Bunching Identification: Density Bounds and Control Groups
The kink response calculation expresses bunching mass as the area under an unobserved counterfactual density. A wide interval with low density can have the same area as a narrow interval with high density. This ambiguity underlies the nonidentification result in (Blomquist et al. 2021). Quantitative density restrictions, such as those in (Bertanha et al. 2023), limit the possible areas and hence the elasticities.
1. Why smoothness is insufficient
In the log-linear kink model, let \(s_0>s_1\) denote log net-of-tax rates and let the elasticity be \(\varepsilon\). The latent interval compressed to the kink has width
The observed bunching mass is \(B=\int_0^Lf(v)\,dv\) after shifting the interval to start at zero. Changing \(\varepsilon\) changes \(L\), but an unrestricted latent density can change within that interval to preserve the same integral and match the observed tails.
Theorem 1. Under the regularity conditions of the nonidentification result, smooth observed tails and positive bunching are compatible with different elasticities and different smooth heterogeneity distributions. Smoothness without a quantitative restriction does not point identify the elasticity. A known Lipschitz bound on the latent density instead restricts the feasible widths and hence gives an identified set for \(\varepsilon\).
The zero-elasticity case needs separate treatment: a smooth continuous heterogeneity model with no behavioral response cannot create positive structural bunching merely by changing a density.
2. Derive the density envelopes
Suppose the endpoint densities are \(a=f(0)\) and \(b=f(L)\), and \(|f(v)-f(u)|\le M|v-u|\). Every feasible density must lie between
and
The endpoint compatibility condition is \(|a-b|\le ML\). Integrating gives necessary area restrictions
For this local filling problem, the area restrictions are sufficient as well. A convex combination of the lower and upper envelopes preserves the endpoints, nonnegativity, and the Lipschitz bound. Its area varies continuously between the two extremes.
3. Read the identified set geometrically
To obtain the most mass at a given width, let the density rise from both endpoints as fast as allowed and meet at a tent. To obtain the least, let it fall as fast as allowed, stopping at zero. Once the lower envelope has a zero middle interval, extending that interval need not add any mass, which can leave the elasticity without a finite upper bound.
The identified set is therefore determined by area compatibility. Conflicting mass and endpoint information can leave no feasible density; intermediate cases bound the width on both sides; some cases allow arbitrarily wide low-density interiors.
The numerical bound \(M\) supplies information that the word “smooth” does not. A control group offers a different source of information by identifying counterfactual probabilities inside the missing interval.
4. Integrating the envelopes when endpoint densities agree
Take \(a=b=h>0\) and \(M>0\). The upper envelope is
so its integral is \(hL+ML^2/4\). The lower envelope is \(\max(0,h-M\min(v,L-v))\). Therefore
In the second case the lower density consists of two endpoint triangles, each of area \(h^2/(2M)\), separated by a zero-density interval. Making that middle interval longer adds no required mass.
The upper-area inequality gives the smallest feasible width,
If \(0<B<h^2/M\), the lower-area inequality also gives
If \(B\geq h^2/M\), no finite upper width bound follows from this local interval problem. Dividing widths by \(s_0-s_1\) yields elasticity bounds. As \(M\downarrow0\), both finite endpoints approach \(B/h\), recovering the constant-density case.
Allowing a steeper slope raises the largest possible mass in a short interval and lowers the required mass in a long one. Weakening the restriction therefore enlarges the identified set.
The unequal-endpoint calculation makes the information in both tails visible. Assume \(a,b\geq0\), \(M>0\), and \(|a-b|\leq ML\). The two lines defining the upper envelope cross at
Integrating \(a+Mv\) on \([0,v_u]\) and \(b+M(L-v)\) on \([v_u,L]\) gives
For the lower envelope, the downward endpoint lines meet at \(v_\ell=(a-b+ML)/(2M)\), at height \((a+b-ML)/2\). If that height is nonnegative, both lines remain above zero at their switch, and
If \(ML\geq a+b\), they hit zero before meeting. The lower area then consists of two disjoint triangles:
At \(ML=a+b\) the two formulas agree. At the smallest compatible width \(ML=|a-b|\), the upper and lower envelopes coincide with the line joining the endpoints. These boundary checks ensure that the algebra describes the stated Lipschitz class.
For each proposed elasticity, substitute \(L=\varepsilon(s_0-s_1)\) into \(I_\ell(L)\leq B\leq I_u(L)\). The observations supply the mass and endpoint densities; the assumed slope bound limits the densities that can connect those endpoints. The inequalities then determine whether the proposed elasticity is feasible.
5. Why the area inequalities are sufficient locally
Let \(I_\ell\) and \(I_u\) be the integrals of the lower and upper envelopes. If \(I_\ell\leq B\leq I_u\) and \(I_\ell<I_u\), choose
This function is nonnegative, has the prescribed endpoint values and integral, and remains \(M\)-Lipschitz by the triangle inequality. If the two areas coincide, their common envelope already has the required area. This proves sufficiency for filling the missing interval subject to the stated endpoint constraints.
This proves a local construction only. A full density must also match the observed tails, integrate to one, and satisfy any imposed global regularity. The paper combines those requirements with the interval bounds to establish sharpness.
6. What control groups add
A control group whose effective local tax incentive does not change can reveal income mass inside the treated group’s missing interval. In the design discussed by (Coles et al. 2022), loss carryforwards help create different effective kink locations under a common statutory schedule. Identification still requires comparability of counterfactual income distributions after the design’s adjustments.
If a comparison identifies the relevant counterfactual cdf \(H_0\), the bunching relation becomes
When \(H_0\) is strictly increasing on the interval, solve
The control group identifies a cdf increment over the missing interval. The structural response formula converts its width to an elasticity. If control and treated groups have different counterfactual distributions after adjustment, the inversion also reflects that composition difference.
7. Conditional heterogeneity as another identifying restriction
The mid-censored log-earnings model writes
Let \(a_j=k-\varepsilon s_j-X'\beta\). Below the kink, \(U<a_0\) and \(y=\varepsilon s_0+X'\beta+U\); at the kink, \(a_0\leq U\leq a_1\); above it, \(U>a_1\) and the other segment is observed. Conditional normality therefore contributes likelihood terms
The kink observations contribute probability mass, not density. Under the uniqueness and specification conditions studied in (Bertanha et al. 2023), this censored model identifies the elasticity; the paper also gives conditions extending the result beyond literal conditional normality. Those conditions are substantive, since an arbitrary misspecified likelihood generally estimates a best approximation rather than the structural parameter.
The observation map is nondecreasing in \(U\), so suitably defined latent conditional quantiles pass through it. Uncensored observations give equalities for the linear index; observations at \(k\) give inequalities. Separating the elasticity from the other coefficients requires sufficient uncensored covariate variation. The atom at the threshold alone cannot provide that rank.
Checking that the envelopes belong to the stated class
A maximum or minimum of finitely many \(M\)-Lipschitz functions is \(M\)-Lipschitz: for a maximum, each \(f_j(x)\le f_j(y)+M|x-y|\le\max_k f_k(y)+M|x-y|\), and reversing \(x,y\) gives the absolute bound. Apply the same argument to negatives for a minimum. It follows that both envelopes obey the slope restriction. Endpoint compatibility makes their values at zero equal to \(a\) and their values at \(L\) equal to \(b\); the lower envelope is no larger than the upper one because each endpoint cone is compatible with the other. Their convex combinations therefore really are admissible fillings, not merely functions with the correct area.
If the fixed tails are already nonnegative, \(M\)-Lipschitz and match those endpoint values, the filling also preserves the global Lipschitz bound. For points on opposite sides of a boundary, insert that boundary and apply the triangle inequality; the lengths add to the distance between the two points. If the tails have total mass \(1-B\), normalization follows as well. These conditions give a complete global extension for the Lipschitz class. Requiring differentiability or a smoother class would impose additional matching conditions and is a different extension problem.