Information Aggregation in Poisson-Elections Mehmet Ekmekci, Boston College Stephan Lauermann, Bonn University November 28, 2016
Abstract The modern Condorcet jury theorem states that under weak conditions, elections will aggregate information when the population is large, in any equilibrium. Our main question is whether the modern Condorcet jury theorem is robust to population uncertainty. We …nd that the modern Condorcet jury theorem holds only if the expected number of voters is independent of the state. If it depends on the state, then additional equilibria exist in which information is not aggregated. The main driving force is that, everything else equal, voters are more likely to be pivotal if the population is small. We show that this result also holds if abstention is allowed and provide conditions under which the additional equilibria are stable.
We are grateful for helpful comments from VJ Krishna and Andrew McLennan. Baris Coskunuzer from the Mathematics Department at Koc University helped with the proof of a critical result, Lemma 9. This work was supported by a grant from the European Research Council (ERC 638115).
1
Introduction
Elections are said to be e¤ective in aggregating information that is dispersed among citizens, for example, about uncertainty regarding future economic prospects, costs and bene…ts of a public good, or the political rami…cations of a trade deal. This belief has been justi…ed by the so called Condorcet Jury theorems (see Ladha (1992)), which asserts that large electorates choose correct outcomes, and in its modern form by Austen-Smith and Banks (1996), Feddersen and Pesendorfer (1997, 1998), Wit (1998), Duggan and Martinelli (2001), and others. Precisely, the modern Condorcet jury theorem states that under weak conditions, in a large voting game with common values, all reasonable (responsive and symmetric) Nash equilibria aggregate information. The Condorcet jury theorem is one motivation for using elections for making collective choices. In its modern form, it provides “a rational choice foundation for the claim that majorities invariably ‘do better’than individuals” (at least, for large electorates); see Austen-Smith and Banks (1996). Most of these earlier contributions assume that the number of voters is deterministic and known. Myerson (1998) observed that the size of the electorate is often uncertain. Importantly, this uncertainty may not be independent of the underlying state of the world. In fact, there are plenty of reasons why the state of the world may be correlated with the expected number of voters: For example, in the case of local elections or referenda, awareness about the election taking place may depend on the perceived economic prospects. Similarly, the awareness of elections may depend on the competency or the motivation of the current o¢ ce holders because of its e¤ect on news coverage and general political engagement. Finally, election turnouts are often subject to manipulation by interested parties who may choose to in‡uence turnout strategically and di¤erently across states; see Ekmekci and Lauermann (2015). This paper studies whether the modern Condorcet Jury Theorem is robust to population uncertainty. To do so, we use the model by Myerson (1998): Voters have to choose among two alternatives. They share common values that depend on an unknown, binary state of nature. The number of voters is Poisson distributed and the mean of the distribution may depend on the state. Each voter observes a private, conditionally independent signal. To start, note that any asymmetry in the expected number of voters itself contains additional information about the state of the world, hence there is one more source of information— in addition to the private signals of the voters— that the electorate could
use to aggregate information. However, as we argue below, even though there is more information that could be used, large electorates may fail to aggregate any information, that is, the modern Condorcet Jury Theorem is not robust to population uncertainty. Because our environment is a common-value environment, we follow McLennan (1998) who showed that in common interest games, every symmetric strategy pro…le that maximizes the social welfare is also a symmetric Nash equilibrium. Our …rst order of business is to extend McLennan (1998)’s theorem for a deterministic population size to our environment where participation is Poisson distributed, and the expected number of voters is state dependent. One notable observation we make is that when the population size is state dependent, the voting game from the point of view of the participants fails to be a common interest game. In other words, an extension of McLennan (1998)’s result can be obtained only when the social welfare function considered in his program is appropriately chosen to include the payo¤s of the nonparticipating voters in the welfare calculation. We use our extension of McLennan (1998) to re-derive the main result from Myerson (1998): If voters have noisy but informative signals about the state of the world, then large electorates in which the population size is state dependent admit at least one symmetric Nash equilibrium that aggregates information, i.e., that chooses the correct outcome with a probability close to one.1 So, as is known from Myerson (1998), one part of the Condorcet Jury theorem survives: Large electorates are able to aggregate information. We show, however, that the second part of the Condorcet Jury theorem fails: there are plausible equilibria that fail to aggregate information when the population is state dependent. In such equilibria, the majority of voters vote as if the state is the one in which there are fewer voters. Therefore, the candidate wins that is preferred in the state in which there are fewer voters. Such equilibria are responsive, and when su¢ ciently informative signals are possible, these equilibria are stable. Thus, our main …nding is that the modern Condorcet Jury theorem holds with population uncertainty if and only if this uncertainty is independent from the state. Otherwise, if the population is statistically state dependent, additional responsive equilibria exist that fail to aggregate information. The key force that helps sustain such equilibria is a “participation curse”. A vote is more likely to change the outcome of the election, i.e., to be pivotal, when there are fewer voters, all else being equal. Therefore, a majority of voters— but not all voters— vote as if 1
In Myerson (1998), this result is proven directly, not using the common interest structure.
2
the state is the one with fewer voters. We then explore whether strategic abstention can help eliminate such “bad” equilibria. Krishna and Morgan (2012) showed that voluntary voting improves on compulsory voting and induces sincere voting outcomes when there are binary signals. In Feddersen and Pesendorfer (1997), abstention allows the uninformed players to participate in a rate that cancels out the e¤ect of partisans who cast their votes in one way independently of their signals. Hence, one may hope strategic abstention to help the electorate “undo” the asymmetry in the population size induced by exogenous factors. However, for the binary signal setting by Krishna and Morgan (2012), we show that allowing abstention does not eliminate responsive equilibria that fail to aggregate information. Somewhat surprisingly, if the asymmetry in the population size is not too big, then in the equilibria that fail to aggregate information, voters with a certain signal are mixing between all three options; they vote with positive probability for both candidates and abstain. This behavior is due to a swing voter’s blessing.2 We also explore abstention with state-dependent participation rates when signals are continuous. In this case, if there is no bound on the informativeness of the signals, then there are always equilibria that fail to aggregate information. Information aggregation fails in our setting because, whenever the expected number of voters depends on the state in a non-trivial way, the probability of being pivotal is di¤erent across states. We explore this general topic also in a companion paper, Ekmekci and Lauermann (2015), where we study a setting in which the number of voters is state-dependent but deterministic. That paper focuses on how the number of voters may endogenously emerge to give rise to equilibria in which information aggregation fails. Here, we consider a di¤erent setting where the number of voters is not deterministic but Poisson distributed in each state. For this case, the present paper provides a comprehensive equilibrium analysis, including stability. Finally, we allow for abstention and characterize equilibrium for voluntary voting with a state-dependent population. The literature has identi…ed other circumstances in which information may fail to aggregate.3 Feddersen and Pesendorfer (1997) show such a failure in an extension (Section 6 of their paper) when the aggregate distribution of preferences remains uncertain conditional on the 2
To the best of our knowledge, this is the …rst paper to identify an equilibrium with a swing voter’s blessing, rather than a swing voter’s curse, in the sense of Feddersen and Pesendorfer (1997). 3 This paragraph is taken from Ekmekci and Lauermann (2015).
3
realized state. Mandler (2012) demonstrates a similar failure if the aggregate distribution of signals remains uncertain. In these settings, the e¤ective state is multi-dimensional. Intuitively, this implies an invertibility problem from the relevant order statistic of the vote shares to payo¤-relevant states. A similar problem is identi…ed by Bhattacharya (2013), who observes the necessity of preference monotonicity for information aggregation. A recent generalization was made by Barelli and Bhattacharya (2013). Gul and Pesendorfer (2009) show that information aggregation fails when there is policy uncertainty. In our setting, conditional on the state, there is no aggregate uncertainty (in the sense that the mean of the Poisson distribution is known), preferences over policies are monotone in the state, and there is no policy uncertainty.
2
Model
The model setup follows Myerson (1998). Voters have to decide between two policies, A and B. There are two states,
and , with prior probability = Pr f! = g ,
with 0 <
< 1 and Pr f! = g = 1
.
The voters have common values: Each voter receives a (von Neumann-Morgenstern) payo¤ of 1 if the policy matches the state, and payo¤ of 0 otherwise. However, voters do not know the realized state. Instead, voters observe noisy signals x 2
[x; x].4 Conditional on the state, the signals are independent and identically distributed. The c.d.f. of the signal distribution is G ( j!). The distribution is atomless and admits a
density. Without loss of generality, signals are ordered so the weak MLRP holds, g (xj ) is weakly decreasing in x. g (xj )
In addition, g (xj!) > 0 for all x 2 (x; x). This, together with G being atomless,
rules out that voters receive perfectly revealing signals with positive probability. nally, for technical convenience, we assume right continuity of limx!x 4
g(xj ) g(xj )
=:
We allow x =
g(xj ) g(xj )
2 R [ f1g and limx!x
1 and x = +1.
4
g(xj ) g(xj )
=:
g(xj ) g(xj )
g( j ) g( j )
Fi-
on (x; x), de…ning
2 R. Signals contain some
information, meaning,
g(xj ) g(xj )
> 1 >
g(xj ) g(xj ) .
Two important special cases are boundedly
informative signals, 1>
g (xj ) g (xj ) >1> > 0, g (xj ) g (xj )
and unboundedly informative signals, g (xj ) g (xj ) = 1 and = 0. g (xj ) g (xj ) The number of voters is Poisson distributed in each state, with an expected number of n = n and n = n; so, the probability that there are t voters in state ! is Pr f~ n = tj!g =
(n! )t e t!
n!
.
The policy is decided by simple majority rule among submitted votes. If there is a tie, then a fair coin ‡ip decides. Abstention is not possible for now. We consider symmetric and pure voting strategies. Given the Poisson setup, symmetry is without loss of generality; see Myerson (1998). A voting strategy is a function a : [x; x] !
[0; 1], with a (x) the probability to vote for A.
Let U (x; W ; a; n) be the expected utility for a voter having signal x who votes for W 2 fA; Bg, given that all other voters use strategy a and the expected number of voters is (n; n) in states
and , respectively. We often drop a and n.
We study voting strategies that form a (Bayesian) Nash equilibrium. A voting strategy a is a Nash equilibrium if and only if U (x; A; a; n) > U (x; B; a; n) implies a (x) = 1 and U (x; A; a; n) < U (x; B; a; n) implies a (x) = 0. To characterize equilibrium, the following is useful. The likelihood ratio of the two states conditional on having signal x and participating is Pr ( jx) = Pr ( jx) 1
n g (xj ) . n g (xj )
Let T denote the event that the number of A and B votes is the same, T
(1) 1 the event
that there is one less A vote than B votes, and T + 1 the event that there is one more A
5
vote. Then, the di¤erence U (x; A; a; n) Pr ( jx) Pr [T
U (x; B; a; n) is equal to
1 1 + Pr [T j ] + Pr [T + 1j ] 2 2 1 1 1j ] + Pr [T j ] + Pr [T + 1j ] 2 2
1j ]
Pr ( jx) Pr [T
Voting A versus voting B changes the payo¤s only in the events T
.
(2)
1, T , and T + 1.
In the …rst event, voting A rather than B increases the probability of A winning from 0 to 1=2, in the second event it increases the probability from 0 to 1, and in the third event it increases the probability from 1=2 to 1. The probability that the decision to vote A versus B turns out to be pivotal is Pr (P iv0 j!) =
1 Pr [T 2
1j!] + Pr [T j!] +
1 Pr [T + 1j!] . 2
It is then easy to see from (1) and (2) that voting for A is a best response for a voter having signal x if and only if (x; a; n) := where
1
1 g (xj ) Pr (P iv0 j ) g (xj ) Pr (P iv0 j )
1;
denotes the critical likelihood ratio.
A strategy a is a cuto¤ strategy if for some x ^, a (x) = 1 if x > x ^ and a (x) = 0 if x
being nonincreasing in x.
Lemma 1. If a strategy forms a Nash equilibrium, it is equivalent to a cuto¤ strategy.5 Our generic notation is x ^ for the strategy: “Vote for A if x 2 (x; x ^), Vote for B if
x 2 (^ x; x).” Abusing notation, let
(x; x ^; n) be the critical likelihood ratio given cuto¤ x ^.
(t; t; n) is continuous in t, then x ^ 2 (x; x) is an interior Nash equilibrium if and only if
If
(^ x; x ^; n) = 1. 5
If the likelihood ratio is constant on some interval, every voting strategy is equivalent to a voting strategy in cuto¤s (because we can reorder votes on that interval). If the likelihood ratio is strictly increasing, every voting strategy is in cuto¤s.
6
3
Welfare Maximization and Nash Equilibria
McLennan (1998) observed that for common interest games, welfare maximizing strategy pro…les are also Nash equilibria. This result extends almost immediately to Poisson games with an in…nite population. However, we …rst need to clarify the welfare function to be maximized, since there are two natural candidates. Let u (i; x ^j!) be the expected payo¤ of a voter conditional on state ! and i voters being
present who vote according to x ^. The expected average surplus of the participating voters given a voting pro…le x ^ is E sur [u; x ^] =
1 X X i Pr f!g Pr f~ n = ij!g u (i; x ^j!) . n!
!2f ; g i=0
The expected payo¤ of a representative agent given x ^ is E rep [u; x ^] =
1 X X
!2f ; g i=0
Pr f~ n = ij!g u (i; x ^j!) .
The expected average surplus weighs the payo¤s in the two states by the population size. The expected payo¤ of a representative agent does not. The …rst welfare criterion is maybe more appropriate if those who do not participate receive payo¤ 0 (and/or do not exist). The second welfare criterion is maybe more appropriate if the non-participating agents are present and receive the same payo¤ u (i; x ^j!), but just don’t vote.
Of these two criteria, only the second corresponds to a game of common interest.
Lemma 2. Any voting strategy x that maximizes the expected payo¤ of a representative agent E rep [u; ] is a Nash equilibrium.
4
Compulsory Voting
Lemma 2 implies that, for all , equilibria exist that aggregate information. Myerson (1998) provides a direct proof of this result. Theorem 1. [Myerson (1998)] For all such that limn!1 E rep [uj^ xn ] = 1.
2 (0; 1), there is a sequence of equilibria f^ xn g1 n=1
7
For this and later proofs, it will be useful to de…ne the median signals x : G (x j ) = 1=2 and x : G (x j ) = 1=2. Since signals contain information, x < x . Proof. Take any x0 2 (x ; x ). Let x ^n maximize the welfare of the representative agent given n. Then, the proposition follows from 1 = lim E rep u; x0
lim E rep [u; x ^n ]
n!1
n!1
1.
The …rst equality is from the weak law of large numbers, the …rst inequality by choice of x ^n , and the last inequality from de…nition of E rep and u (i; x ^j!) 2 f0; 1g. We now study whether all equilibria aggregate information in large elections: How reliably do elections allow voters to make good choices? Is the Condorcet jury theorem robust to population uncertainty? If not, what drives the di¤erence?
4.1
Auxiliary Results
We use the following simple extension of the intermediate value theorem. This lemma deals with the problem that the signal likelihood ratio g (xj ) =g (xj ) is monotone decreasing but may be discontinuous, implying that the critical likelihood ratio
(t; t; n) may be
discontinuous in t as well. Lemma 3. Suppose there are two points a and b such that
(a; a; n)
1
(b; b; n)
(either a < b or a > b). Then, there there exists a Nash equilibrium with cuto¤ x ^ with min fa; bg
x ^
(^ x; x ^; n) = 1.
max fa; bg. If a < b, then there exists a Nash equilibrium cuto¤ with
We will use the following approximation for compulsory voting. Lemma 4. If limn!1 x ^n 2 (x; x), then Pr [P iv0 j ; x ^n ; n] lim n!1 Pr [P iv0 j ; x ^n ; n] 8 p < 1 if lim 2 G (^ xn j ) (1 n!1 = p : 0 if lim 2 G (^ xn j ) (1 n!1
G (^ xn j ))
1 > lim
G (^ xn j ))
1 < lim
8
n!1 n!1
p 2 G (^ xn j ) (1 p 2 G (^ xn j ) (1
G (^ xn j ))
1 ,
G (^ xn j ))
1 .
For
= 1, this simpli…es to ;x ^n ; n]
8 < 1 if
Pr [P iv0 j lim = n!1 Pr [P iv0 j ; x ^n ; n] : 0
if
lim G (^ xn j )
n!1
lim G (^ xn j )
n!1
1 2 1 2
< G (^ xn j ) > G (^ xn j )
1 2 1 2
, .
This lemma follows from standard approximations to pivot probabilities; see Krishna and Morgan (2012). The proof of the lemma is in a separate section in the appendix, where we re-state these general approximations and this and other lemmas for our purposes. The case
= 1 is particularly intuitive. Roughly, the state in which the election is closer to
being tied in expectation becomes arbitrarily more likely conditional on the election being actually tied.
4.2
The Modern Condorcet Jury Theorem
The modern Condorcet theorem states that in large elections, all “reasonable”(symmetric and responsive) equilibria aggregate information. For a deterministic number of voters, this result has been proven by Feddersen and Pesendorfer (1998) and Wit (1998) for binary signals and by Duggan and Martinelli (2001) for a continuum of signals. Krishna and Morgan (2012) prove it for Poisson elections when the expected number of voters is independent of the state,
= 1, and signals are binary. Here, we extend this result to a
continuum of signals. As in Duggan and Martinelli (2001) and Krishna and Morgan (2012), we assume that
1
g (xj ) >1> g (xj ) 1
g (xj ) . g (xj )
(3)
With this assumption, based on their own signal alone, a voter with the strongest signals for
would prefer policy A and a voter with the strongest signal for
would prefer policy
B. The assumption holds if the prior is uniform. The assumption also holds if signals are su¢ ciently informative. Theorem 2. Condorcet Jury Theorem. All equilibria aggregate information if
= 1 (there
is no imbalance) and (3) holds (signals are su¢ ciently informative): For every sequence of Nash equilibrium cuto¤ s fxn g1 n=1 with x
xn
x for all n, limn!1 E rep [uj^ xn ] = 1.
The proof of the theorem is by contradiction. The main step is to show that for any
9
sequence of Nash equilibrium cuto¤s, x < lim x ^n < x ,
(4)
n!1
which then implies that information aggregates by the weak law of large numbers. To show (4), we verify that for any sequence of cuto¤s xn , Pr [P iv0 j ; xn ; n] lim = n!1 Pr [P iv0 j ; xn ; n]
(
1 if x < limn!1 xn 0
if x
limn!1
xn
x , < x.
(5)
This rules out that such sequences are Nash equilibria, of course. A somewhat di¤erent argument deals with the remaining cases xn ! x or xn ! x. Equation (5) follows from the second part of Lemma 4. In particular, if x < limn!1 xn x , then with x0 = limn!1 xn , the MLRP implies that 0 < G x0 j
< G x0 j
1=2.
Therefore, the election is closer to being tied in state — and conditional on being pivotal, a voter becomes almost certain that it is this state. Conversely, if limn!1 xn = x0 is
interior, then Lemma 4 requires that the election is equally close to being tied in both states, meaning, G x0 j
1 2
1 2
= G x0 j
, for otherwise the relative likelihood of
being pivotal explodes. But for the election to be equally close to being tied, it is easy to see that it must be that x < x0 < x — and hence, information is aggregated. For deterministic elections, the analogous result holds only for symmetric and responsive equilibria. For Poisson elections, the symmetry assumption is without loss of generality and there are no non-responsive equilibria; hence, the result is stronger: If
= 1, all equilibria
aggregate information.
4.3
Failure of the Modern Condorcet Jury Theorem
We now show that the modern Condorcet theorem fails if
6= 1.
Theorem 3. Consider a sequence of voting games in which the expected number of participants is (n; n) in states If
and , respectively, and n ! 1.
< 1, then there is a sequence of interior Nash equilibrium cuto¤ s fxn g1 n=1 with
xn 2 (x; x ) for large n, such that B wins in both states with probability converging to one.
10
If
> 1, then there is a sequence of interior Nash equilibrium cuto¤ s fxn g1 n=1 with
xn 2 (x ; x) for large n, such that A wins in both states with probability converging to one.
The proof is in the appendix. The main observations are that lim
n!1
(x ; x ; n) = 1,
(6)
and that for any xR small enough, with x < xR < x , lim
n!1
(xR ; xR ; n) = 0.
(7)
Hence, the intermediate value theorem from Lemma 3 implies that, for all n large enough, there exists some x ^n 2 (xR ; x ) such that
(^ xn ; x ^n ; n) = 1. We verify that
lim x ^n < x , and, hence, B wins with probability converging to one in both states. Again, the critical observations (6) and (7) follow from Lemma 4. For (6), note that for a cuto¤ x ^ = x , the election is tied in state
while B wins with certainty in state .
Thus, it is intuitive that conditional on being pivotal, a voter becomes certain that the state is . Now, consider some xR close to x. Then, with xR small enough, B will win in both states. Moreover, because
< 1, the number of B votes is actually larger in state .
Now, the expected vote di¤erence in state n ((1
G (xR j ))
while the expected vote di¤erence in state n ((1
is G (xR j )) ,
is
G (xR j ))
G (xR j )) .
For xR close enough to x, the expected vote di¤erence is larger in state . Because of the smaller margin of victory, it is therefore intuitive that B is less likely to win in state in state . Thus, conditional on being pivotal, the state is almost certainly
than
for n large,
explaining (7). The fact that there are fewer voters in state
when
< 1 is at the heart of the
aggregation failure. Because the number is smaller, a voter is more likely to be pivotal in 11
that state, and, given that sophisticated voters condition on being pivotal, they tend to support B, even if their signals are strongly in favor of A.
4.4
Stability g(xj ) g(xj ) that6
If signals are unboundedly informative and n, there exists
xnL
with x <
xnL
< xR such
= 1, it is immediate that for any …xed
(xnL ; xnL ; n) > 1.
(8)
Together with (7) and Lemma 3, (8) implies that there exists some equilibrium with x ~n 2
(xnL ; xR ) for all n large enough. Thus, when signals are unboundedly informative, there
are at least two interior equilibria in which information aggregation fails, this one and the previous one with x ^n 2 (xR ; x ). This argument implies also that, when signals are unboundedly informative, there is one “stable” equilibrium that fails to aggregate information by the following argument. From (7) and (8), for all n large enough, ~ (x) =
(x; x; n) cuts 1 from above, at least once
at some point x ~s < xR . If x ~s is the only equilibrium cuto¤ in some neighborhood, then x ~s is an equilibrium cuto¤ that is responsive, and this equilibrium is an expectationally stable equilibrium in the sense of Fey (1997): Intuitively, this equilibrium has the property that if this is the outcome of a dynamic best-response iteration, and, if the process starts in a neighborhood of the equilibrium cuto¤, then the process eventually will converge to the cuto¤. It may be that for some " > 0, and x ~0 < x ~1 , we have ~ (x) > 1 for x 2 (~ x0
"; x ~0 ),
~ (x) = 1 for x 2 [~ x0 ; x ~1 ] and ~ (x) < 1 for x 2 (~ x1 ; x ~1 + "). We call such a set of equilibria
“pseudo-stable,”with singletons being a special case. With this de…nition and the previous observations, it follows immediately that Theorem 4. Suppose g(xj ) g(xj ) = x ^ns such
< 1 and the signal is unboundedly informative,
= 1 and
0. Then, for all n large enough, there is either a stable Nash equilibrium cuto¤
that limn!1 x ^ns < x or a set of pseudo-stable Nash equilibria [^ xn0 ; x ^n1 ] such that
limn!1 x ^n1 < x . 6
g(xj ) g(xj )
This holds because, for any …xed n, the ratio
Pr[P iv0 j ;x;n] Pr[P iv0 j ;x;n]
12
is bounded.
Unstable Equilibria. Applying the same argument to the case with boundedly informative signals, it follows from (6) and (7) that there exists at least one point x ^r 2 (xL ; x ) at which
crosses 1 from below. Thus, when signals are boundedly informative, there
exists at least one equilibria cuto¤ x ^r that is not stable. On the other hand, even though there must be some equilibrium that is not stable, we cannot rule out that there are stable equilibria. In fact, Theorem 4 implies that it is easily possible to construct stable equilibria for boundedly informative signals. Non-Responsive Equilibria. As observed by Myerson (1998), for all
6= 1, if signals
are boundedly informative, then there are also non-responsive equilibria when n is large enough. Consider
< 1 and suppose x ^ = x, so that all voters support B. In that case, the
relative probability of being pivotal is Pr [P iv0 j ; x; n] e = Pr [P iv0 j ; x; n] e
n (1 n (1
+ n) + n)
e
n(1
)1
!n!1 0.
Thus, given that signals are boundedly informative, it is a best response for a voter to vote for B independently of her signal, for n large enough. Non-responsive equilibria are stable.
5
Abstention
We now consider the possibility of abstention or “voluntary voting.” Feddersen and Pesendorfer (1996) noted that voters may have a strict incentive to abstain because of the “swing voter’s curse.” Moreover, abstention increases the expected payo¤ of a representative agent in the best equilibrium relative to compulsory voting. This is an immediate implication of Lemma 2; see also Krishna and Morgan (2012). In addition, Krishna and Morgan (2012) observe that with abstention and a binary signal, there is no longer a con‡ict between voting strategically and voting sincerely that is often present with compulsory voting even when
= 1.7 Thus, abstention may help
eliminate the equilibria that we identi…ed before since these equilibria relied on voters with a strong signal towards state
to nevertheless vote B. Thus, we now ask whether
abstention may help eliminate the bad equilibria. 7
Also, if voting is costly, abstention allows to reduce overall voting costs.
13
With abstention, our generic notation is (y; z) for the strategy: “Vote for A if x 2 (x; y),
Abstain if x 2 (y; z), Vote for B if x 2 (z; x).”We call a voting strategy (y; z) non-responsive if either z = x or y = x, so, either all participants vote B or all vote A. Otherwise, an
equilibrium is responsive. Again, Lemma 2 implies that information aggregation is possible in some equilibria. We state this without proof. Theorem 5. Suppose voting is voluntary. Consider a sequence of voting games in which the expected number of participants is (n; n) in states For all
and , respectively, and n ! 1.
> 0, there exists a sequence of Nash equilibria that aggregates information.
Unboundedly Informative Signals As before, we study whether every sequence of equilibria aggregate information. First, we study the case with unboundedly informative signals. Theorem 6. [Unboundedly Informative Signals.] Suppose voting is voluntary and signals are unboundedly informative. Consider a sequence of voting games in which the expected number of participants is (n; n) in states 1. If
and , respectively, and n ! 1.
< 1, then there is a sequence of responsive Nash equilibria such that B wins in
both states with probability converging to one. 2. If
> 1, then there is a sequence of responsive Nash equilibria such that A wins in
both states with probability converging to one. The proof is in the appendix. The basic idea is this. Consider an auxiliary game (xR ; n) in which voters with signals x
xR > x must vote for B but which otherwise
remains unchanged. By a standard argument, this game has an equilibrium. Then, for n large enough, this equilibrium is also an equilibrium of the original game if xR is small enough, especially xR < x . The critical argument for this proof is that for xR small enough, given any admissible strategy pro…le with y conditional on being pivotal converges to
one.8
z
xR , the probability of state
Thus, voters with signals around xR > x
will optimally vote for B— and hence this restriction does not bind. Thus, this is an equilibrium of the original game. Moreover, the equilibrium is responsive: Since signals are 8
Note that this argument is very similar to the one used in (7) for compulsory voting.
14
unboundedly informative,
g(xj ) g(xj )
= 1. Therefore, for every given n, voters will optimally
vote A for some su¢ ciently small signal. Boundedly Informative Signals
When signals are boundedly informative, unfortunately, the construction via an auxiliary game does not work. If we simply search for an equilibrium with cuto¤s y that equilibrium may turn out to be a non-responsive equilibrium with y = z =
x.9
z
xR ,
We also
cannot use arguments via the intermediate value theorem as in the compulsory voting case, since we are now looking for a two dimensional strategy pro…le. Instead, in the Appendix in Section B, we introduce a new result, Lemma 9, which generalizes the intermediate value theorem in a certain sense to two dimensions. With this result, we can study the case with binary signals, which was also analyzed in Krishna and Morgan (2012). Signals are binary if there is some xB such that the signal likelihood ratios are constant below and above that cuto¤, with
g(xj ) g(xj )
= c 2 (1; 1) for all x 2 [x; xB ) and
g(xj ) g(xj )
x 2 (xB ; x]. Following Krishna and Morgan (2012), we also assume
= c 2 (0; 1) for all
= 1=2, so that (3)
holds.
Theorem 7. [Binary Signals.] Suppose voting is voluntary, signals are binary, and
=
1=2. Consider a sequence of voting games in which the expected number of participants is (n; n) in states
and , respectively, and n ! 1.
1. [Krishna and Morgan (2012)] If
= 1, all sequences of Nash equilibria aggregate
information. 2. If
1 G(xB j ) 1 G(xB j )
< 1 and
6= , there is a sequence of responsive Nash equilibria such
that B wins in both states with probability converging to one. 3. If
> 1 and
G(xB j ) G(xB j )
6= , there is a sequence of responsive Nash equilibria such that
A wins in both states with probability converging to one. In the theorem,
1 G(xB j ) 1 G(xB j )
6=
is a genericity requirement. The construction of the
equilibrium for binary signals may be the main technical contribution of the paper. To the best of our knowledge, the equilibrium construction via the new Lemma 9 is new and may be useful in other contexts as well. 9
Relatedly, we cannot use a …xed point theorem on a restricted strategy space because we cannot ensure that points on the boundary of the space are mapped into the interior.
15
Mixed Equilibrium. With binary signals, the equilibrium turns out to be typically mixed. Consider the case when
< 1 but not too small such that
1 G(xB j ) 1 G(xB j )
< . In this
case, equilibrium may have somewhat surprising properties. In particular, it will be the case that a voter with a signal x < xB (which is a signal indicative of state
), is mixing
between voting for A, voting for B, and abstaining, choosing each of the three actions with strictly positive probability. A necessary condition for this is that there is no swing voters curse. Note that, in contrast to Krishna and Morgan (2012), despite voting being voluntary, the voting strategy is not sincere10 because voters with signal x < xB are voting against their signal. General Signal. We conjecture that Theorem 7 can be extended to all boundedly informative signals. We have veri…ed this for the case in which the signal likelihood ratio g(xj ) g(xj )
is continuous. However, the proof becomes much more involved and we need to
strengthen the Lemma 9. To move even further to general discontinuous signals would require a result analogous to Lemma 9 for certain sets of functions that are discontinuous but monotone along one of the two dimensions. While we believe this to be true, we have not been able to establish it. Compulsory versus Voluntary Voting. In the beginning of this section, we noted that abstention increases the expected payo¤ of a representative agent in the best equilibrium relative to compulsory voting by Lemma 2; see also Krishna and Morgan (2012). However, this is not necessarily the case relative to expected surplus. While we have not found a speci…c example, we believe that there may be parameters for which the expected surplus with compulsory voting is higher than with voluntary voting. Exploring the costs and bene…ts of voluntary voting further will be an interesting avenue for further research.
6
Conclusion
In this paper, we study the set of equilibria of Poisson elections when the expected number of voters is potentially state-dependent. We show that large Poisson elections robustly aggregate information— in the sense that all responsive and symmetric equilibria imply ) ) We say that voting is sincere if x y implies 1 1 g(xj 1 and x z implies 1 1 g(xj 1 (if g(xj ) g(xj ) the strategy pro…le requires voting for either A or B for some signal, then given that signal, voting for A and B is also myopically optimal, that is, if a voter were to be the sole voter and used only the information contained in her signal and participating.) 10
16
the correct choice with probability converging to one— if and only if the expected number of voters is constant across states. If the expected number of voters is di¤erent, then there are additional responsive equilibria that fail to aggregate information. The basic reason for this is that voters are more likely to be pivotal when the electorate is smaller. This leads to equilibria in which voters systematically vote for the policy that is optimal in the state with fewer expected voters. When signals are su¢ ciently informative, these equilibria can be chosen to be stable. Finally, abstention does not eliminate the additional equilibria. For the case with abstention and binary signals, we use a certain “two-dimensional extension” of the intermediate value theorem to construct an equilibrium with novel properties (voters mix between all three options and there is a swing-voters blessing). Our results relate to three contributions: Ekmekci and Lauermann (2015) consider a setting with a deterministic number of voters in each state and compulsory voting and show that information aggregation may fail. Myerson (1998) introduces a model of Poisson elections in which the expected number of voters may be state-dependent and voting is compulsory. He shows that there exist equilibria that aggregate information. Krishna and Morgan (2012) study a model of Poisson elections in which the expected number of voters is independent of the state and abstention is allowed, showing that for the case with binary signals, all equilibria aggregate information.11 Relative to Ekmekci and Lauermann (2015), we consider Poisson elections and allow abstention; relative to Myerson (1998), we show the existence of additional equilibria and allow abstention; and relative to Krishna and Morgan (2012) we allow continuous signals and show that there are additional equilibria when the expected number of voters depends on the state.
A
Appendix
Notation. In the appendix, we denote the expected number of A and B votes in state as B
A
= nG (^ xj ) and
(^ x) = n (1
B
(^ x) = n (1
G (^ xj )). Similarly, for state ,
G (^ xj )). We often drop the arguments from
!
and
x) A (^ !.
= nG (^ xj ),
Sequences and Limits. When taking limits, we mean with respect to subsequences for which a limit exists (in the extended reals). In the context of our proofs, such subsequences can always be found and proving statements for all converging subsequences will be su¢ cient for the desired claims. This convention saves us from introducing notation for layers 11
Their paper considers costly voting, which is the main focus of their analysis.
17
of subsequences. We will not always repeat this quali…er.
A.1
Proofs for Section 3 (Welfare Maximization)
Proof of Lemma 2. We apply the same argument as McLennan (1998), with a slight modi…cation to account for random number of voters . Let u(i; aj!) denote that expected payo¤ of a voter when there are i voters, state is ! and each voter is following the strategy a, and let u(i; a0 j!; a) denote the expected payo¤ of a voter in state ! if he follows the strategy a0 while all
other voters are using the strategy a, and when there are i + 1 voters in total. Finally, let u(a; i; a0 ; jj!) be the expected payo¤ of a voter when i voters are using the strategy a, and j voters are using the strategy a0 . Rede…ne the representative voter’s payo¤ to be E rep [uja] =
X
Prf!g
!
1 X i=0
Pr f~ n = ij!g u (i; aj!) .
In particular, we are interested in lim
E rep [uja]
E rep [uj(1
)a + a0 ]
!0
.
To this end, note the following: Let x be the number of participants who play a and y the number of participants who play a0 , and note that x + y = n ~ . Notice that x has a Poisson distribution with mean n! (1
), and y has a Poisson distribution with mean n! .
Then,
E
rep
[uja]
E
rep
[uj(1
0
)a + a ] =
X !
Pr(y = 0j!) Pr(y = 1j!)
1 X Pr(!)( Pr(~ n = ij!)u(i; aj!) i=0
1 X
Pr(x = ij!)u(i; aj!)
i=0 1 X i=0
Pr(x = ij!)u(i; a0 j!; a)
X
Pr(y = jj!)
j>1
18
1 X i=0
Pr(x = ij!)u(a; i; a0 ; jj!)),
and so E rep [uja]
E rep [uj(1
)a + a0 ] =
X
Pr(!)(
!
Pr(y = 1j!)
1 X
Pr(x = ij!)(u(i + 1; aj!)
i=0
+
X
Pr(y = jj!)
j>1
1 X
u(i; a0 j!; a))
Pr(x = ij!)(u(i + j; aj!)
u(a; i; a0 ; jj!))).
i=0
Consider the term 1X
Pr(!) Pr(y = 1j!)
!
1 X
Pr(x = ij!)(u(i + 1; aj!)
i=0
Because lim
Pr(y = 1j!)
!0
u(i; a0 j!; a)):
= n! ;
and because lim
!0
1 X i=0
0
Pr(x = ij!)(u(i+1; aj!) u(i; a j!; a)) =
1 X i=0
Pr(~ n = ij!)(u(i+1; aj!) u(i; a0 j!; a)),
we have lim
!0
1X
Pr(!) Pr(y = 1j!)
!
1 X
Pr(x = ij!)(u(i + 1; aj!)
i=0
X
Pr(!)n!
!
1 X
u(i; a0 j!; a)) =
Pr(~ n = ij!)(u(i + 1; aj!)
i=0
u(i; a0 j!; a)):
Consider the term 1X
Pr(y = jj!)
j>1
Because
P
j>1 Pr(y
1 X
Pr(x = ij!)(u(i + j; aj!)
u(a; i; a0 ; jj!):
i=0
= jj!) is at the order of
19
2,
and because payo¤s are bounded, we
have lim
!0
1X
Pr(y = jj!)
j>1
1 X
Pr(x = ij!)(u(i + j; aj!)
u(a; i; a0 ; jj!) = 0:
i=0
Therefore, lim
E rep [uja]
E rep [uj(1
)a + a0 ]
!0
=
X
Pr(!)n!
!
1 X i=0
Pr(~ n = ij!)(u(i+1; aj!) u(i; a0 j!; a)).
Suppose now to the contrary of the claim of the Lemma that the strategy a maximizes E rep [uja], and a is not a symmetric Nash equilibrium of the voting game. Then, there is a strategy a0 such that X
Pr(!)n!
!
because
P
!
1 X
Pr(~ n = ij!)(u(i + 1; aj!)
i=0
Pr(!)n!
P1
n i=0 Pr(~
u(i; a0 j!; a)) < 0;
= ij!)u(i; a0 j!; a) is the payo¤ to a voter from following
the strategy a0 when other voters follow strategy a. However, then there is some > 0 such that E rep [uj(1
A.2
)a + a0 ] > E rep [uja], which contradicts that a maximizes E rep [uja].
Proofs for Section 4.1 (Auxiliary Results)
Proof of Lemma 3. (The Generalized Intermediate Value Theorem.) We are done if either the weak MLRP implies (x; a; n)
1 for all x
So, suppose
(a; a; n) = 1 or 1 = (x; a; n)
1 for all x
(b; b; n). To wit, if
(a; a; n) = 1, then
a (and hence voting A is optimal) and
a (and hence voting B is optimal).
(a; a; n) < 1 <
(b; b; n).
Case a > b. Let x ^ = sup fxj (x; x; n) > 1g. By continuity of Pr (P iv0 j ) = Pr (P iv0 j )
on [x; x] and right-continuity of g (xj ) =g (xj ), Also, lim"!0 (x (x; x ^; n)
"; x
1 for all x
"; n)
(^ x; x ^; n) = lim"!0 (x + "; x + "; n)
1. By the MLRP,
(x; x ^; n)
1.
1 for all x < x ^ and
x ^. Hence, x ^ is a Nash equilibrium cuto¤.
Case a < b. Let x ^ = sup fxj (x; x; n) < 1g. As before, by continuity of Pr (P iv0 j ) = Pr (P iv0 j )
and right-continuity of g (xj ) =g (xj ),
(^ x; x ^; n) = lim"!0 (x + "; x + "; n)
tinuity of Pr (P iv0 j ) = Pr (P iv0 j ) and the MLRP, lim"!0 (x By de…nition of x ^, lim"!0 (x previous bound,
"; x
"; n)
1. Hence,
(^ x; x ^; n) = 1.
20
"; x
(^ x; x ^; n)
"; n)
1. By con-
(^ x; x ^; n).
1. Together with the
Recall that nG (^ xj ),
B
A
= nG (^ xj ) and
(^ x) = n (1
B
(^ x) = n (1
G (^ xj )) and similarly
x) A (^
=
G (^ xj )). We approximate the pivotal probabilities. Here 1
and in the following, given any sequence xk k=1 , we say f g for two functions if f ( xk ) limk!1 g xk = 1. To improve readability, we suppress the sequence index k in the follow( ) ing statement. Lemma 5. If
A B
! 1, then Pr [T j ]
Pr [T If
A B
1j ]
p
1 e 2
A
1 e 2
A
B
p
B
! k 2 (0; 1), then Pr [T
e2 A p p 2 2 e2 A p p 2 2
1j ]
e
A
Pr [T j ]
e
A
Pr [T + 1j ]
e
A
B
, A B B
A
A B
B
1=2
.
p I1 2 k B p ; B k p B I0 2 k ; p I1 2 k B p , A k
with I0 : R+ ! R+ a continuous, strictly positive function with limz!1 I0 (z) = I1 (z) = 1,
I0 (0) = 1 and I1 : R+ ! R+ a continuous function that is strictly positive on (0; 1) but limz!0 If
I1 (z) z
= 1=2.
A B
! 0, then Pr [T
1j ]
e
A
B
Pr [T j ]
e
A
B
Pr [T + 1j ]
e
A
B
B,
, A.
Of course, all analogous approximations hold for state , after substituting
W
for
W.
Proof of Lemma 5. The Lemma follows immediately from observations from Krishna and Morgan (2012), namely, 21
Pr [T j ] = e Pr [T
1j ] = e
A
B
A
B
p I0 (2
A B) , 1=2
(9)
p I1 (2
A
A B) ,
(10)
B
where I0 and I1 are modi…ed Bessel functions. The approximations then use properties of the modi…ed Bessel functions, namely, that z pe 2 z
lim
= lim
z!1 I0 (z)
and that
1=2 B A
I1 (z) 1 lim = ) z!0 z 2
z pe 2 z
= 1,
z!1 I1 (z)
I1 2
p
A B
B
p I1 2 = p
A B
! 1.
A B
Now, the approximations follow. Proof of Lemma 4. (Approximation for Compulsory Voting.) Recall that Pr (P iv0 j!) =
1 Pr [T 2
From lim x ^n 2 (x; x), we have 1 Pr [T 2
1j!] + Pr [T j!] +
A B
! 1 and
A B
e
A
e2 A B p p 2 2
p n+2
e p p 2 2
=
p 4 0 < lim p 4
A B A B
2+ 2+
! 1. So, Lemma 5 implies that p
1 1j ] + Pr [T j ] + Pr [T + 1j ] 2
Furthermore, lim x ^n 2 (x; x) implies that
1 Pr [T + 1j!] . 2
q
q
B A
+
B A
+
22
q
q
A B A B
A B
A B
1 2
B
A B
1+
=: K < 1.
1 2
1+
A
+1=2
A
+
B A B
+1=2
+
B
A B
1=2
!
.
1=2
!
This is because with x ^ = lim x ^n , p 4 lim p 4 and
q
2 2 n G (^ xn j
q
B A
+
B A
+
4
A B A B
) (1
= lim p 4 n2 G (^ xn j ) (1 2+
lim 2+ So, Pr [P iv0 j ] lim Pr [P iv0 j ]
q
e = lim e
q
q
A B
2+ =
A B
p n+2 p n+2
p
n 2
= K lim e
p p 4 G (^ xj ) (1 G (^ xj )) = p , 4 G (^ xj ) (1 G (^ xj )) G (^ xn j )) G (^ xn j ))
q
2+
A B A B
(1 G(^ xn j )) G(^ xn j )
+
(1 G(^ xj )) G(^ xj )
+
q
p 4 p 4
A B A B
2+ 2+
q
q
G(^ xn j )(1 G(^ xn j )) 1
q
q
G(^ xn j ) (1 G(^ xn j )) G(^ xj ) (1 G(^ xj ))
B A
+
B A
+ p
2
q
q
.
A B A B
G(^ xn j )(1 G(^ xn j )) 1
.
and the lemma now follows.
A.3
Proofs for Section 4 (Compulsory Voting)
Proof of Theorem 2. (If
= 1, all equilibria aggregate information.)
We show …rst that for any sequence of cuto¤s xn , Pr [P iv0 j ; xn ; n] lim = n!1 Pr [P iv0 j ; xn ; n]
(
1 if x < limn!1 xn 0
if x
limn!1
xn
x , < x.
This rules out that such sequences are Nash equilibria, of course. Consider x limn!1 xn < x. Then, from the MLRP and the fact that signals contain information, 1=2
lim G (^ xn j ) < lim G (^ xn j ) < 1. Now, the claim follows from Lemma 4.
We now rule out equilibria in which xn is close to x. By assumption (3) and rightcontinuity of g ( j ) =g ( j ) at x, there is some xr > x such that for x > xr 1>
g (xj ) . g (xj )
1
Now, if x 2 (xr ; x), then 1=2 < lim G (^ xn j ) < lim G (^ xn j ) < 1 implies that the probability Pr [P iv0 j ; x; n] > Pr [P iv0 j ; x; n]. To see this, consider a …xed voter and suppose the 23
realized number of other voters is m and each of the m other voters supports A with i.i.d. probability G (xj!). If m = 0, then the voter is pivotal in both states with equal likelihood. If m > 0 is even, then the …xed voter a¤ects the election if and only if exactly m 2
other voters support A and B. The probability that exactly
policy is strictly larger in state B since G (xj ) (1
m 2
voters support each
G (xj )) > G (xj ) (1
G (xj )) by
1=2 < lim G (^ xn j ) < lim G (^ xn j ) < 1. If m is odd, then the …xed voter a¤ects the election if and only if she votes A and
for B changes the outcome if
m 1 2 m+1 2
other voters support A and
support A and
m 1 2
m 1 2
support B and vote
support B. With
m 1 2
=: r and
q! := G (xj ), the sum of these two probabilities is
=
2r + 1 (q! )r (1 r 2r + 1 (q! )r (1 r
Again, G (xj ) (1
2r + 1 (q! )r+1 (1 q! )r r+1 2r + 1 q! )r (q! + (1 q! )) = (q! )r (1 r q! )r+1 +
G (xj )) > G (xj ) (1
q! )r .
G (xj )) implies that this probability is higher
in state . Thus, conditional on any realization of the number of other voters (either even or odd), the probability to a¤ect the election is higher in state . Hence, for all x > x > x , Pr [P iv0 j ; x; n] < 1. Pr [P iv0 j ; x; n] Thus, for all n and x > xr ,
1
g (xj ) n Pr [P iv0 j ; x; n] < g (xj ) n Pr [P iv0 j ; x; n] 1
g (xj ) < 1. g (xj )
There can be no equilibrium with a cuto¤ x 2 (xr ; x) for any n. A symmetric argument rules out equilibria with cuto¤s x close to x. Finally, from
= 1, we have for a cuto¤ x that
Pr [P iv0 j ; x; n] Pr [P iv0 j ; x; n]
=
1 2 1 2
=
0+e 0+e
Pr [T Pr [T
1j ] + Pr [T j ] + 12 Pr [T + 1j ] 1j ] + Pr [T j ] + 12 Pr [T + 1j ]
+ 12 e n + 1e 2 n
nn nn
= 1,
which follows because A cannot be behind if the cuto¤ is x (all vote A), a tie occurs only
24
if no voter participates, and A is one ahead if there is exactly one voter. Thus, (x; x; n) =
g (xj g (xj g (xj g (xj
1
=
1
and so by right continuity of g ( j ) =g ( j ) at x,
) n Pr [P iv0 j ; x; n] ) n Pr [P iv0 j ; x; n] ) < 1, ) (x; x; n) < 1 for all x < x. There can be
no equilibrium with cuto¤ x for any n. Similarly, there can be no equilibrium with cuto¤ x for any n. Proof of Theorem 3. (If aggregate information.) We prove the theorem for
6= 1 then there are interior equilibria that do not < 1. The argument for
Suppose signals are boundedly informative, that is,
> 1 is analogous and omitted. g(xj ) g(xj )
We verify that lim
n!1
< 1.
(x ; x ; n) = 1,
(11)
(x; x; n) = 0.
(12)
lim
n!1
From the MLRP and the fact that signals contain information, 1=2 = G (x j ) >
G (x j ) > 0. Now, (11) follows from Lemma 4. For x = x, we have
Pr [P iv0 j ; x; n] Pr [P iv0 j ; x; n]
=
1 2 1 2
=
1 nn + e n + 0 2e 1 n n+e n+0 2e
Pr [T Pr [T
1j ] + Pr [T j ] + 12 Pr [T + 1j ] 1j ] + Pr [T j ] + 12 Pr [T + 1j ] !n!1 0,
where the second equality follows because A is one behind if there is exactly one voter, a tie occurs only if no voter participates, and A cannot be ahead if the cuto¤ is x (all vote B). The limit follows from
< 1. This implies (12).
Given (11) and (12), existence of an interior Nash equilibrium x ^n with
(^ xn ; x ^n ; n) = 1
for all n large enough follows from the generalized intermediate value theorem, Lemma 3. By Lemma 4, the conclusion of (11) also holds if x ^n ! x . Thus, lim x ^n < x , and so B wins with probability converging to one.
25
Suppose signals are unboundedly informative, that is, Since
g(xj ) g(xj )
= 1.
< 1, there exists some xR 2 (x; x ) small enough such that p 2 G (xR j ) (1
G (xR j ))
noting that the left-hand side approaches
n!1
G (xR j ))
1 ,
1 for xR ! x and the right-hand side approaches
. Lemma 4 implies that lim
p 2 G (xR j ) (1
1<
(xR ; xR ; n) = 0.
(13)
Since signals are unboundedly informative, it is immediate that for any given n, there exists xnL with x < xnL < xR such that (xnL ; xnL ; n) > 1. To check: For a …xed n, for x ^ ! x, we have limx^!x
Pr[P iv0 j ;^ x;n] Pr[P iv0 j ;^ x;n]
(14) =
1 e n n+e n +0 2 1 n n+e n +0 e 2
2 (0; 1).
Hence, from signals being unboundedly informative, limx^!x (^ x; x ^; n) = 1 for all n.
Given (13) and (14), existence of a Nash equilibrium x ^n for all n large enough follows
again from Lemma 3.
A.4
Proofs for Section 5 (Abstention)
Notation. In the following, we denote the critical likelihood ratio at the cuto¤ types as A (y; z; n)
=
1
1 g (yj ) Pr [P ivAj ; y; z; n] , g (yj ) Pr [P ivAj ; y; z; n]
1
1 g (zj ) Pr [P ivBj ; y; z; n] . g (zj ) Pr [P ivBj ; y; z; n]
and B
(y; z; n) =
A strategy pro…le (y; z) is an interior Nash equilibrium if x < y 1= A.4.1
A (y; z; n)
=
B
z < x and
(y; z; n) .
Auxiliary Approximation Results
Lemma 6. Suppose abstention is possible and suppose signals are boundedly informative. Pick a sequence of equilibrium cuto¤ s (y n ; z n ). If
A B
! 1 (or, equivalently,
A B
! 1), then there is some K ( ; ) such that
26
lim K ( ; ) 2 (0; 1) and p
Pr [P ivAj ] e ( lim = lim p Pr [P ivAj ] e (
A
p
B
)
A
p
B
)
2
K ( ; ),
2
and similarly there is some K ( ; ) such that lim K ( ; ) 2 (0; 1) and p
Pr [P ivBj ] e ( lim = lim p Pr [P ivBj ] e (
A
p
B
)
A
p
B
)
2
2
K ( ; ).
It follows that under these conditions that for W 2 fA; Bg, Pr [P ivW j ] lim = n!1 Pr [P ivW j ]
(
p
1 if limn!1 0
p
if limn!1
p
A
p
A
B
<
B
>
p p
A A
p p
B
,
B
,
Remark. Observe that for any z < x p
p
A
B
< 0 and
p
A
p
B
< 0.
Hence, p ( =
n
if
p +( p p G (yj ) 1
A
p 1
p
2 B)
G (zj )
A
p
2 B) 2
G (zj )
+
p p G (yj ) >
Thus, it follows from the lemma that Pr [P ivW j ] lim Pr [P ivW j ] 8 p < 1 if limn!1 1 = p : 0 if limn!1 1
p G (yj ) p 1
p 1
G (zj )
G (zj )
2
!
1,
p G (yj ) . (15)
n!1
G (zj ) G (zj )
p p
G (yj ) < limn!1 G (yj ) > limn!1
27
p p
p 1 p 1
G (zj ) G (zj )
p G (yj ) , p G (yj ) .
Proof. We consider W = A. With abstention, we have Pr [P ivAj!] =
1 1 Pr [T j ] + Pr [T 2 2
1j ] .
Suppose signals are boundedly informative. So, from Lemma 5, 1 =
1 2
Pr [P ivAj ] Pr [T j ] + 12 Pr [T 1 2e
n!1
= Let
A
1 2
B
2
pe
Pr [T j
A B
p 2 2 A B 1 ] + 2 Pr [T p 1+ p B
p p 2 B) 1 ep( A p 2 2 2 A B 1 1 2 Pr [T j ] + 2 Pr [T
p p 2 2 K ( ; )= p p 2 2
Then,
1j ]
p
p
Pr [P ivAj ] e ( lim = p Pr [P ivAj ] e (
A
1j ]
A B A B
1+
p
B
)
A
p
B
)
p p
B A
1j ]
.
1+
A
1+
p p p p
B A B
.
A
2
2
K ( ; ).
Given the hypothesis that signals are boundedly informative, it is immediate that lim K ( ; ) 2 (0; 1) . The claim follows. A.4.2
Proof for the Unboundedly Informative Signal Case
Throughout this section, we consider the case with < 1; and signals are unboundedly informative, g (xj ) = 1. g (xj )
28
Let
(xR ; n) be an auxiliary game in which voters with signals x
xR must vote B,
while voters below xR can choose between voting A, B, or abstaining as before. We will show that
(xR ; n) has an equilibrium that satis…es the properties of the theorem and for
which the constraint at xR does not bind. Note that
(xR ; n) has an equilibrium (y; z) by Milgrom and Weber (1985) for all xR .
We use the following Lemma, proven at the end of this section. Lemma 7. Suppose
< 1, abstention is possible, and signals are unboundedly informative.
There exists some xR 2 (x; x ) such that for any sequence (y n ; z n ) with y n
zn
xR ,
Pr [P ivAj ; y n ; z n ; n] Pr [P ivBj ; y n ; z n ; n] = lim = 0. n!1 Pr [P ivBj ; y n ; z n ; n] n!1 Pr [P ivAj ; y n ; z n ; n] lim
Now, suppose that (y n ; z n ) is an equilibrium sequence of satisfy Lemma 7. It cannot be that lim
n!1
B
zn
(xR ; n), for xR chosen to
! xR . Suppose otherwise. If z n ! xR , then
g (z n j ) Pr [P ivBj ] n!1 1 g (z n j ) Pr [P ivBj ] g (xR j ) Pr [P ivBj ] = lim n!1 1 g (xR j ) Pr [P ivBj ] = 0.
(y n ; z n ; n) =
lim
Thus, for any x0 , any voter having a signal in (x0 ; xR ) would have a strict preference to vote for B. Thus, it must be that lim z n < xR . But this implies that all voters with signals in (lim z n ; xR ) prefer voting B to voting A or abstaining. By the MLRP, this implies that in particular all voters with signals x
xR prefer voting B. Hence, the initial restriction of
(xR ; n) relative to the original game does not bind. Therefore, for large n, (y n ; z n ) is also an equilibrium of the original game. Clearly, from lim z n < xR < x , policy B is chosen with probability converging to one. This proves the claim of the theorem. Proof of Lemma 7. There exists a signal xR 2 (x; x ) such that for all y p 1
G (zj )
p p G (yj ) > 29
p 1
z
G (zj )
xR , p G (yj ) .
Hence, it follows from the remark after Lemma 6 that for all y n p
e ( lim p n!1 e (
A
p
B
)
A
p
B
)
zn
xR ,
2
2
= 0.
(16)
Moreover, lim
e e
A
+
A
B
A
B
= 0.
This follows from A
B
+
A
+
B
=
A
B
= n ( G (yj )
B
G (yj ) + (1
G (zj ))
(1
G (zj ))) ,
and ( G (yj )
G (yj ) + (1
G (zj ))
from G (yj ) < G (yj ) (by the MLRP) and necessary by z
(1
(1
G (zj ))) < 0,
G (zj )) < (1
G (zj )) (which is
xR for our choice of xR ).
Case 1. Suppose
A B
! 1. From MLRP, p
Pr [P ivAj ] e ( lim = lim p Pr [P ivAj ] e (
A
p
A
p
A B
! 1. Then, from Lemma 5,
p p 2 2 p p 2 2 2 B) 2
B
)
A B A B
1+ 1+
p p p p
B A B
.
A
If lim y n > x, then we are done because of (16) and the last fractions are bounded. Suppose lim y n = x. Then,
since lim
B B
2 (0; 1) and
p p 2 2 lim p p 2 2
A A
A B A B
p p p p
1+ 1+
B A B
< 1,
A
1.
Similarly,
p
Pr [P ivBj ] e ( lim = lim p Pr [P ivBj ] e (
A
p
A
p
30
p p 2 2 p 2 p 2 2 B) 2
B
)
A B
1+
A B
1+
p p p p
A B A B
,
and
follows from lim
p p
A B
p p 2 2 lim p p 2 2
< 1, lim
Case 2a. Suppose
A B
Then, from Lemma 5,
Pr [P ivBj ] p I1 (2 k) p k
So, if lim
e e
A B
A A
B B
A
2 (0; 1), and
e
A
e
B
A
B
Pr [P ivAj ]
e
A
Pr [P ivBj ]
e
A
A A
0
@I0
0
@I0
B
I0
B
I0
A B
< 1,
A B
1. A B
< 1. This requires y n ! x.
p I1 2 k p 2 k + B k p I1 2 k p p 2 k + A k p
A B
1
A,
1
A,
< 1, we have
p I1 (2 z) , 2 z + B p z p p I1 (2 z) . 2 z + A p z p
< 1 then Pr [P ivAj ] e = lim Pr [P ivAj ] e
A
B
A
B
p ! 0 by (16), I0 2 k > 0,
lim
since lim
1+
= 1 if k = 0. Similarly, from z = lim
lim
since
A B
p p p p
! k < 1 and z = lim
Pr [P ivAj ]
with
B B
A B
1+
= lim
e Pr [P ivBj ] = lim Pr [P ivBj ] e
A
p I0 (2 z) + p I0 2 k +
p I1 (2 k) p k
A
B
A
B
= 0.
31
p I1 (2 z ) p B z p I1 (2 k) p B k
> 0 and lim
p I0 (2 z) + p I0 2 k +
B B
! 0,
< 1. Analogously,
p I1 ( 2 z ) p A z p I1 ( 2 k ) p A k
! 0,
Case 2b. Suppose
A B
! k < 1 and lim
A B
= 1. Then, from Lemma 5, p
p e2
lim
Pr [P ivAj ] e = lim Pr [P ivAj ] e
A
1+ p p p 2 2
A B
B
A
B
A B
!
p I1 (2 k) p B k
p 2 k +
I0
B A
.
Now, observe that p
Moreover, from I0
p 2 k 2 (0; 1), p
1+ p p p 2 2
lim I0
p 2 k +
B A A B
!
p
I1 (2 k) p B k
A
A B
p I1 (2 k) p k
A
B
p
B
2 (0; 1),
lim
p
e2 p A B e2 p 2 A B)
e lim e e p p e ( 2 k = e lim p e ( = 0. 2
p p
)
B
lim
B
A B
2
B A
A B
! 1, and p p
B A
B
=p
A B
1 A B
!1
! 0.
This proves the result. A.4.3
Proof for the Binary Case
Suppose signals are binary and suppose <1 throughout the section. We sometimes refer to a signal x < xB as an “a signal” and a signal x
xB as a “b
signal,” analogously to the notation from existing work on binary signals by Krishna and Morgan (2012) and others. We use the following notation for the signals, r = G (xB j ) and 1 s = 1
r=1
G (xB j ) and 1 32
G (xB j )
s = G (xB j ) ,
so that
r 1 r
1 s s .
>1>
p = q = =
We only consider strategies (y; z) with y
G (min fy; xB g j ) = Pr (Vote Ajx xB ) G (xB j ) G (xB j ) G (min fz; xB g j ) = Pr (Vote Bjx G (xB j ) 1 G (max fz; xB g j ) = Pr (Vote Bjx xB ) . 1 G (xB j )
Noting that the de…nitions above also hold with G(minfy;xB gj ) , G(xB j )
xB . Let
replaced by
xB )
since
etc...
G(minfy;xB gj ) G(xB j )
=
Step 1. Suppose 1
r s
> .
Then, there exist equilibria such that lim np = 1 and lim pr < lim n (1 equilibria,
r). In these
= 1 > p > 0 = q for all n large.
Proof of Step 1. We want to prove the existence of an equilibrium with p 2 (0; 1),
= 1,
and q = 0. Such a strategy pro…le is an equilibrium if 1 Pr [P ivAj s Pr [P ivAj r 1 Pr [P ivAj s Pr [P ivAj r
1 1
Given
] ] ] ]
= 1 and 1 and
r 1 1
1 Pr [P ivBj s Pr [P ivBj r 1 Pr [P ivBj s Pr [P ivBj
] ] ] ]
1, 1.
= 1, and q = 0, (15) becomes
Pr [P ivW j ] lim n!1 Pr [P ivW j ] 8 p < 1 if limn!1 1 = p : 0 if limn!1 1 Fix some
<
1 r s .
(17) r
p
r
p
pr < limn!1 pr > limn!1
p p
p
s
p
s
p p (1 p p (1
s) , s) .
Let p be the p solution to p (1
r)
p
pr =
p 33
p s
p p (1
s) .
(18)
A solution exists by the intermediate value theorem because at p = 0, the left-hand side> right-hand side by 1
r > s , and for p = 1, the left-hand side < 0 < right-hand
side. The solution p satis…es 0 < p < 1. Moreover, the solution also satis…es p r<1
r,
for otherwise the left-hand side would be non-positive while the right-hand side> 0. The p p solution is unique: For this, note that both sides are linear in p with coe¢ cients r > p p 1 s. Thus, p
p (1
r)
p p pr >
p
s
and vice versa for p > p . Pick " > 0 such that 0 < p (1
p p (1
s)
(19)
" < p + " < 1 and (p + ") r <
r). Notice that for all p 2 [p
"; p + "],
A B
= (npr) (n (1
r)) ! 1, validating the
use of the previous approximations. With p as a free variable, de…ne P (p; n) :=
Pr [P ivAj ; p] . Pr [P ivAj ; p]
From (17) and (19), lim P (p
"; n) = 0, and lim P (p + "; n) = 1.
(20)
(A rough intuition may be this: B wins in both states. The margin of victory is necessarily larger in state
and the election closer to being tied in state
election even closer to being tied in state
. Increasing p brings the
. Thus, the probability of state
increases if
we increase p.) Notice that for …xed n, P (p; n) is a continuous function. This and (20) implies that for all large n, there is a pn 2 (p
"; p + ") such that r
1
1 s
P (pn ; n) = 1 for all n.
From (17) and from the de…nition and uniqueness of p , we have pn ! p .
34
(21)
Now, observe that Pr [P ivBj ] Pr [T j ] 1 + = Pr [P ivBj ] Pr [T j ] 1 +
p p p p
A B A B
Pr [T j ] 1 + = Pr [T j ] 1 +
p p p p
p p p p
B A B A
A B A
Pr [P ivAj ] = Pr [P ivAj ]
B
p p p p
A B A
,
(22)
B
using (9) and (10). From, A B A B
=
r (1 r) (1 s) s
s
r
=
(1
r) (1
s)
> 1,
(23)
we have that the second optimality condition for voters with an a signal holds since, r 1
1 Pr [P ivBj ] > 1. s Pr [P ivBj ]
Now, consider voters with a b signal. From the indi¤erence condition of the a voter, Pr [T j ] 1 + lim Pr [T j ] 1 +
p p p p
B A
1
=
B
s r
.
A
Hence, from (22) and (23) Pr [P ivBj ] (1 s) lim = Pr [P ivBj ] r
r
Hence, lim
1
r 1 Pr [P ivBj ] 1 r1 = s Pr [P ivBj ] s
r (1 s
s r) (1
(1 (1
s)
s) s = r) r
=
s
r
1
(1 (1
s) s . r) r
s1 s
r r
< 1.
So, the b voter strictly prefers voting for B to abstaining. Moreover, from MLRP, 1
Therefore,
r 1 Pr [P ivAj ] r 1 Pr [P ivAj ] < = 1. s Pr [P ivAj ] 1 s Pr [P ivAj ]
= 1 is a best response to (pn ; q = 0;
Thus, for large n, (pn ; q = 0;
= 1) for n large enough.
= 1) is an equilibrium pro…le. Finally, from pn ! p
and p solving (18), it must be that pn r <
(1
35
r) = (1
r). This proves the …rst step.
Step 2. Suppose 1> For all n and
>
1
r s
.
(24)
n, there exists an equilibrium pro…le (p ; q ;
= 1 such that for n ! 1 lim rp < lim rq + (1
) with p 2 (0; 1), q 2 (0; 1),
r) .
A strategy pro…le (p; q; ) with p 2 (0; 1), q 2 (0; 1), and r 1 1
1 Pr [P ivAj s Pr [P ivAj r 1 Pr [P ivAj s Pr [P ivAj
] ] ] ]
= 1 and 1 and
r 1 1
= 1 is an equilibrium if
1 Pr [P ivBj s Pr [P ivBj r 1 Pr [P ivBj s Pr [P ivBj
] = 1, ] ] 1. ]
Note that the inequalities are implied by the equalities. De…ne Pr (P ivAj Pr (P ivAj Pr (P ivBj f (p; q) : = Pr (P ivBj g (p; q) : =
for (p; q) 2 [0; 1]2 , p + q
; p; q) , ; p; q) ; p; q) , ; p; q)
1.
The proof uses two lemmas. Lemma 8. There exist n, and p; q; q0 2 [0; 1]3 , with q > q0 , p > 0, q + p rq0 + (1
r) > rp such that for all n
n and p 2 [0; p] ,
r 1 and
1 s
r 1
1, and
f (p; q0 ) > 1,
1 s
f (p; q) < 1,
36
(25)
(26)
So, by continuity of f , for all p 2 [0; p] there exists a solution q~ (p) 2 [q0 ; q] to 1
r 1
s
f (p; q) = 1.
Any such solution satis…es g (0; q~ (0)) > f (0; q~ (0)) ,
(27)
g (p; q~ (p)) < f (p; q~ (p)) .
(28)
and
Proof: Pick some q0 2 (0; 1) with q0 r + (1
r) < (q0 (1
s) + s) ,
which exists by (24) and pick some q 2 (q0 ; 1) with qr + (1 which exists by
r) > (q (1
s) + s) ,
< 1 from (24). Finally, pick p 2 (0; 1 p rq + (1
r)
p p rp > ((1
q) small enough such that
s) q + s)
p
Such p exists since the inequality holds strictly for p = 0. (1
s) p.
Since
((1
s) p, rq0 + (1
r) > rp.
Remark: The choices imply that for all (p; q) 2 [0; p] p and hence
and
(1
p
B
B
>
p
[q0 ; q],
A,
>
p
p
B
p
A
>
p
B
p
A
for q = q0 and any p 2 [0; p] ,
p
B
p
A
<
p
B
p
A
for q = q and any p 2 [0; p] .
A.
Furthermore,
37
(29) s) q + s) >
Proof of Equation (25), 1
r 1
s
f (p; q0 ) > 1 for all 0
p
p.
The proof shows that for any sequence of (p; n), for n ! 1, f (p; q0 ) ! 1.
Case 1:
! 1. Then,
A B
p
Pr [P ivBj ] e ( lim = lim p Pr [P ivBj ] e (
A
p
B
)
A
p
B
)
2
2
K ( ; ),
with lim K ( ; ) 2 (0; 1). We show that p lim ( that is, by choice of p p (1
A
p (1
s) q0 + s
((1 < 1 and (1
B)
2
p (which implies
By choice of q0 ,
and from
p
p
s) p
A B
0 as well. Also
p
B
p
A ),
= 1,
r)
p
rp
2
> 0.
r) ,
s) p < rp.
! k < 1. Suppose k = 0 (the case k 2 (0; 1) is analogous). Then, A
! 0 from
n (1
B
> 0 and lim e
(p p
e (
A
p
A
p
r) ! 1. Similarly,
2 B) 2 B)
r 1 A B
2
s) < r, for all p,
A
p
A
p
)
2
B
)
= 1 (as before). 1
s
f (p; q) < 1.
! 1. Then, from Lemma 6 38
A
B B
A B
! 0. From Lemma 5,
2
B
Proof of Equation (26),
Case 1:
B)
p rq0 + (1
2
p
B +1 B +1
p
>
Pr [P ivBj ] e ( lim = lim p Pr [P ivBj ] e ( since lim
A
s) q0 + s) > rq0 + (1
(1 Case 2:
(
+1 = 1, +1
!
p
Pr [P ivBj ] e ( lim = lim p Pr [P ivBj ] e (
A
p
B
)
A
p
B
)
2
2
K ( ; ),
with lim K ( ; ) 2 (0; 1). We show that p lim ( that is,
p (1
s) q + s
A
p
p (
2 B)
p (1
s) p
2
but this holds by choice of p in (29). Case 2:
A B
A
p
2 B)
p rq + (1
=
1
r)
p
rp
2
<0
! k. This is just as the proof of Case 2 in the proof of Equation (25) and
therefore omitted.
Proof of Equation (27): Any solution q~ (p) with r 1
1 s
f (p; q~ (p)) = 1.
satis…es g (0; q~ (0)) > f (0; q~ (0)) .
(30)
So, suppose p = 0 in the following. A B vote is only pivotal if there is no other vote. An A vote is pivotal is there is either no other vote or one other vote (which would be a B vote). Let m count the number of other B votes. The probability that there is either no voter or 1 voter is Pr (m = 0j ) = e
n(qr+(1 r))
Pr (m = 1j ) = e
n(qr+(1 r))
n (qr + (1
r))
and Pr (m = 0j ) = e
n(q(1 s)+s)
Pr (m = 1j ) = e
n(q(1 s)+s)
39
n (q (1
s) + s)
and so a B vote is pivotal implies n(qr+(1 r))
e Pr [P ivBj ] = Pr [P ivBj ] e
= en(
n(q(1 s)+s)
(q(1 s)+s) (qr+(1 r)))
and Pr [P ivAj ] Pr [P ivAj ]
= =
Let f (q; n)
Pr[P ivBj ] Pr[P ivBj ]
n(qr+(1 r))
e
+e s)+s) + e
e
n(q(1
e e
n(qr+(1 r)) n(q(1
n(q(1 s)+s)
+ (1 r)) n (q (1 s) + s)
1 + n (qr + (1 r)) s)+s) 1 + n (q (1 s) + s)
(recall that we set p (q (1
n(qr+(1 r)) n (qr
0 for this part). Let q 2 (q0 ; q) solve
s) + s) = (qr + (1
r))
Such a solution exists because by our choices at q = q0 , we have s > q0 r + (1 q = q we have s < qr + (1
r).
Then, lim f (q0 ; n) = 1 since at q = q0 , lim n ( (q (1
s) + s)
(qr + (1
r))) = 1,
and lim f (q; n) = 0, since at q = q lim n ( (q (1
s) + s)
(qr + (1
Thus, for all large n, we can …nd a q~n such that 1
r 1
s
f (~ q n ; n) = 1.
Finally, from f (~ q n ; n) < 1,
40
r))) =
1.
r) and for
and
n(qr+(1 r))
Pr [P ivBj ] e = Pr [P ivBj ] e
n(q(1 s)+s)
= en(
(q(1 s)+s) (qr+(1 r)))
,
we get (~ q n (1
s) + s) < (~ q n r + (1
and so with Pr [P ivAj ] e = Pr [P ivAj ] e we have
This and
n(qr+(1 r)) n(q(1 s)+s)
r)) ,
1 + n (qr + (1 r)) , 1 + n (q (1 s) + s)
1 + n (~ q n r + (1 r)) > 1. 1 + n (~ q n (1 s) + s) Pr [P ivAj ] Pr [P ivBj ] 1 + n (~ q n r + (1 r)) = , Pr [P ivAj ] Pr [P ivBj ] 1 + n (~ q n (1 s) + s)
implies Pr [P ivAj ] Pr [P ivBj ] > , Pr [P ivAj ] Pr [P ivBj ] proving what we wanted to show. Remark. This step that fails with
6= 1. If
6= 1, then there may be equilibria in
which p = 0 and q 2 (0; 1), supported by the fact that because when
6= 1, then it does not need to be the case
Pr[P ivAj ] Pr[P ivBj ] Pr[P ivAj ] Pr[P ivBj ] . n that f (~ q ; n) < 1.
This is
Proof Equation (28): Any solution q~ (p) with 1
r 1
s
f (p; q~ (p)) = 1.
satis…es g (p; q~ (p)) < f (p; q~ (p)) . By choice of the boundaries, q~ (p) 2 [q0 ; q] A B
! 1. So,
Pr [P ivBj ] Pr [P ivBj ]
(0; 1) and p 2 (0; 1), implying that
Pr [P ivAj ] Pr [P ivAj ]
41
p p p p
A B A B
,
and inspection shows A
B
B
A
=
q+ rp (1 s) q + s = (1 s) p rq + (1 r) q+
s 1 s 1 r r
> 1,
and hence, Pr [P ivBj ] Pr [P ivAj ] > , Pr [P ivBj ] Pr [P ivAj ] as claimed. QED.
Lemma 9. Continuous Selection. Suppose f : [0; 1]
[0; 1] ! [0; 1] is a function such that
f (0; r) < 1 for all r f (1; r) > 1 for all r f is continuous in (x; r) Then, there exist continuous functions x ^ : [0; 1] ! [0; 1] and r^ : [0; 1] ! [0; 1] such that
r^ (0) = 0, r^ (1) = 1 and
f (^ x (t) ; r^ (t)) = 1 for all t: Proof: See Section B. QED. Proof of Step 2: For all n and
n, there exists an equilibrium pro…le (p ; q ;
) with p 2 (0; 1), q 2 (0; 1),
= 1 such that for n ! 1 lim rp < lim rq + (1
We are looking for (p ; q ) 2 [0; p]
r) .
[q0 ; q] such that
g (p ; q ) = f (p ; q ) = Any such solution will be an equilibrium. Since q aggregate. 42
1
s r
.
q0 and p
p, information fails to
Note that f is a continuous function with f (p; q0 ) < for all p 2 [0; p] and n
1
s r
< f (p; q) ,
n, by (25) and (26) from Lemma 8.
Thus, by the continuous selection lemma, there are continuous functions p^ : [0; 1] !
[0; p] and q^ : [0; 1] ! [q0 ; q] such that p^ (0) = 0 and p^ (1) = p and f (^ p (t) ; q^ (t)) =
1
s r
for all t.
Now, g (^ p (0) ; q^ (0))
f (^ p (0) ; q^ (0)) = g (0; q^ (0))
f (0; q^ (0)) > 0,
g (^ p (1) ; q^ (1))
f (^ p (1) ; q^ (1)) = g (1; q^ (1))
f (1; q^ (1)) < 0;
and
by (27) and (28). By construction, the di¤erence g (p; q) f (p; q) is continuous in (p; q). Hence, g (^ p (t) ; q^ (t)) f (^ p (t) ; q^ (t)) is continuous in t. Therefore, by the intermediate value theorem, there exists some t such that g (^ p (t ) ; q^ (t ))
f (^ p (t ) ; q^ (t )) = 0.
Thus, (p ; q ) = (^ p (t ) ; q^ (t )) solves the equilibrium conditions. Since p^ (t ) ; q^ (t ) 2 [0; p]
B
[q0 ; q], we have lim rp < lim rq + (1
r). This proves Step 2.
Generalized Intermediate Value Theorem
Lemma 9 (Restatement). Continuous Selection. Suppose f : [0; 1] a function such that f (0; r) < 0 for all r f (1; r) > 0 for all r f is continuous in (x; r)
43
[0; 1] ! [ 1; 1] is
Then, there exist continuous functions x ^ : [0; 1] ! [0; 1] and r^ : [0; 1] ! [0; 1] such that
r^ (0) = 0, r^ (1) = 1 and
f (^ x (t) ; r^ (t)) = 0 for all t: For intuition, consider the following related statement. Suppose g : R2 ! R is a
continuous function with g (x0 ) = g0 > 0 for some x0 2 (0; 1)2 and g (x) = 0 for all x2 = [0; 1]2 . Then, for every b 2 (0; g0 ) there exists a loop in [0; 1]2 that goes around x0 such that g (x) R2 .)
b for all x from this loop. (A loop is a continuous map from the unit circle to
Put di¤erently, consider the iso-height lines of some mountain on a map. Then, for
every elevation level that is between the peak of the mountain and the surrounding plains, there is at least one unbroken iso-height line that goes around the peak of the mountain. The lemma makes an analogous statement for the segment of a mountain ridge. The lemma extends the intermediate value theorem in the following sense to twodimensions. Suppose there is another function q : [0; 1]
[0; 1] ! [ 1; 1] that is con-
tinuous and which has the property that q (x; 0) < 0 if f (x; 0) = 0 and q (x; 1) > 0
if f (x; 1) = 0. Then, the lemma implies that there exists a point (x0 ; r0 ) such that f (x0 ; r0 ) = q (x0 ; r0 ) = 0. This is because the lemma implies that q (^ x ( ) ; r^ ( )) is a continuous function with q (^ x (0) ; r^ (0)) < 0 and q (^ x (1) ; r^ (1)) > 0. Proof:12 Let F be the upper contour set of f : F = f(x; r)jf (x; r) > 0g: Notice that F is an open subset of X = [0; 1]
[0; 1] in induced topology, since f is
continuous. E0
Let E be the component of F containing the right edge, YR = f[1; y] j y 2 [0; 1]g. Let
=E
@X. Then, E 0 is an open planar surface. As we can restrict ourselves only to
the end of E 0 containing YR , without loss of generality, we can assume the open surface E 0 has …nite topology, i.e. it has …nite number of ends (it has already genus 0 as it is planar). Therefore, by the classi…cation of surfaces, there exists a compact surface with boundary, which we call S, such that E 0 is homeomorphic to the interior of S, int(S). Let 12
We are thankful for help with the proof of this result to Baris Coskunuzer, Mathematics Department at Koc University.
44
g : int(S) ! E 0 be the homeomorphism (continuous map with continuous inverse). By
inducing the path metric of E 0 , we de…ne a metric on int(S) via g.
Let E be the closure of E 0 in X. Let gb : S ! E be the continuous extension of g. In
particular, for any p 2 @S, let fpn g be a sequence in int(S) with pn ! p. Then qn = g(pn )
de…nes a Cauchy sequence in E 0 as S has the induced metric. Since E is compact, fqn g
is convergent sequence in E, say qn ! q. Then, we naturally de…ne gb(p) = q which is continuous by construction. Notice that the restriction of gb to @S de…nes a continuous and onto map from collection of loops (@S) to E
E 0 , i.e. gb : @S ! (E
E0)
Let xT : inffx 2 [0; 1] j (x; 1) 2 Eg, and xB : inffx 2 [0; 1] j (x; 0) 2 Eg. Note that
xB ; xT 2 (0; 1), because of the intermediate value theorem. Also, because f is continuous, f (xT ; 1) = f (xB ; 0) = 0, and also for any (x; y) 2 (E Let
in E
be the component of @S where gb( )
E0
and gb : @S ! (E
E0)
gb(pT ) = xT and gb(pB ) = xB . As endpoints fpT ; pB g, i.e.
+
[
is not containing YR . Then,
As the arc of . As
+
E), f (x; y) = 0.
YR . Since xT and xB are connected to YR
is onto, then xT ; xB 2 gb( ). Then, let pT ; pB 2
is a loop, there are two closed arcs
=
= gb(
and @
+)
= fpT ; pB g. Let
is a continuous arc in E
+
+
and
in
with with
be the arc where gb(
+)
E with @ = fxT ; xB g.
is homeomorphic to an interval I, gb provides a continuous parametrization
E
E and f = 0 along E
E, the proof follows.
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