Communication Networks A Second Course Jean Walrand Department of EECS University of California at Berkeley

Games: Non-Cooperative One-Shot (Static) Game Nash Equilibrium Cournot Leader-Follower (Stackelberg) Bayes-Cournot Principal-Agent Problem Existence of NE

Static Game Example: Matching pennies HT H1, -1-1, 1 T 1, -1 Actions of Alice Actions of Bob Reward of Alice Reward of Bob One-Shot Game Both players play simultaneously Each player knows both reward functions One-Shot Game Both players play simultaneously Each player knows both reward functions

Static Game Game Example: 2x2 lr tA(t, l), B(t, l)A(t, r), B(t, r) bA(b, l), B(b, l)A(b, r), B(b, r) Actions of Alice Actions of Bob One-Shot Game Both players play simultaneously Each player knows both reward functions One-Shot Game Both players play simultaneously Each player knows both reward functions

Static Game Generally: One-shot game All players play simultaneously, independently Each player knows all the reward functions One-shot game All players play simultaneously, independently Each player knows all the reward functions Reward of player i: Strategy of player i: Goal of player i:

Static Game Example: Matching pennies H: bT: 1 - b H: a1, -1-1, 1 T: 1 - a-1, 11, -1

Static Game Nash Equilibrium In the non-randomized case: In words: No player has an incentive to deviate unilaterally.

Static Game Example: Matching pennies

Static Game Recall: Cournot Duopoly One-shot game Continuous action space Each player knows both reward functions One-shot game Continuous action space Each player knows both reward functions Two firms produce quantity q 1 and q 2 of a product The price is A – q 1 – q 2 For i = 1, 2 the profit of firm i is q i (A – q 1 – q 2 ) - Cq i Two firms produce quantity q 1 and q 2 of a product The price is A – q 1 – q 2 For i = 1, 2 the profit of firm i is q i (A – q 1 – q 2 ) - Cq i

Static Game Nash Equilibrium: Cournot Recall: Profit of firm 2 is q 2 (A – q 1 – q 2 – C) This is maximized for q 2 = (A – q 1 – C)/2 =: (B – q 1 )/2 Recall: Profit of firm 2 is q 2 (A – q 1 – q 2 – C) This is maximized for q 2 = (A – q 1 – C)/2 =: (B – q 1 )/2 Thus, q i = B/3 and reward is u i = B 2 /9.

Static Game Cournot: Cooperation u 2 = q 2 (A – q 1 – q 2 – C) Assume firms cooperate and split the revenues Then they choose q i = q/2 where q maximizes u 1 + u 2 = 2u i = q(A – q – C) = q(B – q) The maximum is q = B/2, so that q i = B/4 and u i = B 2 /8 The firms produce less and make more profit when they cooperate than when they compete. u 2 = q 2 (A – q 1 – q 2 – C) Assume firms cooperate and split the revenues Then they choose q i = q/2 where q maximizes u 1 + u 2 = 2u i = q(A – q – C) = q(B – q) The maximum is q = B/2, so that q i = B/4 and u i = B 2 /8 The firms produce less and make more profit when they cooperate than when they compete.

Static Game Stackelberg: Cournot Recall: u 2 = q 2 (B – q 1 – q 2 ) Assume firm 1 announces q 1 and firm 2 follows Then q 2 = (B – q 1 )/2, so that u 1 = q 1 (B - q 1 )/2 This is maximized by q 1 = B/2 so that q 2 = B/4 Hence u 1 = B 2 /8 and u 2 = B 2 /16 Intuition: Leader (1) has advantage. Recall: u 2 = q 2 (B – q 1 – q 2 ) Assume firm 1 announces q 1 and firm 2 follows Then q 2 = (B – q 1 )/2, so that u 1 = q 1 (B - q 1 )/2 This is maximized by q 1 = B/2 so that q 2 = B/4 Hence u 1 = B 2 /8 and u 2 = B 2 /16 Intuition: Leader (1) has advantage.

Static Game Nash Equilibrium: Bayes Cournot Recall: Profit of firm 2 is u 2 = q 2 (A – q 1 – q 2 – C 2 ) Assume firm i knows C i and the distribution of C 2 - i u 2 is maximized for q 2 = (A – q 1 – C 2 )/2 Similarly, q 1 = (A – q 2 – C 1 )/2 However, no player can solve…. Assume P 2 tries to maximize E[u 2 | q 2, C 2 ] over q 2 E[u 2 | q 2, C 2 ] = q 2 (A – Q 1 – q 2 – C 2 ) where Q 1 = E(q 1 ) This is maximized for q 2 = (A – Q 1 – C 2 )/2 So, Q 2 = (B – Q 1 )/2 with B = A – C, C = E(C 2 ) = E(C 1 ). Hence, Q 1 = Q 2 = Q = B/3, as before. This gives q 2 = (A – B/3 – C 2 )/2 and q 1 = (A – B/3 – C 1 )/2 Then u 2 = (A – B/3 – C 2 )(B/3 + C 1 /2 – C 2 /2)/2 Thus, E(u 2 ) = B 2 /9 +  2 /4 where  2 := var(C i ) Intuition: Produce more when cost is small …. Recall: Profit of firm 2 is u 2 = q 2 (A – q 1 – q 2 – C 2 ) Assume firm i knows C i and the distribution of C 2 - i u 2 is maximized for q 2 = (A – q 1 – C 2 )/2 Similarly, q 1 = (A – q 2 – C 1 )/2 However, no player can solve…. Assume P 2 tries to maximize E[u 2 | q 2, C 2 ] over q 2 E[u 2 | q 2, C 2 ] = q 2 (A – Q 1 – q 2 – C 2 ) where Q 1 = E(q 1 ) This is maximized for q 2 = (A – Q 1 – C 2 )/2 So, Q 2 = (B – Q 1 )/2 with B = A – C, C = E(C 2 ) = E(C 1 ). Hence, Q 1 = Q 2 = Q = B/3, as before. This gives q 2 = (A – B/3 – C 2 )/2 and q 1 = (A – B/3 – C 1 )/2 Then u 2 = (A – B/3 – C 2 )(B/3 + C 1 /2 – C 2 /2)/2 Thus, E(u 2 ) = B 2 /9 +  2 /4 where  2 := var(C i ) Intuition: Produce more when cost is small ….

Static Game Nash Equilibrium: Bayes Cournot vs. Full Information Assume costs C i are random and independent with C = E(C i ),  2 = var(C i ) However, the costs are known by both firms Recall: u 2 = q 2 (A – q 1 – q 2 – C 2 ) This is maximized for q 2 = (A – q 1 – C 2 )/2 Similarly, q 1 = (A – q 2 – C 1 )/2 Solving, we get q 1 = (A + C 2 – 2C 1 )/3 and q 2 = … Hence, u 2 = (A + C 1 – 2C 2 ) 2 /9 so that E(u 2 ) = B 2 /9 + 5  2 /9 where B = A – C In the incomplete information case E(u 2 ) = B 2 /9 +  2 /4 Thus, the cost of lack of information is 11  2 /36 Assume costs C i are random and independent with C = E(C i ),  2 = var(C i ) However, the costs are known by both firms Recall: u 2 = q 2 (A – q 1 – q 2 – C 2 ) This is maximized for q 2 = (A – q 1 – C 2 )/2 Similarly, q 1 = (A – q 2 – C 1 )/2 Solving, we get q 1 = (A + C 2 – 2C 1 )/3 and q 2 = … Hence, u 2 = (A + C 1 – 2C 2 ) 2 /9 so that E(u 2 ) = B 2 /9 + 5  2 /9 where B = A – C In the incomplete information case E(u 2 ) = B 2 /9 +  2 /4 Thus, the cost of lack of information is 11  2 /36

Static Game Stackelberg: Bayes Cournot Recall: u 2 = q 2 (A – q 1 – q 2 – C 2 ) Assume firm 1 announces q 1 Then q 2 = (A – q 1 – C 2 )/2 Hence, u 1 = q 1 (A – q 1 - q 2 – C 1 ) = q 1 (A – q 1 + C 2 – 2C 1 )/2 Note that E[u 1 | q 1, C 1 ] = q 1 (A – q 1 + C – 2C 1 )/2 This is maximized by q 1 = (A + C – 2C 1 )/2 So, u 1 = [B + 2(C – C 1 )][B + 2(C 2 – C 1 )]/8 Thus, E(u 1 ) = B 2 /8 +  2 /2 and E(u 2 ) = B 2 /16 +  2 /2 Intuition: Produce more when cost is small …. If costs are known, E(u 1 ) = B 2 /8 + 5  2 /8 and E(u2) = B 2 /  2 /16 The cost of lack of information is  2 /8 to 1 and 5  2 /16 Recall: u 2 = q 2 (A – q 1 – q 2 – C 2 ) Assume firm 1 announces q 1 Then q 2 = (A – q 1 – C 2 )/2 Hence, u 1 = q 1 (A – q 1 - q 2 – C 1 ) = q 1 (A – q 1 + C 2 – 2C 1 )/2 Note that E[u 1 | q 1, C 1 ] = q 1 (A – q 1 + C – 2C 1 )/2 This is maximized by q 1 = (A + C – 2C 1 )/2 So, u 1 = [B + 2(C – C 1 )][B + 2(C 2 – C 1 )]/8 Thus, E(u 1 ) = B 2 /8 +  2 /2 and E(u 2 ) = B 2 /16 +  2 /2 Intuition: Produce more when cost is small …. If costs are known, E(u 1 ) = B 2 /8 + 5  2 /8 and E(u2) = B 2 /  2 /16 The cost of lack of information is  2 /8 to 1 and 5  2 /16

Static Game Cournot: Summary FullBayes  LF u 1 u 2 B 2 /8 + 5  2 /8 B 2 /  2 /16 B 2 /8 +  2 /2 B 2 /16 +  2 /2  2 /8 5  2 /16 Nash B 2 /9 + 5  2 /9B 2 /9 +  2 /411  2 /36 Coop B 2 /8

Principal-Agent Problem Basic Model Principal Agent Type  Agent Type  Menu of Contracts: (t(  ), q(  )), for  in  Declares a type  ’ Principal’s Utility: S(q) – t S’(.) > 0, S”(.) < 0 S(0) = 0 Agent’s Utility: t – F -  q  = agent’s “type” = Marginal prod. cost Agent can be of two types: Efficient  L or Inefficient  H >  L. Principal does not know the type! Must design good contract. Agent can be of two types: Efficient  L or Inefficient  H >  L. Principal does not know the type! Must design good contract.

Principal-Agent Problem Contract: Bayesian Optimal Contract:

Principal-Agent Problem Bayesian Optimal Contract: U L is the rent that the efficient agent can get by mimicking the inefficient agent. Thus, tradeoff between efficiency and information rent.

Principal-Agent Problem Revelation Principle Principal Agent  (t(  ), q(  )) Principal Agent  (T(m(  )), Q(m(  )))

Existence of Nash Equilibrium Theorem (Nash) Every finite static game has at least one Nash equilibrium (possibly randomized)