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Some Problems from Chapt 13

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Problem 1: A crowded fishery

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Number of boats, and payoff to fishermen Number of Number of Own Own Boats Other People’sPAYOFF Boats

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What is the Nash equilibrium for the stage game for the three fishermen? A)All send one boat. B)All send two boats. C)There is more than one Nash equilibrium for the stage game. D)There are no pure strategy Nash equilibria, but there is a mixed strategy Nash equilibrium for the stage game. E)There are no pure or mixed strategy Nash equilibria for the stage game.

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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.

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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+d 2 +d 3 +…)=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 (d+d 2 +d 3 +…) Grim trigger is bigger if 20<5 (d+d 2 +d 3 +…) This means 20 4/5

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Problem 7 The stage game: Payoff to player 1 is V 1 (x 1,x 2 )=5+x 1 -2x 2 Payoff to player 2 is V 2 (x 1,x 2 )=5+x 2 -2x 1 Strategy set for each player is the interval [1,4] What is a Nash equilibrium for the stage game?

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A)Both players choose 4 B)Both players choose 3 C)Both players choose 2 D)Both players choose 1 E)There is no pure strategy Nash equilibrium.

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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.

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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.

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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+1d 2 +…=9-y+d/1-d) Works out that V(y,y)>V(4,y) if d(8-y)>4

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Problem 2, Chapter 13

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Exploring the problem Note that c, x yields the highest total payoff of 7 for each player. Is this a Nash equilibrium? Why not? What are the Nash equilibria? Can we sustain repeated play of c, x by subgame perfect grim trigger strategies that revert to a not-so-good Nash equilibrium if anyone fails to play c or x?

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Best Responses and the Four Nash Equilbria

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Question 2, Part a When is there a SPNE where: – Player 1 Plays strategy Cdgrim; chooses c so long as all previous play is c,x but moves to d forever if Player 2 ever plays anything but x – Player 2 Plays strategy Xygrim: Choose x so long as all previous play is c,x but moves to y forever if Player 1 ever plays anything but c.

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Checking for SPNE If Player 2 is plays Strategy Xygrim, Player 1’s payoff from playing Cdgrim is 7 in every period so long as the game lasts. Expected payoff from this strategy is 7/(1-d). If Player 1 plays anything other than c at any time, on every later play, Player 2 will play y. Best possibility for Player 1 would be to play b and then d forever. Expected payoff from this strategy is 8 +6/(1-d). Note that once 1 has ticked off 2, 2 will always play w and d is a best response to w. And perpetual w is a best response to perpetual d.

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Comparing Sticking with strategy Cdgrim and continuing to play C is better than any other play if 7/(1-d)>8+6d/1-d) This implies 7>8(1-d) +6d, which implies that d>1/2.

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Other SPNE Grim trigger strategies that revert to other Nash equilibria are also SPNE for sufficiently large d. For example, suppose Player 1 reverts to b forever and 2 reverts to w forever if anyone fail to do c or x. This works if 7/(1-d)>8+3d/(1-d). Equivalently d>1/5.

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Part b of question 2 Don’t worry about this one. It involves an intricate pattern of responses that is hard to follow and in my opinion not worth the effort required to work it out.

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Problem 3 We play the stage game from Problem 2 repeatedly, but only 3 times. Show that some “cooperative behavior” can be sustained in Nash equilibrium. This game has more than one Nash equilibrium and one is better for both than the others. This is what gives us a shot.

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What we learned before. If the stage game has only one Nash equilibrium, then a game consisting of a finite number of repetitions has only one SPNE In this equilibrium, everybody always plays the Nash equilibrium action from the stage game. When there is more than one N.E. for the stage game, we can use the threat of reverting to the worse Nash equlibrium to incentivize good behavior in early rounds.

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Proposed SPNE Player 1: Strategy A1-- Play c in period 1 and c in period 2 if other played x in period 1. Otherwise play b in periods 2 and 3. If Player 2 plays x in periods 1 and 2 plays x in both rounds 1 and 2, then play d in round 3. Player 2: Strategy A2-- Play x in period 1 and x in period 2 if other played c in period 1. Otherwise play w in periods 2 and 3. If Player 1 plays c in periods 1 and 2, play y in period 3.

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Checking that A1,A2 is a SPNE Let’s work backwards. For each possible course of play in first two rounds, the third round is a regular subgame. Play in each of these subgames must be a N.E. One of these subgames occurs where 1 has played c twice and two has played x twice. Strategies A1 and A2 have player 1 play d and two play x in this case. This is a Nash equilibrium. In other subgames for last play, someone has done something other than c or x. In this case, strategies A1 and A2 prescribe b for 1 and w for 2. This is a Nash equilibrium as well. So the A1 and A2 prescribe Nash equilibria for all of the “last play” subgames.

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Best Responses and the Four Nash Equilbria

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Subgames after first play After the first play of the game, there are 25 different regular subgames corresponding to different actions on first play by the players. If on the first play, Player 1 did c and Player 2 did x, then if 1 follows A1 and 2 follows A2, they will play c and x on second round and d and y on third round. They will each get payoff 7+7+6=20.

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Could Player 1 do better in this subgame? The best deviation from strategy A1 for Player 1 would be to play b rather than c at this point Why? If Player 1 plays c on round 1 and b on round 2 and player 2 is playing A2, then Player 2 will play x on rounds 1 and 2 and w on round 3. Best Player 1 can do then is to play b on round 3 and get total payoff 7+8+3=18 Since playing A1 gives him 20>18, A1 prescribes Nash equilibrium play on this subgame.

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What about the other 24 subgames after first round. In the other subgames after the first round, somebody has played something other than c or x. In this case, if Player 2 is playing A2, Player 2 will play w in the next two rounds. If Player 2 is playing w in next two rounds, best response for Player 1 is to play b in next two rounds, which is what Strategy A1 prescribes.

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Conclusion for these subgames We have seen that at all subgames starting after the first round, A1 prescribes best responses to A2. Symmetric reasoning shows that A2 prescribes best responses to A1. Thus we have shown that A1 and A2 prescribe Nash equilibrium play in all regular proper subgames.

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Conclusion for Full Game We still need to show that A1, A2 is a Nash equilibrium for the full game. We saw that payoff to Player 1 from A1 is 20. Suppose Player 1 plays something other than c on first round. Then A2 will have 2 play w in the next two rounds. Best thing other than c for Player 1 on first round is b. After that given that 2 is playing w, playing b is best in the next two rounds for Player 1. So best Player 1 can get by deviating in first round is 8+3+3=14<20.

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Conclusion Symmetric reasoning applies to Player 2. The strategy profile A1, A2 is a subgame perfect Nash equiibrium since the substrategies prescribed in each subgame are Nash equilibria (best responses to each other)

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Problem 4, Ch 13 a) Define a grim-trigger strategy profile. b) Derive conditions whereby this strategy profile is a SPNE. (proposed answer to b: d>3/4)

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Hints for Problem 4 What is a nice outcome for stage game? What is a Nash equilibrium for this game. Define “grim trigger” strategies in which each player does her part of a nice outcome so long as the other does his part, but if either ever does anything else, both revert to the Nash equilibrium forever. Find payoffs from always playing “nice”. Find best you can do by “defecting” from nice play when other is playing the grim trigger.

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Problem 5, Ch 13

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a) Find a SPNE strategy profile that results in an outcome path where both players choose x in every period. Note: x,x is not a Nash equilibrium for stage game, but w,w and z,z are. We see that x,x is better for both than either w,w or z,z. We could construct trigger strategies with either w,w or z,z as the threat. For what values of d is there a SPNE trigger strategy with z,z being the threat?

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Proposed answer to part a With z,z as the reversion “punishment”, We need 6/(1-d)>10+3d/(1-d). This means d>4/7. There is also a SPNE in which the reversion is to w,w for some values of d? For you to figure out: What values of d?

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Part b) Find a SPNE strategy profile that results in an outcome path where players choose x in odd numbered periods and y in even periods. Try strategies. Continue to abide by the rule “play x in odd periods, y in even” so long as nobody has ever violated this rule. If anybody violates the rule, play z forever. Payoff from playing this rule forever is 6+8d+6d 2 +8d 3 +6d 4 +8d 5 +6d 6 +… =6(1+d 2 +d 4 +d 6 +..)+8d(1+d 2 +d 4 …) =6(1+d 2 +(d 2 ) 2 +(d 2 ) 3 +…)+8d(1+d 2 +(d 2 ) 2 +(d 2 ) 3 +…) =6/(1-d 2 )+8d/(1-d 2 )

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Payoff from violating rule Most profitable violation is choose d at start. If other is playing the proposed trigger strategy, Other will play x on first play and violator will get 10 on first play. But ever after, other will play z, and best violator can do is play z. Payoff from doing this is 10+3d/(1-d). Proposed strategy profile is a SPNE if 6/(1-d 2 )+8d/(1-d 2 )>10+3d/(1-d). This is true if 7d 2 +5d>4. We see that the left side of this inequality is increasing in d. We also see that the inequality holds for d=1, but not for d=1/2. (We could solve a quadratic to find exactly which d’s work.)

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Part c) Find a SPNE strategy profile that results in an outcome path in which players choose x in first 10 periods, then always choose z. There ain’t one. Can you see why? Part d) You should be able to show that the one and only grim trigger strategy that does this is one where players revert to z if someone ever deviates form choosing y.

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Problem 7 The stage game: Payoff to player 1 is V 1 (x 1,x 2 )=5+x 1 -2x 2 Payoff to player 2 is V 2 (x 1,x 2 )=5+x 2 -2x 1 Strategy set for each player is the interval [1,4] What is a Nash equilibrium for the stage game?

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A)Both players choose 4 B)Both players choose 3 C)Both players choose 2 D)Both players choose 1 E)There is no pure strategy Nash equilibrium.

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If in the Stage game, the Payoff to player 1 is V 1 (x 1,x 2 )=5+x 1 -2x 2, the Payoff to player 2 is V 2 (x 1,x 2 )=5+x 2 -2x 1, and the strategy set for each player is the interval [1,4], which symmetric strategy profile has highest total payoff? A)x 1 =x 2 =1 B) x 1 =x 2 =2 C) x 1 =x 2 =3 D) x 1 =x 2 =4

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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.

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If both play 2, each gets 5+2-2×2=3 If they both play 2 forever, expected payoff for each is 3(1+d+d 2 +…d n +…)=3/(1-d) Suppose other guy is playing grim trigger “play 2 so long as other guy plays 2. If other ever plays 3, then play 3 forever. If you play 3 on first move and then continue to play 3, you will get 5+3-2×2=4 on first move, then 5+3-2×3=2 forever after. Expected payoff from this is 4+2×d/(1-d). When is 3/(1-d)>4+2d/(1-d)? 3>4(1-d)+2d which implies d>1/2.

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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.

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Problem 8 N doctors share a practice and share all income from it. Doctor can exert effort level, 1,2,…n. Profit of firm is 2(e 1 +e 2 +…e n ) where e i is effort level of Dr. i. Payoff to Dr. i is (1/n) 2(e 1 +e 2 +…e n ) –e i What is a Nash equilibrium? What would be a best cooperative outcome?

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Repeated game with trigger All work at level 10 so long as everyone works at level 10. If anyone ever slacks off, all revert to working at level 1. When does this work? Payoff from following norm is 20-10=10 in each period. Expected total payoff from doing this always is 10/(1-d).

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What if you violate the norm? Best violation would be work at level 1. Payoff then would be (n-1)20/n+(2/n)-1 in first period, then 2 forever after. This is worth (n-1)20/n+(2/n)-1 +2/(1-d). The trigger strategy is a SPNE if 10/(1-d)>(n-1)20/n+(2/n)-1 +2/(1-d)

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