Efficient Influence Functions for the ATT

Why the ATT Has Its Own Influence Function

An average treatment effect on the treated (ATT) asks about the effect of treatment among people who received treatment. This target population differs from the overall population used for the average treatment effect (ATE). Consequently, even when both effects are identified from the same observed data, their efficient influence functions have different weighting structures. Getting those weights right matters: they describe how perturbations in the observed data change the target and determine whether the usual finite-variance, root-$n$ inference framework applies.

Let $A\in\{0,1\}$ denote treatment, $X$ baseline covariates, and $Y$ the observed outcome. Write $Y(1)$ and $Y(0)$ for the potential outcomes under treatment and control. We use independent, identically distributed observations $O=(X,A,Y)$. The consistency condition is $Y=AY(1)+(1-A)Y(0)$. The ATT is

\theta_{\mathrm{ATT}}=E[Y(1)-Y(0)\mid A=1].

Define the treatment probability $e(X)=P(A=1\mid X)$, the treated probability $\pi=P(A=1)$, and the conditional outcome means $m_a(X)=E[Y\mid A=a,X]$, for $a\in\{0,1\}$. These are features of the observed-data distribution.

Identification conditions: Assume consistency, conditional exchangeability $(Y(1),Y(0))\perp A\mid X$, and $0<\pi<1$. To identify the untreated mean for treated people, require $e(X)<1$ for $P(X\mid A=1)$-almost every $X$. Then $$ \theta_{\mathrm{ATT}}=E[m_1(X)-m_0(X)\mid A=1]. $$

The overlap condition here is tailored to the ATT: there must be controls at covariate values represented among treated people. It does not require treated and control observations at every covariate value in the entire population. The condition identifies the target, but by itself it does not guarantee that the influence function has finite variance.

The Efficient Influence Function

In the nonparametric observed-data model, an efficient influence function (EIF), also called the canonical gradient, is a mean-zero, square-integrable function of $O$ whose inner product with any regular score gives the pathwise derivative of the parameter along that score. For the ATT, the formal gradient has a particularly useful representation. To call it an EIF in the usual nonparametric $L_2$ model, we also need it to be square-integrable.

Theorem: Under the identification conditions above and regularity for differentiating conditional means, the formal observed-data gradient for the ATT is $$ \phi_{\mathrm{ATT}}(O) = \frac{A}{\pi}\{Y-m_0(X)-\theta_{\mathrm{ATT}}\} - \frac{1-A}{\pi}\frac{e(X)}{1-e(X)}\{Y-m_0(X)\}. $$ If this function is square-integrable, it is the EIF in the nonparametric $L_2$ model. A sufficient moment condition is $$ E\!\left[A\{Y-m_0(X)-\theta_{\mathrm{ATT}}\}^2\right]<\infty, \qquad E\!\left[(1-A)\left\{\frac{e(X)}{1-e(X)}\right\}^{\!2} \{Y-m_0(X)\}^2\right]<\infty. $$

These conditions explicitly control the second moments of the two nonzero pieces of the gradient. Because their sum is the gradient, finite second moments for both pieces imply $E[\phi_{\mathrm{ATT}}(O)^2]<\infty$. The weighted condition can fail near $e(X)=1$, even if the outcome residuals have finite variance without weighting. Identification and finite-variance inference are therefore distinct requirements.

Deriving the Gradient

A direct derivation makes the role of the weights clear. Write the ATT as a ratio:

\theta_{\mathrm{ATT}} = \frac{N}{\pi}, \qquad N=E[A\{Y-m_0(X)\}], \qquad \pi=E[A].

Indeed, $E[A\{Y-m_0(X)\}]/\pi$ averages the observed treated outcome minus the estimated untreated conditional mean over the treated population. Under the identification conditions, this is the ATT.

Consider a regular one-dimensional perturbation of the observed-data distribution with score $s(O)$, where $E[s(O)]=0$. Holding $m_0$ fixed momentarily, the derivative of $N$ is the covariance of $A\{Y-m_0(X)\}$ with the score. But $m_0$ also changes with the conditional outcome distribution among controls. Denote that derivative by $\dot m_0(X)$. Differentiating the conditional mean gives

\dot m_0(X) = E[\{Y-m_0(X)\}s(O)\mid A=0,X]

where only the conditional-outcome component of the score contributes to this conditional-mean derivative. Thus the full numerator derivative is

\dot N = E[\{A(Y-m_0(X))-N\}s(O)] - E[e(X)\dot m_0(X)].

The second term can be written using observed controls. Conditional on $X$, the probability of being a control is $1-e(X)$, so

E[e(X)\dot m_0(X)] = E\!\left[(1-A)\frac{e(X)}{1-e(X)} \{Y-m_0(X)\}s(O)\right].

The derivative of $\pi=E[A]$ is $E[(A-\pi)s(O)]$. Applying the quotient rule, $\dot\theta_{\mathrm{ATT}}=(\dot N-\theta_{\mathrm{ATT}}\dot\pi)/\pi$, gives the following representation of the derivative:

\dot\theta_{\mathrm{ATT}} = E\!\left[ \left\{ \frac{A}{\pi}(Y-m_0(X)-\theta_{\mathrm{ATT}}) - \frac{1-A}{\pi}\frac{e(X)}{1-e(X)}(Y-m_0(X)) \right\}s(O) \right].

The function in braces is the formal gradient. Both of its terms have mean zero: the treated term does because $E[A\{Y-m_0(X)\}]=\pi\theta_{\mathrm{ATT}}$, and the control residual term does because $E[Y-m_0(X)\mid A=0,X]=0$. In the nonparametric model, if this gradient belongs to $L_2$, it is the canonical gradient and hence the EIF. If the required square-integrability fails, the displayed expression remains the formal derivative representation but is not an EIF in the usual $L_2$ model.

Worked Example: Identifying the ATT From Covariate Strata

Suppose a hypothetical population has two covariate strata, low-risk and high-risk. Among treated people, 75% are low-risk and 25% are high-risk. The observed treated mean outcomes are 12 and 20 units in those strata, respectively; the corresponding control mean outcomes are 5 and 10 units. Assume consistency, exchangeability given stratum, and that both treatment groups occur in each stratum.

The stratum-specific contrasts are $12-5=7$ units and $20-10=10$ units. The ATT averages those contrasts using the covariate distribution among treated people:

\theta_{\mathrm{ATT}} = 0.75(7)+0.25(10) = 5.25+2.50 = 7.75\text{ units}.

The result is 7.75 units for the treated population. The weights are the treated-group proportions, not the proportions in the full population. Under the stated identification assumptions, the observed stratum means identify the relevant potential-outcome means. This calculation identifies the target; it does not by itself establish that an estimator has a finite-variance influence function.

Reading the Two Residual Terms

The treated term is $A/\pi$ times $Y-m_0(X)-\theta_{\mathrm{ATT}}$. For treated observations, it reflects the observed treated outcome relative to the untreated conditional mean, centered by the target ATT. The factor $1/\pi$ rescales contributions from the treated subgroup to the population scale.

The control term uses only controls and weights their untreated-outcome residuals by $e(X)/(1-e(X))$. Controls with covariate values more common among treated people receive greater weight. These residuals account for uncertainty in estimating the untreated conditional mean over the treated covariate distribution. The weight is not simply $1/(1-e(X))$: the target averages over treated covariates, which supplies the factor $e(X)$.

Worked Example: Evaluating Influence-Function Contributions

For a hypothetical data-generating distribution, suppose $\pi=0.40$, $e(X)=0.20$ at the covariate value of interest, $m_0(X)=3$ outcome units, and $\theta_{\mathrm{ATT}}=5$ outcome units. Evaluate the formal gradient for one treated observation with $Y=6$, and for one control observation with $Y=4$.

For the treated observation, $A=1$, so the control term is zero:

\phi_{\mathrm{ATT}} = \frac{1}{0.40}(6-3-5) = 2.5(-2) = -5\text{ units}.

For the control observation, $A=0$, so the treated term is zero. The odds weight is $0.20/(1-0.20)=0.25$:

\phi_{\mathrm{ATT}} = -\frac{1}{0.40}(0.25)(4-3) = -2.5(0.25)(1) = -0.625\text{ units}.

These are the contributions for two particular observations, not estimates of the ATT and not probabilities. The control contribution has smaller magnitude here because the covariate value is less common among treated people than among controls. The expectation of the full gradient over the distribution, unlike either individual contribution, is zero.

How It Differs From the ATE Influence Function

For comparison, the ATE is $\theta_{\mathrm{ATE}}=E[m_1(X)-m_0(X)]$. Under the corresponding identification conditions and suitable square-integrability, its nonparametric EIF is

\phi_{\mathrm{ATE}}(O) = m_1(X)-m_0(X)-\theta_{\mathrm{ATE}} + \frac{A}{e(X)}\{Y-m_1(X)\} - \frac{1-A}{1-e(X)}\{Y-m_0(X)\}.

The difference is not merely a change in the parameter label. The ATE standardizes to the overall covariate distribution and includes a conditional treatment-effect term, $m_1(X)-m_0(X)-\theta_{\mathrm{ATE}}$. Its residual weights use $1/e(X)$ among treated people and $1/\{1-e(X)\}$ among controls. The ATT instead targets the treated covariate distribution: its treated contribution is scaled by $1/\pi$, and its control residual weight is $e(X)/\{1-e(X)\}$, also scaled by $1/\pi$.

Feature	ATT influence function	ATE influence function
Target covariate distribution	Distribution among treated people	Overall population distribution
Treated-outcome residual	Uses $Y-m_0(X)-\theta_{\mathrm{ATT}}$, scaled by $A/\pi$	Uses $Y-m_1(X)$, scaled by $A/e(X)$
Control-outcome residual	Weighted by $(1-A)e(X)/\{1-e(X)\}$, then scaled by $1/\pi$	Weighted by $(1-A)/\{1-e(X)\}$
Conditional effect term	Included through the treated-target centering in its treated term	Explicitly includes $m_1(X)-m_0(X)-\theta_{\mathrm{ATE}}$

This comparison is useful when selecting or checking an influence-function formula: the target population determines the standardization and therefore the weights. Substituting the ATE control weight into the ATT formula, or using the treated-group covariate distribution to describe an ATE, changes the gradient and generally the estimand.

Worked Example: Checking Finite Variance Under Stronger Weighting

Consider a hypothetical setting with constant $X$, $\pi=e(X)=0.40$, $m_0(X)=0$, and $\theta_{\mathrm{ATT}}=2$ outcome units. Suppose the treated outcome has mean 2 and variance 4, while the control outcome has mean 0 and variance 9. These assumptions make both conditional residual means zero and give finite residual variances.

For treated observations, the gradient contribution is $A(Y-2)/0.40$. Its second moment is

E\!\left[\left\{\frac{A(Y-2)}{0.40}\right\}^2\right] = \frac{0.40(4)}{0.40^2} = 10\text{ units}^2.

For controls, the odds weight is $0.40/0.60=2/3$. The second moment of that contribution is

E\!\left[\left\{ \frac{1-A}{0.40}\frac{2}{3}Y \right\}^2\right] = \frac{0.60(2/3)^2(9)}{0.40^2} = 15\text{ units}^2.

The two pieces are never nonzero for the same observation, and each has mean zero, so their cross-product has expectation zero. Thus $E[\phi_{\mathrm{ATT}}^2]=10+15=25\text{ units}^2$, which is finite; the influence-function standard deviation is $5$ units. This finite value supports the usual $L_2$ influence-function framework for this distribution.

Common Pitfall: Confusing Identification With an EIF

A common mistake is to say that consistency, exchangeability, and positivity are enough to establish the EIF formula as an $L_2$ canonical gradient. Those assumptions identify the ATT and yield the formal gradient calculation, but they do not rule out infinite variance. For instance, with constant $X$, $A\sim\mathrm{Bernoulli}(0.5)$ independent of the potential outcomes, $Y(1)=0$, and $Y(0)$ having mean zero but infinite variance, the ATT is identified and equals zero. Yet the formal gradient for controls is $-2Y(0)$, which has infinite variance. It is therefore not an EIF in the usual nonparametric $L_2$ model.

The correction is to check weighted residual second moments as well as identification and overlap. In applications, estimated propensities close to one among treated-like covariate values can make the ATT gradient highly variable. That is not just a numerical inconvenience: it signals that the relevant weighted-integrability condition may be weak or fail.

Key takeaway: The ATT targets the treated covariate distribution. Its formal gradient combines a treated contribution scaled by $1/\pi$ with a control residual weighted by $e(X)/\{1-e(X)\}$ and scaled by $1/\pi$. It is an EIF in the nonparametric $L_2$ model only when the gradient is square-integrable.

Check Your Understanding

Answer each question by referring to the target population and the conditions for the gradient.

State the observed-data identification formula for the ATT and the assumption that lets the control outcome regression identify the untreated mean among treated people.
Why does the ATT control residual receive the weight $e(X)/\{1-e(X)\}$, rather than just $1/\{1-e(X)\}$?
Which two weighted second-moment conditions given in the theorem are sufficient to ensure that the displayed formal gradient is square-integrable?
Describe one important difference between the ATT and ATE influence functions’ treated and control residual weights.
Can the ATT be identified while its formal gradient fails to be an EIF in the usual nonparametric $L_2$ model? Explain briefly.