On Infinite Cheap Talk Equilibria
Sidartha Gordon
yAugust 2010
This paper generalizes the Crawford and Sobel (1982) model of strategic information transmission to situations where not only the magnitude, but also the direction of the Sender’s bias may depend on his type. The results show that the existence of an equilibrium with in…nitely many equilibria is associated with the Sender being more extreme that the Receiver at extreme types, that is with him having anoutward bias. When the bias is globally outward, an in…nite equilibrium exists. Conversely, we show that if there is an equilibrium such that the precision of information revealed around some interior type is in…nite, then the Sender and Receiver must agree at this type, and the Sender’s bias must be locally outward around it. A new …xed-point technique to describe the structure of the equilibrium set in this more general setting is introduced.
Journal of Economic Literature Classi…cation Numbers: D72, D78, D82.
Keywords: Cheap Talk, Strategic Information Transmission, Fixed Points.
Part of this work was undertaken while the author was a postdoctoral researcher at CORE in 2004-2005. The paper has circulated under the title "Informative Cheap Talk Equilibria as Fixed Points" and other titles since october 2004. I am thankful to Oliver Board, Jacques Crémer, Vincent Crawford, Federico Echenique, Peter Esö, Françoise Forges, Rick Harbaugh, Johannes Hörner, Navin Kartik, Kohei Kawamura, Frédéric Koessler, Vijay Krishna, Ming Li, Itai Sher, Joel Sobel, Satoru Takahashi, and Tim van Zandt for helpful comments, conversations, or suggestions. I also thank seminar participants at CORE-UCL, INSEAD, CETC 2006, CEA 2007, the Universities of Cornell, Maastricht, Montréal, Minnesota, Windsor, Western Ontario, Namur, Sabanc¬and the Paris School of Economics for extremely stimulating questions, objections and remarks.
yDepartment of Economics and CIREQ, Université de Montréal, C.P. 6128 succursale
1
Introduction
A growing literature examines how a privately informed agent, the Sender, can in‡uence a decision maker, the Receiver, by supplying relevant, yet unveri…able, in-formation. To in‡uence the decision, all the Sender can do is talk. Talking is free of costs, in the sense that messages do not enter the payo¤ of the players. This problem of cheap talk signalling is interesting when the Sender and the Receiver do not have the same preferences, i.e. when the Sender is biased.
The model of Crawford and Sobel (1982, henceforth CS) is a classic in this lit-erature. The Sender has a type that encodes his relevant information. Both the Sender’s type and the Receiver’s action are unidimensional. Both the Sender and the Receiver have a preferred action that is an increasing function of the Sender’s type. The Sender is biased upwards: for all types, his preferred action is higher than the Receiver’s. Under this assumption, these authors show that information revelation is bounded, in the sense that, in any equilibrium, the set of actions that the Receiver takes, depending on the information he receives, is …nite.
is outward (Theorem 7). In other words, the bias must be outward at least locally around the type at which the information is …nely revealed. This says that it be must downward on a left neighborhood of the agreement type and upward on a right neighborhood of it.
Besides the main insight on the relation between in…nite equilibria and outward biases, the paper provides a detailed analysis of the set of equilibria for games in this larger class. Interestingly, the structure of this set inherits important features of the upward bias case. In particular, we prove that if there is an equilibrium where
> 1 actions are taken by the Receiver, then there are equilibria in which exactly
k actions are taken by the Receiver, where k is any positive integer lower than (Theorem 1): Moreover, the latter are “nested” into the former, in the sense that the boundary points of the -actions equilibrium de…ne bounds within which a k
actions equilibrium (withk < ) necessarily exists (Theorem 1). Similarly, if there is an equilibrium where in…nitely (either countably or uncountably) many actions are taken, then there are equilibria where exactly k actions are taken, where k is any positive integer (Theorem 2): A converse of this result also holds. If for each positive integer k; there is an equilibrium where exactly k actions are taken, then there is an equilibrium where in…nitely many actions are taken (Theorem 2). Sometimes countably, sometimes uncountably, depending on the game.
In another series of results, we provide upper bounds on the amount of information that can be revealed in equilibrium. CS show that whenever the direction of the bias is constant and its magnitude is bounded away from zero, the number of actions the Receiver takes in equilibrium is bounded. We show that this is true whenever the bias’magnitude is bounded away from zero, and whether or not the direction of the bias is constant (Theorem 3). This result is robust in the following sense. In the general case, if the set of types at which the Sender and the Receiver agree is at most countable, then the set of actions the Receiver may take in any equilibrium is at most countable (Theorem 5).
way on the bias direction being the same accross types. If this restriction on the parameters is removed, some of these objects are no longer well-de…ned. Instead, we show that all equilibria where the Receiver takes the same number of actions are the …xed points of a correspondence de…ned on a subset of the Sender’s strategy space. Thus, the set of equilibria is the set of …xed points of a collection of correspondences, which turn out to be de…ned and tractable objects, regardless of the form of the bias. Each of the equilibrium correspondences is monotone, on its domain, for a certain partial order. This monotonicity has strong implications for the structure of the set of the equilibria where the same given number k of actions is taken. In their work, CS introduce a parametric restriction, known as condition M, which implies that this set is a singleton. However, when condition M does not hold, there may be several equilibria where the Receiver takes the same number of actions. We provide an example that satis…es all of the CS assumption except Condition M and has three equilibria with three actions (Example 1). When the Sender has an outward bias, the set of equilibria where a given numberk 2of action is taken is nonempty and has a (complete) lattice structure (Theorem 4). In particular, it has a least and a greatest element. More generally, if the bias is upward at the highest type (this includes both the upward and outward cases), this set may be empty. But if it is not empty, we show that it is an upper-semilattice. In particular, it has a greatest element. We then go on to prove various comparative statics results on this greatestk actions equilibrium. We also provide a simple algorithm that converges monotonically to this equilibrium (Theorem 6).
There are many economic situations where type-dependent bias directions arise naturally. Let us list only a few of them.
First, suppose that the Receiver is a government, and the Sender is an expert, hired by the government to advise it on a one dimensional policy reform from a current status quo a to a new policy a. The expert’s type represents the policy the expert believes the government should take. The government trusts the expert to indicate the direction of the change, i.e. whether a should be greater or lesser than
is more or less reluctant that her to implement large policy changes. To …x ideas, let the type t be distributed in [ 1;1], and the preferred policy of the government under complete information be R(t) = a +t. Instead, the expert would like the government to implement S(t) = a + t; where is a parameter that measures the relative conservatism of the expert in relation to the government . If >1;the expert has an outward bias, because S( 1) < R( 1) < R(1) < S(1). If instead < 1; the expert has an inward bias.
Consider now a second situation. Let the Receiver be a legislature divided in two groups, and let the Sender be an expert, hired to advise it on a one dimensional policy
a. The expert reports to the legislature, which then collectively choose the policy a. Speci…cally, the chosen policy is the outcome of a bargaining game among the two members of the legislature. To …x ideas, let the typetbe any real number in[0;1], and letS(t) = tbe the policy the expert would like the government to implement. Let the preferred policy of one of member1 under complete information beR1(t) = 34+32t,
and let the preferred policy of member2under complete information beR2(t) = 14+32t.
Let the outcome of the legislative bargaining under complete information beR(t) =
R1(t) + (1 )R2(t) = 4 +14 + 32t; where 2 [0;1] is a parameter that measures
the relative bargaining power the …rst group: Here, the expert has an upward bias, with respect to member 1, and a downward bias, with respect to member2. Indeed, we have for allt2[0;1],R1(t)< S(t)< R2(t). But when comparing the expert, and
the legislature’s rule R(t), the bias is upward if < 14; and inward if > 14, since in this case, the inequalities R(0) < S(0) < S(1)< R(1) hold.
of in…nite equilibria in other models that are not strictly speaking special cases of ours, because upward, outward and inward biases apear endogenously in these more complicated games. Beyond the examples and the particular functional form that arises in each of the applications, the aim of the paper is to establish that a general connexion exists between the possibility of in…nite equilibria and an outward bias.
The rest of the paper is organized as follows. Section 2 lays out the model. Section 3 studies …nite informative equilibria. Section 4 studies in…nite equilibria. Section 5 introduces a taxonomy of Sender’s biases, and specializes the results of Sections 3 and 4 to certain special cases. Section 6 provides further results on the structure of the equilibrium set. Section 7 provides a necessary condition for an in…nite equilibrium to exist. Section 8 present the related literature. Section 9 lays out directions for future research.
2
The model
There are two players, the Sender and the Receiver. Only the Sender has payo¤-relevant private information, his type. The Sender observes his type, and sends a message to the Receiver. The Receiver then reads this message, and takes an action. Talking is “cheap”, in the sense that messages do not directly a¤ect payo¤s.
Let T [0;1] be the Sender’s set of types, with typical element t: Let A R be a nonempty set of Receiver’s possible actions, with typical element a: A preference overA is a binary relation that is re‡exive, transitive and complete. The Sender has a preference relation t over A, which depend on his type t. For all a; b 2 A; the
propositiona tb means that the Sender of typet weakly prefers action a to action
b. The corresponding strict preference and indi¤erence relations are denoted by t
and 't. Let denote the family of preferences f tgt2[0;1].
I(t)\I(t0) = ;for all t; t0 2T:A Sender strategy is described by a partition I( ) of
T. For all t; the pool I(t) is the set of Sender types that send the same message as typet under strategyI( ). The encoding of information, i.e. what messages are sent by each of the pools, is irrelevant.
Any strategy I( ) induces an outcome : T ! A such that (t) = R(I(t)):
An equilibrium strategy for (R; ) is a partition I( ) such that for all t; t0 2 T; we
have R(I(t)) t R(I(t0)). Anequilibrium outcome for (R; ); is an outcome that is
induced by some equilibrium strategy for (R; ):1
We now introduce assumptions on the Sender’s preferences. A preference t is single-peaked if it has a unique preferred action S(t) 2 A (its peak) and, among any two distinct actions on the same side of the peak, the one closest the peak is preferred. More precisely, there is an action S(t) 2 A such that for all a; b 2 A
satisfying eithera < b S(t)orS(t) b < a;we haveb ta: We say that the family
is single-peaked if, for all t 2 T; the preference t is single-peaked. Next, we
assume that the Sender’s preference unambiguously shifts in favor of higher actions, as his type increases. The family is single-crossing if higher-types prefer higher actions, i.e. for all s; t 2 T such that s < t, and for all a; b 2 A such that a < b;
we have b s a ) b t a: If the family is single-peaked and single-crossing, then
the function S( ) is nondecreasing on T: Last, we impose regularity on the way the Sender’s preferences change, as the type varies. A strict preference relation should not be reversed by an in…nitesimal change of the type. We say that the family is
type-continuous if, for all a; b2A; the set ft2T :a tbgis closed.
We now turn attention to the Receiver and introduce two assumptions on his re-action function. First, we require that if the Receiver disregards some information when taking his action, then suppressing this information should not a¤ect his deci-sion. The reaction R is consistent if, for all family of disjoint pools T T and all a 2 A satisfying (for all I 2 T, R(I) =a), we have R([I2T I) = a. Second, we
impose regularity on the way the Receiver reacts to interval pools. The reaction R
isrobustif the Receiver reaction to an interval pool that is not a singleton does not
1Besides the more general form of the bias, this setup is slightly more general than CS in other
depend on whether the endpoints of the interval are included in the pool, i.e. for all
s; t 2T such that s < t, we have R([s; t]) =R(]s; t[) =R(]s; t]) =R([s; t[). Abusing notations, for all s; t 2T such that s t, let R(s; t) := R([s; t]):
For any outcome (t)induced by some strategy, letI (t) ft0 2T : (t) = (t0)g
be the outcome partition for : An interval partition is a partition whose pools are all intervals inT (some of them possibly singletons). A well-known result in this type of model is that any equilibrium outcome partition is an interval partition. Indeed, for all t 2T, and any outcome ;we have
R(I (t)) =R(ft0 2T : (t) = (t0)g) =R S
t02T:R(I(t0))= (t)
I((t0))
!
= (t);
where the last equality holds becauseR is consistent. Therefore ( )is also induced by strategyI (t):In particular, any equilibrium outcome partitionI ( )is an equilib-rium strategy for(R; ):Because is single-crossing and by Theorem 2.8.1 in Topkis (1998), this implies that any equilibrium outcome partition is an interval partition, which is our …rst Lemma.
Lemma 1: Let R be consistent and let be single-crossing. Any equilibrium
outcome partition I ( ) for (R; ) is an interval partition.
The rest of the paper is devoted to analyzing which interval partitions are equi-librium outcome partitions.
We now introduce partial orders on vectors and sets of on vectors and monotonicity notions for correspondences. Let m; n be arbitrary positive integers. For any two nonempty subsets X; Y Rm, let X Y if, for all x 2 X, and all y 2 Y, we have
x y. Similarly, let X < Y if, for all x 2 X, and all y 2 Y, we have x < y. Let X Rm and Z
Rn. Let G : X Z be a correspondence such that G(x)
is nonempty for all x 2 X. We say that G is nondecreasing if, for all x y 2 X, we have G(x) G(y). We say that G is increasing if for all x < y 2 X, we have
G(x) < G(y): An equivalent de…nition is that G is nondecreasing (increasing) if all selections from Gare nondecreasing (increasing) functions, in the usual sense.
an important role in our results. Let be the correspondence such that, for alla; b2A
satisfyinga < b, a 0 b and b 1 a; we have (a; b) :=ft 2[0;1] :a'tbg;and for all
a2A satisfyingS(0) a S(1)); we have (a; a) :=ft2[0;1] :S(t) = ag: Let
D :=f(a; b)2A2 : (a < banda 0 b andb 1 a) or (S(0) a=b S(1))g:
Lemma 2: Let be single-peaked, single-crossing and type-continuous.2 Then
is nonempty valued and increasing on D :
Proof. Since is single-crossing and type-continuous, then a 0 b and b 1 a
imply that the set ft 2 [0;1] : a 't bg is a singleton. Similarly, since is
single-peaked, single-crossing, andtype-continuous, the inequalities S(0) a S(1) imply that the set ft 2[0;1] :S(t) =ag is a nonempty closed interval. Therefore (a; b) is a nonempty closed interval for all a b, and is a singleton when a < b. Thus is well-de…ned onD :
We now prove that is increasing on D : Let (a; b) 2 D and (c; d) 2 D ; such that (a; b) (c; d) and (a; b) 6= (c; d): Lets 2 (a; b) and t 2 (c; d): We will prove that s < t. We distinguish three cases. Case 1: c < d. Since s is single-peaked,
then c s d: Since c't d, and is single-crossing, then s < t. Case 2: a < b. Since
t is single-peaked, then b t a. Since b 's a, and is single-crossing, thens < t.
Case 3: a = b < c = d. Since S(s) = a and S(t) = c and S( ) is nondecreasing, therefore s < t, the desired conclusion.
Finally, the following result is useful.
Lemma 3: Let be single-peaked and type-continuous. Then the function S( )
is continuous in t.
Proof. Let t 2 [0;1] and " > 0: Let a := S(t): Since is type-continuous, the set O := fs2[0;1] :a s a "anda s a+"g is open in [0;1]: Therefore it is a
neighborhood of t in [0;1]: For all s 2 O; since s is single-peaked, then S(s) 2
(a "; a+"): ThereforeS( )is continuous at t:
2Lemma 3 would not hold, if we were to replace our de…nition of single-crossing with the standard
strict single-crossing condition that, for alla; b2Asuch thata < b;and alls; t2T such thats < t;
A problem (R; ) is admissible if the Sender preferences are single-peaked, type-continuous and single-crossing; the Receiver reaction R is consistent, robust; and the function (s; t) 7! R(s; t) is increasing. In the remainder of the paper, we restrict attention to admissible problems. We now study the structure of the set of informative equilibria where the Receiver takes …nitely many actions in an admissible problem.
3
Finite informative equilibria
An equilibrium outcome partition is said to be …nite if the Receiver only takes a …nite number of actions. It is well known that any cheap talk game has an equilibrium with a single action, the trivial partition with only one interval[0;1]. Our goal in this section is to describe the structure of the set of …nite informative equilibria, i.e. the set of equilibia where the Receiver takes a …nite number of actions, greater than one. Any partition with 2intervals can be represented by a vector x2T +1 such that
0 =x0 : : : x = 1:LetX be the set of such vectors. For eachl = 1; : : : ; 1;the
typexl is the boundary between the l-th and the (l+ 1)-th intervals of the partition,
ranked in increasing order.
For all 1; an equilibrium vector x 2 X is a vector that represents an equi-librium outcome partition with actions, a -actions equilibrium. It is use-ful to introduce here a -actions equilibrium correspondence. For each 2, let ( ) be the correspondence that maps a vector x in X ; to a set of vectors
(x) 0(x) : : : (x) T +1; where 0(x) x0; (x) x ; and for all
l = 1; : : : ; 1; we have
l(x) (R(xl 1; xl); R(xl; xl+1)):
The domain on which ( )is nonempty-valued is the set D X of vectorsx such that(R(x0; x1); R(x1; x2))2D and (R(x 2; x 1); R(x 1; x ))2D :Indeed, these
two relations, together with the inequalities x0 : : : x ; and the monotonicity of
R( )ensure that, for all l = 1; : : : ; 1; we have (R(xl 1; xl); R(xl; xl 1)) 2D ; i.e.
D could be empty.
For all x 2 D ; we have 0 1(x) : : : 1(x) 1: In addition, for all
x 2 X ; we also have 0 = 0(x) 1(x) and 1(x) (x) = 1; so that for all
x 2 D ; we have (x) X : We will now show that the equilibrium vectors in X
are the …xed-points of the correspondence ( ) in the set D :
Lemma 4: Let 2. A vector x 2 X is an equilibrium vector i¤ x 2 D and
x2 (x): When this is the case and >2, we have x1 < : : : < x 1:
Proof. Since is type-continuous and by Lemma 1, any equilibrium vector x2X
must satisfyx2D and x2 (x): We will now prove that the converse also holds. Let > 2 and let x 2 X be a …xed-point of ( ): To alleviate notations, for all relevant indices l; let Sl := S(xl), let al := R(xl 1; xl). We will prove that S1 <
: : : < S 1. Let H := fh 2 f1; : : : ; 2g:Sh < Sh+1g. Since x 2 X and is single-crossing, we have S0 S1 : : : S and S0 < S : Therefore H 6= ;: Let
h 2 H: Then in particular xh < xh+1: Suppose that h > 1: Since R( ) is increasing,
we have ah < ah+1: Since x is a …xed-point, we have ah 1 'xh 1 ah and ah 'xh ah+1:
Since xh 1 and xh are single-peaked, then Sh 1 ah < Sh < ah+1: In particular,
h 12H:By induction, we obtain that 1; : : : ; h2H: By an identical reasoning, we obtain that h; : : : ; 22H; which proves the claim.
It only remains to prove that the partition represented by x is an equilibrium outcome partition for(R; ). For all l= 1; : : : ; 1;we have al'xl al+1. Since is
single-crossing, this further implies that for all t xl; we have al t al+1; and that
for allt xl; we haveal+1 tal: Thus for all h; l 2 f0; : : : ; g; allt 2]xh 1; xh[; we
haveah tal: Thusx is an equilibrium vector for (R; ).
A set of positive integers is connected to 1 if it is either N or of the form f1; : : : ; Kg: We are now ready to state our …rst main result.
Theorem 1: The set of integers such that there are -actions equilibria is
connected to 1.
The proof of Theorem 1 rests on Lemmas 4, 5, 6, and 7. For all positive integer
m, and all vectors x; z2Rm, we let [x; z] :=
fy2Rm :x y z
Lemma 5: For all 2, the correspondence ( ) is increasing on D : For all
x2D , the set (x) is a closed interval.
Proof. Since R( ) is increasing on f(s; t) 2 [0;1]2 : s t
g, and (by Lemma 2) is increasing on D ; therefore ( ) is increasing on D : Since R( ) is a function and
(a; b) is a closed interval for all (a; b) 2 D , thus (x) is a closed interval for all
x2D .
To state the next result, we need the following de…nitions. A subset L Rm is a lattice if, for each nonempty subset H L, the setfx2L:fxg Hg is nonempty and has a greatest element inL, the in…mum ofH inL, denoted byinfL[H]; and the
set fx2L:fxg Hg is nonempty and has a least element in L, the supremum of
H inL, denoted bysupL[H]:In particular, a nonempty lattice Lhas a least element and a greatest element.3
The set X (for each 2) is a lattice that plays a central role in this paper. Its least element is (0; : : : ;0;1); and its greatest element is (0;1; : : : ;1): For each nonempty subsetH X ;letinf[H] and sup[H] (without subscript) be the in…mum and the supremum ofH inX : For eachl = 0; : : : ; ; thel-th coordinate of inf[H]is the in…mum of the image of H by the projection on the l-th coordinate. Similarly, the l-th coordinate of sup[H] is the supremum of the image of H by the projection on the l-th coordinate.
A subset L X (for some 2) is a sublattice of X if, for each nonempty
H L; we have inf[H] 2 L and sup[H] 2 L: Furthermore, for any 0, and all vectorsx; z 2X , the closed interval[x; z]is a sublattice of X . In particular, for all
2;and allx2D ;the set (x)is a sublattice. Finally, ifLandL0 are sublattices
of X ;then L\L0 is a sublattice of X :
Lemma 6: Let 2. Suppose that there is a nonempty L D ; that is a
sublattice of T +1 such that for all x
2 L, we have (x) L. Then the set of …xed-points of ( ) in L is a nonempty lattice.
3The objects we de…ne here as a lattice, a sublattice, an upper-semilattice (Section 5) and an
Proof. By Lemma 5, ( )is increasing. For allx2D ;the set (x)is a sublattice of
X :The result then follows from Zhou’s (1994) extension of Tarski’s …xed-point the-orem to correspondences. Note that ( )satis…es a stronger monotonicity condition than the one required for Zhou’s result.
For any vector x= (x0; : : : ; xm)2Rm+1, let x j := (x0; : : : ; xj 1; xj+1; : : : ; xm).
Lemma 7: Let 3. Suppose that x2X is a …xed-point of ( ). Then the set
L:= x ( 1); x 1 is a subset of D 1, it is a nonempty sublattice of T , and for all
y2L, we have 1(y) L. Finally, the correspondence 1( ) admits a …xed-point in L.
Proof. Let us …rst verify that L D 1: Since x 2 D ; therefore the relations
(R(x0; x1); R(x1; x2)) 2 D and (R(x 2; x 1); R(x 1; x )) 2 D hold. Because
x0 : : : x and by monotonicity of R( ); then (R(x0; x2); R(x2; x3)) 2 D and
(R(x 3; x 2); R(x 2; x ))2D ; therefore L D 1:
Second,Lis a nonempty sublattice ofX 1:Third, we show that 1(x 1) x 1.
Since x2 (x), then x 0 2 1(x 0). We havex 1 x 0: Since 1( ) is
increas-ing, then 1(x 1) 1(x 0). Therefore 1(x 1) x 0. But since the …rst
coordinate of 1(x 1)isfx0g, and x 1 only di¤ers from x 0 by its …rst coordinate,
which precisely equals 0, therefore 1(x 1) x 1. Fourth, by an identical
reason-ing, we can prove that x ( 1) 1(x ( 1)). From these last two inequalities and
since 1( ) is increasing, we conclude that for all y 2L, we have 1(y) L, the desired conclusion. Lemma 6 ensures then that 1( ) has a …xed-point inL:
Theorem 1 follows immediately from Lemmas 4 and 7.
4
In…nite Equilibria
…nite -actions equilibria is bounded and connected to 1; and there are no in…nite equilibria. This result holds under mild additional assumptions. First, a strict preference relation should not be reversed by an in…nitesimal change of the actions. A preference t is action-continuous, if the set f(a; b)2A:a tbg is closed. We
say that the family is action-continuous if, for all t 2 T; the preference t is
action-continuous. Second, we require that the Receiver’s reaction be continuous in the information he receives. More precisely, we say that the Receiver reaction R( )is
continuous if the mapping (s; t)7!R(s; t) is continuous in the usual sense.4
Theorem2: If there is an in…nite equilibrium, then there also is a -actions equi-librium for any 2N: In addition, if the Sender’s preferences are action-continuous and the Receiver’s reaction is continuous, then the converse holds as well. That is, if there is a -actions equilibrium for any 2N, then there also is at least one in…nite equilibrium.
Proof. The continuity of R( ) and the action-continuity of are needed only to prove the only if implication. We …rst prove the only if implication. (Claims 1 to 6). Let I ( ) be a sequence of equilibrium outcome partitions such that for all = 1;2; : : : ; the equilibriumI ( )has intervals: For all 0;leti : [0;1]![0;1]and
s : [0;1]![0;1]such that for allt 2[0;1]; i (t) = infI (t)ands (t) = supI (t)are respectively the in…mum and the supremum of the pool containingt in the partition
I ( ). Clearly, these functions are both nondecreasing and satisfy for all and all
t2[0;1]; the inequalities i (t) t s (t):
Claim 1: There is a subsequence fng and (unique) nondecreasing functions i( )
and s( )such that in( )converges to i( )andsn( )converges tos( ): Moreover, for all
t2[0;1]; we havei(t) t s(t):
Proof: The functions i ( )and s ( )are all nondecreasing and uniformly bounded on [0;1]. Helly’s Selection Theorem guarantees that a sequence of nondecreasing uniformly bounded functions on [0;1]; has a subsequence which converges to a non-decreasing function. Let fmg denote a sequence and let i : [0;1] ! [0;1] be a nondecreasing function such that im( ) converges to i( ): Next, let fng be a
subse-4Both of these assumptions are only needed for theonly if implication in Theorem 2, the existence
quence fromfmgand lets: [0;1]![0;1]be a nondecreasing function such thatsn( )
converges to s( ): The last inequalities are obvious.k
Let I( ) be a (possibly in…nite) partition of [0;1] into level curves of i( ) +s( ):
Sincei( )+s( )is nondecreasing, each pool inI ( )is an interval, possibly a singleton. Claim 2: The functions i( ) and s( ) are constant on any pool ofI( ).
Proof: Let t; t0 be such that I(t) = I(t0). Then i(t) +s(t) = i(t0) +s(t0) holds.
Since both i( ) and s( ) are nondecreasing, the equality implies that i(t) = i(t0) and
s(t) = s(t0):k
Claim 3: For all t 2[0;1], we have inf[I(t)] = i(t)and sup[I(t)] = s(t):
Proof: By Claim 2, for all t 2 [0;1] and all t0 2 I(t), we have i(t) = i(t0). Since
i(t0) t0, we obtain i(t) t0, for all t0 2 I(t). Therefore i(t) inf[I(t)] for all
t2[0;1]. An identical reasoning proves sup[I(t)] s(t) for all t2[0;1]. Thus for all
t 2 [0;1], we have i(t) inf[I(t)] t sup[I(t)] s(t): For any type t satisfying
i(t) = s(t), all these inequalities hold as equalities and there is nothing more to prove. For the other types, we still need to prove that ]i(t); s(t)[ I(t). Let then t be such that i(t)6=s(t). Let t0 be such that i(t)< t0 < s(t). Let u and v be types such
that i(t) < u < t0 and t0 < v < s(t). Since lim
n1in(t) = i(t) and limn1sn(t) = s(t),
there is a positive integern such that for alln n , we havein(t) uandsn(t) v.
Thus for all n n , we have in(t0) = in(t) and sn(t0) = sn(t). Taking the limit as
n goes to in…nity, we obtain i(t0) = i(t) and s(t0) = s(t). Therefore t0 2 I(t), for all
t0 2]i(t); s(t)[. Therefore ]i(t); s(t)[ I (t). This and the inequalities we obtained in the last paragraph yield the desired conclusion.k
Claim 4: I( )is an equilibrium.
Proof: For all t2[0;1], we havelimn1R(in(t); sn(t)) =R(i(t); s(t)), bycontinuity
of R( ). Since In( ) is an equilibrium, for all t; t0 2[0;1]; we have R(i
n(t); sn(t)) t
R(in(t0); sn(t0)):This relation, the continuity of R( )and the action-continuity of t
imply that for all t; t0 2 [0;1]; we have R(i(t); s(t))
t R(i(t0); s(t0)): Therefore I( )
is an equilibrium.k
It only remains to show that I( ) has an in…nity of intervals.
Claim 5: There is t 2 [0;1] such that i(t ) =t = s(t ). The partition I( ) has an in…nity of intervals.
andsn(un) = in(vn):Letfqgbe a subsequence such thatsq(uq)converges tot 2[0;1]:
Then the sequencesiq(uq)andsq(vq)both converge tot . SinceR( )iscontinuous, we
havelimq1R(iq(uq); sq(uq)) =R(t ; t ). By Lemma 3, the functionS(t)is continuous
and thus limq1S(sq(uq)) = S(t ). For all q, by single-peakedness of the preference
sq(uq), we have
R(iq(uq); sq(uq)) S(sq(uq)) R(iq(vq); sq(vq)):
In the limit where q goes to in…nity, we obtain R(t ; t ) = S(t ): Since Iq( ) is an equilibrium, we have
R(iq(t ); sq(t )) t R(iq(uq); sq(vq)):
Bycontinuity of R( )and action-continuity of t ;we can take the limit asq goes to
in…nity, which yields R(i(t ); s(t )) t R(t ; t ). Since R(t ; t ) = S(t ); we obtain
R(i(t ); s(t )) =R(t ; t ). Since R( ) isincreasing and i(t ) t s(t ), then either we have i(t ) = t = s(t ) or we have i(t ) < t < s(t ): Suppose, by contradiction, that the second case holds, i.e. the inequalities are strict. Let u be a type such that
i(t ) < u < t and let v be a type such that t < v < s(t ): Then there is a positive integer q such that for allq q , we have uq2]u; v[,iq(t )< uand sq(t )> v. Thus
for all q q , we have iq(uq) = iq(t ) < u and sq(uq) = sq(t ) > v. Thus for all
q q , we have sq(vq) iq(uq) > v u > 0, which contradicts that sq(vq) iq(uq)
converges to 0. Therefore i(t ) = t =s(t ).
For all t < t , we have i(t) < t and s(t) t . Since R( ) is increasing, we have R(i(t); s(t)) < R(t ; t ) = S(t ). Therefore S(t ) t R(i(t); s(t)) and thus
s(t) < t . Therefore, if t > 0, the partition I( ) has in…nitely many intervals in a left-neighborhood of t . Similarly, if t < 1, the partition I( ) has in…nitely many intervals in a right-neighborhood of t :k
We now prove the if implication.
Claim 6: If there is an in…nite equilibriu /m, then for all 2;there are two vectors
y; z 2D , such that for allx2[y; z] we have (x) [y; z]:
Proof: If R(0;0) = S(0); let y := 0 : If R(0;0) 6= S(0); then there are t1 <
s(th): Let y := (0; s(t1); : : : ; s(t 1);1): Similarly, if R(1;1) = S(1); let z := 1 : If
R(1;1)6=S(1); then there aret01 < : : : < t0 1 in T; such that s(t0 1) = 1 and for all
h= 2; : : : ; 1; we haves(t0
h 1) =i(t0h): Letz := 0; i(t01); : : : ; i(t0 1);1 : It is then
easy to verify that y and z satisfy the desired condition:k
Claim 6 and Lemma 7 imply that ( )has a …xed point in[y; z];which by Lemma 4 in a -actions equilibrium vector: This ends the proof of the Theorem.
5
A taxonomy of biases
We introduce here a taxonomy of admissible problems, according to the nature of the bias of the Sender versus the Receiver. We then re…ne the results of sections 2 and 3 within some of these categories. Abusing notations, let R(t) R(t; t): This is the reaction of the Receiver when he knows that the type of the Sender is t: One important case occurs when the functionsR(t)orS(t)di¤er by at least some positive constant.
The Sender has a strict biasif there is" >0;such that, for allt2[0;1];we have jR(t) S(t)j ": Under this type of bias, the maximal number of actions taken by the Receiver in an equilibrium is bounded. This result generalizes the CS main result to situations where the bias is strict, but not necessarily upward nor downward for all types.
Theorem 3: Let the bias be strict. Then the set of integers such that there are
-actions equilibria is connected to 1 and bounded, and no equilibrium is in…nite.
Proof. If u and v are actions induced in equilibrium, they satisfy ju vj " (see CS, Lemma 1, for a detailed proof of this fact). Since the set of actions induced in any equilibrium is included in [R(0); R(1)]; this implies that there are at most
(R(1) R(0))= + 1 intervals in any equilibrium: The Theorem is an immediate consequence of this fact and Theorem 1.
is weakly more responsive to the state of the world than the Receiver in extreme situations. This condition is incompatible with a strictly consistent bias.
Outward bias. The Sender has an outward bias if [R(0); R(1)] [S(0); S(1)]:
For all 2, let X X be the set of -actions equilibrium vectors.
Theorem 4: Let the bias be outward. Then, for all 2; the set X is a
nonempty lattice. In addition, if the Sender’s preferences are action-continuous and the Receiver’s reaction is continuous, then there is at least one in…nite equilibrium.
Proof. Let 2:Under outward bias, we have D =X : Moreover, for all x2X ;
we have (x) X : Therefore L := X satis…es the conditions of Lemma 6. The …rst claim in Theorem 4 then follows from Lemma 6. The second claim follows from Lemma 6 and Theorem 2.
We say that the Sender has an inward bias when the locus of Sender’s preferred actions is strictly included in the locus of Receiver’s optimal actions. In other worlds, the Sender is strictly less responsive to the state of the world than the Receiver in extreme situations, i.e. [S(0); S(1)] [R(0); R(1)]:In this case, the maximal number of actions taken in equilibrium may be …nite or in…nite, as we will see in Section 7.
The three conditions of strictly consistent bias, outward bias and inward bias are mutually exclusive. But there are admissible problems that do not belong to any of the three cases. Such problems are such that S(0) R(0) and S(1) R(1) have strictly the same sign but the graphs of R( ) and S( ) are not bounded away from each other (e.g. they cross). Also in this case, the maximal number of actions taken in equilibrium may be …nite or in…nite.
We end this section with a result that demonstrates the robustness of Theorem 3. A typet2T is an agreement type if the reaction of the Receiver when he knows this type coincides with the preferred action of the Sender of this type, i.e. S(t) =R(t)
holds. An action a is induced by type t 2 T at some equilibrium outcome ( ) if
(t) =a: An action is a limit induced action of the equilibrium outcome ( )if it is induced by some type and, in addition, is a limit point of the set of induced actions of the equilibrium ( ); i.e. there is a sequence (an) of actions in A n fa g that
converges to a; and in addition, the sequence (an) is either increasing or decreasing.
Lemma 8: Any limit induced action of an equilibrium outcome ( )is induced by an agreement type.
Proof. Letabe a limit induced action of ( ), and let(an)be a sequence inA nfa g
that converges to a. Without loss of generality, suppose that the sequence (an) is
increasing. For each n; let tn = (an; an+1): Clearly, the sequence (tn) is increasing
and bounded above by 1, and thus converges to some type t : Since the mapping
S( ) is continuous, the sequence (S(tn)) converges to S(t ): By single-peakedness,
we have an S(tn) an+1; for all n 2 N: Therefore S(t ) =a: For all t 2 T such
that R(I(t)) = an; we have t tn t : Therefore an R(tn) R(t ); for all
n 2 N; therefore a R(t ): For all t 2 T such that R(I(t)) = a; we have t t:
ThereforeR(t ) a: In conclusion S(t ) =R(t ) = a; which ends the proof.
Theorem 5: Suppose that the set of agreement types is at most countable. Then
the number of actions taken in any equilibrium outcome is at most countable.
Proof. By Lemma 8, S 1( ) maps the set of limit induced actions of ( ) onto the
set of agreement types. Since the set of agreement types is at most countable, it follows that the set of induced limit actions of ( ) is at most countable. But the set of induced actions of ( )that are not limit induced actions is also at most countable. Therefore the set of induced actions is at most countable, which ends the proof.
6
On equilibria with the same number of actions
CS showed that, when the bias is strict and upward, i.e. there is" >0;such that
R(t) S(t) "for allt 2[0;1];and under a additional restriction on the parameters, known as Condition M, there is at most one equilibrium with a given number of actions.5 When condition M does not hold, the set of such equilibria may not be a singleton. In this section, we provide results on the structure of the set of -actions equilibria, without making this assumption. The main result here is that this set has
5Condition M requires that if t
0; :::; tn and t00; :::; t0n satisfy the conditions R(tk 1; tk) 'tk
R(tk; tk+1) and R t0k 1; t0k 't0
k R t
0
k; t0k+1 for all k 2 f1; :::; n 1g; then t1 < t01 =)
a greatest element, which has interesting comparative statics properties and can be computed using a simple algorithm.
To illustrate, consider the following example of a problem that satis…es all CS assumptions, except that condition M does not hold. It admits three equilibria with three actions.
Example 1: Let R(s; t) = s+2t and US(a; t) = (a S(t))2
; where
S(t) =
8 > > > > < > > > > : t 6 + 13
120 on [0;0:1] 13t
10 1
200 on [0:1;0:35]
t
8 + 15
32 on 0:35; 37 80
t+ 1
80 on
37 80;1 :
The three actions equilibria of the problem de…ned by R(; ) and US(; ) are
repre-sented by vectors such that x2 3(x); i.e. such that x (0; x1; x2;1)and
x1 = S 1
R(0; x1)
2 +
R(x1; x2)
2 ;
x2 = S 1
R(x1; x2)
2 +
R(x2;1)
2 ;
which we can rewrite as
S(x1) =
x1
2 +
x2
4 ;
S(x2) =
x1 4 + x2 2 + 1 4:
There are three such equilibria. They are x0 0;401 ;25;1 ; x00 0;15811;269790;1 and
x000 0;1160;1730;1 :
In Example 1, the set of three actions equilibria contains two equilibria that are not comparable for the partial order on X3; which are x0 and x00; but it also has a
greatest element x000: In a broad set of circumstances, the set of -actions equilibria
has a greatest element that has interesting properties. We say that the Sender has an
occurs when the Sender has a downward bias at 0, i.e. if S(0) R(0). Mirror image results can be obtained in that case, in particular that the set of -actions equilibria has a least element.
To state the next lemma, we need the following de…nitions. A subsetL Rmis an upper-semilatticeif, for any nonempty subsetH L, the setfx2L:fxg Hgis nonempty and has a least element, the supremum ofH inL, denoted bysupL[H]. In particular, a nonempty upper-semilatticeL has a greatest element. A subsetL X
(for some 2) is an upper-subsemilattice of X if, for each nonempty H L;
we havesup[H]2L:
Lemma 9: Let the bias be upward at 1. Let >1 be an integer such thatD 6=;:
Then the set D is an upper-subsemilattice of X : If, in addition, we have X 6= ;, then the set X is an upper-semilattice. In particular, it has a greatest element.
Proof. Since the Sender has an upward bias at 1, then for all x 2 D ; we have
[x;1 ] D : Let H D be nonempty, and let x0 2 H: Clearly sup[H] 2 [x0;1 ]:
Thereforesup[H]2D :Therefore,D is an upper-subsemilattice ofX :
Suppose next that X 6= ;. Let Y be an arbitrary nonempty set of …xed-points of ( ), i.e. Y X : Let y^:= sup[Y]. Since D is an upper-subsemilattice of X ;
then y^2D :Since all elements in Y are …xed-points of ( ), then for ally 2Y, we have y sup[ (y)] sup[ (^y)]: Therefore y^ sup[ (^y)]. Let U := [^y;1 ] D :
For all u2U, we have y^ sup[ (^y)] sup[ (u)] 1 . Thus for allu2U, we have
(x)\U 6= ;. Let Z(x) := (x)\U. Consider the correspondence Z : U U. The set U is a closed interval, therefore it is a nonempty lattice. For all x 2 U, the setZ(x)is also a closed interval included in U, therefore it is a nonempty sublattice of U. Since Z is increasing, we can apply Zhou’s (1994) extension of Tarski’s …xed-point theorem to correspondences. Therefore the set of …xed-…xed-points of Z in U is a nonempty lattice. Let y be the least …xed-point of Z in U. The vector y has the following properties. i) It is a …xed-point of in D , i.e. y 2X . ii) Since y 2U, then y is an upper-bound of Y. iii) Any upper-bound u of Y in X is a …xed-point of Z in U, and therefore y u. Therefore yis the supremum of Y inX , the desired conclusion.
Under the conditions of Lemma 9, the set of vectors that represent -actions equilibria has a greatest element, whenever this set is nonempty. Let the greatest -actions equilibrium be the equilibrium represented by the greatest element of X . The following result shows that, if there is a + 1 actions equilibrium; then the greatest
-actions equilibrium is nested within the greatest + 1 actions equilibrium.
Lemma 10: Let 3:Suppose that the bias is upward at 1. Suppose thatX +1 6=
; (and therefore also X 6=;). Let x be the greatest element in X , and let y be the greatest element in X +1. Then y x y 1.
Proof. First, Lemma 7 ensures that there exists some x 2 X such that y x y 1. Since x x, it follows that y x. Second, let y 2 X +1, let y := (0; x0; : : : ; x )2X +1, and letL:= [y ;1 ]\[y ;1 ]\X +1:ThenL D :This set is
such that for ally 2L, we have +1(y)2L, and it is a nonempty lattice. Therefore it contains a …xed point of +1( ). Thus, there exists y 2 X +1 such that x y 1.
Since y y, we then have x y 1:
We now present a comparative statics result on the greatest -actions equilibrium, for two distinct admissible problems where the Sender has an upward bias at 1.
Corollary 1: Let (R1; 1)and (R2; 2) be two admissible problems. Let 2. Suppose that the Sender has an upward bias at 1 in both of these problems. Suppose that for all x 2 D ; we have inf[ 1(x)] inf[ 2(x)] and sup[ 1(x)] sup[ 2(x)]. Suppose further that problem 1 has a -actions equilibrium: Then problem 2 also has a -actions equilibrium: Let x1 and x2 be respectively the greatest such equilibria for problem 1 and 2. Then x1 x2. If, in addition, for all x
2 D ; we have
1(x)< 2(x), then x1 < x2.
Proof. This result follows directly from Lemma 9 in this paper, and Theorem 2.5.2
by Topkis (1998), which extends Milgrom and Roberts’(1994) Theorem 3 to corre-spondences.
In practice, the following conditions on the primitives(R1; 1)and(R2; 2)imply
that 1( ) 2( ); which is stronger than the joint inequalitiesinf[ 1(x)] inf[ 2(x)]
Sender 2 is more leftist than Sender 1; Receivers are identical.
For all t2[0;1], all a < b 2A; we have [a 1t b])[a 2t b]:
Receiver 2 is more rightist than Receiver 1; Senders are identical.
For all s t2[0;1], we have R1(s; t) R2(s; t):
Our next result relates the existence of in…nite equilibria across two problems. It is assumed that the …rst problem admits an in…nite equilibrium and that the second problem is “more outward” than the …rst, relatively to one of the limit induced actions of the in…nite equilibrium of the …rst problem. When this is the case, we show, the second problem also admits an in…nite equilibrium, with the same limit induced action.
Corollary 2: Let (R1; 1) and (R2; 2) be two admissible problems. Suppose that problem (R1; 1) has an in…nite equilibrium such that there is an action induced
a that is a limit point of the set of induced actions, and let x R 1(a ). Suppose that for all s; t2[0;1]; we have
s t x =) R1(s; t) R2(s; t);
x s t =) R2(s; t) R1(s; t)
and for all a; b2[R1(0;0); R2(1;1)]; we have
a b a =) 1(a; b) 2(a; b);
a a b =) 2(a; b) 1(a; b):
Finally, suppose that x is the unique agreement point of (R2; 2): Then problem
(R2; 2) has an in…nite equilibrium of which a is a limit induced action:
First, we now from Lemma 8, that if a is a limit induced action, then x is an agreement point for(R1; 1):Next, the inequalities imply thatx is also an agreement
point for (R2; 2); but we assume that it is the only one. For any problem (R; )
and any closed interval T0 T; let(R; ; T0) be the problem where the set of types
Proof. Since problem (R1; 1) has an in…nite equilibrium such that a is a limit
induced action, then problem(R1; 1;[0; x ])has an in…nite equilibrium such thata
is a limit induced action as well. Thus problem (R1; 1;[0; x ]) has a -actions
equi-librium, for all 2N:By Corollary 1, this implies that problem(R2; 2;[0; x ])has a
-actions equilibrium for all 2N: By Theorem 2, this implies that (R2; 2;[0; x ])
has an in…nite equilibrium:Similarly, we can show that(R2; 2;[x ;1]) has an in…nite
equilibrium. Sincex is the unique agreement point of both problems (R2; 2;[0; x ])
and (R2; 2;[x ;1]); therefore both of the in…nite equilibria we found for these two
problems havea as the unique limit induced action. Therefore(R2; 2)has an in…nite
equilibrium of whicha is a limit induced action:
We obtained the existence of a greatest -actions equilibrium, as the greatest element of the set of …xed points of the correspondence ( ). It is easy to show that this equilibrium is also the greatest …xed point of the function :x 7!sup[ (x)]in
D (as an application of Corollary 1, for example). In our next result, we provide an algorithm that converges to the greatest -actions equilibrium. The algorithm can be used to compute this equilibrium numerically. Letfxn
g be the sequence of vectors in
D such that x0 = 1 and, for all n 0; we havexn+1 = (xn).
Theorem 6: Suppose that the Sender’s preferences are action-continuous, that
the Receiver’s reaction is continuous, that the bias is upward at 1, and that X 6=;;
for some 2. Then fxn
g converges to the greatest element of X .
Proof. By Lemma 9, the set X has a greatest element x: Since the Sender has an upward bias at 1, we have x1 x0. Since the function ( ) is increasing, this implies that the sequence fxng is nonincreasing. Since fxng is bounded below by
x; therefore it converges to a limit x1 2 [x;1 ] D : Moreover, for all n; we have
x1 xn:Thus, for alln; we have (x1) (xn):Since the sequence (xn)converges
to x1; therefore (x1) x1: Let l be an arbitrary index in f1; : : : ; 1g; and let
l( ) be the l-th coordinate of the function ( ): We will prove that x0l l(x1):
Let n be an arbitrary integer, and let xn
0; : : : ; xn be the coordinates of the vector
xn: Since x1 (xn); thus R(xn
l 1; xnl) x1
l R(x
n
l; xnl+1): Since R( ) is continuous,
then the sequence fR(xn
l 1; xnl)g converges to R(x1l 1; x1l ): Similarly, the sequence
fR(xnl; xnl+1)gconverges toR(x1l ; x1l+1):Since the preference x1
then R(x1l 1; x1l ) x1
l R(x
1
l ; x1l+1); i.e. x1l l(x1): Since this relation holds for all
l = 1; : : : ; 1;thenx1 (x1):Since we also had (x1) x1;then (x1) = x1;
i.e. x1 2X :Since x x1;therefore x1=x; the desired conclusion.
7
Revelation around agreement types
In this section we investigate upon the possibility of full revelation around an agreement type. We assume therefore that there is at least one type x 2 [0;1] and an action a 2R such that
S(x ) = a =R(x ; x )
and we study under what conditions there exists an in…nite equilibrium in a neigh-borhood ofx :We …rst consider the linear-quadratic case, and then the general case.
7.1
Linear-quadratic case
We …rst consider the special case where the Receiver’s reaction and the Sender’s preferred action are a¢ ne functions and the Sender’s preferences can be represented by a quadratic utility function. Let
R(s; t) = (s x ) + (t x ) +a ;
with ; >0and
Us(a; t) = (a [a +d(t x )])2;
with d > 0: Clearly these assumptions ensure that the problem is admissible. If
d= + ; the Sender and the Receiver have the same preferred action at all states, and a fully revealing equilibrium exists. Therefore we restrict attention to the case whered6= + :In this case,x is the unique agreement type:Ifd > + ;the bias is outward. Ifd < + ; the bias is inward.
We now compute equilibria. In equilibrium, a cuto¤ type zh must be indi¤erent
Receiver’s reaction to the information [zh; zh+1]:This implies the arbitrage condition
[a +d(zh x )] [ (zh 1 x ) + (zh x ) +a ]
= [ (zh x ) + (zh+1 x ) +a ] [a +d(zh x )];
which can be rewritten as
zh+1+ ( + 2d)zh+ zh 1 2 ( + d)x = 0;
for all h: A sequence(zh) satis…es this condition if and only if it is of the form zh =
x +xh;where xh is a solution of the homogenous system
xh+1+ ( + 2d)xh+ xh 1 = 0;
for all h: The following lemma is useful.
Lemma 11: Let d6= + :If x = 0;there is an in…nite equilibrium if and only if there is a non-zero monotonic solution (yh)h 1 of the homogenous system, such that
lim
h! 1yh = 0:
If x = 1; there is an in…nite equilibrium if and only if there is a non-zero monotonic solution (xh)h 1 of the homogenous system, such that
lim
h!+1xh = 0:
If x 2 (0;1); there is an in…nite equilibrium i¤ and only i¤ there are two non-zero monotonic solutions (xh)h 1 and (yh)h 1 of the homogenous system, such that
lim
h!+1xh = 0 and h! 1lim yh = 0:
Proof. Suppose that there is an in…nite equilibrium. Sincex is the unique agreement point, the cuto¤s in[0; x )de…ne a sequence(xh)h 1 and the cuto¤s in(x ;1];de…ne
(yh) are such sequence, the sequences x 1 xxh
1 and x + (1 x )
yh
y 1 de…ne the
cuto¤s of an in…nite equilibrium.
Whether the solutions of the homogenous system satisfy the conditions of Lemma 11 depends on the roots of the quadratic polynomial
Q(w) !2+ ( + 2d)!+ ;
whose discriminant is
( + 2d)2 4 = 4
"
d +
2
2
+ +
2
2
+ 2
2#
:
The following lemma summarizes the conditions on ; and d under which there exists an in…nite equilibrium.6 Roughly speaking, for this to occur, it must be that
either(i) the bias is outward, or(ii) the agreement type lies at one of the ends of the type space, the Receiver’s reaction is asymmetric ( 6= ) and more reactive in the direction opposite to the end where the players agree.
Lemma 12: If x 2(0;1); there exists at least one in…nite equilibrium if and only if
d + :
If x 2 f0;1g; there exists at least one in…nite equilibrium if and only if
d +
2 +
p
and
(
x = 0 =)
x = 1 =) :
Proof. Whether there is an in…nite equilibrium or not depends on whether
there are solutions (xh) and (yh) satisfying the conditions of Lemma 11, which in
turn, depends on the sign of :
Case 1: > 0. The polynomial Q has two distinct real solutions and ; such
6The necessary and su¢ cient conditions for existence of an in…nite equilibrium are the same
that < and the general form of a solution of the homogenous system is
xh =a h+b h:
We need to distinguish four subcases, depending on the value of the parameters:
Subcase 1a: d= + : We already pointed out that in this case, there is a fully revealing (thus in…nite) equilibrium, since the Receiver and the Sender have the same preferred action in all states.
Subcase 1b: + < d: Then we have 0< < 1 < and xh = h and yh = h
de…ne solutions of the homogenous system that satisfy the conditions of Lemma 11 for any x 2[0;1]: Therefore there is an in…nite equilibrium for any x 2[0;1]:
Subcase 1c: +2 d < + : We haveQ(0) = >0; Q(1) = 2 ( + d)>0;
and the polynomial Q reaches its minimum at 2d ( + )2 0: Therefore both roots and are positive, but they are either greater than 1 or lower than 1; depending on the value of the parameters: Together, the inequalities >0 and +2 d < +
imply that 6= 12:
If > ; we have 1 < < , and if < ; we have 0 < < < 1: If > ;
the sequence (yh) de…ned by yh = a h+b h with a; b 0; such that (a; b) 6= (0;0)
satis…es the condition for the sequence (yh) in Lemma 11, but no solution of the
homogenous system satis…es the condition for the sequence (xh). Therefore there is
an in…nite equilibrium if and only ifx = 0:In this case, there are even uncountably many in…nite equilibria, one for each pair(a; b)such that a; b 0 and (a; b)6= (0;0)
If instead, < ; the sequence(xh)de…ned byxh =a h+b h with a; b 0;such
that (a; b) 6= (0;0) satis…es the condition for (xh) in Lemma 11, but no solution of
the homogenous system satis…es the condition for the sequence (yh). Therefore, in
this case, there is an in…nite equilibrium if and only if x = 1:In this case, there are even uncountably many in…nite equilibria, one for each pair (a; b) such that a; b 0
and (a; b)6= (0;0):
Observe that when > 0, +2 d < + and x 2 (0;1); there is no in…nite solution.
solution. Then
x1 x0 = a( 1) +b( 1) 0
x2 x1 = a 2 +b 2 = a( 1) +b( 1) 0
x3 x2 = a 3 2 +b 3 2 = 2
" 2
a( 1) +b( 1)
#
0:
Since d +2 and >0; we have <0and <0. Therefore
a( 1) +b( 1) 0
a( 1) +b( 1) 0
2
a( 1) +b( 1) 0:
Since the function y 7! ya( 1) + b( 1) is monotonic, this implies that
a = b = 0; which contradicts that the sequence is non-zero: Therefore, there is no in…nite equilibrium in this case.7
Case 2: <0: Let re i be the complex roots of the polynomialQ. The general solution of the homogenous system is of the form
xh =Arhsin ( h+'):
Since 6= 0; then is not a multiple of ; which implies that any such sequence changes signs in…nitely many times, unless it is the null sequence, withA= 0: There-fore, there is no in…nite equilibrium for any x 2[0;1]in this case.
Case 3: = 0:Let be the double root of the polynomialQ. The general solution of the homogenous system is of the form
xh = (a+bh) h:
If > ; then < 1 and there is an in…nite equilibrium if and only if x = 1: If
< ; then >0and there is an in…nite equilibrium if and only if x = 0:
7.2
General case
We now consider the more general case where the problem is admissible and, in addition, R(; ) is di¤erentiable at the agreement type (x ; x ) and the Sender’s preference are represented by a utility function US(a; t) that is twice di¤erentiable at (a ; x ): Let D be the set of parameters ( ; ; d; x )that satisfy the conditions of Lemma 12. In the general case, we obtain the following necessary conditions for the existence of an in…nite equilibrium for which a is a limit induced action. Roughly speaking, for this to occur, it must be that either (i) the bias is locally outward around the agreement type, or (ii) the agreement type lies at one of the ends of the type space, and the Receiver’s reaction is locally asymmetric around the agreement type and more reactive in the direction opposite to the end where the players agree. This result is thus a partial converse of Theorem 4.
Theorem 7: If there exists an in…nite equilibrium for which a is a limit induced action; then
@R
@s (x ; x ); @R
@t (x ; x );
@2US
@a@t (a ; x ) @2US
@a2 (a ; x )
; x
! 2 D:
Proof. Suppose that the condition of the Theorem does not hold. Since the
complement of D is open, let" >0 such that = @R@s (x ; x ) "; = @R@t (x ; x ) "
and d= @2U S
@a@t(a ;x )
@2U S @a2 (a ;x )
+" satisfy ( ; ; d; x )2 D= . Let
e
R(s; t) = (s x ) + (t x ) +a
and
e
Us(a; t) = (a [a +d(t x )])2:
Since
R(s; t) = R(x ; x ) + (s x )@R
@s (x ; x ) + (t x ) @R
and
(a; b) = (a ; a ) + (a a )d
da (a ; a ) + (b a ) d
db(a ; a ) +o(a a ; b a ):
Moreover,
d
da (a ; a ) =
@2US
@a2 (a ; x ) 2@@a@t2US (a ; x );
then there exists a neighborhood X of x in [0;1]; such that R(s; t) Re(s; t) for all s; t 2 X ; such that s t x and Re(s; t) R(s; t) for all s; t 2 X ; such that x s t. Similarly, there exists a neighborhood of A of (a ; a ); such that
(a; b)<e(a; b); for all a; b2A ; s t a and e(a; b)< (a; b);for all a; b2A ; a a b: Since ( ; ; d; x ) 2 D= , then by Lemma 12, for any closed interval
T X ; that is a neighborhood of x in [0;1]; the problem de…ned by Ues;R; Te
has no in…nite equilibrium. By Corollary 2 and the inequalities stated earlier in this paragraph, this implies that for any closed interval T X ; that is a neighborhood of x in [0;1]; the problem (Us; R; T) has no in…nite equilibrium either. Therefore, the problem de…ned by (Us; R) does not have an in…nite equilibrium for which a is
a limit induced action:
7.3
A counterexample
In Theorem 7 we provided a condition that is necessary for a problem to have an in…nite equilibrium. A plausible conjecture is that a strict version of this condition is also su¢ cient. We now provide a counterexample that shows that this is not true. There are problems that satisfy a strict version of the condition, and that only have a single-action equilibrium.
Example 2: Let R(s; t) = s+2t and US(a; t) = (a S(t))2; where S(t)< 1+24 t
for allt2[0;1]:Letx = 14:Suppose further thatS(x ) = 14 and thatS0(x )>1:The problem de…ned by Us(; ) and R(; ) satis…es the local condition of Theorem 7 in a
strict sense, but it has no two actions equilibrium; as any such equilibrium (0; x1;1) would satisfy
S(x1) =
R(0; x1) +R(x1;1)
2 =
1 + 2x1
which we ruled out by assumption. By Theorem 2, this implies that this problem has no in…nite equilibrium.
8
Related literature
Many models of cheap talk give rise to in…nite equilibria. These can be classi…ed roughly in three groups.
First, a few papers study models that can be directly mapped into special cases of our framework, in the context of a particular application. Stein (1989) uses a unidimensional model, where the Sender is a central bank, and the Receiver is a …nancial market. The equilibrium of this market determines an exchange rate. The central bank has a target exchange rate for today, but the market expect a reversal of the policy tomorrow. As a result, it is less reactive than the central bank would like it to be. Thus, the bias is outward. More recently, Admati and P‡eiderer (2004) and Kawamura (2010) study an extension of CS, where the Sender and the Receiver have asymmetric beliefs over the Sender’s competence. This belief asymmetry generates an e¤ective bias that can be outward, inward, upward or downward, depending on parameters. In each of these papers, the authors are mostly concerned with a speci…c application, assume particular functional forms, and do not seek to identify general conditions on the bias that give rise to in…nite equilibria.
these models has nothing to do with an outward bias, but rather with the presence of a costly signal, or a device that operates as a costly signal.
In a third group of models, the presence of in…nite equilibria can be interpreted us-ing our results, although these models do not map into special cases of our framework. Board, Blume and Kawamura (2008) study a cheap talk model, where the message sent is noise, with some probability. These authors also obtain in…nite equilibria, some countable and some uncountable. This happens because any pool of Senders can mimic noise, by varying the measure of messages it uses. Using a larger measure of messages makes it more likely that the signal is noise, which shifts the receiver reaction towards the center. One can interpret the possibility of in…nite equilibria in the noisy talk model in two ways. On the one hand, one can view the amount of noise mimicked by the sender as a costly signal, as in the papers listed in the second group. However, this cost is endogenous as its e¤ect on each Sender type depends on what types are mimicking the same noise. On the other hand, the situation is as if the Sender has some power over the Receiver’s reaction function. Thus, by mim-icking noise in particular ways, the Sender can make the e¤ective bias “outward”, which then leads to in…nite equilibria. Finally, Morgan and Stocken (2003), Li and Madarasz (2008) and Dimitrakas and Sara…dis (2005) study a CS model where the bias is constant but uncertain. Morgan and Stocken suppose that the Sender may not be biased at all or may have a constant positive bias, Li and Madarasz suppose that the bias can take two values, one positive and one negative, and Dimitrakas and Sara…dis suppose that it is distributed in the interval [0;1]: In all of these papers, the authors …nd in…nite equilibria. The relation between these papers and ours is quite clear. In both of these models, one could replace the two-dimensional type(t; b)
by the sender’s preferred action t+b; which turns it into a one dimensional model. For some parameters of their model, this yields an outward bias, and thus an in…nite equilibrium. This intuition is almost correct but not quite, because the reaction func-tion one obtains after performing this transformafunc-tion is not monotone, which means that these models do not map into special cases of this one.
character-izing the set of equilibria. In contrast, we identify general conditions on the structure of the model, that give rise to in…nite equilibria We also provides a systematic char-acterization of the equilibrium set, while most of these papers chose to concentrate on a few equilibria, to describe the e¤ects the speci…c phenomenon they study (noise, uncertain bias) can have.
Other papers have studied models where the bias is not upward. Melumad and Shibano (1991) study cheap talk signalling, among other mechanisms, and they do not assume that the direction of the bias is constant. But their main focus is on comparing equilibria where either one or two actions are taken, from the point of view of the expected utility of the Sender and the Receiver. Alonso and Matoushek (2008) and Mylovanov and Kovac (2009) study optimal mechanisms in the presence of general biases, when the Receiver can commit to his Reaction. In contrast, we do not assume that the Receiver has any comitment power.
One important contribution that this paper makes is methodological. Following CS, full descriptions of the set of equilibria for cheap talk games of this type are usually obtained through the study of an arbitrage recurrence relation. A -actions equilibrium can be represented by a nondecreasing sequence of boundary types 0 =
x0 : : : x = 1:Such a sequence represents an equilibrium partition, if and only if
it satis…es a recurrence relation, derived from the equilibrium “arbitrage conditions”. This relation links together any three consecutive terms xk 1; xk and xk+1 for all
k = 1; : : : ; 1; and expresses the requirement that the Sender of type xk should
be indi¤erent between the actions induced by the intervals [xk 1; xk] and [xk; xk+1]:
Thus, one can express each of the boundary types as a function of the …rst boundary type x1:An equilibrium is obtained every time one of these functions hits the upper
end of the type space:Studying the equilibria amounts to study these functions. The problem is that when the bias is not upward, these functions are not well-de…ned. As a consequence, these tools cannot be used to describe the equilibrium set in such environments.
