</span><span class=\"mclose\">)]</span><span class=\"mord\">/</span><span class=\"mopen\">(</span><span class=\"mord\">1</span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span><span class=\"mbin\">−</span><span class=\"mspace\" style=\"margin-right:0.2222em;\"></span></span><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:1em;vertical-align:-0.25em;\"></span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.1389em;\">P</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.1389em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span></span></span></span></span></span></span><span class=\"mclose\">)</span></span></span></span>’, used without the condition <span class=\"katex\"><span class=\"katex-mathml\"><math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><msub><mi>P</mi><mi>c</mi></msub><mo>&lt;</mo><mn>1</mn></mrow><annotation encoding=\"application/x-tex\">P_c &lt; 1</annotation></semantics></math></span><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"katex-base\"><span class=\"katex-strut\" style=\"height:0.8333em;vertical-align:-0.15em;\"></span><span class=\"mord\"><span class=\"mord mathnormal\" style=\"margin-right:0.1389em;\">P</span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.1514em;\"><span style=\"top:-2.55em;margin-left:-0.1389em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"></span><span class=\"katex-sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">c</span></span></span></span><span class=\"vlist-s\">​</span></span><span class=\"vlist-r\"><span class=\"vlist\" 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     "location": "p. 678 (PDF p. 2), proof of Proposition 1; the assumptions are in Section 2 on the same page; the related ratio is in the proof of Corollary 1, p. 679 (PDF p. 3)"
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     "body": "proof of Proposition 1, p. 678 (PDF p. 2), and Corollary 1 proof, p. 679. An explicit two-atom randomized full-compliance law has the stated outcome means, null noncomplier and pure-cell events, zero proof denominators and numerators, undetermined noncomplier mean, and valid singleton Proposition 1 intervals.",
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     "body": "<p>Corollary 1, Section 3, p. 678 (PDF p. 2), and its proof, p. 679 (PDF p. 3), under the Section 2 assumptions. The four ordinary finite treatment-assignment cell means and probabilities determine both complier average potential outcomes. They reconstruct the arm means and positive complier probability, the pure-stratum means and masses, the printed noncomplier ratio, both mixture equations, and both displayed complier formulas. Any two models with the same eight cell summaries have the same complier means. The usual measure-theoretic conditions are assumed; only observed cell moments are required, with no moment condition on unused counterfactual outcomes. Positive conditioning events express existence of the ordinary means in the premise.</p>",
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     "body": "Corollary 1, Section 3, p. 678 (PDF p. 2), and its proof, p. 679 (PDF p. 3), under the Section 2 assumptions. The four ordinary finite treatment-assignment cell means and probabilities determine both complier average potential outcomes. They reconstruct the arm means and positive complier probability, the pure-stratum means and masses, the printed noncomplier ratio, both mixture equations, and both displayed complier formulas. Any two models with the same eight cell summaries have the same complier means. The usual measure-theoretic conditions are assumed; only observed cell moments are required, with no moment condition on unused counterfactual outcomes. Positive conditioning events express existence of the ordinary means in the premise.",
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     "body": "<p>Section 2 LATE display, p. 678, and Section 1, p. 677. The population and observed treatment first stages equal the positive complier probability. With the usual measure-theoretic conditions (in particular finite observed arm means), the complier moments exist, the reduced form and both LATE ratios hold, and measurable sample analogues give the Wald IV statistic. Two bounded randomized models share the three reported figures and CACE but have different complier levels. No moments on unused potential outcomes are assumed.</p>",
     "headline": "The ‘LATE theorem’ display of Section 2 (unnumbered), with the Introduction&#x27;s statement that the three reported figures do not identify the average potential outcomes for compliers",
     "location": "p. 678 (PDF p. 2), Section 2, second paragraph; p. 677 (PDF p. 1), Section 1, first paragraph"
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     "body": "Section 2 LATE display, p. 678, and Section 1, p. 677. The population and observed treatment first stages equal the positive complier probability. With the usual measure-theoretic conditions (in particular finite observed arm means), the complier moments exist, the reduced form and both LATE ratios hold, and measurable sample analogues give the Wald IV statistic. Two bounded randomized models share the three reported figures and CACE but have different complier levels. No moments on unused potential outcomes are assumed.",
     "headline": "The ‘LATE theorem’ display of Section 2 (unnumbered), with the Introduction's statement that the three reported figures do not identify the average potential outcomes for compliers",
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 "missing_sources": [
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    "agreement": "not_compared",
    "location": "p. 677 (PDF p. 1), Introduction, second paragraph and footnote 1",
    "method": "I could not compare the reported count of 6 of 20 studies because the study list and coding sheet were unavailable.",
    "status": "blocked",
    "target": "Introduction: reporting survey"
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   {
    "agreement": "agrees_in_part",
    "location": "p. 679 (PDF p. 3), section 4, first paragraph",
    "method": "I translated the source-study outcome construction from Stata. The data reproduce the outcome means to rounding, but give a first stage of about 0.20 and a CACE of 28 points; the reported 0.21 and 29 points agree with the source chapter (P028-N007).",
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    "location": "p. 679 (PDF p. 3), section 4, first paragraph; formula in Proposition 1, p. 678 (PDF p. 2)",
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    "method": "I translated the source-study outcome construction from Stata and recalculated the four summaries. All four agree to rounding when the three unsure or refused control viewing responses are excluded (P028-N007).",
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    "method": "I applied Corollary 1 to the reported outcome means 0.45 and 0.51 and the unrounded take-up rates, cell means and first stage from the data, obtaining 0.4294 and 0.7239. This reconstructs both printed values; the original calculation is unavailable (P028-N007).",
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 },
 "summary": "This letter asks what the usual published summaries of an experiment with noncompliance reveal about compliers. From the outcome mean in each assignment arm and the difference in take-up, it derives sharp bounds on the complier average potential outcomes (Proposition 1) and illustrates them with a debate-viewing experiment. 4 entries are classified as errors (4 Minor), and 2 are clarity notes. The most important error is that the example's point estimates, $0.43$ and $0.72$, need an unstated assumption about 71 missing outcomes.",
 "what_was_checked": "I checked the following statements, which held up within the qualifications below.\n\\begin{itemize}\n\\item The LATE display of Section 2. Under (i)--(iv), $\\mathrm{E}[D \\mid Z = 1] - \\mathrm{E}[D \\mid Z = 0] = P_c$, and $(EY_1 - EY_0)/P_c$ equals the CACE. This follows from the decomposition into always-takers, never-takers and compliers, and it holds exactly in finite models generated at random.\n\\item Equations (1), (2) and (3) and the three displayed inequalities in the proof of Proposition 1, for $0 < P_c < 1$. The misprints are confined to the sentence between (3) and the display (see \\fid{P028-E001}), and the proof omits the case $P_c = 1$ (see \\fid{P028-E002}).\n\\item Both intervals of Proposition 1, as printed. I rederived them by hand. They agree exactly with the set of feasible values in 4000 finite models generated at random, on a grid of 24,167 rational points, and for 6160 compatible triples $(EY_0, EY_1, P_c)$, some with $P_c = 1$.\n\\item The sharpness of Proposition 1. Each endpoint, and each interior point that was tried, is attained by an explicit model that reproduces $(EY_0, EY_1, P_c)$ and in which compliers and never-takers have constant outcomes.\n\\item The remark after Proposition 1 on when the bounds are narrow. With $\\tau = (EY_1 - EY_0)/P_c$, the width of each interval is the minimum of $1$, $1 - |\\tau|$, $EY_0/P_c$, $EY_1/P_c$, $(1 - EY_0)/P_c$, $(1 - EY_1)/P_c$ and $(1 - P_c)/P_c$, and items (i), (ii) and (iii) of the remark correspond to these terms.\n\\item Corollary 1, under its premise that the four cell means exist. Its weighted-subtraction formula holds exactly in 4000 finite models generated at random, with zero-probability cell contributions set to zero for empty cells. The ratio defining $\\mathrm{E}[Y \\mid D(1) = D(0)]$ in its proof requires $P_c < 1$ (see \\fid{P028-E002}).\n\\item Section 4, first paragraph. The reported $0.45$, $0.51$ and $0.21$ give $0.06/0.21 \\approx 0.286$, printed as 29 percentage points, and the intervals $[0, 5/7]$ and $[2/7, 1]$, printed as $[0, 0.71]$ and $[0.29, 1]$. The source chapter reports the difference of 21 points in viewing, the outcome means $0.51$ and $0.45$, and the effect of 29 points (its p.~333 and Tables 12.1 and 12.4).\n\\item Section 4, second paragraph. The four quantities taken from the replication data are reproduced: $80/492 \\approx 0.163$, when three control respondents whose viewing answer was unsure or refused are left out, $320/505 \\approx 0.634$, $40/79 \\approx 0.506$ and $130/294 \\approx 0.442$. The outcome means among respondents whose outcome is recorded are $207/458 \\approx 0.452$ and $240/471 \\approx 0.510$, in agreement with the printed $0.45$ and $0.51$.\n\\item The description of the design in Section 4. The survey instrument and the data agree with it. Respondents were randomly assigned to be asked to watch the debate, and the control group was asked to watch another programme. The treatment $D$ is self-reported viewing, and the outcome $Y$ is a change of opinion about at least one of the two candidates. Noncompliance is two-sided, since 80 of the 495 controls watched and 320 of the 505 encouraged respondents did not.\n\\item The recommendation of Section 5. The quantities $\\mathrm{E}[Y \\mid D = d, Z = z]$ and $\\Pr[D = d \\mid Z = z]$ determine $EY_0$, $EY_1$, $P_c$ and the formula of Corollary 1. For a binary outcome they also determine the joint distribution of $Y$ and $D$ given $Z$.\n\\item The cross-references. Each reference to a numbered equation, proposition or corollary points to the right item. The letter has no tables or figures.\n\\end{itemize}\nI could not obtain the file with the list and coding of the 20 get-out-the-vote studies (footnote 1), so I could not check the count of 6 of 20 in Section 1. No analysis script for Section 4 is available. I reproduced $0.43$ and $0.72$ to rounding using the reported outcome means and the unrounded take-up rates and cell means from the data, with an inferred sample and first stage (see \\fid{P028-N007}). The replication files are the data and Stata code of the source study. Only the 2019 edition of those data, in which age is top-coded, was available. The exclusion restriction is assumed in the example and was not tested. The entries for Angrist, Imbens and Rubin (1996) and for Angrist and Pischke (2009) were not compared with those publications, and the body of Imbens and Rubin (1997) was not read. The letter has no simulations. A statement that has no correction here has not been shown to be correct.",
 "work": {
  "arxiv_id": null,
  "audit_complete": true,
  "authors": [
   "P. M. Aronow",
   "Donald P. Green"
  ],
  "category": "Articles - Theory and Methodology",
  "citation": "P. M. Aronow and Donald P. Green (2013). Sharp Bounds for Complier Average Potential Outcomes in Experiments with Noncompliance and Incomplete Reporting. Statistics and Probability Letters 83(3):677–679.",
  "doi": "10.1016/j.spl.2012.11.012",
  "edition": {
   "pages_in_audited_file": 3,
   "sha256": "c69f86414e9ff27e081e7dc841b732e15abf65dc185c889a33e97dd20789aa50",
   "source_url": "https://doi.org/10.1016/j.spl.2012.11.012",
   "venue": "Statistics and Probability Letters",
   "year": "2013"
  },
  "id": "P028",
  "later_materials": null,
  "title": "Sharp Bounds for Complier Average Potential Outcomes in Experiments with Noncompliance and Incomplete Reporting",
  "type": "Journal article",
  "url": "https://pmaronow.github.io/llm-audit/works/P028/",
  "venue": "Statistics and Probability Letters",
  "year": "2013"
 }
}