Day 75: Bandwidth Selection & Bias Correction

Theory

The trade-off

For a local linear RD estimator with bandwidth $h$ and sample size $n$ ,

\text{Bias} \approx Bh^2, \qquad \operatorname{Var} \approx \frac{V}{nh},

where $B$ depends on the curvature of $E[Y \mid X]$ near the cutoff and $V$ on the outcome variance and the density of $X$ at $c$ .

MSE-optimal bandwidth

Minimising $\text{MSE}(h) = B^2h^4 + V/(nh)$ gives

h_{MSE} = \Big(\frac{V}{4B^2n}\Big)^{1/5} \propto n^{-1/5}.

Data-driven selectors (Imbens–Kalyanaraman; Calonico–Cattaneo–Titiunik, CCT) estimate $B$ (using a pilot local quadratic fit) and $V$ , then plug in. Different bandwidths can be used on each side.

The inference problem

At $h_{MSE}$ , squared bias equals one quarter of the variance, so

\frac{\text{Bias}}{\operatorname{SE}} = \frac12.

A conventional 95% interval $\hat\tau \pm 1.96\,\operatorname{SE}$ then covers only about 92% of the time, and less when bias is underestimated.

Fixes

Undersmoothing: use a smaller bandwidth than $h_{MSE}$ so bias is negligible; the point estimate becomes noisier and the right amount of undersmoothing is unclear.
Robust bias correction (Calonico, Cattaneo and Titiunik, 2014): estimate the bias with a local quadratic fit (pilot bandwidth $b$ ), subtract it, and compute a standard error that includes the variability of the bias estimate. With $b = h$ , the bias-corrected estimator coincides with the local quadratic estimator, and RBC inference is local quadratic inference at bandwidth $h$ .

Current practice: report the local linear estimate at $h_{MSE}$ with the RBC confidence interval; show sensitivity to halving and doubling the bandwidth. Coverage-error-optimal bandwidths (smaller than $h_{MSE}$ ) are an alternative for inference.

Part 1 · Warm-up

Exercise 1 · An optimal bandwidth

With $B = 2$ , $V = 4$ and $n = 1{,}000$ , compute $h_{MSE}$ .

Show solution

Step 1. $h = (4/(4 \times 4 \times 1{,}000))^{1/5} = (1/4{,}000)^{1/5}$ .

Step 2. $\ln 4{,}000 \approx 8.29$ ; divided by 5: 1.66; $e^{-1.66} \approx 0.19$ .

Answer: about 0.19 units of the running variable.

Exercise 2 · More data

If the sample grows 32-fold, how does $h_{MSE}$ change?

Show solution

Step 1. $h \propto n^{-1/5}$ ; $32^{-1/5} = 1/2$ .

Answer: it halves.

Exercise 3 · Coverage at the optimum

With bias equal to half the SE, what is the coverage of a nominal 95% interval?

Show solution

Step 1. The t-statistic is $N(0.5, 1)$ around the truth.

Step 2. Coverage $= P(|Z + 0.5| < 1.96) = \Phi(1.46) - \Phi(-2.46) \approx 0.928 - 0.007 = 0.921$ .

Answer: about 92%.

Exercise 4 · Undersmoothing

To make the bias at most 10% of the SE, by what factor must the bandwidth shrink relative to $h_{MSE}$ ?

Show solution

Step 1. Bias/SE $\propto h^2 \cdot h^{1/2} = h^{5/2}$ .

Step 2. We need the ratio to fall from 0.5 to 0.1: factor 5, so $h$ shrinks by $5^{2/5} \approx 1.90$ .

Answer: use roughly half the MSE-optimal bandwidth; the SE then rises by $\sqrt{1.9} \approx 1.38$ .

Exercise 5 · Ingredients of RBC

List what is estimated in robust bias-corrected inference.

Show solution

Step 1. The local linear estimate $\hat\tau_h$ with bandwidth $h$ .

Step 2. The bias $\hat B$ from a local quadratic fit with pilot bandwidth $b$ .

Step 3. A standard error for $\hat\tau_h - \hat B$ that includes the variance of $\hat B$ and its covariance with $\hat\tau_h$ .

Part 2 · Proofs

Exercise 6 · Deriving the optimal bandwidth

Minimise $B^2h^4 + V/(nh)$ over $h > 0$ .

Show solution

Step 1. Derivative: $4B^2h^3 - V/(nh^2) = 0$ .

Step 2. $h^5 = V/(4B^2n)$ , so $h = (V/(4B^2n))^{1/5}$ .

Step 3. The second derivative $12B^2h^2 + 2V/(nh^3) > 0$ : a minimum. $\blacksquare$

Exercise 7 · Bias is half the SE

Show that at $h_{MSE}$ , $\text{Bias}^2 = \operatorname{Var}/4$ .

Show solution

Step 1. From Exercise 6, $4B^2h^5 = V/n$ , so $B^2h^4 = V/(4nh)$ .

Step 2. The left side is $\text{Bias}^2$ ; the right side is $\operatorname{Var}/4$ . $\blacksquare$

Exercise 8 · The convergence rate

Show that at $h_{MSE}$ the root mean squared error is of order $n^{-2/5}$ , slower than the parametric $n^{-1/2}$ .

Show solution

Step 1. $\text{MSE} = B^2h^4 + V/(nh)$ with $h \propto n^{-1/5}$ .

Step 2. $B^2h^4 \propto n^{-4/5}$ and $V/(nh) \propto n^{-1 + 1/5} = n^{-4/5}$ .

Step 3. $\text{RMSE} \propto n^{-2/5}$ . $\blacksquare$ RD uses only data near the cutoff, so it learns more slowly than an experiment.

Exercise 9 · Why bias correction needs a new SE

Let $\hat\tau_{bc} = \hat\tau - \hat B$ . Show $\operatorname{Var}(\hat\tau_{bc}) = \operatorname{Var}(\hat\tau) + \operatorname{Var}(\hat B) - 2\operatorname{Cov}(\hat\tau, \hat B)$ and explain why using $\operatorname{SE}(\hat\tau)$ alone is wrong.

Show solution

Step 1. Variance of a difference.

Step 2. $\hat B$ is estimated from the same (or overlapping) data with a curvature estimate, which is noisy: $\operatorname{Var}(\hat B)$ is not negligible in large samples when $b$ is of the same order as $h$ .

Answer: ignoring it understates uncertainty; RBC uses the full variance. $\blacksquare$

Exercise 10 · Choosing the bandwidth by eye

Explain why picking the bandwidth that gives the most significant estimate invalidates inference.

Show solution

Step 1. Estimates at different bandwidths are noisy, and some will be significant by chance.

Step 2. Choosing the most significant inflates the probability of rejecting a true null above 5% (specification searching).

Answer: pre-specify a data-driven bandwidth and report sensitivity across bandwidths, not the best one. $\blacksquare$

Part 3 · Applied

Exercise 11 · A scholarship RD bandwidth

A pilot local quadratic fit suggests $B = 0.5$ (enrolment probability per squared score unit, in standardised score units), the variance constant is $V = 2$ , and there are $n = 5{,}000$ students. Compute $h_{MSE}$ , and the bandwidth after a 50% undersmoothing.

Show solution

Step 1. $h = (2/(4 \times 0.25 \times 5{,}000))^{1/5} = (2/5{,}000)^{1/5} = (0.0004)^{0.2}$ .

Step 2. $\ln 0.0004 \approx -7.82$ ; $/5 = -1.56$ ; $e^{-1.56} \approx 0.21$ .

Answer: $h_{MSE} \approx 0.21$ sd of the score; undersmoothed about 0.10.

Exercise 12 · Coverage in a simulation

In the simulation below (true jump 5, curved regression function, 1,000 observations), compare the coverage of conventional local linear intervals with local quadratic (RBC with $b = h$ ) intervals at bandwidths 0.15, 0.3 and 0.5.

Show solution

Step 1. Small $h$ : both cover about 92–93%; the local quadratic interval is wider.

Step 2. At $h = 0.5$ , the local linear estimate is biased (about +0.5) and its interval covers only about 86%.

Step 3. The local quadratic interval keeps coverage near 94% at every bandwidth.

Answer: robust bias correction protects inference when the bandwidth is large enough for bias to matter.

Exercise 13 · A bandwidth robustness table

For an RD on the effect of a pension eligibility age on retirement, outline a robustness table.

Show solution

Step 1. Rows: $h_{MSE}$ , $h_{MSE}/2$ , $2h_{MSE}$ , and the coverage-error-optimal bandwidth.

Step 2. Columns: point estimate, RBC 95% CI, number of observations on each side.

Step 3. Also: local quadratic fits, triangular versus uniform kernel.

Answer: stable estimates across rows support the result; strong sensitivity signals curvature or too few observations near the cutoff.

Exercise 14 · Different bandwidths on each side

Mortality data around a very-low-birth-weight cutoff have many more observations above than below the threshold, and curvature differs. Why might separate bandwidths help?

Show solution

Step 1. Each side's MSE depends on its own density, variance and curvature.

Step 2. Separate bandwidths use more data where it is plentiful and the function is flat, less where it curves.

Answer: lower overall MSE for the jump estimate; CCT-style selectors provide side-specific bandwidths.

Exercise 15 · Few observations near the cutoff

An RD on a school-entry age cutoff has only 60 children within a month of the cutoff. What alternative to continuity-based inference is appropriate?

Show solution

Step 1. Asymptotic local polynomial inference may be unreliable with so few observations.

Step 2. Local randomisation: treat children born within a narrow window as if randomly assigned, check covariate balance, and use Fisher randomisation inference (Day 37).

Answer: a local-randomisation analysis in a small window, reported alongside the continuity-based estimate.

Verify in Python

import numpy as np
from scipy.stats import norm

print("Ex 1 h_MSE:", round((4 / (4 * 4 * 1000)) ** .2, 3), " Ex 11:", round((2 / (4 * .25 * 5000)) ** .2, 3),
      " Ex 3 coverage:", round(norm.cdf(1.96 - .5) - norm.cdf(-1.96 - .5), 3), " Ex 4 factor:", round(5 ** .4, 2))

rng = np.random.default_rng(75)
mu0 = lambda x: 50 + 8 * x + 30 * x ** 2 - 20 * x ** 3
def side(x, y, h, p):
    w = np.clip(1 - np.abs(x) / h, 0, None); m = w > 0
    X = np.column_stack([x[m] ** k for k in range(p + 1)]); W = w[m]
    A = np.linalg.inv(X.T @ (W[:, None] * X)); b = A @ (X.T @ (W * y[m])); r = y[m] - X @ b
    V = A @ (X.T @ ((W ** 2 * r ** 2)[:, None] * X)) @ A                    # heteroskedasticity-robust
    return b[0], V[0, 0]
def rd(x, y, h, p):
    r = x >= 0
    a, va = side(x[r], y[r], h, p); b, vb = side(-x[~r], y[~r], h, p)
    return a - b, np.sqrt(va + vb)
print("  h   LL bias  LL sd  LL cover | LQ bias  LQ sd  LQ cover")
for h in (.15, .3, .5):
    out = {1: [], 2: []}; cov = {1: 0, 2: 0}
    for _ in range(1000):
        x = rng.uniform(-1, 1, 1000); y = mu0(x) + 5 * (x >= 0) + rng.normal(0, 3, 1000)
        for p in (1, 2):
            t, s = rd(x, y, h, p); out[p].append(t); cov[p] += abs(t - 5) < 1.96 * s
    print(f"{h:4.2f}  {np.mean(out[1]) - 5:7.2f}  {np.std(out[1]):5.2f}  {cov[1] / 1000:7.3f} | "
          f"{np.mean(out[2]) - 5:7.2f}  {np.std(out[2]):5.2f}  {cov[2] / 1000:7.3f}")

Key takeaways

The MSE-optimal bandwidth shrinks like n^(−1/5) and leaves a bias equal to half the standard error.
Conventional intervals at that bandwidth under-cover; robust bias-corrected intervals subtract the estimated bias and account for its variance.
Pre-specify a data-driven bandwidth, report RBC intervals, and show sensitivity to the bandwidth.

Day 75. Bandwidth Selection & Bias Correction

Theory

The trade-off

MSE-optimal bandwidth

The inference problem

Fixes

Part 1 · Warm-up

Exercise 1 · An optimal bandwidth

Exercise 2 · More data

Exercise 3 · Coverage at the optimum

Exercise 4 · Undersmoothing

Exercise 5 · Ingredients of RBC

Part 2 · Proofs

Exercise 6 · Deriving the optimal bandwidth

Exercise 7 · Bias is half the SE

Exercise 8 · The convergence rate

Exercise 9 · Why bias correction needs a new SE

Exercise 10 · Choosing the bandwidth by eye

Part 3 · Applied

Exercise 11 · A scholarship RD bandwidth

Exercise 12 · Coverage in a simulation

Exercise 13 · A bandwidth robustness table

Exercise 14 · Different bandwidths on each side

Exercise 15 · Few observations near the cutoff

Verify in Python

Key takeaways

Simulation Lab: Bandwidth & Robust Bias Correction