Imperfect Signaling and the Local Credibility Test

Imperfect Signaling and the Local Credibility Test Hongbin Cai, John Riley and Lixin Ye* November, 2004 Abstract In this paper we study equilibrium refinement in signaling models We propose a Local Credibility Test (LCT) that is somewhat stronger than the Cho and Kreps Intuitive Criterion but weaker than the “strong Intuitive Criterion” of Grossman and Perry Allowing deviations by a pool of “nearby” types, the LCT gives consistent solutions for any positive, though not necessarily perfect, correlation between the signal sender’s true types (e.g., signaling cost) and the value to the signal receiver (e.g., marginal product) It also avoids ruling out reasonable pooling equilibria when separating equilibria not make sense We identify conditions for the LCT to be satisfied in equilibrium for both the finite type case and the continuous type case, and demonstrate that the results are identical as we take the limit of the finite type case We then apply the characterization results to the Spence education signaling model and the Milgrom and Roberts advertising signaling model Intuitively, the conditions for a separating equilibrium to survive our LCT test require that a measure of signaling “effectiveness” is sufficiently high for every type and that the type distribution is not tilted upwards too much *UCLA, UCLA, and Ohio State University We would like to thank In-Koo Cho, David Cooper, Massimo Morelli, James Peck, and seminar participants at Arizona University, Illinois Workshop on Economic Theory, Ohio State University, Penn State University, Rutgers University, UC Riverside, UC Santa Barbara, and Case Western Reserve University, for helpful comments and suggestions All remaining errors are our own Introduction Since the seminal work of Cho and Kreps (1987), various refinement concepts have been proposed to rank different equilibria in signaling games in terms of their “reasonableness” However, the mission is still far from being completed In many applications, signals are “imperfect” in the sense that there is a positive yet imperfect correlation between the signal sender’s true type (e.g., signaling cost) and the signal receiver’s expected value (which then determines her response), see Riley (2001, 2002) Consider a situation in which two of the sender types have a same signaling cost but quite different values to the receiver If these two types not observe their values to the receiver, they are effectively the same type, so the existing refinement concepts, such as the Cho and Kreps Intuitive Criterion, apply in the usual way However, if these two types observe their different values to the receiver, then the Intuitive Criterion is unable to rank equilibria The reason is that if one of the two types likes a deviation, the other also likes it, hence no deviation is credible by a single type This is highly unsatisfactory because the two cases are observationally equivalent The reason for the inconsistent solutions in the above example is that the existing refinement concepts focus on deviations by a single type only and not consider deviations by a pool of types Grossman and Perry (1986a,b), in a bargaining context, propose an equilibrium refinement concept strengthening the Cho-Kreps Intuitive criterion to allow pooling deviations In this paper, in a general signaling model, we weaken slightly the Grossman-Perry Strong Intuitive Criterion, and propose a “Local Credibility Test” (LCT) in which a possible deviation is interpreted as coming from one or more types whose equilibrium actions are nearby We consider only local pooling deviations, first because they seem to us more natural, second because they have much of the power of global pooling deviations, and third because they are more easily analyzed Moreover, the Local Credibility Test does not always rule out pooling equilibria in favor of separating equilibria We will argue that in some situations separating equilibria seem unreasonable while pooling equilibria can be rather appealing By allowing pooling deviations, the LCT avoids ruling out pooling equilibria in such situations Consider a simple two type education-signaling model, in which the high type must take a quite costly signal (e.g., several years of unproductive education) to separate from the low type Now suppose there is only one low type agent in every million high type agents In such a situation separation seems highly unreasonable, because without taking the costly signaling action an agent should not be perceived much differently from being the high type By the LCT, it is easy to see that in any separating equilibrium a pooling deviation to some sufficiently low cost level of the signal is profitable to both types, so no separating equilibrium satisfies the LCT The thrust of our analysis is to derive conditions under which there exist equilibria satisfying the LCT We study a family of continuous type models of which the Spence education signaling and the Milgrom and Roberts advertising signaling are both members We begin by formulating the concept of the LCT for the finite type models first, since the intuition is easier to present Then we consider a discretization of the continuous type model, and take the limit as the discretization becomes finer We characterize conditions under which the Pareto dominant separating equilibrium of the model satisfies the LCT The required conditions are intuitive As long as a measure of signaling “effectiveness” is sufficiently high for every type and the type distribution is not tilted upwards too much, the separating equilibrium can survive our LCT test In the continuous type case, the set of equilibrium signals is dense so that out-ofequilibrium signals can be only found outside the set of equilibrium signals However, thinking of the continuous type case as the limiting case of the finite type case with many close types, it is natural to generalize the concept of the LCT to the continuous type An equilibrium survives the LCT if no deviation-perception pair is credible in the following sense: for any possible deviation signal (on- or off-equilibrium), if it is interpreted as from types of a small neighborhood of the immediate equilibrium type, it is profitable for the types in this neighborhood to deviate to this signal, but unprofitable for types out of this neighborhood to so Another way of thinking about this credibility test in the continuous type case is the following If, for an on-equilibrium signal, there is such a deviation-perception pair, then those nearby types can credibly deviate to the particular on-equilibrium signal by throwing away ε amount of money We derive conditions under which the LCT is satisfied in equilibrium in the continuous type case The conditions are exactly the same as in the limiting finite type case This is satisfactory, because models with continuous types and models with finitely many types are theoretical tools for analyzing the same kind of real world problems Put differently, it would be highly unsatisfactory if an equilibrium refinement concept applies to one case but not the other, or gives different answers for the two cases The paper is structured as follows The next section uses simple examples to illustrate the basic idea of the LCT Then Section presents the general signaling model In Section 4, we formulate the concept of the LCT for the finite type case Then we derive conditions under which the LCT is satisfied by the Pareto dominant separating equilibrium in a discretized continuous type model as the discretization becomes finer Section generalizes the formulation of the LCT to the continuous type case, and shows that the conditions for the LCT are exactly the same as in the discrete type case We discuss an issue of robustness in Section Concluding remarks are in Section Examples A consultant ( si , v j ) has a signaling cost type si and a marginal product of v j , where s1 < s2 < < sn and v1 < v2 < < vm She can signal at level z at a cost of C ( z , si ) We suppose that ∂C ( z , s) < so that a higher type has a lower signaling cost If paid a ∂s wage w her payoff is U ( z, w, si ) = w − C ( z , si ) In a competitive labor market for consultants, her wage will be her marginal product perceived by the market Activity z is a potential signal because the marginal cost of signaling, ∂C ( z , si ) , is a decreasing ∂z function of si The set of signaling cost types is S = {s1 , , sn } , and the set of possible productivity levels is V = {v1 , , vm } The probability of each type, π ( si , v j ) , is common knowledge.1 If π ( si , v j ) = for all i ≠ j , the model reduces to the usual Spence model in which the negative correlation between signaling cost and value to receivers is perfect While we assume π ( si , v j ) > for all i and j , the analysis applies generally Initially we assume that each consultant observes her own signaling cost type but not her productivity We define v( si ) = E{v | si } , and assume that types with lower signaling costs have higher expected productivity , that is, v( s1 ) < v( s2 ) < < v ( sn ) There is a continuum of separating Nash equilibria in this game A Nash separating equilibrium with three signaling cost types is depicted below Each curve is an indifference curve for some signaling cost type A less heavy curve indicates a lower signaling cost type Note that the equilibrium choice for each type si (indicated by a shaded dot) is strictly preferred over the choices of the other types Such an equilibrium fails the Intuitive Criterion first proposed by Cho and Kreps (1987) I1 I2 v( s3 ) I3 v ( s2 ) v( s1 ) zˆ z2 z1 = z3 z Fig: 2-1: Separating Nash Equilibria To see this, suppose an individual chooses the signal zˆ and argues that she is type s2 Is this credible? If the individual is believed, her wage will be bid up to v( s2 ) so she earns the same wage as in the Nash equilibrium but incurs a lower signaling cost As noted by Cho and Kreps (1987), with more than two types, it is necessary to modify their original Intuitive Criterion or it loses much of its power For the modified Intuitive Criterion the question is whether any particular type is uniquely able to benefit from some out-of-equilibrium signal if the signal receivers correctly infer the signaler’s type However ( zˆ, v( s2 )) is strictly worse than ( z1 , v( s1 )) for type s1 and strictly worse than ( z3 , v( s3 )) for type s3 Thus the claim is indeed credible Similar arguments rule out any Nash equilibrium where different signaling cost types are pooled Thus the only equilibrium that satisfies the Intuitive Criterion is the Pareto dominant separating equilibrium (i.e., the Riley outcome) in which each “local upward constraint” is binding Next suppose that each consultant knows both her signaling cost type and her marginal product Again consider the Nash Equilibrium depicted above Suppose in this equilibrium three different types are pooled at each signal level Consider the three types ( s2 , v1 ), ( s2 , v2 ), ( s2 , v3 ) pooled at z2 Suppose a consultant chooses zˆ and claims to be type ( s2 , v3 ) Is this credible? If the claim is believed, the consultant’s wage will rise from v( s2 ) to v3 thus the consultant is indeed better off But any offer that makes type ( s2 , v3 ) better off also make types ( s2 , v1 ) and ( s2 , v2 ) better off, since they have the same signaling cost Thus there is no credible claim that type ( s2 , v3 ) alone can make A similar argument holds for each of the other types Thus any Nash separating equilibrium satisfies the Intuitive Criterion An almost identical argument establishes that any Nash Equilibrium with (partial) pooling satisfies the Intuitive Criterion as well.3 Since all the types with the same signaling cost are observationally equivalent, it seems to us that any argument for ranking the equilibria in the first model (productivity unknown) should also be applicable to the second model as well There is a simple modification to the Intuitive Criterion that achieves this goal A consultant takes an outof-equilibrium action zˆ and argues that she is one of the types who, in the Nash Equilibrium would have chosen z2 This being the case, applying the Bayes Rule, her It can be verified that the Cho and Sobel (1990)’s refinement concept of “divinity”, which is built on the idea of stability of Kohlberg and Mertens (1986) and can be considered as a logic offspring of the Intuitive Criterion, does not have power either in the above example Ramey (1996) extends the Cho and Sobel’s divinity concept to the case of a continuum of types Like the Intuitive Criterion, divinity faces the same problem of distinguishing types ( s2 , v1 ), ( s2 , v2 ), ( s2 , v3 ) to interpret a possible deviation, while these types have the same incentives to deviate Riley (2001) discusses in greater details these and other refinement concepts expected marginal product is v( s2 ) Then once again, the unique Nash Equilibriums satisfying the modified Intuitive Criterion is the Pareto Dominant separating equilibrium We now argue that for some parameter values, the Pareto Dominant separating equilibrium defies common sense Consider the following example Suppose there are two signaling cost types For those with a high signaling cost ( s = s1 ), the cost of signaling is c1 ( z ) with c1 (0) = and c1′ > , and the mean marginal product is 100 For those with a low signaling cost ( s = s2 ) , the signaling cost is c2 ( z ) = (1 − ε )c1 ( z ) and 100 the mean marginal product is 200 The Pareto dominant separating equilibrium is depicted below w 200 100 + ε U ( z , w) = U ( z2 ,200) 100 U1 ( z, w) = U1 (0,100) z Fig 2.2: Separating equilibrium with a gain of The low type must be indifferent between (0, v( s1 )) and the choice of type s2 , that is ( z2 , v( s2 )) Therefore, U1 = 100 = 200 − c1 ( z2 ) and so c1 ( z2 ) = 100 The payoff for type s2 is therefore U = v ( s2 ) − c2 ( z2 ) = 200 − (1 − ε )c1 ( z2 ) = 100 + ε 100 Suppose that only in 100 consultants is of type s1 Then the unconditional mean marginal product is 199 Thus essentially all the social surplus generated by the high types is dissipated by signaling and both types have an income which is approximately half the income that would have in the Nash pooling equilibrium! We believe a better criterion for ranking equilibria should not rule out pooling equilibria in such circumstances We now introduce a further simple modification of the Intuitive Criterion that achieves this outcome Local Credibility Test: Suppose that an out-of-equilibrium signal zˆ is observed and that z − is the largest Nash Equilibrium signal less than zˆ and z + is the smallest Nash Equilibrium signal greater than zˆ , if they exist Let Sˆ be the subset of signaling cost types choosing z − or z + with positive probability For each S ⊂ Sˆ , define v ( S ) = E{v | s ∈ S} Then the equilibrium passes the Local Credibility Test (LCT) if there is no S such that ( zˆ, v ( S )) is strictly preferred over the Nash Equilibrium outcome if and only if s ∈ S Note that if zˆ is smaller (greater) than all equilibrium signals, then z − ( z + ) does not exist and z + ( z − ) is the smallest (largest) equilibrium signal By the above definition, Sˆ is the subset of types choosing z + ( z − ) Also note that by considering a subset of Sˆ to be the singleton set of a single type choosing z + or z − , the definition of the Local Credibility Test allows deviations by single types It follows that only separating equilibria can pass the Local Credibility Test The idea of the Local Credibility Test is weaker than the Strong Intuitive Criterion (SIC) proposed by Grossman and Perry (1986a,b) For any out-of-equilibrium signal zˆ , their criterion considers any subset of types as a potential deviating pool An equilibrium fails the SIC if zˆ is credible for one subset of types Here we restrict attention to local deviations This makes the analysis more tractable and, we believe, more plausible Using the idea of the Local Credibility Test, we show that for the simple consulting example above, the Pareto dominant separating Nash Equilibrium is robust to credible pooling deviations For concreteness, suppose c1 ( z ) = 100 z and c2 ( z ) = 99 z Then in the Pareto dominant separating equilibrium, z1 = and z2 = Consider any convex combination zˆ = (1 − λ ) z1 + λ z2 = λ z2 = λ Then by the definition of the LCT, Sˆ = {s1 , s2 } Consider S = Sˆ The average productivity of these two signaling cost types is 199 For zˆ = λ to be a credible deviation by S , both types must strictly prefer ( zˆ,199) Note that U1 ( zˆ,199) = 199 − 100λ and U ( zˆ,199) = 199 − 99λ , and the equilibrium payoffs are 100 and 101 Thus, for all λ = zˆ < 98 / 99 , both types are indeed better off choosing the out-of-equilibrium zˆ The Pareto dominant separating Nash Equilibrium thus fails the LCT In fact no equilibrium passes the LCT, so the criterion fails to rank the different Nash Equilibria We now show how the LCT can be applied when there are many types and, in the limit, a continuum of types Let the set of signaling cost types be S = {s1 , , sn } , with probabilities { f1 , , f n } where C ( z , si ) = ∑ i f i = Suppose type si has a signaling cost c( z ) , where si +1 − si = δ > We also assume that the expected marginal si product of those of type si have an expected marginal product of v ( si ) = si We seek conditions under which the Pareto dominant separating equilibrium passes the LCT In this equilibrium, the local upward constraints are binding Therefore, as depicted below, those with signaling cost type si −1 are indifferent between ( zi −1 , si −1 ) and ( zi , si ) (The indifference curve is labeled I i −1 ) We construct the signal levels zˆ and zi +1 as follows Choose z so that those with signal type si −1 are indifferent between ( zi , si ) and ( z , 12 si + 12 si +1 ) In the Figure below, these are the points Ci and C Then choose zi +1 so that those with signal cost type si +1 are indifferent between ( z , 12 si + 12 si +1 ) and ( zi +1 , si +1 ) w I i −1 ( zˆ, wˆ ) Ci +1 si +1 s i Ii I i +1 C + 12 si +1 Ci si ( zi −1, si −1 ) zi z zi +1 z Fig 2-3: Applying the LCT with many types In Figure 2.3, these are the points C and Ci +1 We will argue that type si must be indifferent between Ci and Ci +1 as depicted That is, Ci +1 is the efficient separating contract for those with signaling cost type si +1 A type s j consultant is indifferent between ( z , w) and ( z ′, w′) if and only if w− c( z ) c( z ′) = w′ − , sj sj that is, if c( z ′) − c ( z ) = s j ( w′ − w) By construction, a type si −1 consultant is indifferent between ( zi , si ) and ( z , 12 zi + 12 zi +1 ) Appealing to , 10 The above “lower endpoint” problem can be overcome if we modify the model so that there is a significant fixed cost to enter the market If the fixed cost is greater than the maximum profit the lowest type can get, the lowest type and all the types lower than a certain threshold will stay out of the market Then for the lowest type that does enter the market, it cannot choose its complete information optimal price Otherwise, some of the lower types would find it profitable entering the market mimicking it With this type of truncation of the type distribution by entry costs, it is again the case that U < for all the types signaling in the market so Theorem applies Concluding Remarks Except in the special case of perfect correlation between the sender’s true type and the value to the signal receiver, standard refinements (Intuitive Criterion, Divinity, Stability) are not applicable We argue that to have any “bite” at all, a refinement is needed in which the signal receivers take into account the way sender types are distributed We then propose a Local Credibility Test which is somewhat stronger than the Cho and Kreps Intuitive Criterion but milder than the Grossman-Perry Criterion For a class of models which includes the Spence education signaling model and the advertising signaling model, we provide conditions under which the Pareto dominant separating equilibrium satisfies the LCT These conditions are the more likely to be met, (a) the less rapidly the density increases or the more rapidly the density decreases with type, and (b) the more rapidly the marginal cost of signaling decreases with type What is the “right” equilibrium when our conditions are not met? This is a challenging question for which we have no satisfactory answer As can be seen from the simple two-type model depicted in Figure 2.2, our LCT test and the even stronger Grossman and Perry SIC show that the Pareto dominant separating equilibrium is not reasonable But at the same time, all other equilibria are “killed” as well We conjecture that pooling or partial pooling must be a part of any more complete analysis of signaling To make the point as starkly as possible, let p1 be the probability that a sender is low type With p1 = , the high type does not have to signal at all As long as p1 is positive, the high type has to take costly signaling to separate herself from the 30 low type Thus the separating equilibrium has an extreme discontinuity at p = When p is close to zero, the pooling outcome seems more reasonable than the highly inefficient separating equilibrium That is, “reasonable” out-of-equilibrium beliefs not necessarily lead to reasonable outcomes 31 Appendix A: Proofs Proof of Lemma 1: Using sn = s, sn- = s - d, sn+1 = s + d, we rewrite as follows: v ( s, δ )(G ( sn +1 ) − G ( sn −1 )) = [G ( s ) − G ( sn−1 )]s + [G ( sn+1 ) − G ( s )]sn+1 Differentiating with respect to δ on both sides, we have v2 ( s, δ )[G ( sn +1 ) − G ( sn −1 )] + v ( s, δ )[G ′( sn +1 ) + G ′( sn −1 )] = sG ′( sn −1 ) + sn +1G ′( sn +1 ) + G ( sn +1 ) − G ( s ) v22 ( s, d)[G ( sn+1 ) - G ( sn- )] + 2v2 ( s, d)[G ¢( sn+1 ) + G ¢( sn- )] + v ( s, d)[G ¢( sn+1 ) - G ¢( sn- )] =- sG ¢¢( sn- ) + sn+1G ¢¢( sn+1 ) + 2G ¢( sn+1 ) v222 ( s, d)[G ( sn+1 ) - G ( sn- )] + 3v22 ( s, d)[G ¢( sn+1 ) + G ¢( sn- )] + 3v2 ( s, d)[G ¢¢( sn+1 ) - G ¢¢( sn- )] + v ( s, d)[G ¢¢¢( sn+1 ) + G ¢¢¢( sn- )] = sG ¢¢¢( sn- ) + sn+1G ¢¢¢( sn+1 ) + 3G ¢¢( sn+1 ) Letting δ → we obtain: v ( s, δ ) → s; v2 ( s, δ ) → 1/ 2; v22 ( s, δ ) → G "( s ) 2G '( s ) Proof of Lemma 2: First by the continuity of U we have v ( s,0) = s from Differentiating with respect to δ , we have −U1 ( sn −1 , v, y ) + U ( sn −1 , v, y ) U1 ( sn +1 , v, y ) + U ( sn +1 , v, y ) dv dy + U ( sn −1 , v, y ) = −U1 ( sn−1 , sn , zn ) dδ dδ dv dy + U ( sn +1 , v, y ) dδ dδ = U1 ( sn +1 , sn +1 , zn +1 ) + U ( sn +1 , sn +1 , zn +1 ) + U ( sn +1 , sn +1 , zn +1 ) = U ( sn , sn +1 , zn +1 ) + U ( sn , sn +1 , zn +1 ) dzn +1 dδ 32 dzn +1 dδ Q.E.D To save on notation, let z′ = dzn +1 / d δ From we have z ¢=- U ( sn , sn+1 , zn+1 ) U ( sn , sn+1 , zn+1 ) We can obtain the higher order derivatives for zn +1 : z′′ = d zn +1 U (U U − 2U U ) = − 2 33 3 23 dδ U3 d zn +1 U z′′′ = = − 25 6U 32U 232 − 9U 2U 3U 23U 33 + 3U 22U 332 + 3U 2U 32U 233 − U 22U 3U 333  dδ U3 where U = U ( sn , sn +1 , zn +1 ) Write v ¢= dv / d d = v2 ( s, d) and y ¢= dy / d d = y2 ( s, d) From and we have v′ = ∆1 ∆ and y ′ = ∆2 where ∆ ∆ = U ( sn −1 , v, y )U ( sn +1 , v, y ) − U ( sn +1 , v, y )U ( sn −1 , v, y ) ∆1 = U ( sn +1 , v, y )[U1 ( sn −1 , v, y ) − U ( sn −1 , sn , zn )] − U ( sn −1 , v, y )[U ( sn +1 , sn +1 , zn +1 ) − U1 ( sn +1 , v, y ) + U ( sn +1 , sn +1, zn +1 ) + U ( sn +1 , sn +1, zn +1 ) z ′] ∆ = U ( sn −1 , v, y )[U ( sn +1 , sn +1 , zn +1 ) − U1 ( sn+1 , v, y ) + U ( sn+1 , sn +1 , zn +1 ) + U ( sn+1 , sn +1, zn +1 ) z ′] -U ( sn +1 , v, y )[U1 ( sn −1 , v, y ) − U ( sn −1 , sn , zn )] Differentiating the above equations, and using B1, B2 and - , we can derive the following derivatives evaluated at d= : ∆′ = d∆ |δ =0 = 2U 2U13 dδ D ¢¢= ∆1′ = d 2D |d=0 = 4(U13U 23 +U 2U133 ) y ¢ d d2 d ∆1 |δ =0 = 2U 3U13 ×y ′ + 2U 2U13 dδ 33 d ∆1 ′′ ∆1 = |δ =0 = 2(U13 + U 23v′ + U 33 y ′ )U13 ×y ′ + U [−U113 y ′ + ( −U113 + U133 y ′ ) y ′ + U13 y ′′ ] dδ − 2( −U13 + U 23v′ + U 33 y ′ )[U13 ( z ′ − y ′ ) + U 23 z ′ + (U13 + U 23 + U 33 z ′) z ′ + U z ′′] − U 3[U113 ( z ′ − y ′ ) + (U113 + U133 z ′) z ′ + U13 z ′′ − (U113 + U133 y ′ ) y ′ − U13 y ′′ + U 233 z′2 + U 23 z ′′ + (U13 + U 23 + U 33 z ′) z ′′ + (U113 + 2U133 z′ + 2U 233 z ′ + U 333 z ′2 + U 33 z ′′) z ′ + U z ′′′ + (U13 + U 23 + U 33 z ′) z ′′] d ∆2 U 22U13 ′ ∆2 = |δ =0 = −2 − 2U 2U13v′ dδ U3 d 2D ¢ ¢ D = |d=0 = 2U 23 y ¢[(2U13 + 2U 23 +U 33 z ¢) z ¢+U z ¢¢- 2U13 y ¢] dd +U [(3U113 + (3U133 + 3U 233 +U 333 z ¢) z ¢+U 33 z ¢¢) z ¢+ (3U13 + 3U 23 + 2U 33 z ¢) z ¢¢ +U z ¢¢¢- 2U133 y ¢2 - 2U13 y ¢¢] + (U 233 y ¢2 +U 23 y ¢¢)(U +U z ¢) Using L’Hopital’s rule, we have  ∆ ′ 2U 3U13 y ′ + 2U 2U13 U = + y′  v′ = = ∆′ 2U 2U13 U2   U  ′ ∆ 2′ −2U 22U13 − 2U 2U13 y ′ = − − y′  y = ∆′ = 2U 2U13 U3  which implies v′ = v2 ( s,0) = 1/ and y ′ = y2 ( s,0) = −U / 2U Taking second order derivatives, we have d 2v d  ∆1  d ∆1 / d δ − (dv / d δ ) ×(d ∆ / d δ ) =  ÷= dδ dδ  ∆  ∆ d2y d = dδ dδ  ∆  d ∆ / dδ − (dy / dδ ) ×(d ∆ / d δ )  ÷= ∆  ∆  Using L’Hospital’s rule again we have 34 ỡù d 2v D 1ÂÂ- v ÂÂìD Â- v ÂìD  ùù v ( s ,0) = | = d=0 ïï 22 d d2 D¢ ïí ïï ¢¢ ¢¢ ¢ ¢ ¢¢ ïï y ( s,0) = d y | = D - y ×D - y ìD d=0 ùù 22 d d2 D ợ Substituting the expressions of v′ , y ′, ∆′ , ∆′′ , ∆1′′ , ∆ 2′′ derived above into , we can solve for values of v22 ( s,0) as follows: v′′ = v22 ( s,0) = − −U 2U13U 33 + U 2U 3U133 + 2U 3U13U 23 − 2U 32U113 + 4U 3U132 4U 32U13 Q.E.D Proof of Lemma 3: In Figure 4.2, the following conditions hold: U ( s , s , z ) = U ( s , s + δ , z2 )  U ( s + δ , v( s , δ ), z ) = U ( s + δ , s + δ , z2 ) The expected type of s and s + δ is v ( s ,δ ) = sG ( s ) + ( s + δ )[G ( s + δ ) − G ( s )] G( s + δ ) v ( s , δ ) = s Multiplying G ( s + δ ) on Obviously when δ → , we have v ( s ,0) = lim δ →0 both sides of , and differentiating with respect to δ , we have G′( s + δ )v ( s , δ ) + G ( s + δ )v2 ( s , δ ) = G ′( s + δ )( s + δ ) + G ( s + δ ) − G ( s ) Letting δ → we have v2 ( s ,0) = as G ( s ) > Now differentiating once more and letting δ → , we obtain v22 ( s ,0) = 2G ′( s ) G( s ) From we first have v = v ( s ,0) = s by the continuity of U Differentiating with respect to δ , and denote z′ = dz2 / d δ and v′ = dv ( s, δ ) / dδ , we have 35 0 = U ( s, s + δ , z ) + U ( s, s + δ , z ) ×z ′ 2  U1 ( s + δ , s + δ , z2 ) + U ( s + δ , s + δ , z2 ) + U ( s + δ , s + δ , z2 ) ×z ′  = U1 ( s + δ , v, z ) + U ( s + δ , v, z ) ×v ′  Letting δ → we have v′ = v2 ( s,0) = as U > Differentiating once more and evaluating at δ = we obtain 2U13 ( s, s, z ) v ¢¢= v22 ( s,0) =Q.E.D U ( s , s, z ) s′ Proof of Lemma 4: By definition, v ( s, s′) = [G ( s′) − G ( s)] × ∫ xdG ( x) Multiplying s both sides by G ( s′) − G ( s ) and then differentiating by s′ , we have v2 ( s, s′)(G ( s′) − G ( s )) + v ( s, s′)G ′( s′) = s′G ′( s′) Differentiating by s′ again, v22 ( s, s′)(G ( s′) − G ( s )) + 2v2 ( s, s′)G ′( s ') + v ( s, s ′)G ′′( s′) = G ′( s′) + s′G ′′( s′) Setting s′ = s , it follows immediately that v2 ( s, s ) = 1/ Differentiating by s′ again, v222 ( s, s′)(G ( s′) − G ( s )) + 3v22 ( s, s′)G ′( s′) + 3v2 ( s, s′)G ′′( s′) + v ( s, s′)G ′′′( s ′) = 2G ′′( s′) + s′G ′′′( s′) Since v ( s, s) = s and v2 ( s, s ) = 1/ , setting s′ = s we obtain v22 ( s, s ) = G ′′( s ) G′( s ) Q.E.D Proof of Lemma 5: Total differentiating gives U1 ( s¢, v, y )ds ¢+U ( s ¢, v, y )dv +U ( s ¢, v, y )dy = U1 ( s ¢, s ¢, z ( s ¢))ds ¢ U ( s, v, y )dv +U ( s, v, y )dy = Solving the equations, we have dv D = , ds¢ D 36 dy D = ds¢ D D = U ( s¢, v, y )U ( s, v, y ) - U ( s, v, y )U ( s ¢, v, y ), D = U ( s, v, y )[U1 ( s ¢, s ¢, z ( s ¢)) - U1 ( s ¢, v, y )], D =- U ( s, v, y )[U1 ( s ¢, s ¢, z ( s ¢)) - U1 ( s ¢, v, y )] Under Assumption B1, we have dy D U ( s ¢, v, y ) - U1 ( s ¢, s ¢, z ( s ¢)) = = ds¢ D U ( s, v, y ) - U ( s ¢, v, y ) Fix any s, as sÂđ s It must be that v → s, z ( s′) → z ( s ), and y → z ( s ) For the simplicity of notation, write v′( s′) = v2 ( s, s′) and y ′( s′) = y2 ( s, s′) Applying the I’Hopital’s rule, as s′ → s, we get dy ds′ s′→ s U 11 ( s′, v, y ) + U12 ( s ′, v, y )v ′( s′) + U13 ( s′, v, y ) y ′( s′) − U11 ( s′, s ′, z ( s′)) − U12 ( s ′, s ′, z ( s ′)) − U 13 ( s ′, s ′, z ( s ′)) z ′( s ′) U 23 ( s, v, y )v ′( s′) + U 33 ( s, v, y ) y ′( s′) − U 23 ( s′, v, y )v ′( s ′) − U 33 ( s ′, v, y ) y ′( s ′) − U13 ( s ′, v, y ) U ( s′, v, y ) y ′( s′) − U13 ( s′, s′, z ( s′)) z ′( s ′) = lim 13 s ′→ s −U13 ( s′, v, y ) = lim s ′→ s = z′( s ) − dy ds′ s′→ s Hence as s′ → s, dy → 0.5 z′( s ) as long as z′( s ) = − U ( s, s, z ( s )) U ( s, s, z ( s )) is defined ds′ at s , or U ( s, s, z ( s)) ≠ at s Since dv ∆11 ∆11 dy U ( s, v, y ) dy = = =− ds′ ∆ ∆ 21 ds′ U ( s, v, y ) ds′ we have dv U ( s, v, y ) dy U ( s, s, z ( s )) dy = lim − =− lim ′ ′ ds′ s′→ s s → s U ( s, v, y ) ds′ U ( s, s, z ( s )) s →s ds′ = − 0.5 U ( s, s, z ( s )) z′( s ) = 0.5 U ( s, s, z ( s )) for any s such that U ( s, s, z ( s)) ≠ This proves part (i) For part (ii), first note that from z′( s ) = − U ( s, s, z ( s )) U ( s, s, z ( s )) , 37 z′′( s ) = − U 22 U + 2U 23 + U 33 z ′( s ) − z′( s ) 13 U3 U3 From , and by Assumption B1, we have dy − U11 ( s′, s′, z ( s′)) − U13 ( s′, s′, z ( s′)) z ′( s′) ds′ U ( s, v, y ) − U ( s′, v, y ) U 11 ( s′, v, y ) + U13 ( s′, v, y ) d y = ds′2 − U 33 ( s, v, y ) dy dy − U 33 ( s′, v, y ) − U13 ( s′, v, y ) dy ds′ ds′ U ( s, v, y ) − U ( s′, v, y ) ds′ = + U 11 ( s′, v, y ) − U11 ( s′, s′, z ( s′)) U 33 ( s ′, v, y ) − U 33 ( s, v, y )  dy  +  ÷ U ( s, v, y ) − U ( s′, v, y ) U ( s, v, y ) − U ( s′, v, y )  ds ′  dy − U13 ( s′, s′, z ( s′)) z ′( s′) ds′ U ( s, v, y ) − U ( s′, v, y ) 2U13 ( s′, v, y ) Let Li ( s, s′) be the ith term on the right hand side of the above equation For any s such that U ( s, s, z ( s)) ≠ , it can be checked that lim L1 = s′→ s U 113 ( s, s, z ( s )) z ′( s ) U13 ( s, s, z ( s )) lim L2 = − s′→ s U 133 ( s, s, z ( s )) [ z′( s )]2 U13 ( s, s, z ( s )) lim L3 = z′′( s ) − s′→ s d2y ds′2 + 0.5 s′→ s U133 ( s, s, z ( s ))[ z ′( s )]2 U13 ( s, s, z ( s )) Therefore, d2y ds′2 = z′′( s ) + s′→ s U 113 ( s, s, z ( s )) + 0.5U133 ( s, s, z ( s )) z ′( s ) z′( s ) U13 ( s, s, z ( s )) From , and using Assumption B2, we have d 2v U ( s, v, y ) d y U 33 ( s, v, y ) 2U ( s, v, y )U 23 ( s, v, y )   dy  =− − −   ds′ ÷ ds′2 U ( s, v, y ) ds′2  U ( s, v, y ) U 22 ( s, v, y )   38 As s′ → s, we know that d 2v ds′2 =− s′→ s U ( s, s, z ( s )) d y U ( s, s, z ( s )) ds ′2 dv dy → 0.5 and U ( s, v, y ) → 0.5U z ′( s ) = −0.5U So, ds′ ds′ − 0.5 s′→ s U 23 ( s , s, z ( s )) + 0.5U 33 ( s , s, z ( s )) z ′( s ) z′( s) U ( s , s, z ( s )) d y ds′2 s′→ s U + 0.5U 33 z ′( s ) = − 0.5 23 z′( s) z′( s ) U2 −1 U + 2U 23 + U 33 z ′( s ) U 113 +0.5U133 z′( s ) U + 0.5U 33 z ′( s ) = [ 13 ]+ − 0.5 23 z′( s ) U3 U13 U2 U + 2U 23 + U 33 z′( s ) U 113 +0.5U133 z ′( s ) U + 0.5U 33 z ′( s ) = [ 13 z′( s)] + − 0.5 23 z′( s) U2 U13 U2 = U 113 4U13 + 2U 23 U133 U 33 + [ + ]z′( s) + [ z′( s)]2 U13 12 U2 U13 12 U This proves part (ii) Q.E.D Proof of Lemma 6: Multiplying both sides in by G ( s′) and then differentiating by s′ , we have v2 ( s, s′)G ( s′) + v ( s, s′)G ′( s′) = s′G ′( s′) Differentiating by s′ again, we have v22 ( s, s′)G ( s′) + 2v2 ( s, s′)G ′( s ') + v ( s, s′)G ′′( s′) = G ′( s′) + s′G ′′( s′) Setting s′ = s in the two equations above, we prove (i) Total differentiating gives U1 ( s′, v ( s, s′), z ) + U ( s′, v ( s, s′), z ) ×v2 ( s, s ′) = U1 ( s′, s ′, z ( s ′)) Total differentiating once more gives U11 ( s′, v ( s, s′), z ) + U ( s′, v ( s, s′), z ) ×v22 ( s, s′) = U11 ( s′, s′, z ( s′)) + U13 ( s′, s ′, z ( s′)) ×z ′( s ′) U ( s′, s′, z ( s′)) = U11 ( s′, s′, z ( s′)) + U13 ( s′, s′, z ( s′)) ×[ − ] U ( s′, s′, z ( s′)) Taking limits in the two equations above, we get (ii) Q.E.D Proof of Proposition 2: We consider the interior deviation first Following the same notation as in the main text of our paper, we have 39 U ( s, v, y )[U1 ( s ¢, s ¢, z ( s ¢)) - U1 ( s ¢, v, y )] dv D11 = = ds ¢ D U ( s ¢, v, y )U ( s, v, y ) - U ( s, v, y )U ( s ¢, v, y ) - (a / 2b)(1 - a )( s ¢- v) = - (1 - a ) / 2b ×a ( s ¢- s) = s ¢- v s ¢- s dv dv = 1- lim , which implies ¢ s ®s ds ¢ s ®s ds ¢ Using L’Hospital’s rule, we have v2 ( s, s ) = lim ¢ v2 ( s, s ) = 1/ Differentiating with respect to s ¢, we have d v (1- dv / ds ¢)( s ¢- s ) - ( s ¢- v ) = ds¢2 ( s¢- s ) Using L’Hospital’s rule again, we have d 2v - d v / ds ¢2 ( s ¢- s ) d 2v = lim =lim , s Âđs ds  s Âđs 2( sÂ- s ) sÂđs ds Â2 lim which implies v22 ( s, s ) = Therefore, there is no credible interior deviation if ) is strictly concave > G ¢¢( s ) / G ¢( s ) , or if G (× Next consider lower endpoint deviation From Figure 4.2 in the text, we have U ( s ', v , z ) = U ( s ¢, s ¢, z ( s ¢)) Differentiating with respect to s ¢ we have U1 ( s ¢, v , z ) +U ( s ¢, v, z ) ×v2 ( s , s ¢) = U1 ( s ¢, s ¢, z ( s ')) Letting s¢® s , we have v2 ( s , s ) = Differentiating once more time and letting s Âđ s , we have v22 ( s , s ) = a s- c Therefore, there is no lower endpoint deviation if G ( s ) > and a /( s - c) > G ¢( s ) / G ( s ) Q.E.D 40 Appendix B: Application to the reserve price signaling model Cai, Riley and Ye (2003) and Jullien and Mariotti (2003) study the following reserve price signaling model in auction settings A seller of an indivisible good has private information about certain characteristics of the good that the potential bidders not know Let θ ∈ Θ ⊆ R n be the seller’s information The seller’s own valuation of the good is γ s(θ ) , and the common value component of the bidders’ valuations is t (θ ) = s (θ ) , where γ is a positive parameter Bidder i ’s valuation is t + xi , where x i is the private value component that is known to himself only The bidders’ private signals {xi } are i.i.d random variables with a distribution function F (g) and an everywhere positive density function f (g) Suppose the seller uses a sealed bid second price auction to sell the good; and she sets a reserve price r Let sˆ be the perceived type of the seller, i.e., the perceived common value component in bidders’ valuations Using a variable transformation by defining the reserve markup m = r − sˆ , Cai, Riley and Ye (2003) show that the seller’s expected payoff can be expressed as U ( s, sˆ, m) = γ sF(1) (m) + sˆ(1 − F(1) (m)) + B (m) where F(1) (g) is the distribution function of the first order statistics and x B (m) = m( F(2) (m) − F(1) (m)) + ∫ xdF(2) ( x) Thus the model fits into the standard signaling m framework.5 To fit this into our signaling model, we adopt another variable transformation: y = F(1) (m) ∈ [0,1] , and let H ( y ) = B (m( y )) Then the seller’s expected payoff is U ( s, sˆ, y ) = γ sy + sˆ(1 − y ) + H ( y ) It can be verified that Cai, Riley and Ye (2003) study a much more general model with affiliated private signals of the bidders and general valuation functions 41 ′ ( m ) J (m ) F(1) ′ (m ) = − J (m( y )) H ′( y ) = B′( m) y ′(m ) = − F(1) ′ (m ) H ′′( y ) = − J ′( m) F(1) where J ( x ) = x − (1 − F ( x )) / f ( x ) that is assumed to be strictly increasing in x With this transformation, the derivatives of U ( s, sˆ, y ) are U1 ( s, sˆ, y ) = γ y , U ( s, sˆ, y ) = − y, U ( s, sˆ, y ) = γ s − sˆ + H ′( y ) U11 ( s, sˆ, y ) = 0, U12 ( s, sˆ, y ) = 0, U13 ( s, sˆ, y ) = γ U 22 ( s, sˆ, y ) = 0, U 23 ( s, sˆ, y ) = −1, U 33 ( s, sˆ, y ) = H ′′( y ) The standard assumptions and B1 and B2 are all satisfied By the standard results, a separating equilibrium satisfies z′( s ) = − U ( s, s, z ( s )) − z ( s) =− U ( s, s, z ( s )) (γ − 1) s + H ′( z ( s)) We now analyze when this separating equilibrium satisfies the conditions of Theorem We focus on the special case t = γ s, γ ≥ Suppose (1 − γ ) s < J ( x ) so that the separating equilibrium goes through m* ( s ) = x at s Since U ( s, s, z ( s)) = (1 − γ ) s − J ( z ( s )) is decreasing in s , U ( s, s, z ( s)) < (1 − γ ) s − J ( x ) < for all s ∈ [ s, s ] In this model, since U113 ( s, v, y ) = U133 ( s, v, y ) = , condition becomes G ′′( s ) [(4 − 2γ ) z′( s ) + H ′′( z ( s ))( z ′( s )) ] > , 2γ (1 − z ( s )) G′( s ) Condition requires that G ( s ) > and γ G′( s ) > (γ − 1) s + H ′( z ( s)) G ( s ) 42 References Cai, Hongbin, John Riley and Lixin Ye (2003), “Reserve Price Signaling,” Working 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Economics, 87, 355-379 44 ... deviate if the receiver has the perception of v So the LCT is satisfied On the other hand, if the LCT is satisfied, then it must be the case that v ≥ v Otherwise, it is clear from Figures 4.1 and. .. separating equilibria can pass the Local Credibility Test The idea of the Local Credibility Test is weaker than the Strong Intuitive Criterion (SIC) proposed by Grossman and Perry (1986a,b) For any... ∉ S o Then the signal-perception ( yˆ , sˆ) is credible By the definition of the Local Credibility Test given in Section 2, if there does not exist any credible signal-perception, then the separating