Why “Coarsest” Is an Information Claim
A covariate vector may contain many details about each unit, while treatment assignment depends on only some of those details. The propensity score compresses the covariates to a single treatment probability. The important result here is not merely that this compression balances covariates: it is that every balancing score must retain enough information to determine the propensity score.
This gives “coarsest” a precise meaning. It does not mean the propensity score has the fewest possible numerical values in every data set, nor that every other balancing score must be a function of the propensity score. It means that, in terms of information about treatment assignment, any balancing score is at least as informative as the propensity score.
The conditional-independence definition concerns the population distribution. It does not require treatment to have been randomized, and it does not assert that treatment and covariates are marginally independent. Instead, it says that after conditioning on the value of \(b(X)\), knowing the covariates does not further change the probability of treatment.
The Coarsest-Balancing-Score Result
The key step is to express conditional independence in terms of treatment probabilities. Because \(b(X)\) is a function of \(X\), conditioning on both \(X\) and \(b(X)\) gives the same treatment probability as conditioning on \(X\) alone. If \(b(X)\) is balancing, conditional independence also says that this probability equals the probability given \(b(X)\) alone.
The first equality is the definition of the propensity score. The second follows because \(b(X)\) is determined by \(X\); observing it adds no information once \(X\) is known. The third follows from \(A\perp X\mid b(X)\). The final equality defines \(g\) as the treatment probability at a given score value. Conditional probabilities are defined up to sets of probability zero, so the conclusion is an almost-sure statement.
The converse fact helps complete the interpretation: the propensity score is itself a balancing score. Indeed, \(e(X)\) is a function of \(X\), and \(P(A=1\mid X,e(X))=e(X)=P(A=1\mid e(X))\). For binary \(A\), equality of these conditional treatment probabilities gives \(A\perp X\mid e(X)\). Therefore the propensity score is balancing, and every other balancing score determines it.
This is a statement about information, not about the number of labels used to encode a score. Relabeling values does not change what a score distinguishes. Nor does the theorem say that a balancing score contains all information in \(X\), or that it is sufficient for predicting outcomes. It only concerns the covariate information needed to determine the treatment probability.
Worked Example: Grouping Covariate Categories by Treatment Probability
Worked Example: Grouping Covariate Categories by Treatment Probability
Suppose a hypothetical population has a categorical covariate \(X\) with four values. Their population proportions and treatment probabilities are as follows.
| Covariate value | \(P(X=x)\) | \(e(x)=P(A=1\mid X=x)\) | Score \(b(x)\) |
|---|---|---|---|
| North | 0.20 | 0.20 | Low |
| South | 0.30 | 0.20 | Low |
| East | 0.20 | 0.60 | High |
| West | 0.30 | 0.60 | High |
The score \(b(X)\) merges North with South and East with West. In the Low group, the treatment probability is
The same calculation for the High group gives
Thus \(e(X)=g(b(X))\), where \(g(\text{Low})=0.20\) and \(g(\text{High})=0.60\). The grouping is also balancing. For example, within the Low group the treated population masses for North and South are \(0.20(0.20)=0.04\) and \(0.30(0.20)=0.06\). After dividing by the total treated mass \(0.10\), the conditional proportions are 0.40 and 0.60. Among untreated units, the masses are \(0.20(0.80)=0.16\) and \(0.30(0.80)=0.24\); dividing by \(0.40\) again gives 0.40 and 0.60. The covariate distribution is therefore the same in both treatment groups within Low. The corresponding calculation within High gives 0.40 East and 0.60 West in each treatment group.
Here the score is coarser than the full covariate: it does not distinguish North from South, or East from West. Those distinctions are unnecessary for balancing because treatment probability is constant within each score group.
Why Unequal Propensities Cannot Be Merged
For a discrete covariate, the theorem has a direct implication. If a proposed balancing score places two covariate values in the same score group, their propensity scores must be equal. Otherwise, treatment probabilities vary within the group, contradicting the balancing condition.
Worked Example: Detecting a Failed Coarsening
Use the population in the preceding example, but now assign all four covariate values the same score, \(b(X)=\text{One group}\). The treated population masses for North, South, East, and West are respectively \(0.04,\ 0.06,\ 0.12,\ 0.18\). Their sum is \(0.40\), so the conditional covariate proportions among treated units are
The untreated masses are \(0.16,\ 0.24,\ 0.08,\ 0.12\), summing to \(0.60\). Hence
These distributions differ, so the one-group score is not balancing. The theorem shows the same failure through treatment probabilities: a constant score would imply that \(e(X)\) is constant, but the actual propensity scores are 0.20 and 0.60. The covariate values with unequal propensities cannot be merged into a single balancing-score value.
For finite-valued \(X\) and \(b(X)\), the logic can be summarized simply: a score group is balancing precisely when treatment probability is constant across the covariate values in that group, provided both treatment groups occur there. The propensity score records that constant. For more general covariates, the theorem gives the same conclusion through conditional probabilities rather than a finite table.
Worked Example: A Balancing Score Can Keep Extra Detail
Worked Example: A Balancing Score Can Keep Extra Detail
Suppose a hypothetical covariate \(X\) has three categories with population proportions 0.25, 0.35, and 0.40. Their treatment probabilities are 0.30, 0.30, and 0.70. Define a score \(b(X)\) that assigns a separate label to each category: \(b(X)=a\), \(b(X)=b\), or \(b(X)=c\).
Within each score value there is only one covariate category, so the covariate distribution is identical among treated and untreated units in that score group. Thus \(b(X)\) is balancing. Its score-specific treatment probabilities are
The propensity score is recovered by mapping both \(a\) and \(b\) to 0.30 and \(c\) to 0.70. The balancing score \(b(X)\) distinguishes the first two categories even though their propensities match. It is therefore balancing but more detailed than the propensity score for treatment assignment. This example illustrates why the theorem says that \(e(X)\) is a function of \(b(X)\), not that every balancing score must be a function of \(e(X)\).
Assumptions, Diagnostics, and Limits
The result is a mathematical identity under the stated definitions, not a causal-identification theorem. It requires a well-defined binary treatment \(A\), pretreatment covariates \(X\), and a score \(b(X)\) that is genuinely balancing in the population. It does not require an outcome model or an assumption that all confounding has been measured. A balancing score alone does not establish that treatment groups are exchangeable with respect to potential outcomes.
Positivity is a separate consideration. The coarsest-score theorem itself does not require \(0<e(X)<1\): the function identity can still hold when some covariate patterns always receive one treatment. But estimating a treated-versus-untreated contrast within a score group requires both groups to be represented there. A common positivity condition is \(0<e(X)<1\) over the covariate patterns relevant to the target population. Deterministic treatment within a pattern prevents a direct within-pattern comparison, even though \(e(X)\) remains defined.
The ideal result concerns the true propensity score and exact population balance. In applied work, the score is not generally known, and finite samples can show imbalance even when a population-level argument is sound. Useful checks include comparing covariate distributions across treatment groups within score strata and examining whether important covariate categories remain unevenly represented. Persistent imbalance may indicate that the proposed score does not balance in practice, that data are sparse, or that estimates are imprecise. A numerical prediction score is not proof of balance.
The main alternatives clarify the result’s role. Conditioning on the full covariate vector retains all measured baseline detail, but can be unwieldy when \(X\) is high-dimensional. A general balancing score can reduce that detail, but the theorem guarantees it must determine \(e(X)\). The propensity score is therefore a particularly compact balancing summary, not a guarantee that every implementation will achieve balance or that causal assumptions hold. These comparisons concern the balancing property; the choice of an estimation procedure is a separate issue.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- If \(b(X)\) is balancing, write the conditional-probability chain that shows \(e(X)\) is a function of \(b(X)\).
- A discrete covariate has three categories with propensities 0.10, 0.10, and 0.80. Which categories may be combined into one score group without violating the theorem?
- Can a score that assigns a unique value to every covariate pattern be balancing? Explain what the theorem says in that case.
- Does the coarsest-balancing-score result require positivity? What additional issue arises if a score stratum contains only treated units?
Solutions. (1) Since \(b(X)\) is determined by \(X\) and is balancing, \(P(A=1\mid X)=P(A=1\mid X,b(X))=P(A=1\mid b(X))\). The final expression is a function of \(b(X)\), so it equals \(e(X)\) almost surely. (2) The first two categories may be combined because both have propensity 0.10. Neither can be combined with the third, whose propensity is 0.80, if the resulting score is to be balancing. (3) Yes. Each score stratum contains just one covariate pattern, so it is balancing; the propensity is recovered by mapping each score value to that pattern’s treatment probability. (4) No. The function identity does not require positivity. But if a stratum contains only treated units, there is no untreated group within it for a treated-versus-untreated comparison, so positivity for that comparison fails.