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Dynamic Games of Incomplete Information: Exercising the PBE

Dynamic Games of Incomplete Information: Exercising the PBE. APEC 8205: Applied Game Theory. Objectives. Learn How to Find Perfect Bayesian Equilibria (PBE) Finite Types & Actions Continuous Types & Actions. Example 1. (50, 50). U. Player A |1. (40, 25). m. L. R. P=0.4. D. (0, 0).

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Dynamic Games of Incomplete Information: Exercising the PBE

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  1. Dynamic Games of Incomplete Information:Exercising the PBE APEC 8205: Applied Game Theory

  2. Objectives • Learn How to Find Perfect Bayesian Equilibria (PBE) • Finite Types & Actions • Continuous Types & Actions

  3. Example 1 (50, 50) U Player A|1 (40, 25) m L R P=0.4 D (0, 0) Nature Player B (50, 0) 1-P=0.6 U 1- m L R (30, 30) Player A|2 D (0, 50)

  4. How can we solve this game for the PBE? • As with subgame perfection, it is useful to work backwards. • Find B’s best response given R. • Find A’s best response given A’s type & the probability B plays U. • Use Bayes rule to determine  given A’s best responses. • Check  for consistency with B’s best response.

  5. Find B’s best response given R. B’s payoff to choosing U given R: B’s payoff to choosing D given R: B(U|R) >/=/< B(D|R) implies

  6. Graphically sB(U|R) 0.5 m

  7. Find A’s best response given A’s type & the probability B plays U. Type t A’s payoff to choosing R given B(U|R): Type 1 A’s payoff to choosing L:A(L|1) = 40 Type 2 A’s payoff to choosing L:A(L|2) = 30 A(R|1) >/=/< A(L|1) implies A(R|2) >/=/< A(L|2) implies

  8. Graphically sA(R|1)=sA(R|2)=1 0.8 sA(R|1)=[0,1] sA(R|2)=1 sA(R|1)=0 sA(R|2)=1 0.6 sB(U|R) sA(R|1)=0 sA(R|2)=[0,1] sA(R|1)=sA(R|2)=0 0.5

  9. Use Bayes rule to determine  given A’s best responses. = 0.4 = 0.4 sA(R|1)=sA(R|2)=1 0.8 sA(R|1)=[0,1] sA(R|2)=1 = 0 sA(R|1)=0 sA(R|2)=1 = [0,0.4] 0.6 sB(U|R) = [0,1] sA(R|1)=0 sA(R|2)=[0,1] sA(R|1)=sA(R|2)=0 = [0,1] 0.5 = [0,1] m

  10. Check  for consistency with B’s best response. • A(R|1) = A(R|2) = 0, B(U|R) = [0, 0.6], &  = 0.5. • A(R|1) = A(R|2) = 0, B(U|R) = 0, &  < 0.5. Note that all these equilibria are observationally equivalent!

  11. Example 2 (10, 50) (50, 50) U U Player A|1 mR mL L R D P = 1/4 D (30, 0) (0, 0) Player B Nature Player B (0, 0) (10, 0) U 1-P U 1- mR 1- mL L R Player A|2 D D (50, 50) (30, 50)

  12. How can we solve this game for the PBE? • Again, it is useful to work backwards. • Find B’s best response given R & given L. • Find A’s best response given A’s type & the probability B plays U following R & U following L. • Use Bayes rule to determine R & L given A’s best responses. • Check R & L for consistency with B’s best responses.

  13. Find B’s best response given R & given L. B’s payoff to choosing U given action a = R, L: B’s payoff to choosing D given a: B(U|a) >/=/< B(D|a) implies

  14. Graphically mR>1/2 mL<1/2 mR>1/2 mL>1/2 mR>1/2, mL=1/2 1 mR=1/2, mL=1/2 mR=1/2, mL<1/2 sB(U|R) mR=1/2, mL>1/2 0 1 mR<1/2, mL=1/2 mR<1/2 mL<1/2 mR<1/2 mL>1/2 sB(U|L)

  15. Find Type 1 A’s best response given the probability B plays U following R & U following L. Type 1 A’s payoff to choosing R given B(U|R): Type 1 A’s payoff to choosing L given B(U|L): A(R|1) >/=/< A(L|1) implies

  16. Find Type 2 A’s best response given the probability B plays U following R & U following L. Type 2 A’s payoff to choosing R given B(U|R): Type 2 A’s payoff to choosing L given B(U|L): A(R|2) >/=/< A(L|2) implies

  17. Graphically 1 sA(R|1)=1 sA(R|2)=0 sA(R|1)=1 sA(R|2)=1 0.6 sA(R|1) = (0,1) sB(U|R) sA(R|1)=0 sA(R|2)=0 sA(R|1)=0 sA(R|2)=1 sA(R|2) = (0,1) 0.2 0 0.4 1 0.8 sB(U|L)

  18. Use Bayes rule to determine R & L given A’s best responses. mR = 1/ (1 + 3A(R|2)) L= 0 1 sA(R|1)=1 sA(R|2)=0 mR = 1 mL = 0 mR = 1 mL = (1 - A(R|1))/ (1 - A(R|1)) sA(R|1)=1 sA(R|2)=1 0.6 sA(R|1) = (0,1) mR = A(R|1)/ (A(R|1) + 3A(R|2)) L= (1 - A(R|1))/ (4 - A(R|1) - 3A(R|2)) mR = 1/4 mL = [0,1] sB(U|R) sA(R|1)=0 sA(R|2)=0 sA(R|1)=0 sA(R|2)=1 mR = [0,1] mL = 1/4 sA(R|2) = [0,1] mR = A(R|1)/ (A(R|1) + 3) L= 1 0.2 mR = 0 mL = 1 mR = 0 mL = 1/ (4 - 3A(R|2)) 0 0.4 1 0.8 sB(U|L)

  19. Check R & L for consistency with B’s best responses. • A(R|1) = A(R|2) = 0, B(U|R) = B(U|L) = 0, R  ½, and L = ¼. • A(R|1) = 0, A(R|2) = 1, B(U|R) = 0, B(U|L) = 1, R = 0, and L = 1. • A(R|1) = 1, A(R|2) = 0, B(U|R) = 1, B(U|L) = 0, R = 1, and L = 0. • A(R|1) = 0, A(R|2) = 0, B(U|R) = (0, 0.6), B(U|L) = 0, R = ½, and L = ¼.

  20. Example: SLAPP GAME • Players? • Homeowner (H) & Developer (D) • Who does what when? • Nature chooses Homeowner’s value to preserving green space (VH) & Developer’s value to dosing green space (VD). • Homeowner get to choose whether to Fight (F) or Whine (W). • If the homeowner chooses F, H invests effort first in a Stackelberg contest. • If the homeowner chooses W, D invests effort first in a Stackelberg contest. • Who knows what when? • H gets to see its value & D’s value, while D only gets to see its own value. • If H choose F, D gets to see H’seffort before choosing its own. • If H choose W, H gets to see D’seffort before choosing its own. • How are Players Rewarded? • i = Vixi/(xi + xj) – xi for i, j = H, D and i  j.

  21. Characterization of Beliefs • Nature chooses the Homeowner’s value v [v’, v’’] with density f(v) and distribution F(v). • Developer can update it beliefs: • After F: v  RF for fF(v) & FF(v) • After W: v  RW for fW(v) & FW(v)

  22. How can we solve this game using the PBE? • H Leads: • Figure out D’s effort given H’s choice of effort. • Figure out H’s effort & expected reward knowing how D will respond. • D Leads: • Figure out H’s effort given D’s choice of effort. • Figure out D’s effort knowing how H will respond. • Figure out H’s expected reward. • Compare H’s expected rewards to determine when it should choose F & when it should choose W. • Check for consistency of beliefs with H’s and D’s best responses.

  23. H Leads: D’s optimal effort given H’s effort: Which yields the best response:

  24. H Leads: H’s optimal effort given D’s best response: Which yields:

  25. D Leads: H’s optimal effort given D’s effort: Which yields the best response:

  26. D Leads: D’s optimal effort given H’s best response: Which yields:

  27. Compare H’s expected rewards to determine when it should choose F & when it should choose W. UH(W) >/=/< UH(F) as or

  28. Implications • If E(D|W)  1, the homeowner should choose F regardless of its value. • If E(D|W) < 1, • the homeowner should choose W if VD(1 – (1 - E (D|W))0.5)2 > v > VD(1 + (1 - E(D|W))0.5)2,such that RW = (VD(1 – (1 - E (D|W))0.5)2, VD(1 + (1 - E(D|W))0.5)2) and fW(v) = f(v) / (F(VD(1 + (1 - E(D|W))0.5)2) – F(VD(1 – (1 - E (D|W))0.5)2)). • otherwise it should choose F such that RF = {2(xHVD)0.5} and Pr(v = 2(xHVD)0.5) = 1.

  29. Case where 1 < E(rD|W) 0 rH UH(W) - UH(F)

  30. Case where 1 > E(rD|W) rH =1 + (1-E(rD|W))0.5 0 rH 1 rH =1 – (1-E(rD|W))0.5 UH(W) - UH(F)

  31. Comparison of Complete and Incomplete Information Games When 1 > E(rD|W) UH(W) - UH(F) Under Complete Information rH =1 – (1-E(rD|W))0.5 rH =1 + (1-E(rD|W))0.5 0 rH 1 UH(W) - UH(F) Under Incomplete Information

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