Persistence and Snap Decision Making∗ Tomoya Tajika†

June 11, 2018

Behaving consistently is widely observed, which implies that a person clings to his/her initial opinion and ignores future valuable information even if he/she has several chances to state an opinion. Reputation concern explains such behavior. However, in the literature, many studies ignore the case where an expert is allowed to refrain from stating any opinion in the first place. We consider this case in the present study. This possibility enables the sending of only future accurate information. Nevertheless, we find that the expert has an incentive to break silence at the first opportunity and that he/she also maintains his/her initial opinion, which is the driving force behind his/her unnecessary snap decision. Keywords: Reputation, herding, persistency of the initial opinion, snap decision JEL classification: D82, D83, D90

∗ The

author is grateful to Kimiyuki Morita and the participants of the Game Theory Workshop (March 4, 2018, Osaka University of Economics). The usual disclaimer applies. † Institute of Economic Research, Hitotsubashi University, Naka 2-1, Kunitachi, Tokyo, 186-8603, Japan. Email: [email protected]

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1. Introduction Behaving consistently is a widely observed and often valued phenomenon. Politicians who behave inconsistently are criticized for their behavior and economists who predict economic trends that oppose their previous predictions would not be trusted. Indeed, a preference for consistency is a key motivation behind an individual’s behavior (Cialdini, 2006). However, such consistent behavior can lead to inefficient results. For example, consider the following situation. A politician decides to invest in a public project that is likely to generate a profit at first. Over time, however, the project turns out to be likely to make a loss. In this setting, the incentive to behave consistently thus prevents the politician from withdrawing from the investment, which results in an inefficient investment. One might come up with similar examples by replacing politicians with economists or consultants. Many studies in psychology and behavioral economics have aimed to explain this behavior. In social psychology, for instance, Cialdini (2006) summarizes the evidence on and discusses consistent behaviors.1 In the literature, many studies examine reputation concern as a force driving consistent behavior (Li, 2007; Falk and Zimmermann, 2016). In these studies, an expert provides advice to an evaluator. The expert has two chances to provide advice and before each chance, he receives an independent signal whose accuracy positively correlates with the expert’s ability. The second signal is more accurate than the first signal. The expert has reputation concern about his ability. He thus takes care about the evaluator’s assessment of the expert’s ability. In this setting, although giving advice based on the second signal is preferable to the advisee, the expert has an incentive to cling to the first advice that is based on the first signal. To obtain the intuition, if the expert has high ability, which implies that he receives highly accurate signals, he is less likely to receive inconsistent signals. Hence, by pretending to have high ability, the expert behaves consistently. Although this explanation is intuitive, one may argue that if consistent behavior is inevitable and leads to less valuable advice, the expert would remain silent at the initial opportunity in order to conceal the inconsistency of the receiving signals. The present study examines this 1 The

relation to existing studies in economics is discussed in the next section.

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argument and proves that it is generally incorrect. We find that the expert has an incentive to break silence even if it leads to consistent behavior. We refer to the behavior of breaking silence in the first period as snap decision making. The intuition is as follows. By remaining silent in the first period, the evaluator believes that the expert’s decision is based on his second signal. On the contrary, if the expert employs the strategy of behaving consistently, the evaluator believes that the expert’s decision is based on his first signal. As the second signal is more accurate than the first one, the evaluator’s assessment of the expert’s ability based on the decision using the second signal becomes severe regardless of whether the advice matches the situation. Then, the incentive to avoid a severe assessment induces the expert to reveal only less accurate information. In our model, the incentive to behave consistently works as a commitment device to send only the first signal to the evaluator. Therefore, even when the expert can remain silent, there could be unnecessary snap decision making, which leads to an incorrect decision. To avoid such inefficiency, our study suggests that the client should silence the expert in the first period. The remainder of the paper consists of the following sections. In Section 2, we explain our model. Section 3 addresses the case where the expert is forced to provide some advice in each period and provides a sufficient condition that the expert behaves consistently. In Section 4, we analyze the case in which the expert can remain silent in the first period. Section 5 includes several extensions for consistent behavior. We show that our results for consistent behavior are robust even if the expert has arbitrary knowledge about his ability. However, snap decision making does not appear if the expert has no information about his ability. We also examine the case when the true state can change over time. We show that if the changing probability is extreme (i.e., the true state changes surely or does not change surely), consistent behavior remains to be observed. On the contrary, if the changing probability is 50/50, no such behavior can be an equilibrium.

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Relationship to the literature Reputation concern distorts an expert’s decision.2 The relation between reputation concern and persistency in behaviors has been investigated by several studies such as Kanodia, Bushman and Dickhaut (1989); Boot (1992); Prendergast and Stole (1996); Ferreira and Rezende (2007); Li (2007); Falk and Zimmermann (2016). Except for Prendergast and Stole (1996), as in our model, these studies deal with binary signal cases. Prendergast and Stole (1996) deal with a continuous signal case in which the true state and signals are drawn from normal distributions. They show that the reputation-concerned expert becomes conservative over time. Falk and Zimmermann (2017) provide experimental evidence for consistent behavior in a continuous state setting. As in the studies listed above, in our model, we replicate the result in which the expert behaves consistently. In this respect, Li (2007) and Falk and Zimmermann (2016) (hereafter, FZ) are the closest works to our study in the sense that the expert has two chances to recommend a choice and signal accuracy grows over time. Furthermore, FZ provide experimental evidence. In their model, in contrast to ours, there are only two types of signal accuracy. In our study, in contrast to Li’s and FZ’s models (and in most other studies), the set of signal accuracy is an arbitrary non-empty, non-singleton subset of the interval [1/2, 1] and the probability distribution is also arbitrary. In addition, while FZ assume that the evaluator does not observe the realized state, we assume that the evaluator observes it.3 In the examples of politicians and economists, because we can observe the consequences of their behavior, our assumption is more suitable. More importantly, our main contribution is examining the case that the expert can remain silent in the first period and show a result called a snap decision, which is not shown by Li and FZ (and by most other studies).4 This study also relates to the literature on herding behavior when several experts make sequential decisions, because persistency in the expert’s recommendation is seen as herding to his initial recommendation. In a seminal work, Scharfstein and Stein (1990) develop a 2 For

example, see Ottaviani and Sørensen (2006a,b), who provide a model with a general signal structure. assumes that the evaluator observes it. 4 Related to this point, Levy (2004) examines the case that the expert has a choice not to advise. 3 Li

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reputational herding model in which, each expert sequentially makes a decision and is concerned about his reputation. These authors show that in each equilibrium, the second expert follows the decision of the first expert regardless of the second expert’s signal. Among models of reputational herding and anti-herding behaviors, those of Levy (2004) and Sabourian and Sibert (2009) are also similar to ours. In both these models, in contrast to ours, while the expert’s private signal is drawn once, the public signal is also drawn. In Levy’s model, the public signal is drawn before the expert’s decision, while in Sabourian and Sibert’s model, the public signal is drawn after the expert’s decision. Further, after observing the public signal, the expert has the right to change his decision. Levy shows anti-herding to the public signal regardless of the high-ability expert’s signal. Related to our model, Sabourian and Sibert show the persistency of the expert’s decision after observing the public signal regardless of the realization. The results of Falk and Zimmermann (2016); Levy (2004); Sabourian and Sibert (2009) show a (semi) separating equilibrium associated with the expert’s ability. Therefore, the expert must know his ability in their models. However, in our model, most of the results are stated in a pooling equilibrium associated with the expert’s ability. Therefore, it does not matter whether the expert knows his ability or not.5, 6 Indeed, in Section 5, we generalize our result on consistent behavior to the case of an arbitrary information structure about the expert’s knowledge of his ability. This is one of our innovations. This point is important because in an ordinary reputational herding model, the result depends on whether the expert knows the information about his ability or not. Indeed, while in Scharfstein and Stein’s work, the experts do not know their ability, Avery and Chevalier (1999) show that if experts obtain sufficient information about their ability, they anti-herd. Closer to our model, Li (2007) discusses how the result changes regardless of whether the expert knows his ability or not. In contrast to Scharfstein and Stein’s reputational herding model, Banerjee (1992); Bikhchandani, Hirshleifer and Welch (1992) develop statistical herding models in which experts are only 5 Related

to this point, as Levy does, Effinger and Polborn (2001) demonstrate that anti-herding behavior is realized in an equilibrium. In contrast to Levy’s model, however, Effinger and Polborn assume that the experts do not know their ability. 6 In their Appendix, FZ also provide a continuous version of their model, which is a modification of that proposed by Prendergast and Stole (1996). In that model, all states as well as the accuracy of the signals and actions are continuous. FZ show a pooling equilibrium at which decisions are distorted (but not perfectly consistent).

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interested in making the correct choice. As the choices of previous experts also become a signal, the action taken by many experts becomes more attractive, which causes herding behaviors. Belief persistence, a well-known confirmation bias, also relates to our study. This states that once a belief has been formed, people are reluctant to change it (Nickerson, 1998). Rabin and Schrag (1999), for example, provide a model that describes the preference for belief persistence, which also creates persistency in behavior. In our study, unlike their explanation, the belief is (Bayesian) updated, but only the behavior persists. Eyster (2002) and Yariv (2005) also provide models that describe the preference for consistency. In contrast to the studies above, we also address another type of seemingly irrational behavior, namely snap decision making. This is our novelty and the key contribution of the current study. The logic used to explain this behavior can be seen as a kind of self-handicapping strategy— taking a lower-performance action to maintain one’s self-esteem (see Tirole, 2002). In our model, by reporting only the first signal, which yields lower performance, the expert can maintain his reputation related to his ability. Persistent behavior enables the reporting only of the first signal by working as a commitment not to report the second signal. This point is similar to the work of Ferreira and Rezende (2007), who show that information disclosure works as a commitment device to make behavior persistent. In contrast to ours, in the model of Ferreira and Rezende (2007), the benefit of persistent behavior is independent of reputation concern. However, in our model, the benefit of persistent behavior comes from pure reputation concern. In a different setting, Grubb (2011) considers the case where the expert needs to reveal the signal if he provides advice and shows that remaining silent is an equilibrium. This is in contrast to our result.

2. Model The model comprises two players: an expert (sender) and an evaluator (receiver). The expert decides to recommend one of two alternatives R = {x, y}. The state of the world is ω ∈ X = {x, y}. The prior belief that ω = x is true is p = 1/2. The expert has two opportunities to recommend. In opportunity t ∈ {1, 2}, he receives independent signal st ∈ S = {x, y} with

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accuracy θ t (i.e., p(st = ω | ω) = θ t ). The accuracy of the signal grows in terms of the odds θ2 θ1 ratio; that is, the odds ratio of the signal received at the second opportunity is 1−θ = (1+ α) 1−θ , 2 1

where α > 0.7 This assumption means that the expert receives a more accurate signal at his second opportunity. The accuracy growth rate α is common and known to each player, while accuracy at the first opportunity θ 1 , also referred to as the expert’s ability, is the expert’s private information.8 We assume that θ 1 is distributed according to cumulative density function F on set D ⊆ [1/2, 1] and that its associated probability density function is denoted by f .9 Assume that the cardinality of D is no less than 2.10 Let θ¯ := sup D and θ = inf D. Further, assume that θ¯ ∈ D and θ ∈ D. The state of the world is revealed to the evaluator after period 2 and she updates her belief regarding the expert’s ability. For each θ 1 ∈ D, let r(θ 1 ) = (r1 (· : θ 1 ), r2 (·, ·, · : θ 1 )) denote the interim strategy, where r1 (s1 : θ 1 ) : S → R1 = {x, y, œ} and r2 (r1, s1, s2 : θ 1 ) : R1 × S 2 → R2 = {x, y}. That is, r1 is the first report and r2 is the second report. In period 1, not providing a recommendation is allowed, which is denoted by œ. While the first report depends on the ability and first signal s1 , the second report also depends on the first report r1 and second signal s2 . Let β : R1 × R2 × X → ∆(D) denote a belief system regarding the expert’s ability. The expert obtains profit K ⩾ 0 only when his final recommendation coincides with the state of the world. The expert incurs no cost to change his recommendation. More importantly, the expert also gains profit from the evaluator’s assessment about the expert’s ability θ 1 (reputation concern). Precisely, let rt be the recommendation of period t. Then, his (vNM) utility is I(r2 = ω)K + E β [θ 1 | r1, r2, ω], where I is the indicator function: I(E) = 1 when the event E is true and I(E) = 0 otherwise. E β [θ 1 | r1, r2, ω] is the expectation of θ 1 with the evaluator’s updated belief regarding θ 1 . Assume that the expert is risk neutral. 1 θ 2 = (1+α)θ 1+αθ1 . 8 Similar results hold for the case where θ is referred to as ability. 2 9 If D is a countable set, for each x, f (x) is the probability that the expert’s ability θ = x. 1 10 The set D is allowed to be discrete or continuous.

7 Equivalently,

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1st opportunity Expert observes signal s1 ∈ S

Expert recommends r1 ∈ R1

2nd opportunity Expert observes signal s2 ∈ S

Expert recommends r2 ∈ R2

Evaluator observes state ω∈X

Evaluator computes reputation Eβ [θ 1 | r1, r2, ω].

Figure 1: Timeline of the model Figure 1 illustrates the timeline of this model.

Discussion on the specification of the accuracy growth rate We specify the accuracy growth rate as taking the form

θ2 1−θ 2

θ1 = (1 + α) 1−θ . Although this 1

assumption is non-standard, it makes the ability neutral to the choice of whether to change the recommendation if the expert receives an inconsistent signal. To be precise, consider the expert’s optimal choice. As shown later, if the expert receives inconsistent signals, behaving consistently is optimal if and only if 1 θ1 1 − θ2 = ⩾ Q, 1 − θ1 θ2 1+α where Q is a value less than 1. To show that Q < 1, reputation concern plays a crucial role. This is the main source of consistent behavior. If α depends on θ 1 , we may admit a separation with respect to ability, which becomes another channel of the reputation effect. To avoid confusing these two effects, we assume neutrality. In Section 5, we consider the case where α is a function of θ.

3. Equilibrium without the right of silence As the benchmark case, this section examines the case without the right of silence. In each period, the expert needs to make a recommendation. Formally, R1 = R2 = R = {x, y}. Let st ∈ {x, y} be the signal in period t. Let ps1 s2 be the posterior belief of the expert regarding the

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event that x = ω, which is calculated as θ1 θ2 , θ 1 θ 2 + (1 − θ 1 )(1 − θ 2 ) (1 − θ 1 )θ 2 = , θ 1 (1 − θ 2 ) + θ 2 (1 − θ 1 )

θ 1 (1 − θ 2 ) , θ 1 (1 − θ 2 ) + θ 2 (1 − θ 1 ) (1 − θ 1 )(1 − θ 2 ) = . θ 1 θ 2 + (1 − θ 1 )(1 − θ 2 )

px x =

px y =

p yx

py y

Note that p x x > p y x > 1/2 > p x y > p y y . We focus on pure-strategy perfect Bayesian Nash equilibria (PBE) in which the expert recommends the signal that he receives at the first opportunity.

Truthful recommendation Strategy r = (r1, r2 ) is the truthful recommendation if the expert recommends the signal that he received in each period. Precisely, for each θ 1 ∈ D and each history (r1, s1, s2 ) ∈ R × S 2 , r1 (s1 : θ 1 ) = s1 and r2 (r1, s1, s2 : θ 1 ) = s2 . First, we verify whether this strategy is an equilibrium. Let θrTruth be the ex post expected value of the expert’s ability θ 1 with respect to the evaluator’s 1 r2 ω belief after observing recommendations r1, r2 and state realization ω. Superscript Truth indicates that this is the assessment under the truthful recommendation. Suppose that the expert receives signal s1 = x at his first opportunity and recommends it. At the second opportunity, if the truthful recommendation is an equilibrium, the following two inequalities hold: Truth θ Truth px x θ2 xyy − θxxy + K (1 + α) = ⩾ Truth 1 − px x (1 − θ)2 θ x x x − θ Truth xyx + K

(1)

Truth θ Truth px y 1 xyy − θxxy + K . = ⩽ Truth 1 + α 1 − px y θ x x x − θ Truth x yx + K

(2)

The first inequality (1) is the condition that when the expert receives signal s2 = x, r2 = x is optimal. The second inequality (2) is the condition that when he receives signal s2 = y, r2 = y is optimal. To check whether these inequalities are satisfied, we have the following inequality. Truth Truth Truth Lemma 1. For each α ⩾ 0, θ Truth x x x − θ x yx > θ x y y − θ x x y .

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□

Proof. See Appendix A.1. From this calculation, when α = 0, the second inequality (2) becomes Truth θ Truth px y xyy − θxxy + K 1= ⩽ Truth , 1 − px y θ x x x − θ Truth x yx + K

Truth Truth Truth which is violated because θ Truth x x x − θ x yx > θ x y y − θ x x y . Therefore, in the neighborhood of

α = 0, the truthful recommendation is not an equilibrium. Proposition 1. For each K > 0, there exists α¯ > 0 such that for each α < α, ¯ the truthful recommendation is not a PBE. Truth Truth Truth The key to the result is inequality θ Truth x x x − θ x yx > θ x y y − θ x x y , which is rearranged as Truth Truth Truth θ Truth x x x + θ x x y > θ x y y + θ x y x . The left-hand side is the expected reputation gain obtained by

behaving consistently and the right-hand side is that by behaving inconsistently. Under truthful reporting, behaving consistently leads to a higher reputation gain.

Consistent recommendation We now consider another type of strategy, a consistent recommendation. Under this strategy, while the expert truthfully reports his signal in the first period, in the second period, he follows the first period choice irrespective of his second signal. Precisely, for each θ ∈ D, r1 (s1 : θ) = s1 and r2 (r1, s1, s2 : θ) = r1 if r1 (s1 : θ) = s1 .11 On an out-of-the-equilibrium path, that is, if r2 , r1 , we assume that the evaluator believes that the expert recommended his signals truthfully at each opportunity.12 Let us check whether the consistent recommendation is an equilibrium. Consider the second period. From the symmetry of the states, without loss of generality, the expert receives signal s1 = x and thus he chooses r1 = x. Then, as in the previous case, the following inequalities are

11 This 12 FZ

strategy requires consistency only on the equilibrium path. assume the same hypothesis.

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necessary: Cons θ Cons θ2 px x xyy − θxxy + K ⩾ . (1 + α) = Cons 1 − px x (1 − θ)2 θ Cons x x x − θ x yx + K

(3)

Cons θ Cons px y 1 xyy − θxxy + K = ⩾ Cons . 1 + α 1 − px y θ x x x − θ Cons x yx + K

(4)

Here, θrCons is the ex post expected value of the expert’s ability θ 1 under a consistent rec1 r2 ω ommendation.13 Superscript Cons indicates that this is the assessment under the consistent Cons Cons Cons recommendation. If θ Cons x x x − θ x yx > θ x y y − θ x x y , (3) trivially holds. More importantly, when Cons Cons Cons α is sufficiently small, inequality (4) also holds.14 Thus, if θ Cons x x x − θ x yx > θ x y y − θ x x y ,

the consistent recommendation is the optimal choice at the second opportunity. Indeed, the following lemma proves the inequality. Cons Cons Cons Lemma 2. Under any consistent recommendation, θ Cons x x x − θ x y x > θ x y y − θ x x y for each α ⩾ 0.

□

Proof. See Appendix A.1.

Therefore, if inequality (4) holds, a consistent recommendation is optimal in the second period. We can also show that recommending s1 is optimal at the first opportunity. Proposition 2. A consistent recommendation is a PBE if and only if inequality (4) holds. □

Proof. See Appendix A.1. The following is a sufficient condition for inequality (4), which is obvious from Lemma 2.

Lemma 3. For each K > 0, there exists α¯ > 0 such that for each α < α, ¯ inequality (4) holds. According to Proposition 2 and Lemma 3, we have the following. Corollary 1. For each K > 0, there exists α¯ > 0 such that for each α < α, ¯ a consistent recommendation is a PBE. Combining Proposition 1 and Corollary 1, with a sufficiently small α, among the equilibria where the expert recommends truthfully at the first opportunity, the expert behaves consistently. 13 The

formal calculation is given in the proof of Lemma 2. Cons Cons Cons Cons that θ Cons x yx varies by α. According to Lemma 2, θ xxx − θ x yx > θ x y y − θ xx y holds even when α = 0.

14 Note

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Figure 2: V(α), W(α) and 1/(1 + α) Example 1. Suppose that D = {1/2, 3/4, 1} and f (1/2) = a, f (1) = b and f (3/4) = 1 − a − b. Let V(α) :=

Cons θ Cons x y y −θ xx y +K Cons θ Cons xxx −θ x yx +K

and W(α) :=

Truth θ Truth x y y −θ xx y +K Truth θ Truth xxx −θ x yx +K

.15 Then, V(α) and W(α) are drawn in

Figure 2 (K = 1.2, a = b = 0.4). As depicted in Figure 2, when α is smaller than 0.2, inequality (4) is satisfied but (2) is violated. Then, only a consistent recommendation is a PBE.

4. The right to silence The previous section assumes that the expert necessarily recommends some alternative in all periods. However, as accuracy grows, only the second period signal is necessary for the efficient decision. Thus, there is no need for the first period recommendation. Our question is whether the expert remains silent in the first period if he is allowed. To formalize the question, we add the choice to remain silent in the first period. That is, r1 ∈ R1 = {x, y, œ}, where r1 = œ implies that the expert remains silent at the first opportunity. On the contrary, at the second opportunity, the expert must recommend one of the two alternatives, that is r2 ∈ R2 = {x, y}.

15 When

we write a := b, a is defined by b.

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Waiting strategy and a snap decision We focus on the waiting strategy, which requires that r1 = œ and r2 = s2 .16 In this paragraph, we assume that the evaluator has the following out-of-equilibrium belief: if the expert recommends something at the first opportunity, he reports his first signal regardless of his ability. That is, when r1 , œ, the evaluator does not update her belief regarding the expert’s ability but rather believes that the expert receives signal s1 = r1 . The remaining case is discussed in the next subsection. To compare this with the consistent recommendation, we assume inequality (4). In this case, if the expert deviates to report s1 in the first period, from Proposition 2, the consistent recommendation is optimal. We first check the optimality of the second period behavior after remaining silent. In the second period, if r1 = œ, as the second signal is more accurate than the first, the evaluator has no information about the expert’s first signal and believes that the expert recommends his second signal; hence, r2 = s2 is optimal. Consider the first period behavior. We compare two strategies: the waiting strategy and the consistent recommendation. The following lemma characterizes the condition that the waiting strategy is better than consistent recommendations. Lemma 4. There is a concave quadratic function of θ, h(θ : r, β) such that for the expert with θ 1 = θ, the waiting strategy is preferred to consistent recommendations if and only if h(θ : r, β) ⩾ 0. □

Proof. See Appendix A.2.

The function h(θ : r, β) is the difference between the expected utilities gained under the waiting strategy and the consistent recommendation when the expert’s ability is θ. Figure 3 depicts the shape of the h function. We now verify whether the waiting strategy can be an equilibrium. To verify this, from Lemma 4, we only need to check the value of h. In the proof of the following lemma, we can consider another type of waiting strategy such that r1 = œ and r2 = s1 . However, this strategy cannot be a PBE because deviating to r2 = s2 is necessarily profitable. Therefore, if r1 = œ, it must be r2 = s2 in the equilibrium.

16 We

13

h(θ : r, µ) θ 1

1 2

Figure 3: The shape of h(θ : r, µ) can show that h(1 : r, β) < 0, which implies that the expert with ability θ 1 = 1 deviates to recommend a choice in the first period and repeat the first recommendation. Lemma 5. Suppose that inequality (4) holds and θ¯ = 1. Suppose also that the evaluator does not update her belief about whether the expert remains silent. Then, the waiting strategy fails to be a PBE. □

Proof. See Appendix A.2.

The intuition is as follows. If the expert remains silent in the first period, his ability is assessed based on only his second period, while when he speaks in the first period, as the consistent recommendation is employed, his ability is assessed based only on his first signal. With a more accurate signal, if the recommendation is accurate, it is attributed to the growth of the signal but less to ability. On the contrary, if the recommendation is inaccurate, it is attributed to ability more than the case with a less accurate signal. Therefore, if the expert remains silent, his ability is discounted regardless of whether his recommendation is accurate. Thus, even when the expert has the right to silence in the first period, he has an incentive to employ a consistent recommendation. We call this situation a snap decision in the sense that the expert’s decision is based on the immediate (and less accurate) signal. We then verify whether the consistent recommendation strategy is a PBE. Unfortunately, under certain conditions, the consistent recommendation strategy also fails to be a PBE. Lemma 6. Suppose that θ = 1/2 and K > 3. Suppose also that the evaluator does not update her belief about whether the expert remains silent. Then, there exists α¯ such that for each α < α, ¯ the consistent recommendation fails to be a PBE.

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□

Proof. See Appendix A.2.

In the following subsection, by selecting the out-of-equilibrium belief adequately, we show the existence of a PBE that entails a snap decision.

Updating the expert’s ability according to the first period behavior Recall that the net payoff of employing the waiting strategy, h(θ : r, β), is a concave quadratic function of θ. Therefore, waiting is less profitable for the high-ability expert and for the low-ability expert, but not for the medium-ability expert.17 To consider the case that the evaluator updates his belief about the expert’s ability according to the first period behavior, consider the following strategy and belief. • If θ < [θ ∗, θ ∗ ]: r1 = s1 and r2 = r1 when r1 = s1 . • If θ ∈ [θ ∗, θ ∗ ]: r1 = œ and r2 = s2 . • The evaluator believes that if r1 = œ, the expert employs the waiting strategy and if r1 , œ, the expert employs the consistent recommendation strategy. With a sufficiently small α, as the truthful recommendation cannot be a PBE, a possible separating equilibrium such that r1 ∈ {s1, œ} must be the above strategy.18 Therefore, a separating equilibrium associated with ability indicates the pair of the above strategy and belief. Since the difference in the expected utilities between waiting and behaving consistently, denoted by h, is a concave quadratic function, if inequality (4) holds and the waiting strategy is employed by some experts, only this type of strategy is an equilibrium. This type of equilibrium is called tripartite (Figure 4). When θ ∗ = 1/2 or θ ∗ = 1, the tripartite equilibrium degenerates into a bipartite equilibrium (Figure 5). Possible separating equilibria associated with ability must be tripartite or bipartite.

17 This

result relates to Chung and Eső (2009), who consider a signaling model of a reputation-concerned expert. In their model, the expert chooses between informative and uninformative actions and they show that high- and low-ability experts prefer uninformative action. In our model, choosing to wait is a more informative action than not waiting because the expert reveals the second signal, which is more accurate than the first. This may enable the evaluator to estimate the expert’s ability more accurately. 18 For the uniqueness of the PBE such that r = œ, see also footnote 16. 1

15

h(θ : r, µ) r1 = s1

r1 = s1 r1 = œ

θ∗

1 2

θ∗

θ 1

Figure 4: A tripartite equilibrium h(θ : r, µ) θ∗ =

1 2

r1 = œ

r1 = s1 θ∗

1

r1 = s1

θ

1 2

θ∗

h(θ : r, µ) r1 = œ

θ θ∗ = 1

Figure 5: Bipartite equilibria In the second period, for each type that chooses r1 = œ, r2 = s2 is the best response. Suppose that (4) holds. First, by selecting the out-of-equilibrium belief accordingly, both pooling to wait and pooling not to wait (therefore, a consistent recommendation) are equilibria. Proposition 3. (i) The waiting strategy is a PBE. The out-of-equilibrium belief places probability 1 on θ 1 = θ. (ii) Suppose that inequality (4) holds and α is sufficiently small. Then, the consistent recommendation is a PBE. The out-of-equilibrium belief places probability 1 on θ 1 = θ. □

Proof. See Appendix A.2

In each equilibrium, the out-of-equilibrium belief is natural in the following sense. Consider the case that K = 0 and also consider the belief that if r1 , œ, θ 1 = θ. Under the belief, ∂h ∂θ (1

: r, β) = (1 + α)(θ œx x − θ œx y ) + α(θ œx x − θ) > 0. Then, since h is single-peaked, the peak

is greater than 1 and thus for each θ, h(θ : r, β) > h(θ : r, β) > 0. Under the strategy, the expert with θ 1 = θ is most likely to deviate to r1 , œ. Also consider the latter case. That is, if r1 = œ, the evaluator believes that θ 1 = θ. Under this belief,

∂h ∂θ (1/2

: r, β) = α(θ − θ x x x ) + θ x x y − θ x x x < 0. Since h is single-peaked, the peak is less

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than 1/2 and thus 0 > h(θ : r, β) > h(θ : r, β) for each θ. Therefore, the expert with θ 1 = θ is most likely to deviate to r1 = œ. In this sense, the out-of-equilibrium belief considered above is consistent with the spirit of the intuitive criterion (Cho and Kreps, 1987). We now discuss the separating equilibria. In each separating equilibrium, if a type employs a consistent recommendation, the type with sufficiently high ability necessarily reveals his signal to the evaluator in the first period. Lemma 7. There is no bipartite separation equilibrium associated with ability such that θ ∗ = 1. □

Proof. See Appendix A.2.

Intuitively, when θ 1 = 1, the expert does not need the second period signal for his recommendation since he knows the true state in the first period with certainty. Hence, he is only concerned about his reputation. The advantage of recommending a choice in the first period is only reputation concern. Therefore, this advantage vanishes if and only if the expert with θ 1 = 1 prefers to remain silent. This lemma implies that at each separating equilibrium, the efficiency of the final decision is not increasing in ability in the sense that the high-ability expert makes a snap decision, while the medium-ability expert does not. Finally in this section, let us mention the existence of a separating equilibrium associated with the expert’s ability. In fact, in general, with a sufficiently small α, there is no pure-strategy separating equilibrium considered above. Proposition 4. Suppose that K > 3. Then, there exists α¯ such that for each α < α, ¯ no separating equilibrium is associated with the expert’s ability. □

Proof. See Appendix A.2.

Although we assume a large K, this proposition is not trivial in the sense that no type makes a consistent recommendation. Indeed, for each K > 0, with a sufficiently small α, both the consistent recommendation and the waiting strategy are PBEs. By combining Lemmata 5 and 6, and Propositions 3 and 4, under certain conditions, only the strategies shown in Proposition 3 are PBEs.

17

5. Extension Unknown ability In the basic model, we assumed that the expert knows his own ability. In the literature on reputational herding, however, many studies assume that the expert does not know his ability. Our discussion is similar regardless of whether the expert knows his ability. To provide a concise discussion, we modify our game as follows. The expert does not know his ability, but he receives a signal about his ability, denoted by τ. Under receiving ability signal τ, the probability that ability is θ is denoted by f (θ | τ).19 Let T be denoted by the set of ability signals. We impose no restriction on T and ( f (· | τ))τ∈T . Therefore, this includes the known ability case as a special case such that T = D and for each θ ∈ D, f (θ | θ) = 1 and f (θ ′ | θ) = 0 for each θ ′ , θ.20 The information structure also includes the case that the expert obtains no private information about his ability by assuming that |T | = 1. Then, we have the following corollaries, which are generalizations of Proposition 1 and Corollary 1 respectively. The proofs are given in Appendix A.3. Corollary 2. Suppose that the expert receives a signal about his ability. Then, for each K > 0, there exists α¯ > 0 such that for each α < α, ¯ the truthful recommendation is not a PBE. Corollary 3. Suppose that the expert receives a signal about his ability. Then, for each K > 0, there exists α¯ > 0 such that for each α < α, ¯ a consistent recommendation is a PBE. In addition, consider the case that the expert has the option to remain silent. Suppose that the expert receives no information about his ability in the interim stage. Then, in contrast to the known ability case, no snap decision can be observed. Proposition 5. Suppose that the expert receives no information about his ability in the interim stage. Then, if K > 0, for a sufficiently small α, the waiting strategy is better than consistent recommendations in terms of expected utility. □

Proof. See Appendix A.3. 19 If 20 If

D is a continuum, f is the conditional density function of θ. D is a continuum, f (· | θ) is the density of a Dirac measure placing probability 1 on θ.

18

Different accuracy growth rates This subsection considers different accuracy growth rates, which are assumed to be constant in the basic model. In this paragraph, we first assume that the growth rate α is a function of ability θ 1 . Then, as in the basic model, the consistent recommendation is an equilibrium when ˜Cons θ˜Cons 1 xyy − θxxy + K ⩾ Cons , 1 + α(θ 1 ) θ˜x x x − θ˜Cons xyx + K

(5)

where m2 µ − m2 θ˜Cons , θ˜Cons , xxx = xxy = µ 1− µ ∫ θ 2 (1−θ) ∫ f (θ)dθ 1+α(θ)θ , θ˜Cons θ˜Cons xyy = ∫ x y x = ∫ θ(1−θ) f (θ)dθ 1+α(θ)θ

θ 2 (1−θ)(1+α(θ)) f (θ)dθ 1+α(θ)θ . θ(1−θ)(1+α(θ)) f (θ)dθ 1+α(θ)θ

Now, we have ˜Cons ˜Cons ˜Cons θ˜Cons x x x − θ x yx − (θ x y y − θ x x y ) =

1

∫ θ(1−θ) µ(1 − µ) θ(1−θ)(1+α(θ)) f (θ)dθ 1+θα(θ) f (θ)dθ 1+θα(θ) [∫ ( ) ∫ θ(1 − θ)(1 + α(θ)) θ(1 − θ) 1 × f (θ)dθ M − µ(1 − µ)θ f (θ)dθ 1 + θα(θ) 1 + θα(θ) 2 ] ) ( ∫ ∫ θ(1 − θ) θ(1 − θ)(1 + α(θ)) 1 + f (θ)dθ M − µ(1 − µ)θ f (θ)dθ , 1 + θα(θ) 1 + θα(θ) 2

where M = (1 − 2µ)m2 + µ2 . Assume that

∫

1+α(θ) 1+θα(θ) ′

and

1 1+θα(θ)

are weakly decreasing in θ. These

(θ) are guaranteed by assuming that max{−1, αα(θ) − α(θ)(1 + α(θ))} <

θα ′ (θ) α(θ)

for each θ ∈ D. For

example, α(θ) = a/θ b and α(θ) = aθ satisfy the assumption when a ⩾ 0 and b ∈ (0, 1). Then, ˜Cons ˜Cons ˜Cons as in the proof of Lemma 2, we can show that θ˜Cons x x x − θ x y x − (θ x y y − θ x x y ) > 0. If maxθ∈D α(θ) is sufficiently small, (5) holds for each θ ∈ D. Therefore, the consistent recommendation can also be an equilibrium. Another case of the different growth ratio is that although α is independent of θ 1 , α is distributed by a probability density function g on [0, α]. ¯ The result of this case is similar to the

19

previous one. As in the basic model, the consistent recommendation is an equilibrium when ˆCons θˆCons 1 xyy − θxxy + K ⩾ Cons , 1 + α θˆx x x − θˆCons xyx + K

(6)

where m2 µ − m2 θˆCons , θˆCons , xxx = xxy = µ 1− µ ∬ θ 2 (1−θ) 1+αθ f (θ)dθg(α)dα θˆCons , x y x = ∬ θ(1−θ) f (θ)dθg(α)dα 1+αθ

∬ θˆCons xyy = ∬

θ 2 (1−θ)(1+α) f (θ)dθg(α)dα 1+αθ . θ(1−θ)(1+α) f (θ)dθg(α)dα 1+αθ

Each θˆrCons takes a similar form to in the previous case. Therefore, we can show that θˆCons xxx − 1 r2 ω ˆCons ˆCons θˆCons ¯ is sufficiently small, the consistent recommendation is also an x yx − (θ x y y − θ x x y ) > 0. If α equilibrium.

Changing state The basic model of the current study (and that of most existing studies) is that the true state remains constant throughout the game. If the true state changes surely, the situation is virtually the same as in the basic model. Each player knows that the state in the first period is the opposite state in the second period. Then, an expert with high ability is more likely to receive inconsistent signals. In this case, if the expert recommends truthfully in the first period, he has an incentive to recommend the opposite one that he proposed in the first period. On the contrary, if the true state changes with a probability less than 1, the incentives of consistent and inconsistent behaviors are weakened. In particular, if the probability of changing the state is 1/2, the first period signal is useless and therefore the expert has no incentive to behave in such a way. To verify the above discussion, we assume that the true state changes over time with probability q. Formally, let ωt be the true state in period t. The probability that ω1 , ω2 is q. The signal in period t matches the state in the period with probability θ t = Pr(st = ωt ). The evaluator observes r1 , r2 , and ω2 . If the final recommendation matches the second state, the expert is rewarded monetarily.

20

As in the basic model, we first examine the case without the right of silence and focus on the truthful recommendation (i.e., r1 (s1 ) = s1 , r2 (r1, s1, s2 ) = s2 ). We have the following result. Proposition 6.

1. Suppose that q = 1. Then, for each K > 0, there exists α¯ > 0 such that for

each α < α, ¯ the truthful recommendation is not a PBE. If he receives consistent signals, the expert can make a profitable deviation by behaving inconsistently. 2. Suppose that q = 1/2. Then, for each K > 0, the truthful recommendation is a PBE. □

Proof. See Appendix A.3.

If q = 1, we can also show that a type of consistent recommendation is a PBE. The expert recommends the opposite signal that he received in the first period and clings to the first recommendation in the second period.

Competitive recommendation This section extends our model to the two-expert setting. There are two experts {i, j}. These experts share the same accuracy growth rate α but have different abilities, denoted by θi, θ j . Each of them simultaneously recommends at each opportunity t ∈ {1, 2}. Each expert knows his ability but not the other’s. The evaluator does not know the ability of either expert. Let θ kri1r j1ri2r j2 ω = E β [θ k | ri1, r j1, ri2, r j2, ω] be the evaluator’s assessment of expert k ∈ {i, j}’s ability after the state of the world has been revealed. The term rkt is the recommendation of expert k ∈ {i, j} in period t ∈ {1, 2}. Expert k ∈ {i, j} is only concerned about θ kri1r j1ri2r j2 ω . Let µ k , k ∈ {i, j} be the mean and m2k be the second-order moment of expert k’s ability. Now, we investigate the experts’ incentives. Case 1. ri1 = r j1 = x. Consider expert j’s decision. Suppose that s j2 = x. Then, recommending x is dominant. Consider the case that s j2 = y. By updating j’s belief about i’s ability, the expectation of i’s ability is

µix x y :=

θ j (1−θ j )(−αm2i +(1+α)µi ) 1+αθ j θ j (1−θ j )(−αµi +(1+α)) 1+αθ j

21

=

−αm2i + (1 + α)µi , −αµi + (1 + α)

which is independent of θ j . Then, the condition of recommending y is θ j x xri2 y y − θ j x xri2 x y + K µix x y 1 ⩽ . 1 − µix x y 1 + α θ j x xri2 x x − θ j x xri2 yx + K Since i recommends x, x is more likely. This incentive of statistical herding (i.e., term µix x y /(1 − µix x y ), Banerjee 1992; Bikhchandani et al. 1992) strengthens the incentive to make the recommendation consistent. Case 2. ri1 = y, r j1 = x. Consider expert j’s decision. Suppose that s j2 = x. The condition of recommending x is ( ) 1 − µiyx x θj 2 θ j y xri2 y y − θ j yxri2 x y + K (1 + α) ⩾ , µiyx x 1 − θj θ j y xri2 x x − θ j yxri2 y x + K where µiyx x = µix x y . When s j2 = y, the condition of recommending y is 1 − µiy x y 1 θ j yxri2 y y − θ j y xri2 x y + K ⩽ , µiyx y 1 + α θ j yxri2 x x − θ j yxri2 yx + K where µiyx y =

αm2i +µi αµi +1 .

As in the previous case, the incentive of statistical herding (i.e., term (1 − µiri s j1 s j2 )/µis j1 s j2 ) mitigates the incentive to make the recommendations consistent. However, the incentive of statistical herding is also mitigated by reputation concern since the changing opinion inspires the evaluator because of the possibility that the expert has low ability (anti-herding incentive; Avery and Chevalier, 1999; Levy, 2004).21 Therefore, the competition effect becomes ambiguous.

Reputation for the accuracy growth rate Now, consider the case that while θ is fixed, α is unknown to the evaluator, and the expert is concerned about the reputation of α. Denote the probability density function of α by g on R+ .

21 On

the contrary, if the expert does not know his own ability, this effect does not work.

22

Under the truthful recommendation, let αr1r2 ω be the assessment of α. Then, ∫ αx x x = ∫ αx x y = αx yx =

α (1+α)θ 1+αθ g(α)dα

∫

2

(1+α)θ 2 1+αθ g(α)dα ∫ (1−θ)2 α 1+αθ g(α)dα ∫ (1−θ)2 g(α)dα ∫ 1+αθ α 1 g(α)dα ∫ 1+αθ , 1 g(α)dα 1+αθ

= ∫ ∫

= ∫

α (1+α) 1+αθ g(α)dα (1+α) 1+αθ g(α)dα

1 α 1+αθ g(α)dα

1 1+αθ g(α)dα ∫ 1+α α 1+αθ g(α)dα αx y y = ∫ 1+α . g(α)dα 1+αθ

By observing these calculation, we can find that αx x x = αx y y and αx x y = αx yx . Then, reputation is independent of the first period recommendation. Therefore, this problem is reduced to a single period model. Hence, the truthful recommendation is optimal.

6. Conclusion This study investigates the incentive to behave consistently. Although models of consistent behavior are proposed by authors such as Li (2007), Sabourian and Sibert (2009), and FZ, this study not only extends their results, but also shows a novel result, called a snap decision, which is induced from consistent behavior. The combination of consistent behavior and a snap decision makes the expert’s recommendation less valuable. If the expert is a politician, such an inefficient decision could make the population suffer immense damage. As a remedy, our study may suggest that if the expert can obtain a more accurate signal in the future, the client should silence the expert at the first opportunity. Let us point out the limitations of our study, which offer possibilities for future research. One is that the signals and actions are binary. As FZ do, future work must aim to extend our model to a continuous one and investigate the incentive behind making a snap decision. Another is that in this study, reputation concern is only the expectations of others. Ottaviani and Sørensen (2006a) consider a more general form of reputation concern.

23

References Avery, Christopher N. and Judith A. Chevalier (1999) “Herding over the career,” Economic Letters, 63, pp. 327–333. Banerjee, Abhijit (1992) “A Simple Model of Herd Behavior,” Quarterly Journal of Economics, 107(3), pp. 797–817. Bikhchandani, Sushil, David Hirshleifer and Ivo Welch (1992) “A Theory of Fads, Fashion, Custom, and Cultural Change as Informational Cascades,” Journal of Political Economy, 100(5), pp. 992–1026. Boot, Arnoud W. A. (1992) “Why hang on losers? divestitures and takeovers,” Journal of Finance, 47, pp.1401–1423. Cho, In-Koo, and David M. Kreps (1987) “Signaling Games and Stable Equilibria,” Quarterly Journal of Economics, 102(2), pp. 179–221. Chung, Kim-Sau and Péter Eső (2009) “Signaling with career concerns,” Working paper. Cialdini, Robert (2006) Influence: The Psychology of Persuasion, Revised edition, Harper Business. Effinger, Matthias R. and Polborn, Mattias K. (2001) “Herding and Anti-Herding: A Model of Reputational Differentiation,” European Economic Review, 45, pp. 385–403. Eyster, Erik (2002) “Rationalizing the Past: A Taste for Consistency,” Working paper. Falk, Armin and Florian Zimmermann (2016) “Consistency as a Signal of Skills,” Management Science, Articles in Advance. Falk, Armin and Florian Zimmermann (2017) “Information Processing and Commitment,” Economic Journal, Forthcoming. Ferreira, Daniel and Marcelo Resende (2007) “Corporate strategy and information disclosure,” RAND Journal of Economics, 38, pp. 164–184. Grubb, Michael D. “Developing a Reputation for Reticence,” Journal of Economics and Management Strategy, 20, pp.225–268. Kanodia, Chandra, Robert Bushman and John Dickhaut (1989) “Escalation errors and the sunk cost effect: An explanation based on reputation and information asymmetries,” Journal of Accounting Research, 27, pp.59–77. Levy, Gilat (2004) “Anti-herding and strategic consultation,” European Economic Review, 48, pp. 503–525. Li, Wei (2007) “Changing One’s Mind when the Facts Change: Incentives of Experts and the Design of Reporting Protocols,” Review of Economic Studies, 74, pp. 1175–1194. Nickerson, Raymond S. (1998) “Confirmation Bias: A Ubiquitous Phenomenon in Many Guises,” Review of General Psychology, Vol. 2(2), pp. 175–220.

24

Ottaviani, Marco and Peter Norman Sørensen (2006a) “Reputational Cheap Talk,” RAND Journal of Economics, 37(1), pp. 155–175. Ottaviani, Marco and Peter Norman Sørensen (2006b) “Professional advice,” Journal of Economic Theory, 126, pp. 120–142. Prendergast, Canice and Lars Stole (1996) “Impetuous youngsters and jaded oldtimers: Acquiring a reputation for learning,” Journal of Political Economy, 104(6), pp. 1105–1134. Rabin, Matthew and Joel L. Schrag (1999) “First Impressions Matter: A Model of Confirmatory Bias,” Quarterly Journal of Economics, 114(1) pp. 37–82. Sabourian, Hamid and Anne C. Sibert (2009) “Banker Compensation and Confirmation Bias,” Working paper. Scharfstein, David S. and Jeremy C. Stein (1990) “Herd Behavior and Investment,” American Economic Review, 80(3) pp. 465–479. Tirole, Jean (2002) “Rational irrationality: Some economics of self-management,” European Economic Review, 46(4–5), pp. 633–655. Yariv, Leeat (2005) “I’ll See it When I Believe it ? A Simple Model of Cognitive Consistency,” Working paper.

A. Omitted proofs A.1. Proof in Section 3 Proof of Lemma 1. We first note that under the truthful recommendation, the ex post expectation of θ 1 is calculated as θ Truth xxx

=

θ Truth yyy

=

Truth θ Truth xxy = θyyx =

θ Truth xyx Let φ(θ) =

∫

1 1+αθ f (θ) 1 f (θ ′ )dθ ′ 1+αθ ′

=

, µ˜ =

θ Truth xyy ∫

=

∫ ∫ θ 2 θ 2 f (θ)dθ ∫ =∫ θθ 2 f (θ)dθ ∫ (1−θ)2 θ 1+αθ f (θ)dθ ∫ (1−θ)2 1+αθ f (θ)dθ

θ Truth yx x

=

θ Truth yxy

θφ(θ)dθ, m˜ 2 =

∫

θ3 1+αθ θ2 1+αθ

=

f (θ)dθ f (θ)dθ

(A.1) ∫ ∫

θ 2 (1−θ) 1+αθ f (θ)dθ θ(1−θ) 1+αθ f (θ)dθ

.

θ 2 φ(θ)dθ and m˜ 3 =

∫

θ 3 φ(θ)dθ, which are

respectively the first, second, and third moments of θ with density φ. Then, as long as |D| ⩾ 2, for each α ⩾ 0, from the mathematical facts, m˜ 3 µ˜ ⩾ m˜ 22 and m˜ 2 > µ˜ 2 . Note also that because

25

θ ∈ [1/2, 1], m˜ 2 ⩾ m˜ 3 and µ˜ > 1/2. Then, we have22 Truth Truth Truth θ Truth x x x − θ x yx − (θ x y y − θ x x y ) =

m˜ 3 µ˜ − m˜ 22 + (2 µ˜ − 1)m˜ 22 + m˜ 3 (m˜ 2 − µ˜ 2 ) + µ˜ 2 (m˜ 2 − m˜ 3 ) > 0. ( µ˜ − m˜ 2 )(1 − µ˜ − ( µ˜ − m˜ 2 ))m˜ 2 (A.2) □

Proof of Lemma 2. Let µ =

∫

θ f (θ)dθ, m2 =

∫

θ 2 f (θ)dθ, and m3 =

∫

θ 3 f (θ)dθ, which are

the mean, second-order moments, and third-order moments, respectively. Under the consistent recommendation, the ex post expected values of the expert’s ability are θ Cons xxx

=

θ Cons xxy = θ Cons xyx = θ Cons xyy =

∫ θ 2 f (θ)dθ ∫ = mµ2 , θ f (θ)dθ ∫ θ(1−θ) f (θ)dθ 2 ∫ = µ−m 1−µ , (1−θ) f (θ)dθ ∫ ∫ θ 2 (1−θ) θ 2 (1−θ 2 ) f (θ)dθ 1+αθ f (θ)dθ ∫ = ∫ θ(1−θ) , θ(1−θ 2 ) f (θ)dθ f (θ)dθ ∫ ∫ θ1+αθ 2 (1−θ) θθ (1−θ) f (θ)dθ 1+αθ f (θ)dθ ∫ 2 = ∫ θ(1−θ) = θ 2 (1−θ) f (θ)dθ 1+αθ f (θ)dθ

(A.3) θ Cons xyx ,

Cons Cons Cons Cons Cons Cons Cons θ Cons y y x = θ x x y , θ y y y = θ x x x , θ yx x = θ x y y , θ y x y = θ x yx .

Under the consistent recommendation, the difference becomes ∫ θ Cons xxx

−

θ Cons x yx

−

(θ Cons xyy

−

θ Cons xxy )

=

θ(1−θ) 1+αθ ((1

− 2µ)m2 + µ2 − 2µ(1 − µ)θ) f (θ)dθ . ∫ µ(1 − µ) θ(1−θ) f (θ)dθ 1+αθ

To show that the above equation is positive, note that ∫ θ(1 − θ)[(1 − 2µ)m2 + µ2 − 2µ(1 − µ)θ] f (θ)dθ = (µ − m2 )((1 − 2µ)m2 + µ2 ) − 2µ(1 − µ)(m2 − m3 ) = (1 − 2µ)(m3 µ − m22 ) + µ(ξ + (2µ − 1)σ) = (1 − 2µ)(µξ + σ µ2 − σ 2 ) + (µξ + (2µ − 1)σ µ) = (2µ − 1)σ(µ(1 − µ) + σ) + 2(1 − µ)µξ = (1 − µ)µ((2µ − 1)σ + 2ξ) + σ 2 (2µ − 1) > 0, | {z } positive 22 The

calculation is given in Appendix B.1.

26

where σ =

∫

(θ − µ)2 f (θ)dθ = m2 − µ2 and ξ =

∫

(θ − µ)3 f (θ)dθ = m3 − 3µm2 + 2µ3 =

m3 − µm2 − 2µσ, which are the variance and skewness, respectively. The last inequality follows from the fact that µ > 1/2 and ∫ (2µ − 1)σ + 2ξ =

∫

=

(θ − µ)2 (2µ − 1 + 2θ − 2µ) f (θ)dθ (θ − µ)2 (2θ − 1) f (θ)dθ > 0. | {z } positive

2 +µ 2 ˜ Let θ˜ = (1−2µ)m 2µ(1−µ) . Note that (1 − 2µ)m2 + µ − 2µ(1 − µ)θ < 0 if and only if θ > θ. Since ∫ θ(1 − θ)[(1 − 2µ)m2 + µ2 − 2µ(1 − µ)θ] > 0, θ˜ > θ. Then, since 1/(1 + αθ) is non-increasing 2

in θ, ∫

θ(1 − θ) ((1 − 2µ)m2 + µ2 − 2µ(1 − µ)θ) f (θ)dθ 1 + αθ ∫ (θ(1 − θ))((1 − 2µ)m2 + µ2 − 2µ(1 − µ)θ) f (θ)dθ ⩾ > 0, 1 + α θ˜ which shows that ∫ θ Cons xxx

−

θ Cons x yx

−

(θ Cons xyy

−

θ Cons xxy )

=

θ(1−θ) 1+αθ ((1

− 2µ)m2 + µ2 − 2µ(1 − µ)θ) f (θ)dθ > 0. ∫ µ(1 − µ) θ(1−θ) 1+αθ f (θ)dθ

This inequality holds even when α = 0.

□

Proof of Proposition 2. See the proof of Corollary 3. Proposition 2 is a special case of Corol□

lary 3.

A.2. Proofs in Section 4 Proof of Lemma 4. By calculation, the waiting strategy is preferred to consistent recommendations if and only if (θ + αθ 2 )(θ œx x − θ x x x ) + αθ(1 − θ)(θ œx x − θ œx y + K) ⩾ (1 − θ + αθ −

αθ 2 )(θ

27

xxy

− θ œx y ).

(A.4)

The derivation of inequality (A.4) is given in Appendix B.2. Let A(r, β) := θ x x y − θ œx y , B(r, β) := θ œx x − θ x x x and C(r, β) := θ œx x − θ œx y + K. Then, rearranging (A.4) yields23 h(θ : r, β) := αθ 2 (A(r, β) + B(r, β) − C(r, β)) + θ(B(r, β) + αC(r, β) + (1 − α)A(r, β)) − A(r, β) ⩾ 0. As A + B − C = θ x x y − θ x x x − K < 0, h(θ : r, β) is a concave quadratic function.24

□

Proof of Lemma 5. Recall that we assume that the evaluator does not update her belief regarding the expert’s ability when r1 , œ. Then, the assessments by the evaluator are computed as ∫ θ 2 f (θ)dθ θ x x x = ∫ θ f (θ)dθ , ∫ (1+α)θ θ 1+αθ f (θ)dθ θ œx x = ∫ (1+α)θ , 1+αθ f (θ)dθ

∫ θ(1−θ) f (θ)dθ θ x x y = ∫ (1−θ) f (θ)dθ ∫ 1−θ θ 1+αθ f (θ)dθ θ œx y = ∫ 1−θ 1+αθ f (θ)d.θ

(A.5)

Under the waiting strategy and belief system β, functions A, B, and C defined in the proof of Lemma 4 are calculated as follows: ∫

∫

A(r, β) = B(r, β) =

1 µ−m2 1−θ (µ − m − θ(1 − µ)) 1+α 2 1−µ 1+αθ ∫ 1−θ > (1 − µ) ∫ ∫ θ 1+αθ 1 θ(m2 (m − µθ) 2 1+αm /µ 2 ∫ θ − 1+αθ∫ θ <− µ 1+αθ µ 1+αθ

(1 − θ)(µ − m2 − θ(1 − µ)) ∫ 1−θ =0 (1 − µ) 1+αθ

− µθ)

=0

C(r, β) = θ œx x − θ œx y + K > 0. When θ = 1, h(1 : r, µ) = (1 + α)B < 0. Therefore, from Lemma 4, under the belief, if θ¯ = 1, the consistent recommendation is better than the waiting strategy for some θ ∈ D and thus the □

waiting strategy fails to be an equilibrium.

Proof of Lemma 6. We prove this lemma by showing that for a sufficiently small α, h(1/2; r, µ) > 0. When α = 0, h(1/2 : r, µ) = 0. Then, it is sufficient to show that limα→0 hα (1/2 : r, µ) > 0,25 there is no room for confusion, we abuse the notation to drop (r, β). addition, with any probability distribution, θ xxx > θ xx y and θ œxx > θ œx y . 25 By an abuse of the notation, we write h = ∂h for variable z. z ∂z

23 When 24 In

28

which is computed as ( ( )) ∂B ∂ A 1 lim hα (1/2 : r, β) = lim C+2 − <0 α→0 α→0 4 ∂α ∂α limα→0 ∂B ∂α

Note that = ( )2 (µ−m2 ) and limα→0 C = (1−µ)

m22 −µm3 µ2 m2 −µ2 µ(1−µ)

=

(

m2 µ

)2

−

m3 µ ,

limα→0

∂A ∂α

=

(m2 −m3 )(1−µ)−(µ−m2 )2 (1−µ)2

=

(m2 −m3 ) (1−µ)

−

+ K. Note also that | ∂∂αA | < 3/4 and | ∂B ∂α | < 3/4.26

Therefore, if K > 3, limα→0 hα (1/2 : r, β) > 0 and thus the consistent recommendation fails □

to be a PBE.

Proof of Proposition 3. (i) Consider the waiting strategy. Suppose that r1 (θ) = œ for each θ and the evaluator imposes the out-of-equilibrium belief that if r1 , œ, this places probability 1 on θ = θ. Then, since θ x x x = θ x x y = θ, when the expert recommends in the first period, there is no difference in reputation whichever the expert recommends in the second period and the reputation is below that when the expert remains silent. Therefore, we need only to consider the monetary reward. However, since the waiting strategy recommends the best choice with respect to the second signal, the other strategies are never better than the waiting strategy. Therefore, the waiting strategy is optimal for each θ, and thus the pair of the strategy and belief constructs an equilibrium. (ii) Consider the consistent recommendation. Suppose that the expert employs a consistent recommendation irrespective of his ability and the evaluator also has the out-of-equilibrium belief that if r1 = œ, this places probability 1 on θ = θ. Then, since θ œx x = θ œx y = θ, hθ (1/2 : r, β) = α(θ − θ x x y ) + (1 + α)(θ x x y − θ x x x ) < 0.27 Since h is single-peaked, h(θ : r, β) is decreasing in θ ∈ [1/2, 1]. We also have that 1 1 h(1/2 : r, β) = α(θ x x y − θ x x x − K) + (θ x x y − θ x x x + α(K + θ x x y − θ)) − (θ x x y − θ). 4 2 Note that θ x x y − θ x x x < 0 and this does not depend on α. Then, for a small α, h(1/2 : r, β) < 0 and thus h(θ : r, β) < 0 for each θ ∈ D. Hence, the consistent recommendation is better ( )2 ( )2 2 2 −m3 θ ∈ [1/2, 1], we can show that each of mµ2 , mµ3 , m1−µ and µ−m is larger than 1/4 and less than 1. 1−µ The difference is at most 3/4. 27 By an abuse of the notation, we write h = ∂h for variable z. z ∂z

26 Since

29

than waiting for each θ. When inequality (4) holds, as shown in Proposition 2, the consistent □

recommendation is a PBE.

Proof of Lemma 7. Suppose that there is a PBE where θ ∗ = 1. This implies that θ x x y ⩽ θ x x x < θ ∗ < θ œx y ⩽ θ œx x . Therefore, B > 0 and A < 0 and thus h(1 : r, β) > 0 and h(1/2 : r, β) > 0. Since h is a concave quadratic function, h(θ : r, β) > 0 for each θ ∈ [1/2, 1]. This shows that the waiting strategy is preferred to any consistent recommendation for each θ ∈ [1/2, 1], which □

is a contradiction.

Proof of Proposition 4. Suppose by contradiction that for each α, ¯ there exists a separating equilibrium with some α < α. ¯ Without loss of generality, we assume that there exists a separating equilibrium for each α. Consider a sequence of separating equilibria such that limα→0 A + B , 0. First, in this case, with a sufficiently small α, each separating equilibrium is bipartite, that is θ ∗ = 1/2 according to Lemma 7. To see this, limα→0 h(θ : r, β) = θ(A+ B) − A. This fact implies that for each θ, the expert waits if and only if θ(A + B) > A, which is a bipartite equilibrium. From Lemma 7, we state that A+B < 0. Then, in each separating equilibrium, the expert waits if and only if θ <

A A+B .

For the existence of a separating equilibrium, limα→0 h(1 : r, β) = B < 0. Furthermore, the threshold

A A+B

∈ (1/2, 1), which implies that A < 0. However, at this separation equilibrium,

the expert with lower ability remains silent, implying that A = θ x x y − θ œx y > θ ∗ − θ œx y > 0, which is a contradiction. Consider the case that limα→0 A + B = 0. Then, limα→0 h(θ : r, β) = − limα→0 A. Therefore, as long as limα→0 A , 0, there can only be a pooling equilibrium for a sufficiently small α. Now, consider limα→0 A = 0. From the assumption, this fact implies that limα→0 B = 0. If this is a sequence of bipartite separation equilibria, according to Lemma 7, θ ∗ < 1 and θ ∗ = 1/2. This fact implies that θ xr2 ω − θ œr2 ω > ε for each r2, ω ∈ {x, y} and for some ε > 0. This is because while θ œr2 ω is on average below θ ∗ , θ xr2 ω is on average above θ ∗ . Then, it must hold that limα→0 B < 0 and limα→0 A > 0, which is a contradiction. Therefore, the separation equilibrium must be tripartite with a sufficiently small α, that is 1/2 < θ ∗ < θ ∗ < 1. For the existence of a tripartite separation equilibrium for a sufficiently small α, since limα→0 h(θ : r, β) = 0, it must hold that limα→0 hα (1/2 : r, β) < 0, limα→0 hα (θ = 1 : r, β) < 0

30

∫

and limα→0 hα (θ : r, β) > 0 for some θ ∈ [1/2, 1] (see Figure 4). Let µˆ = ∫

mˆ k =

θ ∈[θ∗,θ ∗ ]

∫

θ k f (θ)dθ

θ ∈[θ∗,θ ∗ ]

f (θ)dθ

∫θ ∈[θ∗,θ

∗]

θ f (θ)dθ

θ ∈[θ∗,θ ∗ ]

f (θ)dθ

,

, k ∈ {2, 3}. Then, the following inequality is necessary: ( ( )) 1 ∂B ∂ A lim hα (1/2 : r, β) = lim C+2 − <0 α→0 α→0 4 ∂α ∂α

Note that limα→0 ∂B ∂α = ( )2 ( µ− ˆ mˆ 2 ) and limα→0 C = (1− µ) ˆ

mˆ 22 − µˆ mˆ 3 µˆ 2 mˆ 2 − µˆ 2 µ(1− ˆ µ) ˆ

=

(

mˆ 2 µˆ

)2

mˆ 3 µˆ ,

−

limα→0

∂A ∂α

=

(mˆ 2 −mˆ 3 )(1− µ)−( ˆ µ− ˆ mˆ 2 )2 (1− µ) ˆ2

=

(mˆ 2 −mˆ 3 ) (1− µ) ˆ

−

+ K. Further, | ∂∂αA | < 3/4 and | ∂B ∂α | < 3/4.28

Therefore, if K > 3, limα→0 hα (1/2 : r, β) > 0. This fact implies that with a sufficiently small α, h(1/2 : r, β) > 0, which disproves the existence of a tripartite separating equilibrium.

□

A.3. Proofs in Section 5 Proof of Corollary 2. Consider the second opportunity. Suppose that r1 = s1 = x and s2 = y. Then, the truthful recommendation (i.e., recommending y) is optimal when Truth Truth Truth p x y θ Truth x yx + (1 − p x y )(θ x y y + K) ⩾ p x y (θ x x x + K) + (1 − p x y )θ x x y .

The left-hand side is the expected utility of recommending y and the right-hand side is that of recommending x. Therefore, recommending y is optimal if and only if Truth θ Truth px y xyy − θxxy + K ⩽ Truth . 1 − px y θ x x x − θ Truth x yx + K

Here, p x y is computed as ∫ px y =

1 θ(1 − θ 2 ) f (θ | τ, s1 = x, s2 = y)dθ = . θ(1 − θ 2 ) + θ 2 (1 − θ) 2+α

θ ∈ [1/2, 1], we can show that each of The difference is at most 3/4.

28 Since

(

m ˆ2 µˆ

)2 ,

m ˆ3 m ˆ 2 −m ˆ3 µˆ , 1−µˆ ,

31

( and

µ− ˆ m ˆ2 1−µˆ

)2 is larger than 1/4 and less than 1.

Therefore, recommending y is optimal if and only if Truth θ Truth px y 1 xyy − θxxy + K = ⩽ Truth . 1 − p x y 1 + α θ x x x − θ Truth xyx + K

(A.6)

On the contrary, since the evaluator believes that the expert recommends truthfully, the assessTruth Truth Truth ments are given as (A.1). Therefore, θ Truth x x x − θ x yx − (θ x y y − θ x x y ) is also calculated as (A.2).

Then, with a sufficiently small α, inequality (A.6) is violated.

□

Proof of Corollary 3. Consider the second opportunity. Suppose that r1 = s1 = x. As in the proof of Corollary 2, at the second opportunity, recommending x is optimal when Cons θ Cons px x xyy − θxxy + K ⩾ Cons 1 − px x θ x x x − θ Cons x yx + K

if r2 = x

Cons θ Cons px y xyy − θxxy + K ⩾ Cons 1 − px y θ x x x − θ Cons x yx + K

if r2 = y.

Here, ∫

θθ 2 f (θ | τ, s1 = s2 = x)dθ ⩾ 1/2, θθ 2 + (1 − θ)(1 − θ 2 ) ∫ 1 θ(1 − θ 2 ) = f (θ | τ, s1 = x, s2 = y)dθ = . θ(1 − θ 2 ) + θ 2 (1 − θ) 2+α

px x = px y

Moreover, since the evaluator believes that the expert makes a consistent recommendation, the Cons Cons Cons assessments are given by (A.3). Therefore, θ Cons x x x − θ x yx − (θ x y y − θ x x y ) is calculated as in the

proof of Lemma 2. Since p x x /(1 − p x x ) ⩾ 1 and p x y /(1 − p x y ) = 1/(1 + α) and from Lemma 2, recommending x is optimal if and only if (4) holds. According to Lemma 3, with a sufficiently small α, (4) holds. Consider the first stage. Without loss of generality, we assume that s1 = x. To show that recommending x is optimal at the first opportunity, let r2 (y, x, s2 ) be the continuation strategy when r1 = y. Then, we have the following four cases: 1. r2 (y, x, s2 ) = x. Let S(θ) = θθ 2 + (1 − θ)(1 − θ 2 ), which is the probability that the expert

32

draws the same signals. Then, recommending x is optimal if and only if ∫

[

S(θ)(r x x (θ x x x + K) + (1 − r x x )(θ x x y ))

] + (1 − S(θ))(r x y (θ x x x + K) + (1 − r x y )(θ x x y )) f (θ | τ)dθ ∫ [ > S(θ)(r x x (θ yx x + K) + (1 − r x x )(θ y x y )) ] + (1 − S(θ))(r x y (θ yx x + K) + (1 − r x y )(θ yx y )) f (θ | τ)dθ.

Equivalently, from (A.3), ∫ Cons Cons Cons [θ(θ Cons x x x − θ yx x ) + (1 − θ)(θ x x y − θ yx y )] f (θ | τ)dθ > 0.

Cons Cons Cons Since θ ⩾ 1/2 and (θ Cons x x x − θ yx x ) > (θ yx y − θ x x y ) from Lemma 2,29 the inequality holds.

2. r2 (y, x, s2 ) = y. Recommending x is optimal if and only if ∫

[

Cons S(θ)(r x x (θ Cons x x x + K) + (1 − r x x )(θ x x y ))

] Cons + (1 − S(θ))(r x y (θ Cons x x x + K) + (1 − r x y )(θ x x y )) f (θ | τ)dθ ∫ [ Cons ⩾ S(θ)(r x x (θ Cons y y x ) + (1 − r x x )(θ y y y + K)) ] Cons + (1 − S(θ))(r x y (θ Cons ) + (1 − r )(θ ) + K) f (θ | τ)dθ. x y yyx yyy Equivalently, from (A.3), ∫

[

] Cons Cons Cons θ(θ Cons x x x − θ y y x + K) + (1 − θ)(θ x x y − θ y y y − K) f (θ | τ) ∫ Cons Cons (θ x x x − θ y y x + K) (2θ − 1) f (θ | τ) > 0.

Cons Since θ Cons x x x > θ y y x and θ > 1/2, the above inequality holds.

3. r2 (y, x, x) = x and r2 (y, x, y) = y. Here, consider the truthful recommendation. Recom-

29 Note

Cons Cons Cons that θ Cons yxx = θ x yx = θ x y y = θ yx y .

33

mending x is optimal if and only if ∫

[

Cons S(θ)(r x x (θ Cons x x x + K) + (1 − r x x )(θ x x y ))

] Cons + (1 − S(θ))(r x y (θ Cons x yx ) + (1 − r x y )(θ x y y ) + K) f (θ | τ)dθ ∫ [ ⩾ S(θ)(r x x (θ yx x + K) + (1 − r x x )(θ Cons y x y )) ] Cons + (1 − S(θ))(r x y (θ Cons y y x ) + (1 − r x y )(θ y y y ) + K) f (θ | τ)dθ. Equivalently, from (A.3), ∫ Cons Cons Cons [(θ Cons x x x − θ yx x )θ(2θ 2 − 1) + (1 − θ 2 )(θ yx y − θ x x y )(2θ − 1)] f (θ | τ)dθ > 0.

Cons Cons Cons This inequality holds since θ Cons x x x > θ y x x , θ y x y > θ x x y , θ > 1/2, and θ 2 > 1/2.

If inequality (4) holds, the truthful recommendation is inferior to the consistent recommendation and thus recommending x is optimal. 4. r2 (y, x, x) = y and r2 (y, x, y) = x. In addition, consider the truthful recommendation. Then, recommending x is optimal if and only if ∫

[

Cons S(θ)(r x x (θ Cons x x x + K) + (1 − r x x )(θ x x y ))

] + (1 − S(θ))(r x y (θ x yx ) + (1 − r x y )(θ x y y ) + K) f (θ | τ)dθ ∫ [ Cons ⩾ S(θ)(r x x (θ Cons yx y ) + (1 − r x x )(θ yx y + K)) ] Cons + (1 − S(θ))(r x y (θ Cons y x x + K) + (1 − r x y )(θ y y x )) f (θ | τ)dθ.

34

Equivalently, from (A.3), ∫

[

Cons Cons Cons θθ 2 (θ Cons x x x − θ x yx + K) + (1 − θ)(1 − θ 2 )(θ x x y − θ yx y − K)

] Cons Cons Cons + (1 − θ 2 )θ(θ Cons x yx − θ yx x − K) + θ 2 (1 − θ)(θ x y y − θ y y x + K) f (θ | τ) ⩾ 0 ∫ [ Cons ⇐⇒ θθ 2 (θ Cons x x x − θ x yx ) + Kθ(2θ 2 − 1) ] Cons + (1 − θ)(θ Cons x y y − θ y y x + K)(2θ 2 − 1) f (θ | τ)dθ ⩾ 0. This inequality holds. From the above results, recommending x is optimal and thus the consistent recommendation is □

an equilibrium.

Proof of Proposition 5. As in the derivation of (A.4), the waiting strategy is better than the consistent recommendation if and only if ∫ [θ(θ œx x − θ x x x ) + (θ 2 − θ)(θ œx x − θ œx y + K) − (1 − θ)(θ x x y − θ œx y )] f (θ)dθ ⩾ 0 ∫ αθ(1 − θ) ⇐⇒ H(α, r, β) := µB + C − (1 − µ)A ⩾ 0. 1 + αθ Note that H(0, r, β) = 0 and ∂H(0, r, β) ∂B ∂A = lim µ − (1 − µ) + (µ − m2 )C = (µ − m2 )K > 0, α→0 ∂α ∂α ∂α where ∂B m22 − µm3 , = α→0 ∂α µ2 m2 µ − m2 lim C = − + K. α→0 µ 1− µ

∂ A (m2 − m3 )(1 − µ) − (µ − m2 )2 , = α→0 ∂α (1 − µ)2 lim

lim

Then, for a sufficiently small α, H(0, r, β) > 0, and thus the waiting strategy is a PBE. Proof of Proposition 6.

□

1. To check the statement, without loss of generality, assume that

the expert receives signals s1 = x and s2 . Let ps1 s2 be the probability that ω2 = x. Then,

35

as in the basic model, the following inequalities are necessary: ˜Truth θ˜Truth px x xyy − θxxy + K ⩾ Truth 1 − px x θ˜x x x − θ˜Truth x yx + K Truth Truth θ˜x y y − θ˜x x y + K px y ⩽ Truth . 1 − px y θ˜x x x − θ˜Truth xyx + K

(A.7) (A.8)

Note that ps1 s2 satisfies px y (1 − q)θ 1 (1 − θ 1 ) + q(1 − θ 1 )2 1 = 1 − px y 1+α (1 − q)θ 1 (1 − θ 1 ) + qθ 12 q(1 − θ 1 )θ 1 + (1 − q)θ 12 px x . = (1 + α) 1 − px x q(1 − θ 1 )θ 1 + (1 − q)(1 − θ 1 )2 Here, θ˜rTruth is defined as 1 r2 ω E1 qP1 qP1 + (1 − q)P2 P1 E1 qP1 = qP1 + (1 − q)P0 P1 E0 qP0 = qP0 + (1 − q)P1 P0 qP2 E2 = qP2 + (1 − q)P1 P2

θ˜Truth xxx = θ˜Truth xxy θ˜Truth x yx θ˜Truth xyy

(1 − q)P2 E2 qP1 + (1 − q)P2 P2 (1 − q)P0 E0 + qP1 + (1 − q)P0 P0 (1 − q)P1 E1 + qP0 + (1 − q)P1 P1 (1 − q)P1 E1 + , qP2 + (1 − q)P1 P1 +

where ∫

θ 1 (1 − θ 1 )2 dF, E0 = 1 + αθ ∫ (1 − θ 1 )2 P0 = dF, 1 + αθ

∫

θ 12 (1 − θ 1 ) E1 = dF, 1 + αθ ∫ θ 1 (1 − θ 1 ) P1 = dF, 1 + αθ

If q = 1, px x = 1 + α. 1 − px x

36

∫ E2 =

θ 13

dF, 1 + αθ ∫ θ 12 P2 = dF. 1 + αθ

In addition, for each α ⩾ 0. Truth ˜Truth θ˜Truth θ Truth xyy − θxxy + K x x x − θ x yx + K = > 1. Truth ˜Truth θ˜Truth θ Truth x x x − θ x yx + K xyy − θxxy + K

The latter inequality is shown by Lemma 1. Therefore, for a sufficiently small α, (A.7) is violated and thus the truthful recommendation is not a PBE. 2. If q = 1/2, we can show that px y px x θ 1 (1 + α) 1−θ = >1> = , 1 − px x 1 − θ1 θ(1 + α) 1 − p x y ˜Truth θ˜Truth xyy = θxxx ,

˜Truth θ˜Truth x x y = θ x yx .

The first equality implies that the probability that the second recommendation matches the true state in the second period varies only by the second period signal. The second and third equalities imply that the reputation depends only on whether the second recommendation matches the revealed true state in the second period. Hence, the reputation is independent of the recommendation in the first period. Therefore, this problem is reduced to the single period model and thus the truthful recommendation is optimal. □

37

B. Omitted calculations B.1. Derivation of inequality (A.2)

Truth Truth Truth θ Truth x x x − θ x yx − (θ x y y − θ x x y )

= = = = = = = = =

˜ m˜ 2 − m˜ 3 ) + m˜ 2 ( µ˜ − m˜ 2 )2 (1 − µ˜ − ( µ˜ − m˜ 2 ))(m˜ 3 µ˜ − m˜ 22 ) − m˜ 2 (1 − µ)( ( µ˜ − m˜ 2 )(1 − µ˜ − ( µ˜ − m˜ 2 ))m˜ 2 (1 − µ)( ˜ m˜ 3 µ˜ + m˜ 2 m˜ 3 − 2m˜ 22 ) + ( µ˜ − m˜ 2 )[m˜ 2 ( µ˜ − m˜ 2 ) − (m˜ 3 µ˜ − m˜ 22 )] ( µ˜ − m˜ 2 )(1 − µ˜ − ( µ˜ − m˜ 2 ))m˜ 2 (1 − µ)( ˜ m˜ 3 µ˜ + m˜ 2 m˜ 3 − 2m˜ 22 ) + ( µ˜ − m˜ 2 )[m˜ 2 µ˜ − m˜ 3 µ] ˜ ( µ˜ − m˜ 2 )(1 − µ˜ − ( µ˜ − m˜ 2 ))m˜ 2 (1 − 2 µ˜ + m˜ 2 )m˜ 3 µ˜ + (1 − µ)( ˜ m˜ 2 m˜ 3 − 2m˜ 22 ) + ( µ˜ − m˜ 2 )m˜ 2 µ˜ ( µ˜ − m˜ 2 )(1 − µ˜ − ( µ˜ − m˜ 2 ))m˜ 2 (1 − 2 µ) ˜ m˜ 3 µ˜ + m˜ 2 m˜ 3 − (2 − µ) ˜ m˜ 22 + m˜ 2 µ˜ 2 ( µ˜ − m˜ 2 )(1 − µ˜ − ( µ˜ − m˜ 2 ))m˜ 2 (1 − 2 µ)( ˜ m˜ 3 µ˜ − m˜ 22 ) + m˜ 2 m˜ 3 − m˜ 22 + m˜ 2 µ˜ 2 ( µ˜ − m˜ 2 )(1 − µ˜ − ( µ˜ − m˜ 2 ))m˜ 2 m˜ 3 µ˜ − m˜ 22 − 2m˜ 3 µ˜ 2 + 2 µ˜ m˜ 22 + m˜ 2 m˜ 3 − m˜ 22 + m˜ 2 µ˜ 2 ( µ˜ − m˜ 2 )(1 − µ˜ − ( µ˜ − m˜ 2 ))m˜ 2 2 m˜ 3 µ˜ − m˜ 2 + (2 µ˜ − 1)m˜ 22 + m˜ 2 m˜ 3 − m˜ 3 µ˜ 2 + m˜ 2 µ˜ 2 − m˜ 3 µ˜ 2 ( µ˜ − m˜ 2 )(1 − µ˜ − ( µ˜ − m˜ 2 ))m˜ 2 2 m˜ 3 µ˜ − m˜ 2 + (2 µ˜ − 1)m˜ 22 + m˜ 3 (m˜ 2 − µ˜ 2 ) + µ˜ 2 (m˜ 2 − m˜ 3 ) . ( µ˜ − m˜ 2 )(1 − µ˜ − ( µ˜ − m˜ 2 ))m˜ 2

38

B.2. Derivation of inequality (A.4) The waiting strategy is preferred to consistent recommendations if and only if (θθ 2 + (1 − θ)(1 − θ 2 ))p x x (θ œx x + K) + (θθ 2 + (1 − θ)(1 − θ 2 ))(1 − p x x )(θ œx y ) + (θ(1 − θ 2 ) + (1 − θ)θ 2 )p x y θ œyx + (θ(1 − θ 2 ) + (1 − θ)θ 2 )(1 − p x y )[θ œy y + K] ⩾ [(θθ 2 + (1 − θ)(1 − θ 2 ))p x x + (θ(1 − θ 2 ) + (1 − θ)θ 2 )p x y ](θ x x x + K) + [(θθ 2 + (1 − θ)(1 − θ 2 ))(1 − p x x ) + (θ(1 − θ 2 ) + (1 − θ)θ 2 )(1 − p x y )]θ x x y ⇐⇒ θθ 2 (θ œx x + K) + (1 − θ)(1 − θ 2 )θ œx y + θ(1 − θ 2 )θ œyx + (1 − θ)θ 2 (θ œy y + K) ⩾ θ(θ x x x + K) + (1 − θ)θ x x y ⇐⇒ θθ 2 (θ œx x ) + (1 − θ)(1 − θ 2 )θ œx y + θ(1 − θ 2 )θ œy x + (1 − θ)θ 2 (θ œy y ) + (θ 2 − θ)K ⩾ θ(θ x x x ) + (1 − θ)θ x x y ⇐⇒ θθ œx x + (1 − θ)θ œx y + (θ 2 − θ)(θ œx x − θ œx y + K) ⩾ θ(θ x x x ) + (1 − θ)θ x x y ⇐⇒ θ(θ œx x − θ x x x ) + (θ 2 − θ)(θ œx x − θ œx y + K) ⩾ (1 − θ)(θ x x y − θ œx y ) ⇐⇒ (θ + αθ 2 )(θ œx x − θ x x x ) + αθ(1 − θ)(θ œx x − θ œx y + K) ⩾ (1 − θ + αθ − αθ 2 )(θ x x y − θ œx y ).

The first and second lines are the expected utility of following the waiting strategy and the third and fourth lines are that of following the consistent recommendation. Note that θ 2 − θ = The final line is derived by multiplying both sides by 1 + αθ.

39

αθ(1−θ) 1+αθ .