From a Cutoff Plot to an Estimate
A plot of outcomes against a running variable can suggest a jump at an assignment threshold, but a visual gap is not yet an estimate. To quantify the gap, we approximate the outcome pattern near the cutoff from below and from above, then compare the two fitted values at the cutoff itself. The central idea is to fit the two sides separately: each side may have its own level and its own shape.
This tutorial concerns a sharp regression discontinuity, in which treatment is determined by whether the running variable crosses a cutoff. It focuses on local polynomial estimation of the outcome limits, not on choosing a bandwidth or constructing bias-corrected inference.
Centering the running variable makes the interpretation of the intercept immediate: at the cutoff, \(Z=0\), so every term involving a positive power of \(Z\) is zero. For side \(s\), where \(s\) is either left or right, write the fitted polynomial as
Thus, \(\widehat m_s(0)=\widehat\beta_{s,0}\). The local polynomial estimate of the jump is
The polynomial coefficients are obtained separately on each side. In the simplest version, ordinary least squares minimizes the sum of squared residuals among observations in the local window on that side. More generally, observations can receive nonnegative weights, but they still contribute to distinct left and right fits. The meaning of the intercept contrast does not change.
Why the Intercept Difference Targets the Cutoff Effect
Let \(Y(1)\) and \(Y(0)\) denote the potential outcomes under treatment and no treatment. Under sharp assignment, units just above \(c\) receive treatment and units just below \(c\) do not. The target at the cutoff is the difference between the two potential-outcome means there:
This interpretation relies on assumptions about the data-generating process, not merely on a good-looking plot. Treatment must follow the stated cutoff rule; there must be observations arbitrarily close to the cutoff on both sides; and the conditional mean functions \(E[Y(1)\mid R=r]\) and \(E[Y(0)\mid R=r]\) must be continuous at \(r=c\). In particular, absent treatment, the expected outcome must not itself jump at the cutoff. The usual consistency and no-interference conditions are also understood: observed outcomes correspond to the treatment actually received, and one unit’s treatment does not change another unit’s outcome.
The identification argument is a limit comparison. On the right, observed outcomes near the cutoff reveal treated outcomes; on the left, they reveal untreated outcomes. Continuity allows these nearby means to be extended to the cutoff:
Subtracting the second limit from the first gives \(\tau(c)\). Local polynomials estimate these two limits by fitting outcome patterns near the cutoff and evaluating both fitted curves at \(Z=0\). The polynomial is an approximation to the conditional mean near the threshold; it is not a claim that the outcome relationship is globally polynomial.
Worked Fits
The examples use hypothetical observations and equal weight for every observation inside the stated window. Their small sample sizes make the arithmetic visible; they are not recommendations for sample size or for a particular polynomial degree.
Worked Example: A Local Linear Fit on Each Side
Suppose the cutoff is \(c=50\) points on an eligibility score, and the outcome is a test score measured in points. Within a window of width \(h=2\), imagine these hypothetical observations, with \(z=R-c\):
Left: \((-2,10),(-1,12),(-0.5,13)\). Right: \((0.5,19),(1,20),(2,22)\).
Fit a separate straight line on each side. For the left observations, the points lie exactly on \(Y=14+2z\): at \(z=-2\), the predicted outcome is \(14+2(-2)=10\); at \(z=-1\), it is \(12\); and at \(z=-0.5\), it is \(13\). Thus the left fitted intercept is \(\widehat\beta_{\mathrm{left},0}=14\) points.
For the right observations, the points lie exactly on \(Y=18+2z\): the fitted values at \(z=0.5,1,2\) are \(19,20,22\). The right fitted intercept is \(\widehat\beta_{\mathrm{right},0}=18\) points. Therefore,
The fitted treated-side limit exceeds the fitted untreated-side limit by 4 test-score points. Interpreting that contrast as a treatment effect requires the sharp-assignment and continuity assumptions; the arithmetic alone does not establish them.
Worked Example: Allowing Curvature With Quadratic Fits
Suppose a different hypothetical setting has cutoff \(c=100\) days of age and an outcome measured in milligrams. In a local window, the centered running-variable values and outcomes are left-side pairs \((-3,29),(-2,22),(-1,19)\), and right-side pairs \((1,28),(2,33),(3,40)\). Fit separate quadratic curves.
On the left, the three points are exactly described by \(Y=20+3z+2z^2\). Substitution gives \(29\) at \(z=-3\), \(22\) at \(z=-2\), and \(19\) at \(z=-1\). The fitted left intercept is therefore \(20\) milligrams.
On the right, the points are exactly described by \(Y=25+2z+z^2\): the fitted outcomes at \(z=1,2,3\) are \(28,33,40\). The right intercept is \(25\) milligrams, so the estimated cutoff contrast is \(25-20=5\) milligrams.
The terms involving \(z\) and \(z^2\) allow each side to curve differently away from the threshold. They do not affect the value evaluated at the cutoff, where \(z=0\). A quadratic fit can represent local curvature, but it does not automatically provide a better estimate: its performance depends on the available data and how well a quadratic approximates the conditional mean near the threshold.
Worked Example: Calculating Intercepts When the Data Are Not Collinear
Consider hypothetical outcomes measured in minutes, with cutoff \(c=10\) hours and three observations on each side inside the local window. The left pairs \((z,Y)\) are \((-2,10),(-1,13),(-0.5,13)\). The right pairs are \((0.5,18),(1,21),(2,23)\). Fit an ordinary least-squares line separately on each side.
For a simple linear regression, the slope is \(\widehat\beta_1=\sum_i(z_i-\bar z)(Y_i-\bar Y)/\sum_i(z_i-\bar z)^2\), and the intercept is \(\widehat\beta_0=\bar Y-\widehat\beta_1\bar z\). For the left side, \(\bar z=-3.5/3=-1.1667\) hours and \(\bar Y=36/3=12\) minutes. The sum of centered squared running-variable values is \(1.1667\), and the sum of centered cross-products is \(2.500\). Hence,
For the right side, \(\bar z=3.5/3=1.1667\) hours and \(\bar Y=62/3=20.6667\) minutes. The centered squared-value sum is again \(1.1667\); the centered cross-product sum is \(3.6667\). Thus,
The estimated jump is \(17.00-14.50=2.50\) minutes. The slopes differ, and neither fitted line passes through every observation. The intercept contrast is still obtained by evaluating the two fitted lines at \(z=0\), rather than by comparing observed outcomes at some nonzero distance from the cutoff.
Diagnostics, Alternatives, and Limitations
A local polynomial estimate is only useful if the fitted curves provide credible approximations near the cutoff. Examine the data on each side and the fitted curves together. Sparse observations close to the threshold, large gaps in the running variable, or a curve driven by only one or two distant points make the intercept extrapolation fragile. Residual patterns can also be informative: systematic curvature left over from a linear fit suggests that a straight line may not capture the local pattern adequately. Increasing the polynomial degree is not an automatic fix, because a more flexible curve can become unstable when local data are limited.
The separate fits should not be constrained to share an intercept. Forcing the two sides to have the same intercept rules out the very discontinuity the analysis is designed to estimate. Likewise, a single regression that imposes one smooth relationship across the cutoff can hide a jump unless it explicitly includes a treatment-side indicator and permits the relevant side-specific shape.
Local polynomials are one way to estimate the limits. A simple comparison of average outcomes in bins near the cutoff is easier to describe but depends strongly on which observations fall into each bin and does not model the trend within a bin. A global polynomial uses a wider range of the running variable, but its fitted value at the cutoff can be affected by behavior far from the threshold. Separate local polynomial fits concentrate the approximation near the cutoff and allow different shapes on each side. The introductory material on regression discontinuity explains the identification logic; here, the essential additional choice is how to approximate each nearby outcome curve.
The choice of local window and polynomial degree affects the fitted intercepts. A wider window supplies more data but may include observations whose outcome pattern is less representative of behavior at the cutoff. A higher degree permits more curvature but requires estimating more coefficients. These choices can change the estimate even when the underlying causal assumptions are unchanged. The estimate is also local: it describes the effect at the cutoff, not necessarily an average treatment effect for people far from it.
A Practical Estimation Sequence
Set \(Z=R-c\), so the cutoff is at zero and the polynomial intercept is the fitted cutoff value.
State which observations are used and whether each side is modeled as a constant, line, quadratic, or higher-degree polynomial.
Estimate the coefficients for \(Z<0\) and \(Z\ge0\) using the selected local observations; do not impose a common intercept.
Calculate \(\widehat\beta_{\mathrm{right},0}-\widehat\beta_{\mathrm{left},0}\), report the outcome units, and state that a causal interpretation depends on the RD assumptions.
Check Your Understanding
Answer each question, then compare with the brief solutions.
- If \(c=40\) and a fitted left-side polynomial is \(12-3z+z^2\), what is its fitted value at the cutoff?
- The fitted right and left intercepts are 31 and 27 kilograms. What is the estimated cutoff contrast, and what does its sign mean?
- Why are separate regressions fitted on the two sides rather than requiring a single common intercept?
- Name two conditions needed to interpret the fitted intercept difference as a causal effect at the cutoff.
- What is one limitation of using a higher-degree polynomial in a small local sample?
Solutions. (1) At the cutoff \(z=0\), so the fitted value is \(12\). (2) The contrast is \(31-27=4\) kilograms; the fitted treated-side limit is 4 kilograms higher. (3) A common intercept would force the discontinuity to be zero, disallowing the jump being estimated. (4) Examples include sharp treatment assignment at the cutoff, observations near the cutoff on both sides, and continuity of the potential-outcome conditional means there. (5) The added coefficients can make the fitted curve unstable or overly sensitive to the limited observations.