Lecture VII: Agenda Setting Recommended Reading: Romer & Rosenthal (1978) Baron & Ferejohn (1989)

Lecture VII: Agenda Setting Recall standard median voter theorem: If… 1.Voters i… n (odd) have single-peaked utility functions over a single good, and 2.Any voter can freely place a proposal on the agenda Then, median voter’s position dominates, i.e., is the equilibrium outcome of any sequence of voting

Lecture VII: Agenda Setting In most legislatures, access to agenda control is limited Agenda setter can secure non-median outcomes This result contingent on location of status quo –e.g., given symmetric, single-peaked utility functions (e.g. Euclidean, quadratic), agenda setter can secure p* –Further SQ is from median voter, closer p* can be to agenda setter’s ideal point SQ Agenda setter p*

Lecture VII: Agenda Setting Romer & Rosenthal (1978) generalize this situation Voters’ utility functions single-peaked, but not necessarily symmetric SQ provides different levels of “fallback” utility Main Results: –Median voter’s position no longer dominates –What agenda setter can secure hinges on utility that SQ provides –Absent agenda-setting, cycles may emerge

Lecture VII: Agenda Setting U sq1 V 1 (E) V 2 (E) V 3 (E) U sq2 U sq3 E* If no SQ, E* = median voter’s (2’s) ideal point But if an SQ offers > U sq1, 1 & 3 outvote 2… we could get cycles If U sq2, setter can obtain E* 2 ; Note voter 3 is pivotal If U sq3, setter cannot obtain E* > sq E* 2

Lecture VII: Agenda Setting Baron & Ferejohn (1989) –Generalization of Rubenstein bargaining model to legislative setting where: i.Number of players / bargainers > 2 ii.Majority rule can be used to impose bargains iii.Agenda power a function of recognition and amendment rules (open vs. closed) iv.Choice of these rules is endogenized –Main result: majority rule & closed amendment procedures generate less equitable distributions… but many equilibria are possible.

Lecture VII: Agenda Setting Recall: 1.Ultimatum game: agenda setter’s power to make a take- it-or-leave-it offer secures all of the “pie” 2.Finite Repetition in Rubenstein model: –First-mover advantage remains (and increases in players’ impatience), but.. –Initial agenda setter has to give other player their reservation value

Lecture VII: Agenda Setting The Model: 1.Members {1, …, n}, –Utility increasing in x, risk-neutral –Common discount factor, δ 2.Recognition rule –Member i has probability p i of being recognized –Recognition allows i to propose division of x: x i = (x 1 i, …, x n i ) s.t.  x  1; sq = (0 1, … 0 n ) 3.Amendment rule –Closed or Open 4.Voting rule

Lecture VII: Agenda Setting Closed Rule: Recognition Leg. 1 Proposes x 1 Vote Leg. 2 Proposes x 2 Leg. 3 Proposes x 3 Vote pass {x 1 1, x 2 1, x 3 1 } fail Recognition pass fail pass Recognition {x 1 2, x 2 2, x 3 2 } {x 1 3, x 2 3, x 3 3 }

Lecture VII: Agenda Setting Open Rule: Recognition Leg. 1 Proposes x 1 Put Question Leg. 2 Proposes x 2 Leg. 3 Proposes x 3 pass {x 1 1, x 2 1, x 3 1 } fail Recognition of 2 or 3 x 1 > x 2 Put Question {x 1 1, x 2 1, x 3 1 } Vote on x 1 vs x 2 x 1 < x 2 Recognition of 2 or 3 Vote on x 1 vs x 2

Lecture VII: Agenda Setting An Illustration: –3 legislators under closed rule 1.p i = ( 1 / 3, 1 / 3, 1 / 3 ), x sq = {0, 0, 0} 2.p i = (.4,.4,.2), x sq = {.2,.2,.1} –Case 1: The proposer is indifferent over coalition partners As p i = p j = 1/N, all V i = 1/N x i = {1 – 1 / N, 1 / N, 0}

Lecture VII: Agenda Setting An Illustration: –3 legislators under closed rule 1.p i = ( 1 / 3, 1 / 3, 1 / 3 ), x sq = {0, 0, 0} 2.p i = (.4,.4,.2), x sq = {.2,.2,.1} –Case 2: Both 1 & 2 prefer to coalition with 3 because V 3 < V ~3 Thus 3 knows that she is i) the proposer with p =.2 or ii) a member of the majority coalition with certainty Further, 3 is indifferent between 1 & 2 as coalition partners Thus, 1 & 2 know that they are i) the proposer with p =.4 or ii) a member of the majority coalition with Pr =.5

Lecture VII: Agenda Setting An Illustration: –3 legislators under closed rule 1.p i = ( 1 / 3, 1 / 3, 1 / 3 ), x sq = {0, 0, 0} 2.p i = (.4,.4,.2), x sq = {.2,.2,.1} –Case 2: V 1 = p 1 (.9) + p 2 (0) + p 3 [(½ ×.2) + (½ × 0)] = V 2 = p 2 (.9) + p 1 (0) + p 3 [(½ ×.2) + (½ × 0)] = V 3 = p 3 (.8) + p 1 (.1) + p 2 (.1) =

Lecture VII: Agenda Setting An Illustration: –3 legislators under closed rule 1.p i = ( 1 / 3, 1 / 3, 1 / 3 ), x sq = {0, 0, 0} 2.p i = (.4,.4,.2), x sq = {.2,.2,.1} –Case 2: V 1 =.4(.9) [(½ ×.2) + 0] =.38 V 2 =.4(.9) [(½ ×.2) + 0] =.38 V 3 =.2(.8) +.4(.1) +.4(.1) =.24 –Counterintuitive: legislator 3’s smaller V increases her share of the pie.

Lecture VII: Agenda Setting More generally, given equal recognition probabilities, the expected payoff for any i is, Probability of Recognition × Residual available to proposer Probability of being included in majority coalition × continuation value

Lecture VII: Agenda Setting How does change to open rule affect results? –Initial proposer does not know with certainty who will be recognized after their proposal –Thus, incentive is to propose a distribution that accounts for possibility that all remaining members may have opportunity to present amendments –Distribution becomes increasingly equitable as 1.Players become more patient 2.As N increases (because “insurance coverage” against counter- proposals must be spread more widely)