REVIEW 2 major objections 4 minor 35 references
Placebo Discontinuity Design
T0 review · 2 major / 4 minor · reviewed 2026-08-06 · deepseek-v4-flash
Pith's one-line read A placebo outcome and placebo treatment let regression discontinuity designs recover treatment effects even when the running variable is strategically manipulated.
desk verdict A novel RDD correction via placebo outcomes, but the causal estimand conditions on a null event without a side convention—fix the estimand and it's a solid paper. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The confounding bridge $h(d-d^*, w)$, defined by $\mathbb{E}[Y\mid D=d,U] = \int h(d-d^*, w)\,dP(w\mid D=d,U)$, is the object that carries the argument: it converts unobserved confounding into an integral equation in observed quantities once the placebo treatment $Z$ is used as an instrument. The local instrumental variable estimator approximates the bridge as partially linear, $h(d-d^*,w) = g(d-d^*) + w^{\top}\gamma$, and the finite-sample equivalence $\hat{\tau}_{\mathrm{pdd}} = \hat{\tau}^{y}_{\mathrm{rdd}} - (\hat{\tau}^{w}_{\mathrm{rdd}})^{\top}\hat{\gamma}_{-}$ shows exactly how the placebo outcome's discontinuity adjusts the standard RDD estimate.
What would settle it
In a Monte Carlo study with a known treatment effect, a continuous unobserved confounder, and only a scalar placebo treatment, the completeness condition fails; if the PDD estimate does not concentrate on the true $\tau_0$ while a bounds approach still covers it, the paper's point-identification claim is refuted. Separately, estimating the bridge nonparametrically and testing the partial-linearity restriction would reveal whether $\hat{\tau}_{\mathrm{pdd}}$'s limit is $\tau_0$ or a best-linear approximation.
Extended reading notes
Core claim
The central claim is that the RDD parameter $\tau_0 = \mathbb{E}[Y(1,D,U,\eta_y)-Y(0,D,U,\eta_y)\mid D=d^*]$ is identified even when the distribution of unobserved confounders $U$ jumps at the cutoff, provided expected potential outcomes are continuous conditional on $U$, a placebo outcome $W$ and placebo treatment $Z$ satisfy exclusion and selection conditions, and a confounding bridge $h$ exists and is unique. Theorem 1 shows $\tau_0$ equals the limiting difference of integrals of $h_+$ and $h_-$ against the placebo-outcome distribution at the cutoff, divided by the discontinuity in treatment probability; with completeness (Assumption 6) the bridge is identified from observed data. The proposed local instrumental variable estimator is numerically equal to the standard RDD estimator for the outcome minus the product of the placebo-outcome RDD discontinuity and a weight $\hat{\gamma}_{-}$; the paper proves consistency and, after robust bias correction, asymptotic normality at rate $n^{-2/5}$ at MSE-optimal bandwidths.
Load-bearing premise
The load-bearing premise is that no nonzero function of the unobserved confounder $U$ has conditional mean zero given the running variable and placebo treatment; if this completeness condition fails, multiple bridge functions solve the observed equations and $\tau_0$ is not point identified.
Editorial extensions
If this is right
- Researchers who currently report placebo-outcome discontinuities as evidence that an RDD is invalid can now use those same discontinuities in an adjustment term that recovers the treatment effect.
- When the placebo outcome shows no jump at the cutoff, the estimator numerically reduces to the standard RDD estimate, so the method nests conventional practice.
- The target parameter is the treatment effect on the treated at the cutoff; only the left limit is adjusted, so the estimator does not require estimating a bridge on both sides symmetrically.
- Robust bias-corrected inference is valid at MSE-optimal bandwidths, so confidence intervals can be constructed without undersmoothing.
Reading between the lines
- The point-identification claim rests on a completeness condition that is untestable and will often fail when $U$ is continuous and $Z$ is scalar; in such settings the method likely degrades to partial identification, so its practical scope is narrower than the title suggests.
- The consistency and normal-theory results require the confounding bridge to be exactly partially linear in $W$; if the bridge is nonlinear, the estimator converges to a best-linear approximation of the bridge, and the gap to $\tau_0$ is not recovered by the stated theory.
- Because $\hat{\gamma}_{-}$ is a ratio of covariances, a placebo outcome only weakly related to $U$ will inflate the variance of the adjustment term, mirroring weak-instrument behavior in IV; researchers should report the first-stage strength of $Z$ for $W$.
- The same placebo-corrected template could be applied to other discontinuity designs, such as kink designs or multi-cutoff settings, whenever a pre-treatment proxy for the confounding variable is available.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper proposes a placebo discontinuity design (PDD) for regression discontinuity settings in which the standard continuity of potential outcomes fails because agents manipulate the running variable. The structural model includes an unobserved confounder U, a placebo outcome W, and a placebo treatment Z. Under conditional independence, exclusion, a confounding bridge, and a completeness condition, Theorem 1 expresses the RDD effect tau0 as the limiting difference of bridge integrals divided by the treatment-probability discontinuity. A local linear instrumental-variable estimator is proposed; Proposition 1 shows it equals the standard RDD estimator minus a placebo-discontinuity adjustment term, and Sections 5 and 6 establish consistency and bias-corrected asymptotic normality under a partial-linearity restriction on the bridge. Appendices B and C provide relaxations of the placebo-exogeneity and treatment-assignment assumptions.
Significance. If the identification chain is sound, the PDD is a meaningful extension of the RDD toolkit: it converts placebo tests from diagnostics into corrections, offers point identification in a setting where Gerard et al. (2020) only obtain bounds, and the estimator decomposition is transparent and practically interpretable. The paper is also careful in other respects: the identification lemmas are detailed, the partial-linearity limitation is acknowledged in Section 5.1, and the bias-corrected inference is worked out at the usual RDD rate. The principal caveat, developed below, is that the central formula conditions on D=d* and uses P(w|D=d*) even though the model explicitly permits a discontinuous running-variable density; without a precise definition of this conditional object, the identified quantity is not well-defined.
major comments (2)
- [Sections 3.1-3.3, esp. Lemma 2 and Theorem 1] The estimand tau0 is defined as E[Y(1,D,U,eta_y)-Y(0,D,U,eta_y)|D=d*], and Lemma 2 and Theorem 1 repeatedly use E[.|D=d*] and dP(w|D=d*). The model explicitly allows f(u,eta|d) to be discontinuous at d*, and D is assumed to have a density that is positive but not necessarily continuous. Under a continuous density, {D=d*} is already a null event and the conditional distribution requires a specified version; with a discontinuity, the one-sided limiting conditional distributions generally differ, so the right-hand side of Theorem 1 can take different values depending on whether P(w|D=d*) is read as the right limit, the left limit, or a mixture. The paper does not state a convention. This is not a presentation issue: without defining the estimand through one-sided limits or imposing continuity of the conditioning distribution, the claimed identification is for an object that is not well-defined. I recommend defining tau0 and all bridge integrals in Theorem 1 using explicit one-sided conditional distributions, for example the right-continuous version, and adjusting Lemma 2 and the proof of Theorem 1 accordingly.
- [Proposition 2 and Corollary 1] Proposition 2 characterizes the probability limit of the estimator as tau_y_rdd - (tau_w_rdd)^T gamma_-, where gamma_- is a best local linear approximation. Corollary 1 then asserts consistency for tau0 under equation (1), exact partial linearity of the bridge. The paper does flag this limitation, but the abstract and introduction state consistency without the qualifier, and Section 4.1 introduces the partial-linear class as an approximation rather than a substantive restriction. I recommend that the paper state prominently in the abstract and introduction that the estimator is consistent for tau0 only when the confounding bridge is exactly partially linear in W; otherwise it targets a different, best-linear-approximation object.
minor comments (4)
- [Figure 2 caption] The notes to Figure 2 say 'Figure 1 illustrates the relaxed assumption'; this should refer to Figure 2.
- [Remark 2] The alternative estimand formula for \tilde tau_pdd contains a bare W in the adjustment terms; please clarify whether W denotes the random vector, a sample mean, or an expectation, and define it before use.
- [Theorem B.1] In the sharp-design case of Theorem B.1 the formula keeps the limits as eps down to 0, while the preceding display in the fuzzy case drops them; please reconcile the two displays and state the sharp-design formula consistently.
- [Assumption 7d] The bound M in Assumption 7d is stated as existing for all d in D(epsilon); please state explicitly that M and zeta are uniform in d, which is the form used in the Lyapunov verification in the proof of Theorem 2.
Circularity Check
No significant circularity: the identification and estimation chain is self-contained, with no fitted parameter renamed as a prediction and no load-bearing self-citation.
full rationale
The derivation chain is self-contained. The target tau0 is expressed via Lemma 2 as an average over U of RDD discontinuities conditional on U; Assumption 5 introduces confounding bridge functions h± whose existence is a substantive structural assumption; Lemma 3 shows that any such bridge satisfies an observed integral equation; Lemma 4, together with the completeness condition (Assumption 6, an external rank condition attributed to Newey and Powell and Andrews, not to the authors), recovers uniqueness of the bridge; and Theorem 1 then substitutes the bridge representation into Lemma 2. No equation in the paper sets tau0 equal to a fitted value by construction, and the final estimator is a consistent estimator of the identified functional rather than a prediction extracted from a fitted parameter. The placebo discontinuity is multiplied by a weight gamma_- that is identified from the conditional moment of Lemma 3; it is not forced to equal the RDD estimate. The only notable concern is well-posedness, not circularity: because the model permits f(u,eta_y|d) to be discontinuous at d*, the conditioning event {D=d*} is a null event and P(w|D=d*) in Theorem 1 is version-dependent; this is an identifiability/definitional issue rather than circular reasoning. No self-citations are load-bearing, and the paper explicitly acknowledges the main limitation that consistency requires the partially linear bridge specification in equation (1).
Assumptions & free parameters
free parameters (3)
- gamma_+ and gamma_- (confounding bridge slope coefficients) =
estimated by local IV regression
- g_+(0) and g_-(0) (intercepts of the bridge) =
estimated by local IV regression
- bandwidths h_n and b_n =
data-driven, not fully specified
assumptions (7)
- domain assumption Z is independent of (eta_y, eta_a) given (U,D), eta_w is independent of (D,Z) given U, and Z affects Y only through D; W is only a function of (U, eta_w).
- domain assumption eta_a is independent of (eta_y, U) given D.
- domain assumption E[Y(a,D,U,eta_y)|D=d,U=u] is continuous at d* for all u.
- domain assumption For each side of the cutoff, E[Y|D=d,U] = integral h+/-(d-d*,w) dP(w|D=d,U) for some bounded h.
- domain assumption E{f(U)|D=d,Z=z}=0 for all z implies f(U)=0 almost surely.
- ad hoc to paper h+/-(d-d*,w) = g+/-(d-d*) + w^T gamma+/-.
- standard math Standard smoothness, bounded moments, positive density, and kernel conditions hold.
Cite this review
Pith. "Pith review of Placebo Discontinuity Design." pith.science (2026). https://pith.science/paper/UB5MNY6R
@misc{pith2026250712693,
author = {Pith},
title = {Pith review of: Placebo Discontinuity Design},
year = {2026},
howpublished = {\url{https://pith.science/paper/UB5MNY6R}},
note = {Machine review of arXiv:2507.12693}
}
read the original abstract
Standard regression discontinuity design (RDD) models rely on the continuity of expected potential outcomes at the cutoff. The standard continuity assumption can be violated by strategic manipulation of the running variable, which is realistic when the cutoff is widely known and when the treatment of interest is a social program or government benefit. In this work, we identify the treatment effect despite such a violation, by leveraging a placebo treatment and a placebo outcome. We introduce a local instrumental variable estimator. Our estimator decomposes into two terms: the standard RDD estimator of the target outcome's discontinuity, and a new adjustment term based on the placebo outcome's discontinuity. We show that our estimator is consistent, and we justify a robust bias-corrected inference procedure. Our method expands the applicability of RDD to settings with strategic behavior around the cutoff, which commonly arise in social science.
Figures
Figures from the paper (2 more)
Reference graph
Works this paper leans on
-
[1]
First, we show that𝐴⊥ ⊥𝑍|𝐷,𝑈. We write, P[𝐴=𝑎|𝐷=𝑑,𝑍=𝑧,𝑈 ]=P[𝐴(𝑑,𝜂 𝑎)=𝑎|𝐷=𝑑,𝑍=𝑧,𝑈 ] =P[𝐴(𝑑,𝜂 𝑎)=𝑎|𝐷=𝑑,𝑈 ], where the first line follows from Assumption 2d (exclusion) and the second line follows from Assumption 2b (placebo selection on unobservables). To show that𝑌⊥ ⊥𝑍|𝐷,𝑈, first note that, P[𝑌=𝑦|𝐷=𝑑,𝑍=𝑧,𝑈 ]=P 𝑌(𝐴,𝑑,𝑈,𝜂 𝑦)=𝑦|𝐷=𝑑,𝑍=𝑧,𝑈 =P 𝑌(𝐴,𝑑,𝑈,𝜂 𝑦)=𝑦|𝐷=𝑑...
-
[2]
To show the second claim (𝑊⊥ ⊥𝐷,𝑍|𝑈), we have that, P[𝑊=𝑤|𝐷=𝑑,𝑍=𝑧,𝑈 ]=P[𝑊(𝑈,𝜂 𝑤)=𝑤|𝐷=𝑑,𝑍=𝑧,𝑈 ] =P[𝑊(𝑈,𝜂 𝑤)=𝑤|𝑈 ], where the first line follows from Assumption 2d (exclusion) and the second line follows from Assumption 2b (placebo selection on unobservables)
-
[3]
Similarly, we prove the last claim (𝐴⊥ ⊥𝑈|𝐷): P[𝐴|𝐷=𝑑,𝑈=𝑢 ]=P[𝐴(𝑑,𝜂 𝑎)|𝐷=𝑑,𝑈=𝑢 ] =P[𝐴(𝑑,𝜂 𝑎)|𝐷=𝑑 ]. 34 where the first line follows from Assumption 2d (exclusion) and the second line follows from Assumption 3b (𝜂𝑎⊥ ⊥𝑈|𝐷). □ A.2 Limit Proof of Lemma 2 (limit).Assume a fuzzy design. We proceed in steps
-
[4]
Notice that under Assumption 2 (placebo variables), we are able to write, 𝑌=𝑌(0,𝐷,𝑈,𝜂 𝑦)+𝐴(𝐷,𝜂 𝑎)𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦)
-
[5]
Then, we have that, E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ] =E 𝑌(0,𝑑 ∗+𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗+𝜖,𝑈 −E 𝑌(0,𝑑 ∗−𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗−𝜖,𝑈 +E 𝐴(𝐷,𝜂 𝑎)𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦) |𝐷=𝑑 ∗+𝜖,𝑈 −E 𝐴(𝐷,𝜂 𝑎)𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦) |𝐷=𝑑 ∗−𝜖,𝑈 (I) =E 𝑌(0,𝑑 ∗+𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗+𝜖,𝑈 −E 𝑌(0,𝑑 ∗−𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗−𝜖,𝑈 +E[𝐴(𝐷,𝜂 𝑎)|𝐷=𝑑 ∗+𝜖,𝑈 ]E 𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗+𝜖,𝑈 −E[𝐴(𝐷,𝜂 𝑎)|𝐷=𝑑 ∗−𝜖,𝑈 ]E...
-
[6]
By Assumptions 1 (RDD) and 4 (continuity), we can take the limit of the equality above to show, lim 𝜖↓0 [E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ]] =lim 𝜖↓0 E 𝑌(0,𝑑 ∗+𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗+𝜖,𝑈 −E 𝑌(0,𝑑 ∗−𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗−𝜖,𝑈 +lim 𝜖↓0 [E[𝐴(𝐷,𝜂 𝑎)|𝐷=𝑑 ∗+𝜖]] lim 𝜖↓0 E 𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗+𝜖,𝑈 −lim 𝜖↓0 [E[𝐴(𝐷,𝜂 𝑎)|𝐷=𝑑 ∗−𝜖]] lim 𝜖↓0 E 𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗−𝜖...
-
[7]
First, note that E ∫ ℎ0(𝑑−𝑑 ∗,𝑤)dP (𝑤|𝐷=𝑑,𝑈 ) 𝐷=𝑑 ′ =E ∫ ℎ0(𝑑−𝑑 ∗,𝑤)dP (𝑤|𝑈 ) 𝐷=𝑑 ′ =E ∫ ℎ0(𝑑−𝑑 ∗,𝑤)dP (𝑤|𝐷=𝑑 ′,𝑈) 𝐷=𝑑 ′ = ∫ ℎ0(𝑑−𝑑 ∗,𝑤)dP (𝑤|𝐷=𝑑 ′), where the first and second line follow from Lemma 1b(𝑊⊥ ⊥𝐷|𝑈)and the last line uses the law of iterated expectations
-
[8]
The limits on the right hand side exist due to Assumption 1 (RDD)
Also recall from the proof of Lemma 2 (limit) we showed that, lim 𝜖↓0 [E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ]] =lim 𝜖↓0 [E[𝐴(𝐷,𝜂 𝑎)|𝐷=𝑑 ∗+𝜖]−E[𝐴(𝐷,𝜂 𝑎)|𝐷=𝑑 ∗−𝜖]] ×E 𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗,𝑈 . The limits on the right hand side exist due to Assumption 1 (RDD). Therefore the limit on the left hand side also exists. 38
Show all 35 references
-
[9]
Therefore, we can write the numerator of the expression from Lemma 2 (limit) as, E lim 𝜖↓0 E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ] 𝐷=𝑑 ∗ =E lim 𝜖↓0 ∫ ℎ+(𝜖,𝑤)dP (𝑤|𝐷=𝑑 ∗+𝜖,𝑈 )− ∫ ℎ−(−𝜖,𝑤)dP (𝑤|𝐷=𝑑 ∗−𝜖,𝑈 ) 𝐷=𝑑 ∗ =lim 𝜖↓0 E ∫ ℎ+(𝜖,𝑤)dP (𝑤|𝐷=𝑑 ∗+𝜖,𝑈 )− ∫ ℎ−(−𝜖,𝑤)dP (𝑤|𝐷=𝑑 ∗−𝜖,𝑈 ) 𝐷=𝑑 ∗ =l...
-
[10]
To begin, note thatE[𝑌|𝑑,𝑈 ] is unique
Finally, we argue that the overall expression is unique. To begin, note thatE[𝑌|𝑑,𝑈 ] is unique. Furthermore, in the proof of Lemma 4 (factuals II), we showed that E[𝑌|𝑑,𝑈 ]= ∫ ℎ0(𝑑−𝑑∗,𝑤)dP(𝑤|𝑑,𝑈)almostsurelybyappealingtoAssumption6, so the latter integral is unique. □ B Relax...
-
[11]
42 The limits on the right hand side exist due to Assumption 1 (RDD)
From the proof of Lemma 2 (limit) we showed that, lim 𝜖↓0 [E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ]] =lim 𝜖↓0 [E[𝐴(𝐷,𝜂 𝑎)|𝐷=𝑑 ∗+𝜖]−E[𝐴(𝐷,𝜂 𝑎)|𝐷=𝑑 ∗−𝜖]] ×E 𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗,𝑈 . 42 The limits on the right hand side exist due to Assumption 1 (RDD). Therefore the limit on th...
-
[12]
We can write the numerator of the expression from Lemma 2 (limit) as, E lim 𝜖↓0 {E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ]} 𝐷=𝑑 ∗ =E lim 𝜖↓0 ∫ ℎ+(𝜖,𝑤)dP (𝑤|𝐷=𝑑 ∗+𝜖,𝑈 )− ∫ ℎ−(−𝜖,𝑤)dP (𝑤|𝐷=𝑑 ∗−𝜖,𝑈 ) 𝐷=𝑑 ∗ =E ∫ ℎ+(0,𝑤)dP (𝑤|𝐷=𝑑 ∗,𝑈) 𝐷=𝑑 ∗ −E ∫ ℎ−(0,𝑤)dP (𝑤|𝐷=𝑑 ∗,𝑈) 𝐷=𝑑 ∗ = ∫ ℎ+(0,𝑤)dP (𝑤|𝐷...
-
[13]
To begin, note thatE[𝑌|𝑑,𝑈 ] is unique
Finally, we argue that the overall expression is unique. To begin, note thatE[𝑌|𝑑,𝑈 ] is unique. Furthermore, in the proof of Lemma 4 (factuals II), we showed that E[𝑌|𝑑,𝑈 ]= ∫ ℎ0(𝑑−𝑑∗,𝑤)dP(𝑤|𝑑,𝑈)almostsurelybyappealingtoAssumption6, so the latter integral is unique. □ C Relax...
-
[14]
Notice that under Assumption C.2 (proxy controls), we are able to write, 𝑌=𝑌(0,𝐷,𝑈,𝜂 𝑦)+𝐴(𝐷,𝑈,𝜂 𝑎)𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦) . 46
-
[15]
Using the decomposition above, we can write E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ] =E 𝑌(0,𝑑 ∗+𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗+𝜖,𝑈 −E 𝑌(0,𝑑 ∗−𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗−𝜖,𝑈 +E 𝐴(𝑑∗+𝜖,𝑈,𝜂 𝑎)𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦) |𝐷=𝑑 ∗+𝜖,𝑈 −E 𝐴(𝑑∗−𝜖,𝑈,𝜂 𝑎)𝑌(1,𝐷,𝑈,𝜂 𝑦)−𝑌(0,𝐷,𝑈,𝜂 𝑦) |𝐷=𝑑 ∗−𝜖,𝑈 =E 𝑌(0,𝑑 ∗+𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗+𝜖,𝑈 −E 𝑌(...
-
[16]
Specifically, to go from the second line to the third line, the initial terms cancel, and the later terms factorize
By Assumptions C.1 (RDD) and 4 (continuity), we can take the limit of the equality above to show, lim 𝜖↓0 [E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ]] =lim 𝜖↓0 E 𝑌(0,𝑑 ∗+𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗+𝜖,𝑈 −E 𝑌(0,𝑑 ∗−𝜖,𝑈,𝜂 𝑦)|𝐷=𝑑 ∗−𝜖,𝑈 +𝜏 0 lim 𝜖↓0 [E[𝐴(𝐷,𝑈,𝜂 𝑎)|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝐴(𝐷,𝑈,𝜂 𝑎)|𝐷=𝑑 ∗−𝜖,𝑈 ]] =𝜏 0...
-
[17]
Rearranging and gives 𝜏0= E lim𝜖↓0[E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ]]|𝐷=𝑑 ∗ E lim𝜖↓0[E[𝐴|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝐴|𝐷=𝑑 ∗−𝜖,𝑈 ]]|𝐷=𝑑 ∗
Taking conditional expectations of both sides, we have that E lim 𝜖↓0 [E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ]]|𝐷=𝑑 ∗ =𝜏 0E lim 𝜖↓0 [E[𝐴(𝐷,𝑈,𝜂 𝑎)|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝐴(𝐷,𝑈,𝜂 𝑎)|𝐷=𝑑 ∗−𝜖,𝑈 ]]|𝐷=𝑑 ∗ . Rearranging and gives 𝜏0= E lim𝜖↓0[E[𝑌|𝐷=𝑑 ∗+𝜖,𝑈 ]−E[𝑌|𝐷=𝑑 ∗−𝜖,𝑈 ]]|𝐷=𝑑 ∗ E lim𝜖↓0[E[𝐴|𝐷=𝑑 ∗+...
2001
-
[18]
Using the formula for the inverse of a block matrix, ˆ𝜈+= R⊤ 1 K+R1 R⊤ 1 K+W Z⊤K+R1 Z⊤K+W −1 R⊤ 1 K+Y Z⊤K+Y = (R⊤ 1 K+R1)−1+(R⊤ 1 K+R1)−1R⊤ 1 K+W Q−1 + Z⊤K+R1(R⊤ 1 K+R1)−1 −(R⊤ 1 K+R1)−1R⊤ 1 K+W Q−1 + −Q−1 + Z⊤K+R1(R⊤ 1 K+R1)−1 Q−1 + ...
-
[19]
We can go through the same steps to show thatˆ𝛼−= ˆ𝛽𝑦 −− ˆ𝛽𝑤 − ˆ𝛾−
-
[20]
Plugging these expressions into our estimator forˆ𝜏pdd in Equation (3), we find that, ˆ𝜏pdd= ˆ𝛼+,0+ ˆ𝛽𝑤 +,0 ⊤ ˆ𝛾+− ˆ𝛼−,0+ ˆ𝛽𝑤 +,0 ⊤ ˆ𝛾− = ˆ𝛽𝑦 +,0− ˆ𝛽𝑤 +,0 ⊤ ˆ𝛾++ ˆ𝛽𝑤 +,0 ⊤ ˆ𝛾+− ˆ𝛽𝑦 −,0− ˆ𝛽𝑤 −,0 ⊤ ˆ𝛾−+ ˆ𝛽𝑤 +,0 ⊤ ˆ𝛾− = ˆ𝛽𝑦 +,0− ˆ𝛽𝑦 −,0− ˆ𝛽𝑤 +,0− ˆ𝛽𝑤 −,0 ⊤ ˆ𝛾− = ˆ𝜏𝑦 rdd− ˆ𝜏𝑤 rdd ⊤ ˆ𝛾−
-
[21]
𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖 𝑊⊤ 𝑖 −𝑅⊤ 𝑖,1(R⊤ 1 K−R1)−1R⊤ 1 K−W #−1
Finally, we show the claim thatˆ𝛾− = 1 𝑛 Í𝑛 𝑖=1𝜔𝑖,−{𝑍𝑖(𝑊⊥ 𝑖)⊤} −1 1 𝑛 Í𝑛 𝑖=1𝜔𝑖,−(𝑍𝑖𝑌⊥ 𝑖) . 49 Expanding out our expression forˆ𝛾−, we have that, ˆ𝛾−=Q−1 − Z⊤K−{I−R 1(R⊤ 1 K−R1)−1R⊤ 1 K−}Y = Z⊤K−{I−R 1(R⊤ 1 K−R1)−1R⊤ 1 K−}W −1 Z⊤K−{I−R 1(R⊤ 1 K−R1)−1R⊤ 1 K−}Y = " 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖...
-
[22]
Standard results from e.g
By Proposition 1, ˆ𝜏pdd= ˆ𝛽𝑦 +,0− ˆ𝛽𝑦 −,0− ˆ𝛽𝑤 +,0− ˆ𝛽𝑤 −,0 ⊤ ˆ𝛾−. Standard results from e.g. Hahn et al. (2001, Section 4.1) show that under our conditions, ˆ𝛽𝑦 +,0− ˆ𝛽𝑦 −,0 𝑝 →lim 𝜖↓0 {E[𝑌|𝐷=𝑑 ∗+𝜖]−E[𝑌|𝐷=𝑑 ∗−𝜖]} ˆ𝛽𝑤 +,0− ˆ𝛽𝑤 −,0 𝑝 →lim 𝜖↓0 {E[𝑊|𝐷=𝑑 ∗+𝜖]−E[𝑊|𝐷=𝑑 ∗−𝜖]}. Theref...
2001
-
[23]
Using Lemma E.1 (Q−-limit) and the continuous mapping theorem, we can write that, 𝑛Q−1 − 𝑝 →𝜅−1 lim 𝜖↓0 Cov𝑍𝑖,𝑊⊤ 𝑖 |𝐷=𝑑 ∗−𝜖 −1, 𝜅=𝑓 𝑑,+(𝑑∗) ∫ ∞ 0 𝐾(𝑢)d𝑢
Within the proof of Proposition 1, we have shown that for Q−=Z⊤K− I−R 1(R⊤ 1 K−R1)−1R⊤ 1 K− W, ˆ𝛾−=Q−1 − Z⊤K−{I−R 1(R⊤ 1 K−R1)−1R⊤ 1 K−}Y. Using Lemma E.1 (Q−-limit) and the continuous mapping theorem, we can write that, 𝑛Q−1 − 𝑝 →𝜅−1 lim 𝜖↓0 Cov𝑍𝑖,𝑊⊤ 𝑖 |𝐷=𝑑 ∗−𝜖 −1, 𝜅=𝑓 𝑑,+(𝑑∗...
-
[24]
We verify the claim that𝛾− is the best linear predictor for 𝑌=𝑐+𝑊 ⊤𝛾−+𝜂,lim 𝜖↓0 E[𝜂𝑍𝑖|𝐷 𝑖 =𝑑∗−𝜖]=0,lim 𝜖↓0 E[𝜂|𝐷 𝑖 =𝑑∗−𝜖]=0. Substituting for𝑌in the probability limit gives lim 𝜖↓0 Cov𝑍𝑖,𝑊⊤ 𝑖 |𝐷 𝑖 =𝑑∗−𝜖 −1Cov(𝑍𝑖,𝑌𝑖|𝐷 𝑖 =𝑑∗−𝜖) =lim 𝜖↓0 Cov𝑍𝑖,𝑊⊤ 𝑖 |𝐷 𝑖 =𝑑∗−𝜖 −1 Cov𝑍𝑖,𝑐+𝑊 ⊤ 𝑖 𝛾−+...
-
[25]
Substituting inℎ+(𝑑−𝑑 ∗,𝑤)from Equation (1), lim 𝜖↓0 E[ℎ+(𝜖,𝑊)|𝐷=𝑑 ∗]=𝑔+(0)+E 𝑊⊤|𝐷=𝑑 ∗ 𝛾+ =𝛼+,0+ 𝛽𝑤 +,0 ⊤ 𝛾+
We express the limit as a sum. Substituting inℎ+(𝑑−𝑑 ∗,𝑤)from Equation (1), lim 𝜖↓0 E[ℎ+(𝜖,𝑊)|𝐷=𝑑 ∗]=𝑔+(0)+E 𝑊⊤|𝐷=𝑑 ∗ 𝛾+ =𝛼+,0+ 𝛽𝑤 +,0 ⊤ 𝛾+
-
[26]
By Lemma 3 (factuals) and Equation (1), for all𝑑∈D+(𝜖)recall that we can write, E[𝑌−ℎ +(𝐷−𝑑 ∗,𝑊)|𝐷=𝑑,𝑍=𝑧 ]=E 𝑌−𝑔 +(𝐷𝑖−𝑑∗)−𝑊 ⊤𝛾+|𝐷=𝑑,𝑍=𝑧 =0
It is helpful to characterize a partially linear representation for𝑌. By Lemma 3 (factuals) and Equation (1), for all𝑑∈D+(𝜖)recall that we can write, E[𝑌−ℎ +(𝐷−𝑑 ∗,𝑊)|𝐷=𝑑,𝑍=𝑧 ]=E 𝑌−𝑔 +(𝐷𝑖−𝑑∗)−𝑊 ⊤𝛾+|𝐷=𝑑,𝑍=𝑧 =0. Using this moment condition, it follows that for some𝛿∈[0,𝐷−𝑑 ∗]we ...
-
[27]
Applying the conditional expectationE[·|𝑍,𝐷=𝑑 ∗+𝜖] on both sides, E[𝑌|𝑍,𝐷=𝑑 ∗+𝜖]=𝛼+,0+𝛼+,1𝜖+E 𝑊⊤|𝑍,𝐷=𝑑 ∗+𝜖 𝛾++ 1 2𝑔(2) + (𝛿)𝜖 2
Using the representation in step 2, we now show that𝛼+,0=𝛽 𝑦 +,0−(𝛽 𝑤 +,0)⊤𝛾+. Applying the conditional expectationE[·|𝑍,𝐷=𝑑 ∗+𝜖] on both sides, E[𝑌|𝑍,𝐷=𝑑 ∗+𝜖]=𝛼+,0+𝛼+,1𝜖+E 𝑊⊤|𝑍,𝐷=𝑑 ∗+𝜖 𝛾++ 1 2𝑔(2) + (𝛿)𝜖 2. Further taking the conditional expectationE[·|𝐷=𝑑 ∗+𝜖] on both sides ...
-
[28]
Next, we show thatˆ𝛾+ 𝑝 →𝛾+. As argued in the proof of Proposition 2 (estimand) we have that, ˆ𝛾+ 𝑝 →lim 𝜖↓0 Cov𝑍𝑖,𝑊⊤ 𝑖 |𝐷=𝑑 ∗+𝜖 −1Cov(𝑍𝑖,𝑌𝑖|𝐷 𝑖 =𝑑∗+𝜖) =lim 𝜖↓0 Cov𝑍𝑖,𝑊⊤ 𝑖 |𝐷=𝑑 ∗+𝜖 −1Cov 𝑍𝑖,𝛼+,0+𝛼+,1(𝐷𝑖−𝑑∗)+𝑊 ⊤ 𝑖 𝛾++ ˜𝜂+ 1 2𝑔(2) + (𝛿)(𝐷 𝑖−𝑑∗)2|𝐷 𝑖 =𝑑∗+𝜖 =lim 𝜖↓0 Cov𝑍𝑖,𝑊⊤ 𝑖 |𝐷=...
-
[29]
In the proof of Proposition 1 (estimator decomposition), we showed that ˆ𝛼+,0 = ˆ𝛽𝑦 +,0−( ˆ𝛽𝑤 +,0)⊤ ˆ𝛾+
We also show thatˆ𝛼+,0 𝑝 →𝛼+,0. In the proof of Proposition 1 (estimator decomposition), we showed that ˆ𝛼+,0 = ˆ𝛽𝑦 +,0−( ˆ𝛽𝑤 +,0)⊤ ˆ𝛾+. In step 3, we showed that𝛼+,0=𝛽 𝑦 +,0−(𝛽 𝑤 +,0)⊤𝛾+. Recall that ˆ𝛽𝑦 +,0 𝑝 →𝛽 𝑦 +,0 and ˆ𝛽𝑤 +,0 𝑝 →𝛽 𝑤 +,0 by standard arguments (Hahn et al....
2001
-
[30]
□ E.2 Technical lemma Lemma E.1(Q −-limit).Let Assumptions 7a-c and 8 hold, and assume thatℎ→0and 𝑛ℎ→∞
Collecting results and applying the continuous mapping theorem we have shown that, ˆ𝛼+,0+( ˆ𝛽𝑤 +,0)⊤ ˆ𝛾+ 𝑝 →𝛼+,0+(𝛽 𝑤 +,0)⊤𝛾+ =lim 𝜖↓0 E[ℎ+(𝜖,𝑊)|𝐷=𝑑 ∗] where the convergence uses steps 4 and 5, and the equality uses step 1. □ E.2 Technical lemma Lemma E.1(Q −-limit).Let Assump...
-
[31]
As shown in the proof of Proposition 1, Q−=Z⊤K− I−R 1(R⊤ 1 K−R1)−1R⊤ 1 K− W. Therefore, we have the decomposition 1 𝑛 Q−= 1 𝑛 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖 𝑊⊤ 𝑖 −𝑅⊤ 𝑖,1(R⊤ 1 K−R1)−1R⊤ 1 K−W = 1 𝑛 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖 𝑊⊤ 𝑖 −e⊤ 0(R⊤ 1 K−R1)−1R⊤ 1 K−W − 1 𝑛 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖 𝐷𝑖−𝑑∗ ℎ𝑛 e⊤ 1(R⊤ 1 K−R1)−1R⊤...
-
[32]
(2001, Section 4.1) imply that e⊤ 0(R⊤ 1 K−R1)−1R⊤ 1 K−W 𝑝 →(𝛽 𝑤 −,0)⊤=lim 𝜖↓0 E 𝑊⊤ 𝑖 |𝐷 𝑖 =𝑑∗−𝜖 54 1 ℎ𝑛 e⊤ 1(R⊤ 1 K−R1)−1R⊤ 1 K−W 𝑝 →(𝛽 𝑤 −,1)⊤
Under Assumptions 7a and 8, results from Hahn et al. (2001, Section 4.1) imply that e⊤ 0(R⊤ 1 K−R1)−1R⊤ 1 K−W 𝑝 →(𝛽 𝑤 −,0)⊤=lim 𝜖↓0 E 𝑊⊤ 𝑖 |𝐷 𝑖 =𝑑∗−𝜖 54 1 ℎ𝑛 e⊤ 1(R⊤ 1 K−R1)−1R⊤ 1 K−W 𝑝 →(𝛽 𝑤 −,1)⊤
2001
-
[33]
SetX 𝑖 = 1 ℎ𝑛 1 (𝐷𝑖 <𝑑∗)𝐾 𝐷𝑖−𝑑∗ ℎ𝑛 𝑍𝑖(𝑊𝑖−𝛽 𝑤 −,0)⊤ ∈R 𝑞×𝑞
Here, we show that, 1 𝑛ℎ𝑛 𝑛∑︁ 𝑖=1 1 (𝐷𝑖 <𝑑∗)𝐾 𝐷𝑖−𝑑∗ ℎ𝑛 𝑍𝑖(𝑊𝑖−𝛽 𝑤 −,0)⊤ ! = 1 ℎ𝑛 E 1 (𝐷𝑖 <𝑑∗)𝐾 𝐷𝑖−𝑑∗ ℎ𝑛 𝑍𝑖(𝑊𝑖−𝛽 𝑤 −,0)⊤ +𝑜 𝑝(1). SetX 𝑖 = 1 ℎ𝑛 1 (𝐷𝑖 <𝑑∗)𝐾 𝐷𝑖−𝑑∗ ℎ𝑛 𝑍𝑖(𝑊𝑖−𝛽 𝑤 −,0)⊤ ∈R 𝑞×𝑞. For each componentℓ∈ {(1,1),...,(𝑞,𝑞)}, we show 1 𝑛 Í 𝑖𝑋𝑖,ℓ =E 𝑋𝑖,ℓ +𝑜 𝑝(1) by boundingVar...
-
[34]
We will show that 1 𝑛 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖 𝑊⊤ 𝑖 −e⊤ 0(R⊤ 1 K−R1)−1R⊤ 1 K−W 𝑝 →𝜅lim 𝜖↓0 Cov𝑍𝑖,𝑊⊤ 𝑖 |𝐷 𝑖 =𝑑∗−𝜖
Consider the first term in the decomposition. We will show that 1 𝑛 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖 𝑊⊤ 𝑖 −e⊤ 0(R⊤ 1 K−R1)−1R⊤ 1 K−W 𝑝 →𝜅lim 𝜖↓0 Cov𝑍𝑖,𝑊⊤ 𝑖 |𝐷 𝑖 =𝑑∗−𝜖 . 55 By steps 2 and 3, 1 𝑛 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖 𝑊⊤ 𝑖 −e⊤ 0(R⊤ 1 K−R1)−1R⊤ 1 K−W = 1 𝑛 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖(𝑊𝑖−𝛽 𝑤 −,0)⊤+𝑜 𝑝(1) = 1 𝑛ℎ𝑛 𝑛∑︁ 𝑖...
-
[35]
−”analogues of every object below are defined by replacing each“+
Consider the second term in the decomposition. We will show that 1 𝑛 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖 𝐷𝑖−𝑑∗ ℎ𝑛 e⊤ 1(R⊤ 1 K−R1)−1R⊤ 1 K−W 𝑝 →0. 56 By step 2 and an analagous argument as in step 3, 1 𝑛 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖 𝐷𝑖−𝑑∗ ℎ𝑛 e⊤ 1(R⊤ 1 K−R1)−1R⊤ 1 K−W = 1 𝑛 𝑛∑︁ 𝑖=1 𝜔𝑖,−𝑍𝑖(𝐷𝑖−𝑑∗)(𝛽𝑤 −,1)⊤+𝑜 𝑝(1) =...
2014
Reviewed August 6, 2026 · model on record in the stance chip above.
Discussion (0). Continue with ORCID to comment.