ECO 5346
Sec 001
Klaus Becker
Fall 2013
Homework #8
Solutions
1 There is a rough neighborhood with n  2 residents. Each resident has to decide whether
to engage in the crime of theft. If an individual chooses to be a thief and is not caught by
the police, he receives a payoff of W . If he is caught by the police, his payoff is Z . If he
chooses not to commit theft, he receives a zero payoff. Assume W  0  Z . All n
residents simultaneously decide whether or not to commit theft. The probability of a thief
being caught equals 1/ m where m is the total number of residents who choose to engage
in theft. Thus, the probability of being caught is lower when more crimes are committed
and the police have more crimes to investigate. Find all pure strategy Nash equilibria.
The payoff from being a thief given m  1 other people have also chosen to be thieves is
 m 1
1

W    Z
 m 
m
One Nash equilibrium is when no one chooses to be a thief. Given everyone else chooses not
to turn to crime, a person who chooses to engage in theft is caught for sure so that his payoff
is Z which is less than 0 (the payoff from not being a thief). Now consider an equilibrium in
which m residents choose theft where 1  m  n  1 so that some but not all choose to be a
thief. For it to be optimal for those m residents to be thieves, it must be true that
 m 1 
1

W    Z  0.(SOL5.7.1)
 m 
m
The left-hand side expression is the payoff from being a thief, while 0 is the payoff from
not engaging in theft. For the n  m residents who have avoided a life of crime to find it
optimal to be law-abiding, it must be true that
 m 
 1 

W  
 Z  0.(SOL5.7.2)
 m 1 
 m 1 
Again, the left-hand side expression is the payoff from being a thief but notice now that
there would be m  1 thieves if this resident also chose a life of crime. Let us convince
ourselves that if (SOL5.7.1) is true then (SOL5.7.2) cannot be true.  mm1 W   m1  Z from
(SOL5.7.1) is a weighted average of W, which is positive, and Z, which is negative. If
(SOL5.7.1) is true then that weighted average is non-negative. Now consider
 mm1 W   m11  Z from (SOL5.7.2). It is also a weighted average but notice that it puts more
weight on the positive number - a weight of
than
 mm1 W   m1  Z .
Hence, if
m
m1
 mm1 W   m1  Z
rather than
m 1
m
- and thus must be larger
is non-negative then
 mm1 W   m11  Z
must be positive. But this means (SOL5.7.2) cannot be true. Intuitively, if m residents find
it optimal to engage in theft, then so will m  1 residents since the chances of getting caught
are lower when more people are criminals. From this we conclude that there cannot be a
Nash equilibrium in which some but not all residents are thieves. The only remaining
possibility is that all n residents choose crime. This is an equilibrium if and only if:
 n 1 
1

W    Z  0.
 n 
n
To sum up:

if  nn1 W   1n  Z  0 then there are two Nash equilibria - one with all residents
as criminals and one with none of them as criminals.

If  nn1 W   1n  Z  0 then there is one Nash equilibrium which has all residents
being law-abiding.

2 For the game in Figure 1, find all mixed-strategy Nash equilibria.
Figure 1
a
b
Player 1 c
d
Player 2
y
1,4
2,3
4,6
1,3
x
2,3
5,1
3,7
4,2
z
3,2
1,2
5,4
6,1
First note that c strictly dominates a and y strictly dominates z. Thus, any Nash
equilibrium in mixed strategies must assign zero probability to those dominated strategies.
We can then eliminate them so that the reduced game is
Figure SOL7.6.1
Player 2
x
Player 1
y
b
5,1 2,3
c
3,7 4,6
d
4,2 1,3
For this reduced game, b strictly dominates d so the latter can be deleted. The reduced
game is
Figure SOL7.6.2
Player 2
x
Player 1
y
b 5,1 2,3
c 3,7 4,6
This game has no pure-strategy Nash equilibria. To find the mixed-strategy Nash
equilibria, let p denote the probability that player 1 chooses b and q denote the
probability that player 2 chooses x. The equilibrium conditions ensuring that players want
to randomize are:
q  5  1  q   2  q  3  1  q   4  q  1/ 2
p 1  1  p   7  p  3  1  p   6  p  1/ 3
⇒ (
)
( , ) where p denotes the probability that player 1 chooses b and q denotes
the probability that player 2 chooses x
3. Galileo Galilei is potentially confronted by the Inquisition. To first describe what actually
transpired, Pope Urban II referred Galileo to the Inquisition and he was brought to trial
on April 12, 1633. After verbal persuasion from the Commissary General of the
Inquisition, Galileo confessed that he had gone too far in supporting the Copernican
theory in one of his books (even though he hadn’t). Galileo was then given an
"examination of intention," which involves showing the instruments of torture shown to
the accused. The final hearing by the Inquisition was held on June 22, 1633, at which
time the 69-year old Galileo pleaded for mercy because of his "regrettable state of
physical unwellness." With the threat of torture and imprisonment lurking in the
background, the Inquisitors forced Galileo to "abjure, curse and detest" his work. Galileo
complied in every way and was convicted and sentenced to life imprisonment and
religious penances. Due to his age (and possibly his stature), the sentence was commuted
to house arrest. He was allowed to return to his villa near Florence where he would
remain for the last years of his life. That is history and Figure 2 represents a simple game
theoretic modeling of it.
Figure 2: Galileo Galilei and the Inquisition
Urban VIII
Do not
refer
Refer
Galileo
Urban VIII
Galileo
Inquisitor
3
5
3
Do not
confess
Confess
Inquisitor
5
3
4
Torture
Do not
torture
Galileo
Confess
Do not
confess
4
1
5
2
4
2
1
2
1
(a) Find all of the Nash equilibria.
The strategic form games are shown in Figure SOL8.1.1.
Figure SOL8.1.1
Inquisitor - Torture
Galileo
Urban
DNR
R
C / C C / DNC
3,5,3
3,5,3
5,3,4
5,3,4
DNC / C
3,5,3
DNC / DNC
3,5,3
4,1,5
1,2,1
Inquisitor -Don't Torture
Galileo
C /C
Urban
C / DNC
DNC / C
DNC / DNC
DNR
3,5,3
3,5,3
3,5,3
3,5,3
R
5,3,4
5,3,4
2,4,2
2,4,2
The Nash equilibria are:
(DNR, DNC/DNC, Torture),
(R, C/C, Torture),
(R, C/DNC, Torture),
(DNR,DNC/C, Do not torture),
(DNR, DNC/DNC, Do not torture).
(b) Find all of the subgame perfect Nash equilibria (SPE)
In his last decision node (which is associated with the path Refer-Do not ConfessTorture), Galileo chooses Do not confess. Given this choice, the Inquisitor chooses Do
not torture. At his first decision node (associated with Urban VIII having chosen
Refer), Galileo chooses Do not confess. Finally, using the above result, Urban VIII
chooses Do not refer, as it produces payoff 3, which is greater than payoff 2 from
playing Refer .
Hence the unique subgame perfect Nash equilibrium is:
(DNR, DNC/DNC,Do not torture).
(c) For each Nash equilibrium that is not a subgame perfect Nash equilibrium, explain
why it is not a subgame perfect Nash equilibrium.
There are 4 Nash equilibria which are not subgame perfect Nash equilibria. In
Nash equilibria (DNR,DNC/DNC, Torture) and (R, C/DNC, Torture), the Inquisitor
is making a non-optimal decision by choosing to torture Galileo given Galileo plays
Do not confess in his last decision node. In Nash equilibria (R, C/C, Torture) and
(DNR, DNC/C, Do not torture), Galileo is making a non-optimal decision at his last
decision node. He should play Do not confess instead.
4 President X is in a strategic confrontation with Y’s leader Z. Z’s type determines whether
or not he has weapons of mass destruction (WMD) where the probability he has WMD is
w where 0  w  53 . After learning his type, Z decides whether or not to allow
inspections. If he allows inspections then assume they reveal WMD if Z has them and
does not reveal WMD if he doesn't. If he doesn't allow inspections then uncertainty about
whether he has WMD remains. At that point, President X decides whether or not to
invade. The extensive form is shown in Figure 3.
Figure 3: The WMD game
Nature
No
WMD
WMD
Z
Z
Do not
allow
inspections
Allow
inspections
X
X
Do not
invade Invade
Invade
X
Do not
invade
Do not
allow
inspections
Allow
inspections
Invade
X
Do not
invade Invade
Do not
invade
Z
1
3
2
9
1
4
2
8
X
3
1
3
1
5
9
6
9
Note that X learns Z's type if Z allows inspections but remains in the dark if he does not.
Z's payoffs are such that, regardless of whether he has WMD, his ordering of the
outcomes (from best to worst) are: no inspections and no invasion, inspections and no
invasion, no inspections and invasion, and inspections and invasion. Thus, he prefers not
to allow inspections but is most motivated to avoid an invasion. X's preference ordering
depends very much on whether Z has WMD. If Z has WMD, X wants to invade; if he
does not then he prefers not to invade. Find consistent beliefs for X and values for b and
h , where 0  b  1 and 0  h  1, whereby the following strategy pair is a perfect BayesNash equilibrium:

Z's strategy:
1.
2.
If I have WMD then do not allow inspections.
If I do not have WMD then allow inspections with probability h.

President X’s strategy:
3.
4.
5.
If Z allows inspections and WMD are found then invade.
If Z allows inspections and WMD are not found then do not invade.
If Z does not allow inspections then invade with probability b.
Consider the following beliefs for X: i ) if Z allows inspections and WMD are found then Z
has WMD with probability one; ii) if Z allows inspections and WMD are not found then Z
has WMD with probability zero; and iii) if Z does not allow inspections then Z has WMD
with probability w1 ww (1h ) . Let us begin by showing the consistency of X's beliefs. When
inspections are allowed, it is trivial that beliefs are consistent (they are actually consistent
with the truth, not just Z's strategy). When inspections are not allowed then the posterior
probability of Z having WMD is given by Bayes’ Rule to be:
w 1
w 1  1  w  (1  h)
as with probability w Z has WMD and, in that event, he does not allow inspections with
probability one; while with probability 1 w Z has WMD and, in that event, he does not
allow inspections with probability 1  h. Turning to X's strategy, its optimality is clear when
there are inspections; whether WMD are found or not. When inspections are not allowed, X
is content to randomize (that is, 0 < b < 1 if and only if:



1  w  (1  h)   6
w 1

  3  

 w 1  1  w   (1  h) 
 w 1  1  w   (1  h) 



1  w  (1  h)   9
w 1
 
 1  

 w 1  1  w   (1  h) 
 w 1  1  w   (1  h) 
The left-hand side expression is the expected payoff from invading and the right-hand side
expression is the expected payoff from not invading. Solving this equation for h yields
h  3315ww . Note that 3315ww  0 and 3315ww  1 if and only if 0  w  53 . The latter condition
was assumed.
When he has WMD, it is clearly optimal for Z to not allow inspections. When he does not
have WMD, it is optimal to randomize if and only if:
2b + (1 – b)8 = 4
where he earns a payoff of 4 by allowing inspections - in which case there is no invasion and an expected payoff of 2b + (1 – b)8 from not allowing inspections - where there is an
invasion with probability b . Solving this equation, we find b = 2/3
5 A well-known strategy for sustaining cooperation is Tit-for-Tat. With Tit-for-Tat, a
player starts off with cooperative play and then does whatever the other player did last
period. Tit-for-Tat embodies the idea that "What goes around comes around." For the
Trench Warfare game, it takes the form: In period 1, choose miss. In period t   2  ,
choose miss if the other player chose miss last period and choose kill if the other player
chose kill last period. For the infinitely repeated Trench Warfare game discussed in class,
derive conditions for Tit-for-Tat to be a subgame perfect Nash equilibrium (SPE).
Consider the Allied Soldiers and either period 1 or a period in which both (the
Allied Soldiers and the German Soldiers) chose (miss, miss) in the previous period. Both
players are to choose miss and this'll result in them both choosing miss in the ensuing
period and every period thereafter. The resulting payoff sequence is 4 forever which
has a present value of 4/(1- δ). Alternatively, a player, for example the Allied Soldiers,
could choose kill that'll yield a payoff of 6 in the current period. According to their
strategies, the Allied Soldiers will choose miss and the German Soldiers will choose
kill in the next period. In the period after that, the Allied Soldiers will choose kill and
the German Soldiers will choose miss. They'll keep alternating in their actions in all
periods. The payoff from choosing kill is then:
6 + δ(0) + δ2(6) + δ3(0) + δ4(6) +… = 6/(1- δ2)
Thus, choosing miss is optimal if and only if:
4
6
4
6
1



 4 1     6   
2
1  1 
1   1   1   
2
Now consider a history in which the Allied Soldiers chose miss and the German
Soldiers chose kill in the preceding period. If the Allied Soldiers act according to their
strategy by choosing kill (and the German Soldiers do similarly and choose miss), the
Allied Soldiers’ payoff is 6/(1- δ).
If the Allied Soldiers instead chose miss then the sequence of actions would be (miss,
miss) in the current period and, therefore, also occur in all ensuing periods. The payoff
for that is 4/(1- δ). For it to be optimal for the Allied Soldiers to choose kill when, in the
previous period, they chose miss and the German Soldiers chose kill, it must be true
that:
6
4
6
4
1



 6  4 1      
2
1 
1 
2
1   1    1  
Next consider a history in which the Allied Soldiers chose kill in the previous period and
the German Soldiers chose miss. The Allied Soldiers’ prescribed action of miss is
preferable to choosing kill if and only if:
0   6   20   36 
 2  2  22  32 



2
6
2
 6 

 


2 
 1   1     1  
 1   1 


1
  6  2 1      
2

Finally, we have a history in which both chose kill in the previous period. It is indeed
optimal to choose kill if and only if:
2  2  22  32 
 0   6   20   36 

2
6
1

 
2
1  1 
2
Putting all of these conditions together, Tit-for-Tat is a subgame perfect Nash
equilibrium when:

1
1
1
and     
2
2
2