We study how to compute a PBE by going through the next classic example

EXAMPLE 5.7 (THE BEER-QUICHE GAME, CHO AND KREPS (1987)).

- Player 1 can be weak (w) or strong (s), and let Pr(\(t_1\) = w) = 0.1. The strong type likes beer for breakfast, while the weak type likes quiche.
- Player 1 is ordering his breakfast, and player 2 is watching and contemplating whether to pick a fight with player 1. Player 2 wants to fight with the weak type but walk away from the strong type.
- Player 1 likes to avoid a fight: he gets a payoff of one from the preferred breakfast, and a payoff of two from avoiding the fight.

(1) Represent player 1’s **strategy** \(\sigma_1\) with two parameters:

\(\sigma_1 = (\sigma_1^w(B), \sigma_1^s(B)) = (x, y)\)

\(\sigma_1^w(B)\): for wimpy type, \(\sigma_1^s(B)\): for surly type

(2) Compute player 2’s **Bayesian beliefs** given \(\sigma_1\):

p ≡ \(\mu_2^B(w) = \frac{P_\sigma(a)}{P_\sigma(I_B)} = \frac{0.1x}{0.1x+0.9y} = \frac{x}{x+9y}\)

q ≡ \(\mu_2^Q(w) = \frac{0.1(1-x)}{0.1(1-x) + 0.9(1-y)} = \frac{1-x}{10-x-9y}\)

(3) Find player 2’s **best response** to \(\sigma_1\) given beliefs:

player 2’s BR @ \(I_B = \begin {cases} F & \text{if } p \geq \frac{1}{2} \\ W & \text{if } p < \frac{1}{2} \end {cases}\)

player 2’s BR @ \(I_Q = \begin {cases} F & \text{if } q \geq \frac{1}{2} \\ W & \text{if } q < \frac{1}{2} \end {cases}\)

**(4)** Divide analysis into cases according to player 1’s choice of x and y:

- (x, y) = (1, 1), a
*pooling strategy*in which both types send message B

p = 0.1 but q is unrestricted

If B was sent, W is BR given p = 0.1

Suppose Q was sent and q < 0.5 → W is BR

Not an eqbm ∵ the w-type would then deviate to Q

Suppose Q was sent but q ≥ 0.5 → F is BR

Both types wouldn’t deviate, and thus it is an eqbm

There exists on perfect Bayesian eqbm in pooling strategies:

**⟨(BB, WF), p = 0.1, q ≥ 0.5⟩**

**(4-2)** (x, y) = (1, 0), a separating strategy in which the weak type sends Q but the strong type sends B.

Bayesian beliefs under \(\sigma_1\) = (1, 0) are

p = 1 and q = 0

That is, player 2 exactly infers player 1’s type from the message sent

F is a BR to B and W is a BR to Q

Not an eqbm b/c the wimpy type would deviate to Q

No PBE exists in this case

There is *no* PBE in which player 1 uses the separating strategy BQ.

**(4-3)** (x, y) = (0, 1), another separating strategy

Bayesian beliefs are

p = 0 and q = 1

That is, player 2 exactly infers player 1’s type from the message sent

F is a BR to Q and W is a BR to B

Not an eqbm b/c the wimpy type would deviate to B

No PBE exists in this case

Hence *no* PBE exists in which player employs a separating strategy QB.

**(4-4)** Lastly, we examine (x, y) = (0, 0) in which both types send Q

- q = 0.1 but p is unrestricted
- If Q was sent
**W**is a BR

**⟨(QQ, FW), p ≥ 0.5, q = 0.1⟩**

If p < 1/2, player 2 will walk away whether player 1 orders a Beer or a Quiche

So player 1 has an incentive to debate.

In the weak type, player 2 has no incentive to debate because player 1 gets 3 from Q,

In the strong type, player 1 has an incentive to debate because he gets a payoff of 3 if he orders Beer.

So there is no PBE when p is less than 1/2.

- Reference: Chang-Koo Chi, (39/50) Game Theory and Applications 10 – Computing PBE in signaling games, Jul 14, 2020, https://youtu.be/KrpsV-91YTI