Online Appendix for “Political Kludges”

Proof of Propositions 1a and 1b Focus on Party +1; the calculation for Party −1 is similar. Start with the casep ∈ [p−1, p+1). There, Party+1’s problem is tomaximize the (linear) objective ∂ ∂tu+1 (p(t)) = z+1 d dtp − d dt‖p‖ subject to the constraint (2), which corresponds in ( d dtp, d dt‖p‖)-space to a triangle Conv({v+, v−, vδ}) with vertices v+ = γ ⋅ (1, 1), v− = γ ⋅ (−1, 1), vδ = γ ⋅ (p/‖p‖, −1), whereConv(S) is the convex closure of set S. A linear objective over a simplex is, of course, maximized at one of the vertices of the simplex. Some algebra reveals that vertex v+ is optimal (maximizes the objective) when − p ‖p‖ > 1 − 2 z+1 ; otherwise, vertex v− is optimal. The case p = p+1 is slightly more involved. Here, the objective is no longer linear in ( d dtp, d dt‖p‖); specifically, ∂ ∂tu+1 (p(t)) = z+1| d dtp| − d dt‖p‖. Notice, however, that this objective is linear on each of the half-planes d dt‖p‖ ≤ 0 and on d dt‖p‖ ≥ 0. The intersection of each half-plane with the triangle Conv({v+, v−, vδ}) defines two simplices in ( d dtp, d dt‖p‖)-space over which the objective function is linear: ∂ ∂tu+1 (p(t)) = −z+1 d dtp − d dt‖p‖ over Conv({v+, vm+, vm−}) and ∂ ∂tu+1 (p(t)) = z+1 d dtp − d dt‖p‖ over Conv({v−, vδ, v0−, v0+}) where v0− = γ ⋅ (0, − ‖p‖−|p| ‖p‖+|p|) and v0+ = γ ⋅ (0, 1).