Why Random Assignment Identifies Causal Means
A randomized experiment creates a principled link between observed outcomes and potential outcomes. For each unit, only the outcome under the assignment actually received is observed. Random assignment makes the units assigned to treatment representative, in expectation, of the units’ potential outcomes under treatment; the same logic applies to control. This is an identification argument: it explains what the assignment mechanism lets us learn about causal quantities, not how uncertain a particular estimate is.
Let \(Z_i\) indicate whether unit \(i\) is assigned treatment: \(Z_i=1\) for treatment and \(Z_i=0\) for control. Let \(Y_i(1)\) and \(Y_i(0)\) be that unit’s potential outcomes under treatment and control. Under consistency, the observed outcome is \(Y_i=Z_iY_i(1)+(1-Z_i)Y_i(0)\). The average treatment effect for a population is the mean of the individual effects \(Y_i(1)-Y_i(0)\), or, equivalently, the difference between the population means of \(Y_i(1)\) and \(Y_i(0)\).
The independence condition and positivity do different jobs. Independence ensures that assigned groups are not selected according to their potential outcomes. Positivity ensures that each relevant unit can, under the design, appear in either assignment condition. A known assignment probability is useful for describing the design, but knowing it alone does not guarantee that a raw group average identifies the target mean.
Unbiasedness Under Equal-Probability Assignment
First consider a finite population of \(N\) units, with a fixed number \(n_1\) assigned to treatment and \(n_0=N-n_1\) assigned to control. Under complete random assignment, every subset of \(n_1\) units is equally likely to receive treatment. Each unit therefore has treatment probability \(n_1/N\), and control probability \(n_0/N\). Write the finite-population potential-outcome means as \(\bar{Y}(1)=N^{-1}\sum_{i=1}^{N}Y_i(1)\) and \(\bar{Y}(0)=N^{-1}\sum_{i=1}^{N}Y_i(0)\). The observed treatment and control averages are \(\bar{Y}_{1}=n_1^{-1}\sum_{i=1}^{N}Z_iY_i\) and \(\bar{Y}_{0}=n_0^{-1}\sum_{i=1}^{N}(1-Z_i)Y_i\).
For a treated unit, consistency gives \(Y_i=Y_i(1)\); for a control unit it gives \(Y_i=Y_i(0)\). Thus, the observed averages can also be written using the corresponding potential outcomes. Taking expectations over the random assignment gives the result directly:
Here \(E_Z\) means expectation over the assignment mechanism, holding the finite population’s potential outcomes fixed. The matching control calculation replaces \(Z_i\) with \(1-Z_i\), whose expectation is \(n_0/N\), and yields \(E_Z[\bar{Y}_{0}]=\bar{Y}(0)\). These are design-unbiasedness statements: over repeated assignments under the same design, each group average has the corresponding potential-outcome mean as its expectation. They do not say that a particular realized group average must equal that mean.
Worked Example: Expected Arm Averages in a Small Experiment
Suppose a hypothetical finite population contains four units. The experiment randomly assigns exactly two to treatment and two to control. Their potential outcomes, measured in score points, are:
| Unit | \(Y_i(1)\) | \(Y_i(0)\) |
|---|---|---|
| 1 | 10 | 4 |
| 2 | 12 | 6 |
| 3 | 14 | 8 |
| 4 | 16 | 10 |
The treatment potential-outcome mean is \(\bar{Y}(1)=(10+12+14+16)/4=13\) score points. Each unit has treatment probability \(2/4=0.5\). Using the assignment expectation,
The control potential-outcome mean is \(\bar{Y}(0)=(4+6+8+10)/4=7\) score points. Each unit has control probability \(0.5\), so \(E_Z[\bar{Y}_{0}]=\{0.5(4)+0.5(6)+0.5(8)+0.5(10)\}/2=14/2=7\) score points. The results concern averages across possible assignments; one realized pair of group averages may differ from 13 and 7.
Identification in a Superpopulation
The same argument has a superpopulation version. Suppose units are drawn from a population, treatment is assigned independently of the pair of potential outcomes, and the probability of treatment is \(p\), where \(0<p<1\). Consistency and independence imply that the observed outcome mean among treated units equals the population mean of the treatment potential outcome; similarly, the observed mean among controls equals the mean of the control potential outcome:
The first equality in each line uses consistency for the assignment group in question. The second uses independence of assignment and potential outcomes. Both conditional expectations are defined only when the corresponding assignment has positive probability: \(P(Z=1)=p>0\) and \(P(Z=0)=1-p>0\). The average treatment effect is therefore identified as the difference between these two observed conditional means. Identification does not require that the treatment and control outcomes for the same person both be observed; it requires a design that lets each assignment group represent the relevant potential outcomes in the population.
Worked Example: Reading Observed Means as Potential-Outcome Means
Consider a hypothetical population in which 40% of units have \(Y(1)=20\) points and 60% have \(Y(1)=10\) points. Suppose 25% have \(Y(0)=8\) points and 75% have \(Y(0)=4\) points. Treatment is assigned independently of both potential outcomes with probability \(p=0.4\).
The treatment potential-outcome mean is \(E[Y(1)]=0.40(20)+0.60(10)=8+6=14\) points. Because assignment is independent and \(p>0\), \(E[Y\mid Z=1]=14\) points. The control potential-outcome mean is \(E[Y(0)]=0.25(8)+0.75(4)=2+3=5\) points. Since \(1-p=0.6>0\), \(E[Y\mid Z=0]=5\) points.
Consequently, the population average treatment effect is \(14-5=9\) points. These calculations identify population quantities from the assignment mechanism and the observed conditional outcome distributions. They do not claim that a finite realized sample’s treatment and control means will equal 14 and 5 exactly.
Positivity and the Limits of Raw Averages
Positivity is substantive, not merely a technical convenience. If a unit or a relevant subgroup can never receive one assignment, the data contain no outcomes for that condition in that part of the target population. Random assignment cannot reveal a potential-outcome mean where the design never allows the corresponding outcome to be observed. In a design with covariates or strata, positivity must hold within the relevant covariate groups when the target requires comparisons within those groups.
Equal assignment probabilities also matter for the simple raw-average argument. Under complete random assignment, every unit has the same chance of appearing in each group, so each group is an equally representative sample of the finite population in expectation. If assignment probabilities differ across units, the composition of an observed group can differ from the target population’s composition. Random assignment within groups can still identify group-specific means when positivity holds, but the unadjusted overall group average need not identify the overall population mean.
Worked Example: Unequal Probabilities Can Change Group Composition
Suppose a hypothetical population has 100 units in two equally sized strata. All units in stratum A have treatment potential outcome \(Y(1)=10\) points; all units in stratum B have \(Y(1)=0\) points. The population treatment potential-outcome mean is \(\bar{Y}(1)=\{50(10)+50(0)\}/100=5\) points.
The design randomly selects 40 of the 50 units in A for treatment and 10 of the 50 units in B for treatment. Thus, assignment is random within each stratum, with treatment probabilities \(0.8\) in A and \(0.2\) in B. Both strata satisfy positivity, but the treated group contains 40 units from A and 10 from B. Its observed treatment average is \(\{40(10)+10(0)\}/50=8\) points, not the population mean of 5 points.
The difference occurs because the treated group overrepresents stratum A, whose treatment potential outcomes are higher. Knowing the assignment probabilities does not make this raw average representative of the overall population. The stratum-specific assignment mechanism and the target population’s stratum composition must be taken into account to identify the overall mean.
Worked Example: A Positivity Failure Leaves a Mean Unidentified
Suppose a hypothetical subgroup of 10 units is always assigned treatment, so each has \(\pi_i=1\). Their observed treatment outcomes have an average of 8 points. No control outcome is observed for this subgroup.
Consider two possible sets of missing control potential outcomes, both consistent with the observed data. In the first, the subgroup’s mean \(Y(0)\) is 5 points, making its average treatment effect \(8-5=3\) points. In the second, its mean \(Y(0)\) is 15 points, making its average treatment effect \(8-15=-7\) points. Since the subgroup is never assigned control, both possibilities produce exactly the same observed treatment outcomes.
The subgroup’s average treatment effect is therefore not identified from these data: the observed data cannot distinguish the two counterfactual means. The failure is \(P(Z=0)=0\) for this subgroup. Random assignment elsewhere in the study cannot repair that lack of support.
What the Identification Argument Does—and Does Not—Establish
The argument rests on a clear sequence of conditions. Consistency connects observed outcomes to the potential outcome for the assignment received. Random assignment makes assignment independent of potential outcomes, either for the whole population or conditionally within the design’s relevant groups. Positivity gives each group a nonzero chance of appearing in each required assignment condition. Under equal-probability assignment, the raw treatment and control averages are unbiased for the corresponding finite-population means. In a superpopulation formulation, the corresponding observed conditional means equal the potential-outcome expectations.
These conditions establish identification, not precision. A design can identify a mean while producing a noisy estimate in a small sample. Nor does this argument alone establish a standard error, confidence interval, or test. Those require additional inference calculations. Finally, when assignment probabilities vary, identification of the overall target mean may require accounting for the assignment design and target population composition; the raw group average cannot simply be assumed to work.
- Independence without positivity: even if assignment is unrelated to potential outcomes where both assignments occur, a group with no chance of control provides no observed control outcomes.
- Positivity without independence: both assignments may occur, but if assignment favors units with particular potential outcomes, raw group means need not identify the target means.
- Known probabilities without equal representation: knowing probabilities describes how units entered the assignment groups; it does not imply that an unadjusted average has the target population’s composition.
Check Your Understanding
Use the definitions and arguments above to answer these questions.
- In a finite population of \(N\) units, exactly \(n_1\) are randomly assigned treatment. What is each unit’s treatment probability, and why does the expected treatment average equal \(\bar{Y}(1)\)?
- Which assumption justifies replacing \(E[Y(1)\mid Z=1]\) with \(E[Y(1)]\) in the superpopulation argument?
- A subgroup has treatment probability one. Which potential-outcome mean for that subgroup is not identified by the experiment, and why?
- Explain why knowing unequal assignment probabilities does not, by itself, make a raw treated-group average equal to the overall treatment potential-outcome mean.
- What are the two distinct roles of independence and positivity in the identification argument?