7
Appendix 2: Proofs (For Online Publication)
Proof of Lemma 1. Suppose ((¤1   ¤1  ¤ ) (¤2   ¤2  ¤2 )) is a pure-strategy Nash equilibrium. Then,
 ¤ must satisfy: (1)  ¤ 2 arg max   ((¤  (    ¤0 ) ¤ ) (¤   ¤  ¤ )) and (2)  ¤ = 0 if 0 2
arg max   ((¤  (    ¤0 ) ¤ ) (¤  ¤  ¤ )). 1 2 arg max   ((¤  (    ¤0 ) ¤ ) (¤  ¤  ¤ )) if
and only if ¤ ¡¹
¤ ¸ 0, and 0 2 arg max   ((¤  (   ¤0 ) ¤ ) (¤   ¤  ¤ )) if and only if ¤ ¡¹
¤ ·
¤
¤
¤
0. It follows from (1) and (2) that  = 1 if and and only if  ¡ 
¹  0.
Proof of Lemma 2. Suppose ((¤1   ¤1  ¤ ) (¤2   ¤2  ¤2 )) is a pure-strategy Nash equilibrium. Then,
¤ must satisfy: (1) ¤ 2 arg max  ((   ¤  ¤ ) (¤   ¤  ¤ )) and (2) ¤2 = 0 whenever there
exists  2 arg max  ((  ¤  ¤ ) (¤   ¤  ¤ )) with 2 = 0. It follows from (1) and (2) that

¤1 = 1 and ¤2 = 0 if 1 ¸ 2 , and ¤1 = 0 and ¤2 = 2 if 1  2 , where  = 


³
´
Furthermore, we obtain the following formulas: 1 = (¤1 + (¤1  ¤2 ) ¢  ¤1 ) +1
+2  and 2 =
³
´
(¤2 + (¤1  ¤2 ) ¢  ¤2 ) +1
+2 
 +


If ¤ = 0, ¤ = 0. ¹ = +2 = +2
¸ 0. Thus, ¤ ¡ 
¹ · 0. It follows from Lemma 1
¤
that   = 0. This completes the proof.
P
Proof of Lemma 3. Player ’s self-esteem is given by:  = 2=1  ¤ (¤ ¡ 
¹¤ ) By Lemma 1, we
¤
¤
know that ¤ ¡ 
¹¤  0 if  = 1. It follows that  ¸ 0 and   0 if  1 = 1 or ¤2 = 1. This
proves (1).
P
Recall that  = 2=1 ¤ (¤ ¡ ¹¤ ). If players hold the same values ( ¤ =  ¤ ), it follows that
 =  . Since  ¸ 0,  ¸ 0 when players hold the same values. Since   0 when player
i values academics or music,   0 when the players hold the same values and value academics
or music. This proves (2).
Now suppose players hold di¤erent values ( ¤ 6=  ¤ ). Suppose ¤ = 1. It is easy to show
that  ¤ = 0. What if this were not the case and  ¤ = 1 instead? From Lemma 2, we know
players value at most one activity. So, if  ¤ = ¤ = 1 the players must hold the same values,
which contradicts  ¤ 6=  ¤ . Since we know  ¤ = 0, it follows from Lemma 1 that ¤ ¡ 
¹¤ · 0.

Thus, whenever  ¤ = 1, ¤ ¡ ¹¤ · 0, it follows that  · 0. This proves (3).
Proof of Lemma 4. Suppose the players hold the same values and they value academics or music. It follows from Lemma 3 that the players have strictly positive esteem for one another
(  0). If  = 0+ , the players must interact in equilibrium. We can prove this by contradiction. Suppose the players do not interact in equilibrium (¤1 = ¤2 = 0). In equilibrium,
¤ 2 arg max  ((¤   ¤   ) (¤  ¤  ¤ )). But 0 2
  ((¤   ¤   ) (¤  ¤  ¤ )) since player , in
choosing  = 1 rather than  = 0 gains utility  ¡ . And,  ¡   0 since   0 and
 = 0+ . This proves the …rst part of the lemma.
Suppose one player values academics and the other values music. It follows from Lemma
3 that the players negatively esteem one another ( · 0). The players will not interact in
equilibrium if  = 0+ . Again, we can prove this by contradiction. Suppose the players do
interact. Then, for one of the players (player ), we must have ¤ = 1. In equilibrium, ¤ 2
arg max  ((¤   ¤   ) (¤  ¤  ¤ )). But 0 2
  ((¤   ¤   ) (¤   ¤  ¤ )). If player  chooses
 = 0 rather than  = 1 he gains utility  ¡ (1 ¡  ) . Since,  · 0 and  = 0+ ,
 ¡ (1 ¡  )  0. This proves the second part of the lemma.
45
Proof of Lemma 5. To prove the lemma, it is su¢cient to show two things to be true. Holding
1 and 2 …xed:
(i) If an equilibrium exists for some value of  in which players interact and focus on di¤erent
activities, no equilibrium exists for any value of  in which players focus on the same activity.
(ii) If an equilibrium exists for some value of  in which players interact and hold di¤erent
values, no equilibrium exists for any value of  in which players hold the same values.
Let us assume, without loss of generality, that 2 ¸ 1 
Case 1: 2  
¹
Applying Proposition 2, (i) and (ii) can be restated as follows:
(i) If equilibria of type (3) exist for some , equilibria of types (2), (4), and (5) do not exist for
any values of .
(ii) If equilibria of types (3) or (5) exist for some , equilibria of types (2) and (4) do not exist
for any values of .
1
Showing (i): If an equilibrium of type (3) exists for some , we must have 21 · 14 + 2(+1)
22
and 1  1. Let us suppose this to be the case. In order for an equilibrium of type (2) to exist
2
1
for some , we must have 21 ¸ 1 + (+1)
22 . But, in order for this to hold, 14 + 2(+1)
22 ¸ 21 ¸
2
1 + (+1)
22 , which can never be satis…ed. Thus, equilibria of type (2) do not exist for any values
1
of  In order for an equilibrium of type (4) to exist for some , we must have 21 ¸ 14 + +1
22 .
1
1
1
1
2
2
2
It follows that we must have 4 + +1 2 · 1 · 4 + 2(+1) 2 , which can never be satis…ed. Thus,
equilibria of type (4) do not exist for any values of . In order for an equilibrium of type (5) to
exist for some , we must have 1  21 . But, this contradicts our assumption that 1  1 So,
equilibria of type (5) do not exist for any values of . Thus, (i) holds.
Showing (ii): We have already shown that, if an equilibrium of type (3) exists for some ,
equilibria of types (2) and (4) do not exist for any . If an equilibrium of type (5) exists for some ,
2
1  21  3(+1)
22 . Let us suppose that this condition holds. In order for an equilibrium of type
2
2
2
(2) to exist, we must have 21 ¸ 1+ (+1)
22 . It follows that we must have 3(+1)
22  1 + (+1)
22 ,
4
or ¡ 3(+1)
22  1 (which can never be true). Thus, equilibria of type (2) cannot exist for any
4
. In order for an equilibrium of type (4) to exist, we must have 21  3(+1)
22 . But, this
2
contradicts our assumption that 21  3(+1)
22 . So, equilibria of type (4) do not exist for any .
This shows that (ii) holds.
Case 2: 1  
¹
Applying Proposition A1, (i) and (ii) can be restated as follows:
(i/ii) If equilibria of type (4) exist for some , equilibria of types (1) and (3) do not exist for any
values of .
2
Showing (i/ii): If an equilibrium of type (4) exists for some , we must have 22 ¸ 4 ¡ +1
.
Let us suppose this to be the case. If an equilibrium of type (1) exists for some , we must have
2
2
¡1
2
22 · ¡1
+1 . But, this cannot hold since 2 ¸ 4 ¡ +1 and 4¡ +1  +1  If an equilibrium of type
4
2
(3) exists for some , we must have 22 · +1
. But, again, this cannot hold since 22 ¸ 4 ¡ +1
2
4
and 4 ¡ +1  +1  This establishes (i/ii).
46
Case 3: 
¹  · 1  2 · 
¹
Applying Proposition A2, (i) and (ii) can be restated as follows:
(i) If equilibria of type (7) exist for some , equilibria of types (1) and (2) do not exist for any
values of .
(ii) If equilibria of types (7) and (8) exist for some , equilibria of types (1) and (2) do not exist
for any values of .
Showing (i): If an equilibrium of type (7) exists for some , we must have  · 1. Suppose
2
this is the case. An equilibrium of type (1) only exists if 21 ¸ 1 + +1
22 . But since  · 1
and 2 ¸ 1 , this condition cannot be satis…ed. Therefore, an equilibrium of type (1) cannot
exist. An equilibrium of type (2) only exists if 22 · ¡1
+1 . But, once again, since  · 1 and
2  0, this condition cannot be satis…ed. Therefore, an equilibrium of type (2) cannot exist.
This establishes (i).
Showing (ii): We have already shown that, if an equilibrium of type (7) exists for some ,
equilibria of types (1) and (2) do not exist for any . Existence of an equilibrium of type (8)
also requires  · 1. Therefore, by the same logic, if an equilibrium of type (8) exists for some
, equilibria of types (1) and (2) do not exist for any . This establishes (ii) and completes the
proof.
Proof of Proposition 1. Suppose 2 ¸ 1 and  = 0+ . We will characterize the equilibria that
exist as a function of 1 , 2 , and 
It will be useful, in characterizing the equilibria, to use some shorthand. According to
Lemma 2, we can denote an equilibrium by ((¤1  ¤1 ) ( ¤2  ¤2 )). It is not necessary to specify the e¤ort level, since it can be deduced from Lemma 2. We also know from Lemma 2
that  ¤ 2 f(1 0) (0 1) (0 0)g. Therefore, we will examine all combinations of the following
form to see, under what parameters, they are equilibria of the game: f((1  1 ) (2  2 )) :   2
f(1 0) (0 1) (0 0)g  2 f0 1gg.
To check whether these combinations are equilibria, we will need to consider a variety of
deviations by the players. We will denote deviations by player  by (0  0 ). We do not bother
to specify the e¤ort level of player  when he deviates, since we know the optimal choice of e¤ort.
It is optimal for player , when he deviates, to choose 01 = 1 and 02 = 0 if 1 ¸ 2 ;
0
0
0
it is optimal for
³ player´  to choose 1 = 0 and 2 = ³2 if ´1  2 ; where 1 = (1 +
0
+1
0
(0   ) ¢  1 ) +1
+2  and 2 = (2 + (   ) ¢  2 ) +2 . Player  always focuses e¤ort
on one activity when he deviates and so is below average at one of the activities. Thus, he
will never choose to deviate to  0 = (1 1). Thus, we will only check deviations of the form:
f( 0  0 ) :   2 f(1 0) (0 1) (0 0)g  2 f0 1gg.
We will now systematically check all the combinations. We start with cases where players do
not interact (1 = 2 = 0) and then turn to cases where players do interact.
Case 1: No interaction (1 = 2 = 0)
There are …ve types of combinations to check. We will consider each in turn.
1. 1 = 2 = (1 0)
47
³
´2
+1
2
If the players follow these strategies, 2 = 12 +1
22 ¡ (+2)
2 1 . If player 2 deviates to
+2
³
´2
+1
2
02 = (1 0), 02 = 1, this yields: 20 = 2 +1
22 ¡ 2 (+2)
2 1 . This deviation is pro…table
+2
if 32 ( + 1)22  21 . Suppose this deviation is not pro…table: 32 ( + 1)22 · 21 . If player
2 instead deviates to  02 = (0 0), 02 = 0, this yields: 20 = 0. This deviation is pro…table
if 12 ( + 1)22 · 21 (which is true, since 32 ( + 1)22 · 21 ). In summary, one of these two
deviations will be pro…table. Thus, (1) is never an equilibrium.
2. 1 = (1 0) 2 = (0 1)
³
´2
³
´2
1 +1
2
If the players follow these strategies, they receive: 1 = 12 +1



=
 In
2
2
+2
2 +2
order for this to be an equilibrium, there cannot be any pro…table deviations. There are
ten deviations we need to check.
³
´2
+1
(i) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 = 12 +1
¡ (+2)
2
+2
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
(iii) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 12 +1
max(21  1) ¡ (+2)
2
+2
³
´2
+1
(iv) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 2 +1
¡ 2 (+2)
2
+2
³
´2
+1
(v) Player 1 deviates to 01 = (0 0) and 01 = 1: yields 10 = 12 +1
¡ (+2)
2
+2
³
´2
+1
2
(vi) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 = 12 +1
22 ¡ (+2)
2 1 
+2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
³
´2
+1
2
(viii) Player 2 deviates to 02 = (1 0), 02 = 1: yields 20 = 2 +1
22 ¡ 2 (+2)
2 1 
+2
³
´2
+1
2
(ix) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 12 +1
max(22  1) ¡ (+2)
2 1 
+2
³
´2
+1
2
(x) Player 2 deviates to 02 = (0 0) and 02 = 1: yields 20 = 12 +1
22 ¡ (+2)
2 1 
+2
Observe that (2) will be an equilibrium if and only if (i) through (x) are not pro…table
deviations. This yields ten conditions for (2) to be an equilibrium, but only conditions (iv)
and (viii) are binding:
Condition (iv): 21 ¸
4
+1
Condition (viii): 21 ¸ ( + 1)(22 ¡ 14 )
These conditions can be simultaneously met if  = 0 but not when  ¸ 1. Let us see
why. Since 21 · 22 , we must have: ( + 1)(22 ¡ 14 ) · 22 , or 22 · +1
4 . This implies
2 ¸ 4 , it follows that we must have:
we must also have 21 · +1
.
Since
we
also
have

1
4
+1
4
+1
4
+1
2
·

·

For

¸
1,

,
so
equilibria
of
this type will not exist.
1
+1
4
+1
4
3. 1 = (0 1) 2 = (1 0)
³
´2
³
´2
1 +1
If the players follow these strategies, they receive: 1 = 12 +1


=
22  In
2
+2
2 +2
order for this to be an equilibrium, there cannot be any pro…table deviations. There are
ten deviations we need to check.
48
(i) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 =
1
2
³
+1
+2
´2
21 ¡
+1
2
(+2)2 2 
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
2
(iii) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 
+2
³
´2
+1
2
(iv) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 12 +1
max(21  1) ¡ (+2)
2 2 
+2
³
´2
+1
2
(v) Player 1 deviates to 01 = (0 0) and 01 = 1: yields 10 = 12 +1
21 ¡ (+2)
2 2 
+2
³
´2
+1
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1
¡ (+2)
2
+2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
³
´2
+1
(viii) Player 2 deviates to 02 = (1 0), 02 = 1: yields 20 = 12 +1
22 ¡ (+2)
2
+2
³
´2
+1
(ix) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 2 +1
¡ 2 (+2)
2
+2
³
´2
+1
(x) Player 2 deviates to 02 = (0 0) and 02 = 1: yields 20 = 12 +1
¡ (+2)
2
+2
Observe that (3) will be an equilibrium if and only if (i) through (x) are not pro…table
deviations. This yields ten conditions for (3) to be an equilibrium, but only conditions
(iii)/(v) and (ix) are binding:
Condition (iii)/(v): 22 ¸ ( + 1)(21 ¡ 14 )
Condition (ix): 22 ¸
4
+1
Combining these conditions, we obtain: 22 ¸ max
4. 1 = (0 1) 2 = (0 1)
³
4
+1  (
+ 1)(21 ¡ 14 )
´
³
´2
+1
If the players follow these strategies, 2 = 12 +1
¡ (+2)
If player 2 deviates to
2.
+2
³
´2
+1
02 = (0 1), 02 = 1, this yields: 20 = 2 +1
¡ 2 (+2)
This deviation is always
2.
+2
pro…table. Thus, (4) is never an equilibrium.
5. 1 = (0 0) or  2 = (0 0)
Suppose  = (0 0) and player  chooses some   . Then, when the players follow these
strategies  = 0. We know that player  focuses on at most one activity in equilibrium.
Suppose player  is not focused on activity 1. If player  deviates to 0 = (1 0) 0 = 0,
³
´2
this yields 0 = 12 +1
2  0. Suppose instead player  is not focused on activity 2. If
+2
³
´2
player  deviates to 0 = (0 1) 0 = 0, this yields 0 = 12 +1
 0. So, there is always
+2
a pro…table deviation. Thus, no equilibria of this type exist.
Case 2a: Interaction, initiated by both players (1 = 2 = 1)
Observe that there will be a pro…table deviation for player 1 to  01 =  1  01 = 0 since   0.
Thus, no equilibria exist in which both players initiate interaction.
Case 2b: Interaction, initiated by player 1 only (1 = 1 2 = 0)
49
Observe that, because   0, it will not be pro…table for player 2 to deviate to 02 = 1. Thus,
it will be su¢cient to restrict attention to deviations by player 2 in which 02 = 0.
We know from Lemma 4 that, since  = 0+ , equilibria do not exist in which  = (1 0),
  = (0 1), and players interact. This means that there are three types of combinations to check.
1. 1 = 2 = (1 0)
³
´2
+1
2
If the players follow these strategies, they receive: 1 = 2 +1
21 ¡ 4 (+2)
2 2  2 =
+2
³
´2
+1
2
2 +1
22 ¡ 4 (+2)
2 1  In order for this to be an equilibrium, there cannot be any prof+2
itable deviations. There are seven deviations we need to check.
³
´2
+1
2
(i) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 12 +1
max(21  1) ¡ 2 (+2)
2 2 
+2
³
´2
+1
2
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 = 12 +1
21 ¡ 2 (+2)
2 2 
+2
³
´2
+1
2
(iii) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 12 +1
21 ¡ 2 (+2)
2 2 
+2
³
´2
(iv) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 = 12 +1

+2
(v) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
2
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1
max(22  1) ¡ 2 (+2)
2 1 
+2
³
´2
+1
2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 12 +1
22 ¡ 2 (+2)
2 1 
+2
Observe that (1) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations. This yields seven conditions for (1) to be an equilibrium, but only conditions
(iv), (vi), and (vii) are binding:
Condition (iv): 22  12 ( + 1)(21 ¡ 14 ) (this inequality is strict because player 1 prefers
not to initiate interaction, since  = 0+ ).
Condition (vi): 22 ¸
Condition (vii): 22 ¸
21
+1
+
1
4
2
4 1
3 +1
2. 1 = (0 1) 2 = (0 1)
If the players follow these strategies, they receive: 1 = 2
+1
4 (+2)
2
³
+1
+2
´2
+1
¡4 (+2)
2  2 = 2
³
+1
+2
´2
¡
In order for this to be an equilibrium, there cannot be any pro…table deviations.
There are seven deviations we need to check.
³
´2
+1
(i) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 12 +1
max(21  1) ¡ 2 (+2)
2
+2
³
´2
+1
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 = 12 +1
¡ 2 (+2)
2
+2
³
´2
(iii) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 12 +1
21 
+2
³
´2
+1
(iv) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 = 12 +1
¡ 2 (+2)
2
+2
(v) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
50
(vi) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 =
1
2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 =
1
2
³
³
+1
+2
+1
+2
´2
´2
+1
max(22  1) ¡ 2 (+2)
2
+1
¡ 2 (+2)
2
Observe that (2) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations. This yields seven conditions for (2) to be an equilibrium, but only conditions
(iii) and (vi) are binding:
Condition (iii): 21  4( ¡1
+1 ) (this inequality is strict because player 1 prefers not to
initiate interaction, since  = 0+ ).

Condition (vi): 22 · 4( +1
)
3. 1 = (0 0) or  2 = (0 0)
It is always pro…table for player 1 to deviate to 01 = 0. Suppose
³
´2
+1
2
1 = (0 0) and 2 = (1 0). Then 1 = 12 +1
21 ¡ (+2)
If player 1 deviates to
2 2 .
+2
Suppose  2 = (0 0).
01 = 0, this yields 10 = 0 (which is pro…table when 22 ¸ 12 ( + 1)21 ). If player 1 instead
³
´2
+1
2
deviates to  01 = (1 0), this yields 10 = 4 +1
21 ¡ 2 (+2)
2 2 (which is pro…table when
+2
3
2 (
+ 1)21 ¸ 22 ).
Clearly, one of these deviations will be pro…table. Finally, suppose
³
´2
+1
1 = (0 0) and  2 = (0 1). Then 1 = 12 +1
¡ (+2)
2 . If player 1 instead deviates to
+2
³
´2
+1
01 = (0 1), this yields 10 = 2 +1
¡ 2 (+2)
2 , which is always pro…table. Hence, there is
+2
always a pro…table deviation when  1 = (0 0) or  2 = (0 0).
Case 2c: Interaction, initiated by player 2 only (1 = 0 2 = 1)
Observe that, because   0, it will not be pro…table for player 2 to deviate to 02 = 1. Thus,
it will be su¢cient to restrict attention to deviations by player 2 in which 02 = 0.
We know from Lemma 4 that, since  = 0+ , equilibria do not exist in which  = (1 0),
  = (0 1), and players interact. This means that there are three types of combinations to check.
1. 1 = 2 = (1 0)
³
´2
+1
2
If the players follow these strategies, they receive: 1 = 2 +1
21 ¡ 4 (+2)
2 2  2 =
+2
³
´2
+1
2
2 +1
22 ¡ 4 (+2)
2 1  In order for this to be an equilibrium, there cannot be any prof+2
itable deviations. There are seven deviations we need to check.
³
´2
+1
2
(i) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 = 12 +1
max(21  1) ¡ 2 (+2)
2 2 
+2
³
´2
+1
2
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 12 +1
21 ¡ 2 (+2)
2 2 
+2
³
´2
0
1 +1
+1
0
0
2
(iii) Player 2 deviates to 2 = (0 1), 2 = 1: yields 2 = 2 +2 max(22  1) ¡ 2 (+2)
2 1 
³
´2
+1
2
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = 12 +1
22 ¡ 2 (+2)
2 1 
+2
³
´2
+1
2
(v) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 = 12 +1
22 ¡ 2 (+2)
2 1 
+2
³
´2
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1

+2
51
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
Observe that (1) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations. This yields seven conditions for (1) to be an equilibrium, but only conditions
(i), (ii), and (vi) are binding:
Condition (i): 22 · ( + 1)(21 ¡ 14 )
Condition (ii): 22  34 ( + 1)21 (this inequality is strict because player 1 prefers, all else
equal, not to value activities).
22
Condition (vi): 22  +11 + 14 (this inequality is strict because player 1 prefers not to
initiate interaction, since  = 0+ ).
2. 1 = (0 1) 2 = (0 1)
If the players follow these strategies, they receive: 1 = 2
+1
4 (+2)
2
³
+1
+2
´2
+1
¡4 (+2)
2  2 = 2
³
+1
+2
´2
¡
In order for this to be an equilibrium, there cannot be any pro…table deviations.
There are seven deviations we need to check.
³
´2
+1
(i) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 12 +1
max(21  1) ¡ 2 (+2)
2
+2
³
´2
+1
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 12 +1
¡ 2 (+2)
2
+2
³
´2
+1
(iii) Player 2 deviates to 02 = (1 0), 02 = 1: yields 20 = 12 +1
max(22  1) ¡ 2 (+2)
2
+2
³
´2
+1
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = 12 +1
¡ 2 (+2)
2
+2
³
´2
(v) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 = 12 +1
22 
+2
³
´2
+1
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1
¡ 2 (+2)
2
+2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
Observe that (2) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations. This yields seven conditions for (2) to be an equilibrium, but only condition
(v) is binding:
Condition (v): 22  4( ¡1
+1 ) (this inequality is strict because player 2 prefers not to initiate
interaction, since  = 0+ ).
3. 1 = (0 0) or  2 = (0 0)
It is always pro…table for player 2 to deviate to 02 = 0. Suppose
³
´2
+1
2
2 = (0 0) and 1 = (1 0). Then 2 = 12 +1
22 ¡ (+2)
If player 2 deviates to
2 1 .
+2
Suppose  1 = (0 0).
02 = 0, this yields 20 = 0 (which is pro…table when 21 ¸ 12 ( + 1)22 ). If player 2 instead
³
´2
+1
2
deviates to  02 = (1 0), this yields 20 = 4 +1
22 ¡ 2 (+2)
2 1 (which is pro…table when
+2
3
2 (
+ 1)22 ¸ 21 ).
Clearly, one of these deviations will be pro…table. Finally, suppose
³
´2
+1
2 = (0 0) and  1 = (0 1). Then 2 = 12 +1
¡ (+2)
2 . If player 2 instead deviates to
+2
³
´2
+1
02 = (0 1), this yields 20 = 2 +1
¡ 2 (+2)
2 , which is always pro…table. Hence, there is
+2
always a pro…table deviation when  1 = (0 0) or  2 = (0 0).
52
Case 2: Combined
If we combine Cases 2a, 2b, and 2c we …nd the following.
1. An equilibrium with 1 = 2 = (1 0) and interaction between the players exists if:
(i) 22 
221
+1
+
1
4
(ii) 22 · ( + 1)(21 ¡ 14 )
(iii) 22  34 ( + 1)21
These conditions are the same as those from case 2c, which are weaker than those in case
2b. Intuitively, player 2, who is better at academics than player 1, is more willing to initiate
interaction than player 1.
2. An equilibrium with 1 = 2 = (0 1) and interaction between the players exists if:
(i) 21  4( ¡1
+1 )

(ii) 22 · 4( +1
)
These conditions are the same as those from case 2b, which are weaker than those in case
2c. Intuitively, player 1 is more willing to initiate interaction than player 2 in this instance.
This completes the proof.
Proof of Proposition 2. Suppose 2 ¸ 1 and 2  
¹  . We will characterize the equilibria that
exist as a function of 1 , 2 , , and 
It will be useful, in characterizing the equilibria, to use the same shorthand as we used in
the proof of Proposition 1. Once again, we will examine all combinations of the following
form to see, under what parameters, they are equilibria of the game: f((1  1 ) (2  2 )) :   2
f(1 0) (0 1) (0 0)g  2 f0 1gg. We will …rst examine combinations in which player 2 is a
scholar (2 = (1 0)). Later on, we will show that no equilibria exist in which player 2 is not a
scholar.
Case 1: No interaction (1 = 2 = 0)
Since we assume player 2 is a scholar (2 = (1 0)), we will denote combinations by player 1’s
choice of 1 . There are three combinations to check.
1. 1 = (1 0)
³
´2
+1
2
If the players follow these strategies, they receive: 1 = 12 +1
21 ¡ (+2)
2 2  2 =
+2
³
´2
1 +1
+1
2
22 ¡ (+2)
2 1  In order for this to be an equilibrium, there cannot be any pro…table
2 +2
deviations. There are ten deviations we need to check.
³
´2
(i) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 = 12 +1

+2
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
53
³
´2
+1
2
(iii) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 ¡ 
+2
³
´2
+1
2
(iv) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 12 +1
max(21  1)¡ (+2)
2 2 ¡
+2
³
´2
+1
2
(v) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 = 12 +1
21 ¡ (+2)
2 2 ¡ 
+2
³
´2
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1

+2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
³
´2
+1
2
(viii) Player 2 deviates to 02 = (1 0), 02 = 1: yields 20 = 2 +1
22 ¡ 2 (+2)
2 1 ¡ 
+2
³
´2
+1
2
(ix) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 12 +1
22 ¡ (+2)
2 1 ¡ 
+2
³
´2
+1
2
(x) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = 12 +1
22 ¡ (+2)
2 1 ¡ 
+2
Observe that (1) will be an equilibrium if and only if (i) through (x) are not pro…table
deviations. This yields ten conditions for (1) to be an equilibrium, but only conditions (i)
and (viii) are binding:
2
Condition (i): 21 ¸ 1 + (+1)
22 
³
´2 ³
3 2
Condition (viii):  ¸ +1
+2
2 2 ¡
1
2
+1 1
2. 1 = (0 1)
´
= ¹2 
³
´2
³
´2
1 +1
If the players follow these strategies, they receive: 1 = 12 +1


=
22  In
2
+2
2 +2
order for this to be an equilibrium, there cannot be any pro…table deviations. There are
ten deviations we need to check.
³
´2
+1
2
(i) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 12 +1
21 ¡ (+2)
2 2 
+2
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
2
(iii) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 ¡ 
+2
³
´2
+1
2
(iv) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 12 +1
max(1 21 )¡ (+2)
2 2 ¡
+2
³
´2
+1
2
(v) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 = 12 +1
21 ¡ (+2)
2 2 ¡ 
+2
³
´2
+1
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1
¡ (+2)
2
+2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
³
´2
+1
(viii) Player 2 deviates to 02 = (1 0), 02 = 1: yields 20 = 12 +1
22 ¡ (+2)
2 ¡ 
+2
³
´2
+1
(ix) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 2 +1
¡ 2 (+2)
2 ¡ 
+2
³
´2
+1
(x) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = 12 +1
¡ (+2)
2 ¡ 
+2
54
Observe that (2) will be an equilibrium if and only if (i) through (x) are not pro…table
deviations. This yields ten conditions for (2) to be an equilibrium, but only conditions (i),
(iii), and (viii) are binding:
2
Condition (i): 21 · 1 + +1
22 
³
´2
4
Condition (iii):  ¸ 12 +1
[421 ¡ +1
22 ¡ 1]
+2
³
´2
2
Condition (viii):  ¸ 12 +1
[¡ +1
]
+2
We can combine conditions (iii) and (viii) as follows:
³
´2
³
´
2 ¡ 4 2 ¡ 1 ¡ 2
¹
 ¸ 12 +1
max
4
1
+2
+1 2
+1 = 1 
3. 1 = (0 0)
If the players follow these strategies, 1 = 0. If player 1 deviates to  01 = (0 1), 01 = 0,
³
´2
this yields: 10 = 12 +1
. So, there is always a pro…table deviation. Thus, (3) is never
+2
an equilibrium.
Case 2a: Interaction, initiated by both players (1 = 2 = 1)
Observe that there will be a pro…table deviation to 0 = 0 if   0. If  · 0, deviating
to 0 = 0 is (weakly) unpro…table. Therefore, it will be su¢cient, after imposing the condition
 · 0, to restrict attention to deviations in which 0 = 1. There are three combinations to check.
1. 1 = (1 0)
³
´2
+1
2
If the players follow these strategies, they receive: 1 = 2 +1
21 ¡ 4 (+2)
2 2 ¡ 
+2
³
´2
+1
2
2 = 2 +1
22 ¡ 4 (+2)
2 1 ¡  In order for this to be an equilibrium, there cannot be
+2
any pro…table deviations. There are four deviations we need to check.
(i) Player 1 deviates to  01 = (0 1), 01 = 1: yields 10 =

1
2
³
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 =
1
2
(iii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 =
1
2
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 =
1
2
+1
+2
´2
³
+1
+2
³
+1
+2
³
+1
+2
+1
2
max(21  1)¡2 (+2)
2 2 ¡
´2
+1
2
21 ¡ 2 (+2)
2 2 ¡ 
´2
+1
2
22 ¡ 2 (+2)
2 1 ¡ 
´2
+1
2
22 ¡ 2 (+2)
2 1 ¡ 
Observe that (1) will be an equilibrium if and only if (i) through (iv) are not pro…table
deviations. This yields four conditions for (1) to be an equilibrium, but only conditions (i)
and (ii) are binding:
Condition (i): 21 ¸
1
4
1
2
+1 2
4
2
3(+1) 2 (This
+
Condition (ii): 21 
equal, to not value activities.)
inequality is strict because player 1 prefers, all else
Condition (k):  · 0
55
2. 1 = (0 1)
If 21 ¸ 1, there will always be a pro…table deviation for player 1 to  01 = (0 0) 01 = 1. So
³
´2
we assume 21  1. If the players follow these strategies, they receive: 1 = 12 +1
¡
+2
³
´2
+1
1 +1
+1
2
22 ¡ (+2)
2 ¡  In order for this to be an equilibrium, there
(+2)2 2 ¡  2 = 2 +2
cannot be any pro…table deviations. There are four deviations we need to check.
(i) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 =
1
2
(iii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 2
(iv) Player 2 deviates to
02
= (0 0),
02
= 1: yields
20
=
1
2
³
+1
+2
³
+1
+2
³
³
+1
+2
+1
+2
´2
+1
2
21 ¡ 2 (+2)
2 2 ¡ 
´2
+1
¡ 2 (+2)
2 ¡ 
´2
´2
21 ¡
¡
+1
2
(+2)2 2
+1
(+2)2
¡ 
¡ 
Observe that (2) will be an equilibrium if and only if (i) through (iv) are not pro…table
deviations. This yields four conditions for (2) to be an equilibrium, but only conditions (i)
and (ii) are binding:
Condition (i): 21 ·
1
4
+
1
2
2(+1) 2
Condition (ii): 21  1 (This inequality is strict because player 1 prefers, all else equal, to
not value activities.)
Condition (k):  · 0
3. 1 = (0 0)
If the players follow these strategies, they receive: 1 =
³
´2
1 +1
+1
2
22 ¡ (+2)
2 1 ¡ 
2 +2
1
2
³
+1
+2
´2
21 ¡
+1
2
(+2)2 2
¡  2 =
In order for this to be an equilibrium, there cannot be any pro…table deviations. There are
four deviations we need to check.
³
´2
+1
2
(i) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 ¡ 
+2
³
´2
+1
2
(ii) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 12 +1
max(21  1)¡ (+2)
2 2 ¡
+2
³
´2
(iii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 12 +1
¡ 
+2
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = ¡
Observe that (3) will be an equilibrium if and only if (i) through (iv) are not pro…table
deviations. This yields four conditions for (3) to be an equilibrium, but only conditions (i)
and (ii) are binding:
2
Condition (i): 21  3(+1)
22 (This inequality is strict because player 1 prefers, all else
equal, to not value activities.)
Condition (ii): 21  1 (This inequality is strict because player 1 prefers, all else equal, to
not value activities.)
Condition (k):  · 0
56
Case 2b: Interaction, initiated by player 1 only (1 = 1 2 = 0)
Observe that there will be a pro…table deviation to 02 = 1 if   0. There will not be
a pro…table deviation to 02 = 1 if  ¸ 0. Therefore, it will be su¢cient, after imposing the
condition  ¸ 0, to restrict attention to deviations by player 2 in which 02 = 0. There are three
combinations to check.
1. 1 = (1 0)
³
´2
+1
2
If the players follow these strategies, they receive: 1 = 2 +1
21 ¡ 4 (+2)
2 2 ¡ 
+2
³
´2
+1
2
2 = 2 +1
22 ¡ 4 (+2)
2 1  In order for this to be an equilibrium, there cannot be any
+2
pro…table deviations. There are seven deviations we need to check.
(i) Player 1 deviates to  01 = (0 1), 01 = 1: yields 10 =

1
2
³
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 =
1
2
(iii) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 =
1
2
(iv) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 =
1
2
+1
+2
´2
³
+1
+2
³
+1
+2
³
+1
+2
+1
2
max(21  1)¡2 (+2)
2 2 ¡
´2
+1
2
21 ¡ 2 (+2)
2 2 ¡ 
´2

´2
+1
2
21 ¡ 2 (+2)
2 2 
(v) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
2
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1
22 ¡ 2 (+2)
2 1 
+2
³
´2
+1
2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 12 +1
22 ¡ 2 (+2)
2 1 
+2
Observe that (1) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations. This yields seven conditions for (1) to be an equilibrium, but only conditions
(i), (ii), (iii), and (iv) are binding:
Condition (i): 21 ¸
1
4
1
2
+1 2
4
2
3(+1) 2 (This
+
Condition (ii): 21 
inequality is strict because player 1 prefers, all else
equal, to not value activities.)
³
´2 h
i
3 2
2
2
Condition (iii):  · +1

¡

+2
2 1
+1 2
³
´2
4
Condition (iv):  · +1
(221 ¡ 12 ¡ +1
22 )
+2
Condition (k):  ¸ 0
2. 1 = (0 1)
If 21 ¸ 1, there will always be a pro…table deviation for player 1 to  01 = (0 0) 01 = 1. So,
³
´2
+1
2
assume 21  1. If the players follow these strategies, 1 = 12 +1
¡ (+2)
If
2 2 ¡ .
+2
³
´2
player 1 deviates to  01 = (0 1), 01 = 0, this yields: 10 = 12 +1
. So, there is always a
+2
pro…table deviation. Thus, (2) is never an equilibrium.
57
3. 1 = (0 0)
to  01 = (1 0), 01 = 1, this yields: 10 =
1
2
³
´2
+1
+2
³
´2
2 +1
+2
If the players follow these strategies, 1 =
+1
2
21 ¡ (+2)
2 2 ¡ . If player 1 deviates
+1
2
21 ¡ 2 (+2)
This is a pro…table
2 2 ¡ .
deviation if 32 ( + 1)21  22 . Suppose this is not a pro…table deviation: 32 ( + 1)21 ·
22 . If player 1 instead deviates
to  01 = (0¢0), 01 = 0, this yields: 10 = 0. This is a
¡
+1
1
3
2
2
2
2
pro…table deviation if (+2)
2
1 · 2 , it follows that
2 ( + 1)1 ¡ 2  . Since 2 ( + 1)
¡
¢
1
+1
1
2
2
2
2
2 ( + 1)1 ¡ 2  0. And, since  ¸ 0, it must be true that (+2)2 2 ( + 1)1 ¡ 2  .
Hence, 01 = (0 0), 01 = 0 must indeed be a pro…table deviation. So, it follows that at least
one of the two deviations will be pro…table. So, no equilibrium exists with  1 = (0 0)
Case 2c: Interaction, initiated by player 2 only (1 = 0 2 = 1)
Observe that there will be a pro…table deviation to 01 = 1 if   0. There will not be
a pro…table deviation to 01 = 1 if  ¸ 0. Therefore, it will be su¢cient, after imposing the
condition  ¸ 0, to restrict attention to deviations by player 1 in which 01 = 0. There are three
combinations to check.
1. 1 = (1 0)
³
´2
+1
2
If the players follow these strategies, they receive: 1 = 2 +1
21 ¡ 4 (+2)
2 2  2 =
+2
³
´2
+1
2
2 +1
22 ¡ 4 (+2)
2 1 ¡  In order for this to be an equilibrium, there cannot be any
+2
pro…table deviations. There are seven deviations we need to check.
(i) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 =
1
2
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 =
1
2
(iii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 =
1
2
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 =
1
2
(v) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 =
1
2
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 =
1
2
³
+1
+2
³
+1
+2
³
+1
+2
³
+1
+2
³
+1
+2
³
+1
+2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
´2
+1
2
max(21  1) ¡ 2 (+2)
2 2 
´2
+1
2
22 ¡ 2 (+2)
2 1 ¡ 
´2
´2
´2
´2
+1
2
21 ¡ 2 (+2)
2 2 
+1
2
22 ¡ 2 (+2)
2 1 ¡ 
+1
2
22 ¡ 2 (+2)
2 1 

Observe that (1) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations. This yields seven conditions for (1) to be an equilibrium, but only conditions
(i), (ii), (v), and (vi) are binding:
Condition (i): 21 ¸
1
4
1
2
+1 2
4
2 (This
3(+1) 2
+
Condition (ii): 21 
equal, to not value activities.)
³
´2 ³
3 2
Condition (v):  · +1
+2
2 2 ¡
inequality is strict because player 1 prefers, all else
2
2
+1 1
58
´
Condition (vi):  ·
³
Condition (k):  ¸ 0
+1
+2
´2 ³
222 ¡
1
2
¡
4
2
+1 1
´
2. 1 = (0 1)
If 21 ¸ 1, there will always be a pro…table deviation for player 1 to 01 = (0 0) 01 = 0.
³
´2
+1
So, assume 21  1. If the players follow these strategies, 2 = 12 +1
22 ¡ (+2)
2 ¡ 
+2
³
´2
If player 2 deviates to  02 = (1 0), 02 = 0, this yields: 20 = 12 +1
22 , which is always
+2
pro…table. So, equilibria of this type never exist.
3. 1 = (0 0)
If players follow these strategies, 2 =
1
2
+1
2
(+2)2 1
³
+1
+2
´2
21 ¡
+1
2 ,
(+2)2 2
2 =
1
2
³
+1
+2
´2
22 ¡
¡  In order for this to be an equilibrium, there cannot be any pro…table deviations. There are seven deviations we need to check.
³
´2
+1
2
(i) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 
+2
³
´2
+1
2
(ii) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 12 +1
max(21  1) ¡ (+2)
2 2 
+2
³
´2
(iii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 12 +1
¡ 
+2
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = ¡
³
´2
(v) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 = 12 +1
22 ¡
+2
³
´2
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1

+2
+1
2 
(+2)2 1
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
Observe that (3) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations. This yields seven conditions for (3) to be an equilibrium, but only conditions
(i), (ii), (v), and (vi) are binding:
2
Condition (i): 21  3(+1)
22 (This inequality is strict because player 1 prefers, all else
equal, to not value activities.)
Condition (ii): 21  1 (This inequality is strict because player 1 prefers, all else equal, to
not value activities.)
Condition (v):  · 0
If this condition is combined with condition (k), stated below, we see that we must have
 = 0.
³
´2
1
Condition (vi): +1
( 12 22 ¡ +1
21 ¡ 12 ) ¸ 
+2
Or, using the fact that we must have  = 0, this can be rewritten as: 22 ¸
Condition (k):  ¸ 0
Case 2: Combined
If we combine Cases 2a, 2b, and 2c we …nd the following.
59
2
2
+1 1
+1
1. An equilibrium with 1 = 2 = (1 0) and interaction between the players exists if:
(i) 21 ¸
(ii) 21 
1
4
1
2
+1 2
4
2
3(+1) 2
+
(iii)  · 0 or  ¸ 0,  ·
Since
³
+1
+2
´2 ³
3 2
2 2
be restated as:  ·
¡
³
³
+1
+2
´2 ³
3 2
2 2
¡
2
2
+1 1
´
³
´2 ³
, and  · +1
222 ¡
+2
´
³
´2 ³
´
2
2 ¸ 0 and +1
2 ¡ 1 ¡ 4 2 ¸ 0,

2
2
+1 1
+2
2
+1 1
´2
³
´
+1
3 2
2
2  22 ¡ 1 ¡ 4 2 = 
¹3 .
min

¡

2
+2
2 2
+1 1
2
+1 1
1
2
¡
4
2
+1 1
´
.
condition (iii) can
2. An equilibrium with 1 = (0 1) 2 = (1 0) and interaction between the players exists if:
(i) 21 ·
1
4
+
1
2
2(+1) 2
(ii) 21  1
(iii)  · 0
3. An equilibrium with 1 = (0 0) 2 = (1 0) and interaction between the players exists if:
(i) 1  21 
2
2
3(+1) 2
(ii)  · 0
Additional combinations to consider:
In order to complete the proof, we also need to show that no equilibria exist in which player
2 is not a scholar. There are twelve cases we need to consider.
1. 1 = (1 0),  2 = (0 1) and no interaction (1 = 2 = 0)
³
´2
If both players follow these strategies, 2 = 12 +1
 If player 2 deviates to  02 = (1 0),
+2
³
´2
+1
2
2
2
2
02 = 0, this yields 12 +1
22 ¡ (+2)
2 1 . This deviation is pro…table if 1  2 ¡ +1 1 .
+2
Since 2 ¸ 1 , a su¢cient condition for this deviation to be pro…table is 1  22 ¡
2
or ¡1
+1 2  1, which is true by assumption.
2
2
+1 2 ,
2. 1 = (0 1),  2 = (0 1) and no interaction (1 = 2 = 0)
³
´2
+1
If both players follow these strategies, 2 = 12 +1
¡ (+2)
If player 2 deviates to
2
+2
³
´2
02 = (1 0), 02 = 0, this yields 12 +1
22 . This deviation is pro…table if 22  ¡1
+2
+1 , which
is true by assumption.
3. 1 = (0 0),  2 = (0 1) and no interaction (1 = 2 = 0)
³
´2
If both players follow these strategies, 2 = 12 +1
 If player 2 deviates to  02 = (1 0),
+2
³
´2
02 = 0, this yields 12 +1
22  which is always pro…table.
+2
60
4. 1 = (1 0),  2 = (0 0) and no interaction (1 = 2 = 0)
If both players follow these strategies, 2 = 0 If player 2 deviates to 02 = (0 1), 02 = 0,
³
´2
this yields 12 +1
 which is always pro…table.
+2
5. 1 = (0 1),  2 = (0 0) and no interaction (1 = 2 = 0)
If both players follow these strategies, 2 = 0 If player 2 deviates to 02 = (1 0), 02 = 0,
³
´2
this yields 12 +1
22  which is always pro…table.
+2
6. 1 = (0 0),  2 = (0 0) and no interaction (1 = 2 = 0)
If both players follow these strategies, 2 = 0 If player 2 deviates to 02 = (1 0), 02 = 0,
³
´2
this yields 12 +1
22  which is always pro…table.
+2
7. 1 = (1 0),  2 = (0 1) and interaction (1 = 1 or 2 = 1)
´2
+1
+1
2
22 ¡ (+2)
2 1 ¡ 2 
+2
³
´2
+1
2
2 +1
22 ¡ 2 (+2)
2 1 ¡ 2 .
+2
If 21 ¸ 1. Then, if both players follow these strategies, 2 =
If player 2 deviates to  02 = (1 0), 02 = 2 , this yields 20 =
1
2
³
2
This deviation is pro…table if 22  3(+1)
21 . Since 2 ¸ 1 by assumption, this will indeed
be a pro…table deviation. Now suppose 21  1. Then, it is pro…table for player 1 to
deviate to  01 = (0 0), 01 = 1 . So, a pro…table deviation always exists.
8. 1 = (0 1),  2 = (0 1) and interaction (1 = 1 or 2 = 1)
³
´2
+1
If both players follow these strategies, 2 = 2 +1
¡ 4 (+2)
2 ¡ 2  If player 2 deviates
+2
³
´2
+1
to 02 = (1 0), 02 = 2 , this yields 20 = 12 +1
22 ¡ 2 (+2)
This deviation is
2 ¡ 2 .
+2
pro…table if 22 
4
+1 ,
which is true by assumption.
9. 1 = (0 0),  2 = (0 1) and interaction (1 = 1 or 2 = 1)
³
´2
+1
If both players follow these strategies, 1 = 12 +1
¡ (+2)
2 ¡ 1  If player 1 deviates to
+2
³
´2
+1
01 = (0 1), 01 = 1 , this yields 10 = 2 +1
¡ 2 (+2)
2 ¡ 1 . This deviation is pro…table
+2
if
3
2

1
+1 ,
which is always true.
10. 1 = (1 0),  2 = (0 0) and interaction (1 = 1 or 2 = 1)
³
´2
+1
2
If both players follow these strategies, 2 = 12 +1
22 ¡ (+2)
If player 2
2 1 ¡ 2 
+2
³
´2
+1
2
deviates to  02 = (1 0), 02 = 2 , this yields 20 = 2 +1
22 ¡ 2 (+2)
This
2 1 ¡ 2 .
+2
deviation is pro…table if 22 
a pro…table deviation.
2
2 .
3(+1) 1
Since 2 ¸ 1 by assumption, this will indeed be
11. 1 = (0 1),  2 = (0 0) and interaction (1 = 1 or 2 = 1)
³
´2
+1
If both players follow these strategies, 2 = 12 +1
¡ (+2)
2 ¡ 2  If player 2 deviates
+2
³
´2
+1
to 02 = (1 0), 02 = 2 , this yields 20 = 12 +1
22 ¡ (+2)
This deviation is
2 ¡ 2 .
+2
pro…table if 22  1, which is true by assumption.
61
12. 1 = (0 0),  2 = (0 0) and interaction (1 = 1 or 2 = 1)
If both players follow these strategies, 2 = ¡2  If player 2 deviates to 02 = (1 0),
³
´2
02 = 2 , this yields 20 = 12 +1
22 ¡ 2 . This deviation is always pro…table.
+2
This establishes that no equilibria exist in which player 2 is not a scholar. This completes
the proof.
Proof of Proposition 3. First, let us establish that, in any equilibrium, players value one activity
(except possibly a set of players with 0 mass).
As in the basic model, the returns to e¤ort will be (weakly) greater at one of the activities.
Furthermore, players prefer, all else equal, not to exert e¤ort at activities. It follows that players
will exert e¤ort at – at most – one activity in equilibrium. Players will either be below average
or average at activities when they exert no e¤ort. Since players prefer, all else equal, not to value
activities, it follows that players will either value one activity or zero activities in equilibrium.
Suppose there is an equilibrium in which a fraction  of the players value activity . By a
logic identical to that given in Lemma 5, players who value activity  will interact in equilibrium
with all players who hold the same values (except perhaps a set of measure 0); players who value
activity  will not interact with players who hold di¤erent values (except perhaps a set of measure
0). Following a logic similar to that from Lemma 2, players valuing activity  will choose to exert
e¤ort  +  at activity , and zero e¤ort at other activities. Since a fraction  of the population
exerts e¤ort  +  at activity , and the rest of the population exerts e¤ort 0 at activity , the
average achievement at activity¡  is: 
¹¢ =  ( + ). From this, we conclude that players who
value activity  receive utility: 12 ¡  ( + )2 .
Suppose now that a positive mass of players values no activity in equilibrium. (We will show
that this creates a contradiction). By an identical logic to that given in the proof of Lemma 5,
it follows that such players will not interact with any other player in equilibrium. And, by an
identical logic to that given in Lemma 2, it follows that these players will choose to exert zero
e¤ort at all activities and will receive utility¡ 0. If¢these players deviate and instead choose to
value activity , they would
utility 12 ¡  ( + )2 . Since this deviation cannot be
¡ 1 receive
¢
pro…table, we must have 2 ¡  ( + )2 · 0 for all , which is true if and only if  ¸ 12 for all
. The only way we can have  ¸ 12 for all  is if  = 2 and 1 = 2 = 12 . But, if 1 = 2 = 12 ,
it follows that there cannot be a positive mass of players that value no activity (a contradiction).
Thus, we have established that, in any equilibrium, players values one activity (except possibly
a set of players with
Furthermore, we have established that players who value activity
¡ 1 0 mass).
¢
 receive utility 2 ¡  ( + )2 in equilibrium, where  denotes the fraction of players who
value activity . We will refer to players who value activity  as a group of size  . We will focus
on characterizing the set of equilibria with the property that all groups are of equal size (i.e., for
¹ or 0).
all ,  = 
Case 1: every activity is valued by a positive mass of players in equilibrium.
First, we will examine the possibility that equilibria exist in which every activity is valued by
1
a positive mass of players. Since   0 for all , we must have  = 
for all .
1
Let us check whether it is an equilibrium for all activities to be valued by a fraction 
of the
players. There are two ways in which players can deviate. One way in which a player can deviate
is by joining a di¤erent group. But, since groups are all of the same size, it is not pro…table
to join a di¤erent group. A second way in which a player can deviate is by choosing to value
62
no activity, which yields utility 0.
¡ 1 We1 know
¢ 1 this 2is an unpro…table deviation since players who
belong to groups receive utility 2 ¡  (  + ) ¸ 0. Hence, it is always an equilibrium for
the players to divide into  groups of equal size.
Case 2:    activities are valued by a positive mass of players in equilibrium.
Now, let us examine whether equilibria exist in which    activities are valued by a
1
positive mass of players. We must have  = 
for all activities valued by a positive mass of
players.
1
Let us check whether it is an equilibrium for a fraction 
of the population to value each of
 activities. There are three ways in which players can deviate. One way in which a player
can deviate is by joining a di¤erent group. But, since groups are all of the same size, it is not
pro…table to join a di¤erent group. A second way in which a player can deviate is by choosing
to value no activity, which yields utility 0. A third way in which a player can deviate is by
starting a new group (a group of size 0), which yields utility 12  2 . Observe that this third type of
deviation is always
preferred
to the second type of deviation. This deviation will be unpro…table
¡
¢ 1
1
if and only if: 12 ¡ 
(  + )2  12  2 . The reason this inequality is strict is that a player who
deviates forgoes having to pay a positive (but negligible) cost of initiating interaction with other
players in his group. If  ¸ 1, this deviation
for all . If   1, the condition can
¡ 1 is1 pro…table
¢ 1
1 2
2
be rewritten as:   
¹ where 
¹ solves: 2 ¡ 
¹ (
¹ + ) = 2  .
This completes the proof.
Proof of Proposition A1. Suppose 2 ¸ 1 and 21  
¹  . We will characterize the equilibria that
exist as a function of 1 , 2 , , and 
It will be useful, in characterizing the equilibria, to use the same shorthand as we used in
the proof of Proposition 1. Once again, we will examine all combinations of the following
form to see, under what parameters, they are equilibria of the game: f((1  1 ) (2  2 )) :   2
f(1 0) (0 1) (0 0)g  2 f0 1gg. We will …rst examine combinations in which player 1 is a
musician ( 1 = (0 1)). Later on, we will show that no equilibria exist in which player 1 is not a
musician.
Case 1: No interaction (1 = 2 = 0)
Since we assume player 1 is a musician ( 1 = (0 1)), we will denote combinations by player
2’s choice of  2 . There are three combinations to check.
1. 2 = (0 1)
If the players follow these strategies, they receive: 1 =
+1

(+2)2
1
2
³
+1
+2
´2
+1
¡ (+2)
2  2 =
1
2
³
+1
+2
´2
¡
In order for this to be an equilibrium, there cannot be any pro…table deviations.
There are ten deviations we need to check.
³
´2
(i) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 12 +1
21 
+2
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
(iii) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 2 +1
¡ 2 (+2)
2 ¡ 
+2
³
´2
+1
(iv) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 12 +1
¡ (+2)
2 ¡ 
+2
63
(v) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 =
1
2
(vi) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 =
1
2
³
³
+1
+2
+1
+2
´2
´2
¡
+1
(+2)2
¡ 
22 
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
³
´2
+1
(viii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 2 +1
¡ 2 (+2)
2 ¡ 
+2
³
´2
+1
(ix) Player 2 deviates to  02 = (1 0), 02 = 1: yields 20 = 12 +1
max(22  1) ¡ (+2)
2 ¡ 
+2
³
´2
+1
(x) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = 12 +1
¡ (+2)
2 ¡ 
+2
Observe that (1) will be an equilibrium if and only if (i) through (x) are not pro…table
deviations. This yields ten conditions for (1) to be an equilibrium, but only conditions (iii),
(vi), and (ix) are binding:
³
´2 ³
´
3
1
Condition (iii):  ¸ +1
¡
+2
2
+1
Condition (vi):
¡1
+1
Condition (ix):  ¸
¸ 22
³
´2
1
2
+1
+2
(22 ¡ 1)
We can combine conditions (iii) and (ix) as follows:
³
´2
³
´
3
1
1 2
1
¹
 ¸ +1
max
¡


¡
+2
2
+1 2 2
2 = 1
2. 2 = (1 0)
³
´2
³
´2
1 +1
If the players follow these strategies, they receive: 1 = 12 +1


=
22  In
2
+2
2 +2
order for this to be an equilibrium, there cannot be any pro…table deviations. There are
ten deviations we need to check.
³
´2
+1
(i) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1
¡ (+2)
2
+2
(ii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
³
´2
+1
(iii) Player 2 deviates to  02 = (1 0), 02 = 1: yields 20 = 12 +1
max(1 22 ) ¡ (+2)
2 ¡ 
+2
³
´2
+1
(iv) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 2 +1
¡ 2 (+2)
2 ¡ 
+2
³
´2
+1
(v) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = 12 +1
¡ (+2)
2 ¡ 
+2
³
´2
+1
2
(vi) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 12 +1
21 ¡ (+2)
2 2 
+2
(vii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
2
(viii) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 ¡ 
+2
³
´2
+1
2
(ix) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 12 +1
¡ (+2)
2 2 ¡ 
+2
³
´2
+1
2
(x) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 = 12 +1
21 ¡ (+2)
2 2 ¡ 
+2
64
Observe that (2) will be an equilibrium if and only if (i) through (x) are not pro…table
deviations. This yields ten conditions for (2) to be an equilibrium, but only conditions (i),
(iii), (iv), and (viii) are binding:
Condition (i): 22 ¸
Condition (iii):  ¸
¡1
+1
³
³
+1
+2
´2
1
(¡ +1
)
2
( +1
¡ 12 22 )
³
´2
Condition (viii):  ¸ +1
(221 ¡ 12 ¡
+2
Condition (iv):  ¸
+1
+2
´2
222
+1 )
We can combine conditions (iii), (iv), and (viii) as follows:
³
´2
³
´
2
1
2
1 2
2 ¡ 1 ¡ 22
¹
 ¸ +1
max
¡

¡


2
1
+2
+1 +1
2 2
2
+1 = 2
3. 2 = (0 0)
If the players follow these strategies, 2 = 0. If player 2 deviates to  02 = (1 0), 02 = 0, this
³
´2
yields: 20 = 12 +1
22 . So, there is always a pro…table deviation. Thus, (3) is never an
+2
equilibrium.
Case 2a: Interaction, initiated by both players (1 = 2 = 1)
Observe that there will be a pro…table deviation to 0 = 0 if   0. If  · 0, deviating
to 0 = 0 is (weakly) unpro…table. Therefore, it will be su¢cient, after imposing the condition
 · 0, to restrict attention to deviations in which 0 = 1. There are three combinations to check.
1. 2 = (0 1)
³
´2
+1
If the players follow these strategies, they receive: 1 = 2 +1
¡ 4 (+2)
2 ¡  2 =
+2
³
´2
+1
2 +1
¡4 (+2)
2 ¡ In order for this to be an equilibrium, there cannot be any pro…table
+2
deviations. There are four deviations we need to check.
³
´2
+1
(i) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 12 +1
¡ 2 (+2)
2 ¡ 
+2
³
´2
+1
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 = 12 +1
¡ 2 (+2)
2 ¡ 
+2
³
´2
+1
(iii) Player 2 deviates to  02 = (1 0), 02 = 1: yields 20 = 12 +1
max(22  1) ¡2 (+2)
2 ¡
+2
³
´2
+1
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = 12 +1
¡ 2 (+2)
2 ¡ 
+2
Observe that (1) will be an equilibrium if and only if (i) through (iv) are not pro…table
deviations. Only condition (iii) is binding:
Condition (iii): 22 ·
4
+1
Condition (k):  · 0
65
2. 2 = (1 0)
If 22  1, there will always be a pro…table deviation for player 2 to  02 = (0 0) 02 = 1. So
³
´2
we assume 22 ¸ 1. If the players follow these strategies, they receive: 1 = 12 +1
¡
+2
³
´2
+1
+1
2 ¡  2 = 12 +1
22 ¡ (+2)
2 ¡  In order for this to be an equilibrium, there
+2
(+2)2 2
cannot be any pro…table deviations. There are four deviations we need to check.
(i) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 =
1
2
(iii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 2
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 =
1
2
³
³
³
³
+1
+2
+1
+2
+1
+2
+1
+2
´2
+1
2
21 ¡ 2 (+2)
2 2 ¡ 
´2
+1
¡ 2 (+2)
2 ¡ 
´2
´2
21 ¡
¡
+1
2
(+2)2 2
+1
(+2)2
¡ 
¡ 
Observe that (2) will be an equilibrium if and only if (i) through (iv) are not pro…table
deviations. This yields four conditions for (2) to be an equilibrium, but only condition (iii)
is binding:
Condition (iii): 22 ¸ 4 ¡
2
+1
Condition (k):  · 0
3. 2 = (0 0)
³
´2
+1
If the players follow these strategies, they receive: 1 = 12 +1
¡ (+2)
2 ¡  2 =
+2
³
´2
1 +1
+1
¡ (+2)
If player 2 deviates to  02 = (0 1), 02 = 1, this yields 20 =
2 ¡ 
2 +2
³
´2
+1
2 +1
¡2 (+2)
2 ¡ This deviation is always pro…table. So, this is never an equilibrium.
+2
Case 2b: Interaction, initiated by player 2 only (1 = 0 2 = 1)
Observe that there will be a pro…table deviation to 01 = 1 if   0. There will not be
a pro…table deviation to 01 = 1 if  ¸ 0. Therefore, it will be su¢cient, after imposing the
condition  ¸ 0, to restrict attention to deviations by player 1 in which 01 = 0. There are three
combinations to check.
1. 2 = (0 1)
If the players follow these strategies, they receive: 1 = 2
+1
4 (+2)
2
³
+1
+2
´2
+1
¡4 (+2)
2  2 = 2
³
+1
+2
´2
¡
¡ In order for this to be an equilibrium, there cannot be any pro…table deviations.
There are seven deviations we need to check.
³
´2
+1
(i) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 12 +1
¡ 2 (+2)
2
+2
³
´2
+1
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 12 +1
¡ 2 (+2)
2
+2
³
´2
+1
(iii) Player 2 deviates to  02 = (1 0), 02 = 1: yields 20 = 12 +1
max(22  1) ¡2 (+2)
2 ¡
+2
³
´2
+1
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = 12 +1
¡ 2 (+2)
2 ¡ 
+2
66
(v) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 =
1
2
(vi) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 =
1
2
³
³
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
+1
+2
+1
+2
´2
´2
+1
¡ 2 (+2)
2
22 
Observe that (1) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations. This yields seven conditions for (1) to be an equilibrium, but only conditions
(iii), (v), and (vi) are binding:
Condition (iii):
Condition (v):
Condition (vi):
4
+1
³
+1
+2
³
¸ 22
´2 ³
+1
+2
Condition (k):  ¸ 0
3
2
´2
(2
¡
³
2
+1
¡1
+1
´
´
¸
¡ 12 22 ) ¸ 
2. 2 = (1 0)
If 22  1, there will always be a pro…table deviation for player 2 to  02 = (0 0) 02 = 1. So,
³
´2
+1
assume 22 ¸ 1. If the players follow these strategies, 2 = 12 +1
22 ¡ (+2)
If
2 ¡ .
+2
³
´2
player 2 deviates to  02 = (1 0), 02 = 0, this yields: 20 = 12 +1
22 . So, there is always
+2
a pro…table deviation. Thus, (2) is never an equilibrium.
3. 2 = (0 0)
³
´2
+1
If the players follow these strategies, 2 = 12 +1
¡ (+2)
If player 2 deviates to
2 ¡ .
+2
³
´2
+1
02 = (0 1), 02 = 1, this yields: 20 = 2 +1
¡ 2 (+2)
This is always a pro…table
2 ¡ .
+2
deviation. So, no equilibrium exists with 2 = (0 0)
Case 2c: Interaction, initiated by player 1 only (1 = 1 2 = 0)
Observe that there will be a pro…table deviation to 02 = 1 if   0. There will not be
a pro…table deviation to 02 = 1 if  ¸ 0. Therefore, it will be su¢cient, after imposing the
condition  ¸ 0, to restrict attention to deviations by player 2 in which 02 = 0. There are three
combinations to check.
1. 2 = (0 1)
³
´2
+1
If the players follow these strategies, they receive: 1 = 2 +1
¡ 4 (+2)
2 ¡  2 =
+2
³
´2
+1
2 +1
¡ 4 (+2)
2  In order for this to be an equilibrium, there cannot be any pro…table
+2
deviations. There are seven deviations we need to check.
³
´2
+1
(i) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 12 +1
¡ 2 (+2)
2 ¡ 
+2
³
´2
+1
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 = 12 +1
¡ 2 (+2)
2 ¡ 
+2
³
´2
+1
(iii) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 = 12 +1
¡ 2 (+2)
2
+2
67
(iv) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 =
1
2
³
+1
+2
´2
21 
(v) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
(vi) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 = 12 +1
max(22  1) ¡ 2 (+2)
2
+2
³
´2
+1
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 12 +1
¡ 2 (+2)
2
+2
Observe that (1) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations. This yields seven conditions for (1) to be an equilibrium, but only conditions
(iii), (iv), and (vi) are binding:
³
´2 ³
´
3
2
Condition (iii):  · +1
¡
+2
2
+1
³
´2 ³ ³
´
´
¡1
1 2
Condition (iv):  · +1
2
¡

+2
+1
2 1
Condition (vi):
4
+1
¸ 22
Condition (k):  ¸ 0
2. 2 = (1 0)
If 22  1, there will always be a pro…table deviation for player 2 to 02 = (0 0) 02 = 0.
³
´2
+1
2
So, assume 22 ¸ 1. If the players follow these strategies, 1 = 12 +1
¡ (+2)
2 2 ¡ 
+2
³
´2
If player 1 deviates to  01 = (0 1), 01 = 0, this yields: 10 = 12 +1
, which is always
+2
pro…table. So, equilibria of this type never exist.
3. 2 = (0 0)
³
´2
0
+1
If players follow these strategies, 2 = 12 +1
¡ (+2)
2  If player 2 deviates to  2 = (0 1),
+2
³
´2
+1
02 = 0, this yields: 20 = 2 +1
¡ 2 (+2)
So, equilibria of
2 , which is always pro…table.
+2
this type never exist.
Case 2: Combined
If we combine Cases 2a, 2b, and 2c we …nd the following.
1. An equilibrium with 1 = 2 = (0 1) and interaction between the players exists if:
(i) 22 ·
(ii)  ·
³
4
+1
+1
+2
´2
min
³
3
2
¡
2
+1  2
³
¡1
+1
´
¡ 12 21
´
2. An equilibrium with 1 = (0 1) 2 = (1 0) and interaction between the players exists if:
(i) 22 ¸ 4 ¡
2
+1
(ii)  · 0
Additional combinations to consider:
68
In order to complete the proof, we also need to show that no equilibria exist in which player
1 is not a musician. There are twelve cases we need to consider.
1. 1 = (1 0),  2 = (1 0) and no interaction (1 = 2 = 0)
³
´2
+1
2
If both players follow these strategies, 1 = 12 +1
21 ¡ (+2)
2 2  If player 1 deviates
+2
³
´2
³
´2
1 +1
to 01 = (0 1), 01 = 0, this yields 12 +1
.
This
deviation
is
pro…table
if
21 ¡
+2
2 +2
³
´2
+1
2  12 +1
. Since 1 · 2 , a su¢cient condition for this deviation to be pro…table
+2
(+2)2 2
³
´2
³
´2
+1
1 +1
2
2
is 12 +1

¡


, or ( ¡ 1)21   + 1. This holds by assumption.
1
+2
2 +2
(+2)2 1
2. 1 = (1 0),  2 = (0 1) and no interaction (1 = 2 = 0)
³
´2
³
´2
2 and  = 1 +1
If both players follow these strategies, 1 = 12 +1

 If player 1
2
1
+2
2 +2
³
´2
+1
deviates to  01 = (0 1), 01 = 0, this yields 12 +1
¡ (+2)
This deviation is pro…table
2.
+2
if
¡1
+1
 21 , which is true by assumption.
3. 1 = (1 0),  2 = (0 0) and no interaction (1 = 2 = 0)
If both players follow these strategies, 2 = 0 If player 2 deviates to 02 = (0 1), 02 = 0,
³
´2
this yields 12 +1
 which is always pro…table.
+2
4. 1 = (0 0),  2 = (1 0) and no interaction (1 = 2 = 0)
If both players follow these strategies, 1 = 0 If player 1 deviates to 01 = (0 1), 01 = 0,
³
´2
this yields 12 +1
 which is always pro…table.
+2
5. 1 = (0 0),  2 = (0 1) and no interaction (1 = 2 = 0)
If both players follow these strategies, 1 = 0 If player 1 deviates to 01 = (1 0), 01 = 0,
³
´2
this yields 12 +1
21  which is always pro…table.
+2
6. 1 = (0 0),  2 = (0 0) and no interaction (1 = 2 = 0)
If both players follow these strategies, 1 = 0 If player 1 deviates to 01 = (0 1), 01 = 0,
³
´2
this yields 12 +1
 which is always pro…table.
+2
7. 1 = (1 0),  2 = (1 0) and interaction (1 = 1 or 2 = 1)
³
´2
+1
2
If both players follow these strategies, 1 = 2 +1
21 ¡ 4 (+2)
If player 1
2 2 ¡ 1 
+2
³
´2
+1
2
deviates to 01 = (0 1), 01 = 1 , this yields 10 = 12 +1
¡ 2 (+2)
This
2 2 ¡ 1 .
+2
4
deviation is pro…table if 1  421 ¡ +1
22 . Since 2 ¸ 1 , a su¢cient condition for the
4
deviation to be pro…table is that 1  421 ¡ +1
21 , or 4 ¤ 21   + 1 This is true by
assumption.
8. 1 = (1 0),  2 = (0 1) and interaction (1 = 1 or 2 = 1)
³
´2
+1
If both players follow these strategies, 1 = 12 +1
¡ (+2)
2 ¡ 1  If player 1 deviates
+2
³
´2
+1
to  01 = (0 1), 01 = 1 , this yields 10 = 2 +1
¡ 2 (+2)
2 ¡ 1 . This deviation is always
+2
pro…table.
69
9. 1 = (1 0),  2 = (0 0) and interaction (1 = 1 or 2 = 1)
³
´2
+1
2
If both players follow these strategies, 1 = 12 +1
21 ¡ (+2)
If player 1
2 2 ¡ 1 
+2
³
´2
deviates to  01 = (0 1), 01 = 1 , this yields 10 = 12 +1
¡ 1 . This deviation is
+2
pro…table if 1  21 ¡
pro…table is 1  21 ¡
2
2
+1 2 .
2
2
+1 1 ,
Since 22 ¸ 21 , a su¢cient condition for the deviation to be
2
or 1  ¡1
+1 1 . This is true by assumption.
10. 1 = (0 0),  2 = (1 0) and interaction (1 = 1 or 2 = 1)
³
´2
+1
2
If both players follow these strategies, 1 = 12 +1
21 ¡ (+2)
If player 1
2 2 ¡ 1 
+2
³
´2
+1
2
deviates to  01 = (0 1), 01 = 1 , this yields 10 = 12 +1
¡ (+2)
2 1 ¡ 1 . This deviation
+2
is pro…table if 21  1, which is true by assumption.
11. 1 = (0 0),  2 = (0 1) and interaction (1 = 1 or 2 = 1)
³
´2
+1
If both players follow these strategies, 1 = 12 +1
¡ (+2)
2 ¡ 1  If player 1 deviates to
+2
³
´2
+1
02 = (0 1), 01 = 1 , this yields 10 = 2 +1
¡ 2 (+2)
2 ¡ 1 . This deviation is pro…table
+2
if
3
2

1
+1 ,
which is always true.
12. 1 = (0 0),  2 = (0 0) and interaction (1 = 1 or 2 = 1)
If both players follow these strategies, 1 = ¡1  If player 1 deviates to 01 = (0 1),
³
´2
01 = 1 , this yields 10 = 12 +1
¡ 1 . This deviation is always pro…table.
+2
This establishes that no equilibria exist in which player 1 is not a musician. This completes
the proof.
Proof of Proposition A2. Suppose 2 ¸ 1 , and 
¹  ¸ 1  2 ¸ 
¹  . We will characterize the
equilibria that exist as a function of 1 , 2 , , and 
It will be useful, in characterizing the equilibria, to use the same shorthand as we used in
the proof of Proposition 1. Once again, we will examine all combinations of the following
form to see, under what parameters, they are equilibria of the game: f((1  1 ) (2  2 )) :   2
f(1 0) (0 1) (0 0)g  2 f0 1gg.
Case 1: No interaction (1 = 2 = 0)
There are nine types of combinations to check. We will consider each in turn.
1. 1 = 1 = (1 0)
³
´2
+1
2
If the players follow these strategies, they receive: 1 = 12 +1
21 ¡ (+2)
2 2  2 =
+2
³
´2
1 +1
+1
2
22 ¡ (+2)
2 1  In order for this to be an equilibrium, there cannot be any pro…table
2 +2
deviations. There are ten deviations we need to check.
³
´2
(i) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 = 12 +1

+2
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
70
³
´2
+1
2
(iii) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 ¡ 
+2
³
´2
+1
2
(iv) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 12 +1
max(21  1)¡ (+2)
2 2 ¡
+2
³
´2
+1
2
(v) Player 1 deviates to 01 = (0 0) and 01 = 1: yields 10 = 12 +1
21 ¡ (+2)
2 2 ¡ 
+2
³
´2
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1

+2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
³
´2
+1
2
(viii) Player 2 deviates to 02 = (1 0), 02 = 1: yields 20 = 2 +1
22 ¡ 2 (+2)
2 1 ¡ 
+2
³
´2
+1
2
(ix) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 12 +1
max(22  1)¡ (+2)
2 1 ¡
+2
³
´2
+1
2
(x) Player 2 deviates to 02 = (0 0) and 02 = 1: yields 20 = 12 +1
22 ¡ (+2)
2 1 ¡ 
+2
Observe that (1) will be an equilibrium if and only if (i) through (x) are not pro…table
deviations. This yields ten conditions for (1) to be an equilibrium, but only conditions (i),
(iv), and (viii) are binding:
2
Condition (i): 21 ¸ 1 + +1
22
³
´2
Condition (iv):  ¸ 12 +1
(1 ¡ 21 )
+2
³
´2
1
Condition (viii):  ¸ +1
( 32 22 ¡ +1
21 )
+2
If we combine conditions (iv) and (viii), we obtain:
³
´2
1
 ¸ +1
max( 32 22 ¡ +1
21  12 ¡ 12 21 ) = ¹1
+2
2. 1 = (1 0) 2 = (0 1)
³
´2
³
´2
1 +1
2
If the players follow these strategies, they receive: 1 = 12 +1



=
 In
2
1
+2
2 +2
order for this to be an equilibrium, there cannot be any pro…table deviations. There are
ten deviations we need to check.
³
´2
+1
(i) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 = 12 +1
¡ (+2)
2
+2
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
(iii) Player 1 deviates to 01 = (1 0), 0 = 1: yields 10 =
1
2
(iv) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 2
³
³
+1
+2
+1
+2
´2
´2
max(21  1) ¡
+1
(+2)2
+1
¡ 2 (+2)
2 ¡
³
´2
+1
(v) Player 1 deviates to 01 = (0 0) and 01 = 1: yields 10 = 12 +1
¡ (+2)
2 ¡
+2
³
´2
+1
2
(vi) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 = 12 +1
22 ¡ (+2)
2 1
+2
¡
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
³
´2
+1
2
(viii) Player 2 deviates to  02 = (0 1), 02 = 1: yields 20 = 12 +1
max(22  1)¡ (+2)
2 1 ¡
+2
³
´2
+1
2
(ix) Player 2 deviates to 02 = (1 0), 02 = 1: yields 20 = 2 +1
22 ¡ 2 (+2)
2 1 ¡ 
+2
71
(x) Player 2 deviates to 02 = (0 0) and 02 = 1: yields 20 =
1
2
³
+1
+2
´2
22 ¡
+1
2
(+2)2 1
¡
Observe that (2) will be an equilibrium if and only if (i) through (x) are not pro…table
deviations. This yields ten conditions for (2) to be an equilibrium, but only conditions (i),
(iii), (iv), (vi), and (ix) are binding:
Condition (i): 21 ¸
¡1
+1
³
´2 ³
´
1
Condition (iii):  ¸ ¡ +1
+2
+1
³
´2
2
Condition (iv):  ¸ +1
( +1
¡ 12 21 )
+2
¡ 2
¢
Condition (vi): 21 ¸ +1
2 ¡ 1
2
³
´2
2
Condition (ix):  ¸ +1
(222 ¡ +1
21 ¡ 12 )
+2
If we combine conditions (iii), (vi), and (ix), we obtain:
³
´2
2
2
¡1
 ¸ +1
max( +1
¡ 12 21  222 ¡ +1
21 ¡ 12  +1
) = ¹5
+2
3. 1 = (1 0) 2 = (0 0)
If the players follow these strategies, 2 = 0. If player 2 deviates to  02 = (0 1), 02 = 0,
³
´2
this yields: 20 = 12 +1
. So, there is always a pro…table deviation. Thus, (3) is never
+2
an equilibrium.
4. 1 = 2 = (0 1)
³
´2
+1
If the players follow these strategies, 1 = 2 = 12 +1
¡ (+2)
In order for this to
2.
+2
be an equilibrium, there cannot be any pro…table deviations. There are ten deviations we
need to check.
³
´2
(i) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 12 +1
21
+2
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
(iii) Player 1 deviates to 01 = (1 0), 0 = 1: yields 10 =
1
2
(iv) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 2
³
+1
+2
³
+1
+2
³
+1
+2
´2
max(21  1) ¡
´2
+1
¡ 2 (+2)
2 ¡
´2
+1
(+2)2
¡
+1
¡ 2 (+2)
2 ¡
³
´2
+1
(v) Player 1 deviates to 01 = (0 0) and 01 = 1: yields 10 = 12 +1
¡ (+2)
2 ¡
+2
³
´2
(vi) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 = 12 +1
22
+2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
(viii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 2
(ix) Player 2 deviates to 02 = (1 0), 02 = 1: yields 20 =
³
+1
+2
´2
+1
max(22  1) ¡ (+2)
2 ¡ 
³
´2
+1
(x) Player 2 deviates to 02 = (0 0) and 02 = 1: yields 20 = 12 +1
¡ (+2)
2 ¡
+2
72
1
2
Observe that (4) will be an equilibrium if and only if (i) through (x) are not pro…table
deviations. This yields ten conditions for (4) to be an equilibrium, but only conditions
(iv/viii) and (vi) are binding:
³
´2 ³
´
3
1
¹
Condition (iv/viii):  ¸ +1
¡
+2
2
+1 = 2
Condition (vi): 22 ·
¡1
+1
5. 1 = (0 1) 2 = (1 0)
If the players follow these strategies, 1 =
1
2
³
+1
+2
´2
 2 =
1
2
³
+1
+2
´2
22 .
In order for this to be an equilibrium, there cannot be any pro…table deviations. There are
ten deviations we need to check.
³
´2
+1
2
(i) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 12 +1
21 ¡ (+2)
2 2
+2
(ii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
2
(iii) Player 1 deviates to 01 = (1 0), 0 = 1: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 ¡ 
+2
³
´2
+1
2
(iv) Player 1 deviates to  01 = (0 1), 01 = 1: yields 10 = 12 +1
max(21  1)¡ (+2)
2 2 ¡
+2
³
´2
+1
2
(v) Player 1 deviates to 01 = (0 0) and 01 = 1: yields 10 = 12 +1
21 ¡ (+2)
2 2 ¡ 
+2
³
´2
+1
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1
¡ (+2)
2
+2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
(viii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 2
(ix) Player 2 deviates to 02 = (1 0), 02 = 1: yields 20 =
1
2
³
³
+1
+2
+1
+2
´2
´2
+1
¡ 2 (+2)
2 ¡
+1
max(22  1) ¡ (+2)
2 ¡ 
³
´2
+1
(x) Player 2 deviates to 02 = (0 0) and 02 = 1: yields 20 = 12 +1
¡ (+2)
2 ¡
+2
Observe that (5) will be an equilibrium if and only if (i) through (x) are not pro…table
deviations. This yields ten conditions for (5) to be an equilibrium, but only conditions (i),
(iii), (v), (vi), (viii), and (ix) are binding:
2
Condition (i): 21 · 1 + +1
22
³
´2 ³
´
2 ¡ 2 2 ¡ 1
Condition (iii):  ¸ +1
2
1
2
+2
+1
2
³
´2 ³
´
1 2
1
2¡ 1
Condition (v):   +1

¡

= ¹4 (this inequality is strict because
+2
2 1
+1 2
2
players prefer, all else equal, not to value activities)
Condition (vi): 22 ¸
Condition (viii):  ¸
¡1
+1
³
Condition (ix):  ¸ ¡
+1
+2
³
´2 ³
+1
+2
2
+1
´2 ³
1
+1
¡ 12 22
´
´
If we combine conditions (iii), (viii), and (ix), we obtain:
³
´2
³
´
2
1 2
2 ¡ 2 2 ¡ 1  ¡1
¹
 ¸ +1
max
¡


2
1
+2
+1
2 2
+1 2
2 +1 = 3
73
6. 1 = (0 1) 2 = (0 0)
If the players follow these strategies, 2 = 0. If player 2 deviates to  02 = (1 0), 02 = 0, this
³
´2
yields: 20 = 12 +1
22 . So, there is always a pro…table deviation. Thus, (6) is never an
+2
equilibrium.
7. 1 = (0 0) 2 = (1 0)
If the players follow these strategies, 1 = 0. If player 1 deviates to  01 = (0 1), 0 = 0, this
³
´2
yields: 10 = 12 +1
. So, there is always a pro…table deviation. Thus, (7) is never an
+2
equilibrium.
8. 1 = (0 0) 2 = (0 1)
If the players follow these strategies, 1 = 0. If player 1 deviates to  01 = (1 0), 0 = 0, this
³
´2
yields: 10 = 12 +1
21 . So, there is always a pro…table deviation. Thus, (8) is never an
+2
equilibrium.
9. 1 = 2 = (0 0)
If the players follow these strategies, 1 = 0. If player 1 deviates to  01 = (1 0), 0 = 0, this
³
´2
yields: 10 = 12 +1
21 . So, there is always a pro…table deviation. Thus, (9) is never an
+2
equilibrium.
Case 2a: Interaction, initiated by both players (1 = 2 = 1)
Observe that there will be a pro…table deviation to 0 = 0 if   0. If  · 0, deviating
to 0 = 0 is (weakly) unpro…table. Therefore, it will be su¢cient, after imposing the condition
 · 0, to restrict attention to deviations in which 0 = 1.
There are nine types of combinations to check.
1. 1 = 2 = (1 0)
³
´2
+1
2
If the players follow these strategies, they receive: 1 = 2 +1
21 ¡ 4 (+2)
2 2 ¡ 
+2
³
´2
+1
2
2 = 2 +1
22 ¡ 4 (+2)
2 1 ¡  In order for this to be an equilibrium, there cannot be
+2
any pro…table deviations. There are four deviations we need to check.
(i) Player 1 deviates to  01 = (0 1), 01 = 1: yields 10 =

1
2
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 =
(iii) Player 2 deviates to  02 = (0 1), 02 = 1: yields 20 =

1
2
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 =
³
1
2
´2
³
+1
+2
³
+1
+2
³
1
2
+1
+2
+1
+2
+1
2
max(21  1)¡2 (+2)
2 2 ¡
´2
+1
2
21 ¡ 2 (+2)
2 2 ¡ 
´2
+1
2
22 ¡ 2 (+2)
2 1 ¡ 
´2
+1
2
max(22  1)¡2 (+2)
2 1 ¡
Observe that (1) will be an equilibrium if and only if (i) through (iv) are not pro…table
deviations and  · 0. Only conditions (i) and (ii) are binding:
74
Condition (i): 21 ¸
1
4
1
2
+1 2  Observe that this condition fails to hold when  = 0.
4
2
3(+1) 2 (this inequality is strict because players prefer not to value
+
Condition (ii): 21 
activities, all else equal).
Condition (k):  · 0
2. 1 = (1 0) 2 = (0 1)
There will be a pro…table deviation for player 1 to  01 = (0 0), 01 = 1 if 21 · 1. There will
be a pro…table deviation for player 2 to  02 = (0 0), 02 = 1 if 22 ¸ 1. So, we must have:
21  1 and 22  1. But, this is not possible given that 22 ¸ 21 . Thus, this combination
will never be an equilibrium.
3. 1 = (1 0) 2 = (0 0)
If the players follow these strategies, 2 =
1
2
to 02 = (1 0), 02 = 1, this yields: 20 = 2
whenever 32 22 
1
2
+1 1
³
³
+1
+2
+1
+2
´2
´2
+1
2
22 ¡ (+2)
2 1 ¡ . If player 2 deviates
+1
2
22 ¡ 2 (+2)
2 1 ¡ , which is pro…table
(which always holds). Thus, (3) is never an equilibrium.
4. 1 = 2 = (0 1)
³
´2
+1
If the players follow these strategies, they receive: 1 = 2 = 2 +1
¡ 4 (+2)
2 ¡  In
+2
order for this to be an equilibrium, there cannot be any pro…table deviations. There are
four deviations we need to check.
³
´2
0
1 +1
+1
0
0
(i) Player 1 deviates to  1 = (1 0), 1 = 1: yields 1 = 2 +2 max(21  1) ¡2 (+2)
2 ¡
³
´2
+1
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 = 12 +1
¡ 2 (+2)
2 ¡ 
+2
³
´2
+1
(iii) Player 2 deviates to  02 = (1 0), 02 = 1: yields 20 = 12 +1
max(22  1) ¡2 (+2)
2 ¡
+2
³
´2
+1
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = 12 +1
¡ 2 (+2)
2 ¡ 
+2
Observe that (4) will be an equilibrium if and only if (i) through (iv) are not pro…table
deviations and  · 0. Only condition (iii) is binding:
Condition (iii):
4
+1
¸ 22 . Observe that this condition does not hold when  = 0.
Condition (k):  · 0
5. 1 = (0 1) 2 = (1 0)
There will be a pro…table deviation for player 1 to  01 = (0 0), 01 = 1 if 21 ¸ 1. There will
be a pro…table deviation for player 2 to  02 = (0 0), 02 = 1 if 22 · 1. So, we will assume for
what follows that 21  1 and 22  1. If the players follow these strategies, they receive:
³
´2
³
´2
+1
+1
2 ¡   = 1 +1
1 = 12 +1
¡

22 ¡ (+2)
2
2 ¡  In order for this to be an
+2
2 +2
(+2)2 2
equilibrium, there cannot be any pro…table deviations. There are four deviations we need
to check.
³
´2
+1
2
(i) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 ¡ 
+2
75
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 =
1
2
(iii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 2
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 =
1
2
³
³
³
+1
+2
+1
+2
+1
+2
´2
´2
´2
21 ¡
+1
2
(+2)2 2
¡ 
+1
¡ 2 (+2)
2 ¡ 
¡
+1
(+2)2
¡ 
Observe that (5) will be an equilibrium if and only if (i) through (iv) are not pro…table
deviations and  · 0. Only conditions (i), (ii), and (iii) are binding:
¡
¢
Condition (i): 22 ¸ ( + 1) 221 ¡ 12
Condition (ii): 1  21
Condition (iii): 22 ¸
4+2
+1 .
Condition (k):  · 0
Observe that this condition is violated when  ¸ 2.
6. 1 = (0 1) 2 = (0 0)
³
´2
+1
If the players follow these strategies, 2 = 12 +1
¡ (+2)
2 ¡  If player 2 deviates to
+2
³
´2
+1
02 = (0 1), 02 = 1, this yields: 20 = 2 +1
¡ 2 (+2)
2 ¡ , which is always pro…table.
+2
Thus, (6) is never an equilibrium.
7. 1 = (0 0) 2 = (1 0)
If the players follow these strategies: 1 =
1
2
+1
2 ¡
(+2)2 1
³
+1
+2
´2
+1
2
21 ¡ (+2)
2 2 ¡ , 2 =
1
2
³
+1
+2
´2
22 ¡
In order for this to be an equilibrium, there cannot be any pro…table deviations.
There are four deviations we need to check.
³
´2
+1
2
(i) Player 1 deviates to 01 = (1 0), 01 = 1: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 ¡ 
+2
³
´2
+1
2
(ii) Player 1 deviates to 01 = (0 1), 01 = 1: yields 10 = 12 +1
max(21  1)¡ (+2)
2 2 ¡
+2
³
´2
(iii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 12 +1
¡ 
+2
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = ¡
Observe that (7) will be an equilibrium if and only if (i) through (iv) are not pro…table
deviations and  · 0. This gives us the following conditions:
Condition (i): 22 ¸ 32 ( + 1)21
Condition (ii): 21  1.
violated for  ¸ 2
Condition (iii): 22 ¸ 1 +
Observe that the combination of conditions (i) and (ii) are
2
2
+1 1
Condition (k):  · 0
8. 1 = (0 0) 2 = (0 1)
³
´2
+1
If the players follow these strategies, 1 = 12 +1
¡ (+2)
2 ¡  If player 1 deviates to
+2
³
´2
+1
01 = (0 1), 01 = 1, this yields: 10 = 2 +1
¡ 2 (+2)
2 ¡ , which is always pro…table.
+2
Thus, (8) is never an equilibrium.
76
9. 1 = 2 = (0 0)
If the players follow these strategies, 1 = ¡ If player 1 deviates to  01 = (0 1), 01 = 1,
³
´2
this yields: 10 = 12 +1
¡, which is always pro…table. Thus, (9) is never an equilibrium.
+2
Case 2b: Interaction, initiated by player 1 only (1 = 1 2 = 0)
Observe that there will be a pro…table deviation to 02 = 1 if   0. There will not be
a pro…table deviation to 02 = 1 if  ¸ 0. Therefore, it will be su¢cient, after imposing the
condition  ¸ 0, to restrict attention to deviations in which 02 = 0.
It is useful to observe that, if a particular ( 1   2 ) combination was never an equilibrium in
Case 2a, it will never be an equilibrium in Case 2b. The reason is that there will be an equivalent
pro…table deviation in Case 2b to the one that existed in Case 2a. For this reason, there are only
four types of ( 1   2 ) combinations that need to be checked:
1. 1 = 2 = (1 0)
³
´2
+1
2
If the players follow these strategies, they receive: 1 = 2 +1
21 ¡ 4 (+2)
2 2 ¡ 
+2
³
´2
+1
2
2 = 2 +1
22 ¡ 4 (+2)
2 1  In order for this to be an equilibrium, there cannot be any
+2
pro…table deviations. There are seven types of deviations we need to check.
(i) Player 1 deviates to  01 = (0 1), 01 = 1: yields 10 =

1
2
³
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 =
1
2
(iii) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 =
1
2
(iv) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 =
1
2
+1
+2
´2
³
+1
+2
³
+1
+2
³
+1
+2
+1
2
max(21  1)¡2 (+2)
2 2 ¡
´2
+1
2
21 ¡ 2 (+2)
2 2 ¡ 
´2

´2
+1
2
21 ¡ 2 (+2)
2 2 
(v) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
2
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1
max(22  1) ¡ 2 (+2)
2 1 
+2
³
´2
+1
2
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 12 +1
22 ¡ 2 (+2)
2 1 
+2
Observe that (1) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations and  ¸ 0. Only conditions (i), (ii), (iii), and (iv) are binding:
Condition (i): 21 ¸
1
4
1
2
+1 2
4
2
3(+1) 2 .
+
Condition (ii): 21 
Observe that this condition fails when  = 0
³
´2 ³
´
3 2
2
2 ¸
Condition (iii): +1

¡

+2
2 1
+1 2
³
´2 ³
´
2 ¡ 4 2 ¡ 1 ¸ 
Condition (iv): +1
2
1
+2
+1 2
2
Condition (k):  ¸ 0
77
2. 1 = 2 = (0 1)
³
´2
+1
If the players follow these strategies, they receive: 1 = 2 +1
¡ 4 (+2)
2 ¡  2 =
+2
³
´2
+1
2 +1
¡ 4 (+2)
2  In order for this to be an equilibrium, there cannot be any pro…table
+2
deviations. There are seven types of deviations we need to check.
(i) Player 1 deviates to  01 = (1 0), 01 = 1: yields 10 =
1
2
³
(ii) Player 1 deviates to 01 = (0 0), 01 = 1: yields 10 =
1
2
(iii) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 =
1
2
(iv) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 =
1
2
+1
+2
´2
³
+1
+2
³
+1
+2
³
+1
+2
+1
max(21  1) ¡2 (+2)
2 ¡
´2
´2
´2
+1
¡ 2 (+2)
2 ¡ 
21 
+1
¡ 2 (+2)
2
(v) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 0
³
´2
+1
(vi) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 = 12 +1
max(22  1) ¡ 2 (+2)
2
+2
³
´2
+1
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 12 +1
¡ 2 (+2)
2
+2
Observe that (2) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations and  ¸ 0. Only conditions (iii), (iv), and (vi) are binding:
³
´2 ³ ³
´
´
¡1
1 2
Condition (iii): +1
2
¡

+2
+1
2 1 ¸
³
´2
2
Condition (iv): +1
( 32 ¡ +1
)¸
+2
Condition (vi):
4
+1
Condition (k):  ¸ 0
¸ 22 . Observe that this condition is violated when  = 0.
3. 1 = (0 1) 2 = (1 0)
There will be a pro…table deviation for player 1 to 01 = (0 0), 01 = 1 if 21 ¸ 1. There will
be a pro…table deviation for player 2 to  02 = (0 0), 02 = 2 if 22 · 1. So, we will assume
for what follows that 21  1 and 22  1. If the players follow these strategies, player 1
³
´2
+1
0
2
0
receives: 1 = 12 +1
¡ (+2)
2 2 ¡  Suppose player 1 deviates to  1 = (1 0), 1 = 0.
+2
³
´2
+1
2
This deviation yields 10 = 12 +1
, which is pro…table unless ¡ (+2)
But, it is
2 2 ¸ .
+2
also required that  ¸ 0 for there to be no pro…table deviations. Since these inequalities
cannot simultaneously hold, a pro…table deviations always exists. Thus, (3) is never an
equilibrium.
4. 1 = (0 0) 2 = (1 0)
If the players follow these strategies: 1 =
+1
2
(+2)2 1 
+1
2
2 (+2)
2 2
1
2
³
+1
+2
´2
+1
2
21 ¡ (+2)
2 2 ¡ , 2 =
1
2
Suppose player 1 deviates to 01 = (1 0), 01 = 1, which yields 10 = 2
³
³
+1
+2
+1
+2
´2
´2
22 ¡
21 ¡
¡  This will be a pro…table deviation unless 22 ¸ 32 ( + 1)21 . Now, suppose
³
´2
+1
2
player 1 deviates to 01 = (0 1), 01 = 1, which yields 10 = 12 +1
max(21  1) ¡ (+2)
2 2 ¡
+2
78
. This deviation will be pro…table unless 21  1. The conditions (i) 22 ¸ 32 ( + 1)21 and
(ii) 21  1 imply that 22  32 ( + 1)21 , which is violated for all  ¸ 2. So, no equilibria of
type (4) exist when  ¸ 2.
We will now show that no equilibria of type (4) exist when   2. Suppose player 1
³
´2
+1
2
deviates to  01 = (1 0), 01 = 0. This yields: 10 = 12 +1
21 ¡ (+2)
2 2  This deviation
+2
is pro…table unless  · 0. But, we also need  ¸ 0 (otherwise, there will be a pro…table
deviation). So, the only possible value of  is  = 0. Finally, suppose player 1 deviates
³
´2
to  01 = (0 1), 01 = 0, which yields 10 = 12 +1
 This deviation is pro…table unless
+2
³
´2
³
´2
1 +1
+1
1 +1
2
21 ¡ (+2)
. Combining this with the assumption that  = 0,
2 2 ¡  ¸ 2
2 +2
+2
2
we …nd that it is necessary to have: 21 ¸ +1
22 + 1. But, when   2, this condition is
always violated. Thus, no equilibrium of type (4) exists.
Case 2c: Interaction, initiated by player 2 only (1 = 0 2 = 1)
Observe that there will be a pro…table deviation to 01 = 1 if   0. There will not be
a pro…table deviation to 01 = 1 if  ¸ 0. Therefore, it will be su¢cient, after imposing the
condition  ¸ 0, to restrict attention to deviations in which 01 = 0.
It is useful to observe that, if a particular ( 1   2 ) combination was never an equilibrium in
Case 2a, it will never be an equilibrium in Case 2c. The reason is that there will be an equivalent
pro…table deviation in Case 2c to the one that existed in Case 2a. For this reason, there are only
four types of ( 1   2 ) combinations that need to be checked.
1. 1 = 2 = (1 0)
³
´2
+1
2
If the players follow these strategies, they receive: 1 = 2
21 ¡ 4 (+2)
2 2  2 =
³
´2
+1
2
2 +1
22 ¡ 4 (+2)
2 1 ¡  In order for this to be an equilibrium, there cannot be any
+2
pro…table deviations. There are seven types of deviations we need to check.
(i) Player 2 deviates to  02 = (0 1), 02 = 1: yields 20 =

1
2
³
(ii) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 =
1
2
(iii) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 =
1
2
(iv) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 =
1
2
+1
+2
+1
+2
´2
³
+1
+2
³
+1
+2
³
+1
+2
+1
2
max(22  1)¡2 (+2)
2 1 ¡
´2
+1
2
22 ¡ 2 (+2)
2 1 ¡ 
´2

´2
+1
2
22 ¡ 2 (+2)
2 1 
(v) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
³
´2
+1
2
(vi) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 = 12 +1
max(21  1) ¡ 2 (+2)
2 2 
+2
³
´2
+1
2
(vii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 12 +1
21 ¡ 2 (+2)
2 2 
+2
Observe that (1) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations and  ¸ 0. Only conditions (iii), (iv), (vi), and (vii) are binding:
³
´2 ³
´
3 2
2
2 ¸
Condition (iii): +1

¡

+2
2 2
+1 1
79
³
Condition (iv):
+1
+2
Condition (vi): 21 ¸
Condition (vii):
3 2
4 1
´2 ³
222 ¡
1
4

4
2
+1 1
¡
1
2
+1 2
1
2
+1 2 . Observe
1
2
+
´
¸
that this condition fails to hold when  = 0.
Condition (k):  ¸ 0
2. 1 = 2 = (0 1)
If the players follow these strategies, they receive: 1 = 2
³
+1
+2
1
2
³
´2
+1
¡4 (+2)
2  2 = 2
³
+1
+2
´2
¡
+1
4 (+2)
2 ¡ In order for this to be an equilibrium, there cannot be any pro…table deviations.
There are seven types of deviations we need to check.
(i) Player 2 deviates to  02 = (1 0), 02 = 1: yields 20 =
(ii) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 =
1
2
(iii) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 =
1
2
(iv) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 =
1
2
+1
+2
´2
³
+1
+2
³
+1
+2
³
+1
+2
+1
max(22  1) ¡2 (+2)
2 ¡
´2
´2
´2
+1
¡ 2 (+2)
2 ¡ 
22 
+1
¡ 2 (+2)
2
(v) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
³
´2
+1
(vi) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 12 +1
max(21  1) ¡ 2 (+2)
2
+2
³
´2
+1
(vii) Player 1 deviates to 01 = (0 0), 01 = 0: yields 10 = 12 +1
¡ 2 (+2)
2
+2
Observe that (2) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations and  ¸ 0. Only conditions (i), (iii), and (iv) are binding:
Condition (i):
4
+1
¸ 22 . Observe that this condition is violated when  = 0.
³
´2 ³ ³
´
´
¡1
1 2
Condition (iii): +1
2
¡

+2
+1
2 2 ¸
³
´2
2
Condition (iv): +1
( 32 ¡ +1
)¸
+2
Condition (k):  ¸ 0
3. 1 = (0 1) 2 = (1 0)
There will be a pro…table deviation for player 1 to  01 = (0 0), 01 = 1 if 21 ¸ 1. There
will be a pro…table deviation for player 2 to  02 = (0 0), 02 = 2 if 22 · 1. So, we will
assume for what follows that 21  1 and 22  1. If the players follow these strategies,
³
´2
+1
0
0
player 2 receives: 12 +1
22 ¡ (+2)
2 ¡ Suppose player 2 deviates to 2 = (1 0), 2 = 0.
+2
³
´2
+1
This deviation yields 20 = 12 +1
22 , which is pro…table unless ¡ (+2)
2 ¸ . But, it is
+2
also required that  ¸ 0 for there to be no pro…table deviations. Since these inequalities
cannot simultaneously hold, a pro…table deviations always exists. Thus, (3) is never an
equilibrium.
80
4. 1 = (0 0) 2 = (1 0)
If the players follow these strategies: 1 =
+1
2 ¡
(+2)2 1
1
2
³
+1
+2
´2
21 ¡
+1
2 ,
(+2)2 2
2 =
1
2
³
+1
+2
´2
22 ¡
In order for this to be an equilibrium, there cannot be any pro…table deviations.
There are seven deviations we need to check.
³
´2
+1
2
(i) Player 1 deviates to 01 = (1 0), 01 = 0: yields 10 = 2 +1
21 ¡ 2 (+2)
2 2 
+2
³
´2
+1
2
(ii) Player 1 deviates to 01 = (0 1), 01 = 0: yields 10 = 12 +1
max(21  1) ¡ (+2)
2 2 
+2
³
´2
(iii) Player 2 deviates to 02 = (0 1), 02 = 1: yields 20 = 12 +1
¡ 
+2
(iv) Player 2 deviates to 02 = (0 0), 02 = 1: yields 20 = ¡
³
´2
(v) Player 2 deviates to 02 = (1 0), 02 = 0: yields 20 = 12 +1
22 ¡
+2
³
´2
(vi) Player 2 deviates to 02 = (0 1), 02 = 0: yields 20 = 12 +1

+2
+1
2 
(+2)2 1
(vii) Player 2 deviates to 02 = (0 0), 02 = 0: yields 20 = 0
Observe that (4) will be an equilibrium if and only if (i) through (vii) are not pro…table
deviations and  · 0. Only conditions (i), (ii), (iii), (v), and (vi) will be binding:
Condition (i): 22 ¸ 32 ( + 1)21
Condition (ii): 21  1.
violated for  ¸ 2
Condition (iii): 22 ¸ 1 +
Observe that the combination of conditions (i) and (ii) are
2
2
+1 1
Condition (v):  · 0. Observe that, combined with condition (k), this implies  = 0.
³
´2
³
´2
2 ¡ +1 2 ¡ 1 +1
Condition (vi): 12 +1

¸ 
2
+2
2 +2
(+2)2 1
Given that  = 0, this condition can be rewritten as: 22 ¸ 1 +
2
2
+1 1
Condition (k):  ¸ 0
Case 2: Combined
If we combine Cases 2a, 2b, and 2c, we …nd the following.
1. An equilibrium with 1 = 2 = (1 0) and interaction between the players exists if:
1
4
+ +1
22  21  3(+1)
22
µ ³
´2
³
3 2
(ii)  · max 0 +1
min
+2
2 2 ¡
(i) 21 ¸
1
4
2
2
2
+1 1  22
¡
4
2
+1 1
¡
(iii)  ¸ 1
1
2
´¶
= ¹6
2. An equilibrium with 1 = 2 = (0 1) and interaction between the players exists if:
(i) 22 ·
4
+1
(ii)  · max(0
³
+1
+2
´2
³ ³
´
1 2 3
min 2 ¡1
+1 ¡ 2 1  2 ¡
81
2
+1
´
= ¹7
(iii)  ¸ 1
3. An equilibrium with 1 = (0 1) 2 = (1 0) and interaction between the players exists if:
³
¡ 2 1 ¢´
(i) 22 ¸ max 4+2

(
+
1)
21 ¡ 2 .
+1
(ii) 21  1
(iii)  · 0
Observe that these conditions will be violated when  ¸ 2.
4. An equilibrium with 1 = (0 0) 2 = (1 0) and interaction between the players exists if:
(i) 22 ¸ max( 32 ( + 1)21  1 +
2
2
+1 1 )
(ii) 21  1
(iii)  · 0
Observe that these conditions will be violated when  ¸ 2.
This completes the proof.
82