RDD Flashcards
how do we implement RDD?
identify the cut. There are two ways: local linear regression = y = cons + beta_1(x-c)(1-t) + beta_2(x-c)(t) + beta_3t + u, where beta_1 is slope of control, beta_2 slope on treated, beta_3 is treatment effect. Full discontinuity: higher order poly = y = cons + x + x^2 + x^3 + .. + x^m + t, where t is treatment indicator.
When should we use a strict or fuzzy RDD?
if there is a situation of non-compliance use fuzzy (instrument), producing a TOT. this works because the IV produces an estimate senstive to the exogenous assignment to treatment. BUT AN IV CANNOT SOLVE MANIPULATION.
What are the limitations of RDD?
very low external validity due to LATE right around the cut; no heterogenous effects (if they exist) are detected…if those with X far from C experience different effects these aren’t captured by the LATE
What are the assumptions of RDD?
relationship of x to y is continuous. X is not manipulated to effect who receives treatment (manipulation cannot be observed). There are no other programs or services with the same eligibility rule.
How can we test if f(x) is continuous?
create dummies at the quantiles or deciles of the running variable and see if these dummies are jointly statsig
How can we test if x is manipulated? Under what conditions is manipulation a problem?
we never really know if this happened, but we can examine histograms of the running variable to see if there’s a big dip on one or the other side of the cut. When manipulation is nonrandom then it is a problem.
What are the benefits and costs of using too large a bandwidth?
Benefit is increasing power; cost is that units may become more dissimilar from one another. Plus, if the underlying relationship isn’t linear then it will bias estimates.
how can we choose an optimal bandwidth?
for portion of values of the running variable between two “extremes” (e.g. 10th and 90th percentile) identify a temporary cut and recenter the running variable to that new cut, then run two regressions of y on this newly centered variable - one to the left and one to the right. This procedure produces two intercepts, which we average. this process is repeated for all values of the running variables within the predetermined interval. the bandwidth for which the quantity (mean_y - mean(both intercepts))^2 is minimized represents the optimal bandwidth
how can we interpret IV estimates of TOT?
Y is outcome, Z is assigned to treatment, D is received treatment. Y(Z=1) - Y(Z=0) // D(Z=1) - D(Z=0), where D(Z=1) - D(Z=0) is the proportion of compliers in the sample.
