Matching Weights and the Induced Population

Why the weights define a population

A treatment-effect estimate is not fully described by its numerical value. It also matters for which population the effect is being averaged. In a study where treatment assignment depends on covariates, reweighting the treated and control groups can make them represent the same covariate distribution. The particular weights determine what that shared distribution is—and therefore which average treatment effect is being estimated.

This tutorial focuses on propensity-score matching weights. Despite the name, these are weighting formulas based on the propensity score; they are not the weights produced by every matching algorithm. The earlier material on full matching establishes why a matched design can imply a particular target. Here the target is derived directly from the weighting rule.

Definition: Let $Z$ indicate treatment ($Z=1$ for treated and $Z=0$ for control), let $X$ be the measured pretreatment covariates, and define the propensity score $e(X)=P(Z=1\mid X)$. The matching weight for unit $i$ is $$ w_i = \begin{cases} \frac{\min\{e(X_i),1-e(X_i)\}}{e(X_i)}, & Z_i=1,\\[6pt] \frac{\min\{e(X_i),1-e(X_i)\}}{1-e(X_i)}, & Z_i=0. \end{cases} $$ Equivalently, $w_i=\min\{e(X_i),1-e(X_i)\}/[Z_i e(X_i)+(1-Z_i)(1-e(X_i))]$.

These weights are bounded between 0 and 1. For example, a treated unit with $e(X_i)=0.20$ receives weight $0.20/0.20=1$, while a control unit with that same score receives weight $0.20/0.80=0.25$. The different weights are intentional: each treatment group is being reweighted to represent the same target covariate distribution.

The population induced by matching weights

Let $f(x)$ denote the covariate density in the eligible study population, and write $m(x)=\min\{e(x),1-e(x)\}$. The target covariate density induced by matching weights is

g_{\mathrm{MW}}(x) = \frac{m(x)f(x)}{E[m(X)]}.

Thus, the original covariate distribution is tilted by the factor $m(x)$. Covariate values with propensity scores near 0 or 1 receive little weight; values with scores closer to one-half receive more. The denominator $E[m(X)]$ makes the density integrate to one.

The same target arises in both weighted treatment groups. To see why, the treated covariate density before weighting is proportional to $e(x)f(x)$. Multiplying by the treated weight $m(x)/e(x)$ leaves a quantity proportional to $m(x)f(x)$. For controls, the starting density is proportional to $[1-e(x)]f(x)$; multiplying by $m(x)/[1-e(x)]$ gives the same result.

Key result: Provided $0<E[m(X)]<\infty$, matching weights induce the covariate density $g_{\mathrm{MW}}(x)$ in both treatment groups. The population average treatment effect for that target is $$ \tau_{\mathrm{MW}} = \int \{E[Y(1)\mid X=x]-E[Y(0)\mid X=x]\} g_{\mathrm{MW}}(x)\,dx. $$ It is an average effect for the population tilted by $m(x)$, not automatically the ATE for the original eligible population.

The equality of weighted covariate distributions is a population statement based on the true propensity score. In a finite sample, estimated weights need not make every covariate exactly balanced. The weighted outcome contrast is usually estimated with normalized, or Hájek, weighted means:

\widehat{\tau}_{\mathrm{MW}} = \frac{\sum_{i:Z_i=1}\hat{w}_iY_i}{\sum_{i:Z_i=1}\hat{w}_i} - \frac{\sum_{i:Z_i=0}\hat{w}_iY_i}{\sum_{i:Z_i=0}\hat{w}_i},

where $\hat{w}_i$ is calculated using an estimated propensity score $\hat{e}(X_i)$. Normalizing within each treatment group makes each term a weighted mean on the outcome scale.

Worked Example: Deriving the Target Across Three Covariate Groups

Consider a hypothetical population divided into three covariate groups. Their population proportions are 0.40, 0.35, and 0.25, and their propensity scores are 0.20, 0.50, and 0.80, respectively. Suppose the conditional average treatment effects in those groups are 2, 5, and 9 outcome units. Find the matching-weight target proportions and its average treatment effect.

For the three groups, $m(x)=\min\{e(x),1-e(x)\}$ equals 0.20, 0.50, and 0.20. Multiplying each original group proportion by its matching factor gives unnormalized target masses:

0.40(0.20)=0.080,\qquad 0.35(0.50)=0.175,\qquad 0.25(0.20)=0.050.

The total mass is $0.080+0.175+0.050=0.305$. Dividing each mass by 0.305 gives target proportions 0.2623, 0.5738, and 0.1639. They sum to 1, up to rounding. The matching-weight effect is therefore

\tau_{\mathrm{MW}} = 0.2623(2)+0.5738(5)+0.1639(9) \approx 4.869

The estimated target effect is about 4.869 outcome units for the population represented by these tilted proportions. For comparison, the original-population ATE in this hypothetical example is $0.40(2)+0.35(5)+0.25(9)=4.800$ units. The difference occurs because the two estimands average group effects using different population proportions.

Assumptions for a causal interpretation

The weighting formula describes a target distribution; it does not, by itself, make an outcome contrast causal. For the weighted contrast to identify $\tau_{\mathrm{MW}}$, the following conditions are needed on the target population:

Consistency: each unit’s observed outcome equals its potential outcome under the treatment actually received: $Y=Y(Z)$. Treatment versions must be sufficiently well-defined for this statement to be meaningful.
Conditional exchangeability: $\{Y(1),Y(0)\}$ is independent of $Z$ given $X$. In context, the measured pretreatment covariates must control the relevant confounding.
Positivity on the target support: where $g_{\mathrm{MW}}(x)>0$, both treatment probabilities are positive: $0<e(x)<1$. At covariate values with $e(x)=0$ or $1$, the matching factor is zero, so those values are excluded from this target.
Correct propensity-score information: the true score, or an estimate that consistently captures it, must be used to construct the intended weights. A badly misspecified score can fail to produce the intended target balance.

Under these conditions, conditional exchangeability lets the observed conditional outcome means represent the potential-outcome means: $E[Y\mid Z=z,X=x]=E[Y(z)\mid X=x]$, for $z=0,1$. Combining that identity with the shared weighted covariate distribution yields the matching-weight estimand.

From conditional effects to the weighted contrast

A short derivation makes the identification argument explicit. In the treated group, weighting the joint probability of treatment and covariates gives

w_1(x)P(Z=1\mid X=x)f(x) = \frac{m(x)}{e(x)}e(x)f(x) = m(x)f(x).

For controls, the corresponding expression is

w_0(x)P(Z=0\mid X=x)f(x) = \frac{m(x)}{1-e(x)}[1-e(x)]f(x) = m(x)f(x).

Both weighted groups therefore have the same covariate density after normalization. Their population weighted outcome means are, respectively, the averages of $E[Y(1)\mid X=x]$ and $E[Y(0)\mid X=x]$ over $g_{\mathrm{MW}}(x)$, by consistency and conditional exchangeability. Subtracting the means gives the stated average of the conditional treatment effect. This identifies the target effect; it does not claim that the effect is constant across covariate values.

Worked Example: Calculating a Weighted Outcome Contrast

Suppose a hypothetical sample has two covariate groups, with estimated propensity scores 0.20 and 0.70. In the first group, the treated outcomes are 10 and 12 and the control outcomes are 6 and 8. In the second group, treated outcomes are 15 and 17 and control outcomes are 11 and 13. Calculate the normalized matching-weight contrast.

In the group with score 0.20, treated weights are $0.20/0.20=1$, and control weights are $0.20/0.80=0.25$. In the group with score 0.70, treated weights are $0.30/0.70=3/7$, and control weights are $0.30/0.30=1$.

The treated weighted mean is

\frac{1(10+12)+(3/7)(15+17)}{1+1+3/7+3/7} = \frac{22+96/7}{20/7} = 12.50.

The control weighted mean is

\frac{0.25(6+8)+1(11+13)}{0.25+0.25+1+1} = \frac{3.5+24}{2.5} = 11.00.

The weighted contrast is $12.50-11.00=1.500$ outcome units. These calculations establish the sample’s weighted contrast, not its causal validity on their own. A causal interpretation still requires the assumptions above, and the example does not establish covariate balance beyond the groups and scores specified.

Worked Example: Why the Target Can Differ from the ATE

Suppose a hypothetical eligible population has two equally common covariate groups. Their propensity scores are 0.10 and 0.50, and their conditional treatment effects are 8 and 2 days. Compute the ATE and matching-weight effect.

The ATE gives each group its original proportion of 0.50:

\tau_{\mathrm{ATE}}=0.50(8)+0.50(2)=5.000\text{ days}.

The matching factors are 0.10 and 0.50. The unnormalized target masses are $0.50(0.10)=0.05$ and $0.50(0.50)=0.25$, for a total of 0.30. The normalized target proportions are $0.05/0.30=1/6$ and $0.25/0.30=5/6$. Thus

\tau_{\mathrm{MW}} = \frac{1}{6}(8)+\frac{5}{6}(2) = 3.000\text{ days}.

The estimates differ because the first group, despite having a larger conditional effect, receives less representation in the matching-weight target. The 3.000-day value answers a different population question from the 5.000-day ATE; neither is inherently the correct summary without specifying the target of interest.

Diagnostics, alternatives, and limitations

Useful checks should connect to the target that the weights are intended to create. Inspect propensity-score distributions by treatment group, the range and concentration of the weights, and covariate balance after weighting. The theoretical derivation predicts a common weighted covariate distribution; substantial weighted imbalance in the observed sample is a warning that the estimated weights have not achieved that goal. Also check whether a few observations dominate a weighted mean, even though each raw matching weight is at most 1: normalization can still make a small number of units carry a large share of a group’s total weight.

Matching weights are one choice among several weighting or matching approaches. Inverse-probability weights for the ATE target the original eligible population, while ATT weights target the treated population. Matching weights instead use the minimum of the two treatment probabilities to define their target. Algorithmic matching, including full matching, can induce targets through its matched sets and set-specific weights; those weights need not equal the propensity-score matching weights defined here. These approaches can produce different effect estimates because they may represent different populations, use different units, or achieve different balance.

A common mistake is to say that matching weights estimate the ATE simply because both treated and control observations are weighted. The correction is to derive the induced covariate distribution: it is proportional to $\min\{e(x),1-e(x)\}f(x)$, not generally to $f(x)$. Another mistake is to treat a weighted contrast as automatically causal. Weighting addresses measured covariate representation under the propensity model; it cannot remove confounding from unmeasured variables, repair a violation of consistency, or guarantee good finite-sample balance.

The target also depends on the study’s eligible population and its propensity-score distribution. Restricting eligibility or changing the study population can change $f(x)$, $e(x)$, and therefore $g_{\mathrm{MW}}(x)$. Consequently, an effect estimated with matching weights should be reported with its target population and assumptions, rather than as an unqualified treatment effect.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If $e(x)=0.30$, what are the matching weights for a treated unit and a control unit with those covariates?
A population has two equally common groups with propensity scores 0.20 and 0.60. What are their normalized matching-weight target proportions?
Why does conditional exchangeability matter for interpreting the weighted outcome contrast causally?
True or false: matching weights generally make the weighted contrast equal to the ATE for the original eligible population. Explain.

Solutions. (1) The matching factor is $\min(0.30,0.70)=0.30$. The treated weight is $0.30/0.30=1$; the control weight is $0.30/0.70\approx0.4286$. (2) The matching factors are 0.20 and 0.40. The unnormalized masses are $0.50(0.20)=0.10$ and $0.50(0.40)=0.20$, totaling 0.30, so the normalized target proportions are $1/3$ and $2/3$. (3) Conditional exchangeability allows the observed outcome mean within each treatment and covariate level to represent the corresponding potential-outcome mean. Without it, the weighted difference can retain confounding. (4) False. The induced covariate density is proportional to $\min\{e(x),1-e(x)\}f(x)$, which generally differs from the original density $f(x)$; the matching-weight effect equals the ATE only in special cases where the relevant averages coincide.