Inverse-Probability Weights for the Treated-Group Effect

Why the Treated-Group Effect Needs Its Own Weights

A treatment effect need not be the same for every population. The average treatment effect on the treated (ATT) asks about the average effect for the units that actually received treatment. In an observational study, the untreated units may have a different covariate distribution from the treated units. A direct comparison of their outcomes can therefore compare different populations as well as different treatment states.

Inverse-probability weighting addresses this mismatch by reweighting controls to represent the treated group. For the ATT, treated units retain weight one, while a control unit with covariates $X$ receives a weight based on the odds of treatment, not simply the inverse probability of remaining untreated. The derivation makes clear both why this weight works and which assumptions it requires.

Definition: Let $D$ equal 1 for a treated unit and 0 for a control unit. Let $Y(1)$ and $Y(0)$ denote that unit’s potential outcomes under treatment and control, respectively, and let $Y=D Y(1)+(1-D)Y(0)$ be the observed outcome. The average treatment effect on the treated is $$ \tau_{\mathrm{ATT}}=E[Y(1)-Y(0)\mid D=1]. $$ The propensity score is $e(X)=P(D=1\mid X)$, where $X$ is the set of observed pretreatment covariates.

Identification and the Weighting Identity

The treated units’ average outcome under treatment is observed, by consistency: $Y=Y(1)$ when $D=1$. Their average outcome under control is counterfactual, so it must be identified using assumptions. The central assumption here is conditional exchangeability: given the measured covariates, treatment assignment is independent of the potential outcomes. For identifying the ATT, the relevant part is $Y(0)\perp D\mid X$. We also require that treated units’ covariate values have a positive chance of appearing among controls: $P(D=0\mid X=x)>0$ wherever the treated population has covariate support.

Write $p_1=P(D=1)$ and $p_0=P(D=0)$. Let $f_1(x)$ and $f_0(x)$ denote the covariate distributions among treated units and controls. Bayes’ rule gives the relationship between these distributions and the propensity-score odds:

\frac{f_1(x)}{f_0(x)} = \frac{p_0}{p_1}\frac{e(x)}{1-e(x)}.

Thus the odds $e(x)/(1-e(x))$ adjust the control covariate distribution toward the treated covariate distribution. The constant $p_0/p_1$ does not depend on $x$; it changes the total weight but not the relative representation of different covariate values.

Key identity: For any suitable function $g(X)$, the odds-weighted controls reproduce the treated group’s covariate average after scaling by the treated prevalence: $$ E\left[(1-D)\frac{e(X)}{1-e(X)}g(X)\right] = p_1 E[g(X)\mid D=1]. $$ In particular, under conditional exchangeability and consistency, $g(X)$ can be used to identify the treated group’s mean potential outcome under control.

A short derivation shows why the identity holds. Conditional on $X$, $1-D$ has expectation $1-e(X)$; this cancels the denominator in the odds weight. Also, $E[D\mid X]=e(X)$.

\begin{aligned} E\left[(1-D)\frac{e(X)}{1-e(X)}g(X)\right] &=E\left[E\left\{(1-D)\frac{e(X)}{1-e(X)}g(X)\mid X\right\}\right]\\ &=E[e(X)g(X)]\\ &=E[Dg(X)]\\ &=p_1E[g(X)\mid D=1]. \end{aligned}

To apply this to outcomes, conditional exchangeability implies $E[Y\mid D=0,X]=E[Y(0)\mid X]$. The odds-weighted control outcomes therefore represent the mean of $Y(0)$ over the treated covariate distribution. Subtracting that mean from the observed treated outcome mean identifies the ATT:

\tau_{\mathrm{ATT}} = \frac{E[D Y]}{p_1} - \frac{E\left[(1-D)\dfrac{e(X)}{1-e(X)}Y\right]}{p_1}.

This population expression gives treated units weight $D$, which is one for treated units and zero otherwise, and controls weight $(1-D)e(X)/(1-e(X))$. Both weighted totals are divided by the treated prevalence $p_1$. This common scaling is what makes the contrast target the treated population.

Worked Example: Deriving Odds Weights From Covariate Distributions

Suppose a hypothetical population has a binary covariate $X$, with $P(D=1)=0.4$. Among treated units, 75% have $X=A$ and 25% have $X=B$. Among controls, 25% have $X=A$ and 75% have $X=B$. Find the propensity-score odds in each group and verify that weighting controls reproduces the treated covariate distribution.

For $X=A$, Bayes’ rule gives $e(A)=P(D=1\mid X=A)$. The joint probabilities are $P(D=1,X=A)=0.4(0.75)=0.30$ and $P(D=0,X=A)=0.6(0.25)=0.15$. Thus $e(A)=0.30/(0.30+0.15)=2/3$, and its odds are $e(A)/(1-e(A))=2$.

For $X=B$, the corresponding joint probabilities are $0.4(0.25)=0.10$ for treated units and $0.6(0.75)=0.45$ for controls. Hence $e(B)=0.10/(0.10+0.45)=2/11$, and the odds are $(2/11)/(9/11)=2/9$.

Among controls, the probability masses at $A$ and $B$ are 0.25 and 0.75. After odds weighting, these masses become $0.25(2)=0.50$ and $0.75(2/9)=1/6$. Their total is $0.50+1/6=2/3$. The weighted share at $A$ is $0.50/(2/3)=0.75$; at $B$ it is $(1/6)/(2/3)=0.25$. These are exactly the treated group’s covariate shares.

For example, if the control outcome means at $A$ and $B$ are 5 and 9 units, the treated-population mean of $Y(0)$ is $0.75(5)+0.25(9)=6$ units. If the treated mean outcome is 11 units, the implied ATT is $11-6=5$ units, subject to the identification assumptions.

Sample Estimator and Its Weights

In a sample of $n$ units, let $n_1=\sum_{i=1}^n D_i$ be the number treated and let $\hat e_i$ be an estimated propensity score for unit $i$. Plugging estimated scores into the population expression gives the unnormalized inverse-probability-weighted ATT estimator:

\widehat{\tau}_{\mathrm{ATT}} = \frac{1}{n_1}\sum_{i=1}^{n}D_iY_i - \frac{1}{n_1}\sum_{i=1}^{n}(1-D_i)\frac{\hat e_i}{1-\hat e_i}Y_i.

Every treated unit contributes its observed outcome with weight one. Each control contributes its outcome multiplied by $\hat e_i/(1-\hat e_i)$. The common divisor is $n_1$, so the control term is not, in this formula, divided by the sum of its estimated weights. This form estimates the population identity directly; a finite sample’s total control weight need not equal $n_1$.

The propensity score must be estimated from pretreatment covariates appropriate to the identification argument. The estimator does not make unmeasured confounding disappear: if relevant confounders are omitted, conditional exchangeability may fail even when the fitted scores are accurate for the included covariates. Nor does estimating a propensity score alone establish that the assumptions hold.

Worked Example: Computing an Unnormalized ATT Estimate

Consider hypothetical data on three treated units and three controls. Assume the estimated propensity scores shown below are based on pretreatment covariates, and suppose the identification assumptions are plausible for these covariates. Outcomes are measured in points.

Unit	Status $D$	Observed outcome $Y$	$\hat e$ if control	Control odds weight
T1	1	10	—	1
T2	1	12	—	1
T3	1	8	—	1
C1	0	6	0.50	0.50/0.50 = 1
C2	0	9	0.25	0.25/0.75 = 0.3333
C3	0	5	0.75	0.75/0.25 = 3

The treated mean is $(10+12+8)/3=10$ points. The weighted control outcome total is $1(6)+(0.3333)(9)+3(5)=6+3+15=24$ points. Dividing this total by $n_1=3$ gives 8 points. Therefore,

\widehat{\tau}_{\mathrm{ATT}}=10-\frac{24}{3}=2\text{ points}.

In this sample, the control weights sum to $1+0.3333+3=4.333$, not 3. That difference is possible in a finite sample and does not alter the stated unnormalized estimator. The estimate says that, under its assumptions, the treated units’ mean outcome is estimated to be 2 points higher than their mean outcome would have been under control.

How Weighted Controls Represent the Treated Group

The identity applies not only to the overall covariate distribution but also to functions of the covariates. For example, set $g(X)$ equal to an indicator for membership in a particular age group or hospital. The identity then says that the odds-weighted control count in that category, in expectation, is proportional to the number of treated units in the same category. This is the population reason the weighted controls can stand in for the treated group when estimating its untreated potential outcome.

Worked Example: Recovering Treated Stratum Shares and an ATT

In a hypothetical sample, 40 people are treated: 30 in stratum $A$ and 10 in stratum $B$. There are 20 controls in $A$ and 40 in $B$. For illustration, use the stratum treatment fractions $e(A)=30/(30+20)=0.60$ and $e(B)=10/(10+40)=0.20$ as the propensity scores. The covariate support condition holds in both strata because each contains treated and control units.

The control odds weights are $0.60/0.40=1.5$ in $A$ and $0.20/0.80=0.25$ in $B$. The weighted control counts are $20(1.5)=30$ in $A$ and $40(0.25)=10$ in $B$, matching the treated counts exactly. Thus the weighted control distribution across these strata is the treated distribution.

Suppose the mean control outcomes are 8 units in $A$ and 4 units in $B$, while the treated outcome means are 12 units in $A$ and 11 units in $B$. The treated mean is $(30(12)+10(11))/40=470/40=11.75$ units. The weighted control mean on the treated-group scale is $(30(8)+10(4))/40=280/40=7$ units. The resulting ATT estimate is $11.75-7=4.75$ units.

The weighted control outcomes use the treated group’s stratum shares: $30/40=0.75$ in $A$ and $10/40=0.25$ in $B$. Without the weights, controls would instead be 20/60 in $A$ and 40/60 in $B$, a substantially different mix. The weighted contrast therefore targets the treated group, provided exchangeability within the strata and the other assumptions are credible.

Common Mistakes and Practical Interpretation

Using the wrong control weight. The ATT control weight is $e(X)/(1-e(X))$. A control with $e(X)=0.80$, for example, receives odds weight $0.80/0.20=4$, not $1/(1-0.80)=5$. The latter is a different weighting rule and does not follow from the ATT identity.
Thinking the odds weight makes every control represent one treated person. It is a relative weight that changes how much each control contributes. The distributional identity holds in expectation, and exact matching of weighted counts in a sample is not guaranteed.
Interpreting an estimate as causal without its assumptions. Weighting corrects the measured-covariate distribution in the stated way; it does not by itself justify conditional exchangeability, consistency, or adequate control support.
Changing the divisor without changing the estimator definition. The displayed unnormalized estimator divides the weighted control total by $n_1$, not by the sum of control weights. These are distinct sample calculations and should not be silently interchanged.

The ATT weighting construction matters when the scientific question concerns people who received treatment—for example, the effect among patients who began a therapy or firms that adopted a policy. Its interpretation is tied to that treated population, not automatically to all eligible units. State the target group, the covariates used to form the propensity score, and the assumptions before interpreting the weighted contrast.

Key takeaway: For the ATT, treated units have weight one and controls have odds weight $e(X)/(1-e(X))$. This choice follows from the ratio of treated to control covariate distributions; under conditional exchangeability, consistency, and control support for treated covariate values, the weighted controls represent the treated group’s counterfactual outcomes under control.

Check Your Understanding

Answer each question using the target population and the weight identity, not only the formula.

State the ATT using potential outcomes, and explain which group defines its conditioning event.
For a control unit with propensity score $e(X)=0.40$, calculate its ATT odds weight.
Why does multiplying the control covariate distribution by $e(X)/(1-e(X))$ move it toward the treated covariate distribution?
In the unnormalized sample estimator, why is the weighted control outcome total divided by $n_1$?
Name the assumptions needed to interpret the weighted contrast as a causal ATT, and explain what weighting cannot fix if an important confounder is unmeasured.