- \n
- 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.<\/li>\n\n\n\n
- 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.<\/li>\n\n\n\n
- 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.<\/li>\n<\/ul>\n\n\n\n
![\"\"](\"https:\/\/saraheee.com\/wp-content\/uploads\/2023\/07\/image-35-1024x531.png\")
<\/figure><\/div>\n\n\n(1) Represent player 1’s strategy<\/mark><\/strong> \\(\\sigma_1\\) with two parameters:<\/p>\n\n\n\n
\\(\\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<\/p>\n\n\n\n(2) Compute player 2’s Bayesian beliefs<\/mark><\/strong> given \\(\\sigma_1\\):<\/p>\n\n\n\n
p \u2261 \\(\\mu_2^B(w) = \\frac{P_\\sigma(a)}{P_\\sigma(I_B)} = \\frac{0.1x}{0.1x+0.9y} = \\frac{x}{x+9y}\\)<\/p>\n\n\n\n
q \u2261 \\(\\mu_2^Q(w) = \\frac{0.1(1-x)}{0.1(1-x) + 0.9(1-y)} = \\frac{1-x}{10-x-9y}\\)<\/p>\n\n\n\n
(3) Find player 2’s best response<\/mark><\/strong> to \\(\\sigma_1\\) given beliefs:<\/p>\n\n\n\n
player 2’s BR @ \\(I_B = \\begin {cases}\u00a0F & \\text{if } p \\geq \\frac{1}{2} \\\\\u00a0W & \\text{if } p < \\frac{1}{2} \\end {cases}\\)<\/p>\n\n\n\n
player 2’s BR @ \\(I_Q = \\begin {cases}\u00a0F & \\text{if } q \\geq \\frac{1}{2} \\\\\u00a0W & \\text{if } q < \\frac{1}{2} \\end {cases}\\)<\/p>\n\n\n\n
(4)<\/strong> Divide analysis into cases according to player 1’s choice of x and y:<\/p>\n\n\n\n
- \n
- (x, y) = (1, 1), a pooling strategy<\/em> in which both types send message B<\/li>\n<\/ul>\n\n\n\n
![\"\"](\"https:\/\/saraheee.com\/wp-content\/uploads\/2023\/07\/image-37-1024x797.png\")
<\/figure><\/div>\n\n\np = 0.1 but q is unrestricted
If B was sent, W is BR given p = 0.1<\/p>\n\n\n\nSuppose Q was sent and q < 0.5 \u2192 W is BR
Not an eqbm \u2235 the w-type would then deviate to Q<\/p>\n\n\n\nSuppose Q was sent but q \u2265 0.5 \u2192 F is BR
Both types wouldn’t deviate, and thus it is an eqbm<\/p>\n\n\n\nThere exists on perfect Bayesian eqbm in pooling strategies:<\/p>\n\n\n\n
\u27e8(BB, WF), p = 0.1, q \u2265 0.5\u27e9<\/strong><\/p>\n\n\n\n
(4-2)<\/strong> (x, y) = (1, 0), a separating strategy in which the weak type sends Q but the strong type sends B.<\/p>\n\n\n
\n![\"\"](\"https:\/\/saraheee.com\/wp-content\/uploads\/2023\/07\/image-38-1024x778.png\")
<\/figure><\/div>\n\n\nBayesian beliefs under \\(\\sigma_1\\) = (1, 0) are
p = 1 and q = 0<\/p>\n\n\n\nThat 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<\/p>\n\n\n\nNot an eqbm b\/c the wimpy type would deviate to Q<\/p>\n\n\n\n
No PBE exists in this case<\/p>\n\n\n\n
There is no<\/em> PBE in which player 1 uses the separating strategy BQ.<\/p>\n\n\n\n
(4-3)<\/strong> (x, y) = (0, 1), another separating strategy<\/p>\n\n\n
\n![\"\"](\"https:\/\/saraheee.com\/wp-content\/uploads\/2023\/07\/image-41-1024x815.png\")
<\/figure><\/div>\n\n\nBayesian beliefs are
p = 0 and q = 1<\/p>\n\n\n\nThat 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<\/mark><\/p>\n\n\n\nNot an eqbm b\/c the wimpy type would deviate to B<\/mark><\/p>\n\n\n\n
No PBE exists in this case<\/p>\n\n\n\n
Hence no<\/em> PBE exists in which player employs a separating strategy QB.<\/mark><\/p>\n\n\n\n
<\/p>\n\n\n\n
(4-4)<\/strong> Lastly, we examine (x, y) = (0, 0) in which both types send Q<\/p>\n\n\n
\n![\"\"](\"https:\/\/saraheee.com\/wp-content\/uploads\/2023\/07\/image-42-1024x828.png\")
<\/figure><\/div>\n\n\n- \n
- q = 0.1 but p is unrestricted<\/li>\n\n\n\n
- If Q was sent W<\/mark><\/strong> is a BR<\/li>\n<\/ul>\n\n\n\n
\u27e8(QQ, FW), p \u2265 0.5, q = 0.1\u27e9<\/strong><\/mark><\/p>\n\n\n\n
If p < 1\/2, player 2 will walk away whether player 1 orders a Beer or a Quiche<\/mark><\/p>\n\n\n\n
So player 1 has an incentive to debate.<\/mark>
In the weak type, player 2 has no incentive to debate because player 1 gets 3 from Q,<\/mark>
In the strong type, player 1 has an incentive to debate because he gets a payoff of 3 if he orders Beer.<\/mark><\/p>\n\n\n\nSo there is no PBE when p is less than 1\/2.<\/mark><\/p>\n\n\n\n
<\/p>\n\n\n\n
- (x, y) = (1, 1), a pooling strategy<\/em> in which both types send message B<\/li>\n<\/ul>\n\n\n