Local Randomization Inference Near a Cutoff

Two Ways to Reason Near a Cutoff

Regression discontinuity designs use a running variable and a cutoff to create a sharp contrast in treatment assignment. One way to identify an effect at the cutoff is to assume that conditional expectations of potential outcomes vary continuously there. A different approach is to define a narrow neighborhood around the cutoff and model assignment within that neighborhood as if it came from a randomized experiment. This is called the local randomization approach.

The difference is not just a choice of calculation. Continuity-based identification concerns limits at the cutoff and relies on smoothness as the running variable approaches it. Local randomization instead makes a finite-sample claim about the assignment of the units in a specified neighborhood. If that claim is credible, randomization tests can provide inference for a local effect without fitting a regression function. The key challenge is to justify both the neighborhood and the assignment mechanism.

Definition: Let \(R_i\) be unit \(i\)'s running variable, \(c\) the cutoff, and \(Z_i=1(R_i\ge c)\) the cutoff-based assignment indicator. For a prespecified radius \(h>0\), the cutoff neighborhood is \(W_h=\{i:|R_i-c|\le h\}\). Local randomization treats assignment within \(W_h\), under a stated assignment mechanism, as randomized or as-if randomized conditional on the units included and any fixed features of that mechanism.

The word “local” matters: the claim concerns only units in \(W_h\), not the full study population. Under a sharp RD design, \(Z_i\) is the treatment received. In a fuzzy design, it is an assignment or encouragement and need not equal treatment received; randomization inference about that assignment does not by itself identify a treatment effect for compliers.

Neighborhood, Estimand, and Assignment Mechanism

Choosing \(h\) determines which units are included and therefore changes the population to which the result pertains. A useful neighborhood should be narrow enough that treating units on opposite sides as comparable is plausible, but large enough to include a meaningful number of units and support the proposed randomization inference. The choice should be supported by the assignment context and substantive knowledge, not selected solely because it produces a desired estimate or p-value.

For the units in a fixed neighborhood, write \(Y_i(1)\) and \(Y_i(0)\) for unit \(i\)'s potential outcomes under treatment and control. A natural finite-neighborhood estimand is the average treatment effect among the \(N\) units in the neighborhood:

\tau_{W_h}=\frac{1}{N}\sum_{i\in W_h}\left\{Y_i(1)-Y_i(0)\right\}.

This is an average effect for the included units, not automatically the effect exactly at \(R_i=c\), nor an effect for units outside the neighborhood. One common assignment mechanism is complete randomization: conditional on \(N\) and on exactly \(m\) assigned units, every subset of \(m\) units in \(W_h\) is equally likely to be assigned. Thus each such allocation has probability \(1/\binom{N}{m}\). Other mechanisms, such as independent assignment with a common probability, imply different allocation probabilities and must be analyzed using their own randomization distributions.

In an RD application, assignment is observed to follow the cutoff rule. Local randomization is an additional approximation or design claim: within the selected neighborhood, the observed division into assigned and unassigned units is regarded as one allocation generated by the specified mechanism. It is not enough to note that units are close to the cutoff. If the running variable is deliberately sorted, or if important assignment features make the units on each side systematically different, exchangeability may not be credible.

Worked Example: Defining a Local RD Estimand

Suppose, hypothetically, six applicants have running-variable values \(-0.8,-0.5,-0.2,0.1,0.4,0.7\), the cutoff is \(c=0\), and assignment is \(Z_i=1(R_i\ge0)\). Consider the neighborhood with \(h=0.8\).

All six applicants satisfy \(|R_i-0|\le0.8\), so \(N=6\). The cutoff rule assigns the three applicants with values \(0.1,0.4,0.7\); hence \(m=3\). Under a complete-randomization model conditional on three assigned applicants, there are \(\binom{6}{3}=20\) possible assignment sets, each assigned probability \(1/20\).

The estimand is the average of the six applicants' individual treatment effects, \(\tau_{W_h}=\frac{1}{6}\sum_{i=1}^6\{Y_i(1)-Y_i(0)\}\). It is not a population-wide average and does not become an effect exactly at the cutoff merely because the neighborhood is centered there. The as-if-random assignment claim also needs substantive justification; the arithmetic defining the neighborhood does not establish it.

Estimating an Effect Under Complete Randomization

If assignment is complete randomization within the neighborhood, a simple estimator is the difference in observed mean outcomes between assigned and unassigned units:

\widehat{\tau}_{W_h} = \frac{1}{m}\sum_{i\in W_h:Z_i=1}Y_i - \frac{1}{N-m}\sum_{i\in W_h:Z_i=0}Y_i.

Its design-based justification is straightforward. In repeated allocations under complete randomization, every unit has the same probability of assignment. Averaging the treated-group mean over those allocations gives the finite-neighborhood mean of \(Y_i(1)\); averaging the control-group mean gives the finite-neighborhood mean of \(Y_i(0)\). Subtracting these expectations gives the finite-neighborhood average effect:

E_Z\!\left[\widehat{\tau}_{W_h}\right] = \frac{1}{N}\sum_{i\in W_h}Y_i(1) - \frac{1}{N}\sum_{i\in W_h}Y_i(0) = \tau_{W_h}.

Here \(E_Z\) averages over the specified assignment mechanism while holding the units and their potential outcomes fixed. This unbiasedness statement is about repeated assignments under that mechanism. It does not establish that the mechanism is a credible description of the observed RD assignment.

Worked Example: Calculating a Local Difference in Means

Continue with the hypothetical six applicants in the preceding example. Suppose their observed outcomes, in order of running-variable value, are \(10,11,12,13,14,20\). The three applicants above the cutoff have outcomes \(13,14,20\); the three below have outcomes \(10,11,12\).

The estimated difference is

\widehat{\tau}_{W_h} = \frac{13+14+20}{3} - \frac{10+11+12}{3} = \frac{47}{3}-\frac{33}{3} = \frac{14}{3} \approx 4.667.

The estimate is 4.667 outcome units for this neighborhood. Under the complete-randomization model it estimates the average effect for these six applicants. Without a credible local assignment mechanism, it is only an observed difference between the two sides of the cutoff; the calculation alone does not justify a causal interpretation.

Exact Randomization Tests

A randomization test compares the observed test statistic with the values it would take under assignments permitted by the specified mechanism, assuming a null hypothesis. For a sharp null of no effect, \(H_0:Y_i(1)=Y_i(0)\) for every unit in the neighborhood, all missing potential outcomes are known under the null: each unit's outcome would be the same under either assignment. Outcomes can therefore be held fixed while assignment labels are reassigned according to the mechanism.

For a two-sided test with statistic \(T\), the exact p-value is the probability, under the null and assignment mechanism, of an assignment producing a statistic at least as extreme as the observed one. For complete randomization, this is the number of allowed allocations whose statistic is at least as extreme, divided by the total number of allowed allocations. A p-value is not the probability that the null is true. Its calibration depends on the assignment mechanism and the null being tested.

Worked Example: An Exact Test of the Sharp Null

Use the outcomes \(10,11,12,13,14,20\) and observed assignment in the earlier example, with exactly three of six units assigned. Test the sharp null that assignment has no effect on any applicant's outcome. Use the difference in means as the statistic and a two-sided alternative.

For any allocation, let \(S\) be the sum of the three outcomes assigned to treatment. The total outcome is \(80\), so the difference in means for that allocation is

T=\frac{S}{3}-\frac{80-S}{3}=\frac{2S-80}{3}.

The observed treated sum is \(S=13+14+20=47\), giving \(T_{\text{obs}}=(2(47)-80)/3=14/3\). Among the 20 possible three-unit subsets, this is the largest possible positive difference: the largest treated sum is \(47\). By symmetry, the only allocation with an equally large negative difference is its complement, which has treated sum \(33\). Thus exactly 2 of 20 allocations have \(|T|\ge|T_{\text{obs}}|\), and

p=\frac{2}{20}=0.100.

At significance level \(\alpha=0.05\), this test does not reject the sharp null. The result says that, under the stated null and complete-randomization model, 10.0% of allowed allocations produce a difference at least this extreme. It does not show that there is no effect, and it is not valid if the assumed allocation model is not credible.

Checking Baseline Balance With the Same Logic

The assignment mechanism can also be used to evaluate whether a baseline covariate is unusually imbalanced across the two sides within the selected neighborhood. This is a diagnostic for the plausibility of the local randomization model, not a proof of it. A nonsignificant balance test does not demonstrate exchangeability; a significant imbalance can, however, signal that the proposed model or neighborhood deserves closer scrutiny.

Worked Example: An Exact Covariate-Balance Test

Suppose, hypothetically, the six applicants have a baseline covariate with values \(1,2,3,4,5,6\), and the observed treated group consists of the applicants with values \(4,5,6\). There are still 20 possible assignments of three applicants under complete randomization. The observed treated covariate sum is \(15\), the total is \(21\), and the difference in covariate means is

T_{\text{obs}}=\frac{4+5+6}{3}-\frac{1+2+3}{3}=5-2=3.

For any three-person treated subset with sum \(S\), its covariate difference is \((2S-21)/3\). The largest possible sum is \(15\), giving \(T=3\), and the smallest is \(6\), giving \(T=-3\). These are the only two allocations with absolute difference at least 3. The two-sided exact p-value is therefore \(2/20=0.100\). This provides no evidence of imbalance at the 0.05 level under this test, but it does not establish that all relevant covariates are balanced or that assignment is random.

What the Test Does—and Does Not—Establish

The exactness of a randomization test is conditional on the assignment mechanism being correct. If all \(\binom{N}{m}\) assignments are assumed equally likely, but some units in fact had a much greater chance of being assigned than others, the calculated reference distribution need not be valid. The neighborhood and mechanism should be specified before interpreting the test, and any features on which assignment was conditioned must be reflected in the allowed reallocations.

The sharp null is also important. It specifies every unit's treatment effect, so outcomes under every permitted assignment can be imputed exactly. A test of the sharp null is not generally an exact test of the weaker null that the average treatment effect is zero while individual effects may differ. The two statements are distinct: effects can average to zero and still vary across units. A test or interval for a weaker average-effect claim requires additional methods and assumptions.

Finally, local randomization is not a substitute name for continuity-based identification. Continuity-based RD reasoning identifies a cutoff effect by comparing limiting conditional means as the running variable approaches the cutoff, under continuity assumptions. Local randomization specifies a finite set of units and an assignment distribution for that set. The resulting estimand, assumptions, and inference target are different. A narrow window does not automatically make assignment random, and results from one window need not match results from another.

Common mistake: “The units are close to the cutoff, so their assignments are randomized; therefore the permutation p-value is automatically causal.” Closeness alone does not establish an assignment mechanism. The correction is to define the neighborhood, state precisely which allocations are considered possible and with what probabilities, and justify the as-if-random claim using the assignment context. The randomization test is valid only relative to that model and its null hypothesis.

Define the target neighborhood.
State the cutoff, the radius or other inclusion rule, and the units included. This determines the finite population to which the analysis refers.

Specify the assignment model.
State whether assignments are uniformly allocated subject to a fixed count, independently assigned, or generated by another mechanism. Explain why it is plausible locally.

Choose the estimand and test.
Distinguish the finite-neighborhood effect from a cutoff limit, and state whether the null is sharp or weaker.

Compute and qualify the inference.
Use the allocation probabilities implied by the model, then interpret the estimate and p-value conditional on that model and neighborhood.

Check Your Understanding

Answer these questions using the definitions and examples above.

What population does the finite-neighborhood estimand \(\tau_{W_h}\) describe, and why is it not automatically the effect exactly at the cutoff?
Under complete randomization with \(N=8\) units and \(m=3\) assigned, how many assignments are in the randomization distribution, and what probability does each receive?
Why can a sharp-null randomization test impute all missing outcomes under the null?
What does a two-sided exact p-value of 0.10 mean under a specified assignment mechanism?
Why does a nonsignificant baseline balance test not prove the local randomization assumption?
In one sentence, distinguish the identification basis of local randomization from that of continuity-based RD.