Stabilized Inverse Probability Weights

Why Stabilize Inverse Probability Weights?

Inverse probability weighting (IPW) reweights observed outcomes to estimate a causal effect in a target population. For the average treatment effect (ATE), the unstabilized weight is large when a unit received a treatment that was unlikely given that unit’s covariates. This can create a weighted sample with highly variable weights. Stabilization multiplies each inverse probability weight by the marginal probability of the treatment actually received. The multiplication changes the weights’ scale, often reducing their average magnitude, but it does not change the ATE target when the weighted arm means are calculated appropriately.

Let $A$ be a binary treatment, with $A=1$ for treated and $A=0$ for untreated; let $X$ be the measured pretreatment covariates; and let $Y$ be the observed outcome. Write $Y(1)$ and $Y(0)$ for the potential outcomes under treatment and no treatment, so consistency gives $Y=A Y(1)+(1-A)Y(0)$. Define the propensity score $e(X)=P(A=1\mid X)$ and the marginal treatment probability $p=P(A=1)$. In a sample, $p$ is commonly estimated by the treated fraction $\hat p=n_1/n$, where $n_1$ is the number treated among $n$ units.

Definition: The stabilized ATE weight for unit $i$ is $$ SW_i=\frac{A_i p}{e(X_i)}+\frac{(1-A_i)(1-p)}{1-e(X_i)}. $$ Equivalently, a treated unit receives $p/e(X_i)$, and an untreated unit receives $(1-p)/(1-e(X_i))$. In practice, $e(X_i)$ and $p$ are usually estimated from the data; replacing them by estimates gives estimated stabilized weights.

The numerator is marginal: it depends on treatment status, not on $X_i$. It is not an additional adjustment for confounding. The denominator remains the conditional probability of the treatment actually received, which is the part that corrects for the association between measured covariates and treatment.

Assumptions and the ATE Target

The causal interpretation requires the usual identification conditions. Consistency means the observed outcome equals the potential outcome corresponding to the treatment received. Conditional exchangeability requires $(Y(1),Y(0))\perp A\mid X$: after conditioning on measured pretreatment covariates, treatment assignment carries no further information about the potential outcomes. Positivity requires $0<e(X)<1$ for covariate values in the target population. The potential outcomes must also have finite means for the expectations below to be defined. These conditions, together with a correctly specified propensity score or a consistent estimate of it, support identifying the population ATE $E[Y(1)-Y(0)]$.

The key identity follows by conditioning on $X$. For treated units, conditional exchangeability allows the treated outcome to represent the potential outcome under treatment, and consistency identifies that outcome with $Y$. Thus:

\begin{aligned} E\left[\frac{A pY}{e(X)}\right] &= E\left[E\left\{\left.\frac{A pY}{e(X)}\right|X\right\}\right]\\ &= E\left[\frac{p}{e(X)}E\{AY\mid X\}\right]\\ &= E\left[\frac{p}{e(X)}e(X)E\{Y(1)\mid X\}\right]\\ &=pE[Y(1)]. \end{aligned}

For untreated units, the same argument uses $P(A=0\mid X)=1-e(X)$, giving $E[(1-A)(1-p)Y/\{1-e(X)\}]=(1-p)E[Y(0)]$. Therefore, each stabilized weighted outcome total is scaled by its arm’s marginal probability. Dividing the treated total by $p$, and the untreated total by $1-p$, recovers the respective potential-outcome means in expectation. The scale factors do not replace or redefine the target: both means still refer to the same target population, so their difference is the ATE.

Calculating and Interpreting Stabilized Weights

Worked Example: Computing Weights From Propensity Scores

Consider four hypothetical units with estimated propensity scores $0.20, 0.40, 0.60,$ and $0.80$. Two are treated, so the sample marginal treatment probability is $\hat p=2/4=0.50$. The treated units have scores $0.20$ and $0.60$; the untreated units have scores $0.40$ and $0.80$.

For the treated unit with $e(X)=0.20$, the stabilized weight is $0.50/0.20=2.50$. For the treated unit with $e(X)=0.60$, it is $0.50/0.60=0.8333$. For the untreated unit with $e(X)=0.40$, it is $0.50/(1-0.40)=0.8333$. For the untreated unit with $e(X)=0.80$, it is $0.50/(1-0.80)=2.50$.

The corresponding unstabilized weights are $1/e(X)$ for treated units and $1/\{1-e(X)\}$ for untreated units: $5.00, 1.667, 1.667,$ and $5.00$. Every stabilized weight in this example is one-half of its unstabilized counterpart because $\hat p=1-\hat p=0.50$. The relative weights within each arm are unchanged.

In general, a treated unit’s stabilized weight is $p$ times its unstabilized weight, while an untreated unit’s is $1-p$ times its unstabilized weight. Each arm is multiplied by one common positive constant. That fact explains both the scale change and why normalized arm-specific means are unaffected by stabilization.

Why the Target Does Not Change

A common way to estimate each potential-outcome mean is a Hájek, or normalized weighted mean: divide the weighted outcome sum in an arm by the sum of weights in that arm. Within the treated arm, $p$ appears in every stabilized weight and cancels between numerator and denominator. Within the untreated arm, $1-p$ cancels in the same way.

\frac{\sum_i A_i\{p/e(X_i)\}Y_i} {\sum_i A_i\{p/e(X_i)\}} = \frac{\sum_i A_iY_i/e(X_i)} {\sum_i A_i/e(X_i)}

The analogous equality holds for the untreated mean. Thus the stabilized and unstabilized Hájek estimates are exactly the same for a given set of propensity scores and observed data, apart from numerical rounding. This is an algebraic fact, not a claim that either finite-sample estimate is exactly the population ATE.

For a Horvitz–Thompson calculation, the denominator is handled differently. With known $p$, the treated potential-outcome mean is estimated by the stabilized weighted total divided by $np$; the control mean is the stabilized weighted total divided by $n(1-p)$. These divisions remove the marginal-probability factors. Using the stabilized weighted total divided by $n$ alone estimates a quantity scaled by $p$ or $1-p$, not the potential-outcome mean.

Worked Example: Stabilization Cancels in Normalized Means

Use the four hypothetical units above, and suppose their outcomes are $10$ and $14$ for the two treated units, and $8$ and $6$ for the two untreated units. Their stabilized weights are $2.50, 0.8333, 0.8333,$ and $2.50$, in that order.

The stabilized treated mean is $ \{(2.50)(10)+(0.8333)(14)\}/(2.50+0.8333)=36.6667/3.3333=11.00$. The unstabilized treated weights are $5.00$ and $1.667$, giving $ \{(5.00)(10)+(1.667)(14)\}/(5.00+1.667)=73.338/6.667$, approximately $11.00$; the tiny discrepancy is rounding.

The stabilized untreated mean is $ \{(0.8333)(8)+(2.50)(6)\}/(0.8333+2.50)=21.6664/3.3333$, approximately $6.500$. The unstabilized weights, $1.667$ and $5.00$, produce the same normalized mean. The weighted contrast is therefore approximately $11.00-6.500=4.500$ outcome units. This is a calculation from hypothetical observed data, not by itself evidence that the treatment caused a $4.500$-unit effect; that interpretation depends on the identification assumptions.

Worked Example: Correctly Scaling a Stabilized Horvitz–Thompson Estimate

Suppose a hypothetical sample has $n=4$, with one treated unit, so $\hat p=0.25$. The treated unit has estimated propensity $0.20$ and outcome $12$. The three untreated units have propensity scores $0.30, 0.50,$ and $0.80$, with outcomes $4,8,$ and $6$, respectively.

The treated stabilized weight is $0.25/0.20=1.25$, so the stabilized treated weighted outcome total is $(1.25)(12)=15.00$. The Horvitz–Thompson estimate of the treated potential-outcome mean is $15.00/(n\hat p)=15.00/(4\times0.25)=15.00$. Dividing by $n$ alone would instead give $3.750$, which is the treated weighted total on the wrong scale.

For the untreated units, the stabilized weights are $0.75/0.70=1.0714$, $0.75/0.50=1.500$, and $0.75/0.20=3.750$. Their weighted outcome total is $(1.0714)(4)+(1.500)(8)+(3.750)(6)=38.7856$. The corresponding Horvitz–Thompson estimate of the untreated potential-outcome mean is $38.7856/\{4(0.75)\}=12.9285$. The estimated ATE contrast is $15.00-12.9285=2.0715$ outcome units. The computation illustrates the scale correction; causal interpretation still requires the stated assumptions.

Diagnostics, Comparisons, and Limitations

Stabilization is not a cure for poor overlap. When estimated propensity scores are close to zero for treated units, or close to one for untreated units, the denominators remain small and the corresponding stabilized weights can still be large. Inspect the propensity-score distributions by treatment group and summarize the weight distribution, including its largest values. Also assess covariate balance after weighting: a stabilized formula does not guarantee that the observed weighted covariate distributions are balanced, especially when the propensity model is misspecified or the sample is limited.

The main alternatives have different finite-sample properties and purposes. Unstabilized IPW uses $1/e(X_i)$ and $1/\{1-e(X_i)\}$; stabilized IPW multiplies each arm’s weights by its marginal treatment probability. For a normalized Hájek mean, this multiplication cancels exactly, so stabilization does not alter that estimate. For a Horvitz–Thompson estimator, the stabilized totals must be divided by the corresponding marginal treatment probabilities as shown above. Neither choice of numerator repairs unmeasured confounding or positivity violations.

Common mistake: “The stabilized treated weights are smaller, so they target a smaller treated subpopulation, and the ATE must change.” The correction is that the numerator $p$ is constant across treated units and $1-p$ is constant across untreated units. In normalized arm-specific means, these constants cancel. In Horvitz–Thompson means, the corresponding constants are removed by dividing by $p$ and $1-p$. Stabilization changes scale, not the target estimand.

Stabilized weights can be easier to report and may reduce the raw magnitude of weights, but they do not necessarily reduce every measure of variability, and they do not ensure stable estimates. Extreme weights remain a warning about limited overlap and sensitivity to a small number of observations. Any decision to alter, truncate, or otherwise modify weights changes the estimator and requires separate justification; stabilization alone does not make such a decision.

Practical Construction

Estimate the marginal treatment probability.
For a binary treatment, calculate $\hat p=n_1/n$, or specify the appropriate population marginal probability if it is known.

Obtain propensity scores.
Estimate $e(X_i)=P(A_i=1\mid X_i)$ from pretreatment covariates, and check that each observed treatment assignment has a nonzero estimated probability.

Assign the stabilized weight.
Use $\hat p/\hat e(X_i)$ for treated units and $(1-\hat p)/\{1-\hat e(X_i)\}$ for untreated units.

Choose the estimator and retain its scale.
For normalized arm means, divide weighted outcome sums by the corresponding arm-specific weight sums. For Horvitz–Thompson means, divide stabilized totals by $n\hat p$ and $n(1-\hat p)$, respectively.

Check the weighted data.
Review overlap, extreme weights, and covariate balance, then interpret the contrast only under the identification assumptions.

Check Your Understanding

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

If $\hat p=0.30$ and a treated unit has $\hat e(X_i)=0.15$, what is its stabilized weight?
Why do stabilized and unstabilized Hájek means agree within a treatment arm?
For a stabilized Horvitz–Thompson treated total, what divisor gives the estimated treated potential-outcome mean?
Does multiplying weights by the marginal treatment probability resolve a positivity problem? Explain.

Solutions. (1) The weight is $0.30/0.15=2.00$. (2) The same positive factor $\hat p$ multiplies every treated weight, so it cancels from the weighted numerator and denominator; the untreated factor $1-\hat p$ cancels similarly. (3) Divide by $n\hat p$, not by $n$ alone. (4) No. If the conditional probability of the observed treatment is near zero, the denominator remains near zero and the weight can remain extreme; stabilization does not create overlap or correct unmeasured confounding.