Tutorials › Marginal Structural Models › Why Propensity Model Fit Is Not Balance

Diagnostics and Sensitivity to Misspecification · Tutorial 268 of 501

Why Propensity Model Fit Is Not Balance

Learn why treatment prediction and post-weighting balance are distinct targets, and how the population propensity score connects them.

Advanced 12 min read

What You'll Learn

  • Distinguish treatment-prediction performance from the balance target of inverse-probability weighting.
  • Derive the population balancing identity for a correctly specified propensity score.
  • See how marginal calibration can coexist with substantial covariate imbalance.
  • Calculate how a highly discriminative but miscalibrated model can worsen weighted balance.
  • Compare prediction metrics with direct checks of weighted covariate distributions.
  • Identify assumptions, finite-sample limitations, and common interpretation errors.

Prediction and Balance Are Different Targets

A treatment model predicts treatment assignment from measured covariates. A weighting analysis uses the model’s predicted probabilities to reweight the observed treatment groups. These operations are connected, but they do not ask the same question. Predictive fit asks whether the model assigns probabilities that describe treatment assignment well; balance asks whether weighting has made the measured covariate distributions comparable across treatment groups.

The distinction matters because inverse-probability weights depend on the numerical values of predicted probabilities, not just on how well the model ranks people by treatment likelihood. A model may have a strong likelihood or discrimination measure and still give weights that leave important covariates imbalanced. Conversely, a model can balance covariates well even if it does not classify treatment assignment especially accurately.

Definition: Let \(A\in\{0,1\}\) be treatment and \(L\) a vector of measured pretreatment covariates. The propensity score is \(e(L)=P(A=1\mid L)\). Given a fitted probability \(q(L)\), the unstabilized inverse-probability weight is \(A/q(L)+(1-A)/\{1-q(L)\}\). Covariate balance means that the covariate distributions, or specified features of them, are similar across the weighted treatment groups.

The population balancing claim requires the actual conditional probability \(e(L)\), not merely a fitted probability that performs well on some prediction metric. For a causal interpretation of the resulting weighted comparison, additional assumptions are needed: consistency, conditional exchangeability given \(L\), and positivity. In particular, positivity requires \(0<e(L)<1\) on the covariate support relevant to the target population. For finite-variance weighted summaries, stronger bounds away from zero and one, together with suitable moment conditions, are often useful.

Why the True Propensity Score Balances Covariates

The key mathematical connection is an expectation identity. It applies to any integrable function of the measured covariates, not just to a single covariate mean. The result is about population weighted distributions; it does not promise exact balance in a finite sample.

Theorem: Suppose \(e(L)=P(A=1\mid L)\), \(0<e(L)<1\) almost surely, and \(h(L)\) is integrable. Then
$$ E\!\left[\frac{A h(L)}{e(L)}\right]=E[h(L)] \qquad\text{and}\qquad E\!\left[\frac{(1-A)h(L)}{1-e(L)}\right]=E[h(L)]. $$

Proof. By iterated expectation and the definition of \(e(L)\),

$$ \begin{aligned} E\!\left[\frac{A h(L)}{e(L)}\right] &=E\!\left[E\!\left\{\left.\frac{A h(L)}{e(L)}\right|L\right\}\right]\\ &=E\!\left[\frac{h(L)}{e(L)}E(A\mid L)\right]\\ &=E[h(L)]. \end{aligned} $$

For the untreated group, \(E(1-A\mid L)=1-e(L)\), so the same calculation gives

$$ E\!\left[\frac{(1-A)h(L)}{1-e(L)}\right] =E\!\left[\frac{h(L)}{1-e(L)}E(1-A\mid L)\right] =E[h(L)]. $$

Setting \(h(L)=1\) shows that each group’s expected unnormalized weight is one. Taking \(h(L)\) to be a covariate, an indicator of a covariate-defined set, or another integrable function shows that both weighted groups reproduce the covariate distribution in the full population. In data, estimated probabilities and random sampling make the empirical weighted distributions only approximate this identity. \(\square\)

This identity explains what correct propensity probabilities achieve. It does not say that maximizing a treatment-model likelihood guarantees correct probabilities, nor that any measure of predictive success verifies the identity’s conditions. A fitted model can miss the conditional treatment probabilities even when its overall predictions appear plausible.

Three Examples of Prediction and Balance Diverging

Worked Example: Marginal Calibration but Imbalance

Suppose \(L\) is binary, \(P(L=1)=P(L=0)=0.5\), and the actual treatment probabilities are \(e(1)=0.9\) and \(e(0)=0.1\). A model predicts \(q(L)=0.5\) for everyone. Calculate the treatment prevalence, the weighted covariate means using \(q\), and what happens with the true propensity score.

The population treatment prevalence is

$$ P(A=1)=0.5(0.9)+0.5(0.1)=0.5. $$

Therefore, among everyone assigned the same prediction \(q=0.5\), the population treatment probability is \(0.5\): the prediction is calibrated at that single score value. In a finite sample, however, the observed treatment frequency need not equal \(0.5\).

The fitted weights are \(1/0.5=2\) for treated people and \(1/(1-0.5)=2\) for untreated people. Among treated people, the weighted mass with \(L=1\) is \(P(A=1,L=1)\times2=0.5(0.9)\times2=0.9\). The total treated weighted mass is \(P(A=1)\times2=1\), so the weighted mean of \(L\) is \(0.9\). Among untreated people, the weighted mass with \(L=1\) is \(P(A=0,L=1)\times2=0.5(0.1)\times2=0.1\), and the total is also one. The untreated weighted mean is \(0.1\). Thus the difference in weighted means is \(0.9-0.1=0.8\), despite marginal calibration.

With the true probabilities, the treated weighted mass with \(L=1\) is \(0.5(0.9)/0.9=0.5\), and with \(L=0\) it is \(0.5(0.1)/0.1=0.5\). The untreated weighted masses are \(0.5(0.1)/0.1=0.5\) for \(L=1\) and \(0.5(0.9)/0.9=0.5\) for \(L=0\). Both groups have weighted mean \(L=0.5\), matching the population.

Worked Example: Strong Ranking, Poor Weighted Balance

Keep \(P(L=1)=P(L=0)=0.5\), but let \(e(1)=0.8\) and \(e(0)=0.2\). The fitted model assigns \(q(1)=0.99\) and \(q(0)=0.01\). First assess its discrimination, then calculate the normalized weighted means.

Among treated people, \(P(L=1\mid A=1)=0.8\); among untreated people, \(P(L=1\mid A=0)=0.2\). A treated person’s score exceeds an untreated person’s score when the former has \(L=1\) and the latter has \(L=0\), which occurs with probability \(0.8(0.8)=0.64\). Ties occur when both have \(L=1\) or both have \(L=0\), with total probability \(0.8(0.2)+0.2(0.8)=0.32\). Counting ties as one-half gives an area under the ROC curve of \(0.64+0.5(0.32)=0.80\): the model ranks treatment assignment well.

For the treated group, the weighted mass with \(L=1\) is \(0.5(0.8)/0.99=0.40404\), while that with \(L=0\) is \(0.5(0.2)/0.01=10\). Its normalized weighted mean is therefore \(0.40404/(0.40404+10)\approx0.0388\). For the untreated group, the weighted masses are \(0.5(0.2)/0.01=10\) for \(L=1\) and \(0.5(0.8)/0.99=0.40404\) for \(L=0\). Its normalized weighted mean is \(10/(10+0.40404)\approx0.9612\). The weighted difference is about \(-0.9224\), although the unweighted difference was \(0.8-0.2=0.6\).

The model preserves the ordering of treatment probabilities but gives probabilities that are far from the true values. The resulting inverse weights are correspondingly unsuitable for balancing \(L\). High discrimination alone does not certify the numerical probabilities needed for weighting.

Worked Example: Modest Discrimination, Exact Population Balance

Let \(P(L=1)=P(L=0)=0.5\), with true probabilities \(e(1)=0.55\) and \(e(0)=0.45\). Use these probabilities as the fitted model. Find its discrimination and the weighted means of \(L\).

Among treated people, \(P(L=1\mid A=1)=0.55\), and among controls, \(P(L=0\mid A=0)=0.55\). The area under the ROC curve is \(0.55(0.55)+0.5\{0.55(0.45)+0.45(0.55)\}=0.3025+0.2475=0.55\), a modest ranking result.

The treated weighted mass with \(L=1\) is \(0.5(0.55)/0.55=0.5\), and with \(L=0\) it is \(0.5(0.45)/0.45=0.5\). The untreated weighted mass with \(L=1\) is \(0.5(0.45)/0.45=0.5\), and with \(L=0\) it is \(0.5(0.55)/0.55=0.5\). Each normalized weighted mean is \(0.5\). The model has modest discrimination but, because its probabilities are correct, achieves exact population balance for \(L\).

Why Ranking Metrics Cannot Certify Weighting Probabilities

The second mathematical point formalizes what the examples illustrate. AUC depends on pairwise score ordering, whereas inverse-probability weights depend on probability magnitudes. If \(q\) is a score and \(g\) is strictly increasing, replacing \(q\) by \(g(q)\) leaves every strict ordering unchanged. Ties are unchanged as well. Consequently, the AUC is unchanged.

Result: A strictly increasing transformation of a treatment score preserves its AUC, but generally changes inverse-probability weights. Therefore, AUC alone cannot determine whether those weights balance covariates.

Derivation. For an independent treated score \(Q_1\) and untreated score \(Q_0\), define

$$ \operatorname{AUC}(Q) =P(Q_1>Q_0)+\tfrac12P(Q_1=Q_0). $$

Strict increase gives \(Q_1>Q_0\) exactly when \(g(Q_1)>g(Q_0)\), and \(Q_1=Q_0\) exactly when \(g(Q_1)=g(Q_0)\). Both probabilities in the definition are unchanged. But the treated weight changes from \(1/q\) to \(1/g(q)\), and the untreated weight changes from \(1/(1-q)\) to \(1/\{1-g(q)\}\). Unless the transformation leaves the probabilities unchanged, these weights generally differ. Since weighted covariate summaries use those magnitudes, unchanged AUC does not imply unchanged balance. \(\square\)

A transformation of probabilities need not remain a calibrated probability model. That is precisely why a ranking score and a treatment probability should not be treated as interchangeable: weighting requires probability values that represent the conditional assignment mechanism, not merely a useful ordering.

What to Check and What Can Go Wrong

Treatment-model likelihood, classification accuracy, and AUC summarize prediction from different angles. None is a direct test of post-weighting balance. Balance diagnostics instead compare weighted covariate means, higher moments, or distributions between treatment groups, including across follow-up when the analysis is longitudinal. For a longitudinal treatment history, the weights involve products of conditional treatment probabilities at successive times; errors in those probabilities can compound. The population identity still motivates checking the resulting weighted histories directly.

Quantity examinedWhat it addressesWhat it does not establish
Treatment-model likelihoodHow well the specified probability model fits treatment assignments under its likelihood criterionThat weighted covariate distributions are balanced
AUC or classification accuracyHow well scores rank or classify treatment assignmentThat the predicted probabilities have suitable magnitudes
Calibration summariesWhether predicted probabilities agree with observed frequencies at the level assessedConditional correctness for every relevant covariate pattern, or balance in every covariate feature
Post-weighting balanceWhether specified covariate features or distributions align after weightingThat unmeasured confounding is absent or causal assumptions hold

Calibration itself needs careful interpretation. A prediction can be calibrated on average, as in the first example, while omitting covariate structure that matters for balance. Calibration across coarse groups also only establishes agreement at that resolution; residual differences within a group can remain. Direct balance assessment complements, rather than replaces, examination of the treatment model and the causal assumptions.

A common mistake is to reason, “The model predicts treatment well, so inverse weighting must have removed confounding.” That conclusion does not follow: prediction metrics may emphasize ranking or average fit, while weighting depends on accurate conditional probabilities and overlap. The correction is to inspect post-weighting covariate balance and weight behavior, and to treat poor balance as evidence that the fitted weighting procedure has not achieved its intended measured-covariate target. Good balance, in turn, does not prove that unmeasured confounding is absent or that positivity and exchangeability hold.

The principal limitation is that balance is evaluated for observed covariates and chosen summaries. Two weighted groups can match on means but differ in tails, nonlinear features, or interactions; distributional checks address some of these gaps but cannot establish equality of every feature from finite data. In longitudinal analyses, balance also depends on treatment history and time, so an overall summary can conceal imbalance at particular follow-up times.

Check Your Understanding

For each item, give your answer before reading the solution.

  1. A fitted model has excellent AUC. Does that establish that inverse-probability weighting balances measured covariates? Explain.

    Solution. No. AUC measures ranking, while the weights use probability magnitudes. A model can rank treatment well but give miscalibrated probabilities, as in the second worked example.

  2. If everyone receives a prediction of \(0.5\) and the population treatment prevalence is \(0.5\), what can be said about calibration and balance?

    Solution. The prediction is calibrated at that score value in the population because \(P(A=1\mid q=0.5)=0.5\). This does not establish conditional calibration given covariates or balance after weighting. A finite-sample treatment frequency need not be exactly \(0.5\).

  3. State the population identity that explains why the true propensity score balances an integrable covariate function \(h(L)\) in the treated group.

    Solution. If \(e(L)=P(A=1\mid L)>0\), then \(E[A h(L)/e(L)]=E[E\{A h(L)/e(L)\mid L\}]=E[h(L)]\). The corresponding untreated identity requires \(e(L)<1\).

  4. A model has modest AUC but uses the true conditional treatment probabilities. What does the balancing identity imply, and what does it not imply?

    Solution. The identity implies population balance for integrable functions of the measured covariates, subject to positivity and the stated conditions. It does not guarantee exact finite-sample balance, absence of unmeasured confounding, or validity of the other causal assumptions.

Clinical Biostats