ECO 5341 Signaling Games: Another Example. Saltuk Ozerturk (SMU)
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1 ECO 5341 : Another Example
2 and Perfect Bayesian Equilibrium (PBE) (1,3) (2,4) Right Right (0,0) (1,0) With probability Player 1 is. With probability, Player 1 is. cannot observe P1 s type. However, can observe whether P1 moves or Right and form beliefs about his type.
3 and Perfect Bayesian Equilibrium (PBE) A Perfect Bayesian Equilibrium requires players To maximize their payo s given their beliefs about the type of each player. Whenever possible, these beliefs must be consistent with the actions that players take in the game. In this game P1 s choice of or Right sends a signal to () about his type. Given this signal ( or Right), () forms beliefs about P1 s type and decides to play or.
4 and Perfect Bayesian Equilibrium (PBE) (1,3) (2,4) Right Right (0,0) (1,0) For example, in an equilibrium in which plays and plays Right (Separating Equilibrium), upon observing, () must believe that P1 is with probability 1, whereas upon observing Right () must believe (P1) is Type 2 with probability 1.
5 and Perfect Bayesian Equilibrium (PBE) (1,3) (2,4) Right Right (0,0) (1,0) In an equilibrium in which both and play Right (Pooling Equilibrium), upon observing Right, () must believe that (P1) is with probability and must believe (P1) is with probability.
6 () s optimal actions after observing and Right (1,3) (2,4) Right Right (0,0) (1,0) Let p denote the probability () assigns to (P1) being type 1 upon observing. Hence 1 p is the probability () assigns to (P1) being type 2 upon observing. If () plays upon observing, he gets 3p + 4(1 p) whereas if () plays he gets 1 p: Hence upon observing () always plays.
7 () s optimal actions after observing and Right (1,3) (2,4) Right Right (0,0) (1,0) Let q denote the probability () assigns to (P1) being type 1 upon observing Right. Hence 1 p is the probability () assigns to (P1) being type 2 upon observing Right. If () plays upon observing Right, he gets q whereas if () plays he gets 2(1 q): Hence upon observing Right () plays if and only if and plays if q : q > 2(1 q) ) q > 2 3
8 () s optimal actions after observing and Right (1,3) (2,4) Right Right (0,0) (1,0) Therefore, after observing () always plays and after observing Right () plays if and only if he believes (P1) is with a probability at least 2/3.
9 Separating PBE (1,3) Right (0,0) (2,4) (1,0) Right Is there is a Separating PBE in which plays and plays Right? In this PBE, upon observing Right, () believes that (P1) is with probability 1 and he plays. on observing, () plays. But will deviate from this equilibrium. If Type W plays Right he gets 1 whereas if he plays he would get 2. There is no Separating PBE in which plays and plays Right.
10 Separating PBE (1,3) Right (0,0) (2,4) (1,0) Right Is there is a Separating PBE in which plays Right and plays? In this PBE, upon observing Right, () believes that (P1) is with probability 1 and he plays. on observing, () believes that (P1) is with probability 1 and he plays. and have no incentive to deviate. There is a Separating PBE in which plays Right and plays.
11 Pooling PBE (1,3) Right (0,0) (2,4) (1,0) Right Is there is a Pooling PBE in which both and Type 2 play L? In this PBE, upon observing L, () believes J is with probability and plays. Suppose if () observes R he believes (P1) is with probability q < 2=3 and hence he plays. has no incentive to deviate: If type 1 plays L he gets 1, whereas if he plays R he gets 0. does not deviate either. If type 2 plays L he gets 2, whereas if he plays R he gets 1. This is a PBE!
12 Pooling PBE (1,3) Right (0,0) (2,4) (1,0) Right There is a Pooling PBE in which both and play. In this PBE, upon observing, () believes (P1) is with probability and plays. To sustain this PBE, o equilibrium beliefs of () is such that if he were to observe Right () assigns a probability q < 2=3 that (P1) is and hence he plays. Why do we need () to play if he were to observe R to sustain this PBE?
13 Pooling PBE (1,3) Right (0,0) (2,4) (1,0) Right Is there is a Pooling PBE in which both and Type 2 play R? In this PBE, upon observing R, () believes (P1) is with probability and he plays. Since is not observed we must assign o -equilibrium beliefs for the event that () observes. But note that for any beliefs, () plays after observing. has an incentive to deviate. If type 1 plays L he gets 1, whereas if he plays R he gets 0. No such PBE!
14 Summary So we have one Separating and one Pooling PBE! Pooling PBE: Both and play. In this PBE, upon observing, () believes (P1) is with probability and plays. To sustain this PBE, o equilibrium beliefs of () is such that if he were to observe Right () assigns a probability q < 2=3 that (P1) is and hence he plays. Separating PBE: plays Right and plays. In this PBE, upon observing Right, () believes that (P1) is with probability 1 and he plays. on observing, () believes that (P1) is with probability 1 and he plays.
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