Local Randomization in Regression Discontinuity Designs

Two Ways to Interpret a Regression Discontinuity

In a sharp regression discontinuity (RD), treatment is assigned by whether a running variable $X_i$ is above a cutoff $c$: $D_i=1\{X_i\ge c\}$. The continuity-based interpretation identifies a causal effect at the cutoff by requiring the potential-outcome regression functions to be continuous there. A different interpretation is possible when attention is restricted to a suitably narrow, finite window around $c$: within that window, treatment assignment may be treated as if randomized, conditional on the observed number assigned to treatment.

These are distinct identification strategies. Local randomization does not follow automatically from being close to the cutoff, and it does not mean the running variable itself was randomized. It is a substantive claim that, for the units in a specified window, the observed treatment allocation is plausibly generated by a particular random assignment mechanism. That mechanism—not a generic assumption that “nearby units are similar”—is what supports randomization-based inference.

Definition: A local-randomization RD analysis selects a finite set of units in a window around the cutoff and specifies a probability distribution over treatment assignments within that set. In the common complete-randomization version, conditional on the number $m$ treated among $N$ included units, every allocation with exactly $m$ treated units has probability $1/\binom{N}{m}$. The assignment is assumed to be independent of the units’ potential outcomes, conditional on the analysis window and the specified design.

The Finite Window and Its Assignment Mechanism

A symmetric window is often written as $|X_i-c|\le h$, where $h$ determines which observations enter the local analysis. The window must contain a finite set of units; call its size $N$. The number actually assigned to treatment in that set is $m$. Under complete randomization conditional on $m$, the possible assignments are the $\binom{N}{m}$ subsets of $m$ treated units, each equally likely.

The window is part of the design claim, not just a calculation setting. It should be defined in a way that makes the asserted assignment mechanism plausible, preferably without choosing it to obtain a desired treatment-effect estimate or p-value. A window that is too broad may include units whose treatment status is predictable from meaningful differences in the running variable; a very narrow window may leave too few possible assignments for informative inference. Local randomization therefore requires both a defensible window and a defensible assignment mechanism within it.

The mechanism can differ from complete randomization. For example, assignments might be independent Bernoulli draws with known probabilities, or randomization might be restricted by strata. In each case the randomization distribution must reproduce the mechanism actually assumed. Permuting treatment labels uniformly when the design does not give all allocations equal probability gives the wrong reference distribution.

Worked Example: Specifying a Window and a Fixed-Count Design

Suppose a hypothetical program uses an exam score cutoff of $c=70$. An analyst defines the local window as scores from 68 through 72, inclusive. Six applicants fall in that window, and three received the program. The proposed local-randomization model is complete randomization conditional on exactly three of the six applicants receiving treatment.

There are $\binom{6}{3}=6!/(3!3!)=20$ possible allocations. Under the model, each has probability $1/20=0.05$. The model says that, among these six applicants and conditional on three treated, the observed set of three treated applicants is as likely as any other set of three.

This calculation describes the stipulated assignment mechanism; it does not establish that the mechanism is credible. The analyst must justify why applicants with scores in this window can plausibly be treated as exchangeable with respect to assignment, given the fixed treated count. If, for example, applicants just above 70 were systematically more likely to receive the program because of discretion beyond the stated rule, the uniform allocation model would not represent the assignment process.

Potential Outcomes and the Local Effect

For each unit $i$ in the window, let $Y_i(1)$ be the outcome if treated and $Y_i(0)$ the outcome if untreated. The observed outcome is $Y_i=D_iY_i(1)+(1-D_i)Y_i(0)$. Assume consistency, so the observed outcome equals the potential outcome under the treatment actually received, and assume no interference, so one unit’s treatment does not change another unit’s outcome. The finite-window average treatment effect is $\tau_W=N^{-1}\sum_{i=1}^{N}\{Y_i(1)-Y_i(0)\}$. It describes the average effect for these $N$ units, not automatically an effect for all units at the cutoff or for a broader population.

Under complete randomization conditional on $m$, the difference in observed means is $\widehat{\tau}_W=\bar{Y}_{T}-\bar{Y}_{C}$, where $\bar{Y}_{T}$ and $\bar{Y}_{C}$ are the sample means in the treated and control groups within the window. Its design-based expectation equals $\tau_W$: every unit has probability $m/N$ of treatment, so the expected treated mean is the average of the $Y_i(1)$; every unit has probability $(N-m)/N$ of control, so the expected control mean is the average of the $Y_i(0)$. Subtracting those expectations gives the finite-window effect.

Result: Under complete randomization with $0<m<N$, consistency, and no interference, the difference in means is unbiased over the assignment mechanism for the finite-window average treatment effect: $$ E(\bar{Y}_{T}-\bar{Y}_{C}\mid \text{window},m) = \frac{1}{N}\sum_{i=1}^{N}\{Y_i(1)-Y_i(0)\} = \tau_W. $$ This is a design-based statement: the potential outcomes are treated as fixed, and the expectation is over the possible treatment assignments.

Worked Example: A Difference in Means and Its Design Expectation

Consider four hypothetical applicants in a local window. Their potential outcomes, shown here solely to illustrate the design argument, are $Y(0)=(4,6,8,10)$ and $Y(1)=(6,7,11,12)$. Suppose two applicants are selected for treatment by complete randomization. In the observed allocation, applicants 1 and 3 are treated.

The observed treated outcomes are 6 and 11, so $\bar{Y}_{T}=(6+11)/2=8.5$. The control outcomes are 6 and 10, so $\bar{Y}_{C}=(6+10)/2=8$. Thus $\widehat{\tau}_W=8.5-8=0.5$ outcome units.

The finite-window average effect, using the illustrative potential outcomes, is $\tau_W=[(6-4)+(7-6)+(11-8)+(12-10)]/4=8/4=2$ outcome units. The observed estimate is not equal to this value in the realized allocation. Its unbiasedness is an average over all six equally likely allocations, not a guarantee that this particular estimate equals the finite-window effect.

To see the design expectation directly, each applicant has probability $2/4=1/2$ of being treated. Therefore, the expected treated mean is $(Y_1(1)+Y_2(1)+Y_3(1)+Y_4(1))/4=9$. Each applicant also has probability $1/2$ of being a control, so the expected control mean is $(Y_1(0)+Y_2(0)+Y_3(0)+Y_4(0))/4=7$. Their difference is $9-7=2=\tau_W$.

Randomization Tests Under a Sharp Null

A randomization test compares the observed statistic with the statistics that would result under every assignment allowed by the assumed mechanism. The cleanest exact test uses Fisher’s sharp null hypothesis: $H_0:Y_i(1)=Y_i(0)$ for every unit in the window. Under this null, each unit’s outcome is known under every possible allocation, because treatment would change no unit’s outcome. A statistic such as the difference in means can therefore be recalculated for each permitted allocation.

For a two-sided test, the exact p-value is the fraction of permitted assignments whose statistic is at least as extreme as the observed statistic, typically measured by absolute value. This p-value is exact under the sharp null and the stated assignment mechanism. It is not the probability that the null is true. Nor does rejection or non-rejection alone provide an exact test of the weaker null that the average treatment effect is zero while individual effects may differ. A sharp-null randomization test and inference about an average-effect null are not interchangeable.

Worked Example: An Exact Two-Sided Randomization p-Value

Four hypothetical units have observed outcomes $1,8,2,9$. Exactly two are treated, and the observed treated pair is $\{1,2\}$. Use the difference in means, treated minus control, and test the sharp null that treatment affects no unit.

The observed statistic is $T_{\mathrm{obs}}=(1+8)/2-(2+9)/2=4.5-5.5=-1$. There are six allocations with two treated units. In the order $\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}$, the absolute differences are $1,7,0,0,7,1$, respectively. For example, allocation $\{1,3\}$ gives $(1+2)/2-(8+9)/2=1.5-8.5=-7$, whose absolute value is 7.

Four of the six allocations have an absolute statistic at least as large as $|T_{\mathrm{obs}}|=1$. Therefore, $p=4/6=0.667$. At conventional significance levels this test does not reject the sharp null. The result is conditional on the uniform fixed-count assignment model and tests the sharp null for all four units.

Worked Example: A One-Sided Exact Test With One Treated Unit

In another hypothetical window, three units have observed outcomes $0,1,4$, and exactly one unit is treated. Suppose unit 2, with outcome 1, is observed treated. Test the sharp null against the one-sided alternative that treatment lowers outcomes, using $T=\bar{Y}_{T}-\bar{Y}_{C}$.

The observed statistic is $T_{\mathrm{obs}}=1-(0+4)/2=1-2=-1$. Under the sharp null, the three possible treated units produce statistics $-2.5$, $-1$, and $3.5$. In detail, treating unit 1 gives $0-(1+4)/2=-2.5$; treating unit 2 gives $1-(0+4)/2=-1$; treating unit 3 gives $4-(0+1)/2=3.5$.

For the stated alternative, outcomes lower under treatment correspond to statistics at or below the observed value. Two of the three allocations have $T\le -1$, so the one-sided exact p-value is $2/3=0.667$. The calculation uses the control mean $(0+4)/2=2$ when unit 2 is treated.

What Must Be Believed—and What the Test Does Not Supply

The local-randomization interpretation rests on substantive and design assumptions. The specified window must identify the units to which the claim applies; the assignment probabilities within it must match the actual process; treatment assignment must be independent of the potential outcomes under that mechanism; and consistency and no interference must hold. If assignment probabilities depend on observed features or restrictions, those features and restrictions need to be represented in the assignment distribution.

A deterministic cutoff rule does not itself randomly allocate treatment. Local randomization is an approximation or design-based interpretation applied to a finite window. Its credibility depends on the institutional assignment process and the selected window, not just on a small numerical distance from the cutoff. Evidence that the assumed mechanism is implausible should not be repaired by running a permutation test anyway. In particular, a small p-value calculated under the wrong assignment distribution does not become valid because the arithmetic is exact.

The window effect is also local to the included units. It need not equal a limiting effect exactly at $c$, and it need not generalize to units outside the window. This is a different target from the continuity-based cutoff effect. The distinction matters when reporting results: state the window, the assumed assignment mechanism, the finite-window estimand, and the null tested.

Common pitfall: “The score is close to the cutoff, so treatment was randomized; therefore, any permutation p-value is valid.” Closeness alone does not imply random assignment. The correction is to define a finite window, specify which allocations are possible and their probabilities, and justify why that mechanism is credible for the units in the window. The randomization test is exact only relative to that mechanism and, in the examples above, the sharp null.

Practical Analysis Sequence

Define the local set.
State the cutoff, the rule defining the finite window, and the resulting number $N$ of included units.

Write the assignment mechanism.
Specify the allowed treatment allocations and their probabilities, including whether inference conditions on the treated count $m$.

Name the target and null.
Distinguish the finite-window average treatment effect from a sharp null that states no unit is affected.

Calculate and qualify inference.
Compute the statistic and its randomization distribution under the stated mechanism, then report the window, assumptions, and scope of the conclusion.

Check Your Understanding

Answer each question using the assignment mechanism and estimand relevant to local randomization.

In a window with $N=8$ units and exactly $m=3$ treated under complete randomization, how many treatment allocations are possible, and what probability does each receive?
Why does a fixed-count randomization test require recalculating the statistic only over allocations with the observed treated count?
State the sharp null used for an exact Fisher randomization test, and explain why it is stronger than the null that the finite-window average treatment effect is zero.
Under complete randomization, what does the difference in means estimate in expectation: the effect at the cutoff, or the average effect among the units in the window?
Why is a small distance from the cutoff alone insufficient to justify treating treatment labels as uniformly randomized?