Persuasion when Decision Maker’s Beliefs are

Unknown. Does Speaking Sequentially Help? ∗

Sergii Golovko†

July, 2013

Abstract

I study a model in which perfectly informed, but biased experts consult a

decision maker (DM) whose prior beliefs about the true state of the nature

are unknown. Experts bear a cost of lying and the DM faces a cost of

reading (understanding) experts’ reports. By comparing simultaneous and

sequential communication protocols, I show that the DM benefits in terms

of ex-ante welfare, when experts provide their reports sequentially and the

costs of lying/reading are small enough. The results of this paper also

shed a light on such social phenomenon as partisanship and polarizations

in opinions. A democratic voter is more likely to read a newspaper that is

financed by democrats (partisanship), and after reading this newspaper she

starts to support democrats even more (polarization).

∗I am mostly indebted to Vijay Krishna for constant encouragement and valuable advices.My special gratitude goes to Kalyan Chaterjee and Yu Awaya who invested a lot of their timediscussing this paper on its earlier stages, as well as to Edward Green who bring to my attentionthe conceptual mistake in the previous version of this paper. I also pleased to acknowledge themany useful comments and suggestions of Adam Slawski, Bruno Salcedo, Bruno Sultanum Tex-eira, Neil Wallace, Vaidyanathan Venkateswaran and Nail Kashaev, as well as all the participantsof Edward Green’s and Neil Wallace’s reading groups hold at Penn State.†Department of Economics, The Pennsylvania State University, University Park, PA. 16802,

USA

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1 Introduction

Suppose that during a financial crisis a manager of an international firm has to

decide whether to fire 10, 000 workers. The manager who is aware of the crisis

has some prior beliefs about the right way of how to proceed in this situation.

Fortunately, the manager can ask for advice of the two experts who exactly know

what should be done. Unfortunately for the manager, the experts are biased. One

expert wants to persuade the manager to fire all the workers while the other one

wants the workers to remain employed. What prevents experts from lying and

arguing in favor of their preferable option is the cost of fabricating information

they bear.

With this story in hand, the paper investigates what the optimal communica-

tion protocol between the manager and the experts induces manager to make right

decision. I assume that the communication protocol is constructed by an outsider

who is interested in maximizing the welfare of the manager (which is in my setup

equivalent to the profit of the firm); and I am limiting my attention to two types

of communication protocols: sequential and simultaneous. Reader may wonder:

why the communication protocol could matter in this situation? Suppose the de-

cision maker consults experts sequentially. Then after reading the first report the

decision maker reveals some information about her prior beliefs to the experts.

This information could be used against the decision maker by the second expert

when he is persuading her. However, the latest effect is not present if the decision

maker consults experts simultaneously. Surprisingly, the paper shows that under

some conditions on the costs of reading and lying the decision maker could benefit

in terms of ex-ante welfare, when experts provide their advices sequentially.

Real life situations in which a group of informed people (experts) try to per-

suade a group of uninformed people (decision makers) to change their actions are

not, by far, limited to managerial relationships: political candidates persuade

potential voters during an election campaign, sellers spend millions of dollars

while advertising their products to potential buyers. Additionally, the paper also

sheds light on such social phenomena as partisanship and polarization (in beliefs).

Shortly, the results of the paper suggest the following story: person who supports

democrats is more likely to read newspaper that is financed by democrats (parti-

sanship), and after reading this newspaper she starts to support democrats even

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more strongly (polarization).

Related Literature. The work most closely related to mine is Chen (2007)

and Dziuda (2011). Similarly to this paper, Chen (2007) studies communication

process between biased expert(s) and the decision maker who faces a binary choice

problem. However, her paper different from mine in the way communication is

modeled. Thus, she assumes that information is verifiable and the only way for

experts to manipulate the decision maker’s decision is to hide information. The

main finding of her paper is that in some cases it could be beneficial to consult

biased experts than to consult less biased ones. In a similar framework, but with

multiple pieces of information available to expert, Dziuda (2011) shows that expert

could reveal his unfavorable information to the decision maker.

Also closely related are papers by Kamenica and Gentzkow (2009) and Gentzkow

and Kamenica (2011). Similarly to my paper, these papers study communication

between the decision maker and expert(s) who is(are) interested in persuading the

decision maker. However, these papers propose a model in which expert(s) can

pre-commit to specified strategy. Under some generic conditions, Gentzkow and

Kamenica (2011) shows that there exists a fully revealing equilibrium when the

decision maker has an access to advice of multiple experts. In contrast, the only

equilibrium that exists in my model when we assume that costs of reading and

lying are zero (to make the comparison valid) is so called babbling equilibrium,

when no information is transmitted at all.

The paper also relates to extensive literature on cheap talk1 initiated by the

seminal work of Crawford and Sobel (1982). In the line with this stream of liter-

ature, information in my model is neither verifiable nor contractible. The main

difference of mine paper from cheap talk literature is that information transmission

in my paper is costly while classical cheap talk models assume costless communi-

cation. Probably, the only exceptions are two papers by Chen et al. (2008) and

Kartik (2009) who study costly cheap talk as a selection criteria for equilibrium

in the classical cheap talk models.

While the question of which communication protocol (simultaneous or sequen-

tial) is beneficial for the decision maker is not raised directly, broad cheap talk lit-

erature (Krishna and Morgan, 2001; Battaglini, 2003, 2004; Morgan and Stocken,

2003; Li and Madarasz, 2008) studies one of two communication protocols. Under

1For recent survey on cheap talk literature I refer the interested reader to Sobel (2010).

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some restrictions on parameters Krishna and Morgan (2001); Battaglini (2003,

2004) show that there exists a fully revealing equilibrium when experts provide

their recommendations simultaneously. In contrast, there exist only partially re-

vealing or pooling equilibria when experts are consulted sequentially (Krishna and

Morgan, 2001).

Perhaps surprisingly, this paper shows that when costs of fabricating and inter-

preting information are small enough (the limiting case of which is cheap talk) then

the decision maker benefits with experts talking sequentially. Also the reasoning

for this result to hold is different from cheap talk literature. Mainly, the existence

of the fully revealing equilibrium in cheap talk models explained by properly con-

structed out of equilibrium beliefs, while the main result of this paper is driven by

the ability of the decision maker to credibly signal her private information when

reading the report of the first expert. The idea of the decision maker having some

private information prior to consulting experts is not new. In the context of cheap

talk models, Chen (2009) and de Barreda (2010) show that the decision maker

does not necessarily benefits from having access to private information.

The rest of this paper is organized as follows. Section 2 introduces a model.

Section 3 completely characterizes the equilibrium under both sequential and si-

multaneous communication protocols. Section 4 presents the main result and Sec-

tion 5 concludes. The proofs missing in the main part of the paper are provided

in the Appendix.

2 Model

There are three players: Expert 1, Expert 2 and Decision Maker, hereafter DM.

The state of the nature is θ ∈ {L,R}, Pr(R) = Pr(L) = 1/2. Both experts

observe the true state of the nature perfectly, while DM has a prior belief µ0 that

the true state of the nature is R and this is her private information. The common

prior on µ0 has a distribution function G(·) ∈ C1 and density function g(·).The persuasion takes a form of reports made by each expert. Report is the

binary signal m ∈ {R,L}. Writing report for Expert i is costless if mi = θ, and

requires to pay cost cE otherwise. It is assumed that each expert writes exactly

one report. DM decides which reports to read if at all by paying a cost cDM per

report.

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The DM payoff is: 1{a = θ}−ncDM , where a is an action taking by DM and n

is the number of reports read. That is, DM wants to take an action that matches

the true state of the nature. Expert 1′s payoff equals to 1{a = L}−1{m1 6= θ}·cEand Expert 2′s payoff equals to 1{a = R}−1{m2 6= θ}·cE. All players are assumed

to be the expected utility maximizers.

I allow two forms of communications protocols: simultaneous and sequential.

In the simultaneous communication protocol experts provide their reports simul-

taneously. In the sequential communication protocol, Expert 1 provides his report

first and Expert 2 provides his report only after observing whether DM reads the

first report. Below I discuss the model assumptions and the time line of the game

under the simultaneous and sequential communication protocols. I will use the

phrase Persuasion Game for the game described above.

2.1 Model Assumptions

For the rest of this paper it is assumed that unconditional density function g

satisfies the following assumptions:

Assumption 1. (Symmetry) For all µ0 ∈ [0, 1], g(µ0) = g(1− µ0).

Assumption 2. (Full Support) For all µ0 ∈ (0, 1), g(µ0) > 0.

Assumption 3. (Monotonicity) g is weakly increasing function on (0, 1/2].

Assumption 4. (Concavity) There exists µ1 ∈ [0, 1/2) and µ2 ∈ (1/2, 1] such

that µ0g(µ0) is weakly concave function on [µ1, µ2].

While assumptions 1 and 2 are quite standard in the literature, assumptions 3

and 4 require a little bit of discussion. Formally, density function g(µ) measures

the probability that DM’s beliefs are at a distance µ from a truth (if L is treated

as ”0” and R as ”1”). Then assumptions 1 and 3 imply that statistically, there

is a high probability that DM is more likely to be not well informed about the

true state of the nature. If we assume that g is smooth enough then assumption 4

is directly implied by assumption 1 and 3. The following example provides a full

class of density functions that satisfy all four above assumptions.

Example 1. For all k ≥ 0 let g(µ0) = aµk0(1− µ0)k, where a is chosen in such a

way that1∫0

g(µ0)dµ0 = 1. It is straightforward exercise to show that assumptions

1 − 3 are satisfied. For k = 0, µ0g(µ0) is a linear function and hence, weakly

concave. For k > 0, observe that g ∈ C2, g′(1/2) = 0 and g′′(1/2) < 0. Hence,

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(µ0g(µ0))′ = 2g′(µ0)+µ0g

′′(µ0) ∈ C that is negative if evaluated at 1/2. Therefore,

assumption 4 holds.

Since µ0 is defined as a prior belief that the true state of the nature is R, the

joint density function of θ and µ0, f(·) is defined as:

f(θ, µ0) =

g(µ0)µ0, if θ = R

g(µ0)(1− µ0), if θ = L(1)

Note, that symmetry assumption guarantees that θ is equally likely to take

values L and R. I will write FR(·) and FL(·) to denote conditional distribution

functions on θ = R and θ = L respectively. The corresponding density functions

are fR(·) and fL(·). Then assumption 1 implies that FL(µ0) + FR(1− µ0) = 1 for

all µ0 ∈ [0, 1] and assumption 3 implies that fR is strictly increasing function on

(0, 1/2). Moreover, formular (1) implies that fR(µ0)fL(µ0)

= µ01−µ0 and hence, fR(·)/fL(·)

is strictly decreasing function. Finally, λL and λR denote probability measures

induced by the density functions fL and fR respectively.

2.2 Simultaneous Protocol

The time line of the Persuasion Game with simultaneous communication protocol

proceeds as follows. At stage 0, the true state of the nature θ and DM’s type µ0

are realized. At stage 1, after observing θ, both experts simultaneously choose

the signal they want to report. Finally, at stage 2 DM reads reports, takes action

a ∈ {L,R} and payoffs are realized. Importantly, DM reads reports sequentially.

Say, if DM decides to read report of Expert 2 first then she may condition her

decision whether to read Expert 1’s report on information contained in Expert 2′s

report. The process is diagramed below:

Time0 1 2

Nature chooses

µ0 and θ

Expert 1 and 2

choose report

DM reads reports

and takes action

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2.3 Sequential Protocol

The time line of the Persuasion Game with sequential communication protocol

proceeds as follows. At stage 0, the true state of the nature θ and DM’s type µ0

are realized. At stage 1, after observing θ, Expert 1 chooses the signal he wants

to report. At stage 2 DM decides whether to read Expert 1′s report or do not.

At stage 3, after observing both θ and stage 2 DM decision, Expert 2 chooses

the signal he wants to report. Finally, at stage 4 DM reads reports, takes action

and payoffs are realized. Similarly to simultaneous protocol, DM at stage 4 reads

reports sequentially. Finally, we assume that whenever DM is indifferent between

reading report 1 at stage 2 or 4 she reads report 1 at stage 2. The process is

diagramed below:

Time0 1 2 3 4

Nature

chooses µ0

and θ

Expert 1

chooses

report

ρ(m1|θ)

DM reads

report or

does not

Expert 2

chooses

report

ρ(m2|·)

DM reads

reports

and takes

action

2.4 Strategies and Equilibrium

I adopt the following notation. Set S = {L,R} denotes both state space and space

of actions available to DM. Set M1 = (S × {1})∪ {∅} denotes the set of messages

available to Expert 1. Note, that we do not allow for experts to report nothing

and ∅ is included in set M1 for notational convenience and will be used to denote

the situation when DM does not read report of Expert 1. Define set M2 similarly.

Then set M = M1 ∪M2 is the message space. Finally, I = {∅, 1, 2} is a set of

reports, where ∅ will be used to denote situation when DM does not read any

report.

In the Persuasion Game with simultaneous protocol the strategy for Expert 1

is a mapping ρ1 : S → ∆M1 that specifies the probability ρ1(m1|θ) with which

message m1 is sent by Expert 1 in state θ. Strategy for Expert 2 is defined

similarly. A strategy for DM consists of two parts. The information acquisition

part of DM strategy is a mapping x : [0, 1] ×M → I that specifies the report

x(µ0,m) that DM of type µ0 reads after observing message m ∈ M . The action

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part of DM strategy is a mapping a : [0, 1] ×M1 ×M2 that specifies the action

a(µ0,m1,m2) that DM of type µ0 takes after observing messages m1 and m2. Let

µ(µ0,m1,m2) denote posterior beliefs on θ held by DM after observing messages

m1 and m2. I will write µ(µ0,m1) to denote µ(µ0,m1, ∅) and µ(µ0,m2) to denote

µ(µ0,m2, ∅). Applying Bayes’ rule

µ(µ0,m1,m2) =µ0ρ1(m1|R)ρ2(m2|R)

µ0ρ1(m1|R)ρ2(m2|R) + (1− µ0)ρ2(m1|L)ρ2(m2|L)(2)

for all m1 ∈ M1 and m2 ∈ M2 such that denominator does not equal to zero,

where ρi(∅|θ) = 1 for all i and θ.

Definition 1. Perfect Bayesian equilibrium of the Persuasion Game with simul-

taneous protocol entails as follows: (1) for all messages m1 ∈ M1, m2 ∈ M2 and

DM type µ0, a(µ0,m1,m2) maximizes the DM expected utility given her beliefs

µ(µ0,m1,m2); (2) for all m ∈ M and µ0 ∈ [0, 1], x(µ0,m) maximizes the DM

expected utility given her beliefs µ(µ0,m); (3) the beliefs µ(µ0,m1,m2) are formed

by applying Bayes’ rule (2) whenever possible; (4) given DM strategy (a, x), for all

θ, ρ1(θ) and ρ2(θ) maximize expected utility of Experts 1 and 2 correspondingly.

Next, I will define the equilibrium for the Persuasion Game with sequential

protocol. In the Persuasion Game with sequential protocol a strategy for Expert

1 is a mapping ρ1 : S → ∆M1 that specifies the probability ρ1(m1|θ) with which

message m1 is sent by Expert 1 in state θ. A strategy for Expert 2 is a mapping

ρ2 : S × {1, ∅} → ∆M2 that specifies the probability ρ2(m2|θ, x1) with which

message m2 is sent by Expert 2 in state θ if DM reads the report x1 at stage 2.

A strategy for DM is a triple (x1, x2, a). Information acquisition part of strategy

consists of x1 and x2, where x1 : [0, 1]→ {∅, 1} specifies the report x1(µ0) DM of

type µ0 reads at stage 2 and x2 : [0, 1]×M → I specifies the report x2(µ0,m) that

DM of type µ0 reads at stage 4 after observing message m ∈ M . An action part

of strategy is defined in the same way as for simultaneous protocol.

Let, similarly to simultaneous protocol, µ(µ0, x1,m1,m2) denotes posterior be-

liefs on θ held by DM after reading report x1 at stage 2 and observing messages

m1 and m2. We will write µ(µ0,m1) to denote µ(µ0, x1,m1, ∅) and µ(µ0,m2) to

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denote µ(µ0, x1,m2, ∅). Applying Bayes’ rule

µ(µ0, x1,m1,m2) =µ0ρ1(m1|R)ρ2(m2|R, x1)

µ0ρ1(m1|R)ρ2(m2|R, x1) + (1− µ0)ρ2(m1|L)ρ2(m2|L, x1)(3)

for all x1 ∈ {1, ∅}, m1 ∈ M1 and m2 ∈ M2 such that denominator does not equal

to zero, where ρ1(∅|θ) = 1 and ρ2(∅|θ, x1) = 1 for all θ and x1.

Let σ(µ0|x1) denotes posterior beliefs on µ0 held by Expert 2 after observing

that DM reads report x1 at stage 2. Formally,

σ(µ0|x1) =1(x1(µ0) = x1)∫

1(x1(µ0) = x1)dµ0

(4)

for all µ0 ∈ [0, 1] and x1 ∈ {1, ∅} such that denominator does not equal to zero.

Definition 2. Perfect Bayesian equilibrium of Persuasion Game with simulta-

neous protocol entails as follows: (1) for all messages m1 ∈ M1, m2 ∈ M2 and

DM type µ0, a(µ0,m1,m2) maximizes the DM expected utility given her beliefs

µ(µ0, x1(µ0),m1,m2); (2) for all µ0 ∈ [0, 1], x1(µ0) maximizes the DM utility; (3)

for all m ∈M and µ0 ∈ [0, 1], x2(µ0,m) maximizes the DM expected utility given

her beliefs µ(µ0, x1(µ0),m1,m2); (4) the beliefs µ(µ0, x1,m1,m2) are formed by ap-

plying Bayes’ rule (3) whenever possible; (5) given DM strategy (a, x1, x2), for all

θ, ρ1(θ) maximizes expected utility of Expert 1; (6) given DM strategy (a, x1, x2)

and beliefs σ(µ0|x1), for all θ and x1 ∈ {∅, 1}, ρ2(θ, x1) maximizes expected utility

of Experts 2; and (7) the beliefs σ(µ0|x1) are formed by applying Bayes’ rule (4)

whenever possible.

For the rest of this paper whenever I refer to an equilibrium I implicitly mean

Perfect Bayesian equilibrium.

3 Model Analysis

3.1 Preliminary Results

In this section I outline some preliminary results that hold both for simultaneous

and sequential communication protocols. First thing to note is that DM will

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take action R iff her posterior beliefs µ is greater or equal to one half 2. The next

proposition uses this simple observation to prove that experts never fabricate their

favorable state of the nature on equilibrium path.

Lemma 1. In any equilibrium with either simultaneous or sequential communi-

cation protocol, ρ1(R1|L) = 0 and ρ2(L2|R, ·) = 0.

Proof. I will prove that ρ1(R1|L) = 0. The proof of ρ2(L2|R, ·) = 0 is completely

analogous and hence, omitted. Suppose, by way of contradiction, that ρ1(R1|L) >

0. Note, that all types of DM update their beliefs in the same direction after

observing any particular message. If DM updates her belief up after observing

message R1 (meaning that DM thinks that state R is more likely when she sees

message R1 than when she sees message L1 ) then it is strictly dominant strategy

for Expert 1 to report L1 when the true state of the nature is L and hence,

ρ1(R1|L) = 0. Therefore, DM updates her beliefs down after observing message

R1 or beliefs of DM remain unchanged. Then it is strictly dominant strategy for

Expert 1 to report R1 when the true state of the nature is R as it is costless message

that does not increase the current beliefs of DM. Thus, ρ1(R1|R) = 1. Note also

that ρ1(R1|L) could not be equal to 1 as then report 1 will be uninformative and

will be never requested in equilibrium. Therefore, ρ1(L1|L) > 0 implying that

whenever DM observes message L1 she learns that the true state of the nature

is L and hence, updates her belief to 0. Then DM updates her belief up after

observing message R1. We get a contradiction which ends a proof.

Above lemma simplifies analysis substantially since experts’ strategies are com-

pletely characterized by fabrication probabilities ρ1(L1|R) and ρ2(R2|L, ·). For the

rest of this paper I will use ρ1 to denote ρ1(L1|R), ρ20 to denote ρ2(R2|L, ∅) and

ρ2 to denote both ρ2(R2|L) and ρ2(R2|L, 1). Also Proposition 1 implies that DM

updates her belief down if she reads messages L1 or L2 at any stage of the game,

and up if she reads R1 or R2. Moreover, she learns the true state of the nature

if she reads message R1 or L2. Finally, DM’s decision is completely based on the

last report she read or on her prior beliefs if she reads nothing. In fact, if indepen-

dently of message transmitted in the last report DM will take the same action then

2More precisely DM is indifferent between action R and L if her posterior beliefs µ0 equalsto 1/2. However, since in equilibrium this will be true for measure zero of DM prior beliefs wemay assume without loss of generality that action R is taken in this case.

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DM can take this action without reading the report and economize on the cost of

reading. For instance, if report 1 is the last report read by DM then she should

take action L after observing L2 and action R after observing R2. Hence, DM’s

equilibrium strategy could be completely described by the information acquisition

part.

3.2 Simultaneous Protocol

In this section I completely characterize the set of equilibria for the Persuasion

Game with simultaneous protocol. The first natural question is whether there

exists a truth revealing equilibrium, that is an equilibrium in which ρ1 = ρ2 = 0.

Suppose, that under some conditions on parameters such equilibrium exists. Then

DM will never read two reports, as both reports contain the same information and

reading reports is costly. Consider DM whose type µ0 <12. If DM does not read

any report he takes action L and gets payoff 1− µ0. If DM reads report then she

learns the true state of the nature perfectly and gets utility 1, but also pays the

reading cost cDM > 0 that results in total utility 1 − cDM . Hence, when µ0 <12

DM reads the report iff µ0 ≥ cDM . Similarly, can be shown that DM whose type

µ0 ≥ 12

reads the report iff µ0 ≤ 1 − cDM . Hence, whenever µ0 ∈ [cDM , 1 − cDM ]

the DM reads one of two reports.

Then truth revealing equilibrium exists iff neither Expert 1 nor Expert 2 has

incentive to misreport the true state of the nature. Next lemma states necessary

and sufficient conditions under which truth revealing equilibrium exists.

Lemma 2. There exists truth revealing Equilibrium of the Persuasion Game with

simultaneous communication protocol iff

cE + FR(cDM) ≥ FR(1/2)

Proof. Only if. Suppose that there exists truth revealing Equilibrium. Note,

that if cDM ≥ 1/2 then inequality in the statement of lemma trivially holds.

Therefore, for the rest of the proof assume that cDM < 1/2. Let, Ti denotes the

set of DM types who read report i, i = 1, 2. Note, that T1 ∪ T2 = [cDM , 1− cDM ]

and T1 ∩ T2 = ∅. Then Expert 1 does not have incentive to deviate and send

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message L1 when the true state of the nature is R iff cE ≥ λR(T1). Similarly,

Expert 2 does not have incentive to deviate and send message R2 when the true

state of the nature is L iff cE ≥ λL(T2).

Define sets T1 and T2 in the way that follows: T1 = [cDM , F−1R (cE +FR(cDM))]

and T2 = (1−F−1R (cE +FR(cDM)), 1− cDM ]. It is easy to check that λR(T1) = cE

and λL(T2) = cE. Suppose, by contrary, that inequality in the statement of

proposition does not hold. Then T1 ∩ T2 = ∅ and T1 ∪ T2 is a strict subset of

[cDM , 1− cDM ]. If λL(T1 ∩ T2) = 0 then λL(T2) = λL(T2) as λL(T2) ≤ λL(T2) and

hence, λR(T1) < λR(T1) contradicting to λR(T1) ≤ λR(T1). Therefore, λL(T1 ∩T2) 6= 0. Similarly, λL(T2 ∩ T1) 6= 0. Observe, that3 T1 ∩ T2 > T2 ∩ T1 and taking

into account that fR(µ)/fL(µ) is strictly increasing function we have that:

λR(T1 ∩ T2)λL(T1 ∩ T2)

>λR(T2 ∩ T1)λL(T2 ∩ T1)

.

Then either λR(T1∩ T2) > λR(T2∩ T1) or λL(T1∩ T2) < λL(T2∩ T1). In the former

case:

λR(T1) ≥ λR(T1 ∩ T1) + λR(T1 ∩ T2) > λR(T1 ∩ T1) + λR(T2 ∩ T1) = λR(T1)

contradicting to λR(T1) ≤ λR(T1). In the later case:

λL(T2) ≥ λL(T2 ∩ T1) + λL(T2 ∩ T2) > λL(T1 ∩ T2) + λL(T2 ∩ T2) = λL(T2)

contradicting to λL(T2) ≤ λL(T2). These contradictions ends a proof of ”only if”

part.

If. Suppose, that the inequality in the statement of proposition holds. Define

the strategy of DM in the following way. DM reads report 1 whenever µ0 ∈ T1 =

[cDM , 1/2], reads report 2 whenever µ0 ∈ T2 = [1/2, 1− cDM ] and reads no report

otherwise. Note, that T1 ⊆ T1 and T2 ⊆ T2. Hence, λR(T1) ≤ cE and λL(T2) ≤ cE

implying that experts have no incentive to deviate and misreport the true state

of the nature.

Next, I will derive the necessary and sufficient conditions for existence of an

3I define partial order on the space of all subsets of [cDM , 1 − cDM ] in the following way:A > B iff for all a ∈ A and b ∈ B, a > b.

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equilibrium with fabrications, that is ρ1 > 0 or ρ2 > 0. Note, that there is no

equilibrium in which ρ1 = 0 and ρ2 > 0 as in this case DM will never read report

2 and hence, Expert 2 can economize on the cost of fabrication by reporting L1

when the true state of the nature is L (ρ2 = 0). Similarly, there is no equilibrium

in which ρ1 > 0 and ρ2 = 0. Also there is no equilibrium with ρi = 1, as in

this case report i is uninformative and DM does not read such report. Hence,

ρ1, ρ2 ∈ (0, 1).

I will analyze the DM problem first. Since ρ1 > 0 there exists a positive

measure of DM types who read report 1. Similarly, there exists a positive measure

of DM types who read report 2. The next proposition shows that in equilibrium

DM never reads both reports.

Lemma 3. It is never optimal for the DM to read two reports.

Proof. Suppose, by contrary, that DM reads two reports. By reading two reports

DM can observe one of three pair of messages (L1, L2), (L1, R2) and (R1, R2). Con-

sider the posterior belief of DM µ(L1, R2) and suppose, without loss of generality,

that µ(L1, R2) > 1/2. Clearly, µ(L2) = µ(L1, L2) = 0 and µ(R1) = µ(R1, R2) = 1.

Hence, a(∅, L2) = a(L1, L2) = L and a(R1, ∅) = a(R1, R2) = a(L1, R2) = R.

Therefore, to make a right decision it is enough to know the message contained in

report 2. Then, DM can economize on the cost by reading report 2 alone. We get

a contradiction, which ends the proof.

Lemma 3 tells us that DM either reads one report or does not read any report

at all. Hence, the DM’s problem constitutes in choosing report i from a set I

of possible reports. Let, uSMDM(i) denotes the µ0 type DM’s expected payoff when

she reads report i. Then x(µ0, ∅) = argmaxi∈I

uSMDM(i) is the report DM reads in

equilibrium.

Trivially, if DM reads no reports (i = ∅) then her utility equals to uSMDM(∅) =

max{µ0, 1 − µ0}. Suppose now, that DM reads report 1 in equilibrium, that is

x(µ0, ∅) = 1. Recall that it is optimal to do so only if observing message L1

leads DM to take action L. Then if DM observes message R1 her utility will be

1− cDM as she completely learns θ and if DM observes message L1 her utility will

be 1−µ(L1)− cDM , where µ(L1) is calculated by applying Bayes’ rule (2). Hence,

13

the expected payoff of DM under condition that x(µ0, ∅) = 1 equals to:

uSMDM(1) = Pr(R1) · (1− cDM) + Pr(L1) · (1− µ(L1)− cDM) = 1− µ0ρ1 − cDM .

Similarly, one can show that expected payoff of DM under condition that x(µ0, ∅) =

2 equals to uSMDM(2) = 1− (1− µ0)ρ2 − cDM .

Then the necessary and sufficient conditions that DM of type µ0 reads report

1 are uSMDM(1) ≥ uSMDM(2) and uSMDM(1) ≥ uSMDM(∅). Solving the last two inequal-

ities simultaneously gives us that DM of type µ0 reads report 1 iff µ0 ∈ T1 =

[ cDM

1−ρ1 ,ρ2

ρ1+ρ2). Similarly, could be shown that DM of type µ0 reads report 2 iff

µ0 ∈ T2 = [ ρ2ρ1+ρ2

, 1− cDM

1−ρ2 ]. In all other cases DM reads nothing. The DM acqui-

sition part of equilibrium strategy could be represented by the following figure:

µ00 aSM1∅ aSM12 aSM2∅ 1

∅ 1 2 ∅

Here, aSMij denotes the type of DM who is indifferent between reading report i

and j, aSM1∅ = cDM

1−ρ1 , aSM12 = ρ2ρ1+ρ2

and aSM2∅ = 1− cDM

1−ρ2 .

Once, we describe DM equilibrium strategy we are ready to specify the experts’

equilibrium strategies. Consider, Expert 1. Since ρ1 ∈ (0, 1) Expert 1 should be

indifferent between sending messages L1 and R1 when the true state of the nature

is R. If he sends message R1 his expected utility will increase by the probability

that DM reads report 1 (as DM will take action R after reading message R1 and

L after reading message L1) and decrease by fabrication cost cE1 comparably to

sending message L1. Hence, Expert 1 is indifferent between sending messages L1

and R1 iff the probability that DM reads report 1 conditional on θ = R equals to

the cost of fabrication cE, that is cE = λR(T1). Similarly, Expert 2 is indifferent

between sending message L2 and R2 when the true state of the nature is L iff

cE = λL(T2). Therefore, the fabrication probabilities are determined from the

following system: FR( ρ2ρ1+ρ2

)− FR( cDM

1−ρ1 ) = cE

FR( ρ1ρ1+ρ2

)− FR( cDM

1−ρ2 ) = cE

(5)

It is easy to observe that there is no equilibrium in which ρ1 6= ρ2. Say, if ρ1 > ρ2

then LHS of the first equation is lower than LHS of the second equation implying

14

that both equations could not hold simultaneously. Therefore, ρ1 = ρ2 = ρ∗ =

1 − cDM

F−1R (FR( 1

2)−cE)

. Next proposition summarizes the analysis of the Persuasion

Game with simultaneous communication protocol by stating that equilibrium is

essentially unique, meaning that fabrication probabilities are uniquely determined.

Proposition 1. For all values of parameters cE and cDM there exists essentially

unique equilibrium of the Persuasion Game. In equilibrium experts fabricate with

the same fabrication probabilities ρ1 = ρ2 = ρ∗; ρ∗ = 0 if cE +FR(cDM) ≥ FR(1/2)

and ρ∗ = 1− cDM

F−1R (FR( 1

2)−cE)

otherwise.

Proof. Suppose, that cE+FR(cDM) ≥ FR(1/2). Then according to Lemma 2 there

exists truth telling equilibrium. Moreover, it is easy to verify that (5) has a solution

iff cE + FR(cDM) < FR(1/2). Finally, this solution is ρ∗ = 1− cDM

F−1R (FR( 1

2)−cE)

.

3.3 Sequential Protocol

In this section I characterize the set of equilibria for the Persuasion Game when

experts speak sequentially. Similarly, to simultaneous protocol I first derive condi-

tions under which there is truth revealing equilibrium, that is ρ1 = ρ20 = ρ2 = 0.

Note, that in equilibrium DM of each type reads at most one report and does this

iff µ0 ∈ [cDM , 1− cDM ]. Moreover, as DM that read report 1 gets utility 1− cDMindependently of the time she reads it, she will prefer to do so at stage 2 of the

game. Then truth revealing equilibrium exists when neither Expert 1 nor Expert

2 have incentive to misreport the true state of the nature.

Proposition 2. There is truth revealing Equilibrium of the Persuasion Game with

sequential communication protocol iff

F−1R (cE + FR(cDM)) + F−1R

(cE(FR(cDM) + FL(cDM))

1− cE+ FR(cDM)

)≥ 1 (*)

Proof. See Appendix.

Note, that condition in Proposition 2 is more restrictive than one in Lemma

2. This is due to the fact that under sequential protocol, Expert 2 updates his

posterior beliefs about DM type conditional on whether DM reads report 1 at stage

2 or not. Then, when Expert 1 observes DM not reading report 1 the probability

15

that DM is of type who reads report 2 increases. And hence, it is harder to

prevent Expert 2 from fabricating state θ = L.

Next, I will characterize conditions under which there is an equilibrium with

fabrications (ρ1 > 0 and ρ2 > 0). Similarly to simultaneous protocol, there is no

DM type who reads two reports at stage 4 of the game. However, there could be

an equilibrium in which DM of some type reads two reports: report 1 at stage 2

of the game and report 2 at stage 4. All equilibria with fabrications fall in one of

three types, description of which follows.

Type 1 equilibrium is an equilibrium in which ρ20 = 0 and there is no DM type

who reads both reports. Type 2 equilibrium is an equilibrium in which ρ20 = 0

and there is positive measure of DM types who read both reports. Finally, Type 3

equilibrium is an equilibrium in which ρ20 > 0. Below we will show that for given

values of parameters cE and cDM that do not satisfy condition in Proposition

2 there exists exactly one type of equilibrium with fabrications. Also we will

determine the conditions under which each type of equilibrium exists.

Type 1 Equilibrium. Suppose that ρ20 = 0 and there is no DM type who

reads both reports. Then DM’s problem becomes completely analogous to one of

Persuasion Game with simultaneous protocol. That is DM of type µ0 reads report

1 if µ0 ∈ T1 = [aSQ1∅ , aSQ12 ], reads report 2 if µ0 ∈ T2 = [aSQ12 , a

SQ2∅ ] and does not

read any report otherwise. Here, aSQij denotes the type of DM who is indifferent

between reading report i and j, aSQ1∅ = cDM

1−ρ1 , aSQ12 = ρ2ρ1+ρ2

and aSQ2∅ = 1− cDM

1−ρ2 .

Recall, that Expert 1 is indifferent between fabricating θ = L iff the probability

that DM reads report 1 equals to the fabrication cost cE. Hence, λR(T1) = cE.

Similarly, Expert 2 is indifferent between fabricating θ = R iff the probability

that DM reads report 2 conditional on not reading report 1 equals to cE. Hence,

λL(T2)/(λL(T2) + λL([a1∅, a2∅]C) = cE.

Proposition 3. There exists strictly increasing function g1 : R2++ → R such

that type 1 equilibrium exists iff g1(cDM , cE) ∈ [0, 1] and g1(cDM , cE) > 1 iff cDM

and cE satisfy condition (*). Moreover, the fabrication probabilities ρ1 and ρ2 are

uniquely determined by:FR( ρ2ρ1+ρ2

)− FR( cDM

1−ρ1 ) = cE

(1− cE)FR( ρ1ρ1+ρ2

)− FR( cDM

1−ρ2 ) + cEFR(1− cDM

1−ρ1 ) = cE

(6)

16

Proof. See Appendix.

Corollary 1. Suppose that conditions in Proposition 3 hold, and let ρ∗ denotes

the fabrication probability for the Persuasion Game with simultaneous protocol.

Then ρ1 > ρ∗ and ρ2 > ρ∗.

Proof. See Appendix.

Type 2 and Type 3 Equilibria. Suppose, that in equilibrium there is a positive

measure of DM types who read two reports: report 1 at stage 2 and report 2 at

stage 4. Note, that DM reads report 2 only if she observes message L1 after

reading report 1 as if report 1 contains message R1 the true state of the nature

is learned completely. Then, it is natural to introduce report 3 as a composite

of two reports: report 1 and report 2, where report 2 is read only if report 1

contains message L1. Therefore DM problem constitutes in choosing report from

a set I ∪ {3} = {∅, 1, 2, 3}.Denote uSQDM(µ0, i) the µ0 type DM’s expected payoff when she reads report

i, for all i ∈ I ∪ {3}. With some abuse of notation we will write x(µ0, ∅) =

argmaxi∈I∪{3}

uSQDM(µ0, i) to denote the report DM of type µ0 reads in equilibrium. Then,

similarly to simultaneous communication protocol, we have that uSQDM(µ0, ∅) =

max{µ0, 1− µ0}, uSQDM(µ0, 1) = 1− µ0ρ1 − cDM if x(µ0, ∅) = 1, and uSQDM(µ0, 2) =

1− (1− µ0)ρ2 − cDM if x(µ0, ∅) = 2.

Next, suppose that DM reads report 3 in equilibrium, that is x(µ0, ∅) = 3.

Note, that it is optimal for DM to read report 3 iff µ(L1, L2) < 1/2 < µ(L1, R2).

Hence, the DM’s payoff while reading report 3 equals to 1 − cDM if she observes

message R1, 1−2cDM if she observes pair of messages (L1, L2) and µ(L1, R2)−2cDM

if she observes (L1, R2). Hence, the expected payoff of DM under condition that

x(µ0, ∅) = 3 equals to:

uSQDM(µ0, 3) = Pr(R1)·(1−cDM)+Pr(L1, L2)·(1−2cDM)+Pr(L1, R2)·(µ(L1, R2)−2cDM) =

1− 2cDM − ρ20(1− µ0) + cDMµ0(1− ρ1).

Then the necessary and sufficient conditions that DM of type µ0 reads re-

port 1 are uSQDM(µ0, 1) ≥ uSQDM(µ0, 3), uSQDM(µ0, 1) ≥ uSQDM(µ0, 2) and uSQDM(µ0, 1) ≥uSQDM(µ0, ∅). Solving all three inequalities simultaneously gives us that DM of type

17

µ0 reads report 1 iff µ0 ∈ T1 = [ cDM

1−ρ1 ,ψ

ρ1+ψ], where ψ = (ρ20 + cDM)/(1 − cDM).

Similarly could be shown that DM reads report 2 if µ0 ∈ T2 = ( ψρ1+ψ

, φρ1+φ

] and

reads report 3 if µ0 ∈ T3 = ( φρ1+φ

, 1− cDM

1−ρ2 ], where φ = (ρ2−ρ20−cDM)/cDM . In all

other cases (µ0 ∈ T∅) DM reads nothing. The DM acquisition part of equilibrium

strategy could be represented by the following figure:

µ00 aSQ1∅ aSQ13 aSQ32 aSQ2∅1

∅ 1 3 2 ∅

Here, as usual, aSQij denotes the type of DM who is indifferent between reading

report i and j, aSQ1∅ = cDM

1−ρ1 , aSQ13 = ψρ1+ψ

, aSQ32 = φρ1+φ

and aSQ3∅ = 1 − cDM

1−ρ2 . Since

ρ1 ∈ (0, 1), Expert 1 is indifferent between sending message L1 and R1 at stage 2

of the game when θ = L. This is true iff the probability that DM reads report 1 at

stage 2 equals to the fabrication cost cE, λR(T1) = cE. Similarly, since ρ2 ∈ (0, 1),

the probability that DM reads report 2 equals to the fabrication cost cE. Hence,

λL(T2)/(λL(T∅) + λL(T2)) = cE.

Finally, since ρ20 ∈ [0, 1) the probability that DM reads report 2 after already

reading report 1 is less or equal to cE, and equals to cE if ρ20 6= 0. Hence,

λL(T3)/(λL(T1) + λL(T3)) ≤ cE and ”=” if ρ20 6= 0. Taking into account that

λL(T2)/(λL(T∅) + λL(T2)) = cE we have that λL(T2) + λL(T3) ≤ cE and ”=” if

ρ20 6= 0. Next proposition specifies conditions on parameters under which all three

above conditions are satisfied.

Proposition 4. There exists type 2 or type 3 equilibrium with fabrications iff

g1(cDM , cE) < 0, where function g1 is defined in Proposition 6. Fabrication prob-

abilities are determined by:FR( ψ

ρ1+ψ)− FR( cDM

1−ρ1 ) = cE

FR( ρ1ρ1+ψ

)− FR( cDM

1−ρ2 ) ≤ cE, and = if ρ20 6= 0

(1− cE)FR( ρ1ρ1+φ

)− FR( cDM

1−ρ2 ) + cEFR(1− cDM

1−ρ1 ) = cE

(7)

where ψ = (ρ20 + cDM)/(1 − cDM) and φ = (ρ2 − ρ20 − cDM)/cDM . Moreover, if

2cE + FR(2cDM) ≤ FR(1/2) then only type 3 equilibrium possible.

Proof. See Appendix.

18

Corollary 2. Suppose that conditions in Proposition 4 hold, and let ρ∗ denotes

the fabrication probability for the Persuasion Game with simultaneous protocol. If

ρ20 > 0 and cE ≤ max{(FR(1/2)−FR(µ1))/2, FR(µ2)−FR(1/2)} then ρ2 > ρ∗ > ρ1

and ρ2 − ρ∗ < ρ∗ − ρ1.

Proof. See Appendix.

4 Main Result

Theorem 1. Let function g1 is defined in Proposition 6. Then the following holds:

1. If g1(cDM , cE) ∈ [0, 1], then the DM of every type is weakly better off under

the simultaneous communication protocol.

2. If 2cE+FR(2cDM) ≤ FR(1/2) and cE ≤ max{(FR(1/2)−FR(µ1))/2, FR(µ2)−FR(1/2)}, then the DM is ex-ante better off under the sequential communi-

cation protocol.

Proof. I will prove part 1 of the theorem first. Suppose, that g1(cDM , cE) ∈ [0, 1].

Then according to Proposition 6 there exists type 1 equilibrium with fabrications

in the Persuasion Game with sequential communication protocol. Also according

to Corollary 2 the fabrication probabilities ρ1 and ρ2 satisfy ρ2 > ρ1 > ρ∗. Consider

a DM of type µ0, µ0 ∈ [0, 1] and let i, j ∈ I denote the reports that DM reads

in equilibrium under simultaneous and sequential protocols respectively. Then

uSMDM(i) ≥ uSMDM(j) ≥ uSQDM(j). Hence, DM of type µ0 is weakly better off under

simultaneous protocol.

Next, I will prove part 2 of the theorem. Suppose, that 2cE + FR(2cDM) <

FR(1/2). Then according to Proposition 7 there exists type 3 equilibrium with

fabrications in the Persuasion Game with sequential communication protocol. Also

according to Corollary 3 the fabrication probabilities satisfy ρ2 > ρ∗ > ρ1 and

ρ2−ρ∗ < ρ∗−ρ1. Recall, that under simultaneous communication protocol, type µ0

DM reads report 1 when µ0 ∈ [ cDM

1−ρ∗ , 1/2), reads report 2 when µ0 ∈ [1/2, 1− cDM

1−ρ∗ ]

and reads no report otherwise.

Suppose that under sequential communication protocol the DM uses the same

strategy as under simultaneous communication protocol. To prove part 2 of the

19

theorem it is enough to show that by doing this the DM is better off in ex-

ante terms under sequential communication protocol. Taking into account that

ρ2 − ρ∗ < ρ∗ − ρ1 , for all µ0 ∈ [ cDM

1−ρ∗ , 1/2] we have that

uSQDM(µ0, 1)−uSMDM(µ0, 1) = µ0(ρ∗−ρ1) > µ0(ρ2−ρ∗) = uSMDM(1−µ0, 1)−uSQDM(1−µ0, 1)

and hence

uSQDM(µ0, 1) + uSQDM(1− µ0, 2) > uSMDM(µ0, 1) + uSMDM(1− µ0, 2).

Integrating the last inequality over µ0 ∈ [ cDM

1−ρ∗ , 1/2] and taking into account that

density function g is symmetric around 1/2 gives us the desirable result.

5 Discussion

In this section I will briefly discuss the intuition behind the results presented in the

paper, explain what factors drive the main result, justify assumptions imposed and

point out possible extensions along with directions for future research. Discussion

is logically divided into three parts: results, assumptions and extensions that are

not by no means independent.

Results. The main result of this paper suggests that the DM benefits in terms

of ex-ante welfare, when experts provide their reports sequentially, and the costs of

lying/reading are small enough. Consider, the sequential communication protocol.

Intuitively, when the cost of reading information is small enough some types of the

DM read two reports if doing so provides them with exact information about the

true state of the nature. Then, if the cost of fabricating information is also small,

the second expert has enough incentive to fabricate information after observing

that the DM reads the first report (type 3 equilibrium with fabrications exists).

By reading the first report, the DM credibly signals that she is more likely to

choose the Expert 1’s preferable action. Hence, Expert 2 upon observing that the

DM reads the first report needs to fabricate with lower probability than Expert 1

does. As reports by two experts are substitutes,4 Expert 1 fabricates with lower

probability comparably to the case when experts are consulted simultaneously.

4Even though not perfect, as in order to get access to more precise report provided by Expert2, the DM needs to read the first report

20

Similarly, by not reading the first report, the DM credibly signals that she

is more likely to choose the Expert 2’s preferable action even if she requests no

information. This makes the Expert 2 to fabricate information with higher prob-

ability comparably to the case when experts are consulted simultaneously. Hence,

we have two effects on the DM’s ex-ante welfare. The first effect is positive and

is due to reduction in the fabricating probability by Expert 1, and the second

effect is negative and is due to increase in the fabricating probability by Expert 2.

As Corollary 3 shows, the second effect strictly dominates the first one. On the

other hand, for higher costs of lying/reading DM never reads two reports (type

1 equilibrium with fabrications) and this completely eliminates the first effect.

Consequently, for higher costs of lying/reading the simultaneous communication

protocol is preferable.

In the context of the example, suppose that we are asked to construct a com-

munication protocol between different levels of management prior to knowing the

exact characteristic of each manager. Then, if all such communication interactions

typically involve relatively low costs of manipulating and understanding informa-

tion, the sequential protocol is the correct way of communication between different

levels of management and will produce the highest profit for the firm. If by con-

trary, costs are relatively high, the simultaneous communication protocol is the

one that should be used.

Another interesting result is the one given in the statement of Lemma 3. Basi-

cally, the decision maker is listening to advice of just one expert when experts are

consulted simultaneously. Moreover, decision maker is listening to advice of the

expert whose preferable action she is prone to take even without reading any re-

ports. We call this a partisanship effect. The partisanship effect also persists, even

so in a smaller degree5, when the decision maker consults experts sequentially.

To have a better understanding of the partisanship effect consider the following

example. Suppose that you want to buy a new computer that is up to date and you

are mostly concentrating on choosing between Apple and Dell. Both companies

provide their detailed reports which contain the full description of their newest

model of computer and a comparison with competitors models. You know that

both, Apple and Dell are equally likely to fabricate their reports. Assume also,

5Recall that in both type 2 and type 3 equilibrium with fabrications, some types of DM readtwo reports.

21

you are prone to buy an Apple computer prior to reading reports. You will read

the report provided by Apple as from your prospective this report is more reliable

one. Why will you not read Dell’s report after already reading the Apple’s one?

Well, there is clearly no sense to do this in the case when Apple reports that

the new version of Dell is better. What if Apple’s report argues in favor of Apple

computer? Suppose, that you decided to read Dell’s report. Then the worst choice

you may face is so called ”word against word situation” when Dell will also argue

in favor of itself. However, even in this situation, you will be likely to choose

Apple as you were prone to it prior to reading reports. So, what is the reason of

reading this complicated and most likely, boring report if you anyway decide to

buy Apple?

Let’s go further in our reasoning. Suppose, that you follow my advice and read

just one report provided by Apple. Then it is likely that this report argues in favor

of Apple and you will fall even in more love with an Apple as its again proves its

competitiveness. I call this a polarization effect. To finish our story, suppose that

you finally bought your Apple computer. Then you could be even more prone to

buy Apple products in the future if you have a good experience using computer

purchased. The latest effect, however, is not captured by the current model and

requires its further investigation. To conclude, both partisanship and polarization

effects crucially rely on the assumptions of the model. Thus, strong partisanship

effect (when decision maker is listening to advice of only one expert) may not hold

if instead of binary action space we will allow for discrete or continuous action

space. The discussion of this and other model’s assumptions follow below.

Assumptions. Symmetry is the most crucial assumption that drives the re-

sults. In particular, I assume that experts have access to the same information,

experts bear the same cost of lying and the DM’s prior beliefs distributed symmet-

rically around 1/2. My results may not hold if we impose any kind of asymmetry

in the model. At the same time, while concentrating on the symmetric case we

mitigate the strategic advantages for one of the experts that could arise due to

asymmetry. For instance, if distribution of the DM’s prior beliefs is skewed to

the left or the cost of lying is lower for the Expert 1, then Expert 1 has strategic

advantage against Expert 2.

Another strong assumption is that experts’ utilities are independent of the true

state of the nature. While this is rather standard assumption in the literature on

22

persuasion, it makes the comparison with cheap talk literature outlined in the

introduction less appealing. Also, I assume that experts have access to identical

information. It will be interesting to explore how the results of these paper are

affected by different degree of substitutability in information possessed by two

experts. Moreover, we are limiting ourself to the case of two experts. As long

as symmetry is preserved, the results are unlikely to be affected if we allow the

decision maker to be advised by the board of experts. However, this modification

could produce quite different results if combined with one of other extensions

explained in more details below.

Extensions. The communication protocols presented in this paper are by

no means unique, even though are ones that easily come to mind and are easily

implementable. Also, as pointed out above, I consider the situation when commu-

nication protocol is chosen by the outside person who does not know the decision

maker’s type, or what is alternative, before the decision maker learns her type.

An interesting question for the future research would be to allow decision maker

choosing the communication protocol at the time she already learns her private in-

formation or type. Our analysis suggests that this will provide the decision maker

with additional tool of credibly signaling her type.

While the binary decisions are often observed in real life, checking robustness

of the results to situations where action is either discrete, but not binary, or con-

tinuous is of great interest. Coming back to our managerial example, suppose

that instead of choosing whether to fire workers or not, manager faces a challenge

of how many of them remain to be employed. Similarly, the interest of the first

expert is to fire as many workers as possible, while the goal of the second expert is

an opposite: keep as many workers employed as possible. Potentially, such modi-

fication could be modeled in the standard framework of the cheap talk literature

and could provide an additional insight on why the results of this paper lay in the

sharp contrast with ones implied by cheap talk literature.

Finally, the dynamic version of the model studied in this paper deserves its

special attention. It will be interesting to study the evolution of the decision

maker’s beliefs when she is involved in the sequential rounds of persuasion by

biased experts. Such modification is also going to reflect real life situations more

closely. How the result of Lemma 3 will be modified in this case? Will such

model be able to explain the polarization in opinions? These are quite important

23

questions that require further investigation.

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Appendix

Proof of Proposition 2. Only If. Let, Ti denotes the set of DM’s types who

read report i, i = 1, 2. Note, that T1 ∪ T2 = [cDM , 1 − cDM ] and T1 ∩ T2 = ∅.Similarly to the proof of Lemma 2, Expert 1 does not have incentive to deviate

and send message L1 when the true state of the nature is R iff cE ≥ λR(T1).

Since in equilibrium the DM does not read two reports Expert 2 never fabricates

his report if he observes that DM reads report 1. Suppose that on equilibrium

path DM does not read report 1. Then Expert 2 knows that DM’s type µ0 lies

in the set T2 ∪ [cDM , 1 − cDM ]C . Hence, the probability that Expert 2 faces DM

of type µ0 ∈ T2 conditional on θ = L and the fact that the DM does not read

report 1 equals to λL(T2)/(λL(T2) + λL([cDM , 1 − cDM ]C)). Therefore, Expert 2

does not have incentive to deviate and send message R2 when the true state of the

nature is L iff λL(T2)/(λL(T2) + λL([cDM , 1 − cDM ]C) ≤ cE or what is equivalent

λL(T2) ≤ cE = cEλL([cDM ,1−cDM ]C)1−cE

.

Define sets T1 and T2 in the way that follows: T1 = [cDM , F−1R (cE +FR(cDM))]

and T2 = (1 − F−1R (cE(FL(cDM) + FR(cDM))/(1 − cE) + FR(cDM)), 1 − cDM ]. It

is straightforward exercise to check that λR(T1) = cE and λL(T2) = cE. Suppose,

by contrary, that inequality in the statement of proposition does not hold. Then

T1∩ T2 = ∅ and T1∪ T2 is a strict subset of [cDM , 1− cDM ]. If λL(T1∩ T2) = 0 then

λL(T2) = λL(T2) as λL(T2) ≤ λL(T2) and hence, λR(T1) < λR(T1) contradicting

to λR(T1) ≤ λR(T1). Therefore, λL(T1 ∩ T2) 6= 0. Similarly, λL(T2 ∩ T1) 6= 0.

Observe, that T1 ∩ T2 > T2 ∩ T1 and taking into account that fR(µ)/fL(µ) is

strictly increasing function we have that:

λR(T1 ∩ T2)λL(T1 ∩ T2)

>λR(T2 ∩ T1)λL(T2 ∩ T1)

.

25

Then either λR(T1∩ T2) > λR(T2∩ T1) or λL(T1∩ T2) < λL(T2∩ T1). In the former

case:

λR(T1) ≥ λR(T1 ∩ T1) + λR(T1 ∩ T2) > λR(T1 ∩ T1) + λR(T2 ∩ T1) = λR(T1)

contradicting to λR(T1) ≤ λR(T1). In the later case:

λL(T2) ≥ λL(T2 ∩ T1) + λL(T2 ∩ T2) > λL(T1 ∩ T2) + λL(T2 ∩ T2) = λL(T2)

contradicting to λL(T2) ≤ λL(T2). These contradictions ends a proof of ”only if”

part.

If. Suppose, that the inequality in the statement of proposition holds. Define

the strategy of DM in the following way. DM reads report 1 whenever µ0 ∈T1 = [cDM , F

−1R (cE + FR(cDM))], reads report 2 whenever µ0 ∈ T2 = [cDM , 1 −

cDM ]/T1 and reads no reports otherwise. Note, that T1 = T1 and T2 ⊆ T2. Hence,

λR(T1) = cE and λL(T2) ≤ cE implying that experts have no incentive to deviate

and misreport the true state of the nature.

Proof of Proposition 3. As discussed above there exists type 1 equilibrium with

fabrications iff λR(T1) = cE, λL(T2)/(λL(T2)+λL([ cDM

1−ρ1 , 1−cDM

1−ρ2 ]C)) = cE and there

is no type of DM who reads two reports. Here as before, T1 = [ cDM

1−ρ1 ,ρ2

ρ1+ρ2) and

T2 = [ ρ2ρ1+ρ2

, 1− cDM

1−ρ2 ]. Since λR(T1) = FR( ρ2ρ1+ρ2

)−FR( cDM

1−ρ1 ) condition cE = λR(T1)

is equivalent to the first equation of system (6). Since λL(T2) = FL(1 − cDM

1−ρ2 ) −FL( ρ2

ρ1+ρ2) = FR( ρ1

ρ1+ρ2)− FR( cDM

1−ρ2 ) and λL([ cDM

1−ρ1 , 1−cDM

1−ρ2 ]) = λR([ cDM

1−ρ2 , 1−cDM

1−ρ1 ])

condition λL(T2)/(λL(T2) +λL([ cDM

1−ρ1 , 1−cDM

1−ρ2 ]C)) = cE is equivalent to the second

equation of system (6). Therefore, type 1 equilibrium with fabrications exists iff

the system (5) has a solution and there is no DM type who reads two reports.

First, I will prove that system (6) has a solution ρ1, ρ2 > 0 iff condition in

Proposition 5 does not hold.

Claim 1. System (6) has a solution ρ1, ρ2 > 0 iff condition (*) does not hold.

Proof. Only if. Suppose, that system (6) has a solution ρ1, ρ2 > 0. Then from the

first equation of system (6) it follows that:

FR(ρ2

ρ1 + ρ2) = cE + FR(

cDM1− ρ1

) > cE + FR(cDM),

26

and thus ρ2ρ1+ρ2

> F−1R (cE + FR(cDM)). Similarly, from the second equation of

system (6) it follows that:

FR(ρ1

ρ1 + ρ2) = (FR(

cDM1− ρ2

)+cEFL(cDM

1− ρ1))/(1−cE) > (FR(cDM)+cEFL(cDM))/(1−cE)

and hence, ρ1ρ1+ρ2

> F−1R ((FR(cDM) + cEFL(cDM))/(1− cE)). Therefore,

1 =ρ2

ρ1 + ρ2+

ρ1ρ1 + ρ2

> F−1R (cE1+FR(cDM))+F−1R ((FR(cDM)+cEFL(cDM))/(1−cE)).

If. Suppose, that 1 > F−1R (cE +FR(cDM))+F−1R ((FR(cDM)+cEFL(cDM))/(1−cE)). Define aL = F−1R (cE1+FR(cDM)) and aH = 1−F−1R ((FR(cDM)+cEFL(cDM))/(1−cE)). Note, that a ≥ aL iff FR(a) − FR(cDM) ≥ cE. With some abuse of nota-

tion define ρ1(a) as a solution to FR(a) − FR( cDM

1−ρ1 ) = cE for every a ≥ aL. Such

solution exists as LHS of the equality is continuous function in ρ1 that is lower

than cE if evaluated at ρ1 = 1− cDM and is greater or equal to cE if evaluated at

ρ1 = 0. Moreover, ρ1(a) is strictly increasing continuous function in its argument

as LHS of the equality is strictly increasing continuous function in a and strictly

decreasing continuous function in ρ1. Also it is easy to verify that ρ1(aL) = 0.

Define a ∈ (aL, aH) as a solution to (1− cE)FR(1− a)− FR(cDM) + cEFR(1−cDM

1−ρ1(a)) = cE. a is well defined as LHS of the equality is continuous function

that is greater than cE if evaluated at aL and is less than cE if evaluated at aH6.

With some abuse of notation define ρ2(a) as a solution to (1 − cE)FR(1 − a) −FR( cDM

1−ρ2 )+cEFR(1− cDM

1−ρ1(a)) = cE for every a ∈ [aL, a]. Such solution exists as LHS

of the equality is continuous function in ρ2 that is lower than cE if evaluated at

ρ2 = 1− cDM and is greater or equal to cE if evaluated at ρ2 = 0. Moreover, ρ2(a)

is strictly decreasing continuous function in its argument as LHS of the equality is

strictly decreasing continuous function in both a and ρ2. Also it is easy to verify

that ρ2(aH) = 0.

Finally, define function η : [0, 1]→ [0, 1] as η(a) = ρ2(a)ρ1(a)+ρ2(a)

for all a ∈ [aL, a],

η(a) = 1 if a < aL and η(a) = 0 otherwise. Since ρ1 is increasing and ρ2 is

decreasing continuous functions in a, η(a) is strictly decreasing continuous function

for all a ∈ (aL, a). Moreover, it is easy to verify that η(aL) = 1 and η(aH) = 0

6LHS(aL) = (1 − cE)FR(1 − aL) − FR(cDM ) + cEFR(1 − cDM ) > (1 − cE)FR(1 − aH) −FR(cDM ) + cEFR(1 − cDM ) = cE , LHS(aH) = (1 − cE)FR(1 − aH) − FR(cDM ) + cEFR(1 −cDM

1−ρ1(aH) ) < (1− cE)FR(1− aH)− FR(cDM ) + cEFR(1− cDM ) = cE .

27

implying that η is a continuous function on the whole domain. Therefore, by

Brouwer’s fixed point theorem, there exists a∗ ∈ [0, 1] such that η(a∗) = a∗. Since,

η(a) = 1 > a for all a ≤ aL and η(a) = 0 < a for all a ≥ a, a∗ ∈ (aL, a). Hence,

ρ1(a∗) > 0 and ρ2(a

∗) > 0 solve system (6).

Note, that the proof of ”If” part of claim implies that solution to system (6)

is unique as function η is strictly decreasing function in its argument. Next, I will

show that DM of every type does not read two reports iff ρ2 ≤ cDM

1−cDM.

Claim 2. The DM of every type does not read two reports iff ρ2 ≤ cDM

1−cDM.

Proof. Consider first, the DM of a type µ0, µ0 ≤ aSQ12 . We will show that DM does

not read report 2 after observing the message contained in report 1 iff ρ2 ≤ cDM

1−cDM.

Note, that if DM observes message R1 at stage 2 she learns that θ = R and hence,

she does not read report 2. Suppose, that DM observes message L1 at stage 2.

Then her posterior beliefs that the true state of the nature equals to R is µ(µ0, L1).

Since ρ20 = 0, DM does not read second report iff µ(µ0, L1) ≤ cDM . Applying

Bayes’ rule (3) we have that

µ(µ0, L1) =µ0ρ1

µ0 · ρ1 + (1− µ0) · 1,

and µ(µ0, L1) ≤ cDM iff µ0 ≤ cDM

ρ1(1−cDM )+cDM. The latest is true for all µ0 ≤ aSQ12 iff

ρ2ρ1+ρ2

≤ cDM

ρ1(1−cDM )+cDMor what is equivalent ρ2 ≤ cDM

1−cDM. Note, that in this way

we prove that ρ2 ≤ cDM

1−cDMis a necessary condition for DM of every type to not

read two reports, and sufficient condition for DM of type µ0 ≤ aSQ12 to not read

two reports.

Next, I will prove that DM of a type µ0 > aSQ12 does not read two reports if

ρ2 ≤ cDM

1−cDM. If DM reads two reports (report 1 and 2 if she observes message L1)

her payoff is uSQDM(3) = 1 − cDM − (1 − µ0)cDM − µ0ρ1cDM . If DM reads report

2 her payoff is uSQDM(2) = 1 − (1 − µ0)ρ2 − cDM . It is easy to verify that DM

prefers reading report 2 to reading both reports iff µ0 ≥ (ρ2−cDM )/cDM

ρ1+(ρ2−cDM )/cDM. Since

ρ2 ≤ cDM

1−cDM, (ρ2−cDM )/cDM

ρ1+(ρ2−cDM )/cDM≥ aSQ12 . Then the DM of type µ0, µ0 > aSQ12 will never

read two reports as reading report 2 is a dominant strategy for her.

Finally, I will finish the proof of proposition by defining function g1 in a way

that follows. Let with some abuse abuse of notation ρ1(cDM , cE) and ρ2(cDM , cE)

28

denote a solution to the system (5) if it exists. Define function g1 : R++ →R as g1(cDM , cE) = cDM

1−cDM− ρ2(cDM , cE) if solution to system (5) exists and

g1(cDM , cE) = 1+cDM+cE otherwise. Since condition (*) is satisfied if cDM ≥ 1/2,

g1(cDM , cE) > 1 iff condition (*) is satisfied. Then it follows directly from Claim 1

and Claim 2 that type 1 equilibrium with fabrications exists iff g1(cDM , cE) ∈ [0, 1].

The fact that g1 is strictly increasing function in both arguments follows from

ρ2(cDM , cE) being strictly decreasing in both arguments.

Proof of Corollary 1. Observe, that from the first equation of system (6) it

follows that ρ2ρ1+ρ2

> cDM

1−ρ1 and hence, ρ1ρ1+ρ2

< 1 − cDM

1−ρ1 . Then, from the second

equation of system (6) it follows that:

FR(ρ1

ρ1 + ρ2)− FR(

cDM1− ρ2

) = cE + cE(FR(ρ1

ρ1 + ρ2)− FR(1− cDM

1− ρ1)) < cE.

Note, that the last inequality combining with the first equation of system (6)

implies that ρ2 > ρ1. Hence it is enough to show that ρ1 > ρ∗. It follows from the

first equations of systems (5) and (6) that

FR(ρ2

ρ1 + ρ2)− FR(

cDM1− ρ1

) = FR(1/2)− FR(cDM

1− ρ∗).

Since ρ2 > ρ1,ρ2

ρ1+ρ2> 1

2and therefore, cDM

1−ρ1 >cDM

1−ρ∗ which implies that ρ1 > ρ∗.

Proof of Proposition 4. As discussed above there exists type 2 or type 3

equilibrium with fabrications iff λR(T1) = cE, λL(T2) + λL(T3) ≥ cE and ”=” if

ρ20 6= 0, and λL(T2)/(λL(T∅) + λL(T2)) = cE. Since λR(T1) = λR([ cDM

1−ρ1 ,ρ1

ρ1+ψ]) =

FR( ρ1ρ1+ψ

)−FR( cDM

1−ρ1 ) the first condition is equivalent to the first equation of system

(7). Since λL(T2) +λL(T3) = λL([ ψρ1+ψ

, 1− cDM

1−ρ2 ]) = λR([ cDM

1−ρ2 ,ρ1

ρ1+ψ]) = FR( ρ1

ρ1+ψ)−

FR( cDM

1−ρ2 ) the second condition is equivalent to the second equation of system

(7). Similarly, could be shown that the third condition is equivalent to the third

equation of system (7). Therefore, type 2 or type 3 equilibrium with fabrications

exists iff the system (5) has a solution that satisfies φ > ψ.

Only if. Let (ρ1, ρ2, ρ20) denotes a solution to (7) that satisfies φ > ψ. It is

straightforward exercise to show that φ > ψ implies φ > ρ2 > ψ. Suppose, by

way of contradiction that g1(cDM , cE) ≥ 0. Then either there is no solution to the

29

system FR( ψ′

ρ′1+ψ′ )− FR( cDM

1−ρ′1) = cE

(1− cE)FR(ρ′1

ρ′1+ψ′ )− FR( cDM

1−ψ′ ) + cEFR(1− cDM

1−ρ′1) = cE

(8)

or this solution satisfies ψ′ ≤ cDM

1−cDM. In the former case we have immediate

contradiction with the first and third equations of system (7) as φ > ρ2 > ψ.

Suppose, that solution to above system exists and satisfies ψ′ ≤ cDM

1−cDM. Then

ψ′ ≤ ψ. Comparing the first equation of above system with the first equation of

system (7) one can conclude that ρ1 ≥ ρ′1 and ψ/(ρ1 +ψ) ≥ ψ′/(ρ′1 +ψ′). Finally,

since φ > ψ and ρ2 > ψ the second equation of system (7) implies:

(1− cE)FR(ρ1

ρ1 + ψ)− FR(

cDM1− ψ

) + cEFR(1− cDM1− ρ1

) > cE,

which contradicts to ψ ≥ ψ′, ρ1 ≥ ρ′1, ρ1/(ρ1 + ψ) ≤ ρ′1/(ρ′1 + ψ′) and the second

equation of system (8).

If. Suppose, that g1(cDM , cE) < 0. Then there exists solution to system (8)

and this solution satisfies ψ′ > cDM

1−cDM. Define ρ∗20 > 0 in such a way that for

each ψ′ = ρ20+cDM

1−cDM. Then it follows from the third equation of system (7) and the

second equation of system (8) that if there is a solution to system (5) that satisfies

φ > ψ then ρ20 ≤ ρ∗20.

Next, define function ξ : [0, ρ∗20)→ (0, 1) in such a way that for each ρ20, ξ(ρ20)

denotes the solution to the first equation of system (7) if this equation is treated

as an equation with one unknown ρ1. Similarly, functions ζ : [0, ρ20∗)→ [0, 1] and

η : [0, ρ∗20) → (0, 1) are defined in such a way that for each ρ20, and ρ1 = ξ(ρ20),

ζ(ρ20) denotes the solution to the second equation 7 of system (7) and η(ρ20)

denotes the solution to the third equation of system (7) if these equations are

treated as equations with one unknown ρ2. As next claim shows ξ, ζ and η are

well defined continuous functions.

Claim 3. Functions ξ, ζ and η are well defined continuous functions. Moreover,

ξ is strictly increasing and ζ is strictly decreasing functions in their arguments.

Proof. First, I will prove that ξ is well defined strictly increasing continuous func-

tion in its argument. Fix ρ20 ∈ [0, ρ∗20]. Then LHS of the first equation of system

7More formally, ζ(ρ20) is the solution to the equation that is got from the second inequalityof system (7) when replacing ”≥” by ”=”.

30

(7) is strictly decreasing continuous function in ρ1 that is greater than cE if evalu-

ated at 0 and less that cE if evaluated at 1− cDM . Therefore, for all ρ20 ∈ [0, ρ∗20)

there exists a unique solution ρ1 ∈ (0, 1) to the first equation of system (7) and

hence, ξ is well defined continuous function in ρ20. Also, since LHS of the first

equation of system (7) is strictly increasing in ρ20 and strictly decreasing in ρ1,

ξ is strictly increasing function in its argument. Observe, that ξ(ρ∗20) = ρ∗ and

hence, ξ(ρ20) ≤ ρ∗ for all ρ20 ∈ [0, ρ∗20]. Therefore, ψ ≤ ξ(ρ∗20) for all ρ20 ∈ [0, ρ∗20].

Second, I will prove that ζ is well defined strictly increasing continuous function

in its argument. Again, fix ρ20 ∈ [0, ρ∗20]. Then LHS of the second equation of

system (7) is strictly decreasing continuous function in ρ2 that is greater than cE

if evaluated at 0 (FR( ξ(ρ20)ξ(ρ20)+ψ

) − FR(cDM) > FR( ψξ(ρ20)+ψ

) − FR(cDM) > cE) and

less than cE if evaluated at 1 − cDM . Therefore, for all ρ20 ∈ [0, ρ∗20) there exists

a solution ρ2 ∈ (0, 1) to the second equation of system (7) and hence, ζ is well

defined continuous function in ρ20. Also, since LHS of the second equation of

system (7) is strictly decreasing in ρ20 and ρ2, ζ is strictly decreasing function in

its argument.

Finally, I will prove that η is a well defined continuous function in its argument.

Fixing ρ20 ∈ [0, ρ∗20] note that LHS of the third equation of system (7) is strictly

decreasing continuous function in ρ2 that is greater than cE if evaluated at 0 and

less that cE if evaluated at 1− cDM . Therefore, for all ρ20 ∈ [0, ρ∗20) there exists a

solution ρ2 ∈ (0, 1) to the third equation of system (7) and hence, η is well defined

continuous function in ρ20.

If follows directly from Claim 6 that system (7) is equivalent to the following

system: ρ1 = ξ(ρ20)

ρ2 ≥ ζ(ρ20) and = if ρ20 6= 0

ρ2 = η(ρ20)

(9)

Observe, that for ρ20 = η(ρ∗20), ψ = φ = ρ2 implying that η(ρ∗20) > ζ(ρ∗20).

I claim that there exists ρ20 ∈ [0, ρ∗20) such that (ξ(ρ20), η(ρ20), ρ20) solves (9).

If for some ρ20 ∈ (0, ρ∗20), η(ρ20) = ζ(ρ20) then we are done. Suppose, that

η(ρ20) 6= ζ(ρ20) for all ρ20 ∈ (0, ρ∗20). Then, since η − ζ is a continuous function,

η(0) ≥ ζ(0) and so (ξ(0), η(0), 0) solves (9).

Next, I will prove that if g1(cDM , cE) < 0 then solution to system (9) is unique.

31

Claim 4. Suppose that g1(cDM , cE) < 0. Then there exists a unique solution to

(9).

Proof. Let (ρ1, ρ2, ρ20) is a solution to (9). Suppose, by contrary, that (ρ′1, ρ′2, ρ′20)

is another solution to (9). Let, without loss of generality ρ′20 ≥ ρ20. If ρ′20 = ρ20 6= 0

then ρ′1 = ρ1 and ρ′2 = ρ2 as ξ is strictly increasing and ζ is strictly decreasing

functions. If ρ′20 = ρ20 = 0 then ρ′1 = ρ1 as ξ is strictly increasing function.

Moreover, from the third equation of system (7) it follows that:

(1− cE)FR(ρ1

ρ1 + φ)− FR(

cDM1− ρ2

) = (1− cE)FR(ρ1

ρ1 + φ′)− FR(

cDM1− ρ′2

)

and hence, ρ′2 = ρ2.

Therefore, it should be the case that ρ′20 > ρ20. Then ρ′20 > 0, ρ′1 > ρ1 and

ρ′2 < ρ2 as ξ is strictly increasing and ζ is strictly decreasing functions. ThencDM

1−ρ1 <cDM

1−ρ′1, φρ1+φ

> φ′

ρ′1+φ′ and ψ

ρ1+ψ< ψ′

ρ′1+ψ′ . This implies that T ′3 ⊂ T3 and hence,

λL(T3) > λL(T ′3). Moreover, since λR(T1) = λR(T ′1), T1 < T ′18 and fR(µ)/fL(µ) is

strictly increasing function, λL(T1) > λL(T ′1).

Next, I will prove one technical claim that will be used in the proof of Claim

6 and the proof of Corollary 2.

Claim 5. Suppose that there is type 2 or type 3 equilibrium with fabrications.

Then FR( ρ1ρ1+ψ

)− FR(1/2) < cE and FR(1/2)− FR( ψρ1+ψ

) < cE.

Proof. Since cDM

1−ρ2 < 1/2 the second inequality of system (7) implies:

cE ≥ FR(ρ1

ρ1 + ψ)− FR(

cDM1− ρ2

) > FR(ρ1

ρ1 + ψ)− FR(1/2).

To prove that FR(1/2) − FR( ψρ1+ψ

) < cE observe that fR(µ) < fL(µ) for µ < 1/2

and hence,

FR(1/2)− FR(ψ

ρ1 + ψ) < FL(1/2)− FL(

ψ

ρ1 + ψ) = FR(

ρ1ρ1 + ψ

)− FR(1/2) < cE.

8I define the partial order over the intervals in the following way: [a, b] > [c, d] iff a > c andb > d.

32

Finally, I will provide a proof that 2cE + FR(2cDM) ≤ FR(1/2) is a sufficient

condition for existence of type 3 equilibrium.

Claim 6. Suppose that cE and cDM satisfy 2cE + FR(2cDM) ≤ FR(1/2). Then

there exists type 3 equilibrium.

Proof. First, it is easy to check that if cE and cDM satisfy 2cE + FR(2cDM) ≤FR(1/2) then condition (*) does not hold and hence, there exists equilibrium

with fabrications. Second, 2cE + FR(2cDM) ≤ FR(1/2) implies that cDM < 1/4.

Suppose, by way of contradiction, that there exists type 1 equilibrium with fab-

rications. Then from the first equation of system (6) and Corollary 2 it follows

that

cE > FR(1/2)− FR(cDM

1− ρ2) ≥ FR(1/2)− FR(

cDM1− cDM

1−cDM

) > FR(1/2)− FR(2cDM),

where the last inequality follows from the fact that cDM

1−cDMis increasing in cDM and

cDM < 1/4. We get a contradiction and hence, either type 2 or type 3 equilibrium

with fabrications exists.

Let (ρ1, ρ2, ρ20) solves (7) and suppose, by way of contradiction, that ρ20 = 0.

Then from the first equation of system (7) it follows that ρ1(1−cDM)+cDM < 1−ρ1or what is equivalent ρ1 < (1−cDM)/(2−cDM) < 1/2. Moreover, the first equation

of system (7) implies:

FR(ψ

ρ1 + ψ)− FR(

cDM1− ρ1

) = cE ≤ FR(1/2)− FR(2cDM)− cE

and hence, FR( ψρ1+ψ

) < FR(1/2)− cE which contradicts to FR(1/2)− FR( ψρ1+ψ

) <

cE.

Proof of Corollary 2. Suppose, that ρ20 > 0. Recall, that from the proof of

Proposition 7 we know that ψ < ρ2. Then first two equations of systems (5) and

(7) implies that: FR( ψρ1+ψ

)− FR(12) = FR( cDM

1−ρ1 )− FR( cDM

1−ρ∗ )

FR( ρ1ρ1+ψ

)− FR(12) = FR( cDM

1−ρ2 )− FR( cDM

1−ρ∗ )(10)

33

I will first prove that ρ1 < ρ∗ < ρ2. Note, that ρ1 can not be equal to ψ as in

this case ρ1 = ρ2 = ρ∗ that contradicts to ψ < ρ2. Hence, ψρ1+ψ

is either lower or

greater than 1/2. In the former case ρ1 < ρ∗ < ρ2 and we are done. In the later

case ψ < ρ2 < ρ∗ < ρ1 contradicting to ψρ1+ψ

> 1/2.

Next, I will prove that ρ2 − ρ∗ < ρ∗ − ρ1. It is easy to check that sufficient

condition for ρ2 − ρ∗ < ρ∗ − ρ1 to hold is cDM

1−ρ1 + cDM

1−ρ2 < 2cDM

1−ρ∗ . Define l1 =cDM

1−ρ∗ −cDM

1−ρ1 and l2 = cDM

1−ρ2 −cDM

1−ρ∗ . Then the inequality cDM

1−ρ1 + cDM

1−ρ2 <2cDM

1−ρ∗ could

be rewritten as: l2 < l1. Suppose, by way of contradiction, that l2 ≥ l1 and define

l = 12− ψ

ρ1+ψ. Then from the first equation of system (10) and the fact that fR

is strictly increasing function on (0, 1/2) it follows that l < l1. Summing up two

equations of system (10) and dividing into 2 we have:

FR(1/2 + l) + FR(1/2− l)2

−FR(1/2) =FR( cDM

1−ρ∗ + l2) + FR( cDM

1−ρ∗ − l1)2

−FR(cDM

1− ρ∗).

Since, l2 ≥ l1 > l and FR is a convex function on (0, 1/2), the last equality implies:

FR(1/2 + l) + FR(1/2− l)2

−FR(1/2) >FR( cDM

1−ρ∗ + l) + FR( cDM

1−ρ∗ − l)2

−FR(cDM

1− ρ∗).

(11)

Next thing to note is that it follows from Claim 5 and the first equation of system

(7) that FR(1/2) − FR( cDM

1−ρ1 ) < 2cE and FR( ρ1ρ1+ψ

) − FR(1/2) < cE. The last

inequalities imply that µ1 <cDM

1−ρ1 and µ2 >ρ1

ρ1+ψand therefore, function fR is

concave on ( cDM

1−ρ∗ − l, 1/2 + l). The latest however, contradicts to the inequality

(11). This contradiction ends a proof.

34