Corrections to: Beyond LATE: Estimation of the Average Treatment Effect with an Instrumental Variable

The problem

The sentence ranks two conditions at a fixed covariate profile. The ignorability assumption equates the mean of each potential outcome between the units that received treatment and the units that did not. Assumption 1 equates the mean effect among compliers with the mean effect among all units. To say that Assumption 1 is less restrictive is to say that ignorability implies it. The reason given, that Assumption 1 need not account for baseline selection, shows only that Assumption 1 can hold when ignorability fails. Under two-sided noncompliance the implication fails as well. The treated group mixes compliers with always-takers, and the untreated group mixes compliers with never-takers. In the two populations below, ignorability compares always-takers with never-takers and places no restriction on complier means. Assumption 1 compares compliers with everyone else. The method covers two-sided noncompliance (p. 496), and the Albertson and Lawrence application has it. Someone choosing between ICSW and a method that assumes selection on observables is told that the ICSW assumption is the weaker of the two. It is a different assumption.

Evidence an exact counterexample

I use two exact counterexamples with a single covariate profile. In both, a unit is a complier ($C$), an always-taker ($A$) or a never-taker ($N$) with probabilities $1/2$, $1/4$ and $1/4$, and $Z\sim\mathrm{Bernoulli}(1/2)$ is independent of the type. The five IV assumptions hold, the difference in treatment rates between the arms is $1/2$, and Assumption 3 holds with $\epsilon=1/4$. Half of the treated units and half of the untreated units are compliers.

Note that in population 1 the pair $(Y_0,Y_1)$ is independent of $D$, which is stronger than the displayed equalities. In general, with $\mu_{td}$ the mean of $Y_d$ for type $t$ and $\tau_t=\mu_{t1}-\mu_{t0}$, ignorability holds in these populations if and only if $\mu_{Ad}=\mu_{Nd}$ for $d=0,1$, and Assumption 1 holds if and only if $\tau_C=(\tau_A+\tau_N)/2$.

The implication holds in one case. Suppose that noncompliance is one-sided, $D_0=0$ for every unit, with random assignment and positive probabilities of both assignment arms and compliance at each covariate profile $\mathbf{x}$. If ignorability holds, let $m_d$ be the common value of $\mathrm{E}(Y_d|D,\mathbf{X}=\mathbf{x})$, so that $\mathrm{E}(Y_d|\mathbf{X}=\mathbf{x})=m_d$. The treated units are the compliers assigned to treatment, and I have $m_d=\mathrm{E}(Y_d|D_1>D_0,Z=1,\mathbf{X}=\mathbf{x})=\mathrm{E}(Y_d|D_1>D_0,\mathbf{X}=\mathbf{x})$. Subtracting the identity for $d=0$ from the identity for $d=1$, I obtain Assumption 1.

What survives

Assumption 1 and the consistency argument, which invokes Assumption 1 between equations (10) and (11), are unaffected. The observation that Assumption 1 permits baseline selection and the example of the medical trial also stand. This correction changes no estimate. The ranking is withdrawn. In an application with two-sided noncompliance, the reading of the ICSW estimate as an ATE rests on Assumption 1 itself, and ignorability of the treatment received does not supply it. Whether Assumption 1 holds in either application is not established here.

The problem

Assumption 1 is a condition given the covariates: the mean effect among compliers with profile $\mathbf{x}$ equals the mean effect among all units with profile $\mathbf{x}$. Independence of the effects from compliance, without qualification, is independence in the whole population. It implies that the LATE equals the ATE, and 2SLS is then consistent for the ATE. It does not give the conditional equality, because effects and compliance that are independent overall can be dependent among the units that share a covariate value. ICSW with correct scores then converges to the average over $\mathbf{x}$ of the complier effects, which is not the ATE in general. If the independence is read as conditional on the covariates, Assumption 1 holds and ICSW is consistent, but 2SLS converges to the LATE, which then differs from the ATE in general. Neither of these two readings makes the whole sentence true. The sentence suggests that ICSW cannot hurt when 2SLS already recovers the ATE. The first population below shows that weighting on covariates can introduce a bias.

Evidence an exact counterexample

The evidence is two exact counterexamples and the identities that explain them. I write $\tau=Y_1-Y_0$, $f(\mathbf{x})=\Pr(\mathbf{X}=\mathbf{x})$, $p(\mathbf{x})=\Pr(D_1>D_0|\mathbf{X}=\mathbf{x})$ and $\mu(\mathbf{x})=\mathrm{E}(\tau|D_1>D_0,\mathbf{X}=\mathbf{x})$. Under the IV assumptions, with a discrete covariate and correct scores, the limits of 2SLS and of ICSW are

Assumption 1 implies that $\theta_{\mathrm{ICSW}}$ equals the ATE, $\sum_{\mathbf{x}}f(\mathbf{x})\mathrm{E}(\tau|\mathbf{X}=\mathbf{x})$.

Both populations have a binary covariate $X$, a complier indicator $C$, an effect $\tau\in\{0,1\}$, outcomes $Y_0=0$ and $Y_1=\tau$, and $Z\sim\mathrm{Bernoulli}(1/2)$ independent of everything else. A unit that is not a complier is an always-taker or a never-taker with probability $1/2$ each, independently. In population A, $C$ and $\tau$ are independent $\mathrm{Bernoulli}(1/2)$ variables, and $\Pr(X=1|C,\tau)$ is $3/4$ if $C=\tau=1$ and $1/4$ otherwise. In population B, $X\sim\mathrm{Bernoulli}(1/2)$ and, given $X$, the variables $C$ and $\tau$ are independent with $\Pr(C=1|X)=\Pr(\tau=1|X)$ equal to $1/4$ at $X=0$ and $3/4$ at $X=1$.

In population A, I have $\theta_{\mathrm{ICSW}}=(5/8)(1/4)+(3/8)(3/4)=7/16$. In population B, I have $\theta_{\mathrm{IV}}=(5/16)/(1/2)=5/8$. Assumption 3 holds with $\epsilon=1/3$ in A and $\epsilon=1/8$ in B, and the saturated compliance model of equations (5)–(7) fits both populations, so that the scores are correct and bounded. Solving the IV moment equations in exact arithmetic, with and without $X$ as a regressor, I find the same four limits.

What survives

The consistency result in the appendix, which applies Assumption 1 between equations (10) and (11), is unaffected by this finding, and so are the preceding sentences on a constant effect, the estimator and every reported number. The sufficient condition is withdrawn. The repair adds a restriction on the effects among compliers, which the applications may not satisfy. Whether they satisfy it is not established here.

The problem

The article recommends that the bootstrap repeat the whole procedure, the fit of the compliance score included, and that recommendation is correct. The package says on its help page that it bootstraps “the entire estimation process”, and it does so by reweighting. Each replicate draws exponential weights, multiplies them into the observation weights and calls the fitting routine again. The compliance model is not fitted with those weights. In the one-sided branch the probit is fitted by glm without a weights argument, and the fitted scores are the same in every replicate. In the two-sided branch the weights multiply the likelihood contribution inside the logarithm. This adds a term to the objective that does not depend on the parameters. The set of minimizers is then the same for every vector of positive weights. There are three consequences. The package’s bootstrap standard errors leave out the variation of the first step, contrary to the article and to the help page. Sampling weights or missing-data weights that the user supplies are not used to weight the compliance-model log likelihood, and this can change the estimate itself. A weight of zero makes the two-sided objective infinite at every parameter value. The example on the help page sets the weight to zero for the 9 of its 507 observations with a missing outcome.

Evidence an exact counterexample

The evidence is a derivation for the two-sided objective, an exact counterexample for the one-sided branch, and a run of the package’s code. Let $L_i(\theta)=p_i(\theta)^{D_i}\{1-p_i(\theta)\}^{1-D_i}$, where the code keeps $p_i(\theta)$ in $[10^{-5},1-10^{-5}]$. For positive weights $w_i$, the objective on line 91 is

and the last sum does not depend on $\theta$. The weighted negative log-likelihood is $-\sum_i w_i\log L_i(\theta)$. If some $w_i=0$, then $\log\{L_i(\theta)w_i\}=-\infty$ for every $\theta$.

The counterexample has 16 observations, four in each cell of a binary covariate $W$ and the assignment $Z$. No unit assigned to control is treated. Among the assigned units, one of four is treated at $W=0$ and three of four at $W=1$. Each treated unit has weight $9$ and every other unit has weight $1$. A probit with an intercept and $W$ is saturated, so that the fitted score at each value of $W$ is the proportion treated among the assigned units, weighted when the fit uses the weights. The outcome is $Y=D(1+W)$, and, with an intercept and no other covariates in the IV fit, the weighted IV estimate reduces to $\sum_{Z_i=1}v_iY_i/\sum_{Z_i=1}v_iD_i$ with $v_i$ the weight divided by the score.

Running the unmodified package code on these data, I find scores $(0.25,0.75)$ with and without the weights and a coefficient of $1.5$. With weights = weights added to the call, the coefficient is $1.7$. The first step also matters for the bootstrap. Starting from unit weights, I change the weight of one untreated assigned unit at $W=0$ from $1$ to $t$. The package’s coefficient stays at $3/2$, while refitting the score gives $(t+11)/(t+7)$, with derivative $-1/16$ at $t=1$. In a trace of four bootstrap replicates of the one-sided code on the same $D$, $Z$ and $W$, the scores are identical in the original fit and in every replicate. Each bootstrap coefficient equals the second stage computed with the scores held fixed.

Three points are not established. The size and the direction of the error in the standard errors are unknown. Leaving out the first step removes a variance and a covariance, and the sign of their sum is not determined. In the two-sided branch the weights do enter a regression for starting values, and a numerical optimizer may return different approximate fits from different starts, although the objective has the same minimizers. The default optimizer for the two-sided branch (rgenoud) was not run, and I do not state what the example on the help page returns.

What survives

This package defect does not implicate the article’s mathematical results or reported estimates. The article’s own scripts do not use this package. The one-sided application fits a weighted probit directly and refits it on resampled units. The two-sided application and the bootstrap simulations resample rows and use the helper in the replication files, whose two-sided objective has the weights outside the logarithm. The one-sided branch of that helper has the same misplaced weights, but the article’s one-sided application does not use it. With unit weights and no bootstrap, the package’s compliance models use ordinary unweighted objectives. Package results with nonconstant weights, and all bootstrap standard errors from the package, are to be recomputed after the repair.

The problem

The sentence reads a description of the compliers off the correlations in Table A7 between the estimated compliance score and the covariates: 0.61 for White, 0.38 for Dem and 0.31 for Rep. These numbers are right. They do not measure race or party membership in the six cities. The replication script sets each indicator to 1 when the label of the voter matches and to 0 otherwise, and a blank label counts as 0. Race is blank for all 14,273 voters outside Raleigh. The White indicator is then 1 only for the 3,884 Raleigh voters with the label W, and Raleigh has the highest mean estimated score of the six cities, 0.44 against 0.15 to 0.32. The pooled correlation mostly separates Raleigh from the other cities. Table A7 itself shows that the Raleigh indicator has correlation 0.89 with White and 0.75 with the score, and that the Black indicator is also positively correlated with the score (0.32).

Within Raleigh, the one city where race is recorded, the correlation between the score and the White indicator is $-0.41$. The mean estimated score is 0.432 for the voters recorded as White, 0.487 for those recorded as Black and 0.529 for the 130 voters with another or an unknown label. Among the Raleigh voters assigned to canvassing, 602 of the 1,370 recorded as White were contacted (43.9%), against 160 of the 315 others (50.8%).

Party is blank for every voter in Detroit, Minneapolis and St. Paul and for 489 voters in Columbus. More than 90% of the pooled covariance between the score and each party indicator is covariance between city means. For party the sign does not reverse within Raleigh, where the correlation of the score with the combined Dem/Rep indicator (which includes the code LIB in Dem) is 0.18, against 0.53 in the pooled data.

I conclude that the correlations cited do not support the description of the compliers by race and party. I do not conclude that the description is false. The association between race or party and compliance over the six cities depends on the labels that were not recorded, and these data cannot determine it. The part of the sentence about previous voting is supported.

Evidence an exact counterexample

The evidence is a recalculation from the replication data and an exact counterexample to the inference from recorded labels. The recalculation fits the compliance model of the script. It reproduces the correlations of the compliance score in Table A7 and the twelve point estimates in each column of Table 1 at the printed precision, with the age-missing indicator set before the imputation (see P029-E021). The first table lists the blank labels and the mean score in each city.

In Raleigh, 21 of the 4,660 race labels are unknown or unspecified. The covariance between the score $s$ and an indicator $A$ splits exactly into a part within cities and a part between city means,

where $s_G$ and $A_G$ are the means in the voter’s city and $\bar s$ and $\bar A$ are the overall means.

For White the part within cities is negative, and all of it comes from Raleigh.

For the counterexample, I complete the unobserved quantities in two ways that agree with every recorded label and every observed variable. In both, a voter assigned to canvassing is a complier if contacted, and a control voter, whose status is not observed, is not a complier. I assign the 21 unknown or unspecified Raleigh race labels as not White in both completions. There are 2,551 compliers. In the first completion every voter with a blank race label is not White. Then 3,884 voters are White, 602 of them are compliers, and the compliance rate of the White voters exceeds that of the others by $744791/29225158 \approx 0.025$. In the second completion every voter with a blank race label is White. Then 18,157 voters are White, 2,391 of them are compliers, and the difference is $-131213/1761229 \approx -0.075$. With the estimated scores held fixed, the correlation between the score and race is $0.61$ in the first completion and $-0.37$ in the second. For party, assigning every blank label as outside the two major parties gives a positive association; assigning blank labels as major-party only for noncompliers gives a negative association, with recorded LIB affiliations outside the two major parties in both cases. Both completions agree with the data. This implies that the data do not determine the sign, and they do not show which completion is closer to the truth.

What survives

The correlations of Table A7 are numerically right. The statement that compliers are more likely to have voted previously stands: the score is positively correlated with turnout in 2000 and in 1999 in the pooled data and within cities. The sentence is not an input to any calculation. A change of wording leaves the point estimates of Table 1, the comparison of 5.6 with 6.7 percentage points and the bootstrap as they are. I did not recompute the intervals. The method, the simulations and the application to Albertson and Lawrence (2009) do not depend on the sentence. What changes is the description of the compliers, which must be limited to labels that were recorded.

The problem

The sentence asserts that the SEM conditions of Section 2.1 imply the five assumptions just listed. For the model in equations (1) and (2), with a common effect $\beta_1$, Section 2.1 requires that the correlation between $Z_i$ and $u_i$ tend to zero and that the covariance between $Z_i$ and $D_i$ tend to a nonzero limit. These conditions do not exclude defiers, and a zero correlation does not make $Z_i$ independent of the potential outcomes. In the other direction, the five assumptions let the effect $Y_1-Y_0$ vary across units, and the SEM formulation does not. Neither set of assumptions implies the other. The point that the passage serves is correct: the potential-outcome formulation permits heterogeneous treatment effects.

Evidence an exact counterexample

The evidence is three exact counterexamples. Each is a finite population with binary $Z$ and $D$, no covariates and independently sampled units, in which $D=D_Z$ and $Y=Y_D$.

• The SEM conditions hold and the population contains defiers. A unit is a complier with probability $3/4$ and a defier with probability $1/4$, and $Z\sim\mathrm{Bernoulli}(1/4)$ is independent of the type. I set $u=3/4$ for a defier and $u=-1/4$ for a complier, $Y_0=u$ and $Y_1=u+1$. Then $Y=D+u$ with $\beta_1=1$, and $D=1/4+Z/2+e$ with $\mathrm{E}(e\mid Z)=0$. I have $\operatorname{Var}(Z)=\operatorname{Var}(u)=3/16$, $\operatorname{Cov}(Z,u)=0$ and $\operatorname{Cov}(Z,D)=3/32$. The instrument is independent of all potential outcomes and the exclusion restriction holds, but $\Pr(D_0>D_1)=1/4$.
• The SEM conditions hold and $Z$ is not independent of the potential outcomes. Let $U$ be uniform on $\{-2,-1,1,2\}$, and set $Z=\mathbf{1}\{|U|=2\}$, $D_0=\mathbf{1}\{U>0\}$, $D_1=1$, $Y_0=U$ and $Y_1=U+1$. Then $Y=D+U$, and $D=1/2+Z/2+e$ with $\mathrm{E}(e\mid Z)=0$. I have $\mathrm{E}(U\mid Z)=0$, $\operatorname{Cov}(Z,U)=0$ and $\operatorname{Cov}(Z,D)=1/8$, and there are no defiers. But $\Pr(|Y_0|=2\mid Z=1)=1$ and $\Pr(|Y_0|=2\mid Z=0)=0$.
• The five assumptions hold and the effect is not constant. Let $H$ be uniform on $\{1,2\}$, and let $Z\sim\mathrm{Bernoulli}(1/2)$ be independent of $H$. Every unit is a complier, $Y_0=0$ and $Y_1=H$. The effect is $1$ for half of the units and $2$ for the other half, while a causal reading of equation (1) requires that $Y_1-Y_0=\beta_1$ for every unit.

Note that in the first two populations the treatment is endogenous ($\operatorname{Cov}(D,u)$ is $3/32$ and $1/4$), and the population IV ratio equals the common effect $1$. These populations are not counterexamples to IV estimation under a constant effect.

What survives

Equation (3) is stated under the five assumptions themselves, and Section 3.4 invokes the IV assumptions directly for ICSW. Neither uses the comparison, and no simulation or application result depends on it. Only the comparison changes. The claim that one set of assumptions contains the other is withdrawn.

The problem

The estimator $\hat P_{Ci}$ is the maximum-likelihood estimator of Section 3.1. By equations (5)–(7), the model sets the probability of receiving treatment to $P_{A,C,i}$ for a unit with $Z_i = 1$ and to $P_{A,C,i} P_{A|A,C,i}$ for a unit with $Z_i = 0$. The compliance score is the difference of the two. Since the second probability is at most the first, the model cannot fit a covariate stratum in which the arm $Z = 0$ receives treatment at a higher rate than the arm $Z = 1$. In a saturated model with both arms nonempty in stratum $x$, let $\bar D_{1x}$ and $\bar D_{0x}$ be the treatment-receipt rates in stratum $x$, that is, the shares of units that receive treatment in the arm $Z = 1$ and in the arm $Z = 0$. If $\bar D_{0x} \le \bar D_{1x}$, then the fitted probabilities are these two rates, and the fitted score is $\bar D_{1x} - \bar D_{0x}$. If $\bar D_{0x} > \bar D_{1x}$, then both fitted probabilities equal the pooled rate, and the fitted score is zero. Hence

The difference $\bar D_{1x} - \bar D_{0x}$ is unbiased for the compliance score, and the truncation replaces its negative values by zero. The expectation of $\hat P_{Ci}$ is therefore larger than the compliance score whenever a negative difference has positive probability. This is the case in every finite sample when the stratum contains always-takers and never-takers. The printed sentence is true under one-sided noncompliance, where the arm $Z = 0$ never receives treatment and the truncation cannot bind. Note that the normal CDF plays no role: the formula holds for any $F$ that maps onto $(0,1)$.

Evidence an exact counterexample

The evidence is a derivation of the estimator and an exact counterexample. For the derivation, I fix a stratum with $n_1 > 0$ units in the arm $Z = 1$ and $n_0 > 0$ units in the arm $Z = 0$, of which $k_1$ and $k_0$ receive treatment. I write $p = P_{A,C}$ and $q = P_{A,C} P_{A|A,C}$ for the two receipt probabilities of the model. Up to a constant, the likelihood (7) of the stratum is

and the fitted score is $p - q$ at the maximum. If $k_0/n_0 \le k_1/n_1$, then the unconstrained maximizer $(k_1/n_1, k_0/n_0)$ satisfies the constraint. If $k_0/n_0 > k_1/n_1$, then $k_0 > 0$ and $k_1 < n_1$, so $L$ vanishes on the edges $q = 0$ and $p = 1$. Further, $L$ has no stationary point such that $0 < q < p < 1$, since such a point would be $(k_1/n_1, k_0/n_0)$. The maximum then lies on the diagonal $q = p = t$, where $L(t,t) = t^{k_1 + k_0}(1-t)^{n_1 + n_0 - k_1 - k_0}$ is largest at $t = (k_1 + k_0)/(n_1 + n_0)$. Combining the two cases, I obtain the formula above. A probit model with finite coefficients has $0 < q < p < 1$, and a sample whose maximizer lies on the boundary has a supremum that is not attained. The formula takes the maximum over the closure.

For the counterexample, I take a single stratum in which a unit is an always-taker with probability $1/4$, a complier with probability $1/2$ and a never-taker with probability $1/4$, with no defiers. One of two independently sampled units is assigned to each arm at random. The receipt probabilities are $p_1 = 3/4$ in the arm $Z = 1$ and $p_0 = 1/4$ in the arm $Z = 0$, and the compliance score is $1/2$. The model is correctly specified, with $P_{A,C} = 3/4$ and $P_{A|A,C} = 1/3$.

The expectation of the fitted score is $9/16$, and the bias is $1/16$. The untruncated difference $d_1 - d_0$ has expectation $3/4 - 1/4 = 1/2$. The estimator replaces its value $-1$ in the second row by $0$.

The bias is not special to one unit per arm. Let $T = \bar D_{1x} - \bar D_{0x}$, and let $c$ be the compliance score. Since $\max(T, 0) = T + \max(-T, 0)$ and $\mathbb{E}[T] = c$, I have

and the inequality is strict for all arm sizes whenever $p_0 > 0$ and $p_1 < 1$. With five units in each arm, $p_1 = 1/2$ and $p_0 = 3/10$, I enumerate the 36 pairs of counts and find the expectation $15659/64000 \approx 0.245$, against a compliance score of $1/5$.

Consistency holds. For $c \ge 0$ and every real $t$, I have $|\max(t, 0) - c| \le |t - c|$. Chebyshev’s inequality then implies that, for every $\delta > 0$,

and the right side tends to zero as $n_1$ and $n_0$ tend to infinity. Under one-sided noncompliance, $p_0 = 0$, so $k_0 = 0$ and $\hat P_{Ci} = \bar D_{1x}$, which is unbiased when the arm $Z = 1$ is not empty.

What survives

Assumption 2 requires that $\hat P_{Ci}$ be consistent, not that it be unbiased, and the consistency argument of the Appendix uses the assumption in that form. The estimands, the ICSW estimator and its identification argument do not depend on the sentence. The two applications and the simulations fit parametric models and not saturated ones, and none of their calculations uses the sentence. I did not recompute them for this finding. The phrase “strata means under saturation” on p. 496 needs the same qualification, since the two fitted probabilities are the separate arm means only when $\bar D_{0x} \le \bar D_{1x}$. What changes is the claim that the score is unbiased, which becomes a claim that it is consistent.

The problem

The cut-off $c$ is a quantile of the estimated compliance scores, taken at the level $n^{-\alpha}$. As $n$ grows the level tends to zero, but the quantile need not, since Assumption 3 keeps the true scores above some $\epsilon>0$. The sentence’s conclusion holds for a different reason. The fraction of scores that Winsorization changes is at most $n^{-\alpha}+1/n$, and the weights that it changes stay bounded under conditions of the kind that the article already invokes.

Evidence an exact counterexample

The evidence is an exact counterexample to the limit and a derivation of the corrected statement.

Counterexample. Let the covariate be constant. A unit is an always-taker with probability $1/4$, a complier with probability $1/2$ and a never-taker with probability $1/4$, the assignment $Z\sim\mathrm{Bernoulli}(1/2)$ is independent of the type, and $Y_0=0$, $Y_1=1$. The IV assumptions hold, Assumption 1 holds because the effect is $1$ for every unit, and Assumption 3 holds with $\epsilon=1/4$ because the compliance score is $1/2$. With the exact scores, every quantile of the scores is $1/2$, and $c=1/2$ for every $n$ and every $\alpha$. With the ML estimator of equations (5)–(7), the fitted score is the same for all units and eventually equals the difference between the proportions treated in the two arms, once those proportions are interior and correctly ordered. By the strong law, $c$ converges almost surely to $3/4-1/4=1/2$. No score is below $c$, the weights are unchanged, and the ICSW ratio is $1$, the ATE, whenever its empirical first stage is nonzero. Note that the example contradicts the reason given and not the conclusion.

Derivation. Let $p_n=n^{-\alpha}$, let $c_n$ be the cut-off, and let $m_n$ be the number of scores strictly below $c_n$. For the quantile rule that the code uses, I have $m_n/n\le p_n+1/n$. Suppose that $\max_{i\le n}|\hat{P}_{Ci}-P_{Ci}|\to0$ in probability and that $\mathrm{E}|Y|<\infty$. By Assumption 3 and this convergence, every estimated score exceeds $\epsilon/2$ with probability tending to one. On that event the weights $w_i=1/\hat{P}_{Ci}$ and the Winsorized weights $w_i^*=1/\max(\hat{P}_{Ci},c_n)$ satisfy $0\le w_i-w_i^*\le2/\epsilon$, and they differ for at most $m_n$ units. For any variables with $|V_i|\le K$, I obtain

This bound covers the weighted sums of $\mathbf{1}\{Z_i=z\}$ and $\mathbf{1}\{Z_i=z\}D_i$. For the weighted sums of the outcome, truncating $Y_i$ at a level $M$ and using $\mathrm{E}|Y|<\infty$, I find that the change tends to zero in probability. Since the weighted difference in treatment rates has the positive limit $1/\mathrm{E}\{1/P_C(\mathbf{X})\}$, I conclude that the Winsorized and original estimators have the same probability limit. The bound is of order $n^{-\alpha}$, which for $\alpha=0.275$ is not $o(n^{-1/2})$. The argument does not show that the two estimators have the same first-order distribution, and it does not show that they differ.

What survives

The Winsorization rule, the code, which computes a quantile, and every estimate are unchanged. The appendix does not use the limit $c\to0$, and consistency holds under the conditions stated below. No application estimate and no simulation result is shown to be wrong. Only the reason given in the footnote changes.

The problem

With a plus sign every factor of the product is at least $1$, and the display is not a likelihood. The reduction of equation (7) and the likelihood for binomial regression, which the next sentence names, both have a product. The parameter is also written $\theta_{\mathbf{C}}$ on the left and $\theta_{\mathbf{A,C}}$ on the right. With no always-takers these index the same probability, and one subscript is meant.

Evidence a derivation

The evidence is a derivation. Setting $P_{A|A,C,i}=0$ in equation (7), I obtain the likelihood for unit $i$,

For an assigned unit with $F(\cdot)=3/4$, the printed factor is $5/4$ when $D_i=0$ and $7/4$ when $D_i=1$, in place of $1/4$ and $3/4$. The one-sided likelihood in the replication code is the product.

What survives

Equation (7), the definition of the ML estimate, the sentence on binomial regression and every computed result are unaffected.