Online Appendix: Optimal Retrospective Voting

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1 Online Appendix: Optimal Retrospective Voting Ethan Bueno de Mesquita 1 Amanda Friedenberg 2 The notation and setup will be as in the main text, with the following exceptions: Let x l : Ω R be a random variable, with the interpretation that the Legislator s ideal point is x l (ω) when the true state is ω Ω. Now u l : Ω R {0, 1} R, where u l (ω, p, r) = (p x l (ω)) 2 + rb. We extend the definition of a Bayesian Equilibrium in the obvious way, and similarly the definitions of an Expectationally Optimal and a State-by-State Optimal Equilibrium. Uncertainty About the Legislator s Policy Preferences We begin by pointing out that, under this setup, it is without loss of generality to assume that the Voter knows the Legislator s random variable x l : Ω R. To see this, let x v : Ω R and x l : Ω R be the actual random variables of the Voter and Legislator. Append to the game a set of mappings x l : Ω R, viz. X l, where X l is a metric space. Also, append to the game a (transparent) prior ν (Ω X l ), where marg Ω ν = µ. Take the state space to be Ω X l. Construct the random variable x v : Ω X l R so that x v (ω, x l ) = x v (ω) for all (ω, x l ) Ω X l. 3 Construct the random variable x l : Ω X l R so that x l (ω, x l ) = x l (ω) for all (ω, x l ) Ω X l. Extend the definitions of the payoff functions in the obvious way. game. 4 A Bayesian Equilibrium in this expanded game remains a Bayesian Equilibrium in the original So, if the Voter can design an optimal retrospective voting rule in this expanded game, he can also select the associated equilibrium in the original game. Finally, notice that, if Ω and X l are Polish, then we can also expand the state space to incorporate any uncertainty of the Legislator about the Voter s uncertainty, and so on. (To expand the state space, follow the steps in Brandenburger and Dekel [10, 1993].) With this in mind, we stick to the original setup. Bayesian Equilibrium Begin with a characterization of the set of Bayesian Equilibria: Proposition D1 Fix a Bayesian Equilibrium, viz. (s l, s v ) S l S v, and some state ω Ω. Then one of the following conditions must be satisfied: 1 Department of Political Science, Washington University, 1 Brookings Drive, Campus Box 1063, St. Louis, MO 63130, ebuenode@artsci.wustl.edu. 2 Olin School of Business, Washington University, 1 Brookings Drive, Campus Box 1133, St. Louis, MO 63130, friedenberg@wustl.edu. 3 Notice that x v will fail to be injective, even if x v is injective. 4 Condition 1 follows from the fact that every set in the product σ-algebra has measurable sections. Condition 2 is straight forward. 1

2 1. (a) s l (ω) = x l (ω) (b) s v (p) = 0 for all p [x l (ω) B, x l (ω) + B], 2. there exists B > p 0 where, for all p (x l (ω) p, x l (ω) + p), s v (p) = 0 and (a) s l (ω) {x l (ω) p, x l (ω) + p} (b) s v (s l (ω)) = 1, 3. for all p (x l (ω) B, x l (ω) + B), s v (p) = 0 and (a) s l (ω) {x l (ω), x l (ω) B, x l (ω) + B} (b) s v (s l (ω)) = 1 whenever s l (ω) {x l (ω) B, x l (ω) + B}. The proof follows the proof of Proposition 3.2 in Appendix A. Lemma D1 Fix a Bayesian Equilibrium, viz. (s l, s v ) S l S v. For any state ω Ω, s l (ω) [x l (ω) B, x l (ω) + B]. Proof. Fix some Bayesian Equilibrium (s l, s v ) and a state ω Ω. If x l (ω) B > s l (ω), then u l (x l (ω), s v (x l (ω))) B + B > (s l (ω) x l (ω)) 2 + B u l (s l (ω), s v (s l (ω))), where the first inequality follows from the fact that B > (s l (ω) x l (ω)) and so (s l (ω) x l (ω)) 2 > B. But this contradicts Condition 2 of a Bayesian Equilibrium. Suppose that s l (ω) > x l (ω) + B. Then u l (x l (ω), s v (x l (ω))) B + B > (s l (ω) x l (ω)) 2 + B u l (s l (ω), s v (s l (ω))), where the first inequality follows from the fact that (s l (ω) x l (ω)) > B. Again, this contradicts Condition 2 of a Bayesian Equilibrium. Lemma D2 Fix a strategy for the Voter, viz. s v S v, and a state ω Ω where, for some p [x l (ω) B, x l (ω) + B], s v (p) = 1. Suppose, for every policy p R with s v (p) = 1, that there exists a policy q R with (p x l (ω)) 2 > (q x l (ω)) 2. Then there does not exist a strategy for the Legislator, viz. s l S l, such that (s l, s v ) S l S v is a Bayesian Equilibrium. 2

3 Proof. Fix a state ω Ω. Fix also a strategy for the Voter, viz. s v, that satisfies the stated conditions. Contra hypothesis, suppose that there exists a Bayesian Equilibrium (s l, s v ) S l S v. Notice that, from the statement, we must have that s v (x l (ω)) = 0. (If not, we would be able to find some policy q with 0 > (q x l (ω)) 2, which is impossible.) Suppose that x l (ω) p. Since p x l (ω) B, 0 p x l (ω) B. From this B (p x l (ω)) 2. Analogously, consider the case where p x l (ω). Since x l (ω) + B p, B p x l (ω) 0 and so B (p x l (ω)) 2. So, for any value of p [x l (ω) B, x l (ω) + B], (p x l (ω)) 2 B. We will make use of this fact below. The first step will be to show that s l (ω) = x l (ω). To see this, suppose not. We must have s v (s l (ω)) = 1, else u l (x l (ω), s v (x l (ω))) = 0 > (s l (ω) x l (ω)) 2 = u l (s l (ω), s v (s l (ω))), contradicting Condition 2 of a Bayesian Equilibrium. But then we can find a policy p R with s v (p) = 1 and (s l (ω) x l (ω)) 2 > (p x l (ω)) 2. With this u l (p, s v (p)) = (p x l (ω)) 2 + B > (s l (ω) x l (ω)) 2 + B = u l (s l (ω), s v (s l (ω))), contradicting Condition 2 of a Bayesian Equilibrium. The second and last step is to show that we cannot have s l (ω) = x l (ω). For this, notice that using the statement of the Lemma, we can find some policy p R with s v (p) = 1 and (p x l (ω)) 2 > (p x l (ω)) 2. For this policy, u l (p, s v (p)) = (p x l (ω)) 2 + B > (p x l (ω)) 2 + B B + B = u l (x l (ω), s v (x l (ω))), where the third line follows from the fact established above and the fourth from the fact that s v (x l (ω)) = 0. So, using Condition 2 of a Bayesian Equilibrium, s l (ω) x l (ω). Lemma D3 Fix a Bayesian Equilibrium (s l, s v ) S l S v and a state ω Ω. Suppose that either (i) s v (x l (ω)) = 1 or (ii) for any p R with s v (p) = 1, p R\[x l (ω) B, x l (ω) + B]. Then s l (ω) = x l (ω). Proof. Fix a Bayesian Equilibrium (s l, s v ) and a state ω Ω. First, suppose that the condition of Part (i) is satisfied. Then u l (x l (ω), s v (x l (ω))) = B > (p x l (ω)) 2 + B u l (p, s v (p)), 3

4 for any p R\ {x l (ω)}. By Condition 2 of a Bayesian Equilibrium, s l (ω) = x l (ω). Next, suppose that the condition of part (ii) is satisfied. By Lemma D1, s l (ω) [x l (ω) B, xl (ω) + B]. Fix p [x l (ω) B, x l (ω) + B] with p x l (ω). Then u l (x l (ω), s v (x l (ω))) = 0 > (p x l (ω)) 2 = u l (p, s v (p)). By Condition 2 of a Bayesian Equilibrium, s l (ω) = x l (ω). Lemma D4 Fix a Bayesian Equilibrium (s l, s v ) S l S v and a state ω Ω. Let B > p 0 be a policy with (i) for all p (x l (ω) p, x l (ω) + p), s v (p) = 0 and (ii) s v ({x l (ω) p, x l (ω) + p}) {0}. Then s l (ω) {x l (ω) p, x l (ω) + p} and s v (s l (ω)) = 1. Proof. Fix a Bayesian Equilibrium (s l, s v ) and a state ω Ω. Also, fix some p as in the statement of the Lemma. Throughout this proof, we will suppose that s v (x l (ω) p) = 1. An analogous argument works if s v (x l (ω) + p) = 1. Note, u l (x l (ω) p, s v (x l (ω) p)) = p 2 + B > 0. For p (x l (ω) p, x l (ω) + p), 0 (p x l (ω)) 2 = u l (p, s v (p)). For p R\ [x l (ω) p, x l (ω) + p], either (a) 0 p > (p x l (ω)) or (b) (p x l (ω)) > p 0. In either case, (p x l (ω)) 2 > p 2, so that 0 > p 2 + B > (p x l (ω)) 2 + B u l (p, s v (p)). Taken together with Condition 2 of a Bayesian Equilibrium, these imply that s l (ω) {x l (ω) p, x l (ω) + p}. Finally, suppose that s v (x l (ω) + p) = 0. Then u l (x l (ω) p, s v (x l (ω) p)) = p 2 + B > p 2 = u l (x l (ω) + p, s v (x l (ω) + p)). Again using Condition 2 of a Bayesian Equilibrium, s l (ω) cannot be x l (ω) + p. Lemma D5 Fix a Bayesian Equilibrium (s l, s v ) S l S v and a state ω Ω. Suppose that for all p (x l (ω) B, x l (ω) + B), s v (p) = 0. Then s l (ω) {x l (ω), x l (ω) B, x l (ω) + B}. Moreover, if s l (ω) {x l (ω) B, x l (ω) + B} then s v (s l (ω)) = 1. Proof. Fix a Bayesian Equilibrium (s l, s v ) and a state ω Ω. Note, u l (x l (ω), s v (x l (ω))) = 0. By Lemma D1, s l (ω) [x l (ω) B, x l (ω) + B]. For p (x l (ω) B, x l (ω) + B)\ {x l (ω)}, 0 > (p x l (ω)) 2 = u l (p, s v (p)). 4

5 By Condition 2 of a Bayesian Equilibrium, s l (ω) {x l (ω), x l (ω) B, x l (ω) + B} as required. Now suppose, contra hypothesis, that s l (ω) {x l (ω) B, x l (ω)+ B} but s v (s l (ω)) = 0. Then u l (x l (ω), s v (x l (ω))) = 0 > B = u l (s l (ω), s v (s l (ω))), contradicting Condition 2. Proof of Proposition D1. Immediate from Lemmata D2, D3, D4, and D5. Optimal Retrospective Voting Return to the case where x l (Ω) = {0}. Then Proposition D1 is simply Proposition 3.2. Recall that if the Voter were to choose a strategy for the Legislator, he would choose a constant strategy that corresponds to his expected ideal point, viz. E (x v ). But, when E (x v ) lies within B of the Legislator s ideal point, the Voter can induce the Legislator to choose (i) E (x v ) when his ideal point is positive and (ii) E (x v ) when his ideal point is negative. From the Voter s perspective, this strategy is preferred. (See Lemma A9 in Appendix A.) Now suppose the Legislator s preferences again depend on the state. Fix a policy p that corresponds to an ideal point for the Legislator, i.e., a policy with (x l ) 1 (p) 0. For simplicity, suppose that p + B E (x v ) > p. Proposition D1 suggests that, at any state in (x l ) 1 (p), the Voter can again induce the Legislator to choose (i) E (x v ) when his ideal point is greater than p and (ii) choose 2p E (x v ) when his ideal point is less than x l (ω). At any state contained in (x l ) 1 (p), the Voter would prefer this equilibrium to a constant strategy that selects his ideal point at every state. 5 But this leaves two open questions: Can the Voter induce the Legislator to choose policies as above, for all sets (x l ) 1 (p) with p + B E (x v ) > p? Can the Voter induce the Legislator to choose a policy that she prefers to E (x v ) at any state (i.e., even when E (x v ) does not lie within B of the Legislator s ideal point)? We don t attempt to address these questions here and, instead, leave them for a separate paper. That said, we will conclude with a result that is suggestive of a mildly positive answer, at least in certain circumstances. For this, we need some additional assumptions. We discuss their importance after presenting the result. Assumption D1 The set x l (Ω) is countable. The random variable x l induces a countable partition over the policy space R, where x l (ω) and x l (ω ) are in the same partition member only if x l (ω) = x l (ω ). By Assumption D1, we can find a countable partition, viz. P, so that each partition member, viz. P k, is an interval with (x l ) 1 (P k ). That is, for each partition member P k, there is some state at which the Legislator s ideal point is contained in P k. Moreover, whenever the Legislator s ideal point is contained in this partition member, it is given by the policy p k P k. 5 Do note, we do not show this result. However, we can follow the proof Lemma A9 to establish this fact. 5

6 Assumption D2 The partition P can be chosen so that, for each partition member P k, p k = sup P k + inf P k. 2 Assumption D3 The partition P can be chosen so that, for all ω Ω, x v (ω) P k if and only if x l (ω) P k. Lemma D6 Suppose that x l satisfies Assumption D1 and that the partition P satisfies Assumptions D2 and D3. Let E (x v ) [p 0 B, p 0 + B] and let r l be a constant strategy with r l (Ω) = {E (x v )}. Define Ω k = (x v ) 1 ([p k B, p k + B]), and Ω = k Ω k. There exists a Bayesian Equilibrium, viz. (s l, s v ) S l S v, and a non-empty set of states Φ Ω where u v (ω, s l (ω)) u v (ω, r l (ω)) for all ω Ω u v (ω, s l (ω)) > u v (ω, r l (ω)) for all ω Φ. Remark D1 If, for each k, P k [p k B, p k + B], then Ω = Ω. Proof of Lemma D6. It will be useful to define policies q k. First set q 0 = E (x v ). Let P n P 0 be such that inf P 0 sup P n. If E (x v ) = sup P n, let q n = E (x v ). If E (x v ) > sup P n, choose q n to satisfy the following requirements: { Let q n P n with q n p n and E (x v ) sup P n sup P n q n. Let q n = min q n, p n + } B. Let P n P 0 be such that inf P n sup P 0. If E (x v ) = inf P n, let q n = E (x v ). If inf P n > E (x v ), choose q n to satisfy the { following requirements: Let q n P n with p n q n and inf P n E (x v ) q n inf P n. Let q n = max q n, p n } B. Notice that, by construction, each q k P k. Define (s l, s v ) as follows: For all ω Ω, s l (ω) = q k if and only if x l (ω) P k. Let s v (p) = 1 if and only if p = q k for some k. We now turn to show that (s l, s v ) is a Bayesian Equilibrium. Note that Condition 1 follows from Assumption[ D1 and Lemma A7. For Condition 2, fix some ω Ω with x l (ω) P k. By construction, q k p k, p k + ] B when 0 > k, q k [p k ] B, p k when k > 0, and similarly for q 0. So B (q k p k ) 2 and u l (ω, s l (ω), s v (s l (ω))) = (q k p k ) 2 + B B + B. Fix some p q k. If s v (p) = 0 then, using the above inequality, u l (ω, s l (ω), s v (s l (ω))) 0 u l (ω, p, s v (p)). So, suppose that s v (p) = 1. Then there exists some partition member P j with p = q j. Recall, { then, that q j P j P k (since p q k ). Assume that 0 > k so that q k = min q k, p k + } B. If 6

7 j k, certainly q j q k, so that (using the fact that q k p k ) u l (ω, s l (ω), s v (s l (ω))) = (q k p k ) 2 + B ( q j p k ) 2 + B = ul (ω, p, s v (p)), as required. Suppose then that k > j. Since q k P k, sup P k q k. Using Assumption D2, this implies that 2p k q k inf P k. Now, since k > j, inf P k q j. Notice, then, that, since q k p k, the above inequalities imply that p k 2p k q k q j. As such ( qj p k ) 2 (2pk q k p k ) 2 = (q k p k ) 2. It follows that u l (ω, s l (ω), s v (s l (ω))) = (q k p k ) 2 + B ( q j p k ) 2 + B = ul (ω, p, s v (p)), as required. An analogous argument applies when k 0, so we omit it here. We now need to show that (s l, s v )[ satisfies the desired properties. To show this, fix some state ω Ω with x l (ω) P k and x v (ω) p k B, p k + ] B. If k = 0, then u v (ω, s l (ω)) = (E (x v ) x v (ω)) 2 = u v (ω, r l (ω)), { as required. Suppose that 0 > k. By Assumption D3, min sup P k, p k + } B suppose that s l (ω) = q k. By choice of q k and the fact that sup P k x v (ω), we have x v (ω). First, E (x v ) x v (ω) E (x v ) sup P k sup P k q k x v (ω) q k, where the first and third inequalities are strict whenever sup P k > x v (ω). It follows that u v (ω, s l (ω)) = (q k x v (ω)) 2 (E (x v ) x v (ω)) 2 = u v (ω, r l (ω)), where again the inequality is strict whenever sup P k > x v (ω). Next, suppose that s l (ω) = p k + B qk. Then q k > p k + B x v (ω). So, using the above inequalities we have ( u v (ω, s l (ω)) = p k + ) 2 B x v (ω) > (q k x v (ω)) 2 (E (x v ) x v (ω)) 2 = u v (ω, r l (ω)), 7

8 as required. Notice the role each of the assumptions play in the proof of Lemma D6. Assumption D1 guarantees that the strategy constructed for the Legislator is measurable. Assumption D2 serves to ensure that, when the Voter offers the Legislator incentives within partition member P k, this does not limit his ability to offer incentives in other partition members. This effect can be achieved with the following alternate assumption. 6 Assumption D4 The partition P can be chosen so that, for each partition member P k, [ p k B, p k + ] B P k. Finally, Assumption D3 guarantees that when the Voter offers electoral incentives to choose policy p, the Legislator does not want to choose policy p at states where Voter prefers E (x v ) over p. The extent to which these assumptions can be relaxed is left for future work. 6 To see this, notice that under this alternate assumption q k = p k + B. So, the statement that Assumption D2... implies... 2p k q k inf P k can be replaced by Assumption D4... implies... p k B inf P k. 8