Trimming Rules and Retained Populations

Trimming Defines Who Remains in the Target Population

A trimming rule is not only a data-processing instruction. It defines which units belong to the population being described or analyzed. To make that change precise, represent each unit’s retention status with an indicator and derive the distribution of covariates conditional on retention. The same calculation applies whether the rule is based on a propensity score, a covariate-defined subgroup, or both.

Let $A$ denote treatment assignment, with $A=1$ for treated units and $A=0$ for controls, and let $X$ denote the measured covariates. Write $e(X)=P(A=1\mid X)$ for the propensity score. Let $R=1$ mean that a unit is retained and $R=0$ mean that it is excluded. A trimming rule specified using $X$ is a function of those covariates, so it changes the distribution of $X$ among retained units.

Definition: For a retention rule $g(X)$ taking values 0 or 1, define $R=g(X)$. The retained population is the source population conditional on $R=1$. If $X$ has density $f_X(x)$, its retained density is $$ f_{X\mid R=1}(x)=\frac{g(x)f_X(x)}{P(R=1)} =\frac{g(x)f_X(x)}{\int g(u)f_X(u)\,du}, $$ provided $P(R=1)>0$. For discrete $X$, replace the density and integral with probability mass and a sum.

The numerator preserves the source population’s relative covariate frequencies for units that satisfy the rule and sets the contribution of excluded units to zero. The denominator is the total population probability of satisfying the rule; dividing by it makes the retained distribution integrate or sum to one.

Score-Based and Subgroup-Based Indicators

For fixed score cutoffs $l$ and $u$, where $0\leq l\leq u\leq1$, a symmetric score restriction can be written as

R_{\text{score}}=I\{l\leq e(X)\leq u\}.

Here $I\{\cdot\}$ equals 1 when the condition inside the braces is true and 0 otherwise. The retained covariate density is therefore

f_{X\mid R_{\text{score}}=1}(x) = \frac{I\{l\leq e(x)\leq u\}f_X(x)} {P(l\leq e(X)\leq u)}.

A subgroup restriction is expressed in the same way. If $H(X)=1$ identifies a covariate-defined subgroup of interest, then $R_{\text{subgroup}}=I\{H(X)=1\}$, and the retained population is the source population conditional on belonging to that subgroup. For a rule requiring both the score condition and subgroup membership, multiply the indicators:

R=I\{l\leq e(X)\leq u\}I\{H(X)=1\}.

The multiplication makes $R=1$ only when both conditions hold. More generally, any rule based on covariates can be written as an indicator of its inclusion set. For example, a union of two permitted ranges uses an indicator for membership in either range.

Deriving the Distribution in Each Treatment Arm

The distribution among all retained units is not generally the same as the distribution among retained treated units or retained controls. Even if the same covariate-based rule is applied to everyone, treatment assignment is associated with $X$ whenever $e(X)$ varies. Under a common deterministic rule $R=g(X)$, the treatment-arm distributions follow from conditioning:

f_{X\mid A=a,R=1}(x) = \frac{g(x)P(A=a\mid X=x)f_X(x)} {\int g(u)P(A=a\mid X=u)f_X(u)\,du}.

For treated units, $P(A=1\mid X=x)=e(x)$; for controls, $P(A=0\mid X=x)=1-e(x)$. Thus the treated-arm density has $e(x)$ in its numerator and the control-arm density has $1-e(x)$. This result follows directly from the joint density: $f_{X,A=a,R=1}(x)=g(x)P(A=a\mid X=x)f_X(x)$. Dividing by its integral over $x$, which is $P(A=a,R=1)$, gives the conditional density above.

This distinction matters when describing a retained sample. The overall retained covariate distribution describes the selected target population. The arm-specific distributions describe which covariate values occur among the retained treated units and retained controls. A common retention rule does not make those arm-specific distributions identical.

Worked Example: A Score Threshold With Four Covariate Strata

Suppose a hypothetical population has four covariate strata with probabilities $0.20, 0.30, 0.30,$ and $0.20$. Their respective propensity scores are $0.08, 0.30, 0.70,$ and $0.94$. Apply the rule $0.10\leq e(X)\leq0.90$, then calculate the overall retained distribution and the treated and control distributions among retained units.

The rule excludes the first and fourth strata and retains the second and third. The retention probability is

$$ P(R=1)=0.30+0.30=0.60. $$

The retained population probabilities for strata 2 and 3 are $0.30/0.60=0.5000$ and $0.30/0.60=0.5000$, respectively. The other strata have probability zero in the retained population. These probabilities sum to 1.

For treated units, weight each stratum’s source probability by its propensity score. The retained treated mass is

$$ 0.30(0.30)+0.30(0.70)=0.09+0.21=0.30. $$

Thus the treated-arm distribution across the two retained strata is $0.09/0.30=0.3000$ and $0.21/0.30=0.7000$. For controls, the corresponding mass is

$$ 0.30(1-0.30)+0.30(1-0.70)=0.21+0.09=0.30. $$

The control-arm distribution is therefore $0.21/0.30=0.7000$ in stratum 2 and $0.09/0.30=0.3000$ in stratum 3. The overall retained population is split equally between those strata, even though the retained treated and control arms have different covariate distributions.

Worked Example: Retaining a Covariate-Defined Subgroup

Consider a hypothetical population cross-classified by age band and residence. The source-population probabilities are: younger urban $0.20$, younger rural $0.10$, middle-aged urban $0.30$, middle-aged rural $0.20$, older urban $0.10$, and older rural $0.10$. Suppose the rule retains only the middle-aged subgroup. Find the retained population distribution and, if the scores in the two retained cells are $0.40$ for urban residents and $0.60$ for rural residents, find the distributions in each treatment arm.

The inclusion indicator is $R=I\{\text{middle-aged}\}$. Its probability is $0.30+0.20=0.50$. Within the retained population, the urban probability is $0.30/0.50=0.6000$, and the rural probability is $0.20/0.50=0.4000$. The younger and older cells have retained probability zero.

For treated units, the unnormalized cell probabilities are $0.30(0.40)=0.12$ for urban residents and $0.20(0.60)=0.12$ for rural residents. The total treated mass retained is $0.24$, so each treated cell has probability $0.12/0.24=0.5000$. For controls, the unnormalized probabilities are $0.30(0.60)=0.18$ and $0.20(0.40)=0.08$. Their total is $0.26$, giving urban probability $0.18/0.26=0.6923$ and rural probability $0.08/0.26=0.3077$. All of these distributions are conditional on both the stated treatment arm and retention.

Worked Example: Intersecting a Score Rule and a Subgroup Rule

Let a hypothetical covariate $X$ have a uniform distribution on $[0,1]$, and suppose $e(X)=X$. Retain units with scores between $0.20$ and $0.80$, but only if $X\leq0.60$. Find the retained set, its population probability, its density, and its mean covariate value.

The combined indicator is $R=I\{0.20\leq e(X)\leq0.80\}I\{X\leq0.60\}$. Because $e(X)=X$, both conditions hold exactly when $0.20\leq X\leq0.60$. For a uniform distribution on $[0,1]$, the probability of this interval is its length:

$$ P(R=1)=0.60-0.20=0.40. $$

The retained density is the source density, which is 1, divided by 0.40 on the retained interval, and zero elsewhere. Hence $f_{X\mid R=1}(x)=2.5$ for $0.20\leq x\leq0.60$, and 0 otherwise. Its mean is

E(X\mid R=1)=\int_{0.20}^{0.60}x(2.5)\,dx =2.5\left[\frac{x^2}{2}\right]_{0.20}^{0.60} =2.5(0.18-0.02)=0.4000.

The intersection has a smaller retained set than either rule alone: the score rule permits $X$ up to $0.80$, while the subgroup rule removes values above $0.60$. The denominator must therefore be the probability of the intersection, $0.40$, not the probability of either rule separately.

Using the Rule With Observed Data

In a sample of $n$ units, let $R_i$ be the indicator that unit $i$ satisfies the chosen rule. The empirical retained fraction is $\sum_{i=1}^n R_i/n$. For a covariate category $C$, its empirical proportion among retained units is

\frac{\sum_{i=1}^{n}R_i I\{X_i\in C\}}{\sum_{i=1}^{n}R_i},

provided at least one unit is retained. This is the sample analogue of dividing the source-population probability of being both in $C$ and retained by the probability of retention. To describe treated or control units, include $I\{A_i=a\}$ in both the numerator and denominator. For example, the treated proportion in category $C$ is $\sum_i R_i I\{A_i=1\}I\{X_i\in C\}/\sum_i R_i I\{A_i=1\}$, if the denominator is positive.

If the rule uses an estimated propensity score $\hat e(X)$, its sample indicator is $R_i=I\{l\leq\hat e(X_i)\leq u\}$. The same conditional-distribution calculation applies to the realized rule. In that case, the selected set depends on the fitted score model as well as on the covariates; the formula describes the population or sample selected by that particular rule, not an abstract population selected by a known true score.

For a treatment-specific rule, write $R=I\{A=1\}g_1(X)+I\{A=0\}g_0(X)$, with $g_1$ and $g_0$ taking values 0 or 1. The covariate distribution within arm $a$, conditional on retention, is proportional to $g_a(x)P(A=a\mid X=x)f_X(x)$. If $g_1$ and $g_0$ select different covariate regions, there is not one common retained population with the same inclusion rule for both arms. The two arm-specific distributions should be described separately rather than presented as if they arose from identical eligibility criteria.

Assumptions, Diagnostics, and Common Errors

The distributional identities above require a well-defined source population, a specified retention rule, and positive probability of retention. To use the treatment-arm formulas, the propensity score must represent $P(A=1\mid X)$ in that source population. These calculations do not, by themselves, establish that a causal effect is identified; causal interpretation also depends on the relevant identification assumptions and the target population. If both treatment arms are needed within the retained population, the rule should leave positive treated and control probability there.

Several checks help detect an incorrectly implemented rule. Verify that each $R_i$ is 0 or 1 and that the retained fraction equals the count of retained units divided by the full sample size. Confirm that the retained category probabilities sum to 1. For arm-specific distributions, check that the denominator counts retained units in that arm, not all retained units. For a combined rule, check that units satisfy every required condition. In a sample, a zero denominator means the corresponding conditional distribution cannot be estimated from that sample.

A common mistake is to report the unnormalized retained mass $g(x)f_X(x)$ as if it were a probability distribution. In the four-stratum example, the retained masses $0.30$ and $0.30$ sum to $0.60$, not 1. The correction is to divide each mass by $P(R=1)=0.60$, producing $0.50$ and $0.50$. A second mistake is to use the overall retained distribution as the treated or control distribution. Those distributions require the additional $e(x)$ or $1-e(x)$ factor and their own arm-specific normalizing denominator.

Trimming is one way to define an analysis population. Unlike weighting, which changes units’ contributions without necessarily excluding them, trimming assigns zero inclusion to units outside its rule and renormalizes the remaining covariate distribution. Matching can also lead to a selected analysis sample, but its retained distribution is determined by the matching procedure. These approaches therefore need not target the same population. A limitation of any deterministic restriction is that conclusions based on the retained population do not automatically describe excluded units or the full source population.

Key takeaway: Write the inclusion indicator first, multiply the source covariate distribution by that indicator, and divide by the probability of inclusion. For a treatment-arm distribution, also multiply by the probability of assignment to that arm and normalize within the arm.

Check Your Understanding

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

A source population has probabilities $0.25, 0.50,$ and $0.25$ in three covariate groups. A rule retains the first two groups. What is the retained distribution across the groups?
For a score-based rule $0.15\leq e(X)\leq0.85$, write the retention indicator and state the denominator in the retained covariate density.
Why can the treated and control covariate distributions differ even when the same score rule is applied to both arms?
A rule retains units who are both in subgroup $H(X)=1$ and within the score interval $[l,u]$. Write its indicator.
In a sample, 40 units are retained, including 12 in category $C$. What is the empirical proportion in category $C$ among retained units? Why is $12/n$ not the answer unless the question concerns the full sample?

Solutions. (1) The retention probability is $0.25+0.50=0.75$. The retained probabilities are $0.25/0.75=0.3333$, $0.50/0.75=0.6667$, and 0. (2) $R=I\{0.15\leq e(X)\leq0.85\}$; the denominator is $P(0.15\leq e(X)\leq0.85)=P(R=1)$. (3) The assignment probabilities vary with covariates: treated contributions are weighted by $e(X)$, while control contributions are weighted by $1-e(X)$, followed by separate normalization. (4) $R=I\{l\leq e(X)\leq u\}I\{H(X)=1\}$. (5) The retained proportion is $12/40=0.3000$. Dividing by the full sample size would give the category’s proportion in the full sample, not its proportion conditional on retention.