Augmented Estimation of the Average Treatment Effect on the Treated

Why the Treated Population Needs Its Own Estimand

The average treatment effect on the treated (ATT) asks about the average effect among the people who actually received treatment. It is not the same target as the average treatment effect (ATE): the treated population can have a different distribution of covariates, and therefore a different average effect, from the full population. To estimate the ATT, a useful strategy is to compare treated outcomes with a constructed control group that represents the treated group’s covariate distribution.

The augmented estimator makes that comparison in two linked parts. It uses a regression model for the untreated outcome to adjust outcomes for covariates, and it uses propensity-score odds to reweight controls toward the treated population. Under suitable conditions, either of these two nuisance models can be correct for consistency. That guarantee is conditional on the causal assumptions and support needed to identify the target in the first place.

Definition: Let $A\in\{0,1\}$ denote treatment, $X$ measured pretreatment covariates, and $Y(1)$ and $Y(0)$ the potential outcomes under treatment and no treatment. The average treatment effect on the treated is $$ \tau_{\mathrm{ATT}}=E[Y(1)-Y(0)\mid A=1]. $$ Define the propensity score $e(X)=P(A=1\mid X)$ and the untreated-outcome regression $m_0(X)=E[Y\mid A=0,X]$. The observed outcome is $Y=A Y(1)+(1-A)Y(0)$.

Identification and the Augmented Estimator

The key causal assumption for this estimand is conditional exchangeability of the untreated potential outcome: $Y(0)\perp A\mid X$. It says that, within covariate levels, treated and untreated units have the same mean untreated potential outcome. Along with consistency (the observed outcome equals the potential outcome under the treatment actually received) and no interference between units, this identifies the missing untreated mean for treated units as $m_0(X)$.

A further requirement is overlap for the treated: for covariate values occurring among treated units, $e(X)<1$. There must be a positive chance of observing untreated units at those covariate values. The propensity score may be zero in parts of the covariate space with no treated units; those regions are not part of the treated target. For stable estimation, the relevant propensity odds also need to be sufficiently well behaved, for example bounded or integrable.

Given estimates $\hat e(X_i)$ and $\hat m_0(X_i)$, let $n_1=\sum_{i=1}^n A_i$. The augmented estimator considered here is

\hat\tau_{\mathrm{ATT}} = \frac{1}{n_1}\sum_{i=1}^n A_i\{Y_i-\hat m_0(X_i)\} - \frac{1}{n_1}\sum_{i=1}^n(1-A_i) \frac{\hat e(X_i)}{1-\hat e(X_i)} \{Y_i-\hat m_0(X_i)\}.

The first term is the treated group’s average outcome after subtracting its predicted untreated outcome. The second term corrects that contrast using control-group residuals, weighted by the estimated odds of treatment. Those odds, $e(X)/(1-e(X))$, make controls with covariates common among treated units count more heavily. Both terms are scaled by the number of treated units, not by the number of controls.

This is an unnormalized odds-weighted form: the control residual sum is divided by $n_1$. In a finite sample, the control weights need not sum to $n_1$. That can affect the numerical estimate, but it does not change the consistency argument below when the assumptions and regularity conditions hold.

Worked Example: Calculating the ATT Estimate

Consider hypothetical data with two treated units and three controls. Suppose the untreated-outcome predictions and propensity scores shown below have been obtained from fitted models.

Unit	Treatment	Observed outcome $Y$	$\hat m_0(X)$	$\hat e(X)$
1	1	12	8	0.25
2	1	15	10	0.50
3	0	7	8	0.25
4	0	12	10	0.50
5	0	9	8	0.25

Here $n_1=2$. The treated residual sum is $(12-8)+(15-10)=9$. The control odds are $0.25/0.75=1/3$ for units 3 and 5, and $0.50/0.50=1$ for unit 4. Thus the weighted control residual sum is

\frac{1}{3}(7-8)+1(12-10)+\frac{1}{3}(9-8) =-\frac{1}{3}+2+\frac{1}{3}=2.

Substitution gives

\hat\tau_{\mathrm{ATT}} =\frac{9}{2}-\frac{2}{2} =4.5-1 =3.5.

The estimate is 3.5 outcome units. This calculation alone does not establish a causal effect: its causal interpretation requires the identification assumptions, and its reliability depends on the fitted models and available overlap.

Why Either Nuisance Model Can Be Enough

The estimator’s consistency property follows from two population identities. Write $p_1=P(A=1)>0$, and temporarily use generic working functions $\tilde e(X)$ and $\tilde m_0(X)$ in place of the fitted functions. The population analogue is

\Psi(\tilde e,\tilde m_0) = \frac{E[A\{Y-\tilde m_0(X)\}]}{p_1} - \frac{E[(1-A)\{\tilde e(X)/(1-\tilde e(X))\}\{Y-\tilde m_0(X)\}]}{p_1}.

First suppose the propensity model is correct, so $\tilde e(X)=e(X)$, while the outcome regression can be misspecified. Conditional on $X$, the weighted control contribution satisfies

E\left[(1-A)\frac{e(X)}{1-e(X)}\{Y-\tilde m_0(X)\}\mid X\right] = e(X)E[Y(0)-\tilde m_0(X)\mid A=0,X].

The corresponding treated contribution conditional on $X$ is $e(X)E[Y(1)-\tilde m_0(X)\mid A=1,X]$. Subtracting the control contribution cancels $\tilde m_0(X)$. Conditional exchangeability gives $E[Y(0)\mid A=0,X]=E[Y(0)\mid A=1,X]$, so the difference becomes $e(X)E[Y(1)-Y(0)\mid A=1,X]$. Averaging over $X$ and dividing by $p_1=E[e(X)]$ yields $\tau_{\mathrm{ATT}}$.

Now suppose instead that the outcome regression is correct, $\tilde m_0(X)=m_0(X)$, while the propensity model may be wrong. Given $X$, the control residual has mean zero:

E[Y-m_0(X)\mid A=0,X]=0.

Because the weight $\tilde e(X)/(1-\tilde e(X))$ is a function of $X$, its product with that residual still has conditional expectation zero. The first term becomes $E[A\{Y(1)-m_0(X)\}]/p_1$. By conditional exchangeability, $m_0(X)=E[Y(0)\mid A=1,X]$; hence this term equals $\tau_{\mathrm{ATT}}$. These identities explain the estimator’s double robustness: either the propensity score or the untreated-outcome regression can be correctly specified, provided the causal assumptions hold.

Consistency conditions: A sufficient large-sample argument requires consistency and no interference, conditional exchangeability of $Y(0)$ given $X$, overlap for treated covariate values, and $P(A=1)>0$. In addition, the fitted nuisance functions must converge appropriately, the odds-weighted terms must have finite expectations, and a law of large numbers must apply. If $\hat e$ converges to the true $e$, the outcome model may be misspecified; if $\hat m_0$ converges to the true $m_0$, the propensity model may be misspecified. Double robustness does not protect against failure of the causal assumptions or lack of support.

Worked Example: Correct Propensity Score, Misspecified Outcome Model

Consider a hypothetical population with two covariate strata, low and high. Suppose $P(X=\text{low})=2/3$, $P(X=\text{high})=1/3$, and the true treatment probabilities are $0.25$ and $0.50$, respectively. The treated probability is $(2/3)(0.25)+(1/3)(0.50)=1/3$, and half of the treated population is in each stratum.

Suppose the untreated mean outcomes are 6 in the low stratum and 10 in the high stratum, while treated mean outcomes are 12 and 16, respectively. The true ATT is therefore $\{(12-6)+(16-10)\}/2=6$. Use the misspecified working outcome prediction $\tilde m_0(X)=8$ in both strata.

Among treated units, the average residual is $\{(12-8)+(16-8)\}/2=6$. The control residual means are $6-8=-2$ and $10-8=2$. Correct propensity odds reweight the controls to the treated stratum distribution, which is one-half in each stratum. Thus the weighted control residual average is $\{(-2)+2\}/2=0$. The population augmented contrast is $6-0=6$, exactly the ATT. The example shows why the correct propensity score can compensate for this misspecified outcome prediction.

Worked Example: Correct Outcome Regression, Misspecified Propensity Score

Suppose a hypothetical setting has a correct untreated-outcome regression, but the propensity model is misspecified. In a finite sample, three treated units have residuals $Y_i-\hat m_0(X_i)$ of 2, 4, and 6. Two controls have residuals $-1$ and 2. Suppose the misspecified model assigns both controls $\hat e=0.40$, giving odds $0.40/0.60=2/3$.

The treated residual mean is $(2+4+6)/3=4$. The weighted control residual sum is $(2/3)(-1)+(2/3)(2)=2/3$; divided by $n_1=3$, the adjustment is $2/9=0.2222$. Therefore,

\hat\tau_{\mathrm{ATT}}=4-\frac{2}{9}=3.778.

This finite-sample value need not equal the population ATT. Under a correct outcome regression, control residuals have mean zero conditional on $X$, so weighting them by any well-behaved function of $X$ has population mean zero. The sample residual sum can nevertheless be nonzero, as it is here. Consistency is a large-sample statement, not a promise of exact cancellation in every sample.

Diagnostics, Alternatives, and Limitations

Diagnostics should focus on whether the comparison the estimator needs is plausible. Examine propensity-score distributions separately for treated and control units, and check whether treated covariate profiles have comparable controls. If treated units occupy regions with few or no controls, the estimator relies on extrapolation or on a small number of heavily weighted controls. Inspect the control odds $ \hat e(X_i)/(1-\hat e(X_i)) $, the distribution of weighted control contributions, and whether conclusions change substantially under reasonable model specifications. Balance checks are useful, but good balance on measured covariates does not establish exchangeability or rule out unmeasured confounding.

The main alternatives answer related questions with different vulnerabilities. Outcome regression predicts untreated outcomes for treated units and averages their predicted contrasts; it avoids propensity weighting but can be biased if that regression is wrong. Inverse-probability weighting uses the propensity odds to construct a treated-like control group; it can be consistent with a correct propensity model but may be sensitive to misspecification. The augmented estimator combines these ideas and has the stated consistency protection when either nuisance model is correct. Matching is another way to construct comparable controls, but it is a distinct estimator with its own choices and finite-sample behavior.

The ATT is a causal estimand only for the treated population represented in the data. If some treated covariate patterns have no untreated counterparts, the untreated outcomes for those treated units cannot be identified from the observed data without additional assumptions. Also, having one correct nuisance model does not guarantee accurate finite-sample estimates, valid uncertainty estimates, or robust conclusions when the other model is poorly behaved. The estimator cannot repair unmeasured confounding, incorrect treatment or outcome measurement, or a mismatch between the target population and the data.

Common Mistake

A common mistake is to assume that “doubly robust” means the estimator remains valid if either the causal assumptions or either nuisance model fails. That is not what the property says. It means that, under identification assumptions such as conditional exchangeability and overlap, consistency can hold if either the propensity score model or the untreated-outcome regression is correct. If treated units lack comparable controls, or if important confounders are unmeasured, fitting both nuisance models cannot by itself identify the causal ATT.

Key takeaway: For the ATT, the augmentation uses control-group propensity odds and residuals from an untreated-outcome regression. Its two consistency routes are valuable, but both depend on causal identification, treated-group support, and adequate large-sample behavior.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

In words, which population does the ATT describe, and how does that differ from the ATE?
Why are controls weighted by $e(X)/(1-e(X))$ rather than by $e(X)$ alone?
If the untreated-outcome regression is correct, why can a misspecified propensity model still yield a consistent estimator?
What identification problem arises if treated units have covariate values for which $e(X)=1$?
For a treated residual sum of 18, a weighted control residual sum of 6, and $n_1=4$, calculate the estimator.

Solutions. (1) The ATT is the mean treatment effect among units who received treatment; the ATE averages effects over the full target population. (2) The odds convert the control covariate distribution toward the treated covariate distribution: conditional control mass multiplied by $e(X)/(1-e(X))$ is proportional to treated mass. (3) With a correct $m_0(X)$, the control residual has conditional mean zero given $X$; multiplying it by any suitable function of $X$ preserves that zero conditional expectation, while the treated term identifies the ATT under exchangeability. (4) There are no untreated observations at those treated covariate values, so the missing untreated outcomes for those treated units are not identified by the observed data without additional assumptions. (5) The estimate is $18/4-6/4=4.5-1.5=3.0$ outcome units.