Combining Stratum-Specific Effects for an ATT

Why the Stratum Weights Depend on the Target

Propensity-score stratification first divides units into strata, then compares treated and untreated outcomes within each stratum. Those within-stratum comparisons do not, by themselves, specify an overall treatment effect. To estimate an average treatment effect among the treated, the comparisons must be combined to represent the distribution of strata among treated units. A stratum containing many treated units contributes more to the ATT than a stratum containing few treated units.

This target differs from the average treatment effect (ATE), which averages effects over the target population. The earlier material on combining stratum-specific effects for an ATE uses population stratum proportions. Here the target is specifically the treated group, so the relevant proportions are the treated group’s stratum proportions.

Definition: Let $A$ be the binary treatment indicator, with $A=1$ for treated and $A=0$ for untreated. Let $Y(1)$ and $Y(0)$ be the potential outcomes under treatment and no treatment, respectively. The average treatment effect among the treated is $$ \operatorname{ATT}=E[Y(1)-Y(0)\mid A=1]. $$ For a stratum variable $S$ with strata $k=1,\ldots,K$, define $q_{1k}=P(S=k\mid A=1)$ and $\tau_{1k}=E[Y(1)-Y(0)\mid A=1,S=k]$. Then $\operatorname{ATT}=\sum_{k=1}^{K}q_{1k}\tau_{1k}$.

Identification and the Estimator

The identity above follows by conditioning the ATT on the treated group’s stratum membership. Within a stratum, however, the untreated potential outcomes of treated units are not observed. A comparison group is needed to estimate them. The stratum-specific observed outcome contrast identifies $\tau_{1k}$ if treated and untreated units in that stratum have the same mean untreated potential outcome.

\begin{aligned} \tau_{1k} &=E[Y(1)\mid A=1,S=k]-E[Y(0)\mid A=1,S=k]\\ &=E[Y\mid A=1,S=k]-E[Y\mid A=0,S=k]. \end{aligned}

The second equality uses consistency, so that the observed outcome equals the potential outcome under the received treatment, and the condition $E[Y(0)\mid A=1,S=k]=E[Y(0)\mid A=0,S=k]$. No interference between units is also assumed. Conditional exchangeability of both potential outcomes within strata is a familiar sufficient condition, but for this ATT contrast it is enough to require mean exchangeability for $Y(0)$. In addition, every stratum containing treated units must have untreated comparison units: $P(A=0\mid S=k)>0$ whenever $P(S=k\mid A=1)>0$.

In applications, $S$ is often formed by grouping estimated propensity scores. Exact identification then requires the relevant exchangeability condition to hold within those groups. A propensity score has a balancing property at its exact value; grouping a range of scores does not automatically preserve exact balance. Consequently, the estimator is commonly an approximation whose plausibility should be assessed using covariate balance and comparison support within strata.

For a sample, let $n_{1k}$ and $n_{0k}$ be the treated and untreated counts in stratum $k$, and let $n_1=\sum_k n_{1k}$. Write $\bar Y_{1k}$ and $\bar Y_{0k}$ for their observed outcome means. The estimated treated-group weight is $\hat q_{1k}=n_{1k}/n_1$, and the estimated stratum contrast is $\hat\tau_k=\bar Y_{1k}-\bar Y_{0k}$. Thus:

\widehat{\operatorname{ATT}} =\sum_{k=1}^{K}\frac{n_{1k}}{n_1} \left(\bar Y_{1k}-\bar Y_{0k}\right), \qquad \sum_{k=1}^{K}\frac{n_{1k}}{n_1}=1.

This is a weighted average of the within-stratum contrasts. Its weights are determined by treated counts, not by the total number of treated and untreated units in each stratum. The control count affects how well a control mean can be estimated and whether it exists, but it does not determine how much that stratum represents the treated target.

Worked ATT Calculations

Worked Example: Combining Three Stratum Contrasts

Suppose a hypothetical study groups units into three propensity-score strata. The outcome is a recovery score, measured in points. The following table gives the observed group sizes and means.

Stratum	Treated count $n_{1k}$	Untreated count $n_{0k}$	Treated mean	Untreated mean
Low	40	60	12	10
Middle	40	40	15	12
High	20	20	20	16

There are $n_1=40+40+20=100$ treated units. The stratum contrasts are $12-10=2$, $15-12=3$, and $20-16=4$ points. The treated-group weights are $40/100=0.40$, $40/100=0.40$, and $20/100=0.20$. Therefore,

\widehat{\operatorname{ATT}} =0.40(2)+0.40(3)+0.20(4) =0.80+1.20+0.80 =2.80\text{ points}.

The estimate says that, under the identifying assumptions, treatment increased the mean recovery score among the treated units by an estimated 2.80 points compared with their mean score had they been untreated. Each stratum has both treated and untreated observations, so the calculation has within-stratum comparisons throughout the treated group’s observed distribution. These counts alone do not establish exchangeability; balance and the study design still matter.

Worked Example: Why Equal Stratum Weights Target Something Else

Consider hypothetical treated units in two strata. In the first stratum, 90 treated units have a mean outcome of 7 and untreated units have a mean of 5, giving a contrast of 2 points. In the second, 10 treated units have a mean of 15 and untreated units have a mean of 5, giving a contrast of 10 points. Suppose both strata contain untreated units, and the within-stratum identifying condition is plausible.

The ATT weights are $90/(90+10)=0.90$ and $10/100=0.10$. The estimate is

\widehat{\operatorname{ATT}} =0.90(2)+0.10(10) =1.8+1.0 =2.80\text{ points}.

Giving both strata weight $0.50$ would instead produce $0.50(2)+0.50(10)=6.00$ points. That calculation gives equal importance to the two strata, despite nine times as many treated units being in the first stratum. It does not estimate the ATT for these treated units. The correct ATT weights follow the actual treated-group composition.

Worked Example: A Treated Stratum Without Controls

In another hypothetical study, a low-score stratum contains 30 treated and 25 untreated units. Their outcome means are 8.0 and 6.8 points, respectively, so its estimated contrast is $8.0-6.8=1.2$ points. A high-score stratum contains 10 treated units and no untreated units.

The high stratum represents $10/(30+10)=0.25$ of the treated group, while the low stratum represents $0.75$. The high-stratum treated outcome mean could be observed, but its untreated outcome mean cannot be calculated from these data. If its unobserved stratum-specific ATT is denoted by $\tau_{1,\mathrm{high}}$, the desired overall ATT has the form

\operatorname{ATT} =0.75(1.2)+0.25\tau_{1,\mathrm{high}}.

The first contribution is $0.90$ points, but the second is unknown. The data therefore do not identify the overall ATT without additional assumptions or information. Reporting 1.2 points as the overall ATT would silently drop the 25% of treated units in the high stratum and change the target to treated units in the low stratum. The absence of controls is a support failure, not a reason to assign that stratum zero weight.

Justification, Diagnostics, and Alternatives

The aggregation formula is a direct consequence of the law of total expectation. Partition the treated group by its stratum membership:

\begin{aligned} E[Y(1)-Y(0)\mid A=1] &=\sum_{k=1}^{K}E[Y(1)-Y(0)\mid A=1,S=k]P(S=k\mid A=1)\\ &=\sum_{k=1}^{K}\tau_{1k}q_{1k}. \end{aligned}

This derivation explains why the weights are not a discretionary convention. They are the probabilities of stratum membership within the population named by the estimand, namely the treated group. Replacing $P(S=k\mid A=1)$ with a population stratum proportion changes the averaging population and generally changes the causal estimand.

Before interpreting an estimate, check that treated counts sum to $n_1$ and the calculated ATT weights sum to one. Confirm that every stratum with treated units has untreated comparison units, and inspect whether important pretreatment covariates are similar between treatment groups within each stratum. A large treated share in a stratum with very few controls may also make its control mean unstable. These checks can reveal lack of support or a poor approximation to within-stratum exchangeability; they cannot prove that unmeasured confounding is absent.

Stratification combines a small set of within-stratum outcome contrasts. Propensity-score matching instead constructs comparisons between treated and untreated units according to a matching rule; inverse probability weighting uses weights based on treatment probabilities. These are alternative adjustment strategies, not guarantees of causal identification. The stratified ATT has a transparent target composition, but coarse strata may leave covariate differences within groups, and missing controls in treated strata can prevent estimation of the treated-group target. A reported point estimate alone does not resolve either problem.

Common Mistakes

Using population shares for the ATT. Population stratum proportions target an ATE-style population average. For the ATT, use $n_{1k}/n_1$.
Averaging stratum contrasts equally. Equal weights generally represent neither the treated group nor the population. Use the proportions implied by the estimand.
Dropping a treated stratum with no controls. This changes the target population and does not recover the original ATT. The missing untreated outcome must be addressed through additional defensible information or assumptions.
Treating stratification as proof of exchangeability. Grouping propensity scores does not itself ensure that treated and untreated units are comparable within each group.

Key takeaway: Estimate each stratum’s treated-minus-untreated outcome contrast, then average those contrasts using the treated group’s stratum proportions. The calculation targets the ATT only when within-stratum comparisons are valid and every stratum represented among treated units has adequate untreated comparison support.

Check Your Understanding

Answer each question, then compare your work with the solutions.

Two strata contain 30 and 70 treated units. Their estimated contrasts are 1.5 and 3.0 outcome units. Calculate the stratified ATT.
Why are the untreated counts not the aggregation weights for the ATT?
In a stratum with treated units, no untreated units are observed. What assumption needed for the stratum comparison fails in the observed data, and why is dropping that stratum not a general solution?
State one condition under which a within-stratum outcome contrast identifies the treated units’ stratum-specific effect.

Solutions. (1) The treated weights are $30/100=0.30$ and $70/100=0.70$, so the estimate is $0.30(1.5)+0.70(3.0)=0.45+2.10=2.55$ outcome units. (2) ATT weights represent how treated units are distributed across strata; untreated counts describe the comparison sample, not the target population. (3) There is no observed untreated comparison support in that treated stratum, so its untreated mean cannot be estimated from the data. Dropping it omits treated units and changes the target. (4) For example, within the stratum, the mean untreated potential outcome must be the same for treated and untreated units: $E[Y(0)\mid A=1,S=k]=E[Y(0)\mid A=0,S=k]$, along with consistency.