Problem Set - SOLUTONS 1. Consider the following two-player game: L R T 4, 4 1, 1 B, 3, 3 (a) What is the maxmin strategy profile? What is the value of this game? Note, the question could be solved like we solved the Matching Pennies game in class. We illustrate here a different method of solution. Both are acceptable. Answer: First, note that player has a clear advantage in this game. Start by finding the highest payoff player can guarantee herself. Let p denote the probability that player chooses T in a mixed strategy. To find such a strategy, we find the p that gives player the same expected payoff irrespective of what player chooses: 4p ` p1 pq ˆ p ` p1 pq ˆ 3 ñ 4p p ` p 3 ñ p 1 4 And player can guarantee herself an expected payoff of: 4 ˆ 1 4 ` ˆ 3 4 1 ˆ 1 4 ` 3 ˆ 3 4 5 But can player do better than this? n other words, is it possible for player to guarantee that 5{ is the maximum she will possibly lose? Let s find the strategy for player that guarantees herself an upper bound on what she can lose. Let q denote the probability that player chooses L. As before, we need to find a strategy that gives player the same expected payoff irrespective of what player does: 4q ` p1 qq ˆ p 1q q ` p1 qq ˆ p 3q ñ 4q ` q q 3 ` 1 ñ q 1 And player can guarantee herself a payoff no less than: 4 ˆ 1 1 ˆ 1 ˆ 1 3 ˆ 1 5 Note that the maximum gain that player can guarantee herself is equal to the minimum loss 1
that player can guarantee herself. Therefore, the maxmin strategy profile of this game is 1 `T, 4 ; B, 3 4, 1 `L, ; R, 1, and its value is 5. (b) Without relying on your answer to the previous item, find all Nash equilibria of this game. Answer: Note that none of the players have a strictly dominated strategy, and that there are no pure-strategy Nash equilibria. Recall that in any mixed strategy Nash equilibrium the players have to be indifferent between the possible actions, or otherwise they would have an incentive to deviate. Let ρ denote the probability that player plays T, and let s find the strategy that makes player indifferent between L and R: 4ρ ` p1 ρq ˆ p q ρ ` p1 ρq ˆ p 3q ñ 4ρ ` ρ ρ 3 ` ñ ρ 1 4 Likewise, we need to find the mixed strategy for player that makes player indifferent between T and B. Let σ denote the probability that player plays L, then: 4σ ` p1 σq ˆ 1 σ ` p1 σq ˆ 3 ñ 4σ σ ` σ 3 1 ñ σ 1 Hence, the mixed-strategy Nash equilibrium of this game is `T, 1 4 ; B, 3 4, `L, 1 ; R, 1. (c) Could we have relied on (a) to solve (b)? Answer: Yes. For zero-sum games, the maxmin strategy profile and the Nash equilibrium strategy profile coincide. (d) Now, change player s matrix payoffs according to the positive affine transformation fpxq x 1, to obtain the game above. What are the equilibria of the new game? Hint: Did the players best replies change due to the utility transformation? L R T 7, 4 1, 1 B 3, 5, 3 Answer: The equilibrium of the game does not change. The expected utility theorem states that the (von-neumann-morgenstern) utility function is unique up to positive affine transformations. Hence, the transformation f does not alter player preferences over any mixed strategy profile. (e) Would the answer change, and if so how, if instead we transformed player s payoffs (of the original game) according to gpxq x 1?
Answer: n this case, since g is NOT a positive affine transformation, the equilibrium might actually change. The new game is as follows: L R T 15, 4 0, 1 B 3, 8, 3 Again, let ρ denote the probability that player plays T, and let s find the strategy that makes player indifferent between L and R: 4ρ ` p1 ρq ˆ p q ρ ` p1 ρq ˆ p 3q ñ 4ρ ` ρ ρ 3 ` ñ ρ 1 4. Next we to find the mixed strategy for player that makes player indifferent between T and B. Let σ denote the probability that player plays L, then: 15σ ` p1 σq ˆ 0 3σ ` p1 σq ˆ 8 ñ 15σ ` 5σ 8 ñ σ 5 Hence, the mixed-strategy Nash equilibrium of this game is `T, 1 4 ; B, 3 4, `L, 5 ; R, 3 5, which is different than before. Note, it is s equilibrium strategy that is different even though the s preferences have changed. Why is that?. n the following game, find all equilibria profile. L C R T, 0, 3 4, 7 M 0, 1 1, 4, 7 B 1, 3 3, 9 1, 7 Hint: 1. Use iterated deletion of strictly dominated strategies as discussed in class.. Note that M for player is strictly dominated by p0.5, 0, 0.5q (that is, a 50-50 mix between T and B). Would play M in eq? Answer: Let s start by eliminating strictly dominated strategies, which are never played in equilibrium. For player, p0.5, 0, 0.5q strictly dominates M, since this strategy has an expected payoff of 1.5, if plays L or C, and.5 if plays R. And for player, L is strictly dominated by both C and R. Hence, the game can be simplified to: C R T 0, 3 4, 7 B 3, 9 1, 7 3
There are two pure-strategy Nash equilibria: pt, Rq and pb, Cq. We now look for mixed-strategy equilibria. Let ρ denote the probability that player plays T, and let s find the strategy that makes player indifferent between C and R: 3ρ ` p1 ρq ˆ 9 7ρ ` p1 ρq ˆ 7 ñ 3ρ 9ρ 7 9 ñ ρ 1 3 Next we need to find the mixed strategy for player that makes player indifferent between T and B. Let σ denote the probability that player plays C, then: σ ˆ 0 ` p1 σq ˆ 4 3σ ` p1 σq ˆ 1 ñ 4σ σ 1 4 ñ σ 1 Hence, the mixed-strategy Nash equilibrium of this game is `T, 1 3 ; B, 3, `C, 1 ; R, 1. 3. Consider the standard Cournot model with firms, N t1, u, linear inverse demand P pqq a Q, but with a quasi-fixed cost of p F q if any production takes place. That is @i P N: $ & c q ` p F c i pqq q if q i ą 0 % 0 if q i 0. (a) Write out the best-response correspondences for players 1 and as functions of the other firms production: BR 1 pq q and BR pq 1 q. Keep in mind that a firm would prefer to not produce (and make 0 profit) than make a negative profit. Answer: Each firm will produce if it makes a non-negative profit. Conditional on producing it will ignore the fixed cost and the best response is found by setting the slope of the profit function equal to zero. f π i pq i, q j q pa q i q j cqq i p F q, fixing the other player s strategy to q j, the slope is equal to pa q i q j cq. f we produce a positive quantity the best response is therefore at: And the profit is given by: q i A c q j π i p A c q ˆ j, q j q A A c q j A c qj q j c p F q So the profit is only positive when: pa c q jq 4 p F q pa c q j q p F 4 q ą 0 4
Which is equivalent to: A c q j ą F ô A c F ą q j We can therefore write out the best response function as: $ A c q j & if q ă A c F BR i pq j q 0, A c qj if q A c F % 0 if q ą A c F (b) Draw the graph of BR 1 pq q in the strategy space (like we have done in class) supposing that F P p0, a c q. Make sure to label all axes and points of discontinuity! Answer: See Figure 1 at the end of this document. (c) On a separate graph, draw BR pq 1 q assuming, again, F P p0, a c q. Answer: See Figure at the end. (d) On a separate graph, draw both best-responses under the previous assumptions about F. Also indicate all Nash Equilibria under these assumptions. Answer: See Figure 3 at the end. 4. Prove the indifference principle: Let s be an eq and assume a i, ā i P A i are such that s i pa iq ą 0 and s i pā iq ą 0. Then, u i pa i, s iq u i pā i, s iq Hint: Consult the proof shown in class that strictly dominated strategies are played in eq with probability 0. Answer: To prove this result, we ll assume that it does not hold and show that it implies a contradiction. That is, if u i pa i, s i q u ipā i, s i q, then s cannot be a Nash equilibrium. To see this, suppose that: u i pa i, s iq ą u i pā i, s iq We can then adjust the probabilities on actions a i and ā i in such a way that player i would want to deviate from equilibrium play. n particular, let a i be a generic element of the set A i of actions 5
available to player i, and define the new strategy for player i as: s i pa i q The expected utility of player i from strategy s i u ps i i, s iq ÿ $ s i a i PAi ps i & s i pa i q if a i P A i ta i, ā i u s i pa iq ` s pā i q if a i a i % 0 if a i ā i is then: pa i q ˆ u i pa i, s iq pa i q ` s i pā i qq ˆ u i pa i, s iq ` ÿ a i PAi a i ai,a i ā i u i pa i, s iq Which, by the assumptions we made above, gives a higher utility than s. Therefore, s could not be an equilibrium, a contradiction. (Note: what changes if we assume that u i pa i, s i q ă u ipā i, s i q instead?). 6
q F a c a c F q 1 Figure 1: B 1 pq q, Q1b 7
q F a c a c F q 1 Figure : B pq 1 q, Q1c 8
q F a c Nash Eq a c F q 1 Figure 3: B pq 1 q and B pq 1 q, Q1d 9