Weights for the Average Treatment Effect on the Untreated

Targeting the Untreated Population

A treatment effect depends on which population it describes. The average treatment effect on the untreated (ATU) asks what the average effect would be for units that actually received no treatment. It compares their observed outcomes under no treatment with the outcomes those same units would have had under treatment. Because the treated units provide information about treated outcomes, weighting can reweight them to represent the untreated units’ covariate distribution.

Definition: Let $Z$ be a binary treatment indicator, where $Z=1$ denotes treatment and $Z=0$ denotes no treatment. Let $Y(1)$ and $Y(0)$ be the potential outcomes under treatment and no treatment. The average treatment effect on the untreated is $$ \operatorname{ATU}=E[Y(1)-Y(0)\mid Z=0]. $$ It is the average treatment effect for the population of units with $Z=0$, not the average effect for all units.

The propensity score is $e(X)=P(Z=1\mid X)$, the conditional probability of treatment given the measured covariates $X$. For the ATU, untreated units stay in the target group with weight 1. Treated units receive weight $w(X)=(1-e(X))/e(X)$. These weights are the odds of no treatment versus treatment at a given covariate value.

Assumptions and the Weighting Identity

To interpret the weighted contrast as an ATU, the covariates must be sufficient to account for treatment selection, and there must be treated units who can inform the counterfactual outcomes of the untreated. Specifically, the identification argument uses the following assumptions.

Consistency: the observed outcome is $Y=Y(1)$ when $Z=1$, and $Y=Y(0)$ when $Z=0$. This presumes the treatment and outcome are defined clearly enough that the observed treatment corresponds to the potential outcome being considered.
Conditional exchangeability: $(Y(1),Y(0))$ is independent of $Z$ given $X$. After conditioning on the measured covariates, treatment assignment carries no additional information about the potential outcomes.
Positivity for the untreated target: $e(X)>0$ for covariate values occurring among untreated units. Without treated units at those covariate values, the treated potential outcome for those untreated units cannot be identified from observed data. Since the target units are untreated, their covariate values also have $e(X)<1$ in the population.

The weight formula follows from comparing covariate distributions. By Bayes’ rule, the density of covariates among untreated units is proportional to $P(Z=0\mid X)P(X)$, while the density among treated units is proportional to $P(Z=1\mid X)P(X)$. Their density ratio is

\frac{f(X\mid Z=0)}{f(X\mid Z=1)} = \frac{1-e(X)}{e(X)} \frac{P(Z=1)}{P(Z=0)}.

The factor $P(Z=1)/P(Z=0)$ is constant across covariate values. Thus, weighting treated units by $(1-e(X))/e(X)$ makes their covariate distribution proportional to the untreated distribution; normalizing the weights removes that constant. Untreated units need no reweighting because they already come from the target population.

Conditional exchangeability and consistency then connect this distributional identity to outcomes. At each $X$, treated outcomes have the same conditional mean as $Y(1)$ for untreated units with that $X$. Reweighting those treated outcomes to the untreated covariate distribution therefore identifies $E[Y(1)\mid Z=0]$. The observed untreated outcomes identify $E[Y(0)\mid Z=0]$. Their difference is the ATU.

Identification result: Under consistency, conditional exchangeability, and positivity for the untreated target, $$ \operatorname{ATU} = \frac{E\!\left[Z\frac{1-e(X)}{e(X)}Y\right]} {E\!\left[Z\frac{1-e(X)}{e(X)}\right]} - \frac{E[(1-Z)Y]}{P(Z=0)}. $$ The first term is the treated outcome mean reweighted to represent untreated units; the second is the observed untreated outcome mean.

This is a population identity. In a sample, its direct normalized counterpart divides each group’s weighted outcome total by that group’s total weight. For the treated group the weights are $(1-e(X_i))/e(X_i)$; for the untreated group they are 1. The result is a sample estimate, not a guarantee that the assumptions hold or that the propensity scores are well estimated.

Worked Calculation With Covariate Strata

Worked Example: Reweighting Treated Units to Match Untreated Strata

Hypothetical data. Suppose $X$ has three categories. Each category contains 8 units. The table gives the number of treated and untreated units, the treatment propensity, and the mean outcome in each treatment group. Assume consistency, conditional exchangeability within these categories, and positive treatment probabilities.

Category	Treated count	Untreated count	$e(X)$	Treated mean $Y$	Untreated mean $Y$
A	2	6	0.25	10	4
B	4	4	0.50	14	7
C	6	2	0.75	22	13

The treated-unit weights for categories A, B, and C are respectively $0.75/0.25=3$, $0.50/0.50=1$, and $0.25/0.75=1/3$. Multiplying treated counts by these weights gives $2(3)=6$, $4(1)=4$, and $6(1/3)=2$. Those weighted counts exactly match the untreated counts $6,4,2$, so the reweighted treated group has the same category proportions as the target group.

The weighted treated outcome mean is $$ \frac{2(3)(10)+4(1)(14)+6(1/3)(22)} {2(3)+4(1)+6(1/3)} = \frac{60+56+44}{6+4+2} = \frac{160}{12} \approx 13.3333. $$ The untreated outcome mean is $$ \frac{6(4)+4(7)+2(13)}{6+4+2} = \frac{24+28+26}{12} = \frac{78}{12} =6.5. $$ Therefore, the estimated ATU contrast is $13.3333-6.5\approx6.8333$ outcome units. Under the assumptions, this estimates the average treatment effect for units in the untreated target population.

Worked Calculation With Individual Records

Worked Example: Applying the Weights to a Small Sample

Hypothetical data. Suppose a sample has three untreated units and three treated units. Their propensity scores and observed outcomes are:

Unit	Treatment $Z$	$e(X)$	Observed outcome $Y$	ATU weight
U1	0	0.25	5	1
U2	0	0.50	7	1
U3	0	0.50	9	1
T1	1	0.25	11	3
T2	1	0.50	13	1
T3	1	0.50	17	1

For T1, $w=(1-0.25)/0.25=3$. For T2 and T3, $w=(1-0.50)/0.50=1$. The normalized weighted treated mean is $$ \frac{3(11)+1(13)+1(17)}{3+1+1} = \frac{63}{5} =12.6. $$ The untreated mean, retaining each untreated unit with weight 1, is $$ \frac{5+7+9}{3}=7. $$ The sample ATU contrast is $12.6-7=5.6$ outcome units. It is a weighted comparison targeting the covariate distribution of the observed untreated units; with only six records, it may be imprecise, and the causal interpretation still depends on the assumptions.

Diagnostics, Alternatives, and Limitations

The key diagnostic concern is overlap: whether treated units exist with covariate profiles like those of the untreated target. A low propensity score among a treated unit creates a large ATU weight. Such a unit represents many untreated units with similar covariates, so its outcome can exert substantial influence on the weighted mean.

Worked Example: Recognizing an Extreme ATU Weight

Hypothetical diagnostic calculation. Consider three treated units with propensity scores $0.05$, $0.50$, and $0.90$, and outcomes 20, 10, and 8 outcome units. Their ATU weights are $0.95/0.05=19$, $0.50/0.50=1$, and $0.10/0.90=1/9$. The weighted treated mean is $$ \frac{19(20)+1(10)+(1/9)(8)}{19+1+1/9} = \frac{3518}{181} \approx 19.4365. $$ The first unit contributes $380$ of the $390.8889$ weighted outcome total. Its high influence is a warning that the estimate may be sensitive to a small number of observations. This calculation does not establish that the estimate is biased, but it signals that overlap and sensitivity deserve scrutiny.

Useful checks include inspecting the propensity-score distributions for treated and untreated units, examining the largest weights, and checking whether a small number of treated observations dominate the weighted outcome total. If many untreated units have propensity scores near zero, their corresponding treated counterparts may be absent or represented by very few heavily weighted observations. Weight truncation can reduce variability, but it changes the weighting scheme and can introduce bias; it is not a repair for a failure of positivity.

The target distinguishes these weights from the main alternatives. ATE weights use $1/e(X)$ for treated units and $1/(1-e(X))$ for untreated units to target the full population. ATT weights retain treated units and reweight untreated units by $e(X)/(1-e(X))$, targeting the treated population. ATU weights instead retain untreated units and reweight treated units by $(1-e(X))/e(X)$. These methods answer different questions; one target should not be substituted for another without changing the estimand.

Weighting cannot adjust for unmeasured confounding merely by balancing measured covariates. It also cannot recover an ATU reliably where treatment is effectively absent among the untreated population’s covariate profiles. In practice, propensity scores are estimated, so model misspecification can leave covariate imbalance even when the formula is applied correctly. Finally, a sample weighted contrast is not automatically precise: severe weights can make it highly variable, and a point estimate alone does not communicate that uncertainty.

Common Mistake

Wrong reasoning: “The target is the untreated group, so I should give untreated units weights of $1/(1-e(X))$ and leave treated units unweighted.” That reverses the needed reweighting and uses the ATE-style untreated weight rather than the ATU weight.
Correction: Keep untreated units at weight 1. Reweight treated units by $(1-e(X))/e(X)$, so their covariate distribution represents the untreated target. Then subtract the untreated outcome mean from the weighted treated outcome mean.

Key takeaway: ATU weighting targets units who were untreated: retain their observed outcomes with weight 1, and weight treated observations by the odds of no treatment, $(1-e(X))/e(X)$, to estimate their missing treated-outcome mean.

Check Your Understanding

Answer each question, then compare with the solutions below.

For a unit with propensity score $e(X)=0.20$, calculate the ATU weight for a treated observation.
In one sentence, state the ATU target population and the outcome contrast it describes.
Suppose untreated units occur at covariate values where $e(X)=0$. Which identification condition fails, and why?
For treated outcomes 12 and 18 with propensity scores 0.25 and 0.50, calculate the normalized weighted treated mean. Use the ATU weights.
Explain why an extremely large ATU weight is a diagnostic warning rather than proof that the estimate is biased.

Solutions. (1) $w=(1-0.20)/0.20=4$. (2) The ATU is the average $Y(1)-Y(0)$ among units who actually had $Z=0$. (3) Positivity fails: no treated units occur at those covariate values, so their treated potential outcomes cannot be learned from the treated group. (4) The weights are $3$ and $1$, giving $\{3(12)+1(18)\}/(3+1)=54/4=13.5$ outcome units. (5) A large weight means one or a few treated outcomes may dominate, making the estimate sensitive and potentially imprecise; bias depends on the assumptions, overlap, and model adequacy, not on the weight’s size alone.