Corrections to: Does Regression Produce Representative Estimates of Causal Effects?

The problem

The page defines the contribution of a country as the sum of its weights over all its years, divided by the sum of all the weights in the sample. This is the left side of Expression (9) with $Z_i$ the indicator of the country. The numbers that follow were computed differently: from the mean weight of each country, divided by the sum of the 114 country means. The replication script takes the country mean (tapply(w, jensen.cc$country, mean), line 33) and shades the right panel of Figure 1 with it. Let $p_c$ be the share of country $c$ by sums and $n_c$ its number of observed years. Then the index of country means is

and $q_c=p_c$ for every country only if all countries with positive weight are observed for the same number of years. In this panel $n_c$ ranges from 1 to 22, and the index of country means raises the countries that are observed in few years. Niger, with 4 years, is third on the printed list and 27th by its sum. Under the definition that the page gives, the printed statements change as follows. The twelve named countries hold 47.6% of the weight, which is less than half. Five of them (Niger, Madagascar, Poland, Lesotho and Belarus) are not among the twelve largest contributors, and the order of the other seven changes. The largest 32 countries hold 89.2% of the weight, and 34 are needed to pass 90%. The list of the 14 lowest contributors changes in order and in one member: Namibia replaces China (its tenth name is a separate error, P015-E002). The regional summary and the advice on case selection in the right column were written from the printed list and the map.

Evidence an exact counterexample

I checked the exact recalculation from the authors' data. The regression of regime type on the stated controls was solved in rational arithmetic for all 1,630 country-years and 114 countries, and the residuals satisfy all 123 normal equations exactly. The index of country means reproduces the printed twelve in the printed order, with a share of 50.91%, and it reproduces the printed list of the 14 lowest once its tenth name is corrected (see P015-E002). The shares by country sums are in the last column of the table. Under sums, 10 countries hold 49.7% and 11 hold 53.1%. The printed 90% rounds the mean-based share of 89.71%; it does not assert a minimum of 32 countries to reach at least 90%.

What survives

The finding that few countries carry the estimate survives: 12 of 114 countries hold 56.3% of the weight, 11 pass one half, and 34 pass 90%. The 14 lowest contributors hold 0.011%, inside the printed bound of 0.05%, and Russia, China and Germany each hold less than 0.005%. The aggregation by country takes place after the regression is fitted, so the OLS coefficient and the weights $\hat w_i$ do not change. The left panel of Figure 1, the definition of the weights and the application to Gerber and Huber (2010) are not touched by this error. What changes is the right panel of Figure 1, the two lists, the shares, and the regional summary. Niger, Madagascar, Poland, Lesotho and Belarus leave the twelve, and Chile, Fiji, Zambia, Ecuador and Nepal enter. If the printed numbers were kept, the index would have to be described as a country’s average weight per observed year, which is a different quantity from the contribution that Expression (9) defines.

The problem

The paragraph draws three conclusions from the regression weights. Recall that $w_i = (D_i - \mathrm{E}[D_i \mid X_i])^2$, so that $\mathrm{E}[w_i \mid X_i = x] = \operatorname{Var}[D_i \mid X_i = x]$ by the definition of the conditional variance. One direction is correct. Positivity for two distinct values $d$ and $d'$ at $x$ implies that $\operatorname{Var}[D_i \mid X_i = x] > 0$. Hence positivity fails for every pair at any covariate profile such that this variance is zero in the population. The three conclusions of the paragraph go further, and none of them follows.

First, the paragraph treats a weight of approximately zero as a violation of positivity. A conditional variance that is small and positive is compatible with positivity at every profile, and with a consistent estimator of the representative effect that uses only observed outcomes. A small positive population conditional variance shows that the profile supplies little variation in the treatment and that estimates for it may be imprecise. Second, the paragraph says that positivity holds in the effective sample by construction. At the population level, the effective sample assigns positive weight to the profiles at which the treatment varies. For a binary treatment this is positivity for the pair $(0,1)$. For a treatment with more than two values it means only that $D_i$ takes at least two values at $x$, and these need not be the chosen $d$ and $d'$. The positivity display on the same page is stated for the chosen pair. Both applications have such treatments: a regime score with 21 values in Jensen (2003), and party identification on a five-point scale in Gerber and Huber (2010). Third, the weights used in an application are estimated: they are squared residuals from a fitted regression. An estimated weight that is zero or small does not establish that the conditional variance is zero in the population.

The statement about Jensen (2003), that no consistent, extrapolation-free estimator exists for any $(d,d')$ over the 114 countries, therefore does not follow from the small weights of countries such as Russia and China. Whether the statement is true is a separate question, which I do not settle. In the replication data Russia has two observations, both with regime score 14, and China has seven, all with score 3. These observations are compatible with a violation of positivity, and they do not determine the conditional distribution of the regime score in the population.

Evidence an exact counterexample

The evidence is three exact examples, each computed in rational arithmetic. By the effective population I mean the nominal population reweighted in proportion to $w_i$.

Example A: a small weight with positivity. Let $X_i$ equal $0$ or $1$ with probability $1/2$ each, let $D_i$ be binary with $\Pr[D_i = 1 \mid X_i = 0] = \varepsilon = 10^{-6}$ and $\Pr[D_i = 1 \mid X_i = 1] = 1/2$, and let $Y_i(d) = 2X_i + (1 + X_i)d$. Since the potential outcomes are functions of $X_i$, unconfoundedness holds. The effects are linear in $d$ with $\tau_i = 1 + X_i$, and $\mathrm{E}[D_i \mid X_i] = \varepsilon + (1/2 - \varepsilon)X_i$ is linear in $X_i$. Both treatment values have positive probability at both profiles.

The effective share of the profile $X_i = 0$ is about $4.0 \times 10^{-6}$. The average effect in the nominal population is $3/2$. The population OLS coefficient on $D_i$ is $500000999999/250000999999$, which is the weighted average of Theorem A.1. Let $e(x) = \Pr[D_i = 1 \mid X_i = x]$ and

Computing its first two moments, I find $\mathrm{E}[Z_i] = 3/2$ and $\operatorname{Var}(Z_i) = 2000071/4$. The mean of $n$ independent copies of $Z_i$ is then unbiased for $3/2$ with variance $2000071/(4n)$, and Chebyshev’s inequality implies that it is consistent. Note that this estimator uses only outcomes that are observed under each treatment at each profile. The same construction has positivity for every $\varepsilon$ in $(0, 1/2)$. The effective share of the profile $X_i = 0$ is $\varepsilon(1-\varepsilon)/[\varepsilon(1-\varepsilon) + 1/4]$, which tends to $0$ as $\varepsilon \to 0$. Hence no threshold for “approximately no weight” implies a violation of positivity.

Example B: positive weights, and a chosen pair without support. Let $(X_i, D_i)$ take the values $(0,0)$, $(0,1)$, $(1,0)$ and $(1,2)$ with probability $1/4$ each. Then $\mathrm{E}[D_i \mid X_i] = 1/2 + X_i/2$ is linear in $X_i$.

Every weight is positive and the profile $X_i = 1$ has effective share $4/5$, yet $\Pr[D_i = 1 \mid X_i = 1] = 0$ in the nominal and in the effective population. Positivity for the pair $(0,1)$ fails. I compare two sets of potential outcomes. In the first, $Y_i(d) = d$ for every unit. The second differs from the first only in that $Y_i(1) = 3$ when $X_i = 1$. Both satisfy unconfoundedness. Since $D_i = 1$ never occurs at $X_i = 1$, the observed $(X_i, D_i, Y_i)$ has the same distribution under both. The average of $\tau_i(0,1)$ over the nominal population is $1$ in the first and $2$ in the second; over the effective population it is $1$ and $13/5$. I conclude that no estimator based on the observed data is consistent in both cases.

Example C: an estimated weight of zero. Suppose that $X_i$ and $D_i$ are independent with $\Pr[X_i = 1] = \Pr[D_i = 1] = 1/2$, so that positivity holds and $\operatorname{Var}[D_i \mid X_i = x] = 1/4$ at both profiles. The ordered sample $(X_i, D_i) = (0,0), (0,0), (1,0), (1,1)$ has probability $1/256$. In this sample the regression of $D_i$ on a constant and $X_i$ has coefficients $(0, 1/2)$ and residuals $(0, 0, -1/2, 1/2)$. The two units with $X_i = 0$ receive an estimated weight of exactly zero.

What survives

Theorem A.1, the weights of Expressions (7) and (8), the effective-sample summaries of Expression (9), the simulations and the weighting diagnostics for both applications are unaffected. So is the conclusion that OLS estimates an average effect weighted by the conditional variance of the treatment, which can differ from the average effect in the nominal population. It remains true that zero conditional variance in the population rules out positivity, and that positive conditional variance is positivity for a binary treatment. The paragraph can no longer conclude from near-zero weights that positivity is violated, or that positivity holds in the effective sample when the treatment has more than two values. The statement about Jensen (2003) becomes conditional on a violation of positivity that is shown by other means. No reported number changes. The form of the positivity display for a continuous treatment is a separate point, P015-N011.

The problem

Appendix C derives its result for logistic regression in two steps. The ML estimate solves the estimating equation displayed above. Replacing $\operatorname{logit}^{-1}$ by its first-order Taylor polynomial about $\operatorname{logit}(P_Y)$ then turns this equation into the OLS normal equations, up to the scale factor $c = P_Y(1-P_Y)$. The appendix says that the result holds for probit, Poisson, ordered, multinomial and duration models as well, and that it follows by substituting the predicted mean or inverse link of the model into the estimating equation.

The first step fails in general. For a binary outcome with mean function $\mu$ and linear predictor $\eta_i = W_i\theta$, the likelihood score is $\sum_i W_i\, h(\eta_i)\{Y_i - \mu(\eta_i)\}$ with $h = \mu'/[\mu(1-\mu)]$. For the logit, $h$ is identically $1$ and the printed equation is the likelihood score. For probit, $h(\eta) = \phi(\eta)/[\Phi(\eta)\{1-\Phi(\eta)\}]$ varies with $\eta$ and cannot be taken out of the sum. The substituted equation is then not the probit score, and a root need not be the probit ML estimate. The scores of ordered, multinomial and duration models have other forms again. For example, the Cox partial-likelihood score involves the risk sets, and it cannot be obtained by substituting a scalar inverse link.

The second step changes as well. A first-order expansion of the probit score must expand $h$ along with $\Phi$, which leaves a first-order term that involves the outcome and has no counterpart in the OLS equations. This term vanishes for probit when the expansion point is $P_Y = 1/2$. It can cancel in particular designs, and it can be of higher order under further local assumptions, none of which the appendix states. The appendix therefore establishes the first-order OLS weighting for logistic regression, and the same argument covers the models whose score is a constant multiple of the substituted equation, such as Poisson regression with the log link. For probit and the other listed models the statement is not established by this argument. I do not show that a local version with suitable qualifications fails for those models.

Evidence an exact counterexample

The evidence is a derivation of the score and exact counterexamples.

The score. For a binary outcome with $\Pr[Y_i = 1 \mid W_i] = \mu(\eta_i)$, the log likelihood of one observation is $Y_i \log \mu(\eta_i) + (1 - Y_i)\log\{1 - \mu(\eta_i)\}$. Differentiating in $\eta_i$, I have

For $\mu = \operatorname{logit}^{-1}$, $\mu' = \mu(1-\mu)$ and $h \equiv 1$.

Two parameter values that solve the substituted equation and are not ML estimates. In both examples $W_i = (1 \;\; D_i \;\; X_i)$ with $D_i$ binary, and the cells are the same for $D_i = 0$ and for $D_i = 1$. In the first example $\mu(\eta) = \eta/(1+\eta)$ on $\eta>0$ and $\theta = (1, 0, 1)'$, and the cells $X_i = 0, 1, 2$ have $12$ observations each, with $7$, $6$ and $10$ successes. In the second example the model is probit and $\theta = (0, 0, a)'$ with $a = \Phi^{-1}(3/4)$, and the cells $X_i = -1, 0, 1$ have $8$ observations each, with $3$, $2$ and $7$ successes.

In the probit example the treatment coordinate $2s$ of the score exceeds $13/625$, by a bound obtained in rational arithmetic, and its value is about $0.198$. A point at which the score is not zero is not an ML estimate.

The first-order term. Fix $p$ in $(0,1)$ and let $\eta_0 = \mu^{-1}(p)$, $c = \mu'(\eta_0)$, $h_0 = h(\eta_0)$ and $h_1 = h'(\eta_0)$. Writing $\eta_i = \eta_0 + \delta_i$ and expanding both factors, I obtain

The substitution keeps the first two terms and drops the third, which is of first order in $\delta_i$. Note that $h_1 = 0$ for the logit.

Its effect on the coefficient. Let $X_i$ take the values $-1$, $0$ and $1$ with probability $1/3$ each, let $D_i$ be binary with $\Pr[D_i = 1 \mid X_i] = 1/2 + X_i/4$, and let $\mathrm{E}[Y_i \mid D_i, X_i] = q_0(X_i) + \varepsilon (D_i - 1/2 - X_i/4)$, where $q_0(-1) = q_0(1) = p + 1/16$ and $q_0(0) = p - 1/8$. The mean of $Y_i$ is $p$. Potential outcomes with this conditional mean can be chosen such that unconfoundedness holds; the treatment is binary, and $\mathrm{E}[D_i \mid X_i]$ is linear in $X_i$. The OLS coefficient on $D_i$ is exactly $\varepsilon$. If the substitution were valid, the ML coefficient on $D_i$ would then be $\varepsilon/c$ to first order. Differentiating the equation that sets the likelihood score to zero, I find that the derivative of the ML coefficient at $\varepsilon = 0$ is $1/(c + k/80)$, where $k = h_1/h_0$.

Hence the ML coefficient differs at first order in $\varepsilon$ from the value implied by the substitution. With the same $p$ and with $q_0$ constant, the derivative equals $1/c$. For these models the first-order scale is therefore not a function of the mean of $Y_i$ alone. If $\varepsilon$ is replaced by $\varepsilon\tau(X_i)$ in this example, the relative weights on the effects $\tau(X_i)$ in the derivative still follow $\operatorname{Var}[D_i \mid X_i]$. The example shows that the scaling fails, and it does not show different relative weights.

What survives

Theorem A.1 and Corollary A.1, the formulas for the effective sample under OLS, the comparison with the leverage in Appendix B and the argument about external validity do not use this extension. Both applications compute their weights from least-squares regressions of the treatment on the controls, and none of their reported numbers changes on this account. The logistic score and its first-order reduction to the OLS equations stand, apart from the misprint P015-E010 and the notational point P015-E011. The extension narrows to the models whose score is a constant multiple of the substituted equation. For probit, ordered, multinomial and duration models the same weights are not established by Appendix C and would need a separate argument. The erratum of 2017 does not address this passage.

The problem

“DR Congo” denotes the Democratic Republic of the Congo. In the data it is a separate country with 18 observed years, and it is not among the 14 lowest contributors: counted from the bottom it ranks 63rd by country means and 64th by country sums. The country in tenth place is the Republic of the Congo, with 4 observed years, under either way of aggregating the weights. With DR Congo in the list, the 14 named countries hold about 0.22% of the weight, which is above the bound of 0.05% stated in the preceding sentence.

Evidence an exact counterexample

The evidence is an exact recalculation from the authors' data, cross-checked against the authors' regression in R. With the Republic of the Congo in tenth place, the list is the list of the 14 lowest under the index of country means, which produced the printed figures, and these countries hold 0.0458% of it; the first two entries are tied at zero. The 14 names as printed hold 0.2225% by means and 0.2173% by sums. The authors' file that links the country names of the data to the names of the map also keeps the two countries apart.

What survives

Only the name changes; no weight, coefficient or map does. The bound of 0.05% holds for the correct 14 countries. This correction repairs the list as it was computed. The list by country sums, which is the quantity the text describes, differs further, and P015-E001 gives it.

The problem

The sentence summarizes Model 10 in Table 4 of Jensen (2003), which the preceding sentences name. In that table the coefficient on democracy is 0.021 with three stars, the test statistic beneath it is 2.606, and the legend reads “***$p<.01$”. The coefficient is therefore misquoted in the third decimal, and the significance level is overstated: the source supports $p<.01$, and under a normal reference a statistic of 2.606 has a two-sided $p$ of about 0.009, which is above 0.001.

Evidence a recalculation from the source

This is a check of the cited source: Jensen (2003), International Organization 57(3), p. 604, Table 4, Model 10, in the published version; p. 602 of the same article says that the standard errors are panel-corrected. The replication files agree on the coefficient: fitting OLS with the stated controls on all 1,630 country-years, I find 0.0211. The conventional standard error of that fit gives a two-sided $p$ of 0.043, which follows a different convention from Jensen’s and is not what the sentence quotes.

What survives

The association remains positive and significant at the .01 level in the source. No weight, figure or result of the paper uses these two numbers. Whether the “$p<.05$” quoted on p. 259 for the estimate of Gerber and Huber (2010) matches their table could not be settled, because their Table 4A was not available to me; P015-E004 reports what the replication data show.

The problem

Two descriptions in the paragraph are wrong. First, a one-point move on the party identification scale raises the fitted outcome by 0.077 natural-log points. Since $e^{0.077}-1=0.080$, the implied increase is about 8%. An increase of 20% would need a coefficient of $\log 1.2=0.182$, or a move of about 2.4 points on a scale that runs from $-2$ to $2$. Second, the outcome of the fitted model is the change in the natural log of one plus predicted vacation spending,

where $S_{i,\mathrm{pre}}$ and $S_{i,\mathrm{post}}$ are the predicted spending of respondent $i$ in the two waves. The natural log of the change, $\log(S_{i,\mathrm{post}}-S_{i,\mathrm{pre}})$, is undefined for the 676 respondents whose predicted spending did not change and for the 420 whose predicted spending fell. The words “log change” later in the paragraph, and Table 1, describe the variable correctly.

Evidence an exact counterexample

The evidence is a derivation and a recalculation from the replication files. Let $\hat m(d,x)$ be the fitted linear predictor at party identification $d$ and controls $x$, and let $\hat\beta$ be the coefficient on party identification. Then $\hat m(d+1,x)-\hat m(d,x)=\hat\beta$, which implies that

so the relative increase in the exponentiated fitted outcome is $e^{\hat\beta}-1$. For $x=77/1000$, bounding $e^x$ below by the partial sum $P_7=\sum_{k=0}^{7}x^k/k!$ and above by $P_7+(x^8/8!)/(1-x/9)$ and comparing exact fractions, I obtain

The replication data give a coefficient of 0.07683 on 1,469 observations, hence 7.99% from the unrounded value. In those data the outcome variable is exactly the difference of the two log variables that are stored with it, and each of these equals $\log(1+S)$ to the precision of storage.

Note that the 8% describes the fitted geometric growth factor of $(1+S_{\mathrm{post}})/(1+S_{\mathrm{pre}})$ and not arithmetic mean spending. Consider two data sets of four observations, each with $S_{\mathrm{pre}}=1$ throughout and two observations at each value of a binary $D$. Post-election spending is $(1,1)$ at $D=0$ in both data sets; it is $(1,1)$ at $D=1$ in the first and $(0,3)$ at $D=1$ in the second. The coefficient on $D$ is zero in both, since $[\log(1/2)+\log 2]/2=0$, while mean post-election spending at $D=1$ exceeds that at $D=0$ by 0% in the first and by 50% in the second.

What survives

The coefficient 0.077, the sample of 1,469 and every entry of Table 1 are unaffected: all 76 numerical cells of the table reproduce at the printed precision. The weights and the comparison of the nominal and effective samples do not use the 20%. The printed $p<.05$ is reproduced with standard errors clustered by state (50 clusters, a $t_{49}$ reference, $p=0.0274$). With the ordinary standard errors of OLS, I find $p=0.076$, and I have not been able to check which convention the table of Gerber and Huber (2010) uses.

The problem

The target $\bar\beta$ is a scalar, the average over groups of the treatment coefficient in the Swamy model (10), and Expression (7) is the limit of another scalar, the OLS coefficient on the treatment. The passage therefore says that the estimator pools the quantities of Expression (7), one for each group, with scalar precision weights, and that the effective sample combines those weights with the regression weights within each group. The generalized least squares estimator does not have this form in general. It weights the full vector of OLS coefficients of each group, which holds the intercept, the treatment coefficient and the coefficients on the covariates, by the inverse of the covariance matrix of that vector. The treatment coordinate of the pooled estimate then includes the intercepts and the other coefficients of the groups, with weights that need not vanish, and Expression (7) alone does not characterize it. The words “precision-weighted average” are correct when the weights are matrices that act on the coefficient vectors. What fails is the reduction to the scalar quantities of Expression (7).

Evidence an exact counterexample

The evidence is a derivation of the estimator and an exact counterexample. Let $\theta_g=(\alpha_g,\beta_g,\gamma_g')'$ be the coefficient vector of group $g$, let $Z_g$ be its matrix of regressors, with $Z_g'Z_g$ invertible, and let $\widehat\theta_g$ be its OLS estimate. Suppose that the $\theta_g$ are independent with mean $\mu$ and covariance matrix $S$, independently of the regressors and the errors, and that the errors are uncorrelated within and across groups, with mean zero and variance $\sigma_g^2$ in group $g$ given the regressors. Then $\widehat\theta_g$ has the covariance matrix $\Omega_g=S+\sigma_g^2(Z_g'Z_g)^{-1}$ given the regressors. Let $P_g=\Omega_g^{-1}$. The generalized least squares estimator minimizes $\sum_g(\widehat\theta_g-m)'P_g(\widehat\theta_g-m)$ over $m$, and I obtain

Generalized least squares on the original observations gives the same estimate. Let $e_\beta$ denote the coordinate vector of the treatment. The treatment coordinate of $\widehat\mu$ is the pooled treatment estimate. For given scalars $a_g$, it equals $\sum_ga_g\widehat\beta_g$ for all values of the $\widehat\theta_g$ if and only if $e_\beta'A_g=a_ge_\beta'$ for every $g$.

The counterexample has two groups of three observations, an intercept and a binary treatment, no covariates, $S=I$ and $\sigma_1^2=\sigma_2^2=1$.

The treatment estimates of both groups are zero, and every combination of them with scalar weights is zero. The pooled estimate of the mean coefficient vector is $(68/61,-10/61)'$, and the pooled treatment estimate is $-10/61$. The sample has positive probability under a correctly specified model in which the treatment is assigned at random and the treatment coefficients are independent draws centered on $\bar\beta=0$; under one such model an event producing this sample has probability $19683/134217728$. If each row of the two designs is repeated $r$ times, the pooled treatment estimate is $(-4r-6)/(16r^2+36r+9)$, which is nonzero for every $r$ and tends to zero as $r$ grows.

What survives

Expression (7), Theorem A.1, the comparison with the leverage statistic in Appendix B and the weights computed for the two applications concern OLS and do not use this passage. The distinction in footnote 17 between an average over groups and an average over units is unaffected, and the counterexample does not contradict the consistency of the estimator under a correctly specified model with exogenous treatment: with known covariance matrices its conditional mean is $\mu$, because $\sum_gA_g=I$. The warning that random coefficient models do not by themselves return a representative average of unit-level effects is not refuted. What changes is the account of the effective sample for these models on p. 257, which the introduction previews on p. 252. The effective sample is the stated combination of precision weights and within-group regression weights only under the condition given below, or approximately when the samples within groups are large. No later formula, table or figure uses that account. The check treats the covariance matrices as known and does not address versions of the estimator that estimate them.

The problem

The sentence says that one Taylor expansion of $f(a)=1/[1+\exp(-a)]$, centered at the logit of the mean of $Y_i$, can be carried to any accuracy by taking more terms. That holds only at linear predictors inside the radius of convergence of the series. Since $f$ has poles at the odd multiples of $i\pi$, its Taylor series about a real center $c$ has radius $\sqrt{c^2+\pi^2}$, which equals $\pi$ when $c=0$. A logistic regression can have a design of full rank and a maximum likelihood estimate that is unique and finite, and still have fitted linear predictors that lie farther than this from the center. At such predictors the Taylor polynomials do not approach $f$ as the degree grows. The argument that follows the sentence, and Appendix C, use the first-order term only.

Evidence an exact counterexample

The evidence is an exact counterexample together with a derivation that covers every degree. Consider a logistic regression on an intercept and a binary treatment with 200 observations: 100 have $D_i=0$, of which 1 has $Y_i=1$, and 100 have $D_i=1$, of which 99 have $Y_i=1$. Because both outcomes occur in both cells, the maximum likelihood estimate is unique and finite. It equals $\hat\alpha_\ell=-\log 99$ and $\hat\beta_\ell=2\log 99$, with fitted probabilities $1/100$ and $99/100$ and fitted linear predictors $-\log 99$ and $\log 99$. The mean of $Y_i$ is $1/2$. Then the center of the expansion is $\mathrm{logit}(1/2)=0$ and the radius of convergence is $\pi$. Since $e<11/4$ and $(11/4)^9=2357947691/262144<9801=99^2$, I have $\log 99>9/2>4>\pi$. Both fitted linear predictors lie outside the disk of convergence.

Further, the error is bounded below at every degree. The Taylor series of $f$ at $0$ is

with $b_1=1/4$ and $b_2=1/48$. By the Cauchy–Schwarz inequality, $b_m^2\le b_{m-1}b_{m+1}$ for $m\ge 2$, which implies that $b_{m+1}/b_m\ge b_2/b_1=1/12$ for all $m\ge 1$. For $a\ge 9/2$ the terms $t_m=b_ma^{2m-1}$ therefore satisfy $t_{m+1}/t_m\ge 27/16$. Pairing consecutive terms, I find that at such $a$ every Taylor polynomial with an odd number of nonconstant terms is at least $1/2+a/4\ge 13/8$, and every one with an even number is at most $1/2+a/4-a^3/48\le -35/128$. The fitted probability at $a=\log 99$ is $99/100$, and the constant polynomial equals $1/2$. Hence every Taylor polynomial $T_N$ of $f$ at $0$, of every degree $N\ge 0$, satisfies

and by the symmetry $f(-a)=1-f(a)$ the same bound holds at $-\log 99$. Repeating the data at each value of a binary control $X_i\in\{0,1\}$, I obtain 400 observations, a control coefficient of zero, and the same failure.

What survives

The first-order step of Appendix C is a formal linearization and does not require the series to converge; how accurate it is for given data depends on how far the fitted linear predictors are from the center, which I did not quantify. The OLS weighting result and its derivation in Appendix A, the comparison with leverage in Appendix B and Figure B1, and the two applications, which are fitted by OLS, do not use the sentence. What is lost is the unrestricted promise of arbitrary precision for the nonlinear extension on pp. 256–257, to which the introduction refers on p. 252. The counterexample concerns the single expansion that the sentence specifies; it says nothing about approximations with other centers or by other methods.

The problem

The coefficient of $a - \operatorname{logit}(P_Y)$ in the expansion is the derivative of $\operatorname{logit}^{-1}$ at $\operatorname{logit}(P_Y)$. The printed fraction is not this derivative, because its denominator has $\operatorname{logit}(P_Y)$ where $\exp[\operatorname{logit}(P_Y)]$ belongs. At $P_Y = 1/2$ the printed fraction equals $1$, and the derivative equals $1/4$. The last bracket prints $P_y$ for $P_Y$. The constants $b$ and $c$ that follow the display are those of the correct expansion, and both slips are confined to the display.

Evidence a derivation

The evidence is a derivation and a check of the page. Let $g(a) = \operatorname{logit}^{-1}(a) = [1 + \exp(-a)]^{-1}$. Differentiating, I have