Learning from lazy liars

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(1)PONTIFICIA UNIVERSIDAD CATO LICA DE CHILE INSTITUTO MAGISTER EN. DE ECONOMIA ECONOMIA. TESIS DE GRADO MAGISTER EN ECONOMIA García, Varela, Javiera José Julio, 2019.

(2) PONTIFICIA UNIVERSIDAD CATO LICA DE CHILE INSTITUTO MAGISTER EN. DE ECONOMIA ECONOMIA. LEARNING FROM LAZY LIARS. Javiera José García Varela. Comisión Emilio Depetris Hugo Silva Felipe González Francisco Silva Nicolás Figueroa. Santiago, Julio de 2019.

(3) Learning From Lazy Liars Javiera Garcia Varela. ∗. July 2019. Abstract: This article studies a persuasion problem where biased experts acquires costly information in order to convince a decision maker (DM) to take their preferred action. The information acquired can not be falsified but it can always be concealed. This setting has been studied for preferences that are linear on the DM’s belief, Kartik et al. (2017) showed that adding experts is not necessary good for the DM because of strategic components, the DM might prefer to hire only one highly informed expert than two poorly informed. On this article I study if the timing of the hiring influences the effort decision and show that at least for the first expert, efforts remain as strategic substitutes when hired sequentially. However, I show that if experts are short lived and there is a long term DM, efforts become independent and full learning might be achieved when experts are hired sequentially. Finally I extend the analysis for other preferences.. ∗ Thesis written as a Master student at Department of Economics, Pontificia Universidad Católica de Chile. I would like to thank my thesis advisors Nicolás Figueroa and Francisco Silva. Any errors or omissions are my own responsibility. Comments at: [email protected]. 1.

(4) I. Motivation. Institutions are constantly making decisions and in order to increase the accuracy of those many times experts are solicited. There was a prevalent view on which the decision maker (DM) was always better when more experts were hired, specially if this experts have an agenda and might want to nudge the DM towards their preferred action. Cheap talk and hard evidence models study the problem of an uninformed DM who must rely on experts to take a decision. On both settings, the bias of the experts translates into information loss, hence one attractive solution to this problem is to add experts so that the biases can cancel each other, leading the DM towards the right action. Nonetheless because strategic behavior often appears, it is not obvious that more experts are better. Literature discuss if the informativeness of multiple experts is greater or not and if it differs depending on the bias of the experts (whether they are like or opposed minded) and the timing of the hiring. On cheap talk models when the hiring is simultaneous having two senders apparently leaded to a full revelation equilibrium for any possible bias (Krishna and Morgan 2001b), however Battaglini(2002) shows that after an equilibrium refinement, all equilibria with full revelation are ruled out, hence adding a second expert does not solve the information problem satisfactorily. On the other hand, when experts are sequentially hired, it will depend on the bias of the second expert if the DM’s welfare is greater with 2 experts. If they have the same bias the DM never benefits from hiring a second expert, she will choose the less biased one and ignore the other. Instead if experts have opposed bias, actively consulting both experts always helps the DM, moreover if a “rebutal” stage is allowed, there exists an equilibrium where all information is fully revealed (Krishna and Morgan 2001a). From hard evidence’s side Austen and Smith (1993) propose a model where experts are imperfectly informed about the state and they find that in some cases full revelation is possible even with a single expert but not when two experts are consulted simultaneously, proving that sequential hiring is superior to simultaneous. Another hard evidence paper is Kartik et al (2017) which endogenizes information acquisition and study the impact of adding experts. When it was just a revelation problem (exogenous acquisition of information) a lie could be caught by hiring another expert. Now on one hand, more evidence is needed to convince the DM, but on the other, the DM becomes less responsive to the evidence shown, so incentives for effort are not obvious. The main result of. 2.

(5) this paper is that experts efforts are strategic substitutes, therefore when more experts are hired the total effort that goes into the acquisition of information decreases, hence the decisions are less likely to be accurate. Kartik concludes that sometimes even when there is an opposite bias expert available, the DM would be better without him.. Section II presents Kartik’s model and explains how the expert efforts interacts, particularly why they behave as strategic substitutes. One of the consequences of efforts being substitutes is that the informativeness of any equilibrium of the two sender game is not Blackwell comparable to that of any equilibrium of either one expert game, and so the effects on the DM’s welfare depend on his utility function. However because experts receive independents signals conditional on the real state of the world, we know that in absence of strategic behavior two experts generates more information in Blackwell’s sense than one. Here I show why more effort and higher precision is preferable to any DM. Section III studies for an impartial DM what would happen if experts biases were public information among experts but the DM did not have access to it. This would increase the asymmetry between the experts and the DM. Here I show that each expert hired exerts less effort when the DM does not know their bias and therefore, any Bayesian DM would prefer to know the bias of the experts because that scenario leads to more accurate decisions. When the DM is imperfectly informed, the degree of substitution will decrease with the knowledge of the DM.. On section IV I study Kartik’s problem sequentially. I want to see if Kartik’s results holds for a different timing. First I prove that the DM can not provide higher incentives to the first expert by conditioning her action on the report that he gives her, this is, the first expert would exert the same amount of effort if the DM promise him that if he provides an informative signal no one else will be hired or even that a same biased expert will be hired next, that if she doesn’t promise anything. This because the beliefs after an informative signal are independent of the effort and therefore the problem that the expert face is the same for any of the cases mentioned above. Afterwards I show that the first expert when he knows that a second expert will be hired, always exerts less effort than if he was alone. However for the second sender this is not obvious, e2 (e1 ) is sometimes increasing on e1 and sometimes is decreasing, resembling a Stackelberg game. Section V studies the case of a long term DM and short lived experts. Under this setting hiring the experts sequentially (one on each period) eliminates the strategic component, allowing the study for n experts. To conclude I. 3.

(6) provide evidence that suggest that when several experts are hired, the real state of nature can be revealed and therefore full learning can be attained, which is comparable to the full revelation of the exogenous information models.. Finally section VI address the problem for non linear payoffs. Kartik’s results are strongly associated to the linear preferences, so here I study what happens with other utility functions, if Kartik’s results holds under different preferences. First to illustrate this I present a model where the experts only benefits when the DM takes their preferred action. Here I show that it depends on the DM’s initial belief if she is better with 1 or 2 experts. I provide examples for both cases. Then I present the effort problem for an unspecified utility function F for the experts, and for the first time on this article I give the DM a specific utility function: u(a, ω) = −(a − ω)2 , where a is the action she takes and ω is the real state of the world. I do this in order to show cases where the DM is better with one expert for some particular form of F . This section mainly shows examples on which Kartik’s results holds, this is, that because of strategic components, the DM might prefer to hire only one bias expert, even when an opposite bias expert is available.. II. Model. There is an unknown state of the world ω ∈ {0, 1} with prior probability π ∈ (0, 1) that ω = 1. A Bayesian decision maker (DM) wants to make a decision contingent on the true state but she can not acquire information about the sate by her own, so she relies on experts to collect that information and to inform her. Expert i chooses privately his effort level ei ∈ [0, 1] and obtains as outcome a private signal si ∈ {g, n, b}. With probability ei the expert obtains an informative signal {b, g} which can be verified by a third party if he decides to reveal his signal to the DM, and with probability 1 − ei he obtains an uninformative signal n. The cost of effort Ci (ei ) is given by a differentiable function which is strictly increasing, convex and satisfies Inada conditions. Both experts does this simultaneously.. The precision of the signal is given by p ∈ ( 21 , 1), such that:. 4.

(7) P r[si = g|ei , ω = 1] = P r[si = b|ei , ω = 0] = ei p P r[si = b|ei , ω = 1] = P r[si = g|ei , ω = 0] = ei (1 − p) P r[si = n|ei , ω] = 1 − ei Given that the senders might not receive the same information, even when the investigation was successful (they learned g or b), given any effort, in the absence of strategic behavior, two senders generates more information on Blackwell’s sense than one. After getting the signal si , sender i sends a message mi ∈ {G, B, N } to the DM. They can only send G and B if they receive the signal g or b respectively, but they can always report N (independent of which signal they really got). Senders are potentially biased, therefore they try to manipulate the DM in order to induce her to make the action that they prefer. Preferences are given by β − C1 (e) if the sender is upward biased and (1 − β) − C2 (e) otherwise, where β is the belief of the DM that the state of the world is 1.. To sum, there are three obstacles for learning, first of all is the precision p < 1, then the costly effort that translate into e∗ < 1 and the final obstacle is given by the potential “lies” (information concealing), P(M = N ) > (1 − e∗ ). Let βs1 be the posterior of sender 1 that ω = 1 after receiving s1 , βm1 ≡ π1 the posterior induced to the DM after receiving m1 and βπ1 ,m2 ≡ π2 the belief of the DM after receiving m1 and m2 . Sender 1’s expected payoff after learning s1 is Em2 [π2 | ρ = βs1 ]. Just like on Kartik’s, sender 1 (upwards biased) reports G whenever he receives signal g and N in all the other cases. This is known as sanitization strategies (Shin 1994). The posterior of the DM depends on the conjecture that she has on the effort of the senders (because ei is private information). This conjecture is denoted by êi ∈ [0, 1] and is the same one that the sender −i has on senders i effort. Inada conditions guarantees us an interior solution ei ∈ (0, 1).. Bayes rules implies that for an upward biased sender the update of beliefs are given by: β g = βG =. πp πp + (1 − π)(1 − p). 5.

(8) βb = βB = βN (ê1 ) =. π(1 − p) π(1 − p) + (1 − π)p. π(1 − ê1 p) π(1 − ê1 p) + (1 − π)(1 − ê1 (1 − p)). Where βs is the update of the sender after receiving s and βM is the DM’s update when she gets the message M . Hence for all ê1 ∈ (0, 1) : βb < βN (ê1 ) < βn Hence, when the sender obtains an uninformative signal (obtains signal n), the DM’s update is more pessimistic than the sender’s update. But whenever the DM obtains signal b and conceals it, the DM’s belief is more optimistic than the sender’s. This difference of beliefs is what makes manipulation possible. If effort were observable or disclosure were mandatory, then the unique equilibrium would have e∗1 = e∗2 = 0 (cf. Matthews and Postlewaite, 1985) and both senders’ welfare would be strictly higher.. To avoid repetition I focus on Sender’s 1 incentives and behavior; the analogs for Sender 2 are straightforward. The utility of the upward biased expert hired with a downward biased sender is given by: h i G(e1 , ê1 , ê2 ) ≡ πe1 p(1 − p)ê2 + (1 − π)e1 (1 − p)pê2 πGB (ê1 ê2 ) + h i πe1 p(1 − ê2 + ê2 p) + (1 − π)e1 (1 − p)(1 − pê2 ) πGN (ê1 ê2 ) + h i π(1 − e1 p)(1 − p)ê2 + (1 − π)(1 − e1 + e1 p)pê2 πN B (ê1 ê2 ) + h i π(1 − e1 p)(1 − ê2 + ê2 p) + (1 − π)(1 − e1 + e1 p)(1 − pê2 ) πN N (ê1 ê2 ) − C(e1 ) which gives the following marginal benefit function: M B(ê1 , ê2 ) ≡. η(ê2 )πGB + (λ(π) − η(ê2 ))πGN − h i η(ê2 )πN B + (λ(π) − η(ê2 ))πBB. 6.

(9) such that M B(ê1 , ê2 ) − C(e1 ) = 0 where λ(π) ≡ πp + (1 − p)(1 − π) η(ê2 ) ≡ πp(1 − p)ê2 + (1 − π)(1 − p)pê2 = ê2 p(1 − p) and the final beliefs are given by πGB =. πGN (ê2 ) = πN B (ê1 ) = πN N (ê1 , ê2 ) =. πp(1 − p) = π0 πp(1 − p) + (1 − π)(1 − p)p. πp(1 − ê2 + ê2 p) πp(1 − ê2 + ê2 p) + (1 − π)(1 − p)(1 − ê2 p). π(1 − ê1 p)ê2 (1 − p) π(1 − ê1 p)(1 − p) + (1 − π)(1 − ê1 + ê1 p)pê2. π(1 − ê1 p)((1 − ê2 + ê2 p)) (π(1 − ê1 p)((1 − ê2 + ê2 p)) + (1 − π)(1 − ê1 + ê1 p)(1 − ê2 p). To see that efforts are strategic substitutes we compare M B(ê1 , ê2 ) with M B(ê1 , 0) M B(ê1 , 0) = πGN (ê1 , 0) − πN N (ê1 , 0) λ(π) = πG − πN (ê1 ). η(ê2 ) M B(ê1 , ê2 ) = πGB + λ(π) λ(π). η(ê2 ) 1− λ(π). !. η(ê2 ) − πN B − λ(π). η(ê2 ) 1− λ(π). !. πGN πN N. notice that. P{m2 = B | m1 = G} = πG ê2 (1 − p) + (1 − πG )ê2 p =. 7. η(ê2 ) λ(π).

(10) P{m2 = N | m1 = G} = πN ê2 (1 − p) + (1 − πN )ê2 p >. η(ê2 ) λ(π). therefore πG =. η(ê2 ) η(ê2 ) πGB + (1 − )πGN λ(π) λ(π). but πn = P{m2 = N | m1 = G}πN B +(1− P{m2 = N | m1 = G})πN N. ! η(ê2 ) η(ê2 ) < πN B + 1− πN N λ(π) λ(π). because πN B < πN N we know that e∗1 < ealone , meaning that hiring another experts decreases the 1 incentives for effort and some DM might prefer hiring just one highly informed expert than 2 less informed.. II.a. Effort and Precision. More effort is better. If the DM could hire only one expert she would prefer to have one that exerts more effort, this is, (e1 , 0) >(Blackwell) (e01 , 0) for e1 > e01 . This because the DM could replicate the information provided by an expert that exerts e01 with the information given by e1 . The probability of obtaining each signal conditional on the state when the effort is e is given by:. P(N | ω = 1) = 1 − ep P(G | ω = 1) = ep P(G | ω = 0) = 1 − e(1 − p) P(N | ω = 0) = e(1 − p) If however the DM instead of receiving the expert’s signal with probability 1, receives the “true” signal with probability δ1 =. e0 e. when the real signal is G and with probability δ2 = 1 when N , the. DM would face the same problem that the expert with effort e0 would induce. 8.

(11) The DM receives the signal “good” or ”nothing” depending on the signal that the expert would report. Note that instead of fully revealing the experts message this strategy sometimes reports “good” even when the signal wasn’t g.. P(nothing | ω = 1) = ep(1 − P(good | ω = 1) = ep P(good | ω = 1) = (e − ep)(1 −. e0 ) + 1(1 − ep) = 1 − e0 p e. e0 + 0(1 − ep) = e0 p e. e0 ) + 1(1 − e + ep) = 1 − e0 (1 − p) e. P(nothing | ω = 1) = (e − ep). e0 + 0(1 − e + ep) = e0 (1 − p) e. Because the information structure of e0 can be replicated with e it is straightforward that any DM would prefer e to e0 .. Improving the precision. If the DM could choose the precision of the signals (for example by allowing only some sources of information that are more/less reliable) would she want to increase it? It is not obvious how the agents incentives to acquire information changes with the precision p. On one side if they obtain the desired signal the DM is more convinced that the real state is that one, but whenever they conceal information the DM will be more pessimistic. The DM and the expert’s belief are more alike now, therefore the gap between π B < π N < π n is smaller. I show that more precision actually translate into more effort, improving the DM’s welfare.. Lemma: If the DM could choose the precision she would always choose p = 1 because that maximizes effort since this is increasing on p. max eG 1. G G N G λ(π0 )eG 1 π1 + (1 − λ(π0 )e1 )π1 − C(e1 ). [eG 1]:. 0 G λ(π0 )[π G − π N (êG 1 )] − C (e1 ). 9.

(12) ∂eG 1 G N G 0 = λ0 (π0 )[π G − π N (êG 1 )] + λ(π0 )[π − π (ê1 )] > 0 ∂p The intuition behind this result is straightforward. Each time that the expert conceals information the DM knows that the state is the experts least favorite with probability 1 − e (instead of 1 − ep). Therefore her belief after N is the more pessimistic belief that the DM can form. On the other hand, whenever the sender announces the desired signal, the DM now is more convinced that that is the real state of the world, on the extreme case where p = 1, she knows for a fact after receiving g that the true state is 1. The expert’s maximization problem is now [p − π N ], hence the marginal benefit from acquiring information is greater for the expert when p increases. The question to be asked next is if the DM could choose between improving the precision or hiring another expert what would she choose? How much does the precision has to improve to prefer that over an extra expert. This could be useful when analyzing social welfare because effort is endogenous and costly, hence improving the information process could translate into hiring less experts.. III. Unknown Bias. On this section I sutdy for the symmetric case (π0 = 12 , p1 = p2 , c1 = c2 ), if knowing the bias of the expert benefits the DM or not. I assume that experts are aware of the other experts types, but the DM after receiving N does not know if it comes from an upward or downward biased expert. This would give an opportunity to the experts to deceive the DM because now her belief is closer to what the expert wanted to induce. Hence I want to see if “increasing the payoffs” for the experts translate into a higher effort. To star I will address the upward-downward case when hired simultaneously. For this case the upward biased expert problem is: h i G(e1 , ê1 , ê2 ) ≡ πe1 p(1 − p)ê2 + (1 − π)e1 (1 − p)pê2 πGB (ê1 ê2 ) + h i πe1 p(1 − ê2 + ê2 p) + (1 − π)e1 (1 − p)(1 − pê2 ) πGN (ê1 ê2 ) + h i π(1 − e1 p)(1 − p)ê2 + (1 − π)(1 − e1 + e1 p)pê2 πN B (ê1 ê2 ) + h i π(1 − e1 p)(1 − ê2 + ê2 p) + (1 − π)(1 − e1 + e1 p)(1 − pê2 ) πN N (ê1 ê2 ) − C(e1 ). 10.

(13) therefore M B(ê1 , ê2 ) ≡ η(ê2 )πGB + λ(π)πGN − η(ê2 )πGN − η(ê2 )πN B − λ(π)πN N + η(ê2 )πN N But we know that πnn = π0 because for the symmetric case ê1 = ê2 , and because the DM does not know the bias of the experts πN B = πB and πGN = πG , hence M B(ê1 , ê2 ) ≡ λ(π)(πG − π) Doing the same for the other cases we get the following MB functions:. upward-upward: M B(ê1 , ê2 ) ≡ λ(π)(πG − π) − ê2 [λ(π) − p(1 − p)](2πG − πGG − π) downward-downward: M B(ê1 , ê2 ) ≡ (1 − λ(π))(π − πB ) − ê2 [(1 − λ(π)) − p(1 − p)](π + πBB − 2πB ) This MB functions compared with the ones generated with a DM that knows the experts bias are smaller on every case (recall that for the symmetric case e1 = e2 ). This means that any unbiased DM would prefer to know the bias of the expert before hiring them because this induce the experts to exert higher effort, improving the informativeness of the equilibria.. upward-downward full knowledge: M B(ê1 , ê2 ) ≡ λ(π)(πG − π) + ê2 p(1 − p)[π − πN B ] where [π − πN B ] > 0.. upward-upward full knowledge: M B(ê1 , ê2 ) ≡ λ(π)(πG − π) − ê2 [λ(π) − p(1 − p)](2πG − πGG − π) 11.

(14) where πGN < πG .. downward-downward full knowledge: M B(ê1 , ê2 ) ≡ (1 − λ(π))(π − πB ) − ê2 [(1 − λ(π)) − p(1 − p)](π + πBB − 2πB ) where πB < πBN .. The unknown bias case is Blackwell comparable to the full knowledge, moreover the full knowledge dominates on Blackwell sense the unknown case. We know that for the symmetric case e1 = e2 and I just proved that eunknown < eknown , so we are comparing (eunknown , eunknown ) with 1 1 1 2 (eknown , eknown ) which we know is greater for both components and we know that (e1 , e2 ) >(Blackwell) 1 2 (e01 , e02 ) if e1 > e01 and e2 > e02 . Therefore any Bayesian DM prefers to know the expert type before hiring him.. IV. Kartik Sequentially. The first difference between the sequential and the simultaneous model is that the DM now can choose to hire the next expert or not. After receiving the first signal she might not want to hire another expert for different reasons, for example she could try to incentive the first expert by telling him that if her belief is over some threshold she won’t hire the opposed bias sender.. The new timing of the model is as following: Sender 1 chooses e1 , receives s1 and reports m1 to the DM and to the other sender. The DM updates her belief and decides whether to hire the second sender or not. If she doesn’t then the DM takes the action a∗ (β ≡ π1 ), where π1 is the DM’s belief after receiving message 1, and the payoffs are made. Instead, if she decides to hire the second sender, then the last one will exert effort e2 , obtain s2 and will report m2 . The DM will update her belief to π2 and then take the action a∗ (π2 ). Sender 1’s expected payoff after learning s1 is π1 (1 − P (A)) + P (A)Em2 [π2 | ρ = βs1 ], with P (A) the probability that the DM hires sender 2. Just like on Kartik’s, sender uses sanitization strategies.. 12.

(15) To minimize repetition I show the upward-upward case, the others are straightforward. Solving by backwards induction, we know that the second sender will only report M2 = {G, N } and the belief he faces is π1 , which is the DM’s posterior after receiving sender 1’s report.. The maximization problem for sender 2 is given by: G G G G M ax : (1 − eG 2 )π1,N (ê2 ) + e2 [(π1 p + (1 − π1 )(1 − p))βπ1 ,G + ((1 − π1 )p + π1 (1 − p))π1,N (ê2 )] − C(e2 ) eG 2. Recall that the the probability of obtaining signal g is denoted by λ(π), therefore the optimal effort is the one that satisfies: G G 0 G [eG 2 ] : −π1,N (ê2 ) + [λ(π1 )βπ1 ,G + (1 − λ(π1 ))π1,N (ê2 )] = C (e2 ). Simplifying some terms: 0 G λ(π1 )[βπ1 ,G − π1,N (êG 2 )] = C (e2 ). Then, the second sender faces the same trade-off of the simultaneous case 1 but for a different belief π. One way to see if with sequential timing e∗2 increases is to study how does e changes with π: 0 0 G G −βπ0 1 ,N (êG ∂eG 2 ) + [λ(π1 )βπ1 ,G + (1 − λ(π1 ))βπ1 ,N (ê2 )] + λ(π1 )[βπ1 ,G − π1,N (ê2 )] 2 = G ∂π1 C 00 (e2 ) − λ(π1 )βπ1 ,e2 ,N. (1). for some weak assumptions of C we can show that the denominator of equation (1) is always positive, hence we can focus exclusively on the numerator 0 G λ(π1 )[βπ0 1 ,G − βπ0 1 ,N (êG 2 )] + λ (π1 )[βπ1 ,G − π1,N (ê2 )] ≷ 0. rearranging some terms and denoting βπ1 ,G − π1,N (êG 2 ) by H we get π1. H 0 (π1 ) λ0 (π1 ) ≷ π1 λ(π1 ) H(π1 ). (2). It is easy to note that equation (2) its the comparison between two elasticities. A higher prior π1 will 1 marginal benefit of increasing effort is given by the marginal benefit between obtaining the favourite signal vs announcing N weighted by the probability of obtaining the attractive signal.. 13.

(16) translate into Sender 2 exerting more effort only if the elasticity of λ(π1 ) is grater than the elasticity of H(π1 ). In other words, the effort choice will increase if the probability of obtaining the signal that the sender likes grows faster than the decrease of the benefit of concealing unfavourable information.. Decision Maker’s problem: Because the DM wants to take the action according to the real state of the world her utility is given by: UDM.  1 = −1. if a = ω if a 6= ω. given those preferences, the DM’s strategy is to take action 1 if she believes that state 1 is more likely to happen and action 0 otherwise.  1 a(π) = 0. if π ≥. 1 2. if π <. 1 2. Hence for any posterior π induced by the experts, the utility of the DM is. UDM =.    π ∗ 1 + (1 − π) ∗ (−1)  . if. π>. 1 2.     (1 − π) ∗ 1 + (−1)π. if. π<. 1 2. DM’s hiring decision:. When does the DM benefits from hiring a second sender? A second sender would only be hired if the information that he can provide is relevant, meaning that for some message the DM would change her previous optimal action. If no matter what the second sender reports, the action induced by π1 remains the same, a∗ (π1 ) = a∗ (π2 ) for any m2 ∈ {B, N, G}, then the DM is indifferent between not hiring and hiring another expert. The prior beliefs (π1 ) such that the action of the DM won’t change for any m2 are given by βπ1 ,G ≤. 1 2. or βπ1 ,B ≥ 12 . The beliefs that satisfies this condition are π1 ≤ 1 − p and π1 ≥ p. The two zones that are generated by this beliefs will be denoted as the inaction space. Note that. 14.

(17) the inaction space is reduced when the precision improves (figure 1). On the extreme case that p = 1, the DM would always want to hire another expert.. Figure 1 Sender 1: h i max e1 λ(π)πG + π(1 − e1 p)(1 − p)ê2 + (1 − π)(1 − ê1 + ê1 p)pê2 πN B (ê1 ê2 ) + e1 h i π(1 − e1 p)(1 − ê2 + ê2 p) + (1 − π)(1 − e1 + ê1 p)(1 − pê2 ) πN N (ê1 ê2 ) − C(e1 ) therefore [e∗1 ] :. λ(π)[πG − πN N (ê1 , ê2 )] + ê2 [πN N (ê1 , ê2 ) − πN B (ê1 )] = C 0 (e1 ). and we know that on equilibrium e = ê, then e∗1 is the one that satisfies λ(π)[πG − πN N (e1 , e2 (e1 ))] + ê2 [πN N (e1 , e2 (e1 )) − πN B (e1 )] = C 0 (e1 ) where e2 (e1 ) is the one that solves (1 − λ(πN (ê1 )))[πN N (ê1 , e2 ) − πN B (ê1 )] = C 0 (e2 ) It’s straightforward that for the first experts effort still behaves as strategic substitutes because is the same problem that faces on the simultaneous scenario but for a different e2 . Because the substitution result was true for every e2 < 0 it still holds.. A natural thought is that in order to incentive the first expert the DM could commit to some action 15.

(18) if the former provides an informative signal. Nevertheless this would have no effect on the effort decision whatsoever. The DM could commit to stop hiring or to hire a same bias expert, however the problem that the first expert face under this scheme is the same that would face if an opposite biased expert is hired for sure after him. Recall that the MB for the expert 1 on the upward-downward case is M B(ê1 , ê2 ) ≡. η(ê2 )πGB + (λ(π) − η(ê2 ))πGN − h i η(ê2 ))πN B + (λ(π) − η(ê2 ))πBB. where η(ê2 )πGB + (λ(π) − η(ê2 ))πGN = πG This new “deal” of the DM would only affect the first line of the M B because if the expert reports the uninformative signal he faces the same problem than before (the DM doesn’t want to incentive the message N , therefore the offer it’s only for the informative signal). If no one is hired after message G the MB function does not change: M B(ê1 , ê2 ) ≡. h i λ(π)πG + η(ê2 ))πN B + (λ(π) − η(ê2 ))πBB. If a same bias expert is hired after the message G, M B(ê1 , ê2 ) ≡ h. λ(π)[λ(πG )ê2 πGG + (1 − λ(πG )ê2 )πGN ] i η(ê2 ))πN B + (λ(π) − η(ê2 ))πBB. (3) (4). but [λ(πG )ê2 πGG + (1 − λ(πG )ê2 )πGN ] = πG Hence the DM can not provide efforts incentives conditioning on the messages received.. Sequential vs Simultaneous. Here I study for the symmetric case if hiring sequentially translate into higher effort decisions. I find that e1 is indeed higher for the sequential case but if e2 is higher or not will depend of e∗1 . 16.

(19) I show that e2 (πN (ê1 )) is increasing for πN ∈ [0, π N ) and then decreasing for πN ∈ (π N , 1], and because πN is decreasing on ê1 we know then that for high values of ê1 , specifically ê1 ∈ (e1 , 1), e2 (ê1 ) is increasing on ê1 and for ê1 ∈ (0, e1 ), e2 (ê1 ) is decreasing on ê1 like a Stackelberg game. If C(e) is high enough we can assure this last scenario.. For π = 0.5, C 0 (e) = e we have that e∗2 (πN ). =. 1±. p 1 − 4(1 − λ(πN ))πN (λ(πN ) − p) 2(1 − λ(πN )). which has the following form for a given πN :. It’s straightforward to see that once we have ê1 , e2 can be calculated.. For e1 the problem is the following: e1 = 0.5[p − πN N (e1 )] + e2 (e1 )[πN N (e1 ) − πN B (e1 )]p(1 − p) seq sim seq sim > esim I show now that M B(esim 1 . 1 , e2 (e1 )) − C(e1 ) > 0 implying that e1. 17.

(20) Recall that esim for C 0 (e) = e and π = 0.5 is 1 esim = λ[πG − π] + e2 p(1 − p)[π − πN B (esim 1 1 )] and because we are on the symmetric case we know that e∗1 = e∗2 , then. e. sim. " !# 1 1 − esim p sim = (p − 0.5) + e p(1 − p) 1 − 2 1 − 0.5esim. The expression above depends only of p and we know that. ∂e∗ 1 ∂p. > 0.. sim Now I calculate eseq 2 (e1 ). e2 = (1 − λ(πN (ê1 )))[πN N (ê1 , ê2 ) − πN B (ê1 )]. (5). sim if we replace esim on (5) and then use that e∗2 on the equation for eseq , e2 (esim )). 1 1 we can calculate M B(e. I solved it numerically and obtained that M B seq (esim ) − esim > 0 for every p.. 18.

(21) V. Different Horizons. Lets consider the case where there are multiple periods and the real state of nature reveals at the end of those periods. The DM cares about the action that is taken on every period and she can consult as many experts as she wants on each period. However experts only care about the belief they induce in the period they are consulted. For this scenario the best hiring schedule is to hire only one expert on each period. By doing this the DM no longer face the substitution problem and can take a more accurate decision.. UDM = −(πt − at )2 Usender (at ) = at I study if full learning could be achieved on this scenario, given the strategic components are no longer present. I assume that the DM does not know which type of sender will come next but once she hires him she learns the sender’s type. That scenario gives us the following stochastic kernel:. πt. →.    Probability of obtaining the signal       eG (πt )λ(πt ) the message is G   G G 1 − e (πt ) + (1 − λ(πt ))e (πt ) the message is N      1 − eB (πt ) + λ(πt )eB (πt ) the message is N     B  e (π)(1 − λ(πt )) the message is B.   belief induced      → βπt ,G    → βπt ,NG     → βπt ,NB      → βπt ,B . →. πt+1. The first column is the probability of obtaining each of the signals and the second column is the posterior after that particular signal, where πt+1 can take one of this four values: βπt ,G =. βπt ,NG = βπt ,NB =. πt (1 −. πt p πt p + (1 − πt )(1 − p). πt (1 − eG (πt )p) πt (1 − eG (πt )p) + (1 − πt )(1 − eG (πt )(1 − p)). eB (π. t). +. πt (1 − eB (πt ) + eB (πt )p) B e (πt )p) + (1 − πt )(1 − eB (πt ). βπt ,B =. πt (1 − p) πt (1 − p) + (1 − πt )p. 19. + eB (πt )(1 − p)).

(22) If the real state of nature is 1, then those probabilities actually are:. πt. →.        . eG (πt )p. the message is G. 1 − eG (πt )p. the message is N.   1 − eB (πt )(1 − p)      eB (π )(1 − p) t. the message is N the message is B.   → βπt ,G       →β πt ,NG.  → βπt ,NB       →β. →. πt+1. πt ,B. πt+1 is the same than before because the DM doesn’t know the real state. we know that βπ,B < βπ,NG < βπ,NB < βπ,G We also know that p > λ, hence (1 − p) < (1 − λ) therefore if eG λβπ,G + (1 − eB (1 − λ))βπ,NB + (1 − eG λ)βπ,NG + eB (1 − λ)βπ,B = π then when we replace with the probabilities of each signal when ω = 1 we have that: eG pβπ,G + (1 − eB (1 − p))βπ,NB + (1 − eG p)βπ,NG + eB (1 − p)βπ,B = π + 4 So a priori (before the realization of the state) the expectation of the DM’s belief if he hires another sender is E(πt+1 | πt ) = π but ex post (after the realization of the state) the expectation is E(πt+1 | πt , ω = 1) = π + 4 That the expected beliefs goes up whenever the true state is 1 suggests that full learning could be attained if we had access to several senders. This is equivalent to π + 4 converging to 1 (or zero when ω = 0).. 20.

(23) V.a. Numerical analysis. On this section I simulate the hiring process for some given parameters to see if there is evidence for convergence to the real state. 2. I assume p = 0.752 and a cost function γ e2 with γ = 0.5. For this parameters the optimal effort for each type of agent given π is:. We can see that the effort of each type of expert is symmetric and strictly positive for every π ∈ (0, 1).. Once I have the optimal efforts for each π and knowing the probabilities of obtaining each signal when ω = 1, it can be calculated how many times the belief of the DM ends at each belief π. On Figure 2 it’s shown that as the number of experts hired increases the final belief of the DM collapses to 1, providing evidence for the convergence hypothesis. I did 1000 simulations with parameters p = 0.75 and an initial belief of π = 0.7 and then plotted the resulting distribution of beliefs.. 2 the. results holds for any value of p ∈ ( 12 , 1). 21.

(24) Figure 2.a. Figure 2.b. 22.

(25) Figure 2.c. VI. Non linear utility. The results presented so far are restricted to linear utility functions. It would be interesting to see what happens with different utility functions, if hiring 2 experts could be always better than hiring only one. In order to answer this I solve Kartik’s model sequentially for experts that only care about the action of the DM. On this section I show how for some initial beliefs the DM wants to hire two experts while for another beliefs she might be better off with only one. Afterwards I show that for π = 0.5 this type of experts (“extreme type”) exerts more effort that the one with linear preferences. Next, I present the problem for any utility function F . This is to show that under some specifications of F and a DM’s particular utility function her welfare is greater hiring only one expert, which is what Kartik suggested.. 23.

(26) VI.a. Extreme non linear utility. Utility is given by U upward U downward. = =. 1 2. 1(πT > ) − C(e) 1 − 1(πT >. 1 ) − C(e) 2. where πT is the DM’s final belief. It’s easy to see that when the utility of experts is no longer linear the martingale property does not necessarily holds. We know that λ(π)eG π G + (1 − λ(π)eG )π N = π because Bayesian update follows a martingale, but when the utility is a function of the DM’s belief, λ(π)eG F (π G ) + (1 − λ(π)eG )F (π N ) might be different from π. The timing of the model is as follows: The DM hires the first expert who will exert effort e1 and reveal either {G, N } or {N, B}, then the DM will update her beliefs and decide whether to hire a second expert or not and if so, what bias should it be. If the second sender is hired then he acquires an informative signal with probability e2 and sends message M2 to the DM following sanitization strategies. After receiving M2 the DM updates her belief, take the decision and payoffs are made. Solving by backwards induction we have:. Sender 2. Upward biased: The second experts faces a DM with a belief of π1 which is the belief induced by the first sender and βπ,G , βπ,N are the possible beliefs that he can induce. Unlike the previous section, the sender now only benefits from βπ,G or βπ,N if they are higher than. 1 2,. otherwise his. utility is 0. (1 − eG 2 )1(βπ,N ≥. max eG 2. If βπ,N ≥. 1 2. h 1 1 1 i ) + eG λ(π ) 1 (β ≥ ) + (1 − λ(π )) 1 (β ≥ ) − C(eG 1 π,N 1 π,G 2 2) 2 2 2. then no matter what signal the expert receive his favorite action will be taken. In this. situation the expert will not exert effort because this is costly and he can obtained the preferred action with out putting effort. Likewise if βG ≤. 1 2. the action taken will be his least favorite even. if he obtains favorable evidence and therefore exerting effort is just loss of utility. Taking this two. 24.

(27) conditions, the FOC for the cases where the expert exerts effort is [eG 2]:. λ(π1 ) − C 0 (eG 2)=0. equivalently for the downward bias case the problem is given by: h 1 1 1 i B 1 − (1 − eB 2 )1(βN < ) + e2 (1 − λ(π1 ))1(βB < ) + λ(π1 )1(βN < ) 2 2 2. max eB 2. [eB 2 ]: if C(e) =. e2 2 ,. 1 − λ(π1 ) − C 0 (eB 2 )=0. then e∗G = λ(π1 ). e∗B = 1 − λ(π1 ). Hence the optimal effort for sender 2 is given by:    0   G e2 (π1 ) = λ(π1 )     0. if π1 > if. 1 2.    0   B e2 (π1 ) = (1 − λ(π1 ))     0. 1 2. ≥ π1 ≥ (1 − p). if (1 − p) > π1. if π1 > p if p ≥ π1 ≥ if. 1 2. 1 2. > π1. DM’s Decision:. Knowing that expert 2 does not put effort whenever a∗ (π1 ) is his preferred action, the DM will only hire an upward biased sender when π1 <. DM’s hiring decision.                     . 1 2. and a downward biased sender for π1 > 12 .. if π1 > 1 − p. Nobody Downward biased Upward biased with probability Upward biased Nobody. 25. if. 1 − p ≥ π1 > 1/2. 1 2. if π1 = 1/2 if. 1/2 > π ≥ p if. p > π1.

(28) DM’s action.     . 1. if π2 > 1/2. with probability 0.5 takes action 1.    . 0. if. π = 1/2. if π2 < 1/2. When π1 = p if the downward bias expert is hired and he obtains signal b, then the DM’s posterior will be. 1 2. and therefore she takes action 1 with probability. 1 2. (the same goes for π = p and an. upward biased experts obtaining g).. Sender 1:. For sender 1 the problem is slightly different. We know that for some values of π0 sender 1’s effort will be 0, particularly if he is upward biased and π ∈ (p, 1] sender 1 will exert 0 effort because he knows that even if the downward sender announce B, action 1 will be taken. This is equivalent to not hiring any sender and therefore is not an interesting case to study.. We also know that π1N < 1/2 < π1G , meaning that whenever the upward bias sender obtains good news the second sender hired will be downward bias. If however he gets the bad signal or no signal at all the posterior of the DM will be lower than 1/2 and an upward sender will be hired at t = 2.. The utility of sender 1 then will depend of:. • U = 1 if he obtained a good signal and the second sender did not reverse it. This happens G with probability 1 − (1 − λ(π G ))eB 2 (λ(π )). • U = 0 if he obtained a good signal and the second sender reverse it. This happens with N probability (1 − λ(π G ))eB 2 (λ(π )). • U = 1 if he obtained a bad signal and the next sender reverse the belief. This happens with N probability λ(π B )eG 2 (λ(π )). • U = 0 if he obtained a bad signal and the next sender did not reverse the belief. This happens N with probability 1 − [λ(π B )eG 2 (λ(π ))]. 26.

(29) • U = 1 if he didn’t obtain a signal and the next sender reverse the belief. This happens with N probability λ(π0 )eG 2 (λ(π )). • U = 0 if he didn’t obtain a signal and the next sender does not reverse the belief. This N happens with probability 1 − [λ(π0 )eG 2 (λ(π ))]. The maximization problem then for the first sender (upward biased) is given by:. max eG 1. G N G N B N G N G N G (1−eG 1 )[λ(π0 )e2 (λ(π ))]+e1 λ(π0 )[1−(1−λ(π ))e2 (λ(π ))]+e1 (1−λ(π0 ))[λ(π )e2 (λ(π ))]−C(e1 ). Obviously the first expert effort depends on e2 and we know that this in turn is a non-linear function of π1 , therefore how the effort decision looks for expert 1 depends highly on π0 . Now I will present the problem for two different initial beliefs (π0 ) such that the DM for one of the beliefs wants for sure to hire 2 experts while for the other the decision will depend on her utility function.. For π0 =. (1−p)2 3 p2 +(1−p)2. If only one expert was hired then eialone = 0 for i = {G, B}, this because the downward bias is comfortable with the DM’s decision so he will not exert effort. An upward biased expert would not exert effort either because no matter what signal it’s revealed the DM would still take the undesired action. Now for two experts the only rational choice of experts is {upward, upward}. Given this hiring policy the first expert faces the following problem: max eG 1. G G G λ(π0 )eG 1 [λ(π )e2 ] − C(e1 ). If the second expert is called then his problem is: max eG 2. 3 Perhaps. it is easy to think that π0 =. G λ(π G )eG 2 − C(e2 ). (1−p)2 p2 +(1−p)2. +. 27.

(30) if C 0 (e) = e this gives us the following efforts: e∗1 = λ(π0 )λ(π G )2 =. 4p2 (1 − p)4 p2 + (1 − p)2. e∗1 = λ(π G ) = 2p(1 − p) Because both efforts are strictly positive, the DM is always better hiring 2 experts. This tell us that when the DM has strong inclinations toward one state she benefits from hiring two very motivated experts of the opposite bias. Now lets see a case where the DM might be better with only one expert.. For π0 =. 1 4 2. The one expert problem is: [eG alone ] :. λ(π0 ) − C 0 (eG alone ) = 0. Therefore if C 0 (e) = e, e∗G = λ(π0 ) = 0.5 Two experts: If the first one is upward biased, then the rational choice of experts for the second is downward biased if the former revealed G and an upward biased expert if N was revealed. max eG 1. G G G B G B G G (1 − eG 1 )[λ(π0 )e2 ] + e1 λ(π0 )[1 − (1 − λ(π ))e2 ] + e1 (1 − λ(π0 ))[λ(π )e2 ] − C(e1 ). The second expert effort is given by N eG∗ 2 = λ(π ). eD∗ 2 = 1 − λ(π0 ) Replacing with the corresponding efforts max eG 1. N G G 2 G B N G (1 − eG 1 )[λ(π0 )λ(π )] + e1 λ(π0 )[1 − (1 − λ(π )) ] + e1 (1 − λ(π0 ))[λ(π )λ(π )] − C(e1 ). 4 Again. one might think of π = 0.5 + . 28.

(31) The optimal effort then is given by e∗1 = λ(π0 )(1 − λ(π N )) − p(1 − p)(p − λ(π N )) where e∗1 < e∗alone because 1 − λ(π N ) is less than 1 and p − λ(π N ) is positive. Because the experiments now are no longer Blackwell comparable, only for some DM’s utility function and a range of parameters would be better to hire 2 experts.. Now I compare the “extreme type” effort decisions with the linear expert decisions for a particular initial belief. I show that the extreme type provides higher effort for the first and second sender, therefore the DM would prefer to have this type of experts to hire.. Extreme vs linear:. I will address the problem for π0 =. 1 2. where if the first sender obtains g a downward bias sender. will be hired, and if he obtains b or n another upward sender will be hired. This represents the problem of an “impartial” DM. When π0 =. 1 2. we know that βG = p and therefore is on the limit of the inaction zone. This means. that if the downward sender obtains b the DM’s update will be exactly. 1 2.. I’m going to solve the. problem assuming that if the first sender obtains g the consultation ends at t = 1 and then I will address the case where if π2 =. For π0 =. 1 2. each action is taken with probability 12 .. 1 2. πG = p. πN =. 1−ê1 p 2−ê1. πB = 1 − p. λ(π G ) = 1 − 2p(1 − p). λ(π N ) = π N (2p − 1) + 1 − p. λ(π B ) = 2p(1 − p). 29.

(32) with λ = 12 : max eG 1. 1 G 1 1 N G G B G G B G N (1−eG 1 )[λ( )e2 (λ(π ))]+e1 [λ( )(1−(1−λ(π ))e2 (λ(π ))]+e1 [(1−λ( ))λ(π ))e2 (λ(π ))] 2 2 2. replacing some terms, the maximization problem is: max eG 1. h i eG 1 1 − (π1N (2p − 1) + 1 − p)(1 − 2p(1 − p)) − C(eG 1) 2. the maximization problem for the linear case when hired alone is : lin lin ) λ(π0 )[π1G − π1N ] − C(eG eG 1 1. max G. e1 lin. which we know is greater than esimultaneous . Replacing λ(π0 = 1/2) = 1/2 and π1G (π0 = 1/2) = p 1 we have that max G. e1 lin. lin eG 1 lin [p − π1N ] − C(eG ) 1 2. Therefore experts with non linear utility will exert more effort if [1 − (π1N (2p − 1) + 1 − p)(1 − 2p(1 − p))] > [p − π1N ] This holds only when the following holds p(1 − p) + π1N (1 + p(2p − 1)) And we know this is always true because 1 ≥ p, therefore non linear experts will always exert more effort for π0 = 1/2. Since the effort made by an alone expert with linear utility is greater than is greater than ealone gives us that the one made when hired simultaneously, proving that eextreme 1 1 extreme elin . 1 < e1. What would happen if the DM after receiving G from the first sender and a B from the second (π2 = 1/2) takes the action 1 with probability 0.5?. 30.

(33) Now the utility of getting G in the first period will not be 1, instead it will be G B 1 ∗ (1 − (1 − λ(π G ))eB 2 ) + 0.5 ∗ ((1 − λ(π ))e2 ). that we know it is 1 − 0.5 ∗ (2p(1 − p))2 .. Then if the following holds, experts with non linear utility will exert more effort 1 − 0.5 ∗ (2p(1 − p))2 + 2p(1 − p)2 − π1N (2p − 1)(1 − 2p(1 − p)) > p − π N h i ⇐⇒ p(1 − p)2 + π N 1 + (2p − 1)p > 0 And we know that the last always holds because p ≥ 0.5. Therefore: elinear < eextreme . 1 1. Sender 2:. Extreme: [eG 2]:. λ(π1 ) − C 0 (eG 2)=0. Linear:5 [eG 2]:. λ(π1 )[π G − π N ] − C 0 (eG 2)=0. recall that elinear < eextreme , therefore expert 2 with extreme utility is more likely to face πg 1 1 than expert 2 with linear utility, and because because [π G − π N ] < 1 it’s straightforward that. E(eextreme ) > E(elinear ), therefore: 2 2 (elinear , elinear ) <(Blackwell) (eextreme , eextreme ) 1 2 1 2 Because elinear < eextreme and elinear < eextreme . Hence, any impartial Bayesian DM would prefer 1 1 2 2 to be informed by the extreme type experts. 5 Absent. strategic component, which we know diminish effort.. 31.

(34) VI.b. DM’s welfare when efforts are strategic substitutes. The DM’s welfare is given by an utility function u(a, ω) where a is the action she takes and ω is the real state of nature. So far, DM’s utility has remained general because the results apply for every form of u. For this section I will assume that u(a, ω) = −(a − ω)2 , where it is easy to see that the action a that maximizes the DM’s welfare is a = π. Experts utility depends of the action of the DM, but because a = π the utility is actually a function of the DM’s final belief, F (π). For some specifications of F Kartik’s substitution effect reappears and there are cases on which the DM’s welfare is greater with only one expert. Next I show an example of this.. DM’s utility: UDM =. X. Pi [−πi (1 − πi )2 − (1 − πi )πi2 ]. i. =. −π0 + E(πi2 ). Note that the DM’s utility can be expressed as −π0 + E(π 2 ), therefore to compare the DM’s welfare between 1 or 2 experts we can just pay attention to. E(π2 ). The intuition behind this is that the. DM prefers extreme beliefs because this will likely lead to more accurate decisions.. Effort decision for only one expert. [eG alone ] :. λ(π0 )[F (π G ) − F (π N )] = C 0 (eG 1). while with two experts: G GB [eG ) + (1 − (1 − λ(π G ))e2 )F (π GN ) ] 1 ] : λ(π0 ) [ (1 − λ(π ))e2 F (π. + (1 − λ(π0 )) [ (1 − λ(π B ))e2 F (π N B ) + (1 − (1 − λ(π B ))e2 )F (π N N ) ] − [ (1 − λ(π0 ))e2 F (π N B ) + (1 − (1 − λ(π0 ))e2 )F (π N N ) ] = C 0 (eG 1). 32.

(35) G NG [eB ) − (1 − λ(π G )e1 )F (π N N ) ] 2 ] : λ(π0 ) [ λ(π )e1 F (π. + (1 − λ(π0 )) [λ(π B )e1 F (π BG ) − (1 − λ(π B )e1 )F (π BN ) ] − [λ(π0 )e1 F (π N G ) − (1 − λ(π0 )e1 )F (π N N ) ] = C 0 (eB 2 ) For some utility functions F and cost functions C, UDM (2) < UDM (1) meaning that the DM is better with only one “hard work” expert hired than with two “lazy” experts.. For F (π) = π 2 for the upward bias and (1 − π)2 for the downward, C 0 (e) = 0.3515e, p = 0.85 and π = 0.5 the DM’s welfare is greater with only one expert. For this parameters the alone experts exerts almost full effort (ealone = 1) and because the precision is high this leaves the DM highly informed, while if two experts were hired they would put less effort (e = 0.79), leading to a less informed DM. Hence on this scenario Kartik’s prediction that one expert might be preferred over two holds.. However, if on the previous example the precision was p = 0.6, the DM would be better hiring two experts. The effort of the alone expert would be ealone = 0.1695 while for the 2 experts case efforts would be e1 = e2 = 0.1689. This are indeed substitutes, but because signals are more noisy and the alone experts is less likely to obtain an informative signal than before, the decisions that the DM could make with 2 experts are more accurate and her welfare is higher when 2 are hired.. 33.

(36) References • Austen-Smith, D. (1993). Information and influence: Lobbying for agendas and votes. American Journal of Political Science, 799-833. • Babaioff, M., Feldman, M., Nisan, N. (2009, October). Free-riding and free-labor in combinatorial agency. In International Symposium on Algorithmic Game Theory (pp. 109-121). Springer, Berlin, Heidelberg. • Bhattacharya, S.,. Mukherjee, A. (2013). Strategic information revelation when experts. compete to influence. The RAND Journal of Economics, 44(3), 522-544. • Che, Y. K., Kartik, N. (2009). Opinions as incentives. Journal of Political Economy, 117(5), 815-860. • Grossman, S. J. (1981). The informational role of warranties and private disclosure about product quality. The Journal of Law and Economics, 24(3), 461-483. • Kamenica, E., Gentzkow, M. (2011). Bayesian persuasion. American Economic Review, 101(6), 2590-2615. • Kartik, N., Lee, F. X., Suen, W. (2017). Investment in concealable information by biased experts. The RAND Journal of Economics, 48(1), 24-43. • Krishna, V., Morgan, J. (2001). A model of expertise. The Quarterly Journal of Economics, 116(2), 747-775. • Marx, L. M., Matthews, S. A. (2000). Dynamic voluntary contribution to a public project. The Review of Economic Studies, 67(2), 327-358. • Milgrom, P., Roberts, J. “Relying on the Information of Interested Parties.” RAND Journal of Economics, Vol. 17 (1986), pp. 18–32. • Shin, H. S. (1994). News management and the value of firms. The RAND Journal of Economics, 58-71.. 34.

(37) Appendix 1. Sequential hiring: Sender 2 downward biased B B B B M ax : 1−{(1−eB 2 )π1,N (ê2 )+e2 [((1−π1 )p+(π1 )(1−p))βπ1 ,B +(π1 p+(1−π1 )(1−p))π1,N (ê2 )]}−C(e2 ) eB 2. Optimal effort is given by: B B 0 B [eB 2 ] : π1,N (ê2 ) − [λ(π1 )π1,N (ê2 ) + (1 − λ(π1 ))βπ1 ,B ] = C (e2 ). 0 B ⇐⇒ (1 − λ(π1 ))[π1,N (êB 2 ) − βπ1 ,B ] = C (e2 ). Differentiating by π1 : 0 0 B (1 − λ)(π1,N (êB ∂eB 2 2 ) − βπ1 ,B ) + (1 − λ) (π1,N (ê2 ) − βπ1 ,B ) = ∂π1 C 00 (eB 2 ) − (1 − λ(π1 ))βπ1 ,e2 N. The numerator can be expressed by 0 0 B (1 − λ(π1 ))[βπ0 1 ,N (êB 2 ) − βπ1 ,B ] + (1 − λ(π1 )) [π1,N (ê2 ) − βπ1 ,B ] ≷ 0. 0 βπ0 ,N (êB (1 − λ(π1 ))0 2 ) − βπ1 ,B ≷ 1 (1 − λ(π1 )) π1,N (êB 2 ) − βπ1 ,B. Let I ≡ π1,N (êB 2 ) − βπ1 ,B and multiplying by π1 we obtain: π1. I 0 (π1 ) (1 − λ(π1 ))0 ≷ π1 (1 − λ(π1 )) I(π1 ). Which has the same interpretation as the previous case.. 35.

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