Classes of Games - Signaling Games

1 What we learning • Classes of games: signaling games • Solution concepts: perfect Bayesian equilibrium (PBE), signaling refinements A and B 2 Problems Note: I slightly modified the answer to Problem 5 in Section 5 to make it more clear. Please check the updated section handout online, especially if you did not attend section that week. H A B . No Hire (0, 0) b ir e Jo Jo {2/3} Math School ir e b tMATH H Literature School H b No Hire ir e Jo (-4, 0) H B (1, 2) H A B Jo {1/3} Math School B H b ir e Jo 2 3 No Hire (-4, 0) ir e b . H Jo tLIT H b Literature School ir e Jo No Hire (5, 7) Notice that (1, 2) b ir e (0, 0) (5, 2) R A R (5, 2) (5, 7) Jo b ir e (1, 7) A Problem 1: Stay in Some School (2003 Exam) (1, 7) of potential workers are good in Math. (i) How many pure strategies does S (the worker) have? Give one example. (ii) How many pure strategies does R (the employer) have? Give one example. (iii) What are all of the pure-strategy (Bayesian) Nash equilibria in the game? (iv) What are all of the pure-strategy Perfect Bayesian equilibria in the game? Give a brief intuition for why the set of pure-strategy PBE does or does not eliminate one or more purestrategy Nash equilibria. (v) What are all of the pure-strategy PBE meeting Signaling Refinement A in the game? Give a brief intuition for why the set of pure-strategy PBE meeting Signaling Refinement A does or does not eliminate one or more pure-strategy PBE. 1 The basic ideas in this paper are illustrated by the following simple game. The reader should refer to Figure I throughout. We tell the story of a two-player game with incomplete information concerning one of the two. The first player is called A, and A either is a wimp or is surly. Nature has selected the disposition of A, with probability 0.9 that the A selected is surly. In terms of Figure I, nature has chosen to start the game a t one of the two open dots labeled t , (for game the wimp) andand t , (for the surly type of A). The prior Problem 2: Beer-quiche (Cho Kreps, 1987) The first player is called A, and A either is a wimp (tw ) or is surly (ts ). At the start of the game, A knows his type, and is faced with the choice of what breakfast to have: beer or quiche. After breakfast, A meets with B. B, at this meeting, chooses whether to duel A. (i) What are all of the (Bayesian) Nash equilibria in the beer-quiche game? (ii) What are all of the Perfect Bayesian equilibria in the beer-quiche game? (iii) What are all of the PBE meeting Signaling Refinement A in the beer-quiche game? (This is the weaker of the two signaling refinements, that relies on restricting beliefs based on the sending of some messages by some types being dominated.) (iv) What are all of the PBE meeting Signaling Refinement B in the beer-quiche game? (This is the stronger of the two signaling refinements, equivalent here to Cho-Kreps’s “Intuitive Criterion,” restricting beliefs based on the sending of some messages by some types being dominated by their equilibrium payoffs.) Problem 3: IPO model There is an entrepreneur who would like to take his company public. She may be of the high type, with future profits for the company θ = 2, or of the low type, with future profits for the company θ = 1. The entrepreneur observes her type but the market does not. The market’s prior belief is that θ is equally likely to be high or low. The entrepreneur chooses the fraction q ∈ [0, 1] of the company to sell to the market. The market observes q and forms a belief about the entrepreneur’s type. If the market assigns belief µ(q) that the entrepreneur is of the high type, then the offering price per share is p(q) = µ(q)2 + (1 − µ(q))1 If the entrepreneur’s type is θ, the entrepreneur offers q for sale, and the market pays a per-unit price p, the market’s payoff is θq − pq, while the entrepreneur’s payoff is pq + 2θ (1 − q). Consider first separating equilibria where the high type offers a fraction qH to the public and the low type a fraction qL . Denote the prices corresponding to qH and qL by pH and pL respectively. 2 (i) Show that in any separating PBE, qL = 1, pL = 1. (ii) Derive conditions on qH , pH such that (qH , pH ) could be part of a separating PBE where the entrepreneur chooses qH if she is the high type. (iii) If the fractions (qL , qH ) and corresponding prices (pL , pH ) are part of a separating PBE, what must be true for p(q) for all q ∈ / {qL , qH } in equilibrium? (iv) Is any of the separating PBE ruled out by Signaling Refinement A? Consider now pooling equilibria where both types offer q ∗ and the equilibrium price is p∗ . (v) Show that there exists a pooling PBE in which both types offer q ∗ = 1 and the offering price per share is p∗ = 3/2. (vi) For what other values of q is there a pooling PBE? Characterize these equilibria. (vii) Show that any pooling PBE fails Signaling Refinement B. Problem 4: Semiseparating perfect Bayesian equilibrium A semiseparating (or partially separating/pooling) equilibrium is an equilibrium in which some types of Sender send the same message, while some others send some other messages. A semiseparating equilibrium also arises when mixed strategies are played. For example, if there are two types, a semiseparating equilibrium arises when some messages are sent by both types with positive probability while some other messages are sent by only one of the types with positive probability. Consider the Beer-Quiche game in Problem 2 but suppose that the prior probability that A is a wimp is now 0.9 (and the prior that he is surly is 0.1). Can you find a semiseparating PBE in this modified Beer-Quiche game? 3 3 Answers Problem 1 (i) S has two information sets and two actions available at each information set, so he has a total of 2 × 2 = 4 pure strategies. One example is “Play L if tM AT H , M if tLIT ,” which I denote by LM, meaning that S chooses Literature School if his type is tM AT H and Math School if his type is tLIT . (ii) R has two information sets and three actions available at each information set, so he has a total of 3 × 3 = 9 pure strategies. One example is “Play N if L, B if M,” which I denote by NB, meaning that R chooses No Hire if he observes Literature School and Hire Job B if he observes Math School. (iii) To find the pure-strategy BNE, take each possible strategy for R, find out S’s best response, and check if the original proposed strategy for R is a best response to S’s best response. • Suppose R plays AA. S’s best response is ML. Now is R’s strategy (AA) a best response to ML? Since the two types of S choose different strategies, R knows he is facing tLIT if he observes L. But then his best response is to play B if L (since 7 > 2, 7 > 0). Hence, there is no BNE where R plays AA. • Suppose R plays AN. S’s best response is LL. Is AN a best response for R? Since the two types of S choose the same strategy, R uses his prior beliefs to calculate EU [A] = 7(2/3) + 2(1/3) = 16/3, EU [N ] = 0, EU [B] = 2(2/3) + 7(1/3) = 11/3. So A if L is a best response. And then we are done (since M is off the equilibrium path!). Hence, (LL,AN) is a BNE. • Suppose R plays AB. S’s best response is ML. But then R’s best response is BA. Hence there is no BNE where R plays AB. • Suppose R plays NA. S’s best response is MM. And then using the prior, we can verify that A is a best response to M. Hence (MM,NA) is a BNE. • Suppose R plays NN. S’s best response is ML. But then R’s best response is BA. Hence, there is no BNE where R plays NN. • Suppose R plays NB. S’s best response is MM. But then R’s best response to M is A. Hence there is no BNE where R plays NB. • Suppose R plays BA. S’s best response is ML. R knows he is facing tLIT when he observes L, and tM AT H when he observes M, so R’s best response is BA. Hence (ML,BA) is a BNE. • Suppose R plays BN. S’s best response is LL. But then S’s best response to L is A. Hence there is no BNE where R plays BN. • Suppose R plays BB. S’s best response is ML. But then R’s best response is BA. Hence there is no BNE where R plays BB. In sum, we found three BNE: (LL,AN), (MM,NA), and (ML,BA). (iv) We need to check if for the BNE found above, there is an out-of-equilibrium message. For each such message, we need to check whether there exist beliefs such that R’s strategy is a best response. (Note that we just need to find a single example of beliefs R might have that would justify his strategy as a best response.) • Consider (LL,AN). Both types are playing L, so M is an out-of-equilibrium message. R responds by playing N if M. Are there beliefs that justify R playing N if M? No. Notice 4 that by playing N following M, R obtains a payoff of 0, no matter what type of player S he is facing. But either job gives him at least 2, no matter what type he is facing. Hence, there are no beliefs that R can have that would justify playing N if M, and this BNE is not a PBE. • Consider (MM,NA). Now L is off the equilibrium path. Are there beliefs about types that would justify not hiring if L? No. Not hiring gives R a payoff of 0 no matter what type of S he is facing, while hiring in any job gives R a strictly positive payoff, no matter what type he is facing. Hence, this BNE is not a PBE. • Consider (ML,BA), In this case, there are no out-of-equilibrium messages. Since everything is on the equilibrium path, players are always best responding. Hence, this BNE is a PBE. (v) Signaling Refinement A puts restrictions on what R’s beliefs can be off the equilibrium path. But note that for the unique PBE that we found above, there are no off-the-equilibrium messages (and hence there are no off-the-equilibrium beliefs). Hence, the set of PBE that meet Signaling Refinement A (and B) is the same as the set of PBE. Problem 2 (i) First note that there is no BNE in which the different types of A choose different actions. To see this, suppose that A plays Beer if wimpy and Quiche if surly. Then B must believe that he is facing tw if he observes Beer and ts if he observes Quiche. Then B’s best response is to play Duel if Beer and Don’t duel if Quiche. But then tw would want to deviate to Quiche. Similarly, suppose that A plays Quiche if wimpy and Beer if surly. Then B must believe that he is facing tw if he observes Quiche and ts if he observes Beer. Then B’s best response is to play Duel if Quiche and Don’t duel if Beer. But then tw would want to deviate to Beer. Consider next BNE where both types of A play the same action. Suppose first that A chooses Beer regardless of her type. Then B chooses Don’t Duel if Beer, since B’s belief is that A is surly with (prior) probability 0.9 if Beer. To sustain this as a BNE, we must make sure that no type wants to deviate. Note that the type that could have the most profitable deviation is tw . For this type not to deviate, we need B to choose Duel with probability p > 0.5 upon observing Quiche. Finally, suppose that A chooses Quiche regardless of her type. Then B chooses Don’t Duel if Quiche (since B’s belief is that A is surly with (prior) probability 0.9 if Quiche). To sustain this as a BNE, we must make sure that no type wants to deviate. In this case, the type that could have the most profitable deviation is ts . For this type not to deviate, we need B to choose Duel with probability p > 0.5 upon observing Beer. ∗ , σ ∗ ) with Hence, there are two classes of BNE. The first class is (σA B ∗ σA : Both types of A choose Beer. ∗ : B chooses Don’t duel if Beer and Duel with probability p > 0.5 if Quiche. σB ∗∗ , σ ∗∗ ) with The second class of BNE is (σA B ∗∗ σA : Both types of A choose Quiche. ∗∗ : B chooses Don’t duel if Quiche and Duel with probability p > 0.5 if Beer. σB (ii) The set of PBE is the same as the set of BNE characterized in part (i). To show this, note that for both classes of equilibria, there exist beliefs at the out-of-equilibrium information sets that rationalize B’s dueling. Consider the first class of BNE where both types of A choose Beer. If B believes following Quiche that prob(t = tw ) > 0.5, then playing Duel with probability > 0.5 is an optimal response. (If he believes that prob(t = tw ) = 0.5 if Quiche, then playing 5 Duel with any probability is an optimal response, and if he believes that prob(t = tw ) > 0.5 if Quiche, then playing Duel with probability one is an optimal response.) So these equilibria are perfect Bayesian. Similarly, consider the second class of BNE where both types of A choose Quiche. If B believes following Beer that prob(t = tw ) > 0.5, then playing Duel with probability > 0.5 is an optimal response. So these equilibria are also perfect Bayesian. (iii) The set of PBE that meet Signaling Refinement A is the same as the set of PBE. This refinement does not rule out any PBE since there are no actions that are dominated for some types. The wimpy type prefers Beer to Quiche if B chooses Don’t Duel if Beer and Duel if Quiche, and he prefers Quiche to Beer if B chooses Don’t Duel if Quiche and Duel if Beer. The same is true for the surly type. (iii) The set of PBE that meet Signaling Refinement B is the PBE where both types of A choose Beer, B chooses Don’t Duel if Beer and Duel if Quiche. This refinement rules out the class of PBE where both types of A choose Quiche. We first show that the PBE where both types of A choose Beer survives this refinement. In these equilibria, the out-of-equilibrium message is Quiche. The surly type of A would never deviate to Quiche, since he gets 3 in equilibrium and could get at most 2 by deviating. The wimpy type of A would deviate if B chose Don’t Duel if Quiche, as he would then get a payoff of 3, which is greater than the payoff of 2 he obtains in the equilibrium. Signaling Refinement B then says that B’s belief should be restricted to put no weight on A being surly if Quiche is observed. Then B believes that A is a wimp if Quiche, and hence his best response is to Duel if Quiche. But this supports the equilibrium outcome described above. (More precisely, the PBE that meets Signaling Refinement B is that in which both types of A play Beer, B plays Don’t Duel if Beer, and Duel with probability one if Quiche.) We next show that the PBE where both types of A choose Quiche fail this refinement. In these equilibria, the wimpy type obtains a payoff of 3 at the equilibrium. By deviating to Beer, the best payoff he could obtain is 2. Thus, deviating to Beer is never profitable for this type, no matter what B believes upon observing Beer. But this is different for the surly type. At the equilibrium, this type obtains a payoff of 2. If he deviates to Beer and B believes that he is facing the surly type if he observes Beer, then this type obtains 3 > 2 (since B chooses Don’t Duel). Refinement B then says that we should restrict B’s beliefs to put no weight on the wimpy type of A if he observes Beer. Thus, the only beliefs B could hold if he observes Beer are that A must be surly, and B would not duel. Anticipating this, the surly type of A wants to deviate to Beer. Problem 3 (i) In a separating PBE, the market must believe that θ = 1 when it observes qL . Thus, pL = 1. Then note that it is impossible that qL < 1 in equilibrium, as the low type would want to deviate to qL = 1 to increase his payoff. (ii) In a separating PBE, the market must believe that θ = 2 when it observes qH . Thus, pH = 2. Now for the high type not to want to deviate, it must be that 2qH + (1 − qH ) > 1 which is always true (since q ∈ [0, 1]). And for the low type not to want to deviate, it must be that 1 2qH + (1 − qH ) 6 1 2 6 which holds iff qH 6 1/3. (iii) For all q ∈ / {qL , qH }, p(q) must be such that neither type has incentives to deviate. Since p(q) always satisfies 1 6 p(q) 6 2, no type wants to deviate to q < qH . For qH < q < qL , the following conditions are required respectively for the high type and the low type not to want to deviate 2qH + (1 − qH ) > p(q)q + (1 − q) ⇔ p(q) 6 qH /q + 1 1 1 > p(q)q + (1 − q) ⇔ p(q) 6 1/(2q) + 1/2 2 Thus, the condition for qH < q < qL is p(q) 6 min{qH /q + 1, 1/(2q) + 1/2}. Note that by setting the market’s belief µ(q) = 0 for q ∈ / {qL , qH }, a PBE with qL , qH as described in parts (i) and (ii) can be sustained. (iv) Yes, all the separating PBE with qH < 1/3 are ruled out by Signaling Refinement A. First note that for the high type, there is no fraction q that is dominated. To see this, note that for any q ∈ [0, 1], the highest payoff this type can obtain is maxp Us (H, q, p) = 2q +(1−q) = 1+q, and the lowest payoff he can obtain is minp Us (H, q, p) = 1q +(1−q) = 1. Then, for no fraction q there exists q 0 such that maxp Us (H, q, p) < minp Us (H, q 0 , p). For the low type and any q ∈ [0, 1], the highest payoff he can obtain is maxp Us (L, q, p) = 2q + 1 1 3 1 1 1 2 (1−q) = 2 + 2 q and the lowest payoff he can obtain is minp Us (L, q, p) = 1q+ 2 (1−q) = 2 + 2 q. This gives that all q < 1/3 are dominated for the low type. Hence, Signaling Refinement A says that we should restrict the market’s belief to be µ(q) = 1 if q < 1/3. The price must then be p(q) = 2 for q < 1/3. But then an equilibrium meeting with this refinement cannot have qH < 1/3, as there would be a profitable deviation for the high type. Hence, the only separating PBE that meets Signaling Refinement A is that in which the high type of the entrepreneur offers qH = 1/3, the low type offers qL = 1, and the prices are qH = 2, qL = 1. (v) In a pooling PBE, both types choose the same fraction q ∗ . Thus, the market’s belief is given by the prior, that is, µ(q ∗ ) = 0.5. This gives p∗ = 1 + 1/2 = 3/2. Next, to show that q ∗ = 1 is part of a pooling PBE, suppose the market believes that the entrepreneur is of the low type if it observes q < 1. Then neither type wants to deviate (and the market is best responding to q ∗ with p∗ = 3/2). (vi) As shown in part (v), the price in any pooling PBE must be p∗ = 3/2. For any equilibrium fraction q ∗ , we know that no type will want to deviate to any q < q ∗ if we set the market’s beliefs to be such that µ(q) = 0 for q < q ∗ . But some type may want to deviate to q 0 > q ∗ even if we specify beliefs to be µ(q 0 ) = 0 for q 0 > q ∗ . Indeed, the most tempting deviation is for the low type and it is to offer q = 1. Hence, for no type to have incentives to deviate from q ∗ , the following must hold 1 3/2q ∗ + (1 − q ∗ ) > 1 ⇔ q ∗ > 1/2 2 Therefore, a pooling PBE with q ∗ > 1/2 exists. It is supported by the market believing that the entrepreneur is of the low type for off-equilibrium values of q. (vii) Note that no matter what the market believes upon observing q < 23 q ∗ , the low type would never want to deviate to such q (since 2q+ 21 (1−q) < 3/2q ∗ + 12 (1−q ∗ ) for q < 23 q ∗ ). However, if the market believes that it’s the high type when observing q 0 > 21 q ∗ , then the high type would deviate to q 0 (since 2q 0 + (1 − q 0 ) > 3/2q ∗ + (1 − q ∗ ) for q 0 > 21 q ∗ ). Hence, Signaling Refinement B says that the market must believe that it’s the high type if it observes q ∈ ( 21 q ∗ , 23 q ∗ ). But 7 then the high type has a profitable deviation, and thus all pooling equilibria are ruled out by this refinement. Problem 4 Yes. Consider a PBE where the surly type of player A chooses Beer and the wimpy type mixes between Beer and Quiche with some probability so that if B observes Beer, he finds the two types of A equally likely and thus is indifferent between Duel and Don’t Duel. Hence, if he observes Beer, B mixes between Duel and Don’t duel with some probability so that the wimpy type of A is indifferent between Beer and Quiche, and hence willing to mix between these two pure strategies. Note that these beliefs are derived by using Bayes’ rule. So let p be the probability with which tw plays Beer (and (1 − p) the probability with which tw plays Quiche). Bayes’ rule gives that B’s belief that he is facing tw if he observes Beer is µ(tw ) = 0.9p 0.9p + 0.1 For B to be indifferent between Duel and Don’t duel following Beer, we need µ(tw )1 + (1 − µ(tw ))0 = µ(tw )0 + (1 − µ(tw ))1 which gives µ(tw ) = 1/2. Then µ(tw ) = 1/2 = 0.9p ⇔ p = 1/9 0.9p + 0.1 Now let q be the probability with which B plays Duel if Beer (and (1 − q) the probability with which B plays Don’t Duel if Beer). Note that following Quiche, B knows that he is facing tw , and hence he plays Duel. Hence, for tw to be indifferent between Beer and Quiche we need 0q + 2(1 − q) = 1 ⇔ q = 1/2 Finally, note that given B’s strategy, the best response for the surly type of A is to play Beer with probability one (since 0.5(1) + 0.5(3) > 0). Hence, we found a semiseparating PBE in the modified ∗ , σ ∗ ) given by Beer-Quiche game with (σA B ∗ σA : ts chooses Beer, tw chooses Beer with probability 1/9 and Quiche with probability 8/9. ∗ : B chooses Duel with probability 1/2 and Don’t Duel with probability 1/2 if Beer, and he σB chooses Duel if Quiche. 8