From Local Fits to a Cutoff Estimate
In a sharp regression discontinuity design, treatment changes deterministically at a threshold of a running variable. The causal question is local: how much does the mean outcome change at that threshold? The local linear estimator answers by fitting a line to observations near the cutoff on each side, then subtracting the two fitted values at the cutoff. The central point in its derivation is that the cutoff estimate is an intercept, not the outcome fitted at a typical observation on either side.
Let \(R_i\) be unit \(i\)'s running variable, \(c\) the cutoff, and \(Y_i\) the outcome. Define the centered running variable \(X_i=R_i-c\), so the cutoff is \(X_i=0\). In a sharp design, treatment is \(D_i=1(R_i\ge c)\): units just to the right are treated and units just to the left are untreated.
For a bandwidth \(h>0\), write \(w_i=K(X_i/h)\), where \(K\) is a chosen nonnegative weighting function. The precise shape of \(K\) is a separate design choice; here it is enough that observations with positive weights contribute to the fit. Denote the right-side intercept and slope by \(\hat a_+\) and \(\hat b_+\), and the left-side quantities by \(\hat a_-\) and \(\hat b_-\). The two weighted least-squares fits minimize
The fitted value at the cutoff is \(a_s+b_s(0)=a_s\) on side \(s\). Consequently, the estimator of the outcome jump, with the right side minus the left side as the sign convention, is
This is an estimated discontinuity in the conditional mean outcome. Under the sharp-design identification assumptions below, it estimates the causal effect at the cutoff. Without those assumptions, it remains a fitted difference between two local regressions rather than a causal conclusion.
Deriving the Weighted Intercepts
The derivation can be done separately for either side. For one side, let \(S_0=\sum_i w_i\), \(S_x=\sum_i w_iX_i\), \(S_{xx}=\sum_i w_iX_i^2\), \(S_y=\sum_i w_iY_i\), and \(S_{xy}=\sum_i w_iX_iY_i\), where each sum includes only observations on that side. The weighted sum of squared residuals is \(Q(a,b)=\sum_i w_i(Y_i-a-bX_i)^2\). At its minimum, its derivatives with respect to \(a\) and \(b\) are zero. Dividing the resulting equations by \(-2\) gives the weighted normal equations:
In matrix form, these equations are \(\begin{pmatrix}S_0&S_x\\S_x&S_{xx}\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}S_y\\S_{xy}\end{pmatrix}\). The determinant of the matrix is \(\Delta=S_0S_{xx}-S_x^2\). If \(\Delta>0\), the matrix is invertible, and solving gives:
Applying the intercept formula once to the right-side sums and once to the left-side sums yields \(\hat\tau_{\mathrm{LL}}=\hat a_+-\hat a_-\). The condition \(\Delta>0\) says that there must be positive total weight and weighted variation in \(X_i\) on that side. If all positively weighted observations have the same running-variable value, a line's intercept and slope cannot both be separately determined. Having observations close to the cutoff on both sides is also essential to estimate the two local fits rather than extrapolate one side across an unsupported region.
Worked Example: Exact Local Lines and Their Cutoff Jump
Suppose a hypothetical dataset has equal weights on three observations on each side. The centered running-variable values are \(-3,-2,-1\) on the left and \(0,1,2\) on the right. The left outcomes are \(45/13,58/13,71/13\), and the right outcomes are \(137/13,150/13,163/13\). These numbers lie exactly on a line on each side.
On the left, the mean of \(X\) is \((-3-2-1)/3=-2\), and the mean of \(Y\) is \((45/13+58/13+71/13)/3=58/13\). The left slope is 1, since each one-unit increase in \(X\) increases \(Y\) by \(13/13=1\). Thus the left intercept is \(\hat a_-=58/13-1(-2)=84/13\), or approximately 6.462 outcome units.
On the right, the mean of \(X\) is \((0+1+2)/3=1\), and the mean of \(Y\) is \((137/13+150/13+163/13)/3=150/13\). Its slope is also 1, so the right intercept is \(\hat a_+=150/13-1(1)=137/13\), or approximately 10.54 outcome units.
The estimated jump, using the unrounded intercepts, is \(137/13-84/13=53/13\approx4.077\) outcome units. The fitted difference is positive: the right-side fitted mean at the cutoff is about 4.077 outcome units higher. This calculation alone does not establish a causal effect; the design assumptions must also hold.
Identification Assumptions and the Estimand
In a sharp design, the treatment status changes from zero to one at \(c\). Let \(Y_i(0)\) and \(Y_i(1)\) denote the potential outcomes without and with treatment. The local causal effect of interest is the mean treatment effect at the cutoff, \(E[Y_i(1)-Y_i(0)\mid R_i=c]\), understood through the limits of the potential-outcome means as \(R_i\) approaches \(c\). For the observed outcome jump to identify this effect, the potential-outcome conditional means must be continuous at the cutoff; treatment assignment must follow the sharp threshold rule; and there must be observations, or population support, approaching the cutoff from both sides. These conditions rule out a separate jump in untreated or treated potential outcomes that would otherwise be mistaken for the treatment effect.
The identification argument is brief. Immediately to the right, observed outcomes are \(Y_i(1)\); immediately to the left, they are \(Y_i(0)\). Therefore, the limiting observed-outcome difference is
where continuity supplies the two potential-outcome limits. The right-hand side is the mean causal effect at the cutoff. The local linear estimator approximates the two limits using the intercepts of the separately fitted lines. It is a finite-sample approximation to the identified limits, not an exact identity for arbitrary finite data.
The fit also requires a usable local design: both sides need positively weighted observations and enough variation in \(X_i\) to estimate an intercept and slope. The chosen bandwidth must leave adequate local support. For causal interpretation, the substantive continuity and assignment conditions remain necessary; a well-conditioned regression does not prove them.
Worked Example: Computing Intercepts From Weighted Sums
Consider hypothetical weighted observations summarized separately by side. On the left, the weighted sums are \(S_0=4\), \(S_x=-7\), \(S_{xx}=15\), \(S_y=19\), and \(S_{xy}=-29\). On the right, they are \(S_0=4\), \(S_x=4\), \(S_{xx}=6\), \(S_y=47\), and \(S_{xy}=52\). Assume these sums come from observations with positive weights on both sides and nonzero weighted variation in \(X\).
For the left fit, the determinant is \(4(15)-(-7)^2=60-49=11\). Its intercept is \[ \hat a_-=\frac{15(19)-(-7)(-29)}{11} =\frac{285-203}{11}=\frac{82}{11}\approx7.455. \]
For the right fit, the determinant is \(4(6)-4^2=24-16=8\). Its intercept is \[ \hat a_+=\frac{6(47)-4(52)}{8} =\frac{282-208}{8}=\frac{74}{8}=9.25. \]
The estimated cutoff jump is \(9.25-82/11=79/44\approx1.795\) outcome units. The positive result describes a higher fitted right-side cutoff mean by approximately 1.795 units. Interpreting it as a causal effect requires the sharp-design assumptions, not merely the arithmetic of the weighted regressions.
Why the Two Slopes Are Separate
Allowing a different slope on each side is not a cosmetic choice. The outcome relationship with the running variable may differ to the left and right of the cutoff. A model with a single common slope can force one side's trend onto the other, changing the fitted intercepts and hence the estimated jump.
Worked Example: A Common Slope Changes the Estimated Jump
Suppose a hypothetical dataset has left-side observations \((X,Y)=(-2,6),(-1,8)\) and right-side observations \((0,14),(1,13)\), all with equal weights. Separate lines give a left slope of \((8-6)/[-1-(-2)]=2\) and intercept \(6-2(-2)=10\). The right slope is \((13-14)/(1-0)=-1\) and intercept \(14\). Thus the separate-slope estimate is \(14-10=4\) outcome units.
Now constrain both sides to have a common slope, while retaining a separate intercept for each side. The within-side slope numerator is the sum of the products of deviations from each side's means. On the left, the means are \(\bar X=-1.5\) and \(\bar Y=7\), giving numerator \((-0.5)(-1)+ (0.5)(1)=1\), and denominator \((-0.5)^2+(0.5)^2=0.5\). On the right, the means are \(\bar X=0.5\) and \(\bar Y=13.5\), giving numerator \((-0.5)(0.5)+(0.5)(-0.5)=-0.5\), and denominator \(0.5\).
The common slope is therefore \((1-0.5)/(0.5+0.5)=0.5\). The resulting left intercept is \(7-0.5(-1.5)=7.75\), and the right intercept is \(13.5-0.5(0.5)=13.25\). This restricted model estimates a jump of \(13.25-7.75=5.5\), not 4. The separate local linear estimator avoids imposing that common-slope restriction.
Diagnostics, Alternatives, and Limitations
A calculated estimate can be misleading if the local fits are not supported by the data. Before interpreting the intercept difference, check that observations with positive weights are present on each side, that \(X_i\) varies on each side, and that the fitted lines are based on observations genuinely near the cutoff. A zero or very small determinant signals a singular or poorly conditioned fit. A very small effective set of weighted observations can make the fitted intercept unstable even when the determinant is technically positive.
A plot of the observations and fitted lines can reveal sparse support, influential observations, or a poor local linear approximation. Such a plot is a diagnostic, not proof of continuity or of the absence of sorting around the threshold. The bandwidth and weighting choice matter because they determine which observations influence the local fits. The next steps in a full analysis include examining those choices, assessing sensitivity, and using appropriate uncertainty estimates; the point estimate alone does not quantify uncertainty.
| Approach | What it fits | Key trade-off |
|---|---|---|
| Local linear, separate slopes | A weighted line on each side; subtracts fitted intercepts | Allows different local trends but needs adequate support on each side |
| Local constant | A weighted mean on each side | Simpler fit, but does not adjust for the running variable's local trend |
| Global linear model | A line over a much broader range, often with a treatment indicator | Uses more data but relies more heavily on the chosen global functional form |
The local linear estimator is not automatically reliable because it is called “local.” If the bandwidth includes observations far from the cutoff, a straight line may not approximate the conditional mean well there. If it is too narrow, there may be too little support for stable fits. Nor does a smooth-looking pair of lines establish the causal assumptions. The estimate is local to the cutoff and generally does not describe effects far from it or in other populations.
A common mistake is to subtract the observed averages on the right and left and call the result the local linear estimate. Those averages need not be the fitted values at \(X_i=0\); their difference can reflect different distributions of running-variable values within the neighborhood. The correction is to fit the side-specific regressions and subtract their intercepts. Another mistake is to use the left intercept minus the right intercept without changing the interpretation: under the stated convention, the estimator is right minus left, so reversing the order reverses the sign.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- For one side, suppose \(S_0=5\), \(S_x=-5\), \(S_{xx}=9\), \(S_y=20\), and \(S_{xy}=-25\). What is the weighted intercept?
- Why does the local linear estimator subtract intercepts rather than the two sides' average observed outcomes?
- What does a zero value of \(S_0S_{xx}-S_x^2\) indicate for a side-specific fit?
- Under what condition does the observed limiting outcome jump have a causal interpretation in a sharp design?
Solutions. 1. The determinant is \(5(9)-(-5)^2=45-25=20\), so \(\hat a=[9(20)-(-5)(-25)]/20=(180-125)/20=2.75\) outcome units. 2. The intercept is the fitted mean at the cutoff \(X=0\); observed side averages can be evaluated at different running-variable values and so need not represent the cutoff limits. 3. The weighted regressor matrix is singular: there is no positively weighted variation in \(X\) sufficient to estimate a separate intercept and slope on that side. 4. The assignment must change sharply at the threshold, and the potential-outcome conditional means must be continuous there, with support approaching the cutoff from both sides. Then the limiting observed-outcome difference identifies the mean treatment effect at the cutoff.