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Some  Problems  from  Chapt Some  Problems  from  Chapt

Some Problems from Chapt - PowerPoint Presentation

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Some Problems from Chapt - PPT Presentation

13 Problem 1 A crowded fishery Number of boats and payoff to fishermen Number of Number of Own Own Boats Other Peoples PAYOFF Boats 1 2 ID: 655771

player play strategy nash play player nash strategy payoff equilibrium game stage period trigger players plays playing choose spne grim part subgames

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Slide1

Some Problems from Chapt 13Slide2

Problem 1: A crowded fisherySlide3

Number of boats, and payoff to fishermen

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

20Slide4

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.Slide5

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.Slide6

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 45+ 20(d+d

2

+d

3

+…)

Grim trigger is bigger if

20<

5

(d+d

2

+d

3

+…)

This means 20<5d/(1-d) which implies d>4/5Slide7

Problem 7The 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?Slide8

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.Slide9

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.Slide10

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.Slide11

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)>4Slide12

Problem 2, Chapter 13Slide13

Exploring the problemNote 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?Slide14

Best Responses and the Four Nash

EquilbriaSlide15

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.Slide16

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. Slide17

ComparingSticking 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.Slide18

Other SPNEGrim 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.Slide19

Part b of question 2Don’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.Slide20

Problem 3We 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.Slide21

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. Slide22

Proposed SPNEPlayer 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.Slide23

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

.Slide24

Best Responses and the Four Nash

EquilbriaSlide25

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. Slide26

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

.Slide27

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. Slide28

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

.Slide29

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.Slide30

ConclusionSymmetric 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)Slide31

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)Slide32

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.Slide33

Problem 5,

Ch

13Slide34

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?Slide35

Proposed answer to part aWith

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?Slide36

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+(d2 )3+…)+

8d(1+d2+(d2)2+(d2

)3+…)=6/(1-d2)+8d/(1-d2

)Slide37

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.)Slide38

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.Slide39

Problem 7The 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?Slide40

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.Slide41

If in the Stage game, the Payoff to player 1 is V1(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?

x

1

=x

2

=1

x

1

=x

2

=2

x

1

=x2=3

x1=x2=4Slide42

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.Slide43

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.Slide44

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.Slide45

Problem 8N 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?Slide46

Repeated game with triggerAll 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).Slide47

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)