Doubly Robust Estimation of the ATT

Why the ATT Requires a Different Weighting Argument

The average treatment effect on the treated (ATT) asks how much treatment changed outcomes, on average, for the people who actually received treatment. The comparison is therefore not between the overall treated and control groups without adjustment: it is between treated outcomes and the outcomes those same treated people would have experienced without treatment. An influence-function estimator makes that counterfactual comparison using an outcome regression, a propensity score, or both.

Definition: Let $A$ be a binary treatment indicator, with $A=1$ for treated and $A=0$ for untreated. Let $Y(1)$ and $Y(0)$ denote the potential outcomes under treatment and control, and let $X$ be observed pre-treatment covariates. The ATT is $$ \tau_{\mathrm{ATT}}=E[Y(1)-Y(0)\mid A=1]. $$ Define $e(X)=P(A=1\mid X)$, $m_a(X)=E[Y\mid A=a,X]$ for $a\in\{0,1\}$, and $\pi_1=P(A=1)$.

To identify this estimand from observed data, assume consistency, so $Y=Y(A)$; no interference between units; and conditional exchangeability, $(Y(0),Y(1))\perp A\mid X$. For the ATT, there must also be untreated comparison units over the covariate values found among treated units: $P(A=0\mid X)>0$ for covariate values occurring among treated people. For stable estimation, a stronger overlap condition often requires $e(X)$ to be bounded away from 1 on that treated covariate support. Under these assumptions, $m_0(X)$ identifies the treated units’ mean counterfactual outcome under control, while their observed outcomes identify their mean outcome under treatment.

Identification: Under the assumptions above, $$ \tau_{\mathrm{ATT}} =\frac{E[A\{Y-m_0(X)\}]}{\pi_1} =\frac{E[e(X)\{m_1(X)-m_0(X)\}]}{\pi_1}. $$ The first expression compares observed treated outcomes with the control outcome regression averaged over treated units. The second expresses the same contrast as a treatment-effect regression averaged over the treated covariate distribution.

From the Influence Function to an Estimator

The efficient influence function for the ATT in the nonparametric model can be written using the true nuisance functions as

D_{\mathrm{ATT}}(O) = \frac{A}{\pi_1}\{Y-m_0(X)-\tau_{\mathrm{ATT}}\} - \frac{1-A}{\pi_1}\frac{e(X)}{1-e(X)}\{Y-m_0(X)\}.

Here $O=(X,A,Y)$ denotes one observed unit. The first term centers the treated outcome residual relative to the ATT. The second term uses untreated units to correct the estimate of the treated group’s mean control outcome. Its weight is the propensity odds $e(X)/\{1-e(X)\}$: among controls, units with covariates more common in the treated group receive greater weight.

To obtain an estimator, replace the unknown nuisance functions with estimates $\widehat e(X)$ and $\widehat m_0(X)$, and replace $\pi_1$ with $\widehat\pi_1=n_1/n$, where $n_1=\sum_i A_i$. Solving the empirical influence-function equation for the ATT gives the following estimator.

Estimator: Provided $n_1>0$ and the fitted propensity scores are below 1, $$ \widehat\tau_{\mathrm{ATT,DR}} = \frac{1}{n_1}\sum_{i:A_i=1}\{Y_i-\widehat m_0(X_i)\} - \frac{1}{n_1}\sum_{i:A_i=0} \frac{\widehat e(X_i)}{1-\widehat e(X_i)} \{Y_i-\widehat m_0(X_i)\}. $$ The label “DR” refers to double robustness: under suitable regularity conditions, this estimator is consistent if either the control outcome regression or the propensity score is correctly specified.

Fit the nuisance functions.
Estimate the probability of treatment given covariates, $\widehat e(X)$, and the mean control outcome given covariates, $\widehat m_0(X)$. The outcome model needed here is specifically the control regression.

Predict treated counterfactuals.
For each treated unit, subtract its fitted control outcome from its observed outcome. Average these residuals over the $n_1$ treated units.

Correct with weighted control residuals.
For each control, multiply $Y_i-\widehat m_0(X_i)$ by its fitted propensity odds. Subtract the sum of these weighted residuals, using the same denominator $n_1$.

The denominator is the number treated, not the sum of control weights. This reflects the target population: the ATT is an average over treated units. A normalized alternative can be constructed, but it is a different finite-sample estimator and should not be substituted silently into this formula.

Worked Example: Computing the ATT Estimate

Consider a hypothetical sample with three treated units and four controls. Suppose the fitted control outcome predictions and observed outcomes give treated residuals $Y_i-\widehat m_0(X_i)$ of $4,5,4$. For the controls, suppose the residuals are $1,-1,2,0$, and their fitted propensity scores are exactly $1/3,1/2,2/3,3/5$. These are exact values, not rounded approximations.

The corresponding propensity odds are

\frac{1/3}{1-1/3}=\frac12,\qquad \frac{1/2}{1-1/2}=1,\qquad \frac{2/3}{1-2/3}=2,\qquad \frac{3/5}{1-3/5}=\frac32.

The treated residual sum is $4+5+4=13$. The weighted control residual sum is

\frac12(1)+1(-1)+2(2)+\frac32(0) =\frac12-1+4+0 =3.5.

Thus, with $n_1=3$,

\widehat\tau_{\mathrm{ATT,DR}} =\frac{13-3.5}{3} =\frac{9.5}{3} \approx 3.167.

The estimated ATT is about $3.167$ outcome units. The propensity scores were specified as exact fractions, so using their exact odds gives the displayed result without a discrepancy caused by rounding the scores first.

Why Either Correct Nuisance Model Can Suffice

The robustness claim is most transparent by considering the population version of the estimator. Let $q(X)$ be a candidate propensity function, and let $g(X)$ be a candidate control outcome regression. Suppose $q(X)<1$ on the relevant support. The population contrast corresponding to the estimator is

\Psi(q,g) = \frac{1}{\pi_1} E\left[ A\{Y-g(X)\} - (1-A)\frac{q(X)}{1-q(X)}\{Y-g(X)\} \right].

Under consistency and conditional exchangeability, $E[Y\mid A=1,X]=m_1(X)$ and $E[Y\mid A=0,X]=m_0(X)$. Taking conditional expectations therefore yields

\pi_1\Psi(q,g) = E\left[ e(X)\{m_1(X)-g(X)\} - \{1-e(X)\}\frac{q(X)}{1-q(X)}\{m_0(X)-g(X)\} \right].

If $g=m_0$, the second term is zero, regardless of $q$. The first term becomes $E[e(X)\{m_1(X)-m_0(X)\}]=\pi_1\tau_{\mathrm{ATT}}$. If instead $q=e$, then $\{1-e(X)\}q(X)/\{1-q(X)\}=e(X)$. The terms involving $g(X)$ cancel, leaving $E[e(X)\{m_1(X)-m_0(X)\}]=\pi_1\tau_{\mathrm{ATT}}$. Dividing by $\pi_1$ proves that either correctly specified nuisance function is sufficient for consistency of the population contrast.

This is a consistency result, not a guarantee that every finite-sample estimate is accurate. The fitted propensity odds can be highly variable when scores approach 1, and a badly misspecified nuisance model can still produce unstable estimates even when the other model is correct.

Worked Example: Correct Outcome Regression, Misspecified Propensity

Suppose a hypothetical population has two covariate groups, $L$ and $H$, each with probability $0.5$. The true treatment probabilities are $e(L)=0.2$ and $e(H)=0.8$. The true control and treatment outcome means are $m_0(L)=2$, $m_0(H)=6$, $m_1(L)=5$, and $m_1(H)=10$. The control outcome regression is correct, but suppose the fitted propensity function is misspecified as $q(L)=q(H)=0.5$.

The treated prevalence is

\pi_1=0.5(0.2)+0.5(0.8)=0.5.

The treated-weighted effect numerator is

E[e(X)\{m_1(X)-m_0(X)\}] =0.5(0.2)(5-2)+0.5(0.8)(10-6) =0.3+1.6=1.9.

Because $g=m_0$, the control residual has conditional mean zero in each group. The weighted control residual term therefore has expectation zero even though $q$ is wrong. Hence

\Psi(q,m_0)=\frac{1.9-0}{0.5}=3.8.

The ATT is $3.8$ outcome units. The outcome regression alone suffices for the population estimator to recover it in this example.

Worked Example: Correct Propensity, Misspecified Outcome Regression

Use the same hypothetical population, but now suppose the propensity model is correct and the fitted control regression is misspecified as the constant $g(X)=4$. The treated component of the numerator is

E[e(X)\{m_1(X)-g(X)\}] =0.5(0.2)(5-4)+0.5(0.8)(10-4) =0.1+2.4=2.5.

The expected weighted control residual is

E[e(X)\{m_0(X)-g(X)\}] =0.5(0.2)(2-4)+0.5(0.8)(6-4) =-0.2+0.8=0.6.

Subtracting the weighted control residual and dividing by the treated prevalence gives

\Psi(e,g)=\frac{2.5-0.6}{0.5} =\frac{1.9}{0.5}=3.8.

Thus the correct propensity model recovers the ATT even though the fitted control regression is wrong. The cancellation works because correct propensity odds reweight the control covariate distribution to the treated covariate distribution.

Influence-Function Inference and Its Conditions

When both nuisance functions are consistently estimated, the influence function also describes the estimator’s first-order sampling variation under standard regularity conditions. In particular, overlap must control the propensity odds, the influence function must have finite variance, and nuisance estimation must be sufficiently accurate. With flexible nuisance learners, cross-fitting can help control overfitting bias; a common sufficient type of requirement is that the product of the relevant nuisance estimation errors is $o_p(n^{-1/2})$, together with appropriate norm and moment conditions.

A plug-in estimate of the influence function is

\widehat D_i = \frac{A_i}{\widehat\pi_1} \{Y_i-\widehat m_0(X_i)-\widehat\tau_{\mathrm{ATT,DR}}\} - \frac{1-A_i}{\widehat\pi_1} \frac{\widehat e(X_i)}{1-\widehat e(X_i)} \{Y_i-\widehat m_0(X_i)\}.

When the influence-function approximation is justified, an estimated standard error is

\widehat{SE} = \sqrt{\frac{1}{n^2}\sum_{i=1}^n \widehat D_i^2}.

This standard error is in the same units as the outcome and quantifies estimated sampling uncertainty, not uncertainty about whether the identifying assumptions hold. Importantly, double robustness of the point estimate alone does not automatically justify this standard error if one nuisance model converges to an incorrect limit. In that situation, the estimator’s sampling variation may require additional analysis that accounts for nuisance estimation and the particular model-fitting procedure.

Worked Example: Calculating an Influence-Function Standard Error

Consider a separate hypothetical sample of four units, two treated and two controls. Thus $\widehat\pi_1=2/4=0.5$. The treated residuals $Y_i-\widehat m_0(X_i)$ are $5$ and $7$. For the controls, the fitted propensity scores and residuals are respectively $(0.5,1)$ and $(0.25,-3)$.

The control odds are $1$ and $1/3$, so their weighted residual sum is $1(1)+(1/3)(-3)=0$. The estimate is therefore

\widehat\tau_{\mathrm{ATT,DR}} =\frac{5+7-0}{2}=6.

Using this estimate, the two treated influence-function contributions are $2(5-6)=-2$ and $2(7-6)=2$. The two control contributions are $-2(1)(1)=-2$ and $-2(1/3)(-3)=2$. Consequently,

\widehat{SE} = \frac{\sqrt{(-2)^2+2^2+(-2)^2+2^2}}{4} =\frac{4}{4}=1.

The plug-in influence-function standard error is $1$ outcome unit. This very small sample is used only to show the arithmetic; reliable inference also depends on the regularity and nuisance-estimation conditions described above.

Common Pitfalls and Practical Interpretation

Using the wrong weighting direction. ATT weighting gives controls propensity odds $e(X)/\{1-e(X)\}$, not the inverse odds used in some other targets. The weights make the control covariate distribution resemble that of the treated.
Changing the denominator. In the displayed estimator, both sums are divided by $n_1$. Dividing the control correction by the sum of control weights changes the estimator.
Treating double robustness as protection against confounding. The robustness result concerns misspecification of one of two nuisance models. It does not repair unmeasured confounding, violations of consistency, or a lack of untreated comparators for treated covariate profiles.
Ignoring extreme propensity odds. Scores close to one produce large control weights and can make estimates unstable. Overlap and weight behavior should be examined rather than assuming that double robustness prevents instability.
Assuming a robust estimate guarantees robust inference. Consistency when one nuisance model is correct does not, by itself, validate the usual efficient-influence-function standard error under misspecification of the other model.

Key takeaway: The ATT estimator combines treated outcome residuals with a propensity-odds-weighted correction from controls. Under the identifying assumptions, a correct control outcome regression or a correct propensity score is enough for consistency; stable and valid inference requires additional overlap and nuisance-estimation conditions.

Check Your Understanding

Use the definitions and calculations in this tutorial to answer the following questions.

Why does the ATT estimator use the treated sample size $n_1$ as its denominator?
For a control with fitted propensity score $0.75$, what propensity odds enter the weighted residual correction?
In the population robustness argument, why does a correct control outcome regression make the weighted control residual term have expectation zero?
Using the two-group population in the examples, explain why a correct propensity function cancels the terms involving a misspecified control regression.
Give one reason why consistency under a correct nuisance model does not automatically establish the validity of the usual influence-function standard error.