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
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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)