Why Transform a Propensity Score?
The propensity score is the conditional probability of treatment given baseline covariates. Its balancing property makes it useful to condition on a single number rather than the full covariate vector. Sometimes it is convenient to express that number on another scale: for example, probabilities can be mapped to odds or log-odds. The key question is whether changing the scale changes the information available for balancing.
The answer depends on whether the transformation preserves the score exactly. A one-to-one transformation merely relabels score values; conditioning on the transformed value retains the same information. A many-to-one transformation can merge distinct propensity scores, losing information that was essential to balance.
The earlier material on the balancing property establishes that \(e(X)\) is itself a balancing score. This tutorial asks what happens when we replace it with \(T=g(e(X))\). The question concerns the information encoded by the score, not whether a transformation improves a treatment model or changes the causal assumptions.
The One-to-One Transformation Result
Suppose \(g\) is a measurable function that is one-to-one on the values of \(e(X)\) that can occur, and suppose its inverse on that range is measurable. Then \(T=g(e(X))\) and \(e(X)\) determine each other: \(T\) is computed from \(e(X)\), and \(e(X)=g^{-1}(T)\). Thus conditioning on either score gives the same information.
A short derivation makes the result explicit. Because \(e(X)\) is a function of \(X\), conditioning on both \(X\) and \(e(X)\) gives the same treatment probability as conditioning on \(X\). On the other hand, averaging that probability among units with the same score gives the score itself:
For binary \(A\), equality of these conditional treatment probabilities establishes conditional independence: given the score, treatment probability no longer varies with \(X\). Now, if \(T=g(e(X))\) is one-to-one, \(e(X)=g^{-1}(T)\). Therefore the information in \(T\) and in \(e(X)\) is identical. The same conditional-independence statement holds when conditioning on \(T\).
Strict monotonicity is a convenient way to ensure a one-to-one transformation on an interval, but it is not the essential condition. A transformation need not preserve the order of scores; it must preserve their distinct values so that the original score can be recovered. Conversely, a function that appears one-to-one over its full mathematical domain may fail to be one-to-one over the score values actually used, so the relevant range must be checked.
Worked Example: Log-Odds Retain the Same Balance
Consider a hypothetical population with four baseline-covariate profiles. The profiles have the following population probabilities and propensity scores. Treatment is assigned with the stated probability within each profile.
| Covariate profile | Population probability | Propensity score \(e(X)\) |
|---|---|---|
| L1 | 0.20 | 0.25 |
| L2 | 0.20 | 0.25 |
| H1 | 0.30 | 0.75 |
| H2 | 0.30 | 0.75 |
Use the logit transformation \(g(e)=\log\{e/(1-e)\}\). It is one-to-one for \(0<e<1\), with inverse \(g^{-1}(t)=1/(1+\exp(-t))\). The two score values become \(g(0.25)=\log(1/3)\approx-1.099\) and \(g(0.75)=\log(3)\approx1.099\).
Within the low-score stratum, the conditional population probabilities of L1 and L2 are both \(0.20/(0.20+0.20)=0.50\). For treated units, the probability of profile L1 conditional on being in this stratum and treated is \(0.50(0.25)/0.25=0.50\). For untreated units it is \(0.50(0.75)/0.75=0.50\). The same calculation gives 0.50 for L2 in both groups.
In the high-score stratum, the population probabilities of H1 and H2 conditional on that stratum are both \(0.30/(0.30+0.30)=0.50\). Since both profiles have propensity score 0.75, their conditional probabilities remain 0.50 in both treatment groups: for H1, the treated calculation is \(0.50(0.75)/0.75=0.50\), and the untreated calculation is \(0.50(0.25)/0.25=0.50\).
The logit values label exactly the same two strata as the original scores. Thus the equality of covariate distributions within each stratum holds on the logit scale as well. The calculation illustrates the theorem; the balance does not depend on choosing the probability or log-odds scale.
Why Coarsening Can Fail
A many-to-one function \(h\) can send distinct propensity scores to the same transformed value. Conditioning on \(S=h(e(X))\) then combines units who had different treatment probabilities. In general, those units are weighted differently in the treated and untreated groups, so their covariate distributions need not agree.
For binary treatment, the condition can be stated directly. Given \(X\) and \(S=h(e(X))\), the treatment probability is still \(e(X)\), because \(S\) is already determined by \(X\). Given only \(S\), it is the average \(E[e(X)\mid S]\). For treatment to be independent of \(X\) given \(S\), these probabilities must agree:
Thus balance requires \(e(X)=E[e(X)\mid S]\) with probability one: the coarsened score must still contain enough information to recover the original score. If two distinct score values with positive probability are merged, this condition fails within the merged group. A many-to-one function is not guaranteed to fail in every particular data set—for example, it may merge only values that never occur—but it offers no general preservation result.
Worked Example: Merging Two Score Levels Produces Imbalance
Suppose a hypothetical population has two equally common covariate profiles, \(X=L\) and \(X=H\). Their propensity scores are 0.20 and 0.80, respectively. A coarsening maps both scores to the same value, \(S=0\). Thus every unit is in the same coarsened-score group.
Among treated units in that group, the probability of profile \(L\) is proportional to its population probability times its treatment probability. The two unnormalised quantities are \(0.50(0.20)=0.10\) for L and \(0.50(0.80)=0.40\) for H. Therefore \(P(X=L\mid A=1,S=0)=0.10/(0.10+0.40)=0.20\).
Among untreated units, the corresponding quantities are \(0.50(0.80)=0.40\) for L and \(0.50(0.20)=0.10\) for H. Therefore \(P(X=L\mid A=0,S=0)=0.40/(0.40+0.10)=0.80\).
The two conditional probabilities, 0.20 and 0.80, are not equal. The coarsened score is not balancing, even though it is derived from a balancing score. The failure comes from merging the two distinct treatment probabilities, not from a problem with the original propensity score.
Exact Equality, Numerical Scale, and Diagnostics
An invertible transformation preserves exact score strata: two units have equal original scores if and only if they have equal transformed scores. It does not preserve every numerical feature of the score. In particular, distances, averages, rounding, and calipers depend on the chosen scale. A one-to-one transformation therefore guarantees the population balancing property under exact conditioning, but it does not guarantee identical results from a procedure that uses approximate distances or finite-width bins.
Worked Example: Exact Matching Versus a Fixed Caliper
Suppose two units have propensity scores 0.30 and 0.60. Under the one-to-one transformation \(g(e)=e^2\) on \([0,1]\), their transformed scores are \(0.09\) and \(0.36\). Since the inverse is \(g^{-1}(t)=\sqrt{t}\), no score information is lost.
The units are unequal on both scales, so exact equality gives the same conclusion: they are not an exact score match. But their distance changes from \(|0.60-0.30|=0.30\) to \(|0.36-0.09|=0.27\). With a fixed caliper of 0.28, the pair fails the caliper on the original scale and passes it on the squared scale. Thus invertibility preserves exact conditioning, but a numerical tolerance must be interpreted on the scale on which it is defined.
This distinction matters in diagnostics. If a transformed score is claimed to be one-to-one, check the transformation on the observed score range and verify that numerical operations have not effectively merged distinct values. Rounding log-odds to one decimal place, for example, is a many-to-one operation: several different values may receive the same rounded value. A balance check after such rounding is a check of the rounded groups, not a consequence of the exact transformation theorem.
Choosing and Checking a Score Scale
A useful comparison is between three operations. Keeping the original score retains all its information. Applying an invertible transformation retains that information but can change numerical distances. Coarsening the score may discard information and requires its own balance assessment.
| Operation | Does it retain the original score? | Balance implication |
|---|---|---|
| One-to-one transformation | Yes; recover the score with the inverse | Preserves the exact balancing property |
| Many-to-one coarsening | Not necessarily | May combine different treatment probabilities and fail to balance |
| Rounding or binning | Usually no | Must be evaluated as a coarsening; balance is not guaranteed |
A practical diagnostic is to compare covariate distributions between treatment groups within the score values, strata, or matched sets actually used. For an invertible transformation with exact conditioning, imbalance suggests a practical or implementation issue—such as rounding, limited sample representation, or the use of approximate rather than exact conditioning—not a failure of the population theorem. For a coarsened score, imbalance may be the expected consequence of merging score levels.
There are also limits to the result. The usual logit transformation is finite only when \(0<e(X)<1\); at a propensity score of 0 or 1, the ordinary logit is undefined. One can work with the original probability scale or define an extended transformation with care, but should not silently treat an infinite logit as an ordinary finite score. In addition, the theorem is a statement about the true propensity score. Replacing it with an estimated score does not make the estimated score an exact balancing score merely because an invertible transformation is applied to it; estimated-score balance must be assessed in the data.
A common mistake is to argue that any function of a balancing score must also be balancing because it was calculated from that score. The reasoning confuses being a function of the score with retaining all its information. The correction is to ask whether the original score can be recovered from the transformed value. If yes, conditioning is equivalent; if no, the transformation may pool distinct treatment probabilities, and balance must be checked rather than assumed.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- If \(T=1-e(X)\), is \(T\) a balancing score? State the condition that makes the answer hold.
- A coarsening maps scores 0.10 and 0.90 to the same value. If both score levels occur with positive probability, can balance be guaranteed within that merged group? Explain using conditional treatment probabilities.
- Two scores are 0.20 and 0.50. Under \(g(e)=e^2\), calculate their transformed values and their distance on each scale. What does this imply about a fixed distance threshold?
- Why does rounding an invertible transformation to a fixed number of decimal places require a separate balance check?
Solutions. (1) Yes, because \(g(e)=1-e\) is one-to-one on \([0,1]\), with inverse \(e=1-T\); conditioning on either score retains the same information. (2) No. Within a group containing both values, the treatment probability conditional on \(X\) is either 0.10 or 0.90, while conditional on the merged score it is their weighted average. These probabilities differ whenever both levels have positive conditional probability, so conditional independence does not hold. (3) The transformed values are \(0.20^2=0.04\) and \(0.50^2=0.25\). The original distance is \(0.50-0.20=0.30\), while the transformed distance is \(0.25-0.04=0.21\). A fixed threshold can therefore include a pair on one scale and exclude it on the other. (4) Rounding can map distinct transformed values to the same recorded value. It is therefore many-to-one, so it may combine score levels and no longer has the exact balance guarantee of the unrounded transformation.