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## Some Problems from Chapt 13

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**Problem 1: Chapter 13A FISHERMAN’S CATCH AND PAYOFF**Number of Number of Own Own Boats Other People’s PAYOFF Boats 1 2 25 1 3 20 1 4 15 2 2 45 2 3 35 2 4 20**What is the Nash equilibrium for the stage game for the**three fishermen? • All send one boat. • All send two boats. • There is more than one Nash equilibrium for the stage game. • There are no pure strategy Nash equilibria, but there is a mixed strategy Nash equilibrium for the stage game. • There are no pure or mixed strategy Nash equilibria for the stage game.**Can efficiency be sustained by the Grim Trigger?**• Suppose that the other two fishermen are playing the grim trigger strategy of sending one boat until somebody sends two boats and if anybody ever sends two boats, you send two boats ever after. • If you and the others play the grim trigger strategy, you will always send 1 boat and so will they.**If others are playing grim trigger strategy, would you want**to? • If you play grim trigger, you will always send 1 boat. Your payoff will be 25 in every period. Assume that a fisherman discounts later profits at rate d. Value of this stream is then 25(1+d+d2+d3 +…)=25(1/1-d) • If instead you send 2 boats, you will get payoff of 45 the first time, but only 20 thereafter. • Value of this stream is 45+ 20(d+d2+d3 +…) • Grim trigger is bigger if • 20<5 (d+d2+d3 +…) • This means 20<5d/(1-d) which implies d>4/5**Problem 7**The stage game: • Payoff to player 1 is V1(x1,x2)=5+x1-2x2 • Payoff to player 2 is V2(x1,x2)=5+x2-2x1 • Strategy set for each player is the interval [1,4] What is a Nash equilibrium for the stage game?**What is a Nash equilibrium for the stage game?**• Both players choose 4 • Both players choose 3 • Both players choose 2 • Both players choose 1 • There is no pure strategy Nash equilibrium.**Part b (i)**• If the strategy set is X={2,3}, when is there a subgame perfect Nash equilibrium in which both players always play 2 so long as nobody has ever played anything else. • Compare payoff v(2,2) forever with payoff v(3,2) in first period, then v(3,3) ever after. • That is, compare 3 forever with 4 in the first period and then 2 forever.**Part b(ii) X=[1,4]**• When is there a subgame perfect equilibrium where everybody does y so long as nobody has ever done anything differently and everybody does z>y if anyone ever does anything other than y? • First of all, it must be that z=4. Because actions after a violation must be Nash for stage game. • When is it true that getting V(y,y) forever is better than getting V(4,y) in the first period and then V(4,4) forever.**Comparison**V(y,y) forever is worth V(y,y)/(1-d)=(5-y)/(1-d) V(4,y) and then V(4,4) forever is worth 9-y+1d+1d2+…=9-y+d/1-d) Works out that V(y,y)>V(4,y) if d(8-y)>4