Optimal Retrospective Voting∗
Ethan Bueno de Mesquita†
Amanda Friedenberg‡
First Draft: August 11, 2005
Current Draft: July 12, 2006
Abstract
We argue that optimal retrospective voting can be thought of as an equilibrium selection
criterion—specifically, a criterion that selects so as to maximize voter welfare. Formalized this
way, we go on to provide an informational rationale for retrospective voting. We show that a
Voter can use retrospective voting to extract information from a Legislator with policy expertise.
In particular, when the Voter’s strategy set is sufficiently rich, he can design a retrospective
voting rule that ensures him expected payoffs as high as those he could achieve in the game
where he is a policy expert.
Keywords: Retrospective Voting, Political Economy, Electoral Control, Information Extraction
JEL Classification: D02, D72, D78, D81, D82
∗
We are indebted to Randy Calvert for many conversations related to this project. We have also benefited from
comments by Jim Alt, Heski Bar-Isaac, Adam Brandenburger, Catherine Hafer, Bård Harstad, Dimitri Landa, Motty
Perry, Yuliy Sannikov, Jeroen Swinkels, and seminar participants at Berkely, Columbia, Hebrew University, and
NYU. Bueno de Mesquita thanks the Center for the Study of Rationality, Political Science Department, and Lady
Davis Fellowship, at Hebrew University for their hospitality and support.
†
Department of Political Science, Washington University, 1 Brookings Drive, Campus Box 1063, St. Louis, MO
63130, [email protected].
‡
Olin School of Business, Washington University, 1 Brookings Drive, Campus Box 1133, St. Louis, MO 63130,
[email protected]
Introduction
The political economy literature focuses on two accounts of voter behavior, retrospective and
prospective voting. Retrospective voting is the idea that voters are backward-looking, conditioning
their electoral decisions on politicians’ past performance without concern for their expected future
performance (Key [17, 1966], Fiorina [15, 1981]). For instance, voters may reward politicians for
having achieved economic growth or implemented environmental policies they like, even if such
outcomes do not indicate a higher probability of similar outcomes in the future. Prospective voting
is the idea that voters are forward-looking, conditioning their electoral decisions on expectations
about the future (Kuklinski and West [18, 1981], Lewis-Beck [19, 1988], Lockerbie [21, 1992]). For
instance, voters might support a politician because they believe that while in office, she will provide
economic growth or environmental policies that benefit voters.
There is a long research tradition studying retrospective voting (see, for example, Key [17,
1966], Barro [6, 1973], Fiorina [15, 1981], Ferejohn [14, 1986], Austen-Smith and Banks [5, 1989],
Maskin and Tirole [22, 2004]). Moreover, retrospective voting has been supported empirically—i.e.,
a variety of papers show that voters condition their electoral behavior on the past performance of
politicians. While some of this evidence is consistent with both retrospective and prospective voting
(Lockerbie [20, 1991], Clarke and Stewart [10, 1995]), there is also direct evidence of backwardlooking behavior (Alesina, Londregan and Rosenthal [1, 1993], Norpoth [25, 1996]).
This paper makes two contributions—one conceptual and one substantive. Conceptually, we
suggest that optimal retrospective voting is an equilibrium selection criterion. Substantively, we
provide an informational rationalle for retrospective voting. That is, we show that voters can use
retrospective voting to extract information from politicians with policy expertise.
1
Optimal Retrospective Voting as Equilibrium Selection
What advantage do voters get from backward-looking behavior? An answer, dating back at least to
V.O. Key, is that by threatening not to reelect politicians who perform poorly, voters can introduce
a degree of political accountability or electoral control.1 To borrow from Key (1966, page 77):
The odds are that the electorate as a whole is better able to make a retrospective appraisal of the work of governments than it is to make estimates of the future performance
of nonincumbent candidates. Yet by virtue of the combination of the electorate’s retrospective judgement and the custom of party accountability, the electorate can exert a
prospective influence if not control. Governments must worry, not about the meaning
of past elections, but about their fate at future elections.
1
There is also a literature that shows how past performance can convey information about expected future performance that prospective voters may use (Alesina and Rosenthal [2, 1995], Fearon [13, 1999], Persson and Tabellini
[26, 2000], Ashworth [3, 2005], Besley [8, 2005]). We address that literature, and its critique of retrospective voting,
in Section 3.
1
The idea that voting can create electoral incentives for politicians to represent voter interests
has been formalized by Barro [6, 1973], Ferejohn [14, 1986], Austen-Smith and Banks [5, 1989], and
Maskin and Tirole [22, 2004].2
In order for retrospective voting to serve this purpose, politicians must correctly anticipate the
electoral consequences of their actions. That is, for retrospective voting to be understood as a tool
for voters to control their representatives, it must be part of an equilibrium of a game between
voters and politicians.
Given this, what is optimal retrospective voting? An optimal retrospective voting rule is one
that induces the politician to maximize voter welfare. We formalize this notion of optimality as an
equilibrium selection criterion.
To understand why this is the right way to think about optimality in retrospective voting
games, consider a canonical model of electoral control (between a single politician and voter). The
politician chooses a level of effort, which she finds costly but the voter finds beneficial. Then the
voter observes the politician’s choice, after which he decides whether or not to reelect the politician
(and the game is over).3 The politician likes reelection. The voter’s payoff function does not
(explicitly) depend on whether or not the politician is reelected.
In equilibrium, the politician correctly anticipates the voter’s response to any level of effort.
Consequently, the voting rule adopted by the voter affects the politician’s choice. But, when the
voter actually votes, he is indifferent between reelection rules (i.e., between his strategies). As such,
the game has many equilibria, each described by a voting rule and the effort choice it induces.
Despite the fact that the voter’s payoffs do not depend on whether or not the politician is
reelected, he is still affected by the reelection rule. Different reelection rules may be associated
with different equilibrium choices by the politician. Of course, by the time the voter actually casts
his vote, he cannot change the politician’s equilibrium behavior. But if prior to the start of the
game the voter could choose an equilibrium, he would want to choose one in which his strategy (i.e.,
reelection rule) induces the politician to maximize his (the voter’s) welfare. This is what we mean
by optimal retrospective voting—an equilibrium selection criterion in which the voter’s welfare is
maximized.4
Much of the literature on voting behavior implicitly formalizes optimal retrospective voting as
an equilibrium selection criterion based on voter welfare. One contribution of this paper is to make
the argument explicit. In so doing, we hope to help clarify the role of retrospective voting in the
theory of voter behavior.
2
See Bendor, Kumar and Siegel [7, 2005] for a model of retrospective voting with adaptively rational voters.
Many models study a related game where effort translates into payoffs stochastically. The basic intuition we
discuss in this Section translates directly to such games. (See Section 2 for the relationship between this problem
and the one studied here.)
4
We remind the reader than an equilibrium selection criterion is distinct from an equilibrium refinement and that
a welfare criterion is a classic example of a standard for equilibrium selection. (See, for instance, the discussion in
Myerson [24, 1991], pages 241–242.)
3
2
2
Electoral Control and Information Extraction
We provide a new substantive insight into the study of retrospective voting. In particular, we show
that voters can use retrospective voting both to control politicians and to extract information from
politicians with policy expertise.
In many existing models of retrospective voting, the focus is on giving a politician incentives
to take a costly action: in Barro [6, 1973] and Ferejohn [14, 1986], this is on an “effort” dimension,
and in Austen-Smith and Banks [5, 1989], this is on a “policy” dimension. Any uncertainty is
about the politician’s actions, not about the voter’s preferences over policy outcomes. That is, in
these models the voter knows what he wants (high effort) but not what the politician did. As such,
the optimal retrospective voting rule has features familliar from the canonical model of contracting
under moral hazard.
Yet, in many interesting settings, politicians are policy experts relative to voters. That is, the
politician’s private information is about the likely impact of policies. This type of information may
influence the voter’s preferences over policy outcomes. Here, the voter knows what the politician
did (the policy she chose) but not what he (the voter) wants.5 In this environment, the challenge
in designing an optimal retrospective voting rule is that the voter does not know in advance what
behavior he wants to induce.
One conjecture is that, when the voter is uncertain about his preferences (i.e., his most preferred
policy or “ideal point”), the optimal retrospective voting rule would at best result in the politician
choosing the voter’s expected ideal point. After all, this is the policy the voter himself would
implement, if he were empowered to choose policy directly. We show that the voter can always
design a retrospective voting rule that provides higher payoffs than this. Since the voter could not
have done so well on his own, he must be extracting information from the politician.
The level of information extraction that the voter can achieve depends on how rich his strategy
set is. We allow this richness to vary in two ways.
First, we restrict the voter to pure strategies but allow for the possibility that, after the policy
is chosen, the voter may become informed of the true state. We show that even if the voter only has
a fifty-fifty chance of learning the true state, he can design a retrospective voting rule that ensures
him expected payoffs as high as those he could achieve in a game where he is perfectly informed.
That is, the voter achieves full information extraction.
Second, we allow the voter to employ mixed strategies. Here we show that even if the voter has
no chance of ever learning the true state directly, he can nonetheless design a retrospective voting
rule that extracts full information.
These information extraction results are surprising because they show that the voter’s lack of
information comes at no cost to him. Even when the voter has a significant level uncertainty, he
can still induce the legislator to choose his ideal point in the same set of states he could when
perfectly informed.
5
We thank Motty Perry for suggesting this turn of phrase.
3
We provide tight characterizations of the optimal retrospective voting rule. For instance, we
will see that, in pure strategies, the optimal rule always takes one of two forms, regardless of the
probability that the voter learns the true state.
It is worth noting that Maskin and Tirole [22, 2004] also study a game in which the politician
has private information that influences the voters’ preferences. However, they focus on a different
set of issues and so a different game. There, voting not only serves to control politicians, but also to
select politicians with the same preferences as the voter. If the politician’s preferences are aligned
with the voter’s, the voter’s electoral control and information extraction problems disappear. Here,
we show that in a wide variety of circumstances, voters can solve the problems of electoral control
and information extraction for a given preference ranking of the legislator. Put differently, to solve
these problems, voters need not select a legislator whose preferences are aligned with their own.
3
Is Retrospective Voting Ever Optimal?
We formalize optimal retrospective voting as an equilibrium selection criterion and provide an informational rationalle for retrospective voting. Together, these two points provide a new perspective
on a common critique of restrospective voting.
Critics have argued that, while voters would like to adopt retrospective voting rules that give
politicians optimal incentives, such rules are not credible. At the point when voters actually
vote, their electoral decisions do not affect politicians’ past actions. Thus, rational voters cannot
credibly commit to voting rules for the purpose of providing incentives. Instead, they must be
forward-looking, electing politicians who can be expected to deliver the highest payoffs in the
future (Alesina and Rosenthal [2, 1995], Fearon [13, 1999], Besley [8, 2005]).
We believe that even forward-looking voters can adopt retrospective voting aimed at electoral
control and information extraction. Retrospective and prospective voting are not mutually exclusive. Rather, they are complementary considerations. We justify this claim in two ways.
First, recall that we formalize optimal retrospective voting as an equilibrium selection criterion.
As long as voters are selecting from the set of sequentially rational equilibria, their retrospective
voting rule is credible—there is no commitment problem. Thus, when conceived of in terms of
equilibrium selection, the rationale for retrospective voting is independent of and does not conflict
with sequential rationality.6
Of course, equilibrium selection is interesting only if the game has multiple sequentially rational
equlibria. While many extant models have a unique equilibrium, this is largely due to their adopting
a stylized approach to highlight their substantive results. For example, these models generally
assume that legislative outputs are a simple combination of “ability” and “effort” (Persson and
Tabellini [26, 2000], Ashworth [3, 2005], Besley [8, 2005]). Small changes to such models (e.g.,
adding a policy dimension or richer electoral or institutional environments) can give rise to multiple
6
For a different argument that retrospective and prospective motivations are not mutually exclusive, as well as a
formal model in which both types of motivation arise endogenously, see Snyder and Ting [27, 2005].
4
equilibria, even with forward-looking voters (see, for example, Ashworth and Bueno de Mesquita
[4, 2006]).
Second, beyond the issue of multiplicity, it is not clear that the commitment problem arises
in all interesting voting games. In some settings, it is natural to assume that voters learn about
factors (i.e., their representatives’ ability or trustworthiness) that affect their future payoffs. This is
particularly true in the sort of effort games (without information extraction) that the literature has
tended to focus on. For instance, in models of constituency service or local public goods provision,
a legislator’s individual characteristics are likely to be important. However, other settings do not
naturally suggest this type of learning. This is particularly true in the sort of games with policy
expertise that we focus on. For example, if the voters already know her policy preferences, then
how a legislator votes on purely ideological (environmental policy or gay marriage) or redistributive
issues is unlikely to communicate information about her inherent characteristics.7 In such settings,
observations of legislative action may not affect voters’ beliefs about likely future performance.
Hence, no commitment problem arises.
Given these arguments, we believe that there are good reasons to study retrospective voting
as an independent rationale for why voters condition on observations of past performance. To
this end, we study a simple game of policy choice and elections in which there are no prospective
concerns. (The game ends after the election.) This allows us to focus on how retrospective voting
can be used both to extract information and to discipline legislators, abstracting away from any
prospective incentives.
4
Primitives
There are two players, the Legislator and the Voter. We refer to the Legislator as “she” and the
Voter as “he.” The order of play is as follows.
Nature chooses a state, which determines the
Voter’s policy preferences. The Legislator observes the true state and then chooses a policy. After
this, Nature chooses whether or not to inform the Voter of the true state. The Voter observes the
Legislator’s policy choice and then decides whether or not he will reelect the Legislator.8
The set of states, viz. Ω, is a topological space. Then (Ω, A, µ) is a measure space, where A is
the Borel sigma-algebra and µ is the Voter’s prior. The set of policies is the real line. Write p ∈ R
for a given policy. Nature chooses ι from {i, ni}, where i represents the decision to inform the Voter
of the true state of the world and ni represents the decision not to inform the Voter of this state. It
is ‘transparent’ to the players that Nature chooses the action ‘inform’ with probability π ∈ [0, 1].9
Reelection is a choice r from {0, 1}, where r = 0 represents the Voter’s decision not to reelect the
7
Under our analysis, it is without loss of generality to assume that there is no uncertainty about the Legislator’s
policy preferences. (See the Online Appendix for a formal treatment.) So, even if the Voter is uncertain about the
Legislator’s policy preferences, there is a role for retrospective voting as a selection criterion.
8
We do not explicitly include a challenger. Our game does not include a future in order to focus on the use of
retrospective voting to provide incentives. As such, including a challenger (who may or may not be ex ante identical
to the incumbent) would have no affect on the analysis.
9
By ‘transparent’ we mean what is colloquially referred to as ‘common knowledge.’
5
Legislator. The Legislator’s policy preferences do not vary with the state. In particular, normalize
the Legislator’s ideal point to 0, for every state. (See Section 8.3 for an extension to the case where
the Legislator’s ideal point depends on the state.)
Let xv : Ω → R be a surjective real-valued random variable. The interpretation is that xv (ω)
is the Voter’s ideal point when the true state is ω. Since the mapping is surjective, for any
policy p there exists some state ω such that p is the Voter’s ideal point at ω. The Legislator has
quadratic preferences over policy and seeks reelection. Given a policy p and an ideal point of 0,
the Legislator gains a payoff −p2 . (Our results do not hinge on quadratic payoffs. See Section
8.1.) If reelected, the Legislator receives a payoff of B > 0. Taken together, these imply that the
Legislator’s extensive-form payoff function is ul : R × {0, 1} → R, where
ul (p, r) = −p2 + rB.
For each state, the Voter has quadratic preferences over policy. Formally, the Voter’s extensive-form
payoff function is uv : Ω × R → R, where
uv (ω, p) = − (p − xv (ω))2 .
The extensive form described above induces a strategic form. A strategy for the Legislator is a
map from states to policies, viz. sl : Ω → R. A strategy for the Voter is a map sv : Ω×R×{i, ni} →
{0, 1}, where for every policy p and all states ω, ω 0 ∈ Ω, sv (ω, p, ni) = sv (ω 0 , p, ni). That is, if the
Voter is uninformed, the true state cannot affect his action. Let Sl (resp. Sv ) be the set of pure
strategies for the Legislator (resp. Voter). With this, we can specify strategic-form payoff functions,
viz. Ul : Ω × {i, ni} × Sl × Sv → R and Uv : Ω × Sl → R, with
Ul (ω, ι, sl , sv ) = ul (sl (ω) , sv (ω, sl (ω) , ι)) = −sl (ω)2 + sv (ω, sl (ω) , ι) B
Uv (ω, sl ) = uv (ω, sl (ω)) = − (sl (ω) − xv (ω))2 .
Write Eul (p, sv (ω, p, ι)) for the expected payoffs of the Legislator with respect to ι ∈ {i, ni}, i.e.,
Eul (p, sv (ω, p, ι)) = πul (p, sv (ω, p, i)) + (1 − π) ul (p, sv (ω, p, ni)) .
4.1
Equilibrium
The role of retrospective voting is to induce the Legislator to take actions that are beneficial to
the Voter. That is, if the Voter’s actions are responsive to policy, then the Legislator should
take this into account when choosing policy. This requires that the Legislator correctly anticipate
how the Voter will respond to her choice of policy. Therefore, studying this role of retrospective
voting suggests restricting attention to a solution concept where players’ beliefs are correct, namely
equilibrium.
For now, we will restrict attention to pure-strategy Bayesian Equilibrium. In Section 7, we
6
allow players to make use of behavioral strategies. (From here until Section 7, anytime we refer to
the term Bayesian Equilibrium, we mean a pure-strategy Bayesian Equilibrium.)
Definition 4.1 A pair (s∗l , s∗v ) is a pure-strategy Bayesian Equilibrium if:
(i) s∗l is measurable;
(ii) for each ω ∈ Ω, Eul (s∗l (ω) , s∗v (ω, s∗l (ω) , ι)) ≥ Eul (p, s∗v (ω, p, ι)) for all p ∈ R.
The first requirement is that the Voter must be able to calculate his expected payoffs. For this,
s∗l
must be measurable. Because the Voter makes his reelection decision at the end of the game,
his choice does not directly affect his payoffs. As such, an explicit optimization requirement for the
Voter is omitted.10
The second requirement is that, at each state, the Legislator must choose a
policy that maximizes her expected payoffs, given the Voter’s actual reelection rule.
4.2
Optimal Retrospective Voting
Thus far, we have not considered a non-trivial optimization problem for the Voter. In a Bayesian
Equilibrium, at each state, the Legislator chooses an optimal policy given the actual electoral
strategy s∗v . But, at the time that the Voter makes his reelection decision, he is indifferent among
all strategies. That said, given that in a Bayesian Equilibrium the Legislator correctly anticipates
the Voter’s actual electoral strategy, there is a sense in which the Voter has an optimal reelection
rule. In particular, an optimal retrospective voting rule is one the Voter would choose if he could
survey the Legislator’s best responses for each reelection rule and choose an equilibrium in which
his welfare is maximized.
Put differently, optimal retrospective voting identifies the Voter’s most preferred equilibria from
the set of all Bayesian Equilibria. Notice, for any given strategy of the Voter, the Legislator may
have multiple best responses. Thus, there may be multiple equilibria associated with any given
Voter strategy. Optimal retrospective voting selects a pair of Voter and Legislator equilibrium
strategies that maximizes the Voter’s welfare.
It is important to note that, in equilibrium, the Legislator correctly anticipates the Voter’s
response to policy choice. As such, a feature of an equilibrium analysis (as per Definition 4.1) is
that the Voter’s actual reelection rule influences the Legislator’s behavior. We do not say how
or why players might arrive at playing an equilibrium. So, in particular, we do not require that
the Voter ‘announce’ or ‘commit’ to a strategy before the game is played. Of course, we do not
rule out that ‘announcement’ or ‘commitment’ can lead to equilibrium play. The point is simply
that, in any equilibrium-based solution concept, the Voter’s actual strategy choice influences the
Legislator’s behavior.11
10
For the same reason, we omit a requirement on updating beliefs. Imposing the natural requirement yields an
equivalent definition.
11
This need not be the case for a non-equilibrium solution concept, e.g., rationalizability.
7
Just as we do not say how players arrive at an equilibrium, we do not say how players arrive
at an equilibrium that maximizes the Voter’s welfare. Indeed, one need not view the equilibrium
selection as a behavioral prediction at all. Rather, we are interested in identifying the best possible
outcome the Voter can hope to achieve through retrospective voting. (Again, just as we do not
rule out that that ‘announcement’ or ‘commitment’ may lead to equilibrium play, we do not rule
out that it could lead to a specific equilibrium.)
We need to specify the Voter’s “most preferred” equilibrium. We formalize this with the following two selection criteria:
Definition 4.2 A strategy profile (s∗l , s∗v ) is an Expectationally Optimal Equilibrium if it is
a Bayesian Equilibrium and, for all Bayesian Equilibria (sl , sv ),
Z
Ω
Z
uv (ω, s∗l (ω)) dµ
≥
Ω
uv (ω, sl (ω)) dµ.
Definition 4.3 A strategy profile (s∗l , s∗v ) is a State-by-State Optimal Equilibrium if it is a
Bayesian Equilibrium and there does not exist a Bayesian Equilibrium (sl , sv ) with
uv (ω, sl (ω)) ≥ uv (ω, s∗l (ω))
uv (ω, sl (ω)) >
uv (ω, s∗l (ω))
for all ω ∈ Ω
for some ω ∈ Ω.
A strategy profile (s∗l , s∗v ) is Expectationally Optimal if, among all Bayesian Equilibria, (s∗l , s∗v )
maximizes the Voter’s expected payoffs given the prior µ. Alternatively, if (s∗l , s∗v ) is State-by-State
Optimal, then there does not exist a Bayesian Equilibrium, viz. (sl , sv ), where (i) for every state
ω ∈ Ω, the Voter’s payoff under (sl , sv ) is at least as great as his payoff under (s∗l , s∗v ) and (ii) there
exists a state ω ∈ Ω where the Voter’s payoff under (sl , sv ) is strictly greater than his payoff under
(s∗l , s∗v ). In Section 8.2, we discuss these criteria and explain why they are distinct.
Before continuing, it will be useful to define some loose terminology. Fix two strategy profiles
R
R
(sl , sv ) and (s∗l , s∗v ). If either Ω uv (ω, sl (ω)) dµ > Ω uv (ω, s∗l (ω)) dµ or the strategy profiles satisfy
the conditions in the display of Definition 4.3, we will say “the Voter strictly prefers (sl , sv ) to
(s∗l , s∗v ).”
5
Benchmark Result
Here, we analyze a simple version of the game, one in which it is transparent to the players that
Nature chooses not to inform the Voter (i.e., π = 0). Even in this case, the Voter can extract
information.
This version of the game can be translated into one in which Nature must choose not to inform
the Voter about the true state. Under that specification, the Voter’s strategy need not specify
an action when Nature chooses to inform him. Therefore, we restrict the domain of the Voter’s
strategy sv to be Ω × R × {ni}. Since the true state cannot affect the Voter’s action when he is
uninformed, we can suppress reference to ω and ι = ni. That is, write sv (p) for sv (ω, p, ni).
8
It will be convenient to begin by characterizing the set of Bayesian Equilibria for this game.
Two principles will aid in the characterization. First, by choosing p = 0, the Legislator can assure
herself a payoff of at least zero. Thus, the Voter can never induce the Legislator to choose policies
√
that are further than B from the Legislator’s ideal point. Second, if multiple policies will be
rewarded with reelection, the Legislator never has an incentive to choose any but the policy closest
to her ideal point. Given these facts, the following characterization follows.12
Proposition 5.1 If (sl , sv ) is a Bayesian Equilibrium, then it must take one of the following forms:
(i)
(a) sl (ω) = 0 for all ω ∈ Ω
√ √
(b) sv (p) = 0 for all p ∈ [− B, B],
(ii) there exists B > p ≥ 0 where, for all p ∈ (−p, p), sv (p) = 0 and
(a) sl (ω) ∈ {−p, p} for all ω ∈ Ω
(b) sv (sl (ω)) = 1 for all ω ∈ Ω,
√ √
(iii) for all p ∈ (− B, B), sv (p) = 0 and
√ √
(a) sl (ω) ∈ {0, − B, B} for all ω ∈ Ω
√ √
(b) sv (sl (ω)) = 1 for all sl (ω) ∈ {− B, B}.
Part (i) says that if the Voter does not reward any policies within
√
B of the Legislator’s ideal
point, then the Legislator chooses her ideal point at every state. For Parts (ii)–(iii), consider the
set of policies the Voter rewards with reelection. In particular, let −p and p ≥ 0 be the policies
closest to the Legislator’s ideal point that the Voter rewards with reelection. Part (ii) considers the
√
case where −p and p lie strictly within B of the Legislator’s ideal point. Here, the Legislator’s
√
best response is to choose either of −p or p. Part (iii) considers the case where p = B. Since the
Voter rewards −p or p with reelection, the Legislator is indifferent between choosing these policies
and her ideal point.
5.1
Optimal Retrospective Voting in the Benchmark Case
Were the uninformed Voter empowered to choose policy, maximizing his expected payoffs would
require that he choose his expected ideal point. (See Lemma A6.) However, the Voter cannot
choose policy. Rather, the Legislator is his agent. One conjecture is that this agency relationship
leaves the Voter worse off relative to choosing policy himself. In particular, when the Legislator’s
and Voter’s preferences diverge sufficiently, the Voter cannot offer the Legislator sufficient electoral
incentives to choose his ideal point.
12
The proof of this and all subsequent results can be found in the appendices.
9
− B
− E ( xv )
E ( xv )
0
If xv (ω ) < 0, choose − E ( xv )
B
If xv (ω ) ≥ 0, choose E ( xv )
Figure 5.1: A legislative choice that improves the voter’s expected utility relative to always choosing
his expected ideal point.
But there is another important aspect of this game. In particular, the Legislator is informed of
the true state. As such, while the Voter does not know his ideal point, the Legislator does. The
challenge for the Voter is to use electoral incentives to extract this information from the Legislator
in a way that improves his expected payoffs.
How can the Voter extract such information? Suppose that the Voter could induce the Legislator
to choose his expected ideal point, viz. E (xv ). Refer to Figure 5.1 and suppose that E (xv ) > 0.
Proposition 5.1 suggests that he can also induce the Legislator to choose −E (xv ). Indeed, because
the Legislator’s ideal point is zero, she must be indifferent between (i) choosing E (xv ) and being
reelected and (ii) choosing −E (xv ) and being reelected. This indicates that the Voter can do
better than simply inducing the Legislator to choose his expected ideal point. In particular, there
is an equilibrium where the Legislator chooses E (xv ) when the Voter’s ideal point is positive and
−E (xv ) when the Voter’s ideal point is negative.
More precisely, the Voter will strictly prefer a moderate two-sided voting rule: the Voter reelects
the Legislator if and only if she chooses either −p∗ or p∗ . The Legislator chooses the policy p∗ ≥ 0
when the Voter’s ideal point is positive and otherwise chooses the policy −p∗ . Let Ω+ be the
set of states where the Voter’s ideal point is positive, viz. {ω : xv (ω) ≥ 0}. Under Expectational
Optimality, the Voter selects this voting rule so that p∗ ≥ 0 maximizes
Z
Z
2
−
Ω+
(xv (ω) − p) dµ (ω) −
Ω\Ω+
(xv (ω) + p)2 dµ(ω)
√ √
subject to p ∈ [− B, B].
This policy p∗ will be given by
Z
∗
p =
Ω+
Z
xv (ω) dµ (ω) −
Ω\Ω+
xv (ω) dµ (ω) ≥ 0,
√
√
when this value is in [0, B], and B otherwise. (See Proposition 5.2 below.)
Consider the example in Figure 5.2. Here Ω = R and xv is the identity map. Let µ be the
√ √
uniform distribution on [− B, B]. Since the distribution is symmetric, p∗ is the√ expected value
of the Voter’s ideal point, conditional on the ideal point being positive, i.e., p∗ =
10
B
2 .
Reelect if and only if
- B
- 12 B
1
2
0
1
Choose - 2 B
B
B
Choose
1
2
B
Figure 5.2: A moderate two-sided voting rule.
Reelect if and only if
- 32 B
- B
Choose - B
- 12 B
0
Choose 0
1
2
B
B
Choose
3
2
B
B
Figure 5.3: An extreme two-sided voting rule.
A moderate two-sided voting rule need not be Expectationally Optimal. To see this, continue
to assume that Ω = R and xv is the identity map. Now, let µ be the uniform distribution on
√
√
√
[− 32 B, 32 B]. The policy p∗ = 34 B maximizes the Voter’s expected payoffs given all moderate
√
two-sided voting rules. The Voter’s expected payoffs are − 38 B.
However, the Voter can obtain higher expected payoffs. Consider instead the following extreme
two-sided voting rule, illustrated in Figure 5.3: the Voter reelects the Legislator if and only if she
√
√
chooses a policy of either − B or B. The Legislator (i) chooses her ideal point (zero) when the
√
√
√
Voter’s ideal point is contained in [− 21 B, 21 B], (ii) chooses the policy − B when the Voter’s
√
√
ideal point is strictly less than − 21 B, and (iii) otherwise chooses B. Under this voting rule, the
√
Voter’s expected payoffs are − 14 B. So this voting rule yields strictly higher expected payoffs for
the Voter than any moderate two-sided voting rule.
An extreme two-sided voting rule is indeed a Bayesian Equilibrium. (See Lemma A8.) Intu√ √
√
itively, at any state, the Legislator is choosing a policy among 0, − B, B. The policies − B
√
and B are rewarded with reelection and so yield the Legislator a payoff of zero, at any state. The
policy zero is not rewarded with reelection. So, at any state, the Legislator is indifferent between
choosing any one of these policies.
√
√
Under this voting rule, when the Voter’s ideal point lies in [− 12 B, 12 B] and the Legislator
chooses the policy zero, the Legislator does not get reelected. Thus, there is a sense in which the
Voter punishes the Legislator for choosing the policy he wants her to choose. This is no coincidence.
If the Voter were to reward the Legislator with reelection when she chooses her ideal point, then
11
√
√
the Legislator would never choose the policies − B or B. In order to induce the Legislator to
choose policies close to the Voter’s ideal point when this ideal point is extreme, the Voter cannot
reward the Legislator with reelection for choosing centrist policies.
Similarly, an extreme two-sided voting rule cannot involve the Legislator being rewarded for
√
choosing some policy p ∈ (0, B). If it did, at any state, the Legislator would strictly prefer to
choose this policy to her own ideal point. Thus, in order to induce the Legislator to choose policies
close to the Voter’s ideal point when that ideal point is close to zero, the Voter does not reward
√
any p ∈ (0, B) with reelection.
These examples illustrate two possible types of optimal retrospective voting rules. What differentiates situations in which optimal retrospective voting will involve a moderate versus an extreme
two-sided voting rule? Under the extreme two-sided rule, the policy the Legislator must choose
in order to be reelected is more extreme than under the moderate two-sided rule. In the second
example, the Voter’s beliefs placed greater probability on his ideal point being further from zero.
This is why the Voter favored the extreme two-sided rule.
We now turn to the characterization of the optimal retrospective voting rule in this benchmark
case. It will be useful to first formally define moderate and extreme two-sided voting rules.
Definition 5.1 Fix a policy p∗
Z
∗
p =
Ω+
Z
xv (ω) dµ (ω) −
Ω\Ω+
xv (ω) dµ (ω) .
√
Say (s∗l , s∗v ) is a moderate two-sided voting rule if p∗ ∈ (0, B) and the following are satisfied:
(i) s∗l (ω) = −p∗ whenever xv (ω) < 0, s∗l (ω) = p∗ whenever xv (ω) > 0, and s∗l (ω) ∈ {−p∗ , p∗ }
whenever xv (ω) = 0;
(ii) if p ∈ [−p∗ , p∗ ], then s∗v (p) = 1 if and only if p ∈ {−p∗ , p∗ }.
Definition 5.2 Say (s∗l , s∗v ) is an extreme two-sided voting rule if the following are satisfied:
√
√
√
√
(i) s∗l (ω) = 0 whenever xv (ω) ∈ (− 12 B, 12 B), s∗l (ω) = − B whenever xv (ω) < − 21 B,
√
√
√
√
s∗l (ω) = B whenever xv (ω) > 12 B, s∗l (ω) ∈ {− B, 0} whenever s∗l (ω) = − 12 B, and
√
√
s∗l (ω) ∈ {0, B} whenever s∗l (ω) = 21 B;
√ √
√ √
(ii) if p ∈ [− B, B], then s∗v (p) = 1 if and only if p ∈ {− B, B}.
Proposition 5.2 There exists an Expectationally and State-by-State Optimal Equilibrium. Any
such equilibrium is either a moderate or an extreme two-sided voting rule.
Proposition 5.2 establishes that an optimal retrospective voting rule will take one of two simple
forms, namely a moderate or an extreme two-sided voting rule. When the Voter’s expected ideal
√
point lies within B of zero, he strictly prefers an optimal retrospective voting rule to choosing
12
his expected ideal point at every state. This is one sense in which the Voter can use retrospective
voting to extract information from the Legislator.
But notice that there are only limited circumstances in which the policies chosen coincide with
the Voter’s actual ideal point. (In the case of a moderate two-sided voting rule associated with
policy p∗ , this will be the set of states (xv )−1 ({−p∗ , p∗ }). For an extreme two-sided voting rule, it
√ √
is the set (xv )−1 ({0, − B, B}).)
It will turn out that the Voter can only extract a limited amount of information because his
strategy set is not sufficiently rich. The next two sections enrich the Voter’s strategy set in two
ways.
In Section 6, we allow for the possibility that the Voter may learn the true state after
the Legislator chooses the policy.
information extraction.
Here, we give conditions under which the Voter achieves full
In Section 7, we allow the Voter to make use of behavioral strategies.
Here, we show that—even when there is no chance that the Voter will learn the true state—he can
nonetheless achieve full information extraction.
6
Enriching the Strategy Set: Partially Informed Voter
Contrast the Benchmark specification with the case where it is transparent that Nature always
informs the Voter (i.e., π = 1). Here, the Voter can condition reelection on both the policy and
state. In particular, the Voter can choose a voting rule where, for each state ω, he reelects the
Legislator if and only if she chooses his ideal point xv (ω). When the Voter’s ideal point is within
√
B of zero, such a voting rule will induce the Legislator to choose this ideal point.
In the case where the Voter is fully informed, there is no need to extract information from
the Legislator. The fact that the Voter can induce the Legislator to choose his ideal point does
not reflect information extraction. Rather, it reflects ‘leverage’ over the Legislator due to the fact
that the Voter can condition reelection on his actual ideal point. That is, it reflects a richer set of
strategies.
Now, consider a game in which it is transparent that the Voter learns the true state with some
probability π ∈ (0, 1). Here, the Voter still has some ‘leverage,’ i.e., a richer set of strategies.
But unlike the case where the Voter is fully informed, there is some positive probability that the
Voter will never learn the true state. There are two implications of this. First, while, in principle,
the Voter can condition reelection on the true state, he may not be able to do so in practice—his
‘informed information set’ may not be reached. Second, the Voter would like to use retrospective
voting for both electoral control and information extraction. Can the Voter use the gained ‘leverage’
to extract additional information?
It turns out the answer is yes. To gain an intuition for why, we will begin by looking at a
retrospective voting rule that is analogous to the extreme two-sided rule in the Benchmark case
(π = 0). We will see how the Voter can use his additional ‘leverage’ to alter this rule and improve
his expected payoffs.
13
If
∉ Φ, reelect if and only if
Reelect if ω ∈ Φ
and informed
- B
Choose - B
- πB
0
πB
Choose Voter's Ideal Point
B
Choose
B
Figure 6.1: Modifying an extreme two-sided voting rule.
6.1
An Example
Fix a distribution of µ. Suppose that in the Benchmark specification, an extreme two-sided voting
rule is expectationally optimal, given this distribution. Now, consider a strategy profile, viz. (sl , sv ),
that corresponds to the extreme two-sided voting rule. The strategy sl remains



 0
√
sl (ω) =
− B


 √B
h √
√ i
if xv (ω) ∈ − 12 B, 12 B
√
if xv (ω) < − 12 B
√
if xv (ω) > 12 B.
Whether or not the Voter is informed of the true state, the strategy sv specifies reelection if and
√ √
only if p is contained in {− B, B}. It can be shown that this strategy profile remains a Bayesian
Equilibrium (in the general game). However, it is no longer an optimal retrospective voting rule. In
the event that Nature chooses to inform the Voter, he can condition his reelection decision on the
true state. As such, there is an equilibrium that the Voter strictly prefers to the extreme two-sided
voting rule.
To see this, fix a set of states Φ ⊆ Ω. Consider a voting rule, viz. rv , that satisfies the following
three criteria. First, if the Voter is uninformed, this voting rule agrees with the extreme two-sided
voting rule sv . Second, if the Voter is informed and the state ω is contained in Φ, the Voter reelects
the Legislator if and only if she chooses his ideal point xv (ω). Third, if the Voter is informed but
the true state ω is not contained in Φ, the Voter reelects the Legislator if and only if she chooses a
√ √
policy in {− B, B}. Figure 6.1 illustrates such a voting rule.
√
√
Fix a state in Φ, viz. ω, where the Voter’s ideal point, viz. xv (ω), is not contained in {− B, 0, B}.
At this state, does the given voting rule induce the Legislator to choose the Voter’s ideal point? If
the Legislator chooses the Voter’s true ideal point, her expected payoffs are −xv (ω)2 + πB. The
Legislator will only do so if these payoffs are greater than the expected payoffs from choosing her
own ideal point. This requirement is satisfied if and only if the Voter’s true ideal point lies within
√
πB of zero.
14
- B
- πB
0
πB
B
Choose Voter's Ideal Point
Figure 6.2: Informed incentives.
So, suppose that the Voter chooses Φ to be the set of states where his ideal point is contained
√
√
√
√
in [− πB, πB], i.e., Φ = (xv )−1 ([− πB, πB]). The above arguments suggest the following:
the strategy profile (rl , rv ) is a Bayesian Equilibrium, where (i) rl (ω) = xv (ω) when xv (ω) ∈
√
√
[− πB, πB] and (ii) rl (ω) = sl (ω) otherwise. When the Voter’s true ideal point is contained in
√
√
[− πB, πB]\ {0}, his payoffs under (rl , rv ) are strictly greater than his payoffs under the extreme
two-sided voting rule (sl , sv ). In other cases, the payoffs agree. So, an extreme two-sided rule is no
longer State-by-State Optimal.
6.2
Informed vs. Uninformed Incentives
Return to the example in Section 6.1 above. There we constructed a Bayesian equilibrium, viz.
(rl , rv ), that the Voter strictly prefers to the extreme two-sided voting rule (sl , sv ). We will see
that even this voting rule is not an optimal retrospective voting rule. To see this, we will point to
two types of incentives the strategy rv offers: informed and uninformed incentives.
Begin with informed incentives. Say the Voter offers informed incentives to choose policy p at
a state ω, if the Voter’s strategy takes the following form: if the true state is ω and the Voter is
informed, he reelects the Legislator if and only if she chooses policy p. Returning to the example,
rv offers informed incentives to choose the Voter’s ideal point at any state ω ∈ Φ.
At a given state, can the Voter use informed incentives (alone) to induce the Legislator to choose
his ideal point? If the Voter’s and Legislator’s ideal points are sufficiently divergent—i.e., if the
√
Voter’s ideal point lies further than πB from zero—the answer is no. The argument is the same
as the one given for the strategy profile (rl , rv ), in Section 6.1 above: the Legislator always has
the outside option to choose her ideal point and doing so always gives her expected payoffs that
are at least zero. So if, at the state ω, informed incentives are to be effective by themselves, then
√
−xv (ω)2 + πB ≥ 0. This says that the Voter’s ideal point must lie within πB of zero, if informed
incentives alone are to be effective.
But the good news is that informed incentives are state contingent. As such, they do not conflict
√
with one another. Fix states ω1 , ω2 with xv (ω1 ) , xv (ω2 ) ∈ (0, πB] and xv (ω1 ) > xv (ω2 ). By
offering informed incentives to choose (i) xv (ω1 ) at the state ω1 and (ii) xv (ω2 ) at the state ω2 ,
the Voter can induce the Legislator to choose his respective ideal points at each of these states.
In sum, by using only informed incentives, the Voter can induce the Legislator to choose her
√
√
ideal point if and only if it is contained in [− πB, πB]. This fact is depicted in the shaded region
15
of Figure 6.2.
Contrast this with uninformed incentives. Say the Voter offers uninformed incentives to choose
policy p if the Voter’s strategy takes the following form: if the Voter is uninformed, he reelects the
Legislator if she chooses policy p. Returning to the example, the strategy rv offered uninformed
√
√
incentives to choose the policies − B and B.
Much like informed incentives, uninformed incentives (alone) may not be sufficient to induce
the Legislator to choose the Voter’s ideal point. Again, the Legislator always has the outside option
of choosing her own ideal point, and this option gives her expected payoffs of at least zero. So, for
a given Voter ideal point xv (ω), if uninformed incentives are to be effective by themselves, then
p
−xv (ω)2 + (1 − π) B ≥ 0. Put differently, the Voter’s ideal point must lie within (1 − π) B of
zero.
Unlike informed incentives, uninformed incentives are not state contingent and so may conp
flict with one another. For instance, fix states ω1 , ω2 with xv (ω1 ) , xv (ω2 ) ∈ (0, (1 − π) B] and
xv (ω1 ) > xv (ω2 ). Suppose the Voter only offers uninformed incentives to choose these policies.
Then, at any state, the Legislator strictly prefers xv (ω2 ) over xv (ω1 ). By choosing xv (ω2 ), the
Legislator gets a policy she prefers and does not forgo the benefits of reelection when the Voter
is uninformed. So, the Voter cannot use uninformed incentives (alone) to induce the Legislator to
choose his ideal point at both of these states. That is, because uninformed incentives may conflict
with one another, these incentives are not sufficient to induce the Legislator to choose the Voter’s
p
ideal point whenever it lies within (1 − π) B of zero.
Notice, this conflict need not arise if the Voter offers both (i) informed incentives to choose
xv (ω1 ) at the state ω1 and (ii) uninformed incentives to choose xv (ω1 ). Then, when the Voter’s
ideal point is xv (ω1 ), there is a cost associated with choosing xv (ω2 ). Specifically, by choosing
xv (ω2 ), the Legislator must forgo the benefit of reelection if the Voter is informed.
This idea holds more generally. Using informed incentives in conjunction with uninformed
incentives can eliminate the conflict between uninformed incentives. Moreover, by using both of
these incentives simultaneously, the Voter can induce the Legislator to choose his ideal point even
p
when it lies farther than (1 − π) B from zero.
p
√
To see this, fix states ω1 , ω2 where xv (ω1 ) , xv (ω2 ) ∈ [ (1 − π) B, B] and xv (ω1 ) > xv (ω2 ).
Recall, the Voter cannot use uninformed incentives alone to induce the Legislator to choose xv (ω1 )
when this is his ideal point. Suppose instead the Voter offers both (i) informed incentives to choose
xv (ω1 ) (resp. xv (ω2 )) at the state ω1 (resp. ω2 ) and (ii) uninformed incentives to choose xv (ω1 )
(resp. xv (ω2 )). Not only can the Voter induce the Legislator to choose his ideal point at xv (ω2 )—
p
a policy farther than (1 − π) B from the Legislator’s ideal point—but he can also induce the
Legislator to choose his ideal point at xv (ω1 ). That is, because of their informed component, these
incentives do not conflict with one another.
For instance, if the true state is ω1 and the Legislator chooses the Voter’s ideal point at this
state, her expected payoffs, viz. −xv (ω1 )2 +B, are positive. As such, she has no incentive to choose
her own ideal point. (If she chooses her own ideal point, she is certain not to be reelected.) She
16
- B
- (1-π )B
0
(1-π )B
Choose - B
B
Choose B
Choose Voter's Ideal Point
Figure 6.3: Informed and uninformed incentives together.
also has no incentive to choose xv (ω2 ). If the Voter is uninformed, she is reelected if she chooses
either of xv (ω1 ) or xv (ω2 ). By choosing xv (ω2 ) she gets a policy closer to her ideal point, but
at the cost of forgoing the informed incentives. Because xv (ω2 ) is sufficiently close to xv (ω1 ), the
benefits associated with choosing this more preferred policy are outweighed by the costs of forgoing
the informed incentives. (Indeed, her expected payoffs from xv (ω2 ) are less than or equal to zero.)
In sum, the Voter can use informed incentives in conjunction with uninformed incentives to
p
√
induce the Legislator to choose his ideal point whenever it is contained in [− B, − (1 − π) B] ∪
p
√
[ (1 − π) B, B]. By a similar argument, the Voter can use both informed and uninformed in√
√
centives to induce the Legislator to choose − B (resp. B) whenever his ideal point is less than
√
√
− B (resp. greater than B). These facts are described in Figure 6.3.
6.3
Characterization
The key to characterizing the optimal retrospective voting rule is to notice that we can put Figures
6.2–6.3 together. That is, the incentives associated with each one of these voting rules do not
conflict with one another.
In particular, construct a voting rule that combines the voting rules associated with Figures
6.2–6.3. Here, the Voter provides informed incentives to choose her ideal point whenever it is
contained in
p
p
√
√
√
√
[− B, − (1 − π) B] ∪ [− πB, πB] ∪ [ (1 − π) B, B].
The Voter also offers uninformed incentives to choose policies in
p
p
√
√
[− B, − (1 − π) B] ∪ [ (1 − π) B, B].
This voting rule induces the Legislator to choose the Voter’s ideal point whenever it is contained
in one of the shaded regions of Figures 6.2–6.3. To see this, first fix a state where the Voter’s ideal
p
p
√
√
point lies in [− B, − (1 − π) B] ∪ [ (1 − π) B, B]. At this state, the incentives associated
with this combined voting rule coincide with the incentives associated with Figure 6.3. It follows
that choosing the Voter’s ideal point at this state is indeed optimal. Next, fix a state where
√
√
the Voter’s ideal point lies in [− πB, πB]. At this state, the Legislator has no incentive to
p
p
√
√
choose a policy in [− B, − (1 − π) B] ∪ [ (1 − π) B, B]. While these policies offer uninformed
incentives, they come at the cost of forgoing informed incentives. The uninformed incentives are
17
- B - πB
πB
0
B
Choose Voter's Ideal Point
- B
- (1-π )B
0
(1-π )B
B
Figure 6.4: Full information extraction with π ≥ 12 .
insufficient to induce the Legislator to choose these policies over the Voter’s ideal point. (Indeed,
it is straightforward to check that choosing the Voter’s ideal point gives the Legislator positive
expected payoffs, while choosing a policy in this range gives her expected payoffs that are less than
or equal to zero.)
This says that the Voter can get the Legislator to choose her ideal point both when it is contained
p
√
√
√
in [− πB, πB] (the shaded region in Figure 6.2) and when it is contained in [− B, − (1 − π) B]∪
p
√
[ (1 − π) B, B] (the shaded region in Figure 6.3). The question then is how these Figures fit
together? The answer depends on the probability that the Voter is informed, viz. π.
6.4
Full Information Extraction
Fix π ≥ 12 . This implies that πB ≥ (1 − π) B. So, putting Figures 6.2–6.3 together, we have no
gaps. That is, the Voter can design a retrospective voting rule that induces the Legislator to choose
√ √
his ideal point whenever it is contained in [− B, B]. This is illustrated in Figure 6.4.
Proposition 6.1 Suppose that it is transparent that the Voter will be informed with some probability greater than or equal to one half, i.e., π ≥
State-by-State Optimal Equilibrium, viz.
(s∗l , s∗v ).
1
2.
Then there exists an Expectationally and
Moreover, in any such equilibrium, the Legislator
chooses:
√ √
(i) the Voter’s ideal point, whenever it is contained in [− B, B],
√
√
(ii) the policy − B, whenever the Voter’s ideal point is less than − B, and
√
√
(iii) the policy B, whenever the Voter’s ideal point is greater than B.
Proposition 6.1 is a full information extraction result. It says: suppose that there is at least a
fifty-fifty chance of Nature informing the Voter of the true state. Then the optimal retrospective
voting rule is ‘equivalent’ to the optimal retrospective voting rule when the Voter always knows the
true state of the world.
18
- πB
- B
0
πB
B
Choose Voter's Ideal Point
- B - (1-π )B
0
(1-π )B
B
Figure 6.5: Partial information extraction with π < 12 .
6.5
Partial Information Extraction
Fix π <
1
2.
Putting Figures 6.2–6.3 together, there is now a gap—i.e., the Legislator does not
√ √
choose the Voter’s ideal point whenever it is contained in [− B, B]. The Voter can induce the
Legislator to choose his ideal point when it is contained in the shaded regions of Figure 6.5:
p
p
√
√
√
√
[− B, − (1 − π) B] ∪ [− πB, πB] ∪ [ (1 − π) B, B].
Notice, when it is transparent that Nature does not inform the Voter—that is, when π = 0—this
‘amounts’ to an extreme two-sided voting rule.13 What we have here is a generalized version of an
extreme two-sided voting rule.
In Section 5, we saw that an extreme two-sided voting rule may fail to be Expectationally
Optimal. Under a given distribution µ, the Voter may consider it very likely that his ideal point
√ √
√
√
will be moderate—that is, in [− B, B] but neither too close to − B (resp. B) nor too close
to zero.
If so, a moderate two-sided voting rule may be Expectationally Optimal.
For similar
reasons, a generalized extreme two-sided voting rule may fail to be Expectationally Optimal.
Take the following example: set Ω = R and let xv be the identity map. The Voter’s prior µ is
uniform on
and π =
√
√
√
√
[− 16 2B, − 16 B] ∪ [ 16 B, 16 2B]
1
36 .
Consider the generalized extreme two-sided voting rule associated with Figure 6.5. Here, there
√
√
are exactly two states in the support of µ, viz. − 16 B and 16 B, at which the Legislator chooses
the Voter’s ideal point.
But there is an equilibrium where the Legislator chooses the Voter’s ideal point whenever it
is contained in the support of his distribution. Such an equilibrium is a generalized version of a
moderate two-sided voting rule.
13
We say it ‘amounts’ to an extreme two-sided voting rule because we haven’t specified what the Legislator will
choose outside these ranges. Indeed, once this is specified, taking π = 0 gives an extreme two-sided voting rule.
19
- B
- p2 + π B
-p
0
p
p2 + π B
B
Choose Voter's Ideal Point
Figure 6.6: A generalized moderate two-sided voting rule.
Specifically, suppose that (i) at any state ω in the support of µ, the Voter offers informed
incentives to choose xv (ω) at ω and (ii) the Voter offers uninformed incentives to choose any policy
in the support of µ.
Fix a state ω in the support of µ.
If the Legislator does indeed choose
policy ω, then her expected payoffs are strictly positive, and therefore are higher than from any
other policy. To see this, suppose otherwise, i.e., at the state ω, there is a policy p that offers the
Legislator strictly higher expected payoffs than ω.
Then, the policy p must be associated with
uninformed (but not informed) incentives, so that
−p2 +
Indeed, since p2 ≥
1
36 B,
it follows that
35
B > −ω 2 + B.
36
34
36 B
> −ω 2 + B, or ω 2 >
2
36 B.
This contradicts the fact
that ω is contained in the support of µ.
In sum, there is an equilibrium in which the Legislator chooses the Voter’s ideal point whenever
it is contained in µ.
The Voter achieves this by amending the incentives associated with the
generalized extreme two-sided voting rule. In particular, the Voter changes the range of policies for
which he offers both informed and uninformed incentives—shifting it toward the center. The range
p
√ p
√
√
√
√
√
was previously [− B, (1 − π) B]∪[ (1 − π) B, B], but is now [− 61 2B, 16 B]∪ [ 16 B, 61 2B].
p
This can be done more generally. Refer to Figure 6.6. For any policy p in [0, (1 − π) B],
there is an equilibrium where (i) the Voter offers informed and uninformed incentives in the shaded
p
p
range, i.e., [− p2 + πB, −p] ∪ [p, p2 + πB] and (ii) the Legislator chooses the Voter’s ideal point
p
whenever it is contained in this range. (See Lemma B4.) In Figure 6.5, the policy p is (1 − π) B.
√
In the example above, the policy p is shifted inward to πB.
Shifting p inwards comes at a cost to the Voter. For the intuition, return to the case where
π = 0.
There, we saw an important difference between extreme and moderate two-sided voting
rules.
In particular, under the extreme two-sided voting rule, the Legislator chooses a policy of
zero when the Voter’s ideal point is close to zero. She does not do so under a moderate two-sided
√
voting rule. When the Voter offers incentives to choose policies strictly within B of zero, the
Legislator strictly prefers these policies to her own ideal point. Thus, moving from an extreme to
√
a moderate two-sided voting rule (i.e., shifting p inward from B) comes at the cost of not having
the Legislator choose zero when this is close to the Voter’s ideal point.
20
An analogous issue arises for generalized two-sided voting rules.
Refer to the generalized
extreme two-sided voting rule associated with Figure 6.5. There, the Voter is able to use informed
incentives alone to induce the Legislator to choose his ideal point when it is close to zero. As the
Voter shifts p inward, the range in which these informed incentives are effective is reduced.
√
To see this, return to the example above. There p = 61 B. Fix a state at which the Voter’s
√
√
ideal point ω is contained in (− 16 B, 16 B) and suppose the Voter offers informed incentives to
choose the policy ω at this state.14
Voter’s ideal point are
are
34
36 B.
−ω 2
+
1
36 B,
Then, the Legislator’s expected payoffs from choosing the
√
while her expected payoffs from choosing the policy p = 16 B
So, the Legislator will not choose the Voter’s ideal point at this state.
In Section 6.3, we argued that we can combine the incentives associated with Figures 6.2 and
6.3, i.e., they do not conflict with one another. Here, we see that we may not be able to combine
the incentives associated with Figures 6.2 and 6.6. In particular, giving uninformed incentives for
the policy p (as in Figure 6.6) may conflict with using informed incentives to induce the Legislator
to choose a policy close to zero (as in Figure 6.2).
Fix a state ω. For these incentives not to conflict with one another, the Legislator’s expected
payoffs from choosing p must be less than her expected payoffs from choosing the Voter’s ideal
point at this state. That is,
−xv (ω)2 + πB ≥ −p2 + (1 − π) B.
There is some state that satisfies this condition if and only if p2 ≥ (1 − 2π) B.
Proposition 6.2 Suppose that it is transparent that the Voter will be informed with some probability strictly less than 12 , i.e., π ∈ [0, 12 ). Any State-by-State Optimal Equilibrium takes one of two
forms:
(i) There is a policy p ∈ [0,
p
(1 − 2π) B) such that the Legislator chooses the Voter’s ideal point
p
p
if and only if it is contained in [− p2 + πB, −p] ∪ [p, p2 + πB].
p
p
(ii) There are policies p ∈ [ (1 − 2π) B, (1 − π) B] and q ∈ [0, p) such that the Legislator
p
chooses the Voter’s ideal point if and only if it is contained in [− p2 + πB, −p] ∪ [−q, q] ∪
p
[p, p2 + πB].
Proposition 6.2 says that, when π is strictly less than a half, an optimal retrospective voting
rule must take one of three forms. The first form is a generalized version of a moderate two-sided
voting rule. The second and form is a generalized version of an extreme two-sided voting rule.
In each of these cases, there is an open set of policies X such that the Legislator chooses the
Voter’s true ideal point when it is contained in X. But, again, there is no Bayesian Equilibrium
√ √
in which the Legislator chooses the Voter’s ideal point whenever it is contained in [− B, B].
14
This state is not in the support of the Voter’s prior. The example can be readily amended to include this state
in the support (and retain the properties we discuss).
21
7
Enriching the Strategy Set: Behavioral Strategies
Thus far, we have restricted attention to pure-strategy Bayesian Equilibria. In this Section, we
enrich the Voter’s strategy set by allowing him to use behavioral strategies. (The restriction to
behavioral, rather than mixed, strategies is without loss of generality.)
When it is transparent that Nature will inform the Voter with probability π ≥ 21 , allowing the
Voter this broader choice has no effect. In particular, as seen in Section 6.4, there exists a purestrategy equilibrium in which the Legislator chooses the Voter’s ideal point whenever it is contained
√
within B of zero. Even with the use of behavioral strategies, there do not exist incentives that
√
can induce the Legislator to choose the Voter’s ideal point when it is further than B from zero.
So, the voting rule identified remains Expectationally and State-by-State Optimal, even within this
larger set of equilibria.
Turn to the case where it is transparent that Nature informs the Voter with probability π < 21 .
Now, allowing the Voter to play behavioral strategies does have an effect.15
Proposition 7.1 For all π ∈ [0, 1], there exists a Bayesian equilibrium in behavioral strategies, in
which the Legislator chooses:
√
(i) the Voter’s ideal point, whenever it is contained within B of zero,
√
√
(ii) the policy − B, whenever the Voter’s ideal point is less than − B, and
√
√
(iii) the policy B, whenever the Voter’s ideal point is greater than B.
To understand this result, it will be useful to begin by assuming that the Voter is restricted to
the use of pure strategies and that π ∈ [0, 12 ). Recall that the Voter cannot induce the Legislator to
p
√
choose his ideal point both when his ideal point is a policy p ∈ ( πB, (1 − π) B) and when his
√
ideal point is B. To induce the Legislator to choose policy p, he must offer uninformed incentives.
With these incentives, at any state, the Legislator’s expected payoffs are strictly greater than zero.
√
But then, when the Voter’s true ideal point is B, the Legislator’s expected payoffs from choosing
√
p are strictly greater than zero, while her expected payoffs from choosing B are at most zero.
√
Thus, she would not choose B at this state.
With the use of behavioral strategies, such a conflict need not occur. The Voter can offer the
Legislator uninformed incentives with some probability less than one. In particular, the electoral
rule can have the following features: at any state, if the policy p is chosen and the Voter is
informed, he reelects the Legislator with probability zero. If the policy p is chosen and the Voter
is uninformed, he reelects the Legislator with probability
p2
(1−π)B .
Under this electoral rule, at
every state, the Legislator’s expected payoffs from choosing p are zero. Thus, in equilibrium, the
√
Legislator can still choose B when this is the Voter’s ideal point.
This is again a result on full information extraction. In particular, it says that, when selecting
among the set of Bayesian equilibria in behavioral strategies, the optimal retrospective voting rule
15
We thank Heski Bar-Isaac for suggesting this line of argument.
22
is ‘equivalent’ to the optimal retrospective voting rule when the Voter knows the true state of the
world. This is true even if it is transparent that the Voter has no chance of being informed, i.e.,
if π = 0. The reason is that the Voter has a richer set of strategies and thus has gained ‘leverage.’
8
Discussion
This Section discusses some technical aspects and extensions of the paper.
8.1
The Players’ Payoff Functions
We assume that both players’ policy payoffs are quadratic. This assumption was made for tractability.
All of the results hold in a more general setting. Take the policy space to be a metrizable topological space. We will require that there is a compatible metric that satisfies a certain ‘symmetry’
property. (Lemma A9 suggests what that property must be.) For this metric, a player’s payoff
must be monotonically decreasing as policy moves away from his or her ideal point.
This is a key distinction between our paper and Maskin and Tirole [22, 2004]. In their game,
there are only two actions and so the players’ preferences must violate the ‘symmetry’ property
we require. As a result, their model does not yield the information extraction results on which we
focus.
8.2
Selection Criteria
In Section 4.2, we introduced two selection criteria, namely Expectational Optimality and State-byState Optimality. The two concepts are distinct. The strategy profile (s∗l , s∗v ) may be a State-byState Optimal Equilibrium even though there does not exist a measure under which s∗l maximizes
the Voter’s expected payoffs given all Bayesian Equilibria.16 Conversely, the strategy profile (s∗l , s∗v )
may be Expectationally Optimal even though it is not State-by-State Optimal. This can occur even
if we require the prior to have full support, i.e., if we can find a Bayesian Equilibrium (sl , sv ) where
the non-empty set
{ω ∈ Ω : uv (ω, sl (ω)) > uv (ω, s∗l (ω))}
does not contain a non-empty open subset.
We note that there is a (conceptual) tension between Expectational Optimality and State-byState Optimality. The former requires that the Voter optimize given his prior µ. The latter is a
weak dominance criterion—it asks the Voter to consider all possibilities, even though he may not
be able to do so under his prior µ.
That said, there is a rationale for including both requirements. Individually, each requirement
can be viewed as an ex ante optimality criterion. The Voter may be interested in using an interim
16
This is for the same reason as the well-known fact: a strategy may be purely undominated even though it is not
a best response under any probability measure.
23
optimality criterion, i.e., one where he would not want to revise his selection criterion at any
of his information sets. To the extent that Expectational Optimality is an appropriate ex ante
selection criterion, it is an appropriate selection criterion at the information set where the Voter
is uninformed. Now, consider an information set in which the Voter is informed that the true
state is ω. Here, the Voter would like to use a selection criterion that requires that there does not
exist another Bayesian Equilibrium which yields a strictly higher expected payoff when the Voter
is informed that the true state is ω. Collecting all these information sets together, the Voter would
like to use a weak dominance criterion—namely, State-by-State Optimality.
Many of the arguments in this paper depend only on State-by-State Optimality.17
To see
this, it will be useful to focus the discussion on the case where π = 0. The argument proceeds by
considering a Bayesian Equilibrium, viz. (sl , sv ), that differs structurally from a moderate or an
extreme two-sided voting rule. For instance, suppose that sl is a constant strategy, i.e., specifies the
same policy at every state. We then argue that there is a two-sided Bayesian Equilibrium that (i)
yields the Voter strictly higher payoffs for some non-empty set of states Φ and (ii) otherwise agrees
with the strategy sl . Thus, (sl , sv ) is inconsistent with State-by-State Optimality. Note that, for
the strategy sl , the set Φ corresponds to an open set of ideal points. So, if xv is continuous and
Supp µ = Ω, this also violates Expectational Optimality. But even if we were to assume that xv is
continuous and µ has full-support, Expectational Optimality alone does not give the conclusions
of Proposition 5.2. Suppose that an extreme two-sided voting rule is Expectationally Optimal.
We can find another Bayesian Equilibrium that differs from any extreme two-sided voting rule at
exactly one state. Such an equilibrium will also be Expectationally Optimal, even though it is not
State-by-State Optimal. As such, State-by-State Optimality is important for our analysis.
From the Voter’s perspective, the choice of an optimal voting rule is akin to finding an optimal
mechanism (albeit from a limited set of mechanisms). Dominance has been important in other
areas of mechanism design (Vickery [28, 1961], Chung and Ely [9, 2001], Izmalkov [16, 2004]). It
also has a long history in voting games (Farquharson [12, 1969], Moulin [23, 1979], Dhillon and
Lockwood [11, 2004]).
8.3
The Legislator’s Policy Preferences
Thus far, we have analyzed the case where the Legislator’s policy preferences are independent of
the state. In this subsection, we revisit the Benchmark specification, now allowing the Legislator’s
policy preferences to vary with the state.
Assume π = 0 and consider the follwoing example. If the Voter’s ideal point is contained in
√
R\R+ , the Legislator’s ideal point is p− ∈ [− B, 0). Otherwise, the Legislator’s ideal point is
√
p+ ∈ [0, B]. Consider any Bayesian Equilibrium in the Benchmark specification. We can find
a Bayesian Equilibrium of this new game that (up to a normalization) is equivalent to the initial
17
A notable exception is Lemma A13, which uses Expectational Optimality to pin down the policy p∗ associated
with a Moderate Two-Sided Voting Rule. Expectational Optimality is also important in selecting between a Moderate
and an Extreme Two-Sided Voting Rule.
24
Reelect if and only if
p− − q−
Choose p − − q −
p−
p− + q−
0
p+ − q+
p+
Choose p − + q − Choose p + − q +
p+ + q+
Choose p + + q +
Figure 8.1: The analogue to the Moderate Two-Sided Voting Rule with Two Partition Members.
equilibrium when we restrict (i) the Legislator’s ideal point to be negative (resp. positive) and (ii)
the domain of the Voter’s strategy to be the set of negative (resp. positive) policies. The converse
also holds, i.e., for every Bayesian Equilibrium in the initial game, we can construct an equivalent
Bayesian Equilibrium in this new game.18
With this, the Voter can use a voting rule that is essentially one moderate two-sided voting rule
on the set of negative policies and another moderate two-sided voting rule on the set of positive
policies.19 (See Figure 8.1.) Note that because the Legislator’s and Voter’s ideal points are always
on the same side of zero, the Legislator never has an incentive to choose a policy in the ‘negative
partition member’ when the Voter would like her to choose a policy in the ‘positive partition
√
√
member.’ So, if E (xv ) ∈ [p− − B, p+ + B], there is a Bayesian Equilibrium (sl , sv ) where (i) at
every state, the Voter’s payoffs from sl are at least as high as are his payoffs from E (xv ) and, (ii)
at some states, the Voter’s payoffs from sl are strictly higher than his payoffs from E (xv ).20
The basic set-up here suggests a first step toward generalization. In particular, let P be a
countable partition of the policy space R. Each partition member, viz. Pk , is an interval of some
√
length less than or equal to 2 B. The Legislator’s ideal point at state ω ∈ Ω will be determined
by the random variable xl : Ω → R where
xl (ω) = pk =
sup Pk +inf Pk
2
whenever xv (ω) ∈ Pk .
That is, whenever the Voter’s ideal point lies in the partition member Pk , the Legislator’s ideal
point is the policy pk , the midpoint of the partition member. Notice that this framework maintains
the key assumption that the players’ ideal points always lie in the same partition member. Under
these assumptions, the Voter can again design a voting rule that he strictly prefers to the strategy
where he receives his expected ideal point at every state. (See the Online Appendix for a formal
treatment.)
If the Legislator’s policy preferences are associated with K partition members, in any equilib√
√
The statements in this paragraph hold even when p− + B < 0 or p+ − B > 0.
19
A similar argument can
respect to an Extreme Two-Sided Voting Rule, though the argument is
√ be made with √
more intricate when p− + B > 0 or p+ − B < 0. (See Footnote 18.)
20
Note that the constant strategy associated with E (xv ) may be inconsistent with any Bayesian Equilibrium.
18
25
rium, there are, at most, 2K policies for which the Voter offers electoral incentives that actually
induce the Legislator to choose such policies. As K increases, the Voter can offer more (effective)
electoral incentives. So, if the players’ ideal points always lie in the same partition member, increasing K makes it easier for the Voter to extract information.21 However, increasing K has a
second effect. When the partition members are not required to be symmetric around the Legislator’s ideal point, the incentives a Voter provides the Legislator within one partition member can
conflict with the incentives provided in another partition member.22 Put differently, as K increases,
the informational extraction problem becomes easier, while the discipline problem becomes more
difficult.
One goal of this paper was to provide an informational rationale for retrospective voting. With
this in mind, we focused on the problem of extracting information. We restricted attention to the
case where the informational extraction problem is most difficult, i.e., K = 1. Of course, it is
important to understand the ability of the Voter to extract information in the presence of these
conflicting incentives. It is also important to understand the Voter’s ability to extract information
(under these conflicting incentives) when he can use a richer set of strategies. We leave these
characterizations for future work.
21
In the extreme case, the players’ preferences are perfectly aligned, and the information extraction problem is
trivial.
22
This is why we made the above assumptions about xl . See the Online Appendix for more on these difficulties.
26
Appendix A: Proofs for Section 5
This Appendix provides the proofs for Section 5. As such, throughout this Appendix, we take
the notational conventions outlined at the beginning of that Section. We begin with the proof of
Proposition 5.1. This will require a number of auxiliary results, found in Lemmata A1-A5.
Lemma A1 Fix a strategy sv ∈ Sv . Then there exists some sl ∈ Sl such that (sl , sv ) ∈ Sl × Sv is
a Bayesian Equilibrium if and only if there exists p ∈ R that maximizes ul (·, sv (·)).
Proof. Fix a strategy sv ∈ Sv . First, suppose that there exists p ∈ R that maximizes ul (·, sv (·)).
Set sl : Ω → R so that sl (ω) = p for all ω ∈ Ω. It is immediate that sl is Borel measurable. For all
ω ∈ Ω,
ul (sl (ω) , sv (sl (ω))) ≥ ul (p, sv (p))
for all p ∈ R,
establishing that (sl , sv ) is a Bayesian Equilibrium. Conversely, fix a Bayesian Equilibrium (sl , sv ).
By Condition (ii) of a Bayesian Equilibrium,
ul (sl (ω) , sv (sl (ω))) ≥ ul (p, sv (p))
for all p ∈ R,
holds for all ω ∈ Ω. So, for any given ω ∈ Ω, sl (ω) ∈ R maximizes ul (·, sv (·)), as required.
It will be convenient to introduce a piece of notation. Let P be a binary, complete, and transitive
ordering relation on R × R such that hp, qi ∈ P if and only if p2 ≥ q 2 . Say q is a lower bound
on X ⊆ R with respect to the order relation P if hp, qi ∈ P for all p ∈ X. Say q is the greatest
positive lower bound on X (with respect to the order relation P) if q is a greatest lower bound on
X and q ≥ 0. (So, if X = (−2, −1) then there are two greatest lower bounds, viz. −1 and 1, and
one greatest positive lower bound, 1.)
Fix some strategy sv ∈ Sv . Let inf (sv ) be the greatest positive lower bound on the set of
policies (sv )−1 ({1}) with respect to the order relation P. Notice that inf (sv ) ≥ 0.
Lemma A2 Fix a strategy sv ∈ Sv where, for some p ∈ R, sv (p) = 1 and B ≥ p2 . If sv ({− inf (sv ) ,
inf (sv )}) = {0}, then there does not exist a strategy sl ∈ Sl such that (sl , sv ) ∈ Sl ×Sv is a Bayesian
Equilibrium.
Proof. Fix a strategy sv ∈ Sv that satisfies the conditions in the statement of the Lemma. Notice
that there exists a policy p ∈ R with sv (p) = 1 and B > p2 . To see this, suppose otherwise: using
the statement of the Lemma, there exists some p ∈ R with sv (p) = 1 and B = p2 . Moreover, for
all p ∈ R with p2 > p2 , sv (p) = 0. So, inf (sv ) = |p| contradicting sv (− inf (sv )) = sv (inf (sv )) = 0.
Assume, contra hypothesis, that there exists a strategy sl ∈ Sl where (sl , sv ) is a Bayesian
equilibrium. Fix some ω ∈ Ω and the policy it induces under sl , viz sl (ω). There will be two cases,
the first corresponding to sv (sl (ω)) = 1 and the second corresponding to sv (sl (ω)) = 0.
Case A (sv (sl (ω)) = 1): Since sv (− inf (sv )) = sv (inf (sv )) = 0 and sv (sl (ω)) = 1, it
follows that sl (ω)2 > inf (sv )2 . From this, we know that there must exist some policy p ∈ R with
27
[sl (ω)]2 > p2 > [inf (sv )]2 and sv (p) = 1, else [sl (ω)]2 = [inf (sv )]2 . Now
ul (p, sv (p)) = −p2 + B
> − [sl (ω)]2 + B = ul (sl (ω) , sv (sl (ω))) ,
contradicting Condition (ii) of a Bayesian Equilibrium.
Case B (sv (sl (ω)) = 0): Here
ul (p, sv (p)) = −p2 + B
> −B + B ≥ − [sl (ω)]2 ≥ ul (sl (ω) , sv (sl (ω))) ,
contradicting Condition (ii) of a Bayesian Equilibrium.
Lemma A3 Fix a Bayesian Equilibrium (sl , sv ) ∈ Sl × Sv . Suppose that either (i) sv (0) = 1 or
(ii) for any p ∈ R with sv (p) = 1, p2 > B. Then sl (Ω) = {0}.
Proof. First suppose Condition (i) is satisfied. Then
ul (0, sv (0)) = B > −p2 + B ≥ ul (p, sv (p)) ,
for all p ∈ R\ {0}. By Condition (ii) of a Bayesian Equilibrium, there cannot exist some ω ∈ Ω
with sl (ω) ∈ R\ {0}.
Next, suppose that Condition (ii) is satisfied. Fix some p ∈ R\ {0}. If sv (p) = 1 then p2 > B
so that
ul (0, sv (0)) ≥ −B + B
> −p2 + B = ul (p, sv (p)) .
If sv (p) = 0 then
ul (0, sv (0)) ≥ 0 > −p2 = ul (p, sv (p)) .
Again, using Condition (ii) of a Bayesian Equilibrium, there cannot exist some ω ∈ Ω with sl (ω) ∈
R\ {0}.
Lemma A4 Fix a Bayesian Equilibrium (sl , sv ) ∈ Sl × Sv with B > inf (sv )2 . Then sl (Ω) ⊆
{− inf (sv ) , inf (sv )} and, for all ω ∈ Ω, sv (sl (ω)) = 1.
Proof. Fix a Bayesian Equilibrium (sl , sv ). By Lemma A2 and the fact that B > inf (sv )2 ,
either sv (− inf (sv )) = 1 or sv (inf (sv )) = 1. First, we will show that, for any ω ∈ Ω, sl (ω) ∈
{− inf (sv ) , inf (sv )}. For the purposes of this argument, suppose that sv (− inf (sv )) = 1. (A
corresponding argument will work if sv (inf (sv )) = 1.) Fix p ∈ R\ {− inf (sv ) , inf (sv )}. If sv (p) =
28
1 then p2 > [inf (sv )]2 , so that
ul (− inf (sv ) , sv (− inf (sv ))) = − [inf (sv )]2 + B
> −p2 + B ≥ ul (p, sv (p)) .
If sv (p) = 0 then
ul (− inf (sv ) , sv (− inf (sv ))) = − [inf (sv )]2 + B
> −B + B ≥ ul (p, sv (p)) .
So, by Condition (ii) of a Bayesian Equilibrium, sl (ω) ∈ {− inf (sv ) , inf (sv )}, for any ω ∈ Ω.
Next suppose that, for some ω ∈ Ω, sv (sl (ω)) = 0. Since sl (ω) ∈ {− inf (sv ) , inf (sv )} and
sv ({− inf (sv ) , inf (sv )}) 6= {0}, sv (−sl (ω)) = 1. From this it follows
ul (−sl (ω) , sv (−sl (ω))) = − [sl (ω)]2 + B
> − [sl (ω)]2 = ul (sl (ω) , sv (sl (ω))) ,
contradicting Condition (ii) of a Bayesian Equilibrium.
Lemma A5 Fix a Bayesian Equilibrium (sl , sv ) ∈ Sl × Sv with B = inf (sv )2 . Then sl (Ω) ⊆
√ √
√ √
{0, − B, B}, and sv (sl (ω)) = 1 if sl (ω) ∈ {− B, B}.
Proof. Fix a Bayesian Equilibrium (sl , sv ). First, we will show that, for each ω ∈ Ω, sl (ω) is not
contained in R\ {0, − inf (sv ) , inf (sv )}. Then, we will show that, if sl (ω) ∈ {− inf (sv ) , inf (sv )}
then sv (sl (ω)) = 1.
Fix some policy p ∈ R\ {0, − inf (sv ) , inf (sv )}. If sv (p) = 1 then p2 > B, and so
ul (0, sv (0)) = −B + B
> −p2 + B = ul (p, sv (p)) .
Similarly, if sv (p) = 0 then
ul (0, sv (0)) > −p2 = ul (p, sv (p)) .
¿From this and Condition (ii) of a Bayesian Equilibrium, for all ω ∈ Ω, sl (ω) ∈ {0, − inf (sv ) , inf (sv )}.
Next, suppose that sl (ω) ∈ {− inf (sv ) , inf (sv )} and sv (sl (ω)) = 0. Then
ul (0, sv (0)) > −B = ul (sl (ω) , sv (sl (ω))) ,
contradicting Condition (ii) of a Bayesian Equilibrium. This establishes that if sl (ω) ∈ {− inf (sv ) , inf (sv )}
then sv (sl (ω)) = 1, as required.
Proof of Proposition 5.1. Immediate from Lemmata A2-A3-A4-A5.
29
Next, we turn to the verbal claim that, if the Voter could choose his own policy, then maximizing
his expected utility would require that he always choose his expected ideal point. We formalize this
claim as follows: suppose that the Voter could choose a strategy sl for the Legislator. Unlike the
Legislator’s choice, the Voter’s choice cannot be state-contingent because he does not know the true
state. That is, the Voter can only choose a constant strategy, i.e., a strategy sl with sl (ω) = sl (ω 0 )
for all ω, ω 0 ∈ Ω. With this, it suffices to show the following:
R
Lemma A6 Let E (xv ) = Ω xv (ω) dµ (ω) and let sl : Ω → R be a constant strategy with sl (Ω) =
{E (xv )}. Then, for any constant strategy rl : Ω → R,
Z
Z
uv (ω, sl (ω)) dµ (ω) ≥
Ω
Ω
uv (ω, rl (ω)) dµ (ω) .
Proof. Fix a constant strategy rl : Ω → R with rl (Ω) = {p}. Note
Z
Ω
Z
Z
Z
rl (ω) xv (ω) dµ (ω) −
xv (ω)2 dµ (ω)
Ω
Z
2
2
2
= −p + 2pE (xv ) − E (xv ) + E (xv ) −
xv (ω)2 dµ (ω)
uv (ω, rl (ω)) dµ (ω) = −
2
Ω
rl (ω) dµ (ω) + 2
Ω
Ω
= − (p − E (xv ))2 − var (xv ) ,
where var (xv ) denotes the variance of the random variable xv with respect to µ. Applying this to
sl and an arbitrary constant strategy rl ,
Z
Ω
uv (ω, sl (ω)) dµ (ω) = − var (xv )
Z
2
≥ − (rl (ω) − E (xv )) − var (xv ) =
Ω
uv (ω, rl (ω)) dµ (ω) ,
as required.
We now turn to the proof of Proposition 5.2, which begins with what is known from Proposition
5.1 and then proceeds ‘to rule out’ certain Bayesian Equilibria based on the criteria of Expectational
(Lemma A13) and State-by-State Optimality (A9, A11, and A15). We also establish that Moderate
and Extreme Two-Sided Voting Rules are indeed State-by-State Optimal Equilibria, and at least
one is Expectationally Optimal. Again, this will require a number of auxiliary results, found below.
Lemma A7 Let {P1 , P2 , ..} be a countable partition of R where each partition member Pk is measurable. Also fix some countable {p1 , p2 , ..} ⊆ R. Fix a strategy sl ∈ Sl with sl (ω) = pk whenever
xv (ω) ∈ Pk . Then sl is measurable.
Proof. Fix some measurable set X ⊆ R. Write Xk = X ∩ Pk and note that it is measurable since
Pk is measurable. It suffices to show that each set (sl )−1 (Xk ) is measurable; if so, (sl )−1 (X) is
a countable union of measurable sets and so measurable. If pk ∈
/ Xk then (sl )−1 (Xk ) = ∅ and so
measurable. If pk ∈ Xk then (sl )−1 (Xk ) = (xv )−1 (Xk ). Since xv is measurable, (xv )−1 (Xk ) is a
measurable set, as required.
30
Lemma A8 Fix (sl , sv ) ∈ Sl × Sv satisfying the following properties. The strategy sl ∈ Sl has (i)
√
√
√
√
√
sl (ω) = 0 if xv (ω) ∈ [− 12 B, 12 B], (ii) sl (ω) = − B if − 21 B > xv (ω), and (iii) sl (ω) = B
√ √
√ √
√
if xv (ω) > 12 B. For all p ∈ [− B, B], sv (p) = 1 if and only if p ∈ {− B, B}. Then (sl , sv )
is a Bayesian Equilibrium.
Proof. Condition (i) (of a Bayesian Equilibrium) follows from Lemma A7. Next, fix some ω ∈ Ω.
If sl (ω) = 0, ul (sl (ω) , sv (sl (ω))) = 0. If sl (ω) 6= 0, then ul (sl (ω) , sv (sl (ω))) = −B + B = 0.
√ √
For any p ∈ R\{0, − B, B}, 0 > −p2 = ul (p, sv (p)). This establishes Condition (ii).
Remark A1 The strategy profile (sl , sv ) ∈ Sl × Sv in the statement of Lemma A8 is an Extreme
Two-Sided Voting Rule. An argument analogous to the proof of Lemma A8 establishes that any
Extreme Two-Sided Voting Rule is a Bayesian Equilibrium.
Recall from the text that Ω+ = {ω ∈ Ω : xv (ω) ≥ 0}.
Lemma A9 Fix a Bayesian Equilibrium (rl , rv ) ∈ Sl × Sv where rl is a constant strategy. Then
there is a Bayesian Equilibrium (sl , sv ) ∈ Sl × Sv and a non-empty set of states Φ ⊆ Ω with
uv (ω, sl (ω)) ≥ uv (ω, rl (ω))
for all ω ∈ Ω
uv (ω, sl (ω)) > uv (ω, rl (ω))
for all ω ∈ Φ.
Proof. Fix a Bayesian Equilibrium (rl , rv ) where rl is a constant strategy. We will assume that, for
all ω ∈ Ω, rl (ω) = p ≥ 0. (A corresponding argument will work when p ≤ 0.) It will be convenient
to break up the proof into two cases. The first corresponds to p > 0. The second corresponds to
p = 0.
Case A (p > 0): First notice that rv (p) = 1. If not,
ul (0, rv (0)) > −p2 = ul (p, rv (p)) ,
contradicting Condition (ii) of a Bayesian Equilibrium. Also note that B ≥ p2 . If not,
ul (0, rv (0)) ≥ −B + B
> −p2 + B = ul (p, rv (p)) ,
contradicting Condition (ii) of a Bayesian Equilibrium. Construct (sl , sv ) as follows. Let
(
sl (ω) =
−p
if ω ∈ Ω\Ω+
p
if ω ∈ Ω+ ,
and let sv (p) = 1 if and only if p ∈ {−p, p}. By Lemma A7, sl is measurable. Also notice that, for
any ω ∈ Ω,
ul (sl (ω) , sv (sl (ω))) = −p2 + B ≥ 0,
31
since B > p2 . For any p ∈ R,
(
ul (p, sv (p)) =
−p2 + B
if p ∈ {−p, p}
−p2
otherwise
.
Put together, this establishes Condition (ii) of a Bayesian Equilibrium. As such, (sl , sv ) is a
Bayesian Equilibrium.
Now note that, for any ω ∈ Ω\Ω+ , we have
h
i
uv (ω, sl (ω)) = − p2 + 2pxv (ω) + xv (ω)2
h
i
> − p2 − 2pxv (ω) + xv (ω)2 = uv (ω, rl (ω)) .
Take Φ = Ω\Ω+ and note Φ is non-empty since xv is surjective. Now note that, for ω ∈ Ω+ ,
sl (ω) = rl (ω) and so uv (ω, sl (ω)) = uv (ω, rl (ω)), as required.
Case B (p = 0): Choose (sl , sv ) to be a Bayesian Equilibrium, as in the statement of Lemma
√
√
A8. When xv (ω) ∈ [− 12 B, 21 B], sl (ω) = rl (ω) and so uv (ω, sl (ω)) = uv (ω, rl (ω)). Next,
√
suppose that − 21 B > xv (ω). Here,
h
i
√
uv (ω, sl (ω)) = − B + 2 Bxv (ω) + xv (ω)2
> − [xv (ω)]2 = uv (ω, rl (ω)) ,
√
√
where the second line follows from the fact that − 12 B > xv (ω) and so 0 > B + 2 Bxv (ω).
√
Similarly, for xv (ω) > 12 B,
h
i
√
uv (ω, sl (ω)) = − B − 2 Bxv (ω) + xv (ω)2
> − [xv (ω)]2 = uv (ω, rl (ω)) ,
√
√
where
the second line uses the fact
that xv (ω) > 12 B and so 0 > B − 2 Bxv (ω). Take Φ =
n
o
√
√
ω ∈ Ω : xv (ω) ∈
/ [− 12 B, 12 B] . Since xv is surjective, Φ is non-empty.
√
Lemma A10 Fix p ∈ (0, B) and construct (sl , sv ) ∈ Sl × Sv as follows: Let sl (ω) = −p if
ω ∈ Ω\Ω+ , sl (ω) = p if ω ∈ Ω+ \ (xv )−1 ({0}), and sl (ω) ∈ {−p, p} otherwise. For all p ∈ [−p, p],
sv (p) = 1 if and only if p ∈ {−p, p}. Then (sl , sv ) is a Bayesian Equilibrium.
Proof. Lemma A7 establishes that sl is measurable. To establish Condition (ii) of a Bayesian
Equilibrium, fix some ω ∈ Ω. If p ∈ (−p, p), then
ul (sl (ω) , sv (sl (ω))) = −p2 + B
> −B + B
≥ −p2 = ul (p, sv (p)) .
32
If p ∈
/ (−p, p), then p2 ≥ p2 so that
ul (sl (ω) , sv (sl (ω))) = −p2 + B
≥ −p2 + B ≥ ul (p, sv (p)) ,
as required.
Remark A2 Any Moderate Two-Sided Voting Rule satisfies the conditions of Lemma A10.
√
Lemma A11 Fix p ∈ (0, B) and some associated strategy profile, viz. (sl , sv ) ∈ Sl × Sv , as in
the statement of Lemma A10. Let (rl , rv ) ∈ Sl × Sv be a Bayesian Equilibrium where rv (p) = sv (p)
for all p ∈ [−p, p]. Then
uv (ω, sl (ω)) ≥ uv (ω, rl (ω))
for all ω ∈ Ω.
Moreover, there exists some non-empty set of states Φ ⊆ Ω with
uv (ω, sl (ω)) > uv (ω, rl (ω))
for all ω ∈ Φ
if and only if there exists some ω ∈ (xv )−1 (R\ {0}) with rl (ω) 6= sl (ω).
√
Proof. Fix some p ∈ (0, B). Also fix (sl , sv ) and (rl , rv ) as in the statement of the Lemma. Take
Φ = (xv )−1 (R\ {0}) ∩ {ω ∈ Ω : rl (ω) 6= sl (ω)} .
We need to show that (i) for any ω ∈ Ω, uv (ω, sl (ω)) ≥ uv (ω, rl (ω)), and (ii) ω ∈ Ω if and only
if uv (ω, sl (ω)) > uv (ω, rl (ω)). To do so, we will break the proof into two parts. First, we will
show that if ω ∈ Φ, then uv (ω, sl (ω)) > uv (ω, rl (ω)). Second, we will show that if ω ∈ Ω\Φ, then
uv (ω, sl (ω)) = uv (ω, rl (ω)).
In showing these facts, it is useful to note that rl (Ω) ⊆ {−p, p}. This follows from the fact that
(rl , rv ) is a Bayesian Equilibrium and Lemma A4.
Begin with some ω ∈ Φ. If ω ∈ Ω\Ω+ , then rl (ω) = p. Use the fact that 0 > xv (ω) to get
i
h
uv (ω, sl (ω)) = − p2 + 2pxv (ω) + xv (ω)2
h
i
> − p2 − 2pxv (ω) + xv (ω)2 = uv (ω, rl (ω)) .
Also, if ω ∈ Ω+ \ (xv )−1 ({0}), then rl (ω) = −p. Now, xv (ω) > 0 implies
h
i
uv (ω, sl (ω)) = − p2 − 2pxv (ω) + xv (ω)2
h
i
> − p2 + 2pxv (ω) + xv (ω)2 = uv (ω, rl (ω)) .
33
Taken together, these displays establish that if ω ∈ Φ, then uv (ω, sl (ω)) > uv (ω, rl (ω)). Next, fix
ω ∈ Ω\Φ. Either xv (ω) = 0 or sl (ω) = rl (ω). If the former,
uv (ω, sl (ω)) = −p2 = uv (ω, rl (ω)) .
If the latter, it is immediate that uv (ω, sl (ω)) = uv (ω, rl (ω)).
√
Lemma A12 Fix p ∈ (0, B) and some associated strategy profile, viz. (sl , sv ) ∈ Sl × Sv , as in
the statement of Lemma A10. Then (sl , sv ) is State-by-State Optimal.
Proof. Fix a Bayesian Equilibrium (rl , rv ) ∈ Sl × Sv . We will show that either (i) for all ω ∈ Ω,
uv (ω, sl (ω)) = uv (ω, rl (ω)) or (ii) there exists ω ∈ Ω with uv (ω, sl (ω)) > uv (ω, rl (ω)).
√
Proposition 5.1 tells us that, there exists some q ∈ [0, B] such that rl (Ω) ⊆ {0, −q, q}. We
will make use of this fact below.
Case A: Here q 6= p. Fix some ω ∈ Ω with xv (ω) = p. (Since xv is surjective, such a state
exists.) Then
uv (ω, sl (ω)) = 0 > − (rl (ω) − p)2 = uv (ω, rl (ω)) ,
as required.
Case B: Here q = p. Proposition 5.1 tells us that rl (Ω) ⊆ {−p, p}. If sl (Ω) = rl (Ω) then
certainly we have (i). So suppose there is a state ω ∈ Ω with sl (ω) 6= rl (ω). If the state ω is
contained in Ω\Ω+ then 0 > xv (ω) and rl (ω) = p. It follows that
h
i
uv (ω, sl (ω)) = − p2 + 2pxv (ω) + xv (ω)2
h
i
> − p2 − 2pxv (ω) + xv (ω)2 = uv (ω, rl (ω)) ,
as required. An analogous argument works when the state ω is contained in Ω+ \ (xv )−1 ({0}).
Since xv (ω) > 0 and rl (ω) = −p,
h
i
uv (ω, sl (ω)) = − p2 − 2pxv (ω) + xv (ω)2
h
i
> − p2 + 2pxv (ω) + xv (ω)2 = uv (ω, rl (ω)) ,
as required. Finally, if ω ∈ (xv )−1 ({0}) then certainly uv (ω, sl (ω)) = uv (ω, rl (ω)).
Lemma A13 Define
Z
p=
Ω+
Z
xv (ω) dµ (ω) −
Ω\Ω+
xv (ω) dµ (ω) .
Fix a strategy sl ∈ Sl where sl (ω) = −p if ω ∈ Ω\Ω+ , sl (ω) = p if ω ∈ Ω+ \ (xv )−1 ({0}), and
sl (ω) ∈ {−p, p} otherwise. Let rl ∈ Sl be a strategy where, for some p ≥ 0, rl (ω) = −p if
ω ∈ Ω\Ω+ , rl (ω) = p if ω ∈ Ω+ \ (xv )−1 ({0}), and rl (ω) ∈ {−p, p} otherwise. Then
Z
Ω
Z
uv (ω, sl (ω)) dµ (ω) ≥
34
Ω
uv (ω, rl (ω)) dµ (ω) .
Proof. Note that
Z
Z
Z
2
uv (ω, rl (ω)) dµ (ω) = −
(p − xv (ω)) dµ (ω) −
(−p − xv (ω))2 dµ (ω)
Ω
Ω+
Ω\Ω+
Z h
Z
i
h
i
2
2
=−
p − 2pxv (ω) + xv (ω) dµ (ω) −
p2 + 2pxv (ω) + xv (ω)2 dµ (ω)
Ω+
Ω\Ω+
"Z
# Z
Z
= −p2 + 2p
Ω+
Choosing p ∈ R to maximize
R
Ω uv
xv (ω) dµ (ω) −
p=
Ω+
R
Ω uv
xv (ω) dµ (ω) −
Ω
xv (ω)2 dµ (ω) .
(ω, rl (ω)) dµ (ω) requires
Z
(The second derivative of
Ω\Ω+
Z
xv (ω) dµ (ω) −
Ω\Ω+
xv (ω) dµ (ω) .
(ω, rl (ω)) dµ (ω) is −2, so that this is indeed a maximum.)
Lemma A14 Suppose that
Z
p=
Ω+
Z
xv (ω) dµ (ω) −
Ω\Ω+
xv (ω) dµ (ω) .
Fix a strategy sl ∈ Sl (viz. rl ∈ Sl ) where sl (ω) = −p (resp. rl (ω) = −q) if ω ∈ Ω\Ω+ , sl (ω) = p
(resp. rl (ω) = q) if ω ∈ Ω+ \ (xv )−1 ({0}), and sl (ω) ∈ {−p, p} (resp. rl (ω) ∈ {−q, q}) otherwise.
If p > p > q, then
Z
Z
uv (ω, sl (ω)) dµ (ω) ≥
uv (ω, rl (ω)) dµ (ω) .
Proof. This follows from the proof of Lemma A13: Note that
d
dp
·Z
Ω+
"Z
¸
uv (ω, rl (ω)) dµ(ω) = 2
#
Z
Ω+
xv (ω) dµ (ω) −
Ω\Ω+
xv (ω) dµ (ω) − 2p
is strictly positive whenever
"Z
Ω+
#
Z
xv (ω) dµ (ω) −
Ω\Ω+
xv (ω) dµ (ω) > p,
as required.
Lemma A15 Let (sl , sv ) ∈ Sl × Sv be as in the statement of Lemma A8. Suppose (rl , rv ) ∈ Sl × Sv
√ √
is a Bayesian Equilibrium where rv (p) = sv (p) for all p ∈ [− B, B]. Then
uv (ω, sl (ω)) ≥ uv (ω, rl (ω))
for all ω ∈ Ω.
Moreover, there exists a non-empty set of states Φ ⊆ Ω with
uv (ω, sl (ω)) > uv (ω, rl (ω))
35
for all ω ∈ Φ
if and only if there exists ω ∈ Ω satisfying one of the following properties:
√
√
A. xv (ω) ∈
/ {− 12 B, 12 B} and rl (ω) 6= sl (ω);
√
√
B. xv (ω) = − 12 B and rl (ω) = B;
√
√
C. xv (ω) = 12 B and rl (ω) = − B.
√ √
Proof. Fix a Bayesian Equilibrium (rl , rv ). By Lemma A5, rl (Ω) ⊆ {0, − B, B}. We will
show that (i) for all ω ∈ Ω, uv (ω, sl (ω)) ≥ uv (ω, rl (ω)) and (ii) there exists some ω ∈ Ω with
uv (ω, sl (ω)) > uv (ω, rl (ω)) if and only if there exists some ω ∈ Ω satisfying any one of Conditions
A, B, or C in the statement of the Lemma.
To do so, we will take
½
½
¾
¾
1√ 1√
ΦA = ω ∈ Ω : xv (ω) ∈
/ −
B,
B and rl (ω) 6= sl (ω)
2
2
¾
½
√
1√
B and rl (ω) = B
ΦB = ω ∈ Ω : xv (ω) = −
2
½
¾
√
1√
ΦC = ω ∈ Ω : xv (ω) =
B and rl (ω) = − B .
2
First, we will show that if ω ∈ ΦA ∪ ΦB ∪ ΦC , then uv (ω, sl (ω)) > uv (ω, rl (ω)). Second, we will
show that if ω ∈ Ω\ (ΦA ∪ ΦB ∪ ΦC ), then uv (ω, sl (ω)) = uv (ω, rl (ω)).
√
√
√ √
Case 1A: Fix some ω ∈ ΦA . Here, xv (ω) ∈ (− 21 B, 21 B) if and only if rl (ω) ∈ {− B, B}.
√
If rl (ω) = − B,
uv (ω, sl (ω)) = − [xv (ω)]2
i
h
√
> − B + 2 Bxv (ω) + xv (ω)2 = uv (ω, rl (ω)) ,
√
√
where the inequality follows from the fact that xv (ω) > − 21 B and so B + 2 Bxv (ω) > 0. If
√
rl (ω) = B,
uv (ω, sl (ω)) = − [xv (ω)]2
h
i
√
> − B − 2 Bxv (ω) + xv (ω)2 = uv (ω, rl (ω)) ,
√
√
where the inequality follows from the fact that 12 B > xv (ω) and so B − 2 Bxv (ω) > 0. Next,
√
√
notice that − 21 B > xv (ω) if and only if rl (ω) ∈ {0, B}. If rl (ω) = 0,
h
i
√
uv (ω, sl (ω)) = − B + 2 Bxv (ω) + xv (ω)2
> − [xv (ω)]2 = uv (ω, rl (ω)) ,
36
√
√
where the inequality follows from the fact that − 21 B > xv (ω) and so 0 > B + 2 Bxv (ω). If
√
rl (ω) = B,
h
i
√
uv (ω, sl (ω)) = − B + 2 Bxv (ω) + xv (ω)2
i
h
√
> − B − 2 Bxv (ω) + xv (ω)2 = uv (ω, rl (ω)) ,
where the inequality follows from the fact that xv (ω) is strictly negative.
√
√
Finally, notice that xv (ω) > 21 B if and only if rl (ω) ∈ {0, − B}. If rl (ω) = 0,
i
h
√
uv (ω, sl (ω)) = − B − 2 Bxv (ω) + xv (ω)2
> − [xv (ω)]2 = uv (ω, rl (ω)) ,
where the inequality follows from the fact that xv (ω) >
√
rl (ω) = − B,
1
2
√
√
B and so 0 > B − 2 Bxv (ω). If
h
i
√
uv (ω, sl (ω)) = − B − 2 Bxv (ω) + xv (ω)2
h
i
√
> − B + 2 Bxv (ω) + xv (ω)2 = uv (ω, rl (ω)) ,
where the inequality follows from the fact that xv (ω) is strictly positive.
Case 1B-C: Fix some ω ∈ ΦB ∪ ΦC . In either case,
µ
¶
B
B
= uv (ω, rl (ω)) ,
uv (ω, sl (ω)) = − > − 2B +
4
4
as required.
Case 2: Now fix some ω ∈ Ω\ (ΦA ∪ ΦB ∪ ΦC ). If rl (ω) = sl (ω), then certainly uv (ω, sl (ω)) =
√
uv (ω, rl (ω)). By definition of ΦA ∪ ΦB ∪ ΦC , if rl (ω) 6= sl (ω) then either (i) xv (ω) = − 21 B and
√
√
√
rl (ω) = − B or (ii) xv (ω) = 12 B and rl (ω) = B. In either case,
uv (ω, sl (ω)) = −
B
= uv (ω, rl (ω)) ,
4
as required.
Lemma A16 Any Extreme Two-Sided voting rule is State-by-State Optimal.
Proof. Let (sl , sv ) be an Extreme Two-Sided Voting Rule as in the statement of Lemma A8. It
suffices to show that (sl , sv ) is State-by-State Optimal. (If (ql , qv ) is another Extreme Two-Sided
Voting Rule, then uv (ω, ql (ω)) = uv (ω, sl (ω)) for all ω ∈ Ω. So, if (sl , sv ) is State-by-State
Optimal, (ql , qv ) must also be.)
Fix a Bayesian Equilibrium (rl , rv ) ∈ Sl × Sv . First, we will show that either (i) for all ω ∈ Ω,
uv (ω, sl (ω)) = uv (ω, rl (ω)) or (ii) there exists ω ∈ Ω with uv (ω, sl (ω)) > uv (ω, rl (ω)). Fix some
37
√
state ω ∈ Ω with xv (ω) = − B. We know that such a state exists since xv is surjective. If
sl (ω) 6= rl (ω), then
uv (ω, sl (ω)) = 0 > −(rl (ω) +
√ 2
B) = uv (ω, rl (ω)) ,
√
as required. And, similarly, for a state ω ∈ Ω with xv (ω) = B. So, suppose that sl (ω) = rl (ω)
√ √
√ √
whenever xv (ω) ∈ {− B, B}. Since (rl , rv ) is a Bayesian Equilibrium, for all p ∈ [− B, B],
√ √
√ √
rv (p) = 1 if and only if p ∈ {− B, B}. Put differently, rv (p) = sv (p) whenever p ∈ [− B, B].
Now, by Lemma A15, (i) or (ii) must hold.
Proof of Proposition 5.2. First, we show that any equilibrium that is both Expectationally
Optimal and State-by-State Optimal must take the form of either a Moderate or an Extreme
Two-Sided Voting Rule. Then, we use this fact to show that there exists an Expectationally and
State-by-State Optimal Equilibrium.
Let (s∗l , s∗v ) be an equilibrium that is both Expectationally and State-by-State Optimal. It
must take the form of Part (i), Part (ii), or Part (iii) of Proposition 5.1. Using Lemma A9, (s∗l , s∗v )
cannot take the form of Part (i) in Proposition 5.1. Suppose that it takes the form of Part (ii) in
Proposition 5.1. Then, by Lemmata A11 and A13, (s∗l , s∗v ) is a Moderate Two-Sided Voting Rule.
Finally, suppose that (s∗l , s∗v ) takes the form of Part (iii) in Proposition 5.1. Then, by Lemmata A9
and A15, (s∗l , s∗v ) is an Extreme Two-Sided Voting Rule.
Now, turn to the question of existence. We know that if there exists an Expectationally and
State-by-State Optimal Equilibrium, it must take the form of either a Moderate or an Extreme
Two-Sided Voting Rule. By Lemmata A12 and A16, any such profile is State-by-State Optimal. So
we must only show that either a Moderate or an Extreme Two-Sided Voting rule is Expectationally
Optimal. For this, notice that any two Moderate (resp. Extreme) Two-Sided Voting Rules yield
the same expected payoffs for the Voter. (See Lemmata A11 and A15.)
Define
Z
∗
p =
Ω+
Z
xv (ω) dµ (ω) −
Ω\Ω+
xv (ω) dµ (ω) .
√
If p∗ ∈ (0, B), compare the expected payoffs under any Moderate vs. any Extreme Two-Sided
Voting Rule. Whichever yields higher expected
will be both Expectationally (and State-by³ √ payoffs
´
∗
State) Optimal. Next, suppose that p ∈
/ 0, B . Then, any Extreme Two-Sided Voting Rule is
Expectationally (and State-by-State) Optimal. For p∗ = 0, this follows from Case B in the proof of
√
√
Lemma A9. For p∗ = B, this follows from Lemma A15. For p∗ > B, this follows from Lemmata
A14 and A15.
Appendix B: Results for Section 6
Throughout this Appendix, we will make use of the following results.
√ √
Lemma B1 Fix a Bayesian Equilibrium (sl , sv ) ∈ Sl × Sv . Then sl (Ω) ⊆ [− B, B].
38
Proof. Suppose not. Then there exists some ω ∈ Ω with [sl (ω)]2 > B. At this state,
Eul (0, sv (ω, 0, ι)) ≥ −B + B
> − [sl (ω)]2 + B ≥ Eul (sl (ω) , sv (ω, sl (ω) , ι)) ,
contradicting Condition (ii) of a Bayesian Equilibrium.
Lemma B2 Let sl ∈ Sl be a measurable strategy. Fix some measurable set X ⊆ R and construct
rl ∈ Sl so that
(
rl (ω) =
xv (ω)
if xv (ω) ∈ X
sl (ω)
otherwise.
Then rl is also a measurable strategy.
Proof. Fix some measurable set Y ⊆ R. Then the sets Y ∩X and Y \ (Y ∩ X) are both measurable.
It suffices to show that (rl )−1 (Y ∩ X) and (rl )−1 (Y \ (Y ∩ X)) are measurable. The former comes
from the fact that (rl )−1 (Y ∩ X) = (xv )−1 (Y ∩ X) and xv is measurable. The latter comes from
the fact that (rl )−1 (Y \ (Y ∩ X)) = (sl )−1 (Y \ (Y ∩ X)) and sl is measurable.
We begin with the case of π ≥ 12 . We show that there exists a Bayesian Equilibrium where the
√ √
Legislator chooses the Voter’s ideal point whenever it is contained in [− B, B], and otherwise
√ √
some element of {− B, B} as per the Voter’s preference. We then show that such an equilibrium
is indeed Expectationally and State-by-State Optimal. Lemma B1 is the key to this last step.
Lemma B3 Let π ≥ 12 . Then there exists a Bayesian equilibrium (sl , sv ) ∈ Sl × Sv where:
√ √
(i) If xv (ω) ∈ [− B, B] then sl (ω) = xv (ω);
√
√
(ii) If − B > xv (ω) then sl (ω) = − B;
√
√
(iii) If xv (ω) > B then sl (ω) = B;
Proof. Let sl ∈ Sl be as in the statement of the Lemma. Define sv ∈ Sv as follows: If xv (ω) ∈
√ √
√
√
[− B, B], let sv (ω, p, i) = 1 if and only if p = xv (ω). If − B > xv (ω) (resp. xv (ω) > B),
√
√
let sv (ω, p, i) = 1 if and only if p = − B (resp. p = B). For any ω ∈ Ω, let sv (ω, p, ni) = 1 if
and only if p2 ≥ πB. We will show that (sl , sv ) is indeed a Bayesian Equilibrium.
By Lemmata A7 and B2, the strategy sl ∈ Sl is measurable. To verify that Condition (ii) of
a Bayesian Equilibrium is satisfied, fix some ω ∈ Ω and some policy p ∈ R with p 6= sl (ω). Note
that if πB > p2 , then
0 ≥ −p2 = Eul (p, sv (ω, p, ι)) .
If p2 ≥ πB then
0 ≥ −p2 + πB
≥ −p2 + (1 − π) B = Eul (p, sv (ω, p, ι)) ,
39
where the second inequality comes from the fact that π ≥
1
2.
So, using the above, it suffices to
show that Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ 0. First, consider the case where πB > [xv (ω)]2 . Here
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = −xv (ω)2 + πB > 0,
as required. If [xv (ω)]2 ≥ πB then
(
Eul (sl (ω) , sv (ω, sl (ω) , ι)) =
−B + B = 0
if [xv (ω)]2 ≥ B
−xv (ω)2 + B > 0
if B > [xv (ω)]2 ,
as required.
Proof of Proposition 6.1 . Fix (s∗l , s∗v ) as in the statement of the Proposition. By Lemma B3,
we can choose (s∗l , s∗v ) so that it is indeed a Bayesian Equilibrium. First, we will show that, for
any Bayesian Equilibrium, viz. (rl , rv ), and any ω ∈ Ω we have uv (ω, s∗l (ω)) ≥ uv (ω, rl (ω)). From
this, it immediately follows that (s∗l , s∗v ) is Expectationally and State-by-State Optimal.
√ √
Fix a Bayesian Equilibrium (rl , rv ) and a state ω ∈ Ω. If xv (ω) ∈ [− B, B],
uv (ω, s∗l (ω)) = 0 = arg max {uv (ω, p)} .
p∈R
√
If − B > xv (ω),
√
rl (ω) − xv (ω) ≥ − B − xv (ω) > 0,
√
since rl (ω) ≥ − B (Lemma B1). Then
³ √
´2
uv (ω, s∗l (ω)) = − − B − xv (ω)
≥ − (rl (ω) − xv (ω))2 = uv (ω, rl (ω)) ,
as required. If xv (ω) >
√
B,
0>
since
√
B − xv (ω) ≥ rl (ω) − xv (ω) ,
√
B ≥ rl (ω) (Lemma B1). Then
uv (ω, s∗l (ω)) = −
³√
´2
B − xv (ω)
≥ − (rl (ω) − xv (ω))2 = uv (ω, rl (ω)) ,
as required.
Next, fix an Expectationally and State-by-State Optimal Equilibrium, viz. (sl , sv ). We will
show that it must satisfy the conditions of the Proposition.
ω ∈ Ω,
uv (ω, s∗l (ω))
To see this, recall that, for all
≥ uv (ω, sl (ω)). So, if (sl , sv ) is State-by-State Optimal, we must have that
40
uv (ω, s∗l (ω)) = uv (ω, sl (ω)) for all ω ∈ Ω.
√ √
First, fix ω ∈ Ω with xv (ω) ∈ [− B, B]. Then uv (ω, sl (ω)) = uv (ω, s∗l (ω)) = 0 and
√
√
so sl (ω) = xv (ω). Next fix ω ∈ Ω with − B > xv (ω). By Lemma B1, sl (ω) ≥ − B. If
√
sl (ω) > − B then
√
sl (ω) − xv (ω) > − B − xv (ω) > 0,
and so
³ √
´2
uv (ω, s∗l (ω)) = − − B − xv (ω)
> − (sl (ω) − xv (ω))2 = uv (ω, sl (ω)) ,
√
√
a contradiction. So sl (ω) = − B. Finally, fix ω ∈ Ω with xv (ω) > B. By Lemma B1,
√
√
B ≥ sl (ω). If B > sl (ω) then
0>
√
B − xv (ω) > sl (ω) − xv (ω) ,
so that
uv (ω, s∗l (ω)) = −
³√
´2
B − xv (ω)
> − (sl (ω) − xv (ω))2 = uv (ω, sl (ω)) ,
a contradiction. So sl (ω) =
√
B.
Now we turn to the case where π ∈ (0, 12 ). First, we show a claim discussed in the text.
Lemma B4 Fix some p ∈ [0,
p
Then there exists a Bayesian Equilibrium, viz.
p
p
(sl , sv ) ∈ Sl × Sv , with sl (ω) = xv (ω) whenever xv (ω) ∈ [− p2 + πB, −p] ∪ [p, p2 + πB].
Proof. Fix some p ∈ [0,
p
(1 − π) B].
p
p
(1 − π) B] and define X = [− p2 + πB, −p] ∪ [p, p2 + πB]. Let sl
be a strategy with sl (ω) = xv (ω) if xv (ω) ∈ X and sl (ω) = p otherwise. Construct the strategy
sv so that (i) sv (ω, p, i) = 1 if and only if p = xv (ω) and (ii) sv (ω, p, ni) = 1 if and only if p ∈ X.
We will show that (sl , sv ) is a Bayesian equilibrium.
Condition (i) follows from Lemma B2.
For Condition (ii), begin by fixing some state ω. If xv (ω) ∈ X then
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = −xv (ω)2 + B
≥ −p2 + (1 − π) B,
where the inequality follows from the fact that xv (ω) ∈ X. If xv (ω) ∈ R\X then
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = −p2 + (1 − π) B.
41
Now fix p 6= sl (ω). If p ∈ X then, using the fact that p2 ≥ p2 , we have
Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ −p2 + (1 − π) B
≥ −p2 + (1 − π) B = Eul (p, sv (ω, p, ι)) .
If p ∈ R\X then
Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ −p2 + (1 − π) B
≥ − (1 − π) B + (1 − π) B
≥ −p2
= Eul (p, sv (ω, p, ι)) ,
where the second inequality follows from the fact that (1 − π) B ≥ p2 and the third inequality
follows by construction. This establishes the result.
p
Lemma B4 establishes, for any policy p ∈ [0, (1 − π) B], there exists some Bayesian Equilibrium such that the Legislator chooses the Voter’s ideal point whenever it is contained in
p
p
[− p2 + πB, −p] ∪ [p, p2 + πB].
It does not establish that a State-by-State Optimal equilibrium takes this form. Taken together,
Lemmata B5-B6 will establish this fact.
p
√
Lemma B5 Fix a game with π ∈ (0, 12 ) and some policy p ∈ [ (1 − π) B, B]. Let (rl , rv ) ∈
Sl × Sv be a Bayesian Equilibrium where rv (ω, p, ni) = 0, for all p ∈ (−p, p). Then there exists a
Bayesian Equilibrium (sl , sv ) ∈ Sl × Sv and some set of states Φ ⊆ Ω with
and Φ 6= ∅ whenever p >
uv (ω, sl (ω)) ≥ uv (ω, rl (ω))
for all ω ∈ Ω
uv (ω, sl (ω)) > uv (ω, rl (ω))
for all ω ∈ Φ,
p
(1 − π) B.
Proof. Fix a Bayesian Equilibrium (rl , rv ) as in the statement of the Lemma. By Lemma B1,
√ √
rl (Ω) ⊆ [− B, B]. Suppose that there exists some ω ∈ Ω with p2 > [rl (ω)]2 > πB. Then
rv (ω, rl (ω) , ni) = 0 and so
Eul (0, rv (ω, 0, ι)) ≥ 0
> − [rl (ω)]2 + πB
≥ Eul (rl (ω) , rv (ω, rl (ω) , ι)) ,
contradicting Condition (ii) of a Bayesian Equilibrium. It follows that
√
√
√
√
rl (Ω) ⊆ [− B, −p] ∪ [− πB, πB] ∪ [p, B].
42
We will show that we can construct a Bayesian Equilibrium, viz. (sl , sv ), where (i) uv (ω, sl (ω)) ≥
uv (ω, rl (ω)) for all ω ∈ Ω and (ii) uv (ω, sl (ω)) > uv (ω, rl (ω)) for all ω ∈ Φ. This construcp
tion will be such that Φ 6= ∅ whenever p > (1 − π) B. It will be convenient to define the set
p
p
X = [−p, − (1 − π) B] ∪ [ (1 − π) B, p]. Construct (sl , sv ) as follows: Let
(
sl (ω) =
xv (ω)
if xv (ω) ∈ X
rl (ω)
if xv (ω) ∈ R\X.
If xv (ω) ∈ X, take sv (ω, p, i) = 1 if and only if p = xv (ω). If xv (ω) ∈ R\X, take sv (ω, p, i) = 1 if
and only if p = rl (ω). Finally, for all ω ∈ Ω, take sv (ω, p, ni) = 1 if and only if p2 ≥ (1 − π) B.
We will begin by showing that (sl , sv ) is a Bayesian Equilibrium. Condition (i) follows from
Lemma B2. To establish Condition (ii), fix some ω ∈ Ω and some policy p 6= sl (ω). If (1 − π) B >
p2 then
0 ≥ −p2 = Eul (p, sv (ω, p, ι)) .
If p2 ≥ (1 − π) B then
0 = − (1 − π) B + (1 − π) B
≥ −p2 + (1 − π) B = Eul (p, sv (ω, p, ι)) .
So, it suffices to show that Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ 0.
To see this, first fix xv (ω) ∈ X. Here,
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [xv (ω)]2 + B
≥ −p2 + B
≥ 0,
establishing the desired result. Next, suppose xv (ω) ∈ R\X. Here sl (ω) = rl (ω). Recall that
√
√
√
√
rl (ω) ∈ [− B, −p] ∪ [− πB, πB] ∪ [p, B].
√
√
If rl (ω) ∈ [− πB, πB] then sv (ω, sl (ω) , ni) = 0, so that
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [rl (ω)]2 + πB
≥ Eul (rl (ω) , rv (ω, rl (ω) , ι))
≥ Eul (0, rv (ω, 0, ι))
≥ 0,
43
where the first inequality follows from the fact that rv (ω, rl (ω) , ni) = 0 and the second follows
√
√
from the fact that (rl , rv ) is a Bayesian Equilibrium. If rl (ω) ∈ [− B, −p] ∪ [p, B], then
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [rl (ω)]2 + B
≥ −B + B,
as required.
p
If p = (1 − π) B, take Φ = ∅. If not, take
³
´
p
p
Φ = (xv )−1 (−p, − (1 − π) B] ∪ [ (1 − π) B, p) .
So, when p >
p
(1 − π) B, Φ must be non-empty since xv is surjective. We will show that (sl , sv )
satisfies the desired properties, given this set Φ.
First, fix ω ∈ Φ. Then, rl (ω) 6= xv (ω) since
1
2
> π and
√
√
√
√
rl (ω) ∈ [− B, −p] ∪ [− πB, πB] ∪ [p, B].
With this
uv (ω, sl (ω)) = 0 > uv (ω, rl (ω)) ,
as stated. Next, fix ω ∈ Ω\Φ. If xv (ω) ∈ R\ {−p, p} then sl (ω) = rl (ω). From this, it follows that
uv (ω, sl (ω)) = uv (ω, rl (ω)) .
If xv (ω) ∈ {−p, p} then sl (ω) = xv (ω) so that
uv (ω, sl (ω)) = 0 ≥ uv (ω, rl (ω)) ,
as desired.
Lemma B6 Fix some π > 0 and a Bayesian equilibrium of the associated game, viz. (rl , rv ) ∈
Sl × Sv . Let p ∈ R be the greatest positive lower bound on the
set of policies o
with rv (ω, p, ni) = 1
np
√
2
(for any ω ∈ Ω) given the ordering relation P. Let q = min
p + πB, B . Define Φ so that,
if p ≥ q, Φ = ∅ and otherwise
Φ = (xv )−1 ([−q, −p] ∪ [p, q]) ∩ {ω ∈ Ω : rl (ω) 6= xv (ω)} .
Then there exists a Bayesian Equilibrium, viz. (sl , sv ) ∈ Sl × Sv , with
uv (ω, sl (ω)) ≥ uv (ω, rl (ω))
for all ω ∈ Ω
uv (ω, sl (ω)) > uv (ω, rl (ω))
for all ω ∈ Φ.
44
Proof. Fix a Bayesian equilibrium (rl , rv ) as in the statement of the Lemma. We will show that
we can construct a Bayesian Equilibrium, viz. (sl , sv ), satisfying the desired properties. If p ≥ q,
take (sl , sv ) = (rl , rv ). The result holds trivially. So, for the remainder of the proof, assume that
q > p. It will be convenient to define X = [−q, −p] ∪ [p, q]. Construct (sl , sv ) as follows: Let
(
sl (ω) =
xv (ω)
if xv (ω) ∈ X
rl (ω)
if xv (ω) ∈ R\X.
If xv (ω) ∈ X, let sv (ω, p, i) = 1 if and only if p = xv (ω). If xv (ω) ∈ R\X, let sv (ω, p, i) = 1 if
and only if p = rl (ω). For all ω ∈ Ω, let
(
sv (ω, p, ni) =
0
if p2 > p2
1
if p2 ≥ p2 .
We begin by showing that (sl , sv ) is indeed a Bayesian Equilibrium. Then, we turn to show that
(sl , sv ) satisfies the desired properties.
Condition (i) follows from Lemma B2. We turn to establishing Condition (ii). To do so, it will
be convenient to fix some ω ∈ Ω. First, suppose that xv (ω) ∈ X. Then
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [xv (ω)]2 + B.
Fix p ∈ R with p 6= sl (ω). If p2 > p2 then
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [xv (ω)]2 + B
≥ −B + B
≥ −p2 = Eul (p, sv (ω, p, ι)) ,
where the first inequality follows from the fact that B ≥ q 2 ≥ [xv (ω)]2 . If p2 ≥ p2 ,
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [xv (ω)]2 + B
≥ −q 2 + B
≥ −p2 − πB + B
≥ −p2 + (1 − π) B ≥ Eul (p, sv (ω, p, ι)) ,
where the first inequality follows from the fact that q 2 ≥ [xv (ω)]2 , and the second inequality follows
from the fact that p2 + πB ≥ q 2 .
Now, suppose that xv (ω) ∈ R\X. If p2 > [rl (ω)]2 then rv (ω, rl (ω) , ni) = 0 and so
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [rl (ω)]2 + πB
≥ Eul (rl (ω) , rv (ω, rl (ω) , ι)) .
45
If [rl (ω)]2 ≥ p2 then
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [rl (ω)]2 + B
≥ Eul (rl (ω) , rv (ω, rl (ω) , ι)) .
So, for any xv (ω) ∈ R\X and any policy p ∈ R, we must have
Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ Eul (rl (ω) , rv (ω, rl (ω) , ι))
≥ Eul (p, rv (ω, p, ι)) ,
(B1)
since (rl , rv ) is a Bayesian Equilibrium. In what follows, we will pick p ∈ R with p 6= sl (ω). With
this sv (ω, p, i) = 0. We will divide the argument into three cases.
First suppose p2 > p2 . Then
Eul (0, rv (ω, 0, ι)) ≥ 0
≥ −p2 = Eul (p, sv (ω, p, ι)) ,
so that with Equation B1
Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ Eul (p, sv (ω, p, ι)) ,
as required.
Now fix p2 ≥ p2 . It suffices to show that, whenever p2 > p2 , we mut have
Eul (sl (ω) , sv (ω, sl (ω) , ι)) > Eul (p, sv (ω, p, ι)) .
If so, taking p2 = p2 + ε for ε > 0, we have
Eul (sl (ω) , sv (ω, sl (ω) , ι)) > −p2 − ε + (1 − π) B.
Since this must hold for all ε > 0, we certainly have
Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ −p2 + (1 − π) B = Eul (p, sv (ω, p, ι)) .
So, we turn to establishing the said claim: Fix p2 > p2 . Since p is the greatest positive lower
bound (with respect to the order relation P) on the set of policies that satisfy rv (ω, ·, ni) = 1, there
must exist some policy q (perhaps not distinct from p) with p2 > q 2 ≥ p2 and rv (ω, q, ni) = 1.
Then
Eul (q, rv (ω, q, ι)) ≥ −q 2 + (1 − π) B
> −p2 + (1 − π) B = Eul (p, sv (ω, p, ι)) .
46
With this and Equation B1,
Eul (sl (ω) , sv (ω, sl (ω) , ι)) > Eul (p, sv (ω, p, ι)) .
Now for the second part of the result: namely, showing that (sl , sv ) satisfies the desired properties. First, fix some ω ∈ Φ. Then sl (ω) = xv (ω) and rl (ω) 6= xv (ω), so that uv (ω, sl (ω)) = 0 >
uv (ω, rl (ω)), as stated. Next, fix ω ∈ Ω\Φ. If ω ∈ (xv )−1 (X) then sl (ω) = xv (ω) = rl (ω). If
ω ∈ Ω\ (xv )−1 (X), then sl (ω) = rl (ω). In either case, uv (ω, sl (ω)) = uv (ω, rl (ω)).
Remark B1 Let π ∈ (0, 12 ) and fix a State-by-State Optimal Equilibrium, viz. (s∗l , s∗v ). By Lemma
p
B5, there exists some policy p ∈ [0, (1 − π) B] where sv (ω, p, ni) = 1. With this, q in the statep
ment of Lemma B6 can be taken to be p2 + πB.
Lemma B7 Fix some π ∈ (0, 12 ) and a Bayesian equilibrium of the associated game, viz. (rl , rv ) ∈
Sl × Sv . Let p ∈ R be the greatest positive lower bound on the set of policies with rv (ω, ·, ni) = 1
given the ordering relation P. If p2 ≥ (1 − 2π) B then there exists some policy q ∈ [0, p) and some
set
Φ = (xv )−1 ([−q, q]) ∩ {ω ∈ Ω : rl (ω) 6= xv (ω)}
so that, we can find a Bayesian Equilibrium, viz. (sl , sv ) ∈ Sl × Sv , with
uv (ω, sl (ω)) ≥ uv (ω, rl (ω))
for all ω ∈ Ω
uv (ω, sl (ω)) > uv (ω, rl (ω))
for all ω ∈ Φ.
Proof. Fix a Bayesian Equilibrium (rl , rv ) satisfying the above conditions. Using the argument at
the beginning of the proof of Lemma B5, it follows that
√
√
√
√
rl (Ω) ⊆ [− B, −p] ∪ [− πB, πB] ∪ [p, B].
(We omit repetition of the argument.)
Assume that p2 ≥ (1 − 2π) B. Then p2 − (1 − 2π) B ≥ 0. Let q be a policy with
½q
q = min
Note that since
1
2
p2
¾
√
− (1 − 2π) B, πB .
> π, p > q. Let Φ be as given in the statement of the Lemma. This is well
defined, as q ≥ 0. We want to show that we can construct a Bayesian Equilibrium (sl , sv ) with (i)
uv (ω, sl (ω)) ≥ uv (ω, rl (ω)) for all ω ∈ Ω and (ii) uv (ω, sl (ω)) > uv (ω, rl (ω)) for all ω ∈ Φ. Begin
by constructing sl as
(
sl (ω) =
xv (ω)
if xv (ω) ∈ [−q, q]
rl (ω)
if xv (ω) ∈ R\[−q, q].
47
If xv (ω) ∈ [−q, q], let sv (ω, p, i) = 1 if and only if p = xv (ω). If xv (ω) ∈ R\[−q, q], let sv (ω, p, i) =
1 if and only if p = rl (ω). For all ω ∈ Ω, let
(
sv (ω, p, ni) =
0
if p2 > p2
1
if p2 ≥ p2 .
We begin by showing that (sl , sv ) is indeed a Bayesian Equilibrium. We then turn to show that it
satisfies the desired conditions.
Condition (i) of a Bayesian Equilibrium follows from Lemma B2. We will divide Condition (ii)
into several cases.
Case A: Here xv (ω) ∈ [−q, q].
Fix p 6= xv (ω). When p2 > p2 , we have
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [xv (ω)]2 + πB
≥ −q 2 + πB
≥0
≥ −p2 = Eul (p, sv (ω, p, ι)) ,
where the third line follows from the fact that πB ≥ q 2 . Similarly, when p2 ≥ p2 ,
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [xv (ω)]2 + πB
≥ −q 2 + πB
≥ −p2 + (1 − π) B
≥ −p2 + (1 − π) B ≥ Eul (p, sv (ω, p, ι)) ,
where the third line follows from the fact that p2 − (1 − 2π) B ≥ q 2 .
Case B: Here xv (ω) ∈ R\ [−q, q] and p2 > [rl (ω)]2 .
First, notice that, since (rl , rv ) is a Bayesian Equilibrium, we must have
− [rl (ω)]2 + πB ≥ −p2 + (1 − π) B.
To see this, suppose otherwise, i.e., −p2 + (1 − π) B > − [rl (ω)]2 + πB. Suppose, for every policy
p ∈ R with rv (ω, p, ni) = 1, − [rl (ω)]2 + πB ≥ −p2 + (1 − π) B. Then the greatest lower bound on
the set of policies with rv (ω, ·, ni) = 1 (with respect to the order relation P) is a policy q ∈ R with
q 2 ≥ [rl (ω)]2 + (1 − 2π) B.23 But q 2 > p2 , contradicting p being the greatest positive lower bound
on the set of policies with rv (ω, ·, ni) = 1 (with respect to the order relation P). So, we must have
that there exists some policy p ∈ R with rv (ω, p, ni) = 1, −p2 + (1 − π) B > − [rl (ω)]2 + πB. But
23
Since
1
2
> π, this is well-defined.
48
then
Eul (p, rv (ω, p, ni)) ≥ −p2 + (1 − π) B
> − [rl (ω)]2 + πB ≥ Eul (rl (ω) , rv (ω, rl (ω) , ni)) ,
where the last inequality comes from the fact that rv (ω, rl (ω) , ni) = 0 as p2 > [rl (ω)]2 . This
contradicts (rl , rv ) being a Bayesian Equilibrium.
The takeaway from the above is that
− [rl (ω)]2 + πB ≥ −p2 + (1 − π) B.
It follows that q 2 ≥ [rl (ω)]2 : if q 2 = p2 − (1 − 2π) B this is immediate from the above display. If
p2 − (1 − 2π) B > q 2 then q 2 = πB. Note that πB ≥ [rl (ω)]2 since otherwise
Eul (0, rv (ω, 0, ι)) ≥ −πB + πB
> − [rl (ω)]2 + πB ≥ Eul (rl (ω) , rv (ω, rl (ω) , ι)) ,
contradicting that (rl , rv ) is a Bayesian Equilibrium.
Since q 2 ≥ [rl (ω)]2 ,
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [rl (ω)]2 + πB
≥ −q 2 + πB.
It suffices to show that, for any p ∈ R\{rl (ω)}, −q 2 + πB ≥ Eul (p, sv (ω, p, ι)). If p2 > p2 then
−q 2 + πB ≥ −πB + πB
≥ −p2 = Eul (p, sv (ω, p, ι)) .
If p2 ≥ p2 we have
−q 2 + πB ≥ −p2 + (1 − π) B
≥ −p2 + (1 − π) B ≥ Eul (p, sv (ω, p, ι)) .
Case C: Here xv (ω) ∈ R\[−q, q] and [rl (ω)]2 ≥ p2 . In this case,
Eul (sl (ω) , sv (ω, sl (ω) , ι)) = − [rl (ω)]2 + B
≥ Eul (rl (ω) , rv (ω, rl (ω) , ι)) .
So, using the fact that (rl , rv ) is a Bayesian Equilibrium,
Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ Eul (rl (ω) , rv (ω, rl (ω) , ι)) ≥ Eul (p, rv (ω, p, ι)) ,
49
for all p ∈ R.
Fix p 6= rl (ω). If p2 > p2 , then
Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ Eul (0, rv (ω, 0, ι))
≥ −p2 = Eul (p, sv (ω, p, ι)) ,
as required.
Now suppose p2 ≥ p2 . It suffices to show that, whenever p2 > p2 ,
Eul (sl (ω) , sv (ω, sl (ω) , ι)) > −p2 + (1 − π) B = Eul (p, sv (ω, p, ι)) .
If so, then taking p2 = p2 + ε for any ε > 0,
Eul (sl (ω) , sv (ω, sl (ω) , ι)) > −p2 − ε + (1 − π) B = Eul (p, sv (ω, p, ι)) .
So,
Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ −p2 + (1 − π) B = Eul (p, sv (ω, p, ι)) .
We then turn to establishing the claim: Fix p2 > p2 and suppose, contra hypothesis, that
Eul (p, sv (ω, p, ι)) = −p2 + (1 − π) B ≥ Eul (sl (ω) , sv (ω, sl (ω) , ι)) .
Note, there exists some policy q with p2 > q 2 ≥ p2 and rv (ω, q, ni) = 1.
(If not, this would
contradict p being a greatest lower bound on the set of policies with rv (ω, ·, ni) = 1.) Then, using
the fact that Eul (sl (ω) , sv (ω, sl (ω) , ι)) ≥ Eul (q, rv (ω, q, ι)) as established above, we have
Eul (q, rv (ω, q, ι)) = −q 2 + (1 − π) B
> −p2 + (1 − π) B
≥ Eul (sl (ω) , sv (ω, sl (ω) , ι))
≥ Eul (q, rv (ω, q, ι)) ,
a contradiction.
With this, we have established that (sl , sv ) is a Bayesian Equilibrium. We now show that (sl , sv )
satisfies the desired properties. To see this, first fix ω ∈ Φ. Then sl (ω) = xv (ω) and rl (ω) 6= xv (ω),
so that uv (ω, sl (ω)) = 0 > uv (ω, rl (ω)). For ω ∈ Ω\Φ, we must have sl (ω) = rl (ω) so that
uv (ω, sl (ω)) = uv (ω, rl (ω)).
p
√
Remark B2 In the statement of Lemma B7, we can take q = min{ p2 − (1 − 2π) B, πB}.
Lemma B8 Fix a Bayesian Equilibrium, viz.
(sl , sv ) ∈ Sl × Sv , where, for some policy p,
sv (ω, p, ni) = 1 and p ∈ sl (Ω). Let p be the greatest lower bound on these policies, with respect to
P. If, for some state ω, p2 > [sl (ω)]2 = [xv (ω)]2 then min{p2 − (1 − 2π) B, πB} ≥ [xv (ω)]2 .
50
Proof. Fix a Bayesian Equilibrium, viz. (sl , sv ), satisfying the conditions of the Lemma. Fix also
a state ω with p > sl (ω) = xv (ω). Then, by condition (ii) of a Bayesian Equilibrium,
− [xv (ω)]2 + πB ≥ Eul (xv (ω) , sv (ω, xv (ω) , ι))
≥ Eul (p, sv (ω, p, ι))
≥ −p2 + (1 − π) B,
from which it follows that p2 − (1 − 2π) B ≥ [xv (ω)]2 . Similarly,
− [xv (ω)]2 + πB ≥ Eul (xv (ω) , sv (ω, xv (ω) , ι))
≥ Eul (0, sv (ω, 0, ι))
≥ 0,
from which it follows that πB ≥ [xv (ω)]2 .
Corollary B1 In the statement of Lemma B8, suppose that (1 − 2π) B > p2 . Then, for each ω,
[sl (ω)]2 ≥ p2 .
Lemma B9 Fix a Bayesian Equilibrium with sv (ω, p, ni) = 1 for some policy p. If sl (ω) = xv (ω)
then p2 + πB ≥ [xv (ω)]2 .
Proof. Fix a Bayesian Equilibrium, viz. (sl , sv ), satisfying the conditions of the Lemma. Then
− [xv (ω)]2 + B ≥ Eul (xv (ω) , sv (ω, xv (ω) , ι))
≥ Eul (p, sv (ω, p, ι))
≥ −p2 + (1 − π) B,
where the second line follows from condition (ii) of a Bayesian Equilibrium. The result now follows
immediately.
Proof of Proposition 6.2. Fix a State-by-State Optimal Equilibrium, viz. (s∗l , s∗v ). By Lemma
p
B5-B6, there exists some policy p ∈ [0, (1 − π) B] such that (i) p is the greatest positive lower
bound on the set of policies with s∗v (ω, p, ni) = 1 (given the ordering relation P) and (ii) s∗l (ω) =
xv (ω) whenever ω is contained in
p
p
(xv )−1 ([− p2 + πB, −p] ∪ [p, p2 + πB]).
If (1 − 2π) B > p2 the the result follows from Corollary B1 and Lemma B9. If p2 ≥ (1 − 2π) B
then the result follows from Lemmata B7-B8-B9 (and Remark B2).
51
Appendix C: Voter Randomization
In this Appendix, we suppose that each player can make use of a randomization device. We
formalize this by allowing players to choose behavioral strategies.24
Retrospective voting is a
method of equilibrium selection from among the set of equilibria in behavioral strategies.
A behavioral strategy for the Legislator, denoted by bl , will specify a measures in ∆ (R), one
for each information set of the Legislator. Write bl (ω) for the distribution associated with the
behavioral strategy bl at the information set where the Legislator learns the true state is ω ∈ Ω.
Analogously, a behavioral strategy for the Voter, denoted by bv , will specify a measure contained
in ∆ ({0, 1}), one for each of the Voter’s information sets. Write bv (ω, p) for the measure specified
by bv at the information set where the true state is ω ∈ Ω, the Legislator chooses policy p ∈ R, and
the Voter is informed. Write bv (p) for the measure specified by bv at the information set where
the Legislator chooses policy p ∈ R, and the Voter is uninformed. Let Bl (resp. Bv ) be the set of
behavioral strategies of the Legislator (resp. Voter).
Fix a behavioral strategy for the Voter, viz. bv ∈ Bv . When the true state is ω ∈ Ω and the
Legislator chooses policy p ∈ R, write Eul (p, bv [ω, p]) for the expected utility of the Legislator,
where the expectation is taken with respect to bv and ι ∈ {i, ni}, i.e.,
Eul (p, bv [ω, p]) = −p2 + [bv (ω, p) (r = 1)] ∗ πB + [bv (p) (r = 1)] ∗ (1 − π) B.
Now fix a behavioral strategy for the Legislator, viz. bl ∈ Bl . When the true state is ω ∈ Ω, write
Euv (ω, bl (ω)) for the expected utility of the Voter, where the expectation is taken with respect to
bl , i.e.
Z
Euv (ω, bl (ω)) = −
p∈R
(p − xv (ω))2 dbl (ω) (p) .
A Bayesian Equilibrium is now simply an equilibrium in behavioral strategies. That is, (bl , bv ) ∈
Bl × Bv is a Bayesian Equilibrium if, for each ω ∈ Ω, Eul (bl (ω) , bv [ω, bl (ω)]) ≥ Eul (p, bv [ω, p])
for all p ∈ R.25 With this, we can extend the Definitions 4.2-4.3 to behavioral strategies, in the
natural way.
Lemma C1 Let π ∈ [0, 12 ). Then there exists an equilibrium in behavioral strategies, viz. (bl , bv ) ∈
Bl × Bv , with:
√ √
(i) If xv (ω) ∈ [− B, B] then bl (ω) (xv (ω)) = 1;
√
√
(ii) If − B > xv (ω) then bl (ω) (− B) = 1;
√
√
(iii) If xv (ω) > B then bl (ω) ( B) = 1.
24
If we allowed players to make use of mixed strategies, the results in this Appendix would not change.
Since the Legislator’s strategy is no longer a map, there is no need to impose a measurability requirement. This
is why we restrict attention to behavioral strategies (see Footnote 24).
25
52
Proof. Let bl ∈ Bl be as in the statement of the Lemma. Define bv ∈ Bv as follows: when
√ √
xv (ω) ∈ [− B, B], let
(
1 if p = xv (ω)
bv (ω, p) (r = 1) =
0 if p 6= xv (ω).
√
√
√
When − B > xv (ω) (resp. xv (ω) > B), let bv (ω, p) (r = 1) = 1 if and only if p = − B (resp.
√
p = B). Let

if πB ≥ p2

 0 2
p
bv (p) (r = 1) =
if (1 − π) B > p2 > πB
(1−π)B


1
if p2 ≥ (1 − π) B.
Notice that for any p ∈ R with (1 − π) B > p2 > πB,
p2
(1−π)B
∈ (0, 1). So, bv (p) can indeed be
taken to be a probability measure.
We now show that (bl , bv ) is a Bayesian Equilibrium. To do so, fix some ω ∈ Ω and some policy
√ √
p∈
/ Supp bl (ω). Note, for any p ∈ [− B, B] ∩ (Ω\ Supp bl (ω)), p 6= xv (ω). So, if πB > p2 , then
0 ≥ −p2 = Eul (p, bv [ω, p]) .
If (1 − π) B > p2 > πB, then
0 = −p2 +
p2
(1 − π) B = Eul (p, bv [ω, p]) .
(1 − π) B
If p2 ≥ (1 − π) B, then
0 = − (1 − π) B + (1 − π) B
≥ −p2 + (1 − π) B = Eul (p, bv [ω, p]) .
So, it suffices to show that Eul (bl (ω) , bv [ω, bl (ω)]) ≥ 0, for all ω ∈ Ω.
First, when πB ≥ [xv (ω)]2 ,
Eul (bl (ω) , bv [ω, bl (ω)]) = − [xv (ω)]2 + πB ≥ 0,
as required. Second, suppose that (1 − π) B > [xv (ω)]2 > πB. Here,
[xv (ω)]2
Eul (bl (ω) , bv [ω, bl (ω)]) = − [xv (ω)] + πB +
(1 − π) B
(1 − π) B
2
= πB,
53
satisfying the desired inequality. Third, suppose that B ≥ [xv (ω)]2 ≥ (1 − π) B. Now
Eul (bl (ω) , bv [ω, bl (ω)]) = − [xv (ω)]2 + B
≥ −B + B,
as required. Last, suppose that [xv (ω)]2 > B. Here,
Eul (bl (ω) , bv [ω, bl (ω)]) = −B + B,
as desired.
Lemma C2 Fix an equilibrium, viz. (bl , bv ) ∈ Bl × Bv , in behavioral strategies. For any ω ∈ Ω
√ √
and any X ⊆ R\[− B, B], bl (ω) (X) = 0.
The proof of Lemma C2 is essentially the proof of Lemma B1, so we omit it here.
Proposition C1 Let π ∈ [0, 12 ). Then there exists an Expectationally and State-by-State Optimal
Equilibrium, viz. (b∗l , b∗v ) ∈ Bl × Bv . Moreover, this equilibrium takes the following form:
√ √
(i) If xv (ω) ∈ [− B, B], then the Legislator chooses xv (ω) with probability 1;
√
√
(ii) If − B > xv (ω), then the Legislator chooses − B with probability 1;
√
√
(iii) If xv (ω) > B, then the Legislator chooses B with probability 1.
Proof. Let (b∗l , b∗v ) be a behavioral strategy as in the statement of the Proposition. By Lemma
C1, we can choose (b∗l , b∗v ) so that this is indeed an equilibrium.
First, we will show that for any equilibrium in behavioral strategies, viz. (bl , bv ), and any
state ω ∈ Ω, we must have Euv (ω, b∗l (ω)) ≥ Euv (ω, bl (ω)). From this, it follows that (b∗l , b∗v ) is
Expectationally Optimal and State-by-State Optimal.
√ √
Fix a Bayesian Equilibrium (bl , bv ) and a state ω ∈ Ω. If xv (ω) ∈ [− B, B],
Euv (ω, b∗l (ω)) = 0 = arg max {uv (ω, p)} ,
p∈R
√
so certainly 0 ≥ Euv (ω, bl (ω)). Next suppose − B > xv (ω). Let X − ⊆ R be the set of policies
√
p with − B > p. By Lemma C2, bl (ω) (X − ) = 0. For any p ∈
/ X −,
√
p − xv (ω) ≥ − B − xv (ω) > 0.
It follows that
´2
³ √
Euv (ω, b∗l (ω)) = − − B − xv (ω)
Z
≥−
(p − xv (ω))2 dbl (ω) (p) = Euv (ω, bl (ω)) .
p∈R
54
Finally, suppose that xv (ω) >
√
√
B. Let X + ⊆ R be the set of policies p with p > B. By Lemma
C2, bl (ω) (X + ) = 0. For any p ∈
/ X +,
0>
√
B − xv (ω) ≥ p − xv (ω) .
Using this, we have
³√
´2
B − xv (ω)
Z
≥−
(p − xv (ω))2 dbl (ω) (p) = Euv (ω, bl (ω)) .
Euv (ω, b∗l (ω)) = −
p∈R
Now we fix an Expectationally and State-by-State Optimal Equilibrium, viz. (bl , bv ), and show
that it satisfies the three conditions of the Proposition. For this, recall that, for all ω ∈ Ω, we must
have Euv (ω, b∗l (ω)) ≥ Euv (ω, bl (ω)). Since (bl , bv ) is State-by-State Optimal, Euv (ω, b∗l (ω)) =
Euv (ω, bl (ω)) for all ω ∈ Ω.
√ √
First, fix some ω ∈ Ω with xv (ω) ∈ [− B, B]. Here Euv (ω, bl (ω)) = Euv (ω, b∗l (ω)) = 0 and
√
so bl (ω) (xv (ω)) = 1. Next, fix some ω ∈ Ω with − B > xv (ω). By Lemma C2, if bl (X) > 0 then
√ √
√
there exists Y ⊆ X ∩ [− B, B] with bl (Y ) = bl (X) > 0. For any policy p ∈ Y \{− B}, we must
have
√
p − xv (ω) > − B − xv (ω) > 0.
√
So, if bl (ω) (Y ) > 0 and − B ∈
/ Y,
´2
³ √
Euv (ω, b∗l (ω)) = − − B − xv (ω)
Z
>−
(p − xv (ω))2 dbl (ω) (p) = Euv (ω, bl (ω)) .
p∈R
√ √
√
As such, for any Y ⊆ [− B, B] with bl (ω) (Y ) > 0 we have − B ∈ Y . Now note that, if
√
√ √
bl (ω) (− B) = 1, since otherwise there must exist some Y ⊆ [− B, B] with bl (ω) (Y ) > 0 and
√
√
√
− B∈
/ Y . A corresponding argument establishes that, if xv (ω) > B then bl (ω) ( B) = 1.
Proof of Proposition 7.1. This follows immediately from Lemma C2, Proposition 6.1, and
Proposition C1.
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