Why Compare These Estimates?
Inverse-probability weighting and propensity-score subclassification can both estimate an average treatment effect (ATE), but they do not necessarily produce the same number in a finite sample. Comparing them is useful because the difference can reveal how much the result depends on using individual-level propensity scores versus averages within coarser strata. It can also expose differences in the population weights the calculations apply.
The comparison is meaningful only when the estimand and analysis population are held fixed. Here the target is the ATE for the study population: the average difference between each unit’s potential outcome under treatment and its potential outcome under control. Assume a binary treatment \(A_i\), observed outcome \(Y_i\), pretreatment covariates \(X_i\), and propensity score \(e(X_i)=P(A_i=1\mid X_i)\). The observed outcome is \(Y_i=A_iY_i(1)+(1-A_i)Y_i(0)\).
For both estimates to have a causal interpretation, the relevant identifying assumptions are consistency, conditional exchangeability \(Y(a)\perp A\mid X\) for \(a=0,1\), and positivity \(0<e(X_i)<1\) for covariate values in the target population. Consistency connects observed outcomes to potential outcomes; exchangeability says that, conditional on the measured covariates, treatment assignment is independent of the potential outcomes; positivity requires a possibility of either treatment at each relevant covariate value. The propensity-score model must also estimate scores adequately for the weighting procedure. Subclassification requires its strata to control confounding sufficiently well; grouping scores into strata does not guarantee that covariates are balanced within them.
Two Estimators and Their Target Weights
Let \(S_i=k\) indicate that unit \(i\) belongs to stratum \(k\). Write \(n_k\) for the number of units in stratum \(k\), \(n_{1k}\) and \(n_{0k}\) for its treated and control counts, and \(\bar{Y}_{1k}\) and \(\bar{Y}_{0k}\) for the corresponding observed outcome means. With all strata represented in both treatment groups, the usual sample ATE subclassification estimate is
The stratum effect is weighted by \(n_k/n\), the fraction of the study population in that stratum. Thus, this estimator targets the study population’s stratum distribution, not an equally weighted average across strata unless the strata happen to have equal sizes.
One common ATE inverse-probability-weighted estimator is the Horvitz–Thompson form:
In this formula, each treated outcome receives weight \(1/\hat{e}(X_i)\), and each control outcome receives weight \(1/[1-\hat{e}(X_i)]\). Division by \(n\) puts the weighted contrast on the study-population scale. A normalized, or Hájek, version instead divides the weighted treated and control outcome totals by their respective total weights before taking the difference. The two versions can differ in finite samples, so a comparison should state which version is used.
The weighting formula applies the target-population weight at the individual level. Subclassification first forms an outcome contrast in each stratum, then applies the stratum’s population share. This distinction matters when estimated scores vary substantially within strata, when strata have unequal sizes, or when treatment effects differ across strata.
Worked Comparison on One Hypothetical Dataset
Worked Example: Different Individual Scores, Different Estimates
Consider eight hypothetical units, divided into two propensity-score strata. Outcomes are measured in points. Each stratum contains two treated and two control units. The displayed scores are hypothetical estimated propensity scores; the treatment and outcome columns show what was observed.
| Stratum | Treatment \(A_i\) | Estimated score \(\hat{e}(X_i)\) | Observed outcome \(Y_i\) (points) |
|---|---|---|---|
| Low | 1 | 0.30 | 10 |
| Low | 1 | 0.40 | 12 |
| Low | 0 | 0.20 | 6 |
| Low | 0 | 0.30 | 7 |
| High | 1 | 0.70 | 18 |
| High | 1 | 0.80 | 20 |
| High | 0 | 0.60 | 10 |
| High | 0 | 0.70 | 12 |
First calculate the subclassification estimate. In the low stratum, the treated mean is \((10+12)/2=11\) points and the control mean is \((6+7)/2=6.5\) points, so the difference is \(4.5\) points. In the high stratum, the corresponding means are \(19\) and \(11\) points, giving a difference of \(8\) points. Since each stratum contains four of the eight units, each receives weight \(4/8=0.5\):
Now use the Horvitz–Thompson IPW estimator with the individual scores shown. The weighted treated total is \(10/0.30+12/0.40+18/0.70+20/0.80=114.0476\) points. The weighted control total is \(6/0.80+7/0.70+10/0.40+12/0.30=82.5\) points. Therefore,
Both calculations use the same eight units and aim at the study-population ATE, but they yield different estimates: \(6.25\) points by subclassification and \(3.943\) points by IPW. The subclassification calculation uses two within-stratum contrasts and equal stratum weights. IPW uses different weights for individual observations, including larger weights for observations whose estimated treatment probability is lower in their observed group. The difference is not, by itself, evidence that one method is correct and the other is wrong. It is a reason to inspect score behavior, covariate balance, and the details of the analysis.
When the Estimates Coincide
A useful special case gives a direct algebraic link between these methods. Suppose each unit in stratum \(k\) receives the same estimated score, and that score is the observed treatment fraction in that stratum: \(\hat{e}_i=n_{1k}/n_k\) for units in stratum \(k\). Then the Horvitz–Thompson IPW estimate is exactly the subclassification estimate.
To see why, consider the treated contribution from stratum \(k\). There are \(n_{1k}\) treated units, so dividing each treated outcome by \(n_{1k}/n_k\) multiplies their total by \(n_k/n_{1k}\). Dividing the resulting sum by the full sample size \(n\) gives \(n_k/n\) times the treated mean. The same reasoning applies to the control contribution, using \(1-\hat{e}_i=n_{0k}/n_k\). Subtracting these two terms and summing across strata produces the subclassification formula:
Taking the difference for each stratum and summing gives \(\hat{\tau}_{\mathrm{IPW}}=\hat{\tau}_{\mathrm{sub}}\). This is an algebraic identity under the stated score choice, not a general guarantee that independently fitted individual-level scores will reproduce subclassification. The identity also applies to the unnormalized Horvitz–Thompson estimator as written; normalized weighted means need not have the same finite-sample equality.
Worked Example: Confirming the Equality With Stratum-Level Scores
Return to the same hypothetical data. Each stratum has two treated and two control units, so its treatment fraction is \(2/4=0.5\). Assign every unit in each stratum a score of \(0.5\). The low-stratum contrast remains \(4.5\) points and the high-stratum contrast remains \(8\) points. The IPW calculation can now be grouped by stratum:
This equals the subclassification estimate because the common score within each stratum equals that stratum’s observed treatment fraction. The scores in this calculation are deliberately different from the individual scores in the previous calculation: the example illustrates the equality condition, not a claim that the two score specifications are interchangeable in an actual analysis.
Unequal Stratum Sizes Determine the Target Weights
The stratum weights are part of the estimand calculation, not a cosmetic choice. For the study-population ATE, a stratum containing 75% of the study units receives three times the weight of one containing 25%, regardless of how many strata are used. Equal weighting across strata would instead target an artificial population that gives every stratum the same share.
Worked Example: Applying the Study-Population Shares
Suppose a hypothetical study has two strata. The first contains 75% of the units and has a treated-minus-control outcome difference of \(2\) points. The second contains 25% and has a difference of \(10\) points. The ATE subclassification estimate is
Giving each stratum equal weight would instead calculate \(0.5(2)+0.5(10)=6.0\) points. That is not the same study-population ATE calculation: it overweights the smaller stratum relative to its share of the study population. Under the stratum-level score condition derived above, ATE IPW also gives \(4.0\) points. This example demonstrates why a difference in estimates can reflect different target weights, not just different arithmetic.
Diagnostics, Interpretation, and Limitations
When estimates disagree, begin by checking that both calculations use the same outcome scale, analysis sample, target population, and estimand. Then inspect the mechanisms that can create differences:
- Within-stratum balance: Subclassification controls confounding only to the extent that relevant covariates are balanced within each stratum. Similar scores alone do not prove that every covariate is balanced.
- Individual weights: IPW can be strongly affected by observations with large inverse-probability weights. Check the range and distribution of weights and whether a small number of observations dominate the weighted totals.
- Stratum contributions: Report each stratum’s size, treated and control counts, estimated contrast, and contribution to the overall estimate. A large change in one stratum can explain much of the difference.
- Score specification: Compare the score estimates used for weighting with the score groupings used for subclassification. Coarse strata discard within-stratum score variation; weighting retains it.
- Uncertainty: A difference between point estimates does not establish that the methods estimate different causal effects. Compare uncertainty using an inference procedure appropriate to each estimator; if using a bootstrap, repeat the propensity-score estimation and analysis steps within each resample.
Treatment-effect heterogeneity can make the role of weights especially visible. If stratum-specific effects differ, changing the weights changes their average. Conversely, even when effects are constant, finite-sample variation, imperfect score estimation, and residual confounding within strata can lead to differing estimates. Neither estimate should be selected merely because it is larger, smaller, or more appealing.
Both approaches have limitations. Subclassification can leave residual confounding within strata and its estimate can depend on how strata are formed. IPW uses individual scores but can be sensitive to extreme weights and model misspecification. In either case, causal interpretation still relies on the identifying assumptions; computation alone cannot establish exchangeability or positivity. The introductory material on propensity-score subclassification and ATE weighting supplies the procedures, while this comparison emphasizes that their weights encode how the target population is represented.
Common Mistake
Correction: They can target the same ATE without being algebraically identical in a finite sample. Exact equality holds under the particular stratum-level score condition shown above for the Horvitz–Thompson form. With individual scores, the estimators apply different finite-sample weights and may differ. First verify the formulas, sample, and target weights; then investigate balance, score behavior, and uncertainty before drawing conclusions.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A study has stratum shares 0.60 and 0.40, with within-stratum differences of 3 and 8 outcome points. What is the subclassification estimate?
- In a stratum with \(n_k=20\), \(n_{1k}=8\), and \(n_{0k}=12\), what common score in that stratum makes the Horvitz–Thompson contribution align with subclassification?
- Why can estimates differ when both methods use the same sample and aim at the ATE?
- Does exact agreement between IPW and subclassification establish that exchangeability holds? Explain.
- If one stratum contains 80% of the study population, should it receive weight 0.5 in the ATE subclassification estimate just because there are two strata?
Solutions. (1) The estimate is \(0.60(3)+0.40(8)=1.8+3.2=5.0\) outcome points. (2) The common score is \(n_{1k}/n_k=8/20=0.40\); its complement is \(0.60=12/20\). (3) Individual-level IPW scores and stratum-level contrasts apply different finite-sample weights; coarse strata also discard within-stratum score variation. (4) No. Agreement is a numerical property of the estimates and does not verify conditional exchangeability, which is an identifying assumption about treatment assignment and potential outcomes. (5) No. For the study-population ATE, that stratum receives weight 0.80; the other receives 0.20. Equal weighting would represent a different target distribution.