CCompeting Persuaders in Zero-Sum Games ∗ Dilip Ravindran † and Zhihan Cui ‡ August 20, 2020
Abstract
We study a Bayesian Persuasion game with multiple senders employing condi-tionally independent experiments. Senders have zero-sum preferences over whatinformation is revealed. We characterize when a set of states cannot be pooled inany equilibrium, and in particular, when the state is (fully) revealed in every equi-librium. The state must be fully revealed in every equilibrium if and only if senderutility functions are sufficiently nonlinear. In the binary-state case, the state is fullyrevealed in every equilibrium if and only if some sender has nontrivial preferences.Our takeaway is that ‘most’ zero-sum sender preferences result in full revelation. ∗ We thank Navin Kartik, Elliot Lipnowski, Alessandro Pavan, Jacopo Perego, and seminar participantsfor very helpful comments. † Department of Economics, Columbia University. Email: [email protected] ‡ School of International and Public Affairs, Columbia University. Email: [email protected] a r X i v : . [ ec on . T H ] A ug Introduction
A key question in the economics of persuasion is the effect of competition on informationprovision. While it has been shown that information disclosure increases with competitionin some settings (Battaglini (2002); Milgrom and Roberts (1986); Shin (1998)), in otherscompetition has the opposite effect (Emons and Fluet (2019); Kartik et al. (2017)). Inthis paper we address the question by modelling two or more senders persuading one (ormultiple) receiver(s) about an unknown state. The senders influence the receiver’s beliefsby disclosing information in the manner of Bayesian Persuasion. Unlike existing work,our senders simultaneously choose conditionally independent experiments; the receiverobserves these experiments and their realizations and updates her belief. To fix ideas,consider competing lobbyists commissioning reports to persuade a politician (or entirelegislature) to vote yes/no on a climate change bill. Here the state may be whether climatechange is a threat; the politician would only like the bill to pass if it is while lobbyistshave differing interests in it passing. Our analysis applies equally well to prosecutor anddefense attorneys persuading a judge or media outlets persuading voters.We consider an environment where senders are maximally-competitive —the senders’payoffs are zero-sum functions of the receiver’s posterior. This assumption is naturalfor the lobbyist example as lobbyists may only care about the probability the bill issuccessfully passed, which varies with the politician’s posterior. Our question is: howdoes competition affect how much information is revealed in equilibrium and how doesthis change with the number of senders?There is always an equilibrium of this game in which all senders fully reveal the state.Our main result is that typically the state is fully revealed in every equilibrium. We findthat —under mild technical assumptions —when sender utility functions are sufficientlynonlinear (in particular are nonlinear on every edge of the simplex) then regardless of thenumber of senders the state is fully revealed in every equilibrium. If utility functionsare sufficiently linear, there are equilibria in which the receiver does not always learn thestate. Two implications are worth mentioning. In the binary-state case, the state is fullyrevealed in every equilibrium if and only if some sender has nontrivial preferences. If thereceiver chooses among a finite set of actions, then generically the receiver learns enoughto take her first best action; furthermore, the state is fully revealed in every equilibriumif and only if the receiver prefers a different action in every state. The every quanitifier implies, by standard arguments, that if the conditions for full revelation aremet for zero-sum utilities, then for utilities close to those all equilibria are almost fully revealing. i canguarantee a payoff strictly larger than from full revelation whenever her opponent j doesnot fully reveal the state. Sender i can do this by choosing an experiment that ensuresthe posterior will fall in regions she has an ‘advantage’ and not in regions her opponenthas an advantage.This idea extends to arbitrary finite state spaces and more than two senders. Givenchoices of experiments for each sender, we say that a set of states is not pooled if thereceiver never assigns positive probability to all of them. We show that a subset of statesis not pooled in every equilibrium if and only if conditional on the receiver learning thestate is in this subset, some sender has strict preferences over what further informationto reveal. For instance, a pair of states is not pooled in all equilibria if any only if somesender’s utility is nonlinear on the edge of the simplex between those two states. Itfollows that nonlinearity on every edge is necessary and sufficient for full revelation inevery equilibrium.
Related Literature.
This paper relates most closely to work in the multiplesender Bayesian Persuasion literature, most notably Gentzkow and Kamenica (2016) andGentzkow and Kamenica (2017) (henceforth GK). These authors’ main finding is thatmore competition leads to no less equilibrium information. GK (2017), like us, find thatin zero-sum games sufficient nonlinearity in sender preferences ensures full revelation inevery equilibrium. However, both GK papers make a strong and arguably unrealisticassumption that senders can choose from a set of ‘Blackwell connected’ experiments thatallow for arbitrary correlation in senders’ experiments’. In contrast we study the case ofconditionally independent experiments; conditional independence is a common assump-tion in information economics. We discuss the distinction in Section 5 and show that, butfor a few additional assumptions we make, our results generalize that of GK (2017).Boleslavsky and Cotton (2018) and Au and Kawai (2020) study two senders per-suading a receiver. However their setups are substantially different from ours becauseeach sender can only reveal information about part of the state (their own type); as aconsequence, they find unique non-fully revealing equilibria. Li and Norman (2018a), Liand Norman (2018b), and Wu (2017) consider Bayesian Persuasion with multiple sendersmoving sequentially. Finally, in a concurrent paper, Dworczak and Pavan (2020) (hence- i.e. the line joining degenerate beliefs on the two states. There is a state ω ∈ Ω = { , ..., N } . All agents have a common prior belief on ω with fullsupport π ∈ int∆(Ω). There are M > , ..., M , who persuade a receiver. Fix a set of signals S with | S | = | ∆(Ω) | . The game starts with each sender i simultaneously choosing a set S i ⊂ S , | S i | < ∞ , and an experiment Π i : Ω → ∆( S i ).Each Π i gives the probability of the receiver receiving each signal in S i conditional oneach state. As | S i | < ∞ , senders may only choose finite signal experiments; we relax thisassumption in Supplementary Appendix B. Implicit in this definition of experiments isthat senders’ experiments are independent conditional on the state.The receiver observes the choices of Π , ..., Π M (and implicitly S , ..., S M ). Then,the state is realized (but not observed by the receiver) and signals from each of the M experiments, s ∈ S , ..., s M ∈ S M , are realized and observed by the receiver. The receiveris Bayesian and updates his belief on ω to some posterior β ∈ ∆(Ω). Senders receive theirpayoffs and the game ends.Senders’ payoffs depend only on the receiver’s posterior belief β . Each sender i has apiecewise analytic utility function u i : ∆(Ω) → R . Crucially, we assume senders’ payoffsare zero-sum: u ( β ) + ... + u M ( β ) = 0 for all β ∈ ∆(Ω). For any state l = 1 , ..., N let δ l ∈ ∆(Ω) represent the belief that puts probability 1 on state l . Due to the structure of As we do not explicity model the receiver acting, the model allows for any number of receivers. All equilibria with the finite signal restriction are equilibria without it. That is, each u i is defined by a finite partition of ∆(Ω) into convex sets and a real analytic functionfor each element of the partition. Note this restriction is not necessary; see Section 5 for discussion. This could represent the reduced form of a game where the receiver chooses an action a ∈ A afterobserving experiment realizations. The receiver has preferences u r ( a, ω ) and the senders may also havestate dependent preferences { u i ( a, ω ) } i which are zero-sum: (cid:80) i u i ( a, ω ) = 0 for all a ∈ A, ω ∈ Ω. u i ( δ l ) = 0 for all senders i = 1 , ..., M and all states l = 1 , ..., N . A strategy profile is a choice of experiment for each sender (Π , ..., Π M ). Let U i (Π , ..., Π M ) = E Π ,..., Π M [ u i ( β )] be sender i ’s ex-ante expected utility from (Π , ..., Π M ); the expectationis over experiment realizations, of which β is a function. Senders choose experiments tomaximize their ex-ante expected utility. Interim Beliefs.
Instead of thinking of sender i picking Π i , it is easier to think of i choosing a distribution over the receiver’s interim beliefs. For any i and choice of Π i , letΓ i ∈ ∆(Ω) be the random variable representing the receiver’s belief on ω if she observesonly the realization of Π i , s i ∈ S . Γ i represents the interim belief of the receiver aftershe observes information from Π i but before viewing the realizations from { Π j } j (cid:54) = i andupdating to her posterior belief. Following Kamenica and Gentzkow (2011), it is without loss for us to recast thechoice of experiment of each sender i as a selection of a Bayes-plausible distribution ofthe interim beliefs, Γ i , the experiment induces. As we have restricted senders to pickingfinite signal experiments, a pure strategy for sender i is a selection of a Bayes-plausible Γ i with finite support. Henceforth, when we use Γ i it implicit that this random variable isBayes-plausible and has finite support. A strategy profile is a vector (Γ , ..., Γ M ). Fixingany strategy profile and sender i , let Γ − i denote the experiment induced by observingrealizations { Γ j } j (cid:54) = i .There are two benchmark experiments to consider. We say Γ i is fully revealing, orΓ i = Γ F R , if Pr(Γ i = δ l ) = π l ∀ l ∈ Ω. If any sender chooses a fully revealing experiment thereceiver learns the state with certainty. The second benchmark is the fully uniformativeexperiment which we will denote Γ U ; Γ i = Γ U if Pr(Γ i = π ) = 1. Equilibrium.
A Nash Equilibrium of this game is a vector of random variables(Γ , ..., Γ M ) such that no sender i can strictly improve her ex-ante expected utility, U i (Γ , ..., Γ M ),by deviating.There is a trivial NE of this game: (Γ F R , ..., Γ F R ). All senders are left indifferentacross all experiment choices as the state will be fully revealed by other senders’ ex- The normalization entails adding an affine function to each u i . Though it may change senders’preferences over posterior beliefs, it will not change preferences over strategy profiles. As the receiver is Bayesian, the order in which she views signal realizations does not matter. A random variable Γ is Bayes-plausible if E [Γ] = π . Note that any finite mixture of pure strategies is also a pure strategy. every equilibrium.
First we derive the main results for the two-sender binary-state case. The intuition willextend to the general case.
Figure 1:
Example of u (blue) and u (red). u ( β ) = β for β < . u ( β ) = 1 − β for β ≥ .
6. Sender 1’s preferences are those in Kamenica and Gentzkow (2011)’s leading examplewith discontinuity at 0 . u (0) = u (1) = 0. Let Ω = { , } . A belief here is a scalar representing the probability the state is ω = 1. Figure 1 shows an example of sender preferences. A sender i ’s strategy is achoice of interim belief random variable Γ i ∈ [0 , , Γ chosen, thereceiver’s posterior belief can be written as a function of the interim beliefs realized fromboth experiments. If Γ = x and Γ = y , then the posterior is: β ( x, y ) = (1 − π ) xyxy − πx − πy + π (1)6ote that β (1 , y ) = β ( x,
1) = 1 and β (0 , y ) = β ( x,
0) = 0; if either interim belieffully reveals the state, the other is irrelevant. Note β (0 ,
1) and β (1 ,
0) are not well definedbut this is not an issue as it is impossible for one sender to fully reveal ω = 0 while theother reveals ω = 1.For either sender i , given any strategy Γ i , consider the distribution of Γ i conditionalon Γ j = x ( j (cid:54) = i ): P r (Γ i = y | Γ j = x ). P r (Γ i = y | Γ j = x ) can be constructed by takingthe signal structure Π i that corresponds to Γ i and deriving the distribution over interimbeliefs it induces if x and not π was the receiver’s prior.Given an opponent choice of Γ j , define W i ( x ) as sender i ’s expected payoff conditionalon generating Γ i = x . For a fixed Γ W ( x ) is written: W ( x ) = (cid:88) y ∈ supp [Γ ] u ( β ( x, y )) P r (Γ = y | Γ = x ) dy (2)Note that W (1) = W (0) = W (0) = W (1) = 0; if either players’ experimentgenerates a fully revealing interim belief then the other experiment is irrelevant. Twospecial cases are important. When Γ = Γ F R then, regardless of u or the prior, W ( x ) = 0for all x . This is because Γ will reveal the state to be 0 or 1; any interim belief sender1 produces can only affect the relative probability of these events, both of which yield u = 0. Meanwhile, when Γ = Γ U , then W ( x ) = u ( x ) as x will be the receiver’sposterior. The result below will be useful.
Lemma 1.
In any equilibrium (Γ , Γ ) :(1) U (Γ , Γ ) = U (Γ , Γ ) = 0 .(2) W ( x ) ≤ and W ( x ) ≤ for all x ∈ [0 , . We provide a formal proof in Supplementary Appendix A, but the logic is straight-forward. Property (1) follows from the game being zero-sum and the observation thateach sender i can guarantee a payoff U i = 0 by fully revealing the state. While property(1) says that sender equilibrium payoffs equal those from full revelation, it does not say7hat we must have full revelation in equilibrium: for instance if the u and u are linear,any (Γ , Γ ) constitute an equilibrium.Property (2) holds because any violation leads to a contradiction of (1). Fix any Γ j such that sender i (cid:54) = j has W i ( x ) > x . We can find a Γ i with support only on { x, , } ; as i gets strictly positive expected utility whenever x is realized and 0 otherwise, U i (Γ i , Γ j ) >
0. Hence such Γ j cannot be played in equilibrium. The main result for the two-sender binary-state case relies on Lemma 1. We showthat if utility functions are nonlinear, then in equilibrium at least one sender i must chooseΓ i = Γ F R , or else W j ( x ) will violate property (2). Proposition 1.
The state is fully revealed in every equilibrium if and only if u is non-linear. The ‘only if’ direction is trivial —if u is linear then senders are indifferent betweenall strategy profiles. Hence the result can be restated as:There is full relevation in every equilibrium ⇐⇒ ∃ (Γ , Γ ) , (Γ (cid:48) , Γ (cid:48) ) and a sender i with U i (Γ , Γ ) (cid:54) = U i (Γ (cid:48) , Γ (cid:48) ) . We discuss the intuition and sketch the proof but leave details to SupplementaryAppendix A. In the rest of this section we discuss the ‘if’ direction: u nonlinear = ⇒ full revelation in every equilibrium.The idea can be seen using the example in Figure 1 with any prior. Note that forall r ∈ [0 . , u ( β ) > β ∈ [ r, r . Suppose for contradictionthat sender 2 plays a non-fully revealing strategy Γ in some equilibrium. As Γ (cid:54) = Γ F R ,Pr(0 < Γ < >
0; let y = min supp [Γ ] \ { , } ∈ (0 ,
1) be in the smallest interior beliefin the support of Γ . Using the definition of β ( x, y ), define x by β (x , y) = r . Conditionalon Γ = x ∈ [x , β ( x, y ) ∈ [ r,
1) for all interior y in Γ ’s support. But then for all x ∈ [x ,
1) we have: As an example consider u in Figure 1 with Γ = Γ U . Then W (0 .
6) = 0 . U (Γ , Γ ) > putting support on { , . , } . ( x ) = u ( β ( x, (cid:124) (cid:123)(cid:122) (cid:125) =0 P r (Γ = 0 | Γ = x ) + u ( β ( x, (cid:124) (cid:123)(cid:122) (cid:125) =0 P r (Γ = 1 | Γ = x )+ (cid:88) y ∈ supp [Γ ] \{ , } u ( β ( x, y )) (cid:124) (cid:123)(cid:122) (cid:125) > P r (Γ = y | Γ = x ) (cid:124) (cid:123)(cid:122) (cid:125) > > . This contradicts Lemma 1 property (2).The broader intuition is as follows. We say a sender i has an advantage on anysubset of [0 ,
1] on which u i is strictly positive; for instance in the example, sender 1 hasan advantage on [0 . , While senders would like the receiver’s posterior to fall intheir regions of advantage with high probability, neither sender’s experiment unilaterallycontrols the posterior. However, in the example sender 1 has an advantage at the end ofthe unit interval, [ r, that is interior with positive probabilitythen sender 1 can find extreme enough interim beliefs x ≥ x guaranteeing that, conditionalon x being realized and Γ being interior, the receiver’s posterior is in [ r, fully reveals the state both senders get utility 0, and so overall sender 1 gets a strictlypositive expected payoff from generating an interim belief x ∈ [x , u in the example. As β → u approaches u (1) = 0 from above, as is the case in the example, we can findan r ∈ (0 ,
1) such that sender 1 has an advantage on [ r, u , we canreplicate the same argument to show that any Γ (cid:54) = Γ F R cannot be played in equilibrium.If u approaches 0 from below as β → u must approach from above, and so wemust have Γ = Γ F R in any equilibrium. The same argument applies whenever u or u approach 0 from above as β → u , u are piecewise analytic, there is only one other case to consider: u , u nonlinear and u ( β ) = u ( β ) = 0 for all β in some neighborhoods of both 0 and 1. Heretoo there will be a sender with a region of advantage closest to the ends of [0 ,
1] who canfind a violation of Lemma 1 property (2) whenever her opponent does not fully reveal thestate. Let r = sup { β ∈ [0 ,
1] : u ( β ) (cid:54) = 0 } be the supremum of posteriors at which u , u are nonlinear (note r < u ( r ) > u ( r − ) > We use the word advantage because Lemma 1 tells us that both senders will get ex-ante expectedutility 0 in equilibrium. Any posteriors that yield strictly better utility than this for a sender are relativelyadvantageous to that sender. u or u ). Suppose Γ (cid:54) = Γ F R . Defining y andx as before, if u ( r ) >
0, then W (x) = u ( r ) P r (Γ = y | Γ = x) >
0. If u ( r − ) >
0, then W (x − (cid:15) ) > (cid:15) > Now we apply the logic from the previous section to N ≥ M ≥ T ≥ , ..., Γ T , let β (Γ , ..., Γ T ) be the receiver’s posterior beliefafter observing all T realizations. First note Lemma 1 extends to this setting (the logicis the same).For any strategy profile (Γ , ..., Γ M ) and subset of states Ω (cid:48) ⊆ Ω, we say Ω (cid:48) is notpooled if Pr( β l (Γ , ..., Γ M ) > ∀ l ∈ Ω (cid:48) ) = 0 (otherwise, Ω (cid:48) is pooled ). When Ω (cid:48) is notpooled, the receiver will always be able to rule out at least one of the states in the set.For any Ω (cid:48) ⊆ Ω, let ∆(Ω (cid:48) ) = { γ ∈ ∆(Ω) : (cid:80) l ∈ Ω (cid:48) γ l = 1 } be the subset of the simplexassigning probability 1 to ω ∈ Ω (cid:48) . Note that for two states l, k , ∆( { l, k } ) is the edge ofthe simplex between δ l and δ k . Hence Proposition 1 can be restated as: states 0 , u i is nonlinear on ∆( { , } ). Proposition2 generalizes Proposition 1 and characterizes when any subset of states is not pooled inevery equilibrium. Proposition 2.
Consider any Ω (cid:48) ⊆ Ω . No equilibrium pools Ω (cid:48) if and only if for somesender i u i is nonlinear on ∆(Ω (cid:48) ) . Suppose the receiver learns that ω ∈ Ω (cid:48) ⊆ Ω. Conditional on this event, if u i islinear on ∆(Ω (cid:48) ) for all i then all senders are indifferent across all additional informationthat can be revealed. Meanwhile, if for some i u i is nonlinear on ∆(Ω (cid:48) ), then there issome additional experiment that i either strictly prefers or disprefers to not providingany additional information. Proposition 2 says that conditional on the receiver learningthat ω ∈ Ω (cid:48) , some sender having strict preferences over revealing additional informationcharacterizes Ω (cid:48) being not pooled in every equilibrium. One important implication is thatrevealing no information, the strategy profile (Γ U , ..., Γ U ), is an equilibrium if and only ifall senders are indifferent across all strategy profiles.We give the broad idea of the proof here and give a detailed proof sketch in theAppendix. The full proof is in Supplementary Appendix A. The ‘only if’ direction isstraightforward; if all u i are linear on ∆(Ω (cid:48) ) then all senders fully revealing the state10henever ω ∈ Ω \ Ω (cid:48) and revealing no further information whenever ω ∈ Ω (cid:48) is a non-fullyrevealing equilibrium.Now for the ‘if’ direction. For any Ω (cid:48) ⊆ Ω and sender i , fixing an opponent strategyΓ − i consider W i ( x ) on ∆(Ω (cid:48) ). First note that Γ i = x ∈ ∆(Ω (cid:48) ) = ⇒ β ( x, Γ − i ) ∈ ∆(Ω (cid:48) )w.p. 1. Generating an interim belief in ∆(Ω (cid:48) ) tells the receiver that ω ∈ Ω (cid:48) , ensuringthe posterior is also on this set. Conditional on x ∈ ∆(Ω (cid:48) ), the only information Γ − i can convey is relative probabilities of states in Ω (cid:48) . When evaluating W i ( x ) on ∆(Ω (cid:48) ),sender i can treat Γ − i as an experiment just about states in Ω (cid:48) . If some u i is nonaffineon ∆(Ω (cid:48) ) we can apply a similar argument to Proposition 1. Firstly, at least one senderhas an advantage somewhere on ∆(Ω (cid:48) ). We can find some sender j such that wheneverPr(Γ − j = y s.t. y l > ∀ l ∈ Ω (cid:48) ) > − j pools Ω (cid:48) ), W j ( x (cid:48) ) > x (cid:48) ∈ ∆(Ω (cid:48) ).Like in Proposition 1, x (cid:48) will be extreme enough —close enough to the boundaries of ∆(Ω (cid:48) )—to ensure that whenever Γ − j assigns positive probability to all states in Ω (cid:48) , β ( x (cid:48) , Γ − j )falls in a region of j ’s advantage with positive probability. Otherwise, β ( x (cid:48) , Γ − j ) will fallwhere j gets 0 utility. Hence W j ( x (cid:48) ) >
0, violating Lemma 1 and implying that Γ − j mustnot pool Ω (cid:48) in equilibrium.Whenever no pair of states can be pooled in any equilibrium, the state is fully revealedall equilibria. Applying Proposition 2 to every pair of states: Theorem 1.
The state is fully revealed in every equilibrium if and only if for every pairof states l and k there exists sender i with u i nonlinear on ∆( l, k ) . This is an immediate corollary of Proposition 2. Theorem 1 shows that preferencesbeing sufficiently nonlinear characterizes all equilibria being fully revealing. The statemay not be fully revealed only if all senders have linear preferences on an edge of thesimplex. Linearity along any edge for any sender, let alone all senders, is knife-edge andso for typical sender preferences the state is fully revealed in all equilibria.Thus far, senders’ preferences have been defined over posterior beliefs. Theorem 1yields a clean result when preferences can be microfounded by modeling a single receiverchoosing from a finite set of actions. Suppose after observing all experiment realizationsthe receiver takes an action a and receives a payoff u r ( a, ω ) while each sender i getspayoff u i ( a, ω ). We make the generic assumption that no agent is indifferent between anyactions at any state and assume the receiver uses a fixed tie-breaking rule when indifferentbetween actions.In the space of posteriors, senders have piecewise linear utility functions. Thesefunctions are linear on subset ∆(Ω (cid:48) ) if and only if the receiver has the same best action11t every state in Ω (cid:48) . By Proposition 2, Ω (cid:48) ⊆ Ω is not pooled in every equilibrium if andonly if the receiver has different best actions at some states in Ω (cid:48) . This implies a versionof Theorem 1: the state is fully revealed in every equilibrium if and only if the receiverhas a different best action at every state . Further, when states are pooled in equilibrium,additional information would not change the receiver’s action. Hence the receiver alwayslearns enough take her first best action . See Supplementary Appendix B for formal details.
Robustness to zero-sumness.
Using standard upper hemicontinuity arguments, wecan show Theorem 1 is robust.
Result.
Suppose senders’ utility functions converge to zero-sum and utilities aresufficiently nonlinear in the limit. Whenever convergent, the information revealed alongany sequence of equilibria converges to full revelation. Robustness is one reason we focus on conditions for full revelation in all equilibria.Note that if the limiting preferences are linear on every edge of the simplex, it is still pos-sible for the information revealed in all equilibria to converge to full revelation. Examplesare provided and the result is formalized in Supplementary Appendix B.
Private Information.
Suppose when the game begins each sender receives a privatesignal. We assume these signals are bounded and realized from finite signal conditionallyindependent experiments. In equilibrium, senders could potentially signal their privateinformation through their choice of experiment. However, for ‘most’ sender preferencesthe takeaway from Theorem 1 remains the same. Result.
Suppose senders receive private signals before the game. In all but a knife-edge case of preferences, the state is fully revealed in every equilibrium.
The knife-edge case includes all senders having linear utility on every edge of thesimplex. Details are in Supplementary Appendix B but the logic is similar to the baselinemodel.
Experiments without finite signals.
We have focused on finite signal equilibria Convergence for utilities is in the sup norm. For information, the notion is convergence in distributionof the receiver’s posterior. They induce beliefs bounded away from the simplex’s boundaries. This result is most similar to that of Dworczakand Pavan (2020).
Piecewise analytic utility.
Our assumption that utility functions are piecewiseanalytic is not necessary for Theorem 1. We just need to rule out pathological utilityfunctions that, under our normalization, take values oscillating infinitely about 0 on anyedge of the simplex.
Comparison to Gentzkow and Kamenica (2017).
GK (2017) obtains a verysimilar result to ours in a different setting. They consider a game in which senders areallowed to arbitrarily correlate their signal realizations and show that nonlinearity onevery edge of the simplex guarantees full revelation in every equilibrium. Our conditonalindependence assumption makes this paper different for three reasons. For applications,conditional independence corresponds to senders conducting investigations independently,which is an important benchmark. Secondly, our results still go through if senders areallowed to correlate (to any extent) their experiments’ realizations. This is because ourproofs primarily relied on senders being able to deviate to experiments which are con-ditionally independent of their opponents’. Even when senders are allowed to correlatesignal realizations, this is still possible.
Result:
Suppose each sender’s strategy space contains only finite signal experimentsand includes every finite signal conditionally independent experiment. Then Propositions1 and 2 and Theorem 1 hold (regardless of what correlated experiments are also available).
This result combined with Theorem 1 nests the zero-sum result of GK (2017) withthe caveats that we assume piecewise analytic utility functions and restrict strategy spacesto finite signal experiments.Finally it’s important to note the forces that deliver the full revelation results in bothpapers are different. In GK (2017), ability to correlate experiments gives each sender muchmore control over the posterior. Given any Γ − i played by her opponents, a sender i canplay a different experiment for each realization of Γ − i . Senders’ ability to manipulatethe posterior belief by belief makes Proposition 2, and hence Theorem 1, much easier toprove. In our setting senders have less control over posteriors; the strongest tool a senderhas is using extreme interim beliefs to ensure poteriors are similarly extreme. The sufficient condition is the same as that in our private information result. iming and Relationship to Dworczak and Pavan (2020). See Section 1 fora description of DP (2020). One difference between our model and DP’s is timing. Intheir model with conditionally independent experiments, nature can condition her choiceof experiment on the persuader’s choice. This is similar to a version of our model with twosenders moving sequentially. In a sequential version of our model, senders 1 , ..., M movein order observing all previous experiment choices (but not realizations); we are interestedin subgame perfect Nash Equilibria (SPNE) of this game. Note that for each simultaneousgame there are multiple corresponding sequential games, one for each ordering of senders.The following result helps clarify the relationship between the our baseline model and asequential version.
Result.
If for u , ..., u M there is full revelation in every SPNE of the sequential gamewith the senders moving in some order, then there is full revelation in every equilibriumof the simultaneous game. We prove the result Supplementary Appendix B and also show that the conversedoes not hold. The result shows that we are guaranteed full revelation in equilibriumfor a (weakly) larger set of sender preferences with simultaneity than with sequentiality.This is in line with Norman and Li (2018a) and Wu (2017), which show that simultaneouspersuasion cannot generate less information than sequential. Differences in timing are one reason our full revelation results are starker than thoseof DP. Our results are also stronger due to our focus on equilibria in finite signals. Finally,while our results concern the total information revealed by multiple senders, their resultsemphasize only the information revealed by a single persuader (less so their opponent—nature). Formally, our condition for a set Ω (cid:48) ⊆ Ω to be not pooled in every equilibriumis equivalent to the persuader in DP’s model having a unique optimal strategy of notpooling Ω (cid:48) or nature minimizing her payoff by not pooling Ω (cid:48) . Proof Sketch of Proposition 2.
The ‘only if’ direction is straightforward. The approach to proving the ‘if’ directionis similar to that of Proposition 1. Suppose some u j is nonlinear on ∆(Ω (cid:48) ). Fix a strategy However both papers allow senders to correlate experiments arbitrarily. Wu (2017) also considerszero-sum games, but only shows existence of a fully revealing equilibrium. , ..., Γ M ) that pools Ω (cid:48) and let Γ be the experiment induced by observing therealizations of all M experiments; we show, via violation of Lemma 1 property (1) (gener-alized to N ≥ , M ≥ i and an interim belief x such that E [ u i ( β (x , Γ))] >
0. As in the proof of Lemma1 property (2), i can then construct an experiment Γ (cid:48) i with support on { δ , ..., δ N , x } andobtain a strictly positive payoff by playing Γ (cid:48) i in addition to, conditionally independently,Γ i . In the binary-state case i was a sender who had an advantage at points closest tothe extremes of the [0 ,
1] interval. This idea is simply generalized when N and/or Ω (cid:48) arelarger.Let ∆ int (Ω (cid:48) ) = { γ ∈ ∆(Ω (cid:48) ) : γ l > ∀ l ∈ Ω (cid:48) } ; this is the set of beliefs in ∆(Ω (cid:48) ) whosesupport is Ω (cid:48) . We first make some simple observations (proofs are trivial and omitted). Observation 1.
Conditional on an interim belief x ∈ ∆ int (Ω (cid:48) ) being realized, thefollowing hold. (1) β ( x, Γ) ∈ ∆(Ω (cid:48) ) w.p 1 as x has ruled out states Ω \ Ω (cid:48) . (2) β ( x, Γ) ∈ ∆ int (Ω (cid:48) ) if and only if Γ assigns positive probability to all states in Ω (cid:48) . (3) Realizationsof Γ that assign probability 0 to Ω (cid:48) occur w.p. 0.The case of | Ω (cid:48) | = 2 is simple. Conditional on interim beliefs x ∈ ∆ int ( { l, k } ), β ( x, Γ) ∈ ∆( { l, k } ) (Observation 1) and so we are in a binary-state world. We canidentify sender i and x similarly Proposition 1 —the extension from two to M > | Ω (cid:48) | >
2, first consider an example with N = 3and Ω (cid:48) = Ω. Let A = { γ ∈ ∆(Ω) : u j ( γ ) > j } be the set of posteriors at whichsome sender has an advantage and let cl ( A ) be its closure. Note that by zero-sumness, atall posteriors outside A all senders get 0 utility. Suppose cl ( A ) is as shown in Figure 2. Note we only need Proposition 2 with | Ω (cid:48) | = 2 to prove Theorem 1. igure 2 It is convenient to represent each belief γ ∈ ∆(Ω) by two ratios: r ( γ ) = γ γ and r ( γ ) = γ γ + γ . As in Figure 2 cl ( A ) does not touch the boundaries of ∆(Ω), we canfind ¯ r = max γ ∈ cl ( A ) r ( γ ) with 0 < ¯ r < ∞ ; the dotted line shows points γ ∈ ∆(Ω) with r ( γ ) = ¯ r . Amongst the points in argmax γ ∈ cl ( A ) r ( γ ), we can then identify a uniquepoint, ¯ β ∈ ∆ int (Ω), with the largest r ratio (see Figure 2). We will find x ∗ ∈ ∆ int (Ω)conditional on which the posterior will either fall outside of cl ( A ) (yielding payoff 0) orwill equal ¯ β . A sender who has an advantage at or near ¯ β will hence be able to get strictlypositive utility by generating an interim belief at, or near, x ∗ .Let Z = { γ ∈ supp [Γ] : γ , γ , γ > } (note Z is nonempty). Conditional ongenerating x ∈ ∆ int (Ω), all beliefs in the support of Γ that are not in Z can be ignoredas they will result in posteriors on the boundaries of ∆(Ω) and hence, in the example,yield utility 0. Consider interim beliefs x ∈ ∆( { , } ); for all y ∈ Z , as x → δ , β ( x, y ) → δ = ⇒ r ( β ( x, y )) → ∞ and as x → δ , β ( x, y ) → δ = ⇒ r ( β ( x, y )) →
0. Thefiniteness of Z , continuity of Bayesian updating, and intermediate value theorem togetherimply that there exists x (cid:48) ∈ ∆( { , } ) such that min y ∈ Z r ( β ( x (cid:48) , y )) = ¯ r . Next considertaking a convex combination of x (cid:48) and δ : λδ +(1 − λ ) x (cid:48) ; as λ → β ( λδ +(1 − λ ) x (cid:48) , y ) → δ for all y ∈ Z . While r ( λδ + (1 − λ ) x (cid:48) ) changes in λ , r ( λδ + (1 − λ ) x (cid:48) ) does not as λ does not affect the relative probabilities assigned to states 2 and 3. As a consequence, λ also does not affect r ( β ( λδ + (1 − λ ) x (cid:48) , y )) for any y ∈ Z . Hence by the intermediate These are not well defined for all γ , but this won’t be a problem in our example. λ ∗ ∈ (0 ,
1) and x ∗ = λ ∗ δ + (1 − λ ∗ ) x (cid:48) such that:(1) min y ∈ Z r ( β ( x ∗ , y )) = ¯ r (2) min y (cid:48) ∈ argmin y ∈ Z r ( β ( x ∗ ,y )) r ( β ( x ∗ , y (cid:48) )) = r ( ¯ β ) (3)Hence for all y ∈ Z , we have either β ( x ∗ , y ) = ¯ β or β ( x ∗ , y ) (cid:54)∈ cl ( A ). Posteriors forwhich the latter is true will fall either to the left of the dotted line in Figure 2 or on thedotted line and (strictly) above ¯ β . If ¯ β ∈ A , then set x = x ∗ and E [ u i ( β (x , Γ))] > i . If ¯ β (cid:54)∈ A , then there exists some β (cid:48) ∈ A close to ¯ β and x close to x ∗ such thateither β (x , y ) = β (cid:48) or β (x , y ) (cid:54)∈ cl ( A ) for all y ∈ Z .The steps in proving the existence of x generalize beyond this example. Due toObservation 1, the same logic applies for N > (cid:48) = { , , } whenever cl ( A ) ∩ ∆(Ω (cid:48) ) ⊂ ∆ int (Ω (cid:48) ) (i.e. cl ( A ) does not touch the boundaries of ∆(Ω (cid:48) )). If | Ω (cid:48) | = K > cl ( A ) ∩ ∆(Ω (cid:48) ) ⊂ ∆ int (Ω (cid:48) ) we can extend the ratio representation of a belief γ to ratios r ( γ ) , ..., r K − ( γ ) and iterate the same procedure to find ¯ r , ..., ¯ r K − and ¯ β . x ∗ will now have to satisfy K − cl ( A ) ∩ ∆(Ω (cid:48) ) (cid:54)⊂ ∆ int (Ω (cid:48) ) requires additional details and we leave it to SupplementaryAppendix A. References
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A Supplementary Appendix A: Proofs
Definitions and Facts.
The following definitions and facts are used in both Supple-mentary Appendix A and B.Let P be the set of Bayes-plausible finite support elements of ∆(∆(Ω)). It will beconvenient to talk about a strategy for sender i as a choice of interim belief Γ i withprobability mass function p i ∈ P (in the text of the paper we did not introduce notationfor the distribution of Γ i ).For any strategy profile (Γ , ..., Γ M ) and subset of senders S ⊆ { , ..., M } , let therandom variable Γ S be the receiver’s belief after observing realizations of { Γ j } j ∈ S but notthe realizations of { Γ j } j (cid:54)∈ S ; let p S be it’s probability mass function and p S ( ·| ω = k ) be its18robability mass function conditional on the state being k . Let Γ − S and p − S be the sameobjects for the complementary set of senders.For any disjoint subsets of senders S, S (cid:48) ⊂ { , ..., M } and any fixed strategy profile(Γ , ..., Γ M ), let p S (cid:48) ( ·| x ) be the probability mass function of Γ S (cid:48) conditional on Γ S = x . p S (cid:48) ( y | x ) = (cid:88) k ∈ Ω p S (cid:48) ( y | ω = k, x ) P r ( ω = k | Γ S = x ) = (cid:88) k ∈ Ω p S (cid:48) ( y | ω = k ) x k = (cid:88) k ∈ Ω P r ( ω = k | y ) p S (cid:48) ( y ) P r ( ω = k ) x k = (cid:88) k ∈ Ω x k y k p S (cid:48) ( y ) π k (A.1)Where the second equality comes from conditional independence of Γ S and Γ S (cid:48) . ClaimA.1 tells us that conditional on Γ S , with probability 1 Γ S (cid:48) assigns positive probability toat least one state that Γ S assigns positive probability to (i.e. Γ S (cid:48) cannot contradict Γ S ).This is a simple implication of Bayesian updating. Claim A.1.
For any disjoint subset of senders
S, S (cid:48) , Γ S , Γ S (cid:48) , and x ∈ ∆(Ω) : p S (cid:48) ( y | x ) = 0 for all y s.t. y l = 0 for all l ∈ Ω for which x l > . Further, there exists y ∈ supp [Γ S (cid:48) ] such that p s (cid:48) ( y | x ) > .Proof. The first statement, that p S (cid:48) ( y | x ) = 0 for all y such that y l = 0 for all l for which x l >
0, follows immediately from Equation A.1. The second statement follows from Bayes-plausibility of Γ S (cid:48) . For every l such that x l >
0, as π l >
0, there exists y ∈ supp [Γ S (cid:48) ] with y l ≥ π l >
0; by Equation A.1, p ( y | x ) > y .Let β l ( x , ..., x M ) = P r ( ω = l | x , ..., x M ) be the receiver’s posterior belief that ω = l after observing experiment realizations Γ = x , ..., Γ M = x M . By Bayes rule: β l ( x , ..., x M ) = P r (Γ = x , ..., Γ M = x M | ω = l ) P r ( ω = l ) P r (Γ = x , ..., Γ M = x M )= Π Mi =1 p i ( x i | ω = l ) π l (cid:80) Nk =1 Π Mi =1 p i ( x i | ω = k ) P r ( ω = k ) = Π Mi =1 P r ( ω = l | x i ) p i ( x i ) P r ( ω = l ) π l (cid:80) Nk =1 Π Mi =1 P r ( ω = k | x i ) p i ( x i ) P r ( ω = k ) π k = Π Mi =1 x il /π M − l (cid:80) Nk =1 Π Mi =1 x ik /π M − k (A.2)Where the second equality uses the conditional independence of Γ , ..., Γ M . Note that β l is not well defined when for each state k ∈ Ω there exists sender j with x jk = 0. However19t is straightforward to see by applying Claim A.1 that such a realization of (Γ , ..., Γ M )occurs with zero probability; after viewing the realizations of any number of experiments,the Bayesian receiver will have a well defined posterior w.p. 1.For any strategy profile (Γ , ..., Γ M ) and disjoint sets of senders S , ..., S T , we similarlydefine define the receiver’s posterior as a function of interim belief realizations from eachexperiment: { Γ S s = y S s } s =1 ,...,T . β l ( y S , ..., y S T ) = Π Ts =1 y S s l /π T − l (cid:80) Nk =1 Π Ts =1 y S s k /π T − k (A.3)Note: Γ S ∪ ... ∪ S T = β (Γ S , ..., Γ S T ), as both define the receiver’s belief after observingrealizations of Γ S , ..., Γ S T .Claim A.2 shows that if any subset of experiments in a strategy profile generate aninterim belief in ∆(Ω (cid:48) ) then the posterior will fall in ∆(Ω (cid:48) ) w.p. 1. Claim A.2.
For any strategy profile (Γ , ..., Γ M ) , disjoint subsets of senders S , ..., S T ,and states Ω (cid:48) ⊆ Ω , if Γ S ∈ ∆(Ω (cid:48) ) then β (Γ S , ..., Γ S T ) ∈ ∆(Ω (cid:48) ) w.p. .Proof. This can be seen from the definition of β ( y S , ..., y S T ) which implies β l ( y S , ..., y S T ) =0 for all l (cid:54)∈ Ω (cid:48) . After observing Γ S ∈ ∆(Ω (cid:48) ), the receiver updates to an interim beliefassigning 0 probability to all states outside of Ω (cid:48) . No additional information can changethis. Claim A.3.
For any strategy profile (Γ , ..., Γ M ) , Ω (cid:48) ⊆ Ω , and any subsets of senders S :If Γ S does not pool Ω (cid:48) then (Γ , ..., Γ M ) does not either.Proof. Let S (cid:48) = { , ..., M } \ { S } . As Γ S does not pool Ω (cid:48) , then P r (Γ S = y : s.t. y l > ∀ l ∈ Ω (cid:48) ) = 0. If Γ S = y , Γ S (cid:48) = y (cid:48) , then by Equation A.3, if y l = 0 then β l ( y, y (cid:48) ) = 0.Hence as w.p. 1 Γ S assigns 0 probability to at least one state in Ω (cid:48) , β (Γ S , Γ S (cid:48) ) does aswell and so (Γ , ..., Γ M ) does not pool Ω (cid:48) . A.1 Section 2
Normalization of utility functions.
Here we show that we can normalize u i ( δ l ) = 0 forall i = 1 , ..., N , l = 1 , ..., M without changing senders’ preferences over strategy profilesor the zero-sumness of the game. 20uppose senders have utility functions u (cid:48) , ..., u (cid:48) M with u (cid:48) + ... + u (cid:48) M = 0 for all β .For i = 1 , .., M let α i : ∆(Ω) → R be the affine function α i ( β ) = − (cid:80) l β l u i ( δ l ). Foreach i , define the function u i : ∆(Ω) → R as u i = u (cid:48) i + α i . Then u i ( δ l ) = 0 for all i , l = 1 , ..., N . Note that utility function u i preserves the same preferences over strat-egy profiles as u (cid:48) i , as for any strategy profile (Γ , ..., . Γ M ), E p ,...,p M [ u i ( β (Γ , ..., Γ M ))] = E p ,...,p M [ u (cid:48) i ( β (Γ , ..., Γ M ))] − (cid:80) l π l u i ( δ l ) - the latter term is a constant. Finally note that α ( β ) + ... + α M ( β ) = 0 for all β ∈ ∆(Ω), so u + .... + u M = 0. A.2 Section 3
Lemma 1. General Case:
In any equilibrium (Γ , ..., Γ M ): (1) U i (Γ , ..., Γ M ) = 0 for i = 1 , ..., M and (2) W i ( x ) ≤ x ∈ ∆(Ω) and i = 1 , ..., M . Proof.
We prove (1) first. First note that as the functions { u i } i =1 ,...,M are zero-sum, so are { U i } i . To see this, fix any (Γ (cid:48) , ..., Γ (cid:48) M ) and let Γ (cid:48) be the random variable representing thereceiver’s posterior after viewing all M experiment realizations and p (cid:48) be its pmf. Then (cid:80) Mi =1 U i (Γ (cid:48) , ..., Γ (cid:48) M ) = (cid:80) i (cid:80) β ∈ supp [Γ (cid:48) ] u i ( β ) p (cid:48) ( β ) = (cid:80) β ∈ supp [Γ (cid:48) ] p (cid:48) ( β ) (cid:80) i u i ( β ) = 0. Nextnote that any sender i choosing Γ i = Γ F R yields U i (Γ F R , Γ − i ) = 0 for all Γ − i . Hence inany equilibrium (Γ , ..., Γ M ), each sender gets U i (Γ , ..., Γ M ) ≥
0. Finally no sender canhave U i (Γ , ..., Γ M ) > U j (Γ , ..., Γ M ) < j (cid:54) = i .For (2) we prove the contrapositive. Fix any sender i and opponents’ strategy profileΓ − i such that W i ( x ) > x ∈ ∆(Ω); we will show Γ − i cannot be played inequilibrium. Consider the strategy Γ (cid:48) i with distribution p (cid:48) i and support only on x and { δ l } l =1 ,...,N . Set p (cid:48) i ( x ) > π l − x l p (cid:48) i ( x ) > l (such avalue exists as π l > l ). Bayes-plausibility implies we must have: p (cid:48) i ( δ l ) = π l − x l p (cid:48) i ( x ) > l (as the support of Γ (cid:48) i is { x, δ , ..., δ M } ). Then U i (Γ (cid:48) i , Γ − i ) = W i ( x ) p (cid:48) i ( x ) + (cid:80) l u i ( δ l ) p (cid:48) i ( δ l ) >
0. Property (1) of the lemma implies Γ − i cannot be playedin equilibrium; hence W i ( x ) ≤ i in any equilibrium. Proposition 1.
Proposition 1 is implied by Proposition 2, proven in the next section.However, as the proving Proposition 1 is much simpler than Proposition 2, we provide aproof here for exposition.
Proof.
The ‘only if’ direction is trivial. If u is linear so is u . Under our normalizationof u (0) = u (1) = 0, this implies that u ( β ) = u ( β ) = 0 for all β ∈ [0 , , Γ ) is an equilibrium.21ow for the ‘if’ direction. Suppose u (and hence u ) are nonlinear. Let q = sup { β ∈ [0 ,
1] : u ( β ) (cid:54) = 0 } be the supremum of posteriors at which u , u are nonlinear. We provethe result in two cases. Case 1: q = 1 . If q = 1, then by the piecewise analycity of u , u , there exists r < u ( β ) > u ( β ) < β ∈ [ r, u ( β ) < u ( β ) > u ( β ) > β ∈ [ r, in someequilibrium. As Γ (cid:54) = Γ F R , Pr(0 < Γ < >
0; let y = min supp [Γ ] \ { , } ∈ (0 ,
1) bein the smallest interior belief in the support of Γ . Using the definition of β ( x, y ), definex by β (x , y) = r . Conditional on Γ = x ∈ [x , β ( x, y ) ∈ [ r,
1) for all interior y in Γ ’ssupport. But then for all x ∈ [x ,
1) we have: W ( x ) = u ( β ( x, (cid:124) (cid:123)(cid:122) (cid:125) =0 P r (Γ = 0 | Γ = x ) + u ( β ( x, (cid:124) (cid:123)(cid:122) (cid:125) =0 P r (Γ = 1 | Γ = x )+ (cid:88) y ∈ supp [Γ ] \{ , } u ( β ( x, y )) (cid:124) (cid:123)(cid:122) (cid:125) > P r (Γ = y | Γ = x ) (cid:124) (cid:123)(cid:122) (cid:125) > > . This contradicts Lemma 1 property (2) and hence Γ = Γ F R in all equilibria.
Case 2: q < . We break this case into two subcases.First suppose u ( q ) (cid:54) = 0. WLOG assume u ( q ) > u ( q ) > (cid:54) = Γ F R in some equilibrium. Then let r = q and define y , x as before.Again Lemma 1 prpoerty (2) is violated as: W (x) = u ( β ( x, (cid:124) (cid:123)(cid:122) (cid:125) =0 P r (Γ = 0 | Γ = x) + u ( β ( x, (cid:124) (cid:123)(cid:122) (cid:125) =0 P r (Γ = 1 | Γ = x)+ (cid:88) y ∈ supp [Γ ] \{ , , y } u ( β (x , y )) (cid:124) (cid:123)(cid:122) (cid:125) =0 P r (Γ = y | Γ = x) (cid:124) (cid:123)(cid:122) (cid:125) > > u ( β (x , y)) (cid:124) (cid:123)(cid:122) (cid:125) = u ( r ) > P r (Γ = y | Γ = x) (cid:124) (cid:123)(cid:122) (cid:125) > > . Now suppose u ( q ) = u ( q ) = 0. Then by piecewise analycity of utilities either u ( q − ) > u ( q − ) >
0. WLOG assume u ( q − ) > (cid:54) = Γ F R in some equilibrium. Define y as before. There exists r < q and x such that22 > r, q ) and β (x , y) = r . Then we have W (x) >
0, violating Lemma 1property (2).
A.3 Proof of Proposition 2
A.3.1 ‘Only if ’ direction.
Suppose for some Ω (cid:48) ⊆ Ω all senders have linear utilities on ∆(Ω (cid:48) ). Let x (cid:48) ∈ ∆(Ω (cid:48) ) with x (cid:48) k = π k (cid:80) n ∈ Ω (cid:48) π n ∀ k ∈ Ω (cid:48) . Consider the experiment Γ (cid:48) with P r (Γ (cid:48) = δ l ) = π l for all l (cid:54)∈ Ω (cid:48) and P r (Γ (cid:48) = x (cid:48) ) = (cid:80) n ∈ Ω (cid:48) π n . Γ (cid:48) is Bayes-plausible and has finite support. The strategyprofile (Γ (cid:48) , ..., Γ (cid:48) ) is an non-fully revealing equilibrium. To see this consider a sender i ’sincentive to deviate. If ω ∈ Ω (cid:48) then Γ − i ∈ ∆(Ω (cid:48) ) = ⇒ β (Γ , ..., Γ M ) ∈ ∆(Ω (cid:48) ) w.p. 1(Claim 2); as u i is linear on ∆(Ω (cid:48) ), i has no profitable deviation conditional on ω ∈ Ω (cid:48) .Conditional on ω (cid:54)∈ Ω (cid:48) , Γ − i fully reveals the state and no deviation from i can changethis. A.3.2 ‘If ’ direction.
Let ∆ int (Ω (cid:48) ) = { γ ∈ ∆(Ω (cid:48) ) : γ l > ∀ l ∈ Ω (cid:48) } ; this is the set of beliefs in ∆(Ω (cid:48) ) whosesupport is Ω (cid:48) .We first prove the result for the case of | Ω (cid:48) | = 2. This case is simpler than the caseof | Ω (cid:48) | > | Ω (cid:48) | = 2. Proof for | Ω (cid:48) | = 2 . Proof.
WLOG let Ω (cid:48) = { , } . Suppose some sender i has u i nonlinear on ∆( { , } ). Foreach sender j (cid:48) let r (cid:48) j (cid:48) = sup { t ∈ [0 ,
1] : u j (cid:48) ( tδ + (1 − t ) δ ) > } . Let r (cid:48) = max j (cid:48) =1 ,...,M r (cid:48) j (cid:48) and j ∈ argmax j (cid:48) =1 ,...,M r (cid:48) j (cid:48) . As u i is nonlinear there exists γ ∈ ∆ int ( { , } ) with u i ( γ ) (cid:54) = 0.If u i ( γ ) < u i (cid:48) ( γ ) > i (cid:48) (zero-sumness); otherwise u i ( γ ) > r (cid:48) exists and is > j = 1. We prove the ‘if’ direction in 2 cases. Case 1: r (cid:48) = 1 . u is piecewise real analytic and so ∆( { , } ) can be partitioned into23ntervals each of which u is real analytic on. Each γ ∈ ∆( { , } ) can be represented byscalar γ —how close it is to δ . For some a ∈ [0 , u is real analytic on an interval { γ ∈ ∆( { , } ) : γ ∈ ( a, } (this a ∈ [0 ,
1) is not unique; any selection will do). This impliesthat there are a finite number of points (possibly zero) on { γ ∈ ∆( { , } ) : γ ∈ ( a, } atwhich u = 0. As r (cid:48) = 1, there exists r ∈ ( a,
1) such that u ( γ ) > γ ∈ ∆( { , } )s.t. γ ∈ [ r,
1) (again, this r will not be unique; any selection will do).Suppose, for contradiction, that some equilibrium (Γ , ..., Γ M ) pools { , } . Then wemust must have P r (Γ − = y s.t. y , y > >
0, i.e. Γ − pools { , } , by Claim A.3.Let Z = { y ∈ supp [Γ − ] : y , y > } ; Z is nonempty. For any x ∈ ∆ int ( { , } ) and y ∈ Z , we have p − ( y | x ) > β ( x, y ) ∈ ∆ int ( { , } ) (by Claim A.2 and equation A.2).Note that as x → δ we have β ( x, y ) → δ = ⇒ β ( x, y ) → x → δ we have β ( x, y ) → δ = ⇒ β ( x, y ) →
1. As Z is finite, this implies min y ∈ Z β ( x, y ) goes to 0 as x → δ and min y ∈ Z β ( x, y ) goes to 1 as x → δ . As for all y ∈ Z β ( x, y ) is continuous in x for x ∈ ∆ int ( { , } ), min y ∈ Z β ( x, y ) is also continuous in x for x ∈ ∆ int ( { , } ). By theintermediate value theorem there exists x ∈ ∆ int ( { , } ) such that min y ∈ Z β (x , y ) = r .Note that by equation A.2, β (x , y ) = δ for all y ∈ supp [Γ − ] with y > y = 0;similarly β (x , y ) = δ for all y ∈ supp [Γ − ] with y > y = 0. Finally by Claim A.1, p − ( y | x) = 0 for all y ∈ supp [Γ − ] with y = y = 0.Putting this together: W (x) = (cid:88) y ∈ supp [Γ − ] y =0 ,y > u ( δ ) (cid:124) (cid:123)(cid:122) (cid:125) =0 p − ( y | x) + (cid:88) y ∈ supp [Γ − ] y =0 ,y > u ( δ ) (cid:124) (cid:123)(cid:122) (cid:125) =0 p − ( y | x)+ (cid:88) y ∈ Z u ( β (x , y )) (cid:124) (cid:123)(cid:122) (cid:125) > p − ( y | x) (cid:124) (cid:123)(cid:122) (cid:125) > > { , } . Case 2: r (cid:48) < . First, if u ( r (cid:48) ) >
0, then set r = r (cid:48) and derive x just as in Case 1.As in Case 1, we have W (x) >
0, violating Lemma 1 property (2). Next if u ( r (cid:48) ) < i (cid:54) = 1 must have u i ( r (cid:48) ) > i to 1and repeat the same argument.Next assume u ( r (cid:48) ) = 0. Now for some a ∈ [0 , r (cid:48) ), u is real analytic on an interval24 γ ∈ ∆( { , } ) : γ ∈ ( a, r (cid:48) ) } (this a ∈ [0 , r (cid:48) ) is not unique; any selection will do). Thisimplies that there are a finite number of points (possibly zero) on { γ ∈ ∆( { , } ) : γ ∈ ( a, r (cid:48) ) } at which u = 0. This implies that there exists r ∈ ( a, r (cid:48) ) such that u ( γ ) > γ ∈ ∆( { , } ) with γ ∈ [ r, r (cid:48) ) (again, this r will not be unique; any selection will do).Note that u ( γ ) ≥ γ ∈ ∆( { , } ) with γ ≥ r .Suppose, for contradiction, that in some equilibrium (Γ , ..., Γ M ) pools { , } . Wefollow identical steps in defining Z and x. Note that for all y ∈ Z , β k (x , y ) ∈ [ r,
1) = ⇒ u ( β (x , y )) ≥
0. By the definition of x, there exists y ∈ Z such that β (x , y) = r = ⇒ u ( β (x , y)) >
0. For all y ∈ supp [Γ − ] \ Z either β (x , y ) ∈ { δ , δ } or p − ( y | x) >
0. Hence W (x) >
0, violating Lemma 1 property (2).Now we proceed with the analysis for | Ω (cid:48) | ≥ General Analysis.
Suppose some u j is nonlinear on ∆(Ω (cid:48) ). Fix a strategy profile (Γ , ..., Γ M ) that poolsΩ (cid:48) ; let Γ be the experiment induced by observing the realizations of all M experimentsand p be its probability mass function. We show, via violation of Lemma 1 property (1),that (Γ , ..., Γ M ) is not an equilibrium. To do this it sufficies to identify a sender j and aninterim belief x such that when Γ , ..., Γ M are played, conditional on generating interimbelief x (from an experiment played additionally and conditionally independently to Γ j ) j gets a strictly positive expected payoff: : E [ u i ( β (x , Γ))] >
0. As in the proof of Lemma1 property (2), j can then construct an experiment Γ (cid:48) j with support on { δ , ..., δ N , x } andobtain a strictly positive payoff by playing Γ (cid:48) j in addition to, conditionally independently,Γ j . The remainder of the proof shows that such a sender j and x exist.For i = 1 , ..., M let A i = { γ ∈ ∆(Ω) : u i ( γ ) > } and D i = { γ ∈ ∆(Ω) : u i ( γ ) < } be the sets of posteriors at which i has an advantage and disadvantage respectively. Let A = ∪ i A i be the union of these advantage sets (also equal to the union of disadvantagesets as utilities are zero-sum) and cl ( A ) be its closure.We say a subset of states Θ ⊆ Ω ( | Θ | >
1) is minimal if A ∩ Θ (cid:54) = ∅ and Θ (cid:48) ∩ A = ∅ for all Θ (cid:48) ⊂ Θ. Note that if A is empty, then there are no minimal subsets. Meanwhile if A is nonempty, any subset of states that intersects A (i.e. any set Θ for which some u i isnonlinear on ∆(Θ)) is either minimal or has a minimal subset: Claim A.4.
Every subset Θ ⊆ Ω for which u i (for some i ) is nonlinear on ∆(Θ) is eitherminimal or has a subset Θ (cid:48) that is minimal. roof. If for some i u i is nonlinear on ∆(Θ), then A ∩ ∆(Θ) (cid:54) = ∅ . Either Θ is minimal, orthere exists a subset Θ (cid:48) ⊂ Θ that intersects A . Now set Θ = Θ (cid:48) and repeat this processuntil Θ is minimal; it must be minimal at some point because Ω is finite and states areremoved from Θ each iteration.By Claim A.4, it is sufficient to prove Proposition 2 for minimal Ω (cid:48) alone. If allminimal sets cannot be pooled in equilbirium, then any set on which there are nonlinearsender preferences cannot be pooled, as all such sets have a minimal subset. Henceforthwe assume Ω (cid:48) is minimal. Let | Ω (cid:48) | = K ≤ N and WLOG let Ω (cid:48) = { , ..., K } .It is convenient for us to represent any belief γ ∈ ∆(Ω (cid:48) ) by the ratios ( r ( γ ) , ..., r K − ( γ )) ∈ ( R + ∪ {∞} ) K − , where for k = 1 , ..., K −
1: (1) r k ( γ ) = γ k − (cid:80) kl =1 γ l when 1 − (cid:80) kl =1 γ l isnonzero, (2) when this doesn’t hold r k ( γ ) = ∞ if γ k > r k ( γ ) = 0 if γ k = 0. We callthis the ratio representation of γ . The ratio r k ( γ ) tells us the ratio of probability massassigned to state k by γ to the mass assigned to states k + 1 , ..., K . Lemma A.1.
Note for any γ, γ (cid:48) ∈ ∆(Ω (cid:48) ) we have r k ( γ ) = r k ( γ (cid:48) ) for all k = 1 , ..., K − if and only if γ = γ (cid:48) ; that is, ratio representations for beliefs in ∆(Ω (cid:48) ) are unique.Proof. The ‘if’ direction is trivial; we prove the ‘only if’ direction as follows. First suppose r k ( γ ) < ∞ for all k = 1 , ..., K −
1. This implies that 1 − (cid:80) kl =1 γ l is nonzero for all k (or else, let k (cid:48) be the minimum k for which 1 − (cid:80) kl =1 γ l = 0; but then we must have γ k (cid:48) > ⇒ r k (cid:48) ( γ ) = ∞ —contradiction). But then from its definition, r uniquely pinsdown γ ( γ = r ( γ )1+ r ( γ ) ), after which r pins down γ , ..., r K − pins down γ K − , and γ K is pinned down by 1 = (cid:80) Kl =1 γ l ). Now suppose r k (cid:48)(cid:48) ( γ ) = ∞ for some k (cid:48)(cid:48) . Note that thisimplies γ k = 0 for all k > k (cid:48)(cid:48) ; further, 1 − (cid:80) kl =1 γ l > k < k (cid:48)(cid:48) and hence r k ( γ ) < ∞ for all k < k (cid:48)(cid:48) . Then γ , ..., γ k (cid:48)(cid:48) − are uniquely pinned down by using the definitions of r ( γ ) , ..., r k (cid:48)(cid:48) − ( γ ) (just as in the previous case). γ k (cid:48)(cid:48) is pinned down by 1 = (cid:80) Kl =1 γ l .The the continuity of r k ( γ ) on part of ∆(Ω (cid:48) ) will be useful later: Claim A.5.
For k = 1 ..., K − , r k ( γ ) is continuous in γ for γ ∈ ∆( { k, ..., K } ) \ { δ k } .Proof. r k ( γ ) = γ k − (cid:80) kl =1 γ k . As γ ∈ ∆( { k, ..., K } ), the denominator is strictly positive when γ k < r k ( γ ) is continuous in γ on this domain.The following simple results will be useful.26 emma A.2. Suppose
K > . For any < L < K , let x ∈ ∆( { L, ..., K } ) , x (cid:48) ∈ ∆( { , ..., L − } ) and y ∈ ∆(Ω) . If β ( x, y ) is well defined, then r k ( β ( λx (cid:48) + (1 − λ ) x, y )) = r k ( β ( x, y )) for all k = L, ..., K − and λ ∈ [0 , .Proof. For any n ∈ Ω, β n ( λx (cid:48) + (1 − λ ) x, y ) = ( λx (cid:48) n +(1 − λ ) x n ) y n π n (cid:80) Nn (cid:48) =1 ( λx (cid:48) n (cid:48) +(1 − λ ) x n (cid:48) ) y n (cid:48) π n (cid:48) For k ≥ L : r k ( β (( λx (cid:48) + (1 − λ ) x ) , y )) = β k (( λx (cid:48) + (1 − λ ) x ) , y ) (cid:80) Nn = k +1 β n (( λx (cid:48) + (1 − λ ) x ) , y ) = ( λx (cid:48) k + (1 − λ ) x k ) y k /π k (cid:80) Nn = k +1 ( λx (cid:48) n + (1 − λ ) x n ) y n /π n = (1 − λ ) x k y k /π k (1 − λ ) (cid:80) Nn = k +1 x n y n /π n = x k y k /π k (cid:80) Nn = k +1 x n y n /π n whenever the denominator is nonzero; when the denominator is nonzero, this expres-sion is equal to r k ( β ( x, y )). When the denominator and numerator are zero, r k ( β ( λx (cid:48) +(1 − λ ) x, y )) = r k ( β ( x, y )) = 0 and when the denominator is zero and the numerator isnonzero, r k ( β ( λx (cid:48) + (1 − λ ) x, y )) = r k ( β ( x, y )) = ∞ . Claim A.6.
Suppose
K > . For any < L < K , let x ∈ ∆( { L, ..., K } ) , x (cid:48) ∈ ∆( { , ..., L − } ) and y ∈ ∆(Ω) . If β ( x, y ) and β ( x (cid:48) , y ) are well defined then β k ( λx (cid:48) + (1 − λ ) x, y ) = λβ k ( x (cid:48) , y ) + (1 − λ ) β k ( x, y ) for all λ ∈ [0 , , k = 1 , ..., N .Proof. Simple algebra.Let Z = { z ∈ supp [Γ] : z n > n ∈ Ω (cid:48) } . Note Z is nonempty as (Γ , ..., Γ M )pools Ω (cid:48) . For k = 1 , ..., K − x ∈ ∆ int (Ω (cid:48) ) define M x ( k ) recursively starting with k = K − M x ( K −
1) = argmin z ∈ Z r K − ( β ( x, z )) (A.4) From its definition, one can see β ( x, y ) is only not well defined when y assigns probability 0 to everystate that x assigns strictly positive probability to. x ( K −
1) is nonempty as Z is. For k < K −
1, let M x ( k ) = argmin z ∈ M x ( k +1) r k ( β ( x, z ));these sets are nonempty for all k . For k = 1 , ..., K − m x ( k ) by: pick z ∈ M x ( k )and let m x ( k ) = r k ( β ( x, z )). m x ( k ) is well defined for all k . M x ( K −
1) gives the set of realizations of Γ that, conditional on interim belief x beingrealized from a different experiment, would induce the lowest r K − ratio of posteriorsamong those in Z . m x ( K −
1) gives the value of this lowest r K − ratio. M x ( K −
2) givesthe subset of M x ( K −
1) that would result in lowest r K − ratio of posteriors conditionalon x being realized and m x ( K −
2) gives this value, etc.Note any z ∈ M x (1) must satisfy: r k ( β ( x, z )) = m x ( k ) for all k = 1 , ..., K − β ( x, y ) ∈ ∆(Ω (cid:48) ) ( x ∈ ∆ int (Ω (cid:48) ) and Claim A.2), by Lemma A.1, ratios m x (1) , ..., m x ( K −
1) uniquely pin down the value of β ( x, z ) for all z ∈ M x (1). If we have | M x (1) | >
1, thismeans that multiple realizations of Γ, z (cid:54) = z (cid:48) , produce the same posterior conditional on x . This is possible when x assigns probability 0 to states z, z (cid:48) do not — z and z (cid:48) differingon these states may not affect the posterior.Using the objects introduced above, we finish proving Proposition 2 in two cases. Inthe first case, cl ( A ) ∩ Ω (cid:48)(cid:48) = ∅ for all Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) . This case include the example in the mainAppendix in the text of paper; the same logic generalizes. The second case to consider is cl ( A ) ∩ Ω (cid:48)(cid:48) (cid:54) = ∅ for some Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) . Case 1: cl ( A ) ∩ Ω (cid:48)(cid:48) = ∅ for all Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) .Lemma A.3. Suppose cl ( A ) ∩ ∆(Ω (cid:48)(cid:48) ) = ∅ for all Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) . Then there exists x ∗ ∈ ∆(Ω (cid:48) ) and ¯ β ∈ cl ( A ) such that for all y ∈ Z either: (1) β ( x ∗ , y ) (cid:54)∈ cl ( A ) or (2) β ( x ∗ , y ) = ¯ β .Proof. Define the point ¯ β ∈ cl ( A ) ∩ ∆(Ω (cid:48) ) as follows. Let E ( K −
1) = argmax γ ∈ cl ( A ) ∩ ∆(Ω (cid:48) ) r K − ( γ )and e ( K −
1) = max γ ∈ cl ( A ) ∩ ∆(Ω (cid:48) ) r K − ( γ ). For k = 1 , ..., K −
2, let E ( k ) = argmax γ ∈ E ( k +1) r k ( γ )and e ( k ) = max γ ∈ E ( k +1) r k ( γ ). Note that as cl ( A ) ∩ ∆(Ω (cid:48)(cid:48) ) = ∅ for all Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) , we have0 < e ( k ) < ∞ for all k = 1 , ..., K −
1. Further, by Lemma A.1, | E (1) | = 1 as γ ∈ E (1)must satisfy r k ( γ ) = e ( k ) for all k = 1 , ..., K −
1. Let ¯ β be the unique element in E (1).Consider x ∈ ∆ int ( { K − , K } ). Note that as x → δ K , β ( x, y ) → δ K = ⇒ r K − ( β ( x, y )) → y ∈ Z . Similarly as x → δ K − , β ( x, y ) → δ K − = ⇒ r K − ( β ( x, y )) → ∞ for all y ∈ Z . By the finiteness of Z , x → δ K = ⇒ min y ∈ Z r K − ( β ( x, y )) →
28 and x → δ K − = ⇒ min y ∈ Z r K − ( β ( x, y )) → ∞ . The continiuity of β ( x, y ) in x ,continuity of r K − in β ( x, y ) (Claim A.5), and finiteness of Z together imply the con-tinuity of min y ∈ Z r K − ( β ( x, y )) in x . By the intermediate value theorem, there exists x (cid:48) ∈ ∆ int ( { K − , K } ) with min y ∈ Z r K − ( β ( x (cid:48) , y )) = e ( K − m x (cid:48) ( K −
1) = e ( K − x (cid:48) ∈ ∆( { k (cid:48) + 1 , ..., K } ) such that for all k = k (cid:48) + 1 , ..., K − m x (cid:48) ( k ) = e ( k ). We find x (cid:48)(cid:48) ∈ ∆( { k (cid:48) , ..., K } ) with m x (cid:48)(cid:48) ( k ) = e ( k ) for all k = k (cid:48) , ..., K − x (cid:48) ( λ ) = λδ k (cid:48) + (1 − λ ) x (cid:48) for λ ∈ (0 , λ → β ( x (cid:48) ( λ ) , y ) → δ k (cid:48) = ⇒ r k (cid:48) ( β ( x (cid:48) ( λ ) , y )) → ∞ for all y ∈ Z and as λ → β ( x (cid:48) ( λ ) , y ) → x (cid:48) = ⇒ r k (cid:48) ( β ( x (cid:48) ( λ ) , y )) → y ∈ Z . For all λ ∈ [0 , y ∈ Z , k = k (cid:48) + 1 , ..., K − r k ( β ( x (cid:48) ( λ ) , y )) = r k ( β ( x (cid:48) , y ))by Lemma A.2; hence changing λ will leave m x (cid:48) ( k ) = e ( k ) for k = k (cid:48) + 1 , ..., K −
1. Byfiniteness of M x (cid:48) ( k (cid:48) + 1), continuity of β ( x (cid:48) ( λ ) , y ) for all y ∈ M x (cid:48) ( k (cid:48) + 1), continuity of r k (cid:48) in β ( x (cid:48) ( λ ) , y ) for all y ∈ M x (cid:48) ( k (cid:48) + 1), and in the intermediate value theorem, there exists λ ∗ ∈ (0 ,
1) and x (cid:48)(cid:48) = λ ∗ δ k (cid:48) + (1 − λ ∗ ) x (cid:48) ∈ ∆( { k (cid:48) , ..., K } ) such that m x (cid:48)(cid:48) ( k (cid:48) ) = e ( k (cid:48) ). Then m x (cid:48)(cid:48) ( k ) = e ( k ) for all k = k (cid:48) , ..., K − k (cid:48) = 1, by equation A.5 we find x ∗ ∈ ∆(Ω (cid:48) ) with, for all y ∈ M x ∗ (1) and k = 1 , ..., K − r k ( β ( x ∗ , y )) = e ( k ). Hence for all y ∈ M x ∗ (1), β ( x ∗ , y ) = ¯ β . Meanwhile for all y (cid:54)∈ M x ∗ (1), there exists 1 ≤ k (cid:48) ≤ K − r k ( β ( x ∗ , y )) = e ( k ) for all k > k (cid:48) and r k (cid:48) ( β ( x ∗ ) , y ) > e ( k (cid:48) ); this implies (by definitionof e ( k (cid:48) )) that β ( x ∗ , y ) (cid:54)∈ cl ( A ).The following result follows from Lemma A.3 almost immediately. Lemma A.4.
Suppose cl ( A ) ∩ ∆(Ω (cid:48)(cid:48) ) = ∅ for all Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) . Then there exists x ∈ ∆(Ω (cid:48) ) and β (cid:48) ∈ A such that for all y ∈ Z either: (1) β ( x , y ) (cid:54)∈ cl ( A ) or (2) β ( x , y ) = β (cid:48) .Proof. Find x ∗ and ¯ β as per Lemma A.3. If ¯ β ∈ A then set x = x ∗ and β (cid:48) = ¯ β and we’redone.Assume now ¯ β (cid:54)∈ A . As Z finite, cl ( A ) is closed, and β ( x, y ) is continuous in x forall y ∈ Z , there exists (cid:15) > N (cid:15) = { x ∈ ∆(Ω (cid:48) ) : | x − x ∗ | < (cid:15) } such that for all x ∈ N (cid:15) and y ∈ Z , β ( x ∗ , y ) (cid:54)∈ cl ( A ) = ⇒ β ( x, y ) (cid:54)∈ cl ( A ).By equation A.2, for any γ ∈ ∆(Ω (cid:48) ) and y ∈ Z there exists x ∈ ∆(Ω (cid:48) ) such that β ( x, y ) = γ . Also by equation A.2, if β ( x, y ) = β ( x, y (cid:48) ) for some x ∈ ∆ int (Ω (cid:48) ) and y, y (cid:48) ∈ Z , then β ( x (cid:48) , y ) = β ( x (cid:48) , y (cid:48) ) for all x (cid:48) ∈ ∆ int (Ω (cid:48) ).29s ¯ β ∈ cl ( A ), there is a sequence of beliefs in A converging to ¯ β . By continuity of β ( x, y ) in x for all y ∈ Z and the facts in the previous paragraph, there exists β (cid:48) ∈ A close to ¯ β and x ∈ N (cid:15) such that for all y ∈ Z , β ( x ∗ , y ) = ¯ β = ⇒ β (x , y ) = β (cid:48) .Hence we have either β (x , y ) = β (cid:48) ∈ A or β (x , y ) (cid:54)∈ cl ( A ) for all y ∈ Z .This next Lemma wraps up the proof of the ‘if’ direction of Proposition 2 for thecase that cl ( A ) ∩ ∆(Ω (cid:48)(cid:48) ) = ∅ for all Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) : Lemma A.5.
Suppose cl ( A ) ∩ ∆(Ω (cid:48)(cid:48) ) = ∅ for all Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) . Then there exists a sender j and x ∈ ∆ int (Ω (cid:48) ) with E [ u j ( β ( x , Γ))] > .Proof. Find β (cid:48) and x as per Lemma A.4.Note that for all y ∈ supp [Γ] \ Z , one of the following holds. (1) y l = 0 for all l ∈ Ω (cid:48) .(2) y l = 0 for some l ∈ Ω (cid:48) but y k > k ∈ Ω (cid:48) . For y in case (1), by Claim A.1 p ( y | x) = 0. For y in case (2), equation A.2 implies β (x , y ) ∈ Ω (cid:48)(cid:48) for some Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) ; as Ω (cid:48) is minimal we have u i ( β (x , y )) = 0 for all i .For all y ∈ Z such that β (x , y ) (cid:54) = β (cid:48) , u i ( β (x , y )) = 0 for all i .Let j be some sender with u j ( β (cid:48) ) >
0. Then E [ u j ( β (x , Γ))] = u j ( β (cid:48) ) P r (Γ ∈ { y ∈ Z : β (x , y ) = β (cid:48) }| x) > Case 2: cl ( A ) ∩ Ω (cid:48)(cid:48) (cid:54) = ∅ for some Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) . Note that it is still the case that Ω (cid:48) is minimal, so A ∩ Ω (cid:48)(cid:48) = ∅ for all Ω (cid:48)(cid:48) (cid:40) Ω (cid:48) .Let Ω (cid:48)(cid:48) ∈ argmin { Ω (cid:48)(cid:48)(cid:48) ⊂ Ω (cid:48) : cl ( A ) ∩ Ω (cid:48)(cid:48)(cid:48) (cid:54) = ∅} | Ω (cid:48)(cid:48)(cid:48) | be one of the smallest subsets of Ω (cid:48) thatintersects cl ( A ). Note that 1 ≤ | Ω (cid:48)(cid:48) | < | Ω (cid:48) | . Define Z , and all other necessary objects, asin the previous case. Then note that cl ( A ) ∩ Θ = ∅ for all Θ (cid:40) Ω (cid:48)(cid:48) and so by an identicalargument to Lemma A.3 we can find ¯ β ∈ cl ( A ) ∩ ∆ int (Ω (cid:48)(cid:48) ) and x ∗ ∈ ∆ int (Ω (cid:48)(cid:48) ) such thatfor all y ∈ Z either β ( x ∗ , y ) = ¯ β or β ( x ∗ , y ) (cid:54)∈ cl ( A ). Then by a similar argument toLemma A.4, we can find x ∈ ∆ int (Ω (cid:48) ) close to x ∗ and β (cid:48) ∈ A close to ¯ β such that either β (x , y ) = β (cid:48) or β (x , y ) (cid:54)∈ cl ( A ) for all y ∈ Z .Following the same argument as in Lemma A.5, all y ∈ supp [Γ] \ Z either occur with30robability 0 conditional on x or result in a posterior outside of ∆ int (Ω (cid:48) ) conditional onx (yielding 0 utility by minimality of Ω (cid:48) ). Letting j be some sender with u j ( β (cid:48) ) >
0, wehave E [ u j ( β (x , Γ))] > A.3.3 Additional Claim
In the proof of Proposition 2 given, when utilities are nonlinear on some ∆(Ω (cid:48) ) and(Γ , ..., Γ M ) pooled Ω (cid:48) , we were able to find a sender j who could take advantage of Ω (cid:48) being pooled and find x conditonal on which she gets strictly positive expected utility.Identifying this sender j did not depend on the strategy profile (Γ , ..., Γ M ). Hence, theclaim below (which is useful for further results in Supplementary Appendix B) holds. Claim A.7.
Suppose for some sender i and Ω (cid:48) ⊆ Ω that u i is nonlinear on ∆(Ω (cid:48) ) . Thenthere exists a sender j such that for any (Γ , ..., Γ M ) that pools Ω (cid:48) j can find some x suchthat E [ u j ( β ( x , Γ , ..., Γ M ))] > . A.4 Proof of Theorem 1
Theorem 1.
Proof. ‘If’ direction. Proposition 2 implies that if for every l, k some u i is nonlinear on∆( { l, k } ), then in any equilibrium (Γ , ..., Γ M ), w.p. 1 β n (Γ , ..., Γ M ) > n ∈ Ω. Hence w.p. 1 we must have β n (Γ , ..., Γ M ) = 1 for some n ∈ Ω; the state is fullyrevealed.‘Only if’ direction. Suppose for some l, k ∈ Ω, u i is linear along ∆( { l, k } ) for all i .Then the equilibrium construction in the Proposition 2 ‘only if’ direction is a non-fullyrevealing equilibrium. 31 Supplementary Appendix B: Extensions
B.1 Single Receiver with Finite Actions
Setup.
There is a single receiver. Let A = { a , ..., a A } be a finite set of actions; the gameis as in the baseline model except after observing realizations of Γ , ..., Γ M the receiverpicks an action a ∈ A after which the players get payoffs. The reciever’s utility functionis u r : A × Ω → R and senders’ utility functions are u i : A × Ω → R for i = 1 , ..., M .Sender’s preferences are zero-sum: for all a ∈ A , ω ∈ Ω, (cid:80) Mi =1 u i ( a, ω ) = 0. We make theassumption that for any a (cid:54) = a (cid:48) ∈ A , ω ∈ Ω, u r ( a, ω ) (cid:54) = u r ( a (cid:48) , ω ) and u i ( a, ω ) (cid:54) = u i ( a (cid:48) , ω )for all i ∈ { , ..., M } ; this is generically true. The equilibrium concept is Perfect BayesianEquilibrium (PBE).For any β ∈ ∆(Ω) let A ∗ ( β ) = argmax a ∈A (cid:80) Nl =1 u r ( a, l ) β l . In any PBE, after signalsare realized and the receiver updates to posterior belief β ∈ ∆(Ω), the receiver takes anaction a ∈ A ∗ ( β ) that maximizes expected utility (sequential rationality). Let R indiff = { β ∈ ∆(Ω) : | A ∗ ( β ) | > } be the set of posteriors at which the receiver is indifferentbetween multiple best actions. We assume that at posteriors in β ∈ R indiff the receiverbreaks ties by choosing the lowest indexed action in A ∗ ( β ). Hence in equilibrium thereciever takes action a ∗ ( β ) = argmin a b ∈ A ∗ ( β ) b ; this is well defined and single valued foreach β ∈ ∆(Ω).For i = 1 , ..., M define v i : ∆(Ω) → R as sender i ’s expected utility from any posterior β ∈ ∆(Ω) in a PBE following the specified tie breaking rule: v i ( β ) = (cid:80) Nl =1 u i ( a ∗ ( β ) , l ) β l .First we show that in any PBE satisfying this tie-breaking rule, v i ’s are zero-sum andpiecewise analytic. This will allow us to directly apply our previous results to them. Claim B.1.
In any PBE in which the receiver chooses action a ∗ ( β ) = argmin a b ∈ A ∗ ( β ) b after signals are realized, v , ..., v M are each piecewise analytic and are zero-sum.Proof. Zero-sumness is trivial. For all β , (cid:80) Mi =1 v i ( β ) = (cid:80) i (cid:80) Nl =1 u i ( a ∗ ( β ) , l ) β l = (cid:80) Nl =1 β l (cid:80) i u i ( a ∗ ( β ) , l ) = 0.Now we show piecewise analyicity. For each a b ∈ A , let ∆ a b = { β ∈ ∆(Ω) : a b = a ∗ ( β ) } . Note that for every a b ∈ A the set ∆ a b is convex; for any β, β (cid:48) ∈ ∆ a b we can see a ∗ ( λβ + (1 − λ ) β (cid:48) ) = a b for all λ ∈ (0 ,
1) as follows. For any b (cid:48) < b we have:32 (cid:88) l =1 u r ( a b , l ) β l > N (cid:88) l =1 u r ( a b (cid:48) , l ) β lN (cid:88) l =1 u r ( a b , l ) β (cid:48) l > N (cid:88) l =1 u r ( a b (cid:48) , l ) β (cid:48) l = ⇒ N (cid:88) l =1 u r ( a b , l )( λβ l + (1 − λ ) β (cid:48) l ) > N (cid:88) l =1 u r ( a b (cid:48) , l )( λβ l + (1 − λ ) β (cid:48) l )For any b (cid:48) > b the first two inequalities must hold weakly which implies the thirdholds weakly. So at ( λβ l + (1 − λ ) β (cid:48) l ) a b yields weakly higher utility for the receiver thanall higher indexed actions. Together this implies a ∗ ( λβ + (1 − λ ) β (cid:48) ) = a b and hence ∆ a b isconvex. Note that { ∆ a b } b =1 ,...,A partition ∆(Ω); hence they form a partition of ∆(Ω) intoconvex sets. On each ∆ a b , each v i = (cid:80) Nl =1 u i ( a b , l ) β l ( i = 1 , ..., M ) is linear in β . Henceeach v i is piecewise linear on ∆(Ω). So { v i } i =1 ,...,M are piecewise analytic.Lemma B.1 below shows that all v i ’s are linear on ∆(Ω (cid:48) ) if and only if the receiverprefers the same action at all states in Ω (cid:48) . Lemma B.1.
Fix Ω (cid:48) ⊆ Ω , | Ω (cid:48) | > . v , ..., v M are all linear on ∆(Ω (cid:48) ) if and only if a ∗ ( δ l ) = a ∗ ( δ k ) for all l, k ∈ Ω (cid:48) .Proof. We prove the ‘only if’ direction by proving the contrapositive. Consider any states l (cid:54) = k ∈ Ω (cid:48) . Suppose a ∗ ( δ l ) (cid:54) = a ∗ ( δ k ). For β ∈ ∆( { l, k } ) ⊆ ∆(Ω (cid:48) ), the receiver hasexpected utility u r ( a, l ) β l + u r ( a, k ) β k . Note by our assumption that no agent is indifferentbetween any two actions at any state, we must have that u r ( a ∗ ( δ l ) , l ) > u r ( a (cid:48) , l ) for all a (cid:48) (cid:54) = a ∗ ( δ l ) and u r ( a (cid:48) , k ) < u r ( a ∗ ( δ k ) , k ) for all a (cid:48) (cid:54) = a ∗ ( δ k ). By continuity, this impliesthat there exists 1 > β uk > β lk > β ∈ ∆( { l, k } ) if: (1) β k < β lk then a ∗ ( β ) = a ∗ ( δ l ) and (2) β k > β uk then a ∗ ( β ) = a ∗ ( δ k ). For β ∈ ∆( { l, k } ) and i ∈ { , ..., M } , v i ( β ) = β k u i ( a ∗ ( β ) , k ) + (1 − β k ) u i ( a ∗ ( β ) , l ) = u i ( a ∗ ( β ) , l ) + β k ( u i ( a ∗ ( β ) , k ) − u i ( a ∗ ( β ) , l )).When β k > β uk , v i is linear in β k with slope ( u i ( a ∗ ( δ k ) , k ) − u i ( a ∗ ( δ k ) , l )). When β k < β lk , v i is linear in β k with slope ( u i ( a ∗ ( δ l ) , k ) − u i ( a ∗ ( δ l ) , l )). These slopes are different because: u i ( a ∗ ( δ k ) , k ) − u i ( a ∗ ( δ l ) , k ) > > u i ( a ∗ ( δ k ) , l ) − u i ( a ∗ ( δ l ) , l )= ⇒ u i ( a ∗ ( δ k ) , k ) − u i ( a ∗ ( δ k ) , l ) (cid:54) = u i ( a ∗ ( δ l ) , k ) − u i ( a ∗ ( δ l ) , l )33ence v i is nonlinear along ∆( { l, k } ) and hence nonlinear on ∆(Ω (cid:48) ) ⊇ ∆( { l, k } ).‘If’ direction. Suppose for all l ∈ Ω (cid:48) , a ∗ ( δ l ) = a . Note that ∆(Ω (cid:48) ) is the convex hullof { δ l : l ∈ Ω (cid:48) } . Then, by convexity of the set ∆ a (see proof of Claim B.1), we must have∆(Ω (cid:48) ) ⊆ ∆ a . For all i ∈ { , ..., M } v i is linear on ∆ a (proof of Claim B.1) and hence∆(Ω (cid:48) ).Finally we prove our main results for the finite action model. Proposition B.1 isimportant as it says that even when the receiver does not fully learn the state, they learnadequately —that is, enough that further learning would not influence their action. Proposition B.1.
In any PBE in which the receiver chooses action a ∗ ( β ) = argmin a b ∈ A ∗ ( β ) b , the receiver takes their first best action w.p. .Proof. Fix a PBE and consider all posteriors induced by the equilibrium experiments withpositive probability. At any posterior β ∈ { δ , ..., δ N } , the receiver clearly takes their firstbest action. Now consider any posterior that occurs with positive probability that doesnot fully reveal the state. At such a β , there exists Ω (cid:48) ⊆ Ω with | Ω (cid:48) | > β l > l ∈ Ω (cid:48) . But then Ω (cid:48) is being pooled in equilibrium which implies (Proposition 2) that v , ..., v M are all linear on ∆(Ω (cid:48) ) which implies (Lemma B.1) that there exists a ∈ A suchthat a ∗ ( δ l ) = a for all l ∈ Ω (cid:48) . Note that ω ∈ Ω (cid:48) (all other states are ruled out by β ). Byconvexity of ∆ a (proof of Claim B.1), a ∗ ( β ) = a = a ∗ ( δ ω ). Hence, the receiver’s ex-postpayoff is always their first best payoff: u r ( a ∗ ( ω ) , ω ).The finite action characterization of all equilibria being fully revealing follows imme-diately from Theorem 1 and Lemma B.1. Corollary B.1.
The state is fully revealed in every PBE in which the receiver choosesaction a ∗ ( β ) = argmin a b ∈ A ∗ ( β ) b if and only if for every pair of states l and k a ∗ ( δ l ) (cid:54) = a ∗ ( δ k ) .Proof. ‘If’ direction. If for a pair of states l and k , a ∗ ( δ l ) (cid:54) = a ∗ ( δ k ), then by Lemma B.1some v i is nonlinear (and hence nonlinear) on ∆( { l, k } ). If this is true for every pair ofstates, then along every edge of the simplex we have nonlinearity of some sender’s utilityfunction which implies (Theorem 1) full revelation in every equilibrium.‘Only if’ direction. If a ∗ ( δ l ) = a ∗ ( δ k ) for any pair of states l and k , then LemmaB.1 implies that all v i ’s are linear on ∆( { l, k } ). This implies (Theorem 1) that there arenon-fully revealing equilibria. 34 .2 Robustness Here we consider the robustness of our results to the assumption that preferences arezero-sum. Consider a game identical to the baseline model but with utility functions u , ..., u M ( u i : ∆(Ω) → R ) that need not be zero-sum. We assume that all utilities arepiecewise analytic and make the normalization u i ( δ l ) = 0 for all senders i and states l .We adopt notation from the baseline model whenever it obviously carries over.Before presenting the robustness results, which concern the information revealed inequilibrium as preferenes approach zero-sum, we discuss what we can say in this moregeneral setting. As in the baseline model, we have no issues with equilibrium existence;for any u , ...u M there is a fully revealing equilibrium (Γ F R , ..., Γ F R ). Of course our results in the paper, starting with Lemma 1, rely on the zero-sumnessof preferences and do not generalize to this setting. While Lemma 1 says senders mustget their full revelation payoff in every equilibrium of a zero-sum game, this is no longertrue when preferences may be nonzero-sum. As an example, suppose all senders have thesame preferences; then there will be an equilibrium in which one sender plays a singlesender optimal experiment (a Bayesian Persuasion, or BP, solution) and all others playΓ U . This BP solution may not fully reveal the state and may yield the senders strictlylarger payoffs than the 0 payoff from full revelation. More generally, when preferencesare not zero-sum, agreement between senders (even if it isn’t complete agreement) mayallow them to play experiments that yield them strictly positive payoffs in equilibrium.For any γ ∈ ∆(Ω) and u , ..., u M , let (cid:80) i u i ( γ ) be the total surplus shared amongsenders when the receiver’s posterior is γ . Note this surplus is 0 when γ ∈ { δ , ..., δ N } .Let M S ( u , ..., u M ) = sup γ ∈ ∆(Ω) (cid:80) Mi =1 u i ( γ ) be the supremum of the total surplus sendersget at any posterior. Note M S ( u , ..., u M ) = 0 when the game is zero-sum. While wecannot pin down equilibrium payoffs as we did in Lemma 1, we can upperbound them foreach sender. While senders may get above utility 0 in equilibrium, none can attain utilityhigher than the maximum surplus: Lemma B.2.
In any equilibrium (Γ , ..., Γ M ) , for all senders i = 1 , ..., M : ≤ U i (Γ , ..., Γ M ) ≤ M S ( u , ..., u M ) . The reason this is an equilibrium is the same —no sender’s experiment is pivotal when others arefull revealing the state. In fact whenever there exists a posterior γ (cid:48) that yields the senders utility larger than 0, then all BPsolutions yield utility larger than 0. To see this, note that there exists an experiment that puts supportonly on γ (cid:48) and δ , ..., δ N (such a construction is shown in the proof of Lemma 1). This experiment yieldsall senders utility strictly greater than 0. roof. The first inequality, 0 ≤ U i (Γ , ..., Γ M ), follows from the fact that each sender canalways fully reveal the state and obtain payoff 0.Note that for all strategy profiles (Γ (cid:48) , ..., Γ (cid:48) M ) (with distributions p (cid:48) , ..., p (cid:48) M ), we have: (cid:80) i U i (Γ , ..., Γ M ) = (cid:80) i E p (cid:48) ,...,p (cid:48) M [ u i ( β (Γ , ..., Γ M ))] = E p (cid:48) ,...,p (cid:48) M [ (cid:80) i u i ( β (Γ , ..., Γ M ))] ≤ E p (cid:48) ,...,p (cid:48) M [ M S ( u , ..., u M )] = M S ( u , ..., u M ).Suppose for contradiction that for equilibrium (Γ , ..., Γ M ) and sender i , U i (Γ , ..., Γ M ) >M S ( u , ..., u M ). Then the previous paragraph implies that there exists sender j with U j (Γ , ..., Γ M ) <
0. But then j can profitably deviate to Γ j = Γ F R . Contradiction.Lemma B.2 implies that when
M S ( u , ..., u M ) = 0, all senders get payoff 0 in ev-ery equilibrium. In one special case, when only fully revealing posteriors generate thismaximum surplus, all equilibria must be fully revealing: Corollary B.2.
If for all γ (cid:48) (cid:54)∈ { δ , ..., δ M } (cid:80) i u i ( γ (cid:48) ) < , then the state is fully revealedin every equilibrium.Proof. Note that
M S ( u , ..., u M ) = 0. For any strategy profile (Γ , ..., Γ M ) that is fully re-vealing, U i (Γ , ..., Γ M ) = 0 for all i . For (Γ , ..., Γ M ) that is not fully revealing, (cid:80) i U i (Γ , ..., Γ M ) <
0, as with strictly positive probability the surplus at the posterior is strictly less than 0(when the state is not fully revealed). This implies that for each non-fully revealing(Γ , ..., Γ M ), there is a sender j such that U j (Γ , ..., Γ M ) <
0. Hence, by Lemma B.2, thestate is fully revealed in equilibrium.The logic of Corollary B.2 is similar to that of Proposition 1 of Gentzkow and Ka-menica (2016). If sender surplus is uniquely maximized at fully revealing posteriors, anynon-fully revealing strategy profile leaves at least one sender strictly worse off than fullrevelation, which is always an available strategy.We now turn our attention to what we can say as preferences approach zero-sum.We consider convergence of utilities under the sup norm. A sequence of utility functions { u k } ∞ k =1 converges to a function u , or u k → u , if lim k →∞ sup γ ∈ ∆(Ω) | u k ( γ ) − u ( γ ) | = 0. Fora sequence of profiles of utility functions { ( u k , ..., u kM ) } ∞ k =1 (for notational convenience wewill drop the limits), ( u k , ..., u kM ) → ( u , ..., u M ) if u ki → u i for i = 1 , ..., M .For a sequence of strategies/experiments { Γ k } ∞ k =1 , with each Γ k distributed accordingto pmf p k , we say Γ k → Γ (where Γ has distribution p ), or Γ k converges in distribution to36, if for all γ ∈ ∆(Ω), lim k →∞ p k ( γ ) = p ( γ ). For any strategy profile (Γ , ..., Γ M ), let therandom variable Γ(Γ , ..., Γ M ) denote the receiver’s posterior after observing realizationsof all Γ , ..., Γ M (i.e. the experiment induced by combining all M senders’ experiments).The following result says that as utilities converge to zero-sum sufficiently nonlinearfunctions, the information revealed along any sequence of equilibria, whenever convergent,converges to full revelation. Proposition B.2.
Fix a sequence of games with utilities { ( u k , ..., u kM ) } with ( u k , ..., u kM ) → ( u , ..., u M ) and (cid:80) i u i ( γ ) = 0 for all γ ∈ ∆(Ω) . For each k let (Γ k , ..., Γ kM ) be an equilib-rium of game ( u k , ..., u kM ) . Suppose for every pair of states l, k ∈ Ω there exists an i with u i nonlinear on ∆( { l, k } ) . Then if Γ(Γ k , ..., Γ kM ) → Γ , Γ = Γ
F R . Proposition B.2 is important, as indicates that Theorem 1 does not qualitatively relyon the knife-edge assumption of zero-sum preferences. As utilities get close to zero-sumand sufficiently nonlinear, the information revealed in every equilibrium (if it converges)gets close to full revelation. Before proving the result, we prove a lemma which showsProposition 2 is similarly robust.
Lemma B.3.
Fix a sequence of games with utilities { ( u k , ..., u kM ) } with ( u k , ..., u kM ) → ( u , ..., u M ) and (cid:80) i u i ( γ ) = 0 for all γ ∈ ∆(Ω) . For each k let (Γ k , ..., Γ kM ) be an equilib-rium for ( u k , ..., u kM ) . Suppose for some Ω (cid:48) ⊆ Ω and i , u i is nonlinear on ∆(Ω (cid:48) ) . Then if Γ(Γ k , ..., Γ kM ) → Γ , Γ does not pool Ω (cid:48) .Proof. For strategy profile (Γ (cid:48) , ..., Γ (cid:48) M ) (with distributions p (cid:48) , ..., p (cid:48) M ), let U ki (Γ (cid:48) , ..., Γ (cid:48) M ) = E p (cid:48) ,...,p (cid:48) M [ u ki ( β )] be sender i ’s expected utility when she has preferences u ki .Let the distribution of Γ be p and for any k let the distribution of Γ(Γ k , ..., Γ kM ) be p k .Note that as Γ(Γ k , ..., Γ kM ) has finite support for every k , Γ must also finite support (by thedefinition of convergence in distribution). Define T = supp [Γ] ∪ ( ∪ ∞ k =1 supp [Γ(Γ k , ... Γ kM )]);note by the finiteness of all terms in the union, T is countable.Suppose for contradiction that some u i is nonlinear on Ω (cid:48) and Γ pools Ω (cid:48) . By ClaimA.7, there exists a sender j such who can find experiment Γ j (with distribution p j ) suchthat in the limiting (zero-sum) game ( u , ..., u M ), U j (Γ j , Γ) = c > k , M S ( u k , ..., u kM ) = sup γ (cid:80) i u ki ( γ ) = sup γ (cid:80) i u i ( γ ) + ( u ki ( γ ) − u i ( γ )) ≤ sup γ (cid:80) i u i ( γ ) + | u ki ( γ ) − u i ( γ ) | . As ( u k , ..., u kM ) → ( u , ..., u M ), the second term goes to0 for all i as k → ∞ and hence M S ( u k , ..., u kM ) → M S ( u , ..., u M ) = 0. This implies37here exists K s.t. ∀ k > K , M S ( u k , ..., u kM ) < c/
2. By Lemma B.2, for all k > K , U kj (Γ kj , Γ k − j ) < c/ j playing experiment Γ kj as well as (conditionally independently) playingΓ j , while her opponents’ play Γ k − j . The expected payoff that j gets from this can bewritten U kj (Γ(Γ j , Γ kj ) , Γ k − j ) = U kj (Γ j , Γ(Γ k , ..., Γ kM )), as it does not affect j ’s payoff if sheplays Γ kj or her opponents’ do. As T ⊃ supp [Γ(Γ k , ..., Γ kM )] for each k , we can write U kj (Γ j , Γ(Γ k , ..., Γ kM )) = (cid:80) x ∈ supp [Γ j ] (cid:80) y ∈ supp [ T ] u kj ( β ( x, y )) p k ( y | x ) p j ( x ). Then:lim k →∞ U kj (Γ j , Γ(Γ k , ..., Γ kM )) = (cid:88) x ∈ supp [Γ j ] (cid:88) y ∈ supp [ T ] lim k →∞ u kj ( β ( x, y )) p k ( y | x ) p j ( x )For any y ∈ T , lim k →∞ u kj ( β ( x, y )) = u j ( β ( x, y )) and lim k →∞ p k ( y | x ) = p ( y | x ) (bydefinition of p k ( y | x ) and convergence in distribution). Hence:lim k →∞ U kj (Γ j , Γ(Γ k , ..., Γ kM )) = (cid:88) x ∈ supp [Γ j ] (cid:88) y ∈ supp [ T ] u j ( β ( x, y )) p ( y | x ) p j ( x ) = U j (Γ j , Γ) = c This implies that there exists K (cid:48) such that ∀ k > K (cid:48) , U kj (Γ j , Γ(Γ k , ..., Γ kM )) > c/ K (cid:48)(cid:48) = max { K, K (cid:48) } . For all k > K (cid:48)(cid:48) we have: U kj (Γ kj , Γ k − j ) < c/ U kj (Γ j , Γ(Γ k , ..., Γ kM )) >c/
2. But for all k > K (cid:48)(cid:48) , this contradicts that (Γ k , ..., Γ kM ) is an equilibrium as j has aprofitable deviation of Γ(Γ j , Γ kj ).We now prove Proposition B.2. Proof.
Suppose not. Then Γ(Γ k , ..., Γ kM ) → Γ (cid:54) = Γ F R . Then Γ pools some set of states Ω (cid:48) with | Ω (cid:48) | ≥
2; let l, k ∈ Ω (cid:48) . There is a sender with u i nonlinear on ∆( { l, k } ); hence u i isnonlinear on ∆(Ω (cid:48) ). But then by Lemma B.3 Γ must not pool Ω (cid:48) . Contradiction.Proposition 4 relates to the standard results on the upper hemicontinuity of the setof equilibria (although here we are not concerned with the set of equilibrium actionsthemselves but instead the set of information that could be revealed in equilibrium). Asis also standard, we do not have the corresponding lower hemicontinuity properties. Inparticular, it is possible for there to be non-fully revealing equilibria in the limit, but onlyfully revealing equilibria along the sequence. It is not hard to come up with examples ofthis; we provide a one here. 38 xample 1. Suppose Ω = { , } and there are two senders 1 and 2. Consider the sequenceof utility functions { ( u k , u k ) } with u k ( β ) = − k for all β ∈ [0 , \ { , } , u k (0) = u k (1) = 0,and u ( β ) = 0 for all β ∈ [0 , u as u ( β ) = 0 for all β . Thennote ( u k , u k ) → ( u, u ). Proposition 1 states that in the game ( u, u ), there are non-fullyrevealing equilibria (any strategy profile is an equilibrium). Note for any k , the game( u k , u k ), u k ( β ) + u k ( β ) < β (cid:54)∈ { , } . Hence by Corollary B.2, all equilibria of( u k , u k ) are fully revealing for all k . B.3 Infinite Signal Experiments
So far we have restricted senders to choosing experiments with a finite number of signals,or equivalently interim belief distributions p i ∈ P that have finite support. In this sectionwe demonstrate that our takeaway from the finite signal results —that for typical senderpreferences we have full revelation in every equilibrium —extends when senders can choosefrom a more general set of experiments. Senders now choose any experiments Π : Ω → ∆( S ) (with no restrictions on the signal space). Again, we recast choices of experiments aschoices of interim belief distributions. For technical convenience we restrict our attentionto senders choosing interim belief distributions that can be written as the sum of absolutelycontinuous and discrete distributions. We will call the space of pure strategies, or theset of Bayes-plausible distributions that satisfies this requirement G . Formally G = { g ∈ ∆(∆(Ω)) : E g [Γ i ] = π, g = g c + g d for some abs. cont. and discrete (respectively)measures g c , g d ∈ ∆(∆(Ω)) } . Each g ∈ G is a generalized density. A strategy for sender i is a choice of random variable Γ i with generalized density g i ∈ G . Preferences overstrategy profiles for a sender i are given by U i (Γ , ..., Γ M ) = E g ,...,g M [ u i ( β )]. The gameand equilibrium concept are otherwise identical to the finite signal model (including ournormalization of the u i ’s). Remark:
Note that our equilibrium analysis under both the finite signal restrictionand under the technical restriction above can be seen as equilibrium selections. Any finitesignal equilibrium will also be an equilibrium when senders are allowed to pick strategiesfrom G and any equilibria with strategies selected from G will be equilibria in a gamewhere senders can pick any distribution in ∆(∆(Ω)).We make one additional technical assumption on utility functions: As ∆(Ω) ⊂ R N , by the Lebesgue Decomposition Theorem this only rules senders choosing distribu-tions with singular continuous components. g is the density function g c on intervals where the distribution is absolutely continuous and theprobability mass function g d everywhere else. ssumption B.1. For each l ∈ Ω and i ∈ { , ..., M } , u i is real analytic in some neigh-borhood of δ l . Note that Assumption B.1 only rules out piecewise analytic utility functions for whichsome δ l lies on the boundary between different pieces. For any states l, k let v l,k ∈ R N bethe vector from δ l to δ k . For any sender i let ∇ v l,k u i ( · ) be the directional derivative of u i moving along v l,k . Note for some i ∇ v l,k u i ( · ) may not be well defined at some pointson ∆(Ω); but under Assumption 7.1 for all i and all l it is a well defined continuousfunction in some neighborhood of δ l . With this assumption we can define Condition B.1.Condition B.1 concerns the shape of utility functions on an edge of the simplex and willbe sufficient for a pair of states l and k to not be pooled in every equilibrium. Definition B.1.
For any states l, k and sender i we say that u i satisfies Condition B.1on ∆( { l, k } ) if either ∇ v l,k u i ( δ l ) (cid:54) = u i ( δ k ) − u i ( δ l ), ∇ v l,k u i ( δ k ) (cid:54) = u i ( δ k ) − u i ( δ l ), or both. For example, utility functions that look like Figure 7.2 along edge ∆( { l, k } ) do notsatisfy Condition B.1; nor does a utility function that is linear along that edge. Functionsthat look this those in Figure 1 do satisfy Condition B.1. Figure B.3
Proposition B.3.1.
For any pair of states l and k , if there exists a sender i with u i satisfying Condition B.1 on ∆( { l, k } ) , then l and k are not pooled in every equilibrium. v l,kk = 1, v l,kl = − v l,kn = 0 for all n (cid:54) = l, k . Under the normalization we made, u i ( δ k ) − u i ( δ l ) = 0.
40e prove Proposition B.3.1 after redefining some objects for this setting. For anysender i and strategy profile for all opponents { Γ j } j (cid:54) = i , we define W i ( x ) for x ∈ ∆(Ω)analogously to the finite signal case. W i ( x ) = (cid:90) ∆(Ω) u i ( β ( x, y )) p − i ( y | x ) dy (B.1)For any vector v ∈ R N and y ∈ ∆(Ω), let ∇ v p − i ( y | x ) and ∇ v β k ( x, y ) (for any state k ) be the directional derivates of these two functions with respect to x along vector v .Note the following. ∇ v p − i ( y | x ) = ∇ v N (cid:88) l =1 x l y l p − i ( y ) π l = N (cid:88) l =1 v l y l p − i ( y ) π l (B.2) ∇ v β k ( x, y ) = ∇ v x k y k π k (cid:80) Nl =1 x l y l π l = − N (cid:88) l =1 v l y l π l x k y k π k ( (cid:80) Nn =1 x n y n π n ) + v k y k π k (cid:80) Nl =1 x l y l /π l (B.3)For any states l, k ∈ { , ..., N } let ∆ int ( { l, k } ) = ∆( { l, k } ) \ { δ l , δ k } be the nondegen-erate beliefs in ∆( { l, k } ). Let ∆ ( { l, k } ) = { β ∈ ∆(Ω) : β l , β k = 0 } and ∆ ( { l, k } ) = { β ∈ ∆(Ω) : β l + β k < β l and/or β k (cid:54) = 0 } . Note { δ l , δ k }∪ ∆ int ( { l, k } ) ∪ ∆ ( { l, k } ) ∪ ∆ ( { l, k } ) =∆(Ω) and all four sets are disjoint.We now prove Proposition B.3.1 Proof.
First note Lemma 1 still holds in this context by an identical proof. In any equi-librium (Γ , ..., Γ M ) all senders i have: U i (Γ , ..., Γ M ) = 0 and W i ( x ) ≤ x ∈ ∆(Ω).WLOG let l = 1, k = 2. For notational convenience let v = v , . Suppose u i satisfies Condition B.1 on ∆( { , } ) for some i . WLOG we consider the case ∇ v u i ( δ ) (cid:54) = u i ( δ ) − u i ( δ ) = 0. Then by zero-sumness (derivatives must also be zero-sum wherethey exist for all senders) there exists a sender j with ∇ v u j ( δ ) = c <
0. We will provethat Γ − j must not pool { l, k } ; this clearly implies the Proposition B.3.1 as by Claim A.3,(Γ , ..., Γ M ) also will not pool { l, k } .Suppose for contradiction there is an equilibrium (Γ , ..., Γ M ) such that Γ − j pools { l, k } . For x ∈ ∆(Ω), by the product rule and the partition of ∆(Ω) into { δ , δ } , ∆ int ( { , } ),∆ ( { , } ) , ∆ ( { , } ): 41 v W j ( x ) = ∇ v ( u j ( β ( x, δ )) p − j ( δ | x ) + u j ( β ( x, δ )) ∇ v p − j ( δ | x )+ ∇ v ( u j ( β ( x, δ )) p − j ( δ | x ) + u j ( β ( x, δ )) ∇ v p − j ( δ | x )+ (cid:90) ∆ int ( { , } ) ∇ v u j ( β ( x, y )) p − j ( y | x ) dy + (cid:90) ∆ int ( { , } ) u j ( β ( x, y )) ∇ v p − j ( y | x ) dy + (cid:90) ∆ ( { , } ) ∇ v u j ( β ( x, y )) p − j ( y | x ) dy + (cid:90) ∆ ( { , } ) u j ( β ( x, y )) ∇ v p − j ( y | x ) dy + (cid:90) ∆ ( { , } ) ∇ v u j ( β ( x, y )) p − j ( y | x ) dy + (cid:90) ∆ ( { , } ) u j ( β ( x, y )) ∇ v p − j ( y | x ) dy though this may not be well defined for some x . Note: ∇ v ( u j ( β ( x, y )) p − j ( δ | x )) = N (cid:88) k =1 ∂u j ( β ( x, y )) ∂β k ∇ v β k ( x, y ) p − j ( y | x )(again this may not be well defined for some x ). Now consider x ∈ ∆( { , } ). Forsuch x , with some algebra: ∇ v β k ( x, y ) p − j ( y | x ) = ( y /π − y /π ) x k y k π k x y π + x y π + v k y k π k (B.4)One can check that for y ∈ ∆ ( { , } ) and for y ∈ { δ , δ } this expression is 0 forall k = 1 , ..., N . This tells us that the 1st, 3rd, and 7th terms of ∇ v W j ( x ) are 0 for x ∈ ∆( { , } ). Note that for y ∈ ∆( { , } ) , ∇ v p − i ( y | x ) = 0, and so the 8th term is also0. We consider the limit of ∇ v W j ( x ) for x ∈ ∆( { , } ) as x → δ . We show this limitexists and is negative.lim x → δ ∇ v W j ( x ) = lim x → δ u j ( β ( x, δ )) ∇ v p − j ( δ | x ) + lim x → δ u j ( β ( x, δ )) ∇ v p − j ( δ | x )+ (cid:90) ∆ int ( { , } ) N (cid:88) k =1 lim x → δ ∂u j ( β ( x, y )) ∂β k ∇ v β k ( x, y ) p − j ( y | x ) dy + (cid:90) ∆ int ( { , } ) lim x → δ u j ( β ( x, y )) ∇ v p − j ( y | x ) dy + (cid:90) ∆ ( { , } ) N (cid:88) k =1 lim x → δ ∂u j ( β ( x, y )) ∂β k ∇ v β k ( x, y ) p − j ( y | x ) dy + (cid:90) ∆ ( { , } ) lim x → δ u j ( β ( x, y )) ∇ v p − j ( y | x ) dy
42e evaluate this term by term. The first term is 0 as ∇ v p − j ( δ | x ) is finite anddoes not depend on x and lim x → δ β ( x, δ ) = δ which implies (by continuity of u j in aneighborhood of δ ) lim x → δ u j ( β ( x, δ )) = 0. The second term is also 0 as ∇ v p − j ( δ | x )is finite and does not depend on x and lim x → δ β ( x, δ ) = δ (by L’Hopital’s rule) whichimplies (by continuity of u j in a neighborhood of δ ) lim x → δ u j ( β ( x, δ )) = 0.The fourth term is also 0 as for all y ∈ ∆ int ( { , } ), lim x → δ β ( x, y ) = δ and u j ( δ ) =0 while ∇ v p − j ( y | x ) is finite and does not depend on x . The sixth term is also 0 for thefollowing reason. For y ∈ ∆ ( { , } ) with y >
0, lim x → δ β ( x, y ) = δ ; as u j ( δ ) = 0, theterms inside the integral are 0 when y >
0. For y ∈ ∆ ( { , } ) with y = 0, we musthave y >
0; then by L’Hopital’s rule lim x → δ β ( x, y ) = δ and u j ( δ ) = 0 meaning termsinside the interal are 0 when y = 0.Using (B.4) note that ∇ v β k ( x, y ) p − j ( y | x ) = 0 for all k (cid:54) = 1 , k x k = v k = 0. For y ∈ ∆( { , } ) int we have: lim x → δ ∇ v β ( x, y ) p − j ( y | x ) = − y π p − j ( y ) andlim x → δ ∇ v β ( x, y ) p − j ( y | x ) = y π p − j ( y ). The same holds for y ∈ ∆ ( { , } ) with y > y ∈ ∆ ( { , } ) with y = 0, we have: lim x → δ ∇ v β ( x, y ) p − j ( y | x ) = lim x → δ ∇ v β ( x, y ) p − j ( y | x ) =0 (applying L’Hopital’s rule).Putting this together:lim x → δ ∇ v W j ( x ) = (cid:90) ∆ int ( { , } ) ∪{ y ∈ ∆ ( { , } ): y > } [ ∂u j ( δ ) ∂β − ∂u j ( δ ) ∂β ] y π p − j ( y ) dy = ∇ v u j ( δ ) (cid:90) ∆ int ( { , } ) ∪{ y ∈ ∆ ( { , } ): y > } y π p − j ( y ) dy = c (cid:90) ∆ int ( { , } ) ∪{ y ∈ ∆ ( { , } ): y > } y π p − j ( y ) dy (B.5)As Γ − j pools Ω (cid:48) , there exists y ∈ supp [Γ] for which y , y >
0. Such a y must fallinside the set (∆ int ( { , } ) ∪ { y ∈ ∆ ( { , } ) : y > } ) (any point in the completementof this set assigns probability 0 to at least one of states 1 , y ∈ (∆ int ( { , } ) ∪ { y ∈ ∆ ( { , } ) : y > } ) for which p j ( y ) > y >
0. As y (cid:48) ≥ y (cid:48) ∈ ∆(Ω), the integral on the righthand side of equation B.5 is strictly positive.As c <
0, we have lim x → δ ∇ v W j ( x ) < W j ( δ ) = 0, this implies that for some x ∗ ∈ ∆( { , } ) close enough to δ , wemust have W j ( x ∗ ) >
0, contradicting Lemma 1.Condition B.1 holding on each edge for some sender is a sufficient condition for full43evelation in any equilibrium.
Theorem B.3.1.
If for every pair of states l and k there exists a sender i such that u i satisfies Condition B.1 on ∆( { l, k } ) then the state is fully revealed in every equilibriumin which senders choose experiments from G .Proof. If for every pair l, k there is a sender with u i satisfying Condition B.1 on ∆( { l, k } ),then by Proposition B.3.1, no pair of states is pooled in any equilibrium. Hence w.p. 1the posterior assigns positive probability to only 1 state and hence must fully reveal thatstate.Note that a function not satisfying Condition B.1 along a given edge is knife-edge—it requires a particular directional derivative to take a certain value at two points. Ifno sender has a utility function satisfying Condition B.1 along an edge, this is even moreparticular. Hence we ‘typically’ expect Condition B.1 to be satified on each edge forsome u i and so Theorem B.3.1 says we should typically expect fully revelation in everyequilibrium. B.4 Privately Informed Senders
Consider our baseline model with one modification: each sender receives a private signalbefore the game. For simplicity, senders’ private signals are realizations of finite signalexperiments that are conditionally (on ω ) independent across senders. We think of theexperiments in terms of the beliefs they induce. Formally each sender i draws a privatebelief b i ∈ ∆(Ω) with b i ∼ B i ∈ P , | supp [ b i ] | < ∞ , and E [ b i ] = π (Bayes-plausibility).The distributions { B i } Ni =1 are conditionally independent. We make one more assumption:that for each Ω (cid:48) (cid:40) Ω, supp [ b i ] ∩ ∆(Ω (cid:48) ) = ∅ for all i ; no sender’s private information rulesout any states. A pure strategy for sender i is a mapping from private beliefs (or types) to choicesof finite signal experiments: σ i : supp [ b i ] → P . As before, we use Γ i to denote the interimbelief produced by sender i ’s experiment. i chooses the distribution of Γ i , p i ∈ P afterobserving her own type. Importantly, we define Γ i to be the interim belief the receiverholds after viewing realization of i ’s experiment but without updating her belief on Ω fromobserving the choice of p i (we formalize this updating in the next paragraph). Hence Γ i is These assumptions are not necessary. Having already made the assumption of finite signals, this assumption is equivalent saying signalsare bounded. i ’s experiment ignoring information from signalling. A purestrategy profile is a vector ( σ , ..., σ M ).The receiver’s posterior belief β is a function of the signal realizations she observesas well as the experiment choices she observes. The receiver will form beliefs about each { b i } i =1 ,...,M independently via a belief function µ i : ∆(∆(Ω)) → ∆( supp [ b i ]) ( i = 1 , ..., M )which maps choices of experiment to a belief on the sender’s type. For an experimentchoice of p i by sender i , let µ i ( p i )[ b ] denote the probability the receiver assigns to b i = b under belief function µ i . For i = 1 , ..., M let τ i ( µ i , p i ) ∈ ∆(Ω) be the receiver’s belief on ω given belief function µ i after observing experiment choice p i from sender i but not itssignal realization or any other senders’ experiment choices are realizations. Then for eachstate l ∈ Ω: τ il ( µ i , p i ) = (cid:88) b ∈ supp [ b i ] P r ( ω = l | b i = b ) P r ( b i = b | p i ) = (cid:88) b ∈ supp [ b i ] b l µ i ( p i )[ b ] (B.6)For any sender i let α i (Γ i , p i , µ i ) ∈ ∆(Ω) be the random variable representing thereceiver’s interim belief given µ i after observing just the choice p i and the realizationof Γ i . α i hence captures the receiver’s belief after taking into account all information—signalling and otherwise —from sender i . For any l ∈ Ω: α il (Γ i , p i , µ i ) = τ il ( µ i , p i )Γ i,l /π l (cid:80) Nk =1 τ ik ( µ i , p i )Γ i,k /π k (B.7)For fixed { µ i } i =1 ,...,M , after observing { p i } and realizations { Γ i } , the receiver updatesby Bayes rule to a posterior belief for each l ∈ Ω: β l ( { Γ i } , { p i } , { µ i } ) = Π Mi =1 α il (Γ i , p i , µ i ) /π M − l (cid:80) Nk =1 Π Ms =1 α ik (Γ i , p i , µ i ) /π M − k (B.8)A PBE (in pure strategies) is a strategy profile ( σ , ..., σ M ) and a set of belief functions( µ , ..., µ M ) satisfying two conditions. First, no sender i can strictly gain from deviatingfrom σ i ( b i ) for any b i ∈ supp [ b i ]: While other senders’ experiment choices and the realizations of (Γ , ..., Γ M ) will affect the receiver’sbelief about each b i , as players’ types and experiment realizations are conditionally independent theywill only affect the receiver’s beliefs through learning about ω . This updating will hence not affect thereceiver’s belief on ω , which is all that players care about. The functions { µ i } are what is important forevaluating senders’ payoffs. i ∈ { , ..., M } , b i ∈ supp [ b i ] : E { B j } j (cid:54) = i [ E σ i ( b i ) , { p j } j (cid:54) = i [ u i ( β )] | b i ] ≥ E { B j } j (cid:54) = i [ E p (cid:48) i , { p j } j (cid:54) = i [ u i ( β )] | b i ] for all p (cid:48) i ∈ P (B.9)where the receiver’s posterior β is formed using (B.8). It is important to note thatsender i may not know { p j } j (cid:54) = i but forms beliefs about these given her own private infor-mation to evaluate expected utility.Second, beliefs must follow Bayes rule on path: ∀ i ∈ { , ..., M } and p ∈ P s.t. ∃ b i ∈ supp [ b i ] with σ i ( b i ) = p : ∀ b ∈ supp [ b i ], µ i ( p )[ b ] = σ ( b )= p B i ( b ) (cid:80) b (cid:48) ∈ supp [ b i ] σ ( b (cid:48) )= p B i ( b (cid:48) ) (B.10)The result below is Lemma 1 adapted to this setting with private information. Lemma B.4.
Take any equilibrium ( σ , ..., σ M ) , ( µ , ..., µ M ) . For any sender i and b ∈ supp [ b i ] , conditional on b i = b sender i gets expected utility .Proof. Fix any equilibrium. No sender i can get expected utility strictly less than 0conditional on any b i = b ∈ supp [ b i ], as playing the fully revealing experiment guaranteesexpected utility 0. This means each sender i ’s expected utility unconditional on type, (cid:88) b (cid:48) ∈ supp [ b i ] E { B j } j (cid:54) = i [ E σ i ( b (cid:48) ) , { p j } j (cid:54) = i [ u i ( β )]] B i ( b (cid:48) ) (B.11)is weakly positive. As the game is zero-sum, the sum of all senders’ unconditionalexpected utilities must be 0; as each of these payoffs is weakly positive, it must be EquationB.11 is equal to 0 for each i . But then as each term in the summation of Equation B.11is weakly positive, sender i ’s expected utility conditional on b i = b cannot be strictlypositive, and hence it must be 0.We redefine state pooling in the game with private information as follows. An equi-librium ( σ , ..., σ M ), ( µ , ..., µ M ) does not pool a set of states Ω (cid:48) ⊆ Ω if
P r ( β l > ∀ l ∈ Ω (cid:48) ) = 0. Otherwise, the equilibrium pools Ω (cid:48) . The following Lemma is useful for the mainresults; it says an equilibrium pools Ω (cid:48) if and only if α i does for every sender i .46 emma B.5. An equilibrium ( σ , ..., σ M ) , ( µ , ..., µ M ) pools Ω (cid:48) ⊆ Ω if and only if for all i P r ( α il (Γ i , p i , µ i ) > ∀ l ∈ Ω (cid:48) ) > .Proof. ‘If’ direction. If for all i P r ( α il (Γ i , p i , µ i ) > ∀ l ∈ Ω (cid:48) ), then as α i is conditionally(on state) across senders (as b i and Γ i are), P r ( α il (Γ i , p i , µ i ) > ∀ l ∈ Ω (cid:48) ∀ i = 1 , ..., M ) >
0. By equation B.8,
P r ( β l > ∀ l ∈ Ω (cid:48) ) > i , P r ( α il (Γ i , p i , µ i ) > ∀ l ∈ Ω (cid:48) ) = 0, then byequation B.8, β l = 0 for some l ∈ Ω (cid:48) w.p. 1.We now provide a sufficient condition for a pair of states to be not pooled in everyequilibrium. The sufficient condition is the same as that in Section B.3 and will lead tothe same sufficient condition for full revelation in all equilibria. We note as before thatthis condition is satisfied for all but a knife-edge case of sender preferences. As in SectionB.3 we make the mild technical assumption that all sender utilities are real analytic insome neighborhood of δ l for all states l (Assumption B.1). Lemma B.6.
Suppose Assumption B.1 holds. Consider any pair of states l and k . { l, k } is not pooled in every equilibrium if for some i u i satisfies Condition B.1 on ∆( { l, k } ) .Proof. WLOG let l = 1, k = 2. For notational convenience let v = v , . Suppose u i satisfies Condition B.1 on ∆( { , } ) for some i . WLOG we consider the case ∇ v u i ( δ ) (cid:54) = u i ( δ ) − u i ( δ ) = 0. Then by zero-sumness (derivatives must also be zero-sum where theyexist for all senders) there exists a sender j with ∇ v u j ( δ ) = c <
0. This implies thereexists r ∈ (0 ,
1) such that for all γ ∈ ∆( { , } ) with γ > r , u j ( γ ) >
0. In other words, j has a region of advantage along ∆( { , } ) close to δ .Suppose for contradiction there is an equilibrium ( σ , ..., σ M ), ( µ , ..., µ M ) that pools { l, k } . For each j (cid:48) (cid:54) = j , let Λ j (cid:48) = α j (cid:48) (Γ j (cid:48) , p j (cid:48) , µ j (cid:48) ). By Lemma B.5, for all j (cid:48) (cid:54) = j , Λ j (cid:48) pools { , } . Note Λ j (cid:48) is also a random variable (where randomness is over p j (cid:48) and therealization of Γ j (cid:48) ) with finite support (due to finite support of b j (cid:48) and Γ j (cid:48) ). If all senders j (cid:48) (cid:54) = j follow the equilibrium play, Λ j (cid:48) is also Bayes-plausible (with mean π ); this isbecause Γ j (cid:48) Bayes-plausible, any learning the receiver does about b j (cid:48) must follow Bayesrule on path, and the distribution of b j (cid:48) has mean π . Let Λ − j denote the interim beliefinduced by viewing realizations of all { Λ j (cid:48) } j (cid:48) (cid:54) = j ; note this experiment also pools { , } ;this is also a finite signal Bayes-plausible experiment.We will now find a profitable deviation for sender j . This deviation will take the formof an experiment p (cid:48) j , which we will construct, that generates strictly positive expected47tility no matter what j ’s type is.We can rewrite β l for l ∈ Ω (from equation B.8) conditional on this deviation p (cid:48) j as: β l (Λ − j , α j (Γ j , p (cid:48) j , µ j )) = Λ − j,l α jl (Γ j , p (cid:48) j , µ j ) /π l (cid:80) Nk =1 Λ − j,k α jk (Γ j , p (cid:48) j , µ j ) /π k (B.12)It is also useful to write down the probability distribution of Λ − j conditional on α j (Γ j , p (cid:48) j , µ j ): P r (Λ − j = y | α j (Γ j , p (cid:48) j , µ j )) = N (cid:88) l =1 α jl (Γ j , p (cid:48) j , µ j ) y l π l (B.13)Let Y = { y ∈ supp [Λ − j ] : y , y > } ; note this set is nonempty as Λ − j pools { , } .Let Y = { y ∈ supp [Λ − j ] : y , y = 0 } , Y = { y ∈ supp [Λ − j ] : y > , y = 0 } , and Y = { y ∈ supp [Λ − j ] : y = 0 , y > } . These sets partition the support of Λ − j .Consider j generating interim belief Γ j = x ∈ ∆( { , } ) \ { δ , δ } . As in Proposition2’s proof, we will try and find such an x conditional on which j gets a strictly positiveexpected payoff. We will then construct p (cid:48) j which assigns positive probability to x . Notethat P r ( α j (Γ j , p (cid:48) j , µ j ) ∈ ∆( { l, k } ) \ { δ , δ }| Γ j = x ) = 1. This can be seen from thedefinition of α j (equation B.7) and is a consequence of every private belief b ∈ supp [ b j ]having b n > n ∈ Ω which implies τ jl ( µ j , p (cid:48) j ) > l ∈ Ω. This impliesthat
P r (Λ − j ∈ Y | Γ j = x ∈ ∆( { , } ) \ { δ , δ } ) = 0 (equation B.13). Also note thatconditional on Γ j = x ∈ ∆( { , } ) \ { δ , δ } , β = δ when Λ − j ∈ Y and β = δ whenΛ − j ∈ Y (see equation B.12); these posteriors both yield utility 0.Finally, note that conditional Γ j = x ∈ ∆( { , } ) \ { δ , δ } , we have P r (Λ − j ∈ Y ) > α j ∈ ∆( { l, k } ) \ { δ , δ } ). Note that for each y ∈ Y and α j (Γ j , p (cid:48) j , µ j ) ∈ ∆( { , } ) \ { δ , δ } , β k ( y, α j (Γ j , p (cid:48) j , µ j )) is continuous in α j (Γ j , p (cid:48) j , µ j ), β k ( y, α j (Γ j , p (cid:48) j , µ j )) → α j (Γ j , p (cid:48) j , µ j ) →
1, and β k ( y, α j (Γ j , p (cid:48) j , µ j )) → α j (Γ j , p (cid:48) j , µ j ) →
0. As Y is finite, the function min y ∈ Y β k ( y, α j (Γ j , p (cid:48) j , µ j )) is also continuous in its secondargument for α j (Γ j , p (cid:48) j , µ j ) ∈ ∆( { , } ) \ { δ , δ } and goes to 0 or 1 as α j (Γ j , p (cid:48) j , µ j ) goes to0 or 1 respectively. By the intermediate value theorem, there exists a α r ∈ (0 ,
1) such thatmin y ∈ Y β k ( y, α r ) = r . When 1 > α j (Γ j , p (cid:48) j , µ j ) > α r , we have β k ( y, α j (Γ j , p (cid:48) j , µ j )) ∈ ( r, y ∈ Y .Note that for Γ j = x ∈ ∆( { , } ) \{ δ , δ } , we have α j ( x, p (cid:48) j , µ j ) = τ j ( µ j ,p (cid:48) j ) x /π τ j ( µ j ,p (cid:48) j ) x /π + τ j ( µ j ,p (cid:48) j ) x /π .48e can rewrite this as: α j ( x, p (cid:48) j , µ j ) = τj µj,p (cid:48) j ) τj µj,p (cid:48) j ) x /π x /π + τj µj,p (cid:48) j ) τj µj,p (cid:48) j ) x /π . Note that as private beliefs havefinite support and cannot rule out any state, for all beliefs the receiver may hold about j ’stype, µ ∈ ∆( supp [ b j ]), the corresponding belief τ this induces on Ω ( τ l = (cid:80) b ∈ supp [ b j ] ) b l µ [ b ])must have τ τ ≥ d > d ∈ (0 , α j ( x, p (cid:48) j , µ j ) = τ j ( µ j ,p (cid:48) j ) τ j ( µ j ,p (cid:48) j ) x /π x /π + τ j ( µ j ,p (cid:48) j ) τ j ( µ j ,p (cid:48) j ) x /π ≥ dx /π x /π + dx /π (B.14) α j ( x, p (cid:48) j , µ j ) is continuous in x on ∆( { , } ) \ { δ , δ } and will also fall in ∆( { , } ) \{ δ , δ } . By equation B.14, as x → α j ( x, p (cid:48) j , µ j ) → regardless of what beliefs thereceiver holds. Hence there exist x ∗ ∈ ∆( { , } ) \{ δ , δ } such that 1 > α j ( x ∗ , p (cid:48) j , µ j ) > α r .Hence we have β k ( y, α j ( x ∗ , p (cid:48) j , µ j )) ∈ ( r,
1) for all y ∈ Y . Conditional on x ∗ , j getsexpected utility: (cid:88) y ∈ Y u j ( δ ) (cid:124) (cid:123)(cid:122) (cid:125) =0 P r (Λ − j = y | α j ( x ∗ , p (cid:48) j , µ j )) + (cid:88) y ∈ Y u j ( δ ) (cid:124) (cid:123)(cid:122) (cid:125) =0 P r (Λ − j = y | α j ( x ∗ , p (cid:48) j , µ j ))+ (cid:88) y ∈ Y u j ( β ( y, α j ( x ∗ , p (cid:48) j , µ j ))) (cid:124) (cid:123)(cid:122) (cid:125) > P r (Λ − j = y | α j ( x ∗ , p (cid:48) j , µ j )) (cid:124) (cid:123)(cid:122) (cid:125) > > j can construct a stratgy p (cid:48) j which assigns positive probability only to x ∗ and { δ , ..., δ N } . p (cid:48) j yields j strictly positiveexpected utility conditional on x ∗ being realized and 0 utility otherwise. Hence LemmaB.4 is violated. Contradiction. Hence no equilibrium can pool { , } .Lemma B.6 implies that Condition B.1 being satisfied by some u i on each edge ofthe simplex is sufficient for full revelation in all equilibria. It is worth noting again thatis sufficient condition is satisfied for all but a knife-edge case of sender preferences. Proposition B.3.
Suppose Assumption B.1 holds. The state is fully revealed in everyequilibrium if for all pairs of states l and k , there is some u i that satisfies Condition B.1on ∆( { l, k } ) . roof. Argument is identical to Theorem B.3.1’s proof.
B.5 Sequential Moving Senders
Consider a sequential version of our baseline model. Senders 1 , ..., M move in order,observing all previous experiment choices (but not realizations); we are interested in purestrategy subgame perfect Nash Equilibria (henceworth just SPNE) of this game. Notethat for a simultaneous game, there are multiple corresponding sequential games, one foreach ordering of senders.A few facts easily carry over from the simultaneous case. First, all senders must getutility 0 in equilibrium (as the game is zero-sum and anyone can fully reveal the state).Second, full revelation can be supported as an SPNE outcome in a game with sendersmoving in any order. We can construct such an equilibrium with all senders playing Γ
F R on path and playing any sequentially rational strategies off path. No sender after the firsthas an incentive to deviate if those upstream from them have not (as the receiver willthe learn the state from upstream senders). The first sender cannot strictly gain fromdeviating, as a strict gain would imply a strict loss for a downstream sender; sequentialrationality rules this out as the downstream sender can fully reveal the state to avoid aloss. As for the simultaneous game, the interesting question is when all equilibria (hereSPNE) are fully revealing. The following results and discussion clarify the relationshipbetween the our simultaneous model and a sequential version.
Proposition B.5.1.
If for u , ..., u M there is full revelation in every SPNE of the sequen-tial game with the senders moving in some order, then there is full revelation in everyequilibrium of the simultaneous game.Proof. We prove the following statement, from which the result follows: if there exists anon-fully revealing equilibrium in the simultaneous game, then, for any order of senders,there exists a non-fully revealing SPNE in the sequential game.Choose any ordering of senders 1 , ..., M . Consider any non-fully revealing equilibriumof the simultaneous game, (Γ , ..., Γ M ) and let Γ be the experiment induced by observingthe realizations of Γ , ..., Γ M . In the sequential game, consider the following strategyprofile: (1) sender 1 plays Γ. (2) each sender i = 2 , ..., M plays Γ U if all previous sendershaven’t deviated and play some seqentially rational strategies otherwise. We will showthis is an SPNE. By Lemma 1 all senders get utility 0 from following perscribed play as Γis the information revealed in an equilibrium of the simultaneous game and no additional50nformation is revealed. First note sender 1 has no strict incentive to deviate as anyprofitable deviation would give some downstream sender strictly negative utility. This isnot possible along any path of play in an SPNE as this downstream sender can alwaysfully reveal the state.If sender 1 plays Γ, senders 2 , ..., M have no incentive to deviate for the followingreason. Consider any deviation Γ (cid:48) j (cid:54) = Γ U for sender j , 2 ≤ j ≤ M . This deviation leads to apath of play producing information from Γ as well as additional conditionally independentexperiments. Suppose, for contradiction, this deviation yields j strictly positive expectedutility. But then j has a profitable deviation from the simulatenous game equilibrium(Γ , ..., Γ M ); if j unilaterally plays these additional conditionally independent experimentsin addition to Γ j , she gets strictly positive utility.Proposition B.5.1 implies that the set of (zero-sum) utility functions under whichthere is full revelation in all equilibria in the simultaneous game contains the set underwhich for some order of senders there is full revelation in all SPNE of the sequential game.The converse is not true. It is possible for there to be full revelation in every equilibriumof the simultaneous game but, for every order of senders, non-fully revealing SPNE insequential game. The following example demonstrates this. Example 2.
There are two senders 1 , , ,
3. Assume that π < π < π (this is not necessary, but eases exposition; the assumption rules outany two states having equal prior probabilities but is otherwise without loss). Suppose u ((1 / , / , u ((1 / , , / u ((1 / , / , u ((1 / , , / − u ((0 , / , / u ((0 , / , / −
1. At all other γ ∈ ∆(Ω), u ( γ ) = u ( γ ) = 0. Sender 1 has an advantage at single points along edges ∆( { , } ) and ∆( { , } )and sender 2 has a single advantage on edge ∆( { , } ); on the rest of the simplex, neithersender has an advantage. Figure B.4 summarizes this. By Theorem 1, the state is fullyrevealed in all equilibria of the simultaneous game. However we will show that regardlessof the order senders move in, there is always a non-fully revealing SPNE.51 igure B.4 First suppose sender 1 plays first, then sender 2. Consider the sender 1 playing Γ s.t. P r (Γ = δ ) = π and P r (Γ = ( π π + π , π π + π , − π (this distribution satisfiesBayes-plausibility). Suppose sender 2 plays Γ = Γ U whenever sender 1 plays this andfollowing a deviation plays any sequentially rational Γ . Note the posterior when followingperscribed play will either be δ or ( π π + π , π π + π , / , / , ω = 3 the receiver learns the stateand conditional on ω ∈ { , } the posterior will lie on ∆( { , } ), on which sender 2 hasno points of advantage, w.p. 1 (by Claim A.2). This is a non-fully revealing equilibrium.If sender 2 plays first, we can construct an analogous non-fully revealing equilibriumwith Γ s.t. P r (Γ = δ ) = π and P r (Γ = (0 , π π + π , π π + π )) = 1 − π . Sender 1 playsΓ = Γ U on path and any sequentially rational Γ otherwise.In the simultaneous version of Example 2, the state is fully revealed in every equilib-rium for the following reasons. States 2 and 3 must not be pooled in equilibrium becauseif not sender 2, who has an advantage along ∆( { , } ), could find an experiment yielding astrictly positive payoff (violating Lemma 1). More precisely, Γ cannot pool states { , } ,because if not sender 2 can gain a strictly positive payoff. Similarly, Γ cannot pool { , } or { , } or sender 1 take advantage. Each sender is ‘responsible’ for not pooling somestates in equilibrium because their opponent has an advantage.More generally, when the state must be fully revealed in every equilibrium of asimultaneous game, for every subset of states Ω (cid:48) there is some sender j who could takeadvantage of Γ − j pooling Ω (cid:48) . This is shown by Claim A.7 (in the proof of Prosition 2);sender j must have an advantage somewhere on ∆(Ω). When sender j moves last in thesequential game, then by the same argument, all upstream senders must (collectively)not pool Ω (cid:48) in any SPNE. For example, when sender 2 moved second in the example,sender 1 did not pool { , } . However, as the last moving sender may only be able to takeadvantage of some subsets of states being pooled, we need not get full revelation. If thereis a sender j who can take advantage of Ω (cid:48) being pooled for any Ω (cid:48) ∈ Ω s.t. | Ω (cid:48) | > Proposition B.5.2.
If for utility functions u , ..., u M a set of states Ω (cid:48) ⊆ Ω is not pooledin every equilibrium of the simultaneous game, then, for some ordering of senders, Ω (cid:48) isnot pooled in every SPNE of the sequential game.Proof. Suppose Ω (cid:48) ⊆ Ω (cid:48) is not pooled in every equilibrium of the simultaneous game.Then by Proposition 2, u i ( γ ) > i and some γ ∈ ∆(Ω (cid:48) ). By Claim A.7,there exists a sender j such that for every Γ − j that pools Ω (cid:48) , there exists a Γ j such that U j (Γ j , Γ − j ) >
0. Consider any ordering of senders with j moving last. Then as all sendersmust get utility 0, in any SPNE senders upstream from j (collectively) do not pool Ω (cid:48) (orelse j ’s best response yields strictly positive utility). Hence, by Lemma A.3, all SPNE donot pool Ω (cid:48)(cid:48)