Why Assignment and Treatment Receipt Are Different
In an experiment, researchers may randomize an offer, encouragement, or assignment without forcing every unit to receive the treatment. For example, some people offered a program may not participate, while some not offered it may obtain similar services elsewhere. The randomized variable is the assignment; actual treatment receipt is a separate variable. Keeping them distinct is essential for describing how units respond to assignment.
For each unit, potential outcomes can describe both the outcome that would occur under each assignment and whether treatment would be received under each assignment. This tutorial focuses on the latter. A unit’s compliance type is determined by its pair of potential treatment-receipt values, not by the single receipt value observed in the experiment.
The equality \(D_i=D_i(Z_i)\) is the consistency relation: the observed receipt is the potential receipt corresponding to the assignment actually received. As in the potential-outcomes framework, the potential receipts refer to well-defined assignment conditions, and one unit’s assignment does not change another unit’s potential receipt. The pair \((D_i(0),D_i(1))\) describes a unit before assignment; it does not mean that both potential receipts are observed.
The Four Compliance Types
There are four possible pairs because each of the two potential receipt values is either 0 or 1. The order matters: the first value describes receipt under assignment 0, and the second describes receipt under assignment 1.
| Potential receipt pair | Compliance type | Meaning |
|---|---|---|
| \((0,1)\) | Complier | Receives treatment if assigned 1, but not if assigned 0 |
| \((1,1)\) | Always-taker | Receives treatment under either assignment |
| \((0,0)\) | Never-taker | Does not receive treatment under either assignment |
| \((1,0)\) | Defier | Receives treatment if assigned 0, but not if assigned 1 |
These labels summarize how receipt responds to assignment. They do not describe whether a unit’s outcome improves, whether a unit follows a researcher’s instructions in every respect, or whether treatment receipt is beneficial. For instance, a complier’s label only says that receipt changes from 0 to 1 when assignment changes from 0 to 1.
Worked Example: Classifying Potential Receipt Pairs
Suppose a hypothetical experiment has four units whose potential receipts are known for illustration. Classify each unit.
| Unit | \(D_i(0)\) | \(D_i(1)\) | Pair | Type |
|---|---|---|---|---|
| A | 0 | 1 | \((0,1)\) | Complier |
| B | 1 | 1 | \((1,1)\) | Always-taker |
| C | 0 | 0 | \((0,0)\) | Never-taker |
| D | 1 | 0 | \((1,0)\) | Defier |
For Unit A, the first entry is 0 and the second is 1, so receipt follows assignment: this unit is a complier. Unit D has the reverse pattern, so it is a defier. In an actual experiment, however, each unit receives only one assignment. The table is a hypothetical illustration of the definition, not a claim that both potential receipts can be directly recorded for each participant.
What an Observed Assignment–Receipt Pair Reveals
Only one potential receipt is observed for a unit: \(D_i(0)\) if \(Z_i=0\), or \(D_i(1)\) if \(Z_i=1\). Therefore, an observed assignment and receipt usually narrow down the possible compliance types but do not identify the individual type.
| Observed assignment \(Z_i\) | Observed receipt \(D_i\) | Compatible types |
|---|---|---|
| 0 | 0 | Complier or never-taker |
| 0 | 1 | Always-taker or defier |
| 1 | 0 | Never-taker or defier |
| 1 | 1 | Complier or always-taker |
Worked Example: Interpreting an Observed Receipt
In a hypothetical randomized encouragement experiment, a participant is assigned \(Z_i=1\) and is observed to receive treatment, \(D_i=1\). Which compliance types remain possible?
Because the participant was assigned 1, the observation reveals \(D_i(1)=1\). It does not reveal \(D_i(0)\). If \(D_i(0)=0\), the potential receipt pair is \((0,1)\), so the participant is a complier. If \(D_i(0)=1\), the pair is \((1,1)\), so the participant is an always-taker. Both types are compatible with the observed data.
It would be incorrect to label this participant a complier merely because the participant received treatment when assigned 1. Always-takers do so as well. A single observed assignment–receipt pair does not reveal the missing potential receipt.
Receipt Rates and Compliance-Type Shares
Although an individual type is generally not observed, randomized assignment allows the receipt rate in each assignment arm to inform us about population type shares. Define \(p_1=P(D_i=1\mid Z_i=1)\) and \(p_0=P(D_i=1\mid Z_i=0)\). Under randomized assignment, the units assigned to each arm represent the corresponding distribution of potential receipts, subject to the assignment design and sampling uncertainty.
In the population, a unit receives treatment under assignment 1 if it is a complier or an always-taker. Under assignment 0, it receives treatment if it is an always-taker or a defier. If \(\pi_C,\pi_A,\pi_N,\pi_D\) denote the shares of compliers, always-takers, never-takers, and defiers, respectively, then:
The two receipt rates do not, in general, determine all four type shares. The following derivation shows exactly what they do determine: bounds on the shares, rather than a unique value for each.
Write \(\pi_A=a\). The rate equations then require \(\pi_C=p_1-a\) and \(\pi_D=p_0-a\).
Substituting those expressions into the equation that the four shares sum to 1 gives \(\pi_N=1-p_1-p_0+a\).
Each share must be at least 0. Thus \(a\geq 0\), \(a\geq p_1+p_0-1\), \(a\leq p_1\), and \(a\leq p_0\). Hence \(\max(0,p_1+p_0-1)\leq a\leq\min(p_1,p_0)\).
Since \(\pi_C=p_1-a\), its smallest value occurs at the largest allowed \(a\), and its largest value at the smallest allowed \(a\). Therefore \(\max(0,p_1-p_0)\leq\pi_C\leq\min(p_1,1-p_0)\).
This calculation assumes the receipt rates describe the relevant population margins. In a finite randomized experiment, the observed arm-specific rates are estimates of those margins, not necessarily their exact values. Random assignment does not reveal both potential receipts for any one participant; it lets the two assignment arms provide information about different margins across units.
Worked Example: Bounding the Complier Share
Suppose a hypothetical randomized study has 100 units assigned to each arm. In the assignment-1 arm, 70 receive treatment; in the assignment-0 arm, 30 receive treatment. The observed receipt rates are
Using these rates as estimates of the population margins, the share of compliers is bounded by
Thus these rates alone are compatible with a complier share ranging from 40% to 70%. The other type shares are not uniquely determined either. For example, the always-taker share \(a\) can range from \(\max(0,0.70+0.30-1)=0\) to \(\min(0.70,0.30)=0.30\). At \(a=0.10\), the implied shares are \(\pi_C=0.60\), \(\pi_D=0.20\), and \(\pi_N=0.10\); together with \(\pi_A=0.10\), they sum to 1 and reproduce both receipt rates. At \(a=0.30\), the implied shares are \(\pi_C=0.40\), \(\pi_D=0\), and \(\pi_N=0.30\), also consistent with the rates.
What an Additional No-Defiers Assumption Changes
Sometimes an analysis assumes that no unit would reduce treatment receipt when assigned to the offer rather than the comparison condition. In the notation here, this means \(\pi_D=0\). This is an additional assumption about the potential receipt pairs; it is not guaranteed by random assignment and cannot be verified for every individual from the observed data.
With \(\pi_D=0\), the assignment-0 receipt rate satisfies \(p_0=\pi_A\), while \(p_1=\pi_C+\pi_A\). The type shares are then determined by the two rates:
For the hypothetical rates above, this assumption gives estimated shares of 40% compliers, 30% always-takers, 30% never-takers, and 0% defiers. This is one particular decomposition of the observed rates; without the no-defiers assumption, the earlier bounds show that other decompositions are possible. Whether the assumption is substantively defensible requires considering how assignment could affect receipt in the study setting.
Common Mistake: Treating Observed Receipt as Type
A common error is to classify every participant who received treatment in the assignment-1 arm as a complier, or every participant who received treatment in the assignment-0 arm as an always-taker. That reasoning uses only one observed potential receipt and silently assumes a value for the unobserved one. The correction is to list the compatible pairs: \(Z_i=1,D_i=1\) is consistent with a complier or an always-taker, while \(Z_i=0,D_i=1\) is consistent with an always-taker or a defier.
A second error is to treat randomization as if it made individual compliance types observable. Randomization makes assignment independent of the potential receipts, but it does not assign both conditions to the same unit. It supports comparisons of receipt rates across arms; it does not directly show a participant’s two potential receipts. When reporting type shares, distinguish observed sample rates, estimates of population rates, and conclusions that require additional assumptions.
Check Your Understanding
Use the ordered pair convention: the first potential receipt is under assignment 0 and the second is under assignment 1.
- What compliance type corresponds to the potential receipt pair \((1,0)\)?
- A unit is assigned \(Z_i=0\) and has observed receipt \(D_i=0\). Which types are compatible with that observation?
- Why does observing \(Z_i=1,D_i=1\) not establish that a unit is a complier?
- If the assignment-1 receipt rate is 0.65 and the assignment-0 receipt rate is 0.25, what bounds do these rates imply for the complier share?
- What additional assumption would make the four population type shares determined by the two assignment-specific receipt rates?