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The Balancing Property and Estimation · Tutorial 118 of 699

Known Versus Estimated Propensity Scores

Learn why the true propensity score balances covariates in the population, while an estimated score need not balance them exactly in the observed sample.

Intermediate 10 min read

What You'll Learn

  • State the population balancing property of the true propensity score.
  • Derive why covariate distributions agree within a true-score stratum.
  • Distinguish population balance from balance in a realized sample.
  • Explain how model error and data-dependent estimation can leave fitted-score imbalance.
  • Diagnose imbalance without treating a fitted score as proof of balance.
  • Compare estimated scores with using the true score or directly checking covariates.

Why Known and Estimated Scores Must Be Distinguished

A propensity score is defined from the treatment assignment mechanism, not from a particular fitted model. Its population balancing result is exact: among units with the same true score, treatment status is independent of measured baseline covariates. In practice, however, the true treatment mechanism is usually unknown, so analysts estimate scores from data. The resulting fitted values are not automatically the true scores, and even a well-behaved fitting procedure does not promise exact balance in every finite sample.

Keeping those claims separate matters when interpreting matching, weighting, or other procedures that use propensity scores. A balance theorem about a population quantity is not a guarantee about the covariate counts in a particular data set. Balance must be assessed in the data and in the sample actually used for analysis.

Definition: Let \(A\) be a binary treatment indicator, with \(A=1\) denoting treatment, and let \(X\) denote measured pretreatment covariates. The true propensity score is \(e(X)=P(A=1\mid X)\). A fitted propensity score for unit \(i\) is an estimated value \(\hat e_i=\hat e(X_i)\) produced using a model and data. The true score is a feature of the population assignment mechanism; fitted scores are estimates and can depend on the observed sample.

The true-score balancing property concerns the conditional distribution of \(X\) among treated and untreated units at a fixed score value. It is not a claim that treated and untreated units have identical covariates overall. The score can vary across the population, and covariate distributions can differ substantially before conditioning on it. For the causal use of the score, positivity—\(0<e(X)<1\) on the covariate values of interest—is also important. The balancing identity itself is about treatment assignment and covariates; causal interpretations additionally require causal identification assumptions such as conditional exchangeability.

Why the True Score Balances Covariates

For any covariate-defined set \(B\), consider units whose true score equals \(s\), where \(0<s<1\). On this set, the probability of treatment is \(s\) for every covariate value. The probability of both belonging to \(B\) and receiving treatment, conditional on the score, is therefore \(s\) times the probability of belonging to \(B\). Dividing by the treatment probability within the score stratum gives the treated covariate distribution. The same calculation for controls cancels the factor \(1-s\).

$$ \begin{aligned} P(X\in B\mid A=1,e(X)=s) &=\frac{P(A=1,X\in B\mid e(X)=s)}{P(A=1\mid e(X)=s)}\\ &=\frac{s\,P(X\in B\mid e(X)=s)}{s}\\ &=P(X\in B\mid e(X)=s),\\[4pt] P(X\in B\mid A=0,e(X)=s) &=\frac{(1-s)\,P(X\in B\mid e(X)=s)}{1-s}\\ &=P(X\in B\mid e(X)=s). \end{aligned} $$

Since these equalities hold for every covariate-defined set \(B\), they state that \(A\) and \(X\) are independent conditional on the true score: \(A\perp X\mid e(X)\). The argument is a population calculation. It does not say that each realized sample will have precisely equal covariate frequencies within a score stratum.

Worked Example: Balance Within a True-Score Stratum

Suppose a hypothetical population has three covariate categories, \(U\), \(V\), and \(W\), with probabilities \(P(U)=0.20\), \(P(V)=0.30\), and \(P(W)=0.50\). Its true scores are \(e(U)=0.20\), \(e(V)=0.20\), and \(e(W)=0.80\). Find the proportion in category \(U\) among treated and untreated units in the score stratum \(e(X)=0.20\).

Within this stratum, the total population probability is \(0.20+0.30=0.50\). The proportion in category \(U\), before separating by treatment, is \(0.20/0.50=0.40\).

Among the whole population, the probability of being in \(U\) and treated is \(0.20(0.20)=0.04\). The probability of being in the low-score stratum and treated is \(0.50(0.20)=0.10\). Thus, \(P(U\mid A=1,e(X)=0.20)=0.04/0.10=0.40\).

For controls, the corresponding joint probability is \(0.20(1-0.20)=0.16\), and the probability of being in the low-score stratum and untreated is \(0.50(1-0.20)=0.40\). Hence, \(P(U\mid A=0,e(X)=0.20)=0.16/0.40=0.40\). The two conditional proportions agree, as the true-score property requires.

Why a True Score Does Not Guarantee Sample Balance

The preceding result describes probabilities in a population. A finite data set is one realization of treatment assignments and covariates. Random variation can leave treated and untreated units with noticeably different covariate distributions even when every unit has the same true score. If a procedure further restricts the sample—for example, by retaining only some units for matching—the balance of that selected sample is a separate empirical question.

Worked Example: Random Imbalance Despite a Known Score

Consider a hypothetical sample of 10 treated and 10 untreated units. Suppose \(X\) is binary and the true score is \(e(X)=0.50\) for both values of \(X\). In one possible realization, 8 of the 10 treated units have \(X=1\), while 2 of the 10 untreated units have \(X=1\).

The observed proportions with \(X=1\) are \(8/10=0.80\) among treated units and \(2/10=0.20\) among controls. Their difference is \(0.80-0.20=0.60\), or 60 percentage points.

There is no contradiction with the balancing property. At the population level, \(P(X=1\mid A=1)=P(X=1\mid A=0)=P(X=1)\), because the true score is constant. The displayed counts are a possible finite-sample fluctuation around that equality. A known true score gives the population identity; it does not erase sampling variation.

Why Estimated Scores May Not Balance

An estimated score can differ from the true score for several reasons. The model may omit relevant covariates, impose an inadequate functional form, or be unstable because the sample contains little information in some covariate regions. Even if the model is appropriate, estimated parameters vary from sample to sample. In addition, an in-sample fitted score is calculated using data that include the observed treatment assignments. The fixed-score population argument cannot simply be applied to these data-dependent values as if they were known population probabilities.

Under suitable conditions, a correctly specified and consistently estimated model can make fitted scores approach the true scores as sample size grows. That is an asymptotic statement, not an exact finite-sample guarantee. It does not imply that every fitted-score stratum has identical treated and control covariate distributions, nor that every matching or weighting implementation achieves balance. The quality of the result depends on the model, the data, and the subsequent analysis.

Worked Example: A Common Fitted Score Can Conceal Imbalance

Suppose a hypothetical data set has 20 units. Among the 10 units with \(X=A\), 8 are treated and 2 are controls. Among the 10 units with \(X=B\), 2 are treated and 8 are controls. An analyst uses an intercept-only treatment model, which assigns the same fitted score to every unit. The overall treated proportion is \(10/20=0.50\), so the common fitted score is \(0.50\).

All units therefore have the same fitted score, but their covariates are not balanced. Among treated units, the proportion with \(X=A\) is \(8/10=0.80\). Among controls, it is \(2/10=0.20\). The difference is \(0.80-0.20=0.60\).

The equal fitted values do not establish balance: the model has compressed two covariate groups into one score value. If the true treatment probabilities differ by \(X\), this simple model also fails to represent the assignment mechanism. In either case, inspecting the fitted score alone would miss the observed covariate imbalance.

Diagnostics, Comparisons, and Limitations

Balance diagnostics should be applied to the groups that will actually be compared, after any matching, weighting, or restriction. Examine each important baseline covariate directly, not just the distribution of fitted scores. For a binary covariate, compare group proportions and their difference. For a quantitative covariate, compare means and inspect distributional summaries or plots; equal means alone do not rule out differences in spread or shape. Standardized differences can help compare imbalances across covariates measured on different scales, but no single threshold proves that all confounding has been removed.

The main alternatives answer different questions. If the true score were known, the population balancing identity would be exact, but the observed sample could still fluctuate. Using an estimated score makes the method practicable, but introduces model and estimation uncertainty and requires empirical diagnostics. Direct covariate adjustment does not rely on a score to summarize \(X\), but it has its own modeling and design choices. None of these approaches makes unmeasured covariates balanced merely by construction.

1
Clarify which score is being used.
Distinguish the population quantity \(e(X)\) from the fitted values \(\hat e_i\), and record how the fitted values were obtained.
2
Check covariates in the analysis sample.
Compare treated and control distributions for the baseline covariates relevant to the design, rather than relying on similar score distributions alone.
3
Investigate poor balance.
Look for model misspecification, sparse covariate regions, lack of overlap, and restrictions that leave unlike groups. Revise the design and reassess rather than assuming the score has solved the problem.

A common mistake is to reason, “The propensity score balances covariates by definition, so the fitted scores must have made my sample balanced.” The first clause applies to the true score in the population; the conclusion does not follow for estimated scores in a finite sample. Similar fitted-score distributions are not proof of covariate balance, and covariate balance by itself is not proof that all causal assumptions hold. Diagnostics can reveal problems in measured covariates, but cannot verify the absence of unmeasured confounding.

Key takeaway: The true propensity score has an exact population balancing property. Estimating a score does not automatically transfer that exact property to finite-sample fitted values. Assess covariate balance in the analysis sample and interpret it alongside the modeling assumptions and causal assumptions.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. In one sentence, distinguish the true propensity score from a fitted propensity score.
  2. In a true-score stratum with \(e(X)=0.30\), suppose \(P(X\in B\mid e(X)=0.30)=0.40\). What is \(P(X\in B\mid A=1,e(X)=0.30)\), under the balancing property?
  3. A sample has a known true score of \(0.50\) for every unit, but observed proportions with a binary covariate are 0.65 among treated units and 0.45 among controls. Is this logically incompatible with the balancing property? Why?
  4. An intercept-only fitted model assigns every unit a score of 0.40. Does that establish balance on a measured binary covariate? What should be checked?

Solutions: (1) The true score is \(P(A=1\mid X)\) in the population; a fitted score is an estimate produced from a model and data. (2) It is \(0.40\), because the true score makes \(A\) independent of \(X\) within that score stratum. (3) No. The identity is a population statement, and finite samples can show random imbalance; the observed difference is \(0.65-0.45=0.20\). (4) No. A common fitted value does not establish balance; compare the binary covariate proportions in the treated and control groups in the analysis sample.

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