Weights for the Average Treatment Effect on the Treated

Targeting an Effect for the Treated

An inverse-probability weighting analysis depends on which population the effect is meant to describe. The average treatment effect on the treated (ATT) asks about the effect of treatment for units that actually received treatment. The treated units therefore already represent the target population. Weighting is needed to make the observed controls represent the treated units’ covariate distribution, so their outcomes can help estimate what the treated units would have experienced without treatment.

Definition: Let $A$ be a binary treatment indicator, with $A=1$ for treated units and $A=0$ for controls. Let $Y(1)$ and $Y(0)$ be the potential outcomes under treatment and control, and let $X$ denote pretreatment covariates. The average treatment effect on the treated is $$ \operatorname{ATT}=E[Y(1)-Y(0)\mid A=1]. $$ The propensity score is $e(X)=P(A=1\mid X)$. ATT weighting assigns weight $1$ to each treated unit and weight $e(X)/(1-e(X))$ to each control unit.

The definition separates the estimand from the method of estimation. The ATT is a causal effect for the treated population; it is not automatically the effect in the full population. Weighting is one way to estimate it when the required causal assumptions are plausible and the propensity score is adequately specified or known.

Why the Control Weight Is the Propensity-Score Odds

Within covariate values $X=x$, treated units occur with probability $e(x)$, whereas controls occur with probability $1-e(x)$. A control unit in that covariate group must therefore be upweighted in proportion to the treated-to-control ratio, $e(x)/(1-e(x))$. Treated units need no corresponding adjustment: they already come from the population whose effect is the target.

The following derivation makes the distribution-matching argument precise. Assume consistency, so the observed outcome is $Y=Y(A)$; no interference between units; conditional exchangeability, $(Y(1),Y(0))\mathbin{\perp} A\mid X$; and positivity on the treated covariate support, $0<e(X)<1$. For identification of the ATT, conditional exchangeability of $Y(0)$ alone is sufficient, but the stronger condition stated here is a familiar sufficient assumption. Positivity means that covariate profiles occurring among treated units also have a chance of occurring among controls.

\begin{aligned} E\left[(1-A)\frac{e(X)}{1-e(X)}Y\mid X\right] &=(1-e(X))\frac{e(X)}{1-e(X)}E[Y\mid A=0,X]\\ &=e(X)E[Y(0)\mid X]. \end{aligned}

The first equality uses $P(A=0\mid X)=1-e(X)$; the second uses consistency and exchangeability to identify the control outcome mean with $E[Y(0)\mid X]$. Also, $E[(1-A)e(X)/(1-e(X))\mid X]=e(X)$. Averaging over $X$, the weighted control outcome total and weight total are therefore proportional to the corresponding potential-outcome total and population total among treated units:

\begin{aligned} E\left[(1-A)\frac{e(X)}{1-e(X)}Y\right] &=E[e(X)E[Y(0)\mid X]] =E[AY(0)],\\ E\left[(1-A)\frac{e(X)}{1-e(X)}\right] &=E[e(X)]=P(A=1). \end{aligned}

Since $E[AY(0)]/P(A=1)=E[Y(0)\mid A=1]$, the normalized weighted control mean identifies the treated population’s mean outcome under control. By consistency, the observed treated mean identifies $E[Y(1)\mid A=1]$. Subtracting the two gives the ATT. This argument also shows why controls must be reweighted toward the treated covariate distribution, rather than toward the full population.

Calculating ATT Weights and the Weighted Contrast

In a finite sample, write $e_i=e(X_i)$. Give each treated unit weight $1$, and give each control unit weight $e_i/(1-e_i)$. The treated outcome mean is the ordinary mean among treated units. The weighted control outcome mean is the sum of control outcomes multiplied by their weights, divided by the sum of those weights. The sample ATT estimate is their difference:

\widehat{\operatorname{ATT}} = \frac{\sum_i A_iY_i}{\sum_i A_i} - \frac{\sum_i(1-A_i)\dfrac{e_i}{1-e_i}Y_i} {\sum_i(1-A_i)\dfrac{e_i}{1-e_i}}.

The denominator in the second term makes it a weighted mean of control outcomes. It does not change the relative control weights: multiplying every control weight by the same positive constant leaves that weighted mean unchanged. In practice the propensity score is usually estimated, so the result also depends on whether the model estimates the relevant treatment probabilities adequately.

Worked Example: Converting Propensity Scores to ATT Weights

Suppose three hypothetical control units have propensity scores $0.20$, $0.50$, and $0.80$. Their ATT weights are their propensity-score odds.

For the first control, $0.20/(1-0.20)=0.20/0.80=0.25$. For the second, $0.50/(1-0.50)=1.00$. For the third, $0.80/(1-0.80)=4.00$. Each treated unit, by contrast, receives weight $1$.

The control with propensity score $0.80$ receives a larger weight because controls with covariates associated with a high probability of treatment are relatively scarce compared with treated units. That does not mean this particular control is four times as important as a treated unit for every purpose; it means its outcome contributes four times as much as a control with weight $1$ to the weighted control total before that total is divided by the sum of control weights.

Worked Example: A Stratified ATT Calculation

Consider a hypothetical dataset with two covariate strata, low and high risk. Suppose the propensity scores are $0.50$ in the low-risk stratum and $0.25$ in the high-risk stratum. The observed outcomes are:

Stratum	Treated outcomes	Control outcomes	Propensity score
Low risk	8, 10, 12	4, 6, 8	0.50
High risk	16	10, 12, 14	0.25

Assume these propensity scores are the relevant probabilities for the covariate strata. The assumptions needed for a causal ATT include consistency, no interference, exchangeability within strata, and positivity. Positivity holds in both strata because each has a propensity score strictly between zero and one. The exchangeability assumption cannot be verified from this table; it must be justified by the study design and measured covariates.

The treated outcome mean is $(8+10+12+16)/4=46/4=11.5$. The low-risk control weight is $0.50/(1-0.50)=1$, and the high-risk control weight is $0.25/(1-0.25)=1/3$. The sum of control weights is $3(1)+3(1/3)=4$. The weighted control outcome total is $(4+6+8)(1)+(10+12+14)(1/3)=18+12=30$. Thus the weighted control mean is $30/4=7.5$, and the estimated ATT is $11.5-7.5=4.0$ outcome units.

The weighting has made the control stratum totals match the treated totals: the treated group contains three low-risk units and one high-risk unit, while the weighted controls contribute total weight $3$ from the low-risk stratum and $1$ from the high-risk stratum. The estimated contrast is therefore for the treated covariate mix, not the unweighted control mix.

Diagnostics and Failure Modes

A propensity-score model can produce weights without producing a credible ATT estimate. Diagnostics should ask whether the weighted controls resemble the treated units on the covariates used to estimate treatment assignment, and whether the estimate depends excessively on a small number of controls. These checks do not prove that confounding has been removed: balance in measured covariates cannot establish balance in unmeasured covariates.

Worked Example: An Extreme Control Weight

In a hypothetical sample, one control has propensity score $0.90$, and three other controls each have propensity score $0.20$. Their ATT weights are $9$ and $0.25$, respectively. Suppose the corresponding control outcomes are $20$, $4$, $5$, and $6$ units.

The total control weight is $9+3(0.25)=9.75$. The weighted outcome total is $9(20)+0.25(4+5+6)=180+3.75=183.75$. The weighted control mean is $183.75/9.75\approx18.85$ outcome units. The high-score control contributes $9/9.75\approx92.31\%$ of the total control weight.

This calculation is not, by itself, proof that the estimate is invalid. It is a warning that the counterfactual mean is highly dependent on one control. A useful concentration summary is the effective sample size, defined here for weights $w_i$ as $(\sum_i w_i)^2/\sum_i w_i^2$. It is approximately $9.75^2/[9^2+3(0.25^2)]=1.171$ control units, despite there being four observed controls. The example also illustrates why extreme weights and covariate balance should both be examined; a visually balanced weighted sample can still have high uncertainty if only a few observations carry substantial weight.

If $e(X)=1$ for a covariate profile among treated units, there are no controls with that profile under the assumed treatment mechanism, and the odds weight is undefined. Even when scores are just below one, the weights can become very large and unstable. A practical assessment therefore includes the distribution of estimated propensity scores by treatment group, the largest control weights, the contribution of the largest weights to the weighted mean, and covariate balance after weighting. Poor balance may indicate a misspecified propensity model or inadequate overlap. Extreme weights indicate that the estimate may be sensitive to a few observations. Neither problem can be fixed merely by reporting the formula.

How ATT Weighting Differs From Other Approaches

Approach	Target or role	Key distinction
ATT weighting	Effect among treated units	Treated weights are 1; control weights are $e(X)/(1-e(X))$.
ATE weighting	Effect in the full target population	Uses inverse-probability treatment weights for both treatment groups; its target differs from the ATT.
Matching	Depends on the matched sample and design	Selects or pairs comparison units rather than using every control with odds weights.
Outcome regression	Depends on the modeled outcome contrast and target averaging	Models outcomes conditional on covariates instead of balancing controls through propensity-score weights.

These approaches are not interchangeable just because they use the same covariates. The estimand, identifying assumptions, and sensitivity to model choices should be considered together. ATT weighting specifically uses the observed treated covariate distribution as its target. Its strengths are a direct weighting interpretation and retention of treated units; its limitations include dependence on measured-confounder exchangeability, adequate propensity-score estimation, and control availability across the treated covariate support. Weighting does not turn an observational comparison into a randomized experiment.

Common Mistakes

Using the ATE control weight for an ATT. The reasoning “inverse probability weights always use $1/(1-e)$ for controls” confuses the ATE construction with ATT weighting. For the ATT, the control weight is $e/(1-e)$, while treated units retain weight $1$.
Interpreting a large weight as a large causal effect. A weight describes how much a unit contributes to the weighted comparison, not the size of its treatment effect.
Ignoring the target population. The ATT describes treated units. It need not equal the ATE, even if both are estimated without sampling error.
Treating balance as proof of no confounding. Balance diagnostics concern observed covariates; they cannot show that all relevant confounders were measured or adequately handled.
Proceeding when treated profiles have no controls. An undefined or extreme odds weight signals a positivity problem, not a harmless arithmetic inconvenience.

Key takeaway: For the ATT, keep treated units at weight $1$ and weight controls by the propensity-score odds $e(X)/(1-e(X))$. The resulting weighted control mean targets the treated units’ mean outcome under control only under the causal assumptions and overlap needed for identification.

Check Your Understanding

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

A control unit has propensity score $0.60$. What is its ATT weight?
Why do treated units receive weight $1$ in ATT weighting?
In a treated covariate stratum with propensity score $0.40$, what is the control weight, and what overlap condition does this score satisfy?
A weighted control mean is 6.2 units and the treated mean is 9.1 units. What is the estimated ATT, and which population does it describe?
Why is a control weight that is extremely large a reason for caution?

Solutions. (1) $0.60/(1-0.60)=0.60/0.40=1.5$. (2) The treated units already constitute the target population for the ATT, so they do not need reweighting to represent themselves. (3) The weight is $0.40/(1-0.40)=0.40/0.60\approx0.667$. The score satisfies positivity for that stratum because it is strictly between zero and one, so both treatment states have positive probability. (4) The estimate is $9.1-6.2=2.9$ outcome units, for the treated population. (5) It can let one or a few controls dominate the weighted counterfactual mean, making the estimate unstable and sensitive to those observations.