From a Distance Table to Subclasses
A distance table gives a number for each treated-control pair: a smaller number usually indicates a closer pair under the chosen distance measure. Full matching uses these pairwise relationships to form subclasses that can contain one or more treated units and one or more control units. The practical task is to decide which pairwise links define the subclasses, then check that the resulting sets are valid.
The distance table alone does not uniquely specify a full matching. A construction also depends on which links are allowed and selected. A link may be ruled out by a prespecified restriction, such as a caliper; even among allowed links, the analyst must identify an allocation. The checks in this tutorial concern the structure of that allocation, not whether a particular allocation is optimal.
The word “connected” matters: a subclass does not require every treated unit to be directly linked to every control unit in that subclass. For example, links from two treated units to the same control can connect all three into one subclass. Conversely, being close to a unit in the table does not place a unit in a subclass unless the corresponding link is actually selected.
A Reliable Construction and Audit
Begin by identifying the treated and control units, and confirm what the distance measure represents. If the design specifies an eligibility rule, apply it to the table before choosing links. Then list the selected links explicitly. The connected groups formed by those links are the candidate subclasses. Finally, audit both the groups and the units not assigned to them.
Apply any stated restriction to each treated-control distance. Do not treat a small distance as an automatic assignment.
Record which eligible treated-control pairs are used to connect units. A unit can have several selected links.
Units joined directly or through a chain of selected links belong to the same subclass.
Check that every subclass has both treatment groups, every included unit appears once, and each included unit has a selected link.
List isolated or otherwise excluded units separately. Do not silently count them as matched.
A short graph argument justifies why this procedure produces mixed subclasses. In a connected component containing at least one edge, that edge joins a treated vertex to a control vertex, so both treatment groups occur in the component. Every other vertex in the component is joined to that edge’s endpoints by a path, and every link on the path also joins opposite treatment groups. Thus the component contains both groups. A component with no edge is an isolated unit, not a mixed subclass. Distinct connected components share no vertices, so their subclasses do not duplicate units.
This argument establishes structural validity, not statistical adequacy. Distances must be defined sensibly for the study, and the matching design must be assessed on its own merits. A valid partition does not by itself establish exchangeability, eliminate unmeasured confounding, or guarantee good covariate balance.
Worked Example: Building Two Subclasses From a Distance Table
Suppose a hypothetical design has treated units \(T_1,T_2,T_3\) and control units \(C_1,C_2,C_3,C_4\). The entries below are distances. For this example, a supplied eligibility rule permits links with distance at most \(0.50\).
| \(C_1\) | \(C_2\) | \(C_3\) | \(C_4\) | |
|---|---|---|---|---|
| \(T_1\) | 0.12 | 0.72 | 0.55 | 0.62 |
| \(T_2\) | 0.16 | 0.68 | 0.48 | 0.59 |
| \(T_3\) | 0.51 | 0.61 | 0.11 | 0.14 |
The eligible pairs include \(T_1\)-\(C_1\), \(T_2\)-\(C_1\), \(T_2\)-\(C_3\), \(T_3\)-\(C_3\), and \(T_3\)-\(C_4\), because their distances are no greater than \(0.50\). Choose the following links for this feasible construction: \(T_1\)-\(C_1\), \(T_2\)-\(C_1\), \(T_3\)-\(C_3\), and \(T_3\)-\(C_4\). The eligible link \(T_2\)-\(C_3\) is not selected; eligibility does not require selection.
The first two selected links connect \(T_1,T_2,C_1\), giving subclass \(\{T_1,T_2,C_1\}\). The other two connect \(T_3,C_3,C_4\), giving subclass \(\{T_3,C_3,C_4\}\). The first subclass has two treated units and one control; the second has one treated unit and two controls. Both are mixed.
Count the units explicitly: \(2+1=3\) units in the first subclass and \(1+2=3\) in the second, for \(3+3=6\) included units. The full table has seven units, so \(C_2\) is excluded. It has no eligible link under the stated rule. The audit is therefore: two valid mixed subclasses, all six included units have at least one selected link, and \(C_2\) is explicitly excluded. This construction is feasible; the example does not claim that it minimizes any full-matching objective.
How to Check the Construction
A useful audit is more than checking that the subclass labels look plausible. Write down the membership of each subclass, count treated and control units separately, and compare the union of those sets with the original unit list. If a unit appears twice, the proposed sets are not a partition. If a unit appears in no set, determine whether it is intentionally excluded or was accidentally omitted.
The degree of a unit in the selected-link graph is the number of selected links incident to it. An included unit must have degree at least one. Degree one is sufficient for inclusion: for example, one control can connect two treated units if each treated unit is linked to that control. A unit can also have degree greater than one. Multiple links do not create multiple copies of that unit; they simply connect it to other members of its subclass.
Worked Example: Checking a Subclass With Unequal Group Sizes
Consider hypothetical selected links \(T_1\)-\(C_1\), \(T_1\)-\(C_2\), \(T_2\)-\(C_2\), and \(T_2\)-\(C_3\). They connect the five units \(T_1,T_2,C_1,C_2,C_3\) into one subclass.
The treated count is \(n_T=2\), and the control count is \(n_C=3\), so the subclass size is \(n_T+n_C=2+3=5\). Both counts are positive, so the subclass is mixed. The degrees are \(2\) for \(T_1\), \(2\) for \(T_2\), \(1\) for \(C_1\), \(2\) for \(C_2\), and \(1\) for \(C_3\). Every unit has positive degree, and each appears once in the subclass membership list.
This is not a collection of three independent pairs: the links share units, and the connected component is one subclass. Describing it as three pairs would count \(C_2\) twice and obscure the actual structure. The correct record is one subclass with two treated and three controls.
When the Audit Finds a Problem
Consider a second hypothetical distance table with three treated units and two controls. Suppose the design supplies the same \(0.50\) eligibility threshold.
| \(C_1\) | \(C_2\) | |
|---|---|---|
| \(T_1\) | 0.18 | 0.74 |
| \(T_2\) | 0.24 | 0.81 |
| \(T_3\) | 0.63 | 0.77 |
Worked Example: Identifying Isolated Units
The only eligible links are \(T_1\)-\(C_1\) and \(T_2\)-\(C_1\), since \(0.18\leq0.50\) and \(0.24\leq0.50\); the other four distances exceed \(0.50\). Selecting both eligible links gives the mixed subclass \(\{T_1,T_2,C_1\}\), with two treated units and one control.
The remaining units \(T_3\) and \(C_2\) have degree zero. The included count is \(2+1=3\), while the original count is \(3+2=5\); the difference \(5-3=2\) agrees with the two isolated units. Neither isolated unit belongs to a mixed subclass in this construction. It would be incorrect to list \(\{T_3\}\) or \(\{C_2\}\) as a full-matching subclass because each set contains only one treatment group.
This example also shows why the distance table and the selected-link list should both be retained. The table explains which links were available under the rule; the link list makes the actual construction auditable. If a unit is isolated, the analyst must report it and decide how to handle it under the study design. A structural check cannot make an ineligible pair eligible.
Feasibility, Alternatives, and Limitations
A distance table can support several different matching structures. Pair matching restricts each matched set to one treated and one control unit. Variable-ratio matching allows different numbers of controls across treated units, while full matching permits subclasses with one or more units from each group. These structures answer different design needs. In full matching, a connected subclass may have a larger group on either side, but it still must contain both groups.
A nearest-neighbor choice is a local rule for selecting close partners; it does not, by itself, certify that all proposed subclasses form a valid full-matching partition. Exact matching and coarsened exact matching define matches through covariate agreement or categories rather than simply reading a distance table. Whatever the approach, the chosen structure should be checked against its own inclusion rules and the study’s balance goals.
The construction here is a feasibility exercise: it shows how a given set of links yields mixed subclasses and how to verify membership. It does not establish that the links are globally optimal, that another feasible construction would be worse, or that the resulting covariates are balanced. In particular, adding a link between two existing components can merge them into a larger subclass, but whether that change is desirable depends on the design criterion and distances. Merely increasing the number of links is not a proof of improvement.
Common diagnostic warnings include an isolated unit mistakenly listed as matched, an all-treated or all-control group labeled as a subclass, repeated membership across sets, and a count that does not reconcile with the original sample. Also check that distances are on the intended scale and that any eligibility rule was applied consistently. Even a perfectly audited construction addresses only the matching structure; it cannot rule out unmeasured confounding or guarantee a causal interpretation.
Check Your Understanding
Answer each question by auditing the links and membership, not just by inspecting the distance values.
- Selected links are \(T_1\)-\(C_1\) and \(T_2\)-\(C_1\). What is the subclass, and how many units from each treatment group does it contain?
- A proposed subclass contains \(T_1,T_2,C_1,C_2\), with selected links \(T_1\)-\(C_1\) and \(T_2\)-\(C_2\). Is it connected as one subclass? Explain.
- There are four treated and three control units. The listed subclasses contain three treated and three controls, with no repeated members. How many units are unaccounted for, and what must be checked next?
- Why is a unit with a small distance to a control not necessarily included in a constructed subclass?
Solutions. (1) Both links share \(C_1\), so they form the connected subclass \(\{T_1,T_2,C_1\}\), containing two treated units and one control. (2) No. The two selected links make two separate components, \(\{T_1,C_1\}\) and \(\{T_2,C_2\}\). The proposed four-unit set is not connected as one subclass unless an additional selected link joins the components. (3) Seven units were present and six are listed, so one is unaccounted for. Check whether it is intentionally excluded or omitted; the unit counts and membership list must reconcile with the original units. (4) Inclusion depends on the selected links and their connected components, not merely on a distance appearing in the table. A small distance may be ineligible under a design restriction or may simply not be selected.