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Game Theory
Solutions to Problem Set 4
1 Hotellingís model
1.1 Two vendors
Consider a strategy proÖle (s 1 ; s 2 ) with s 1 6 = s 2 : Suppose s 1 < s 2 : In this case, it is proÖtable to for player 1 to deviate and choose a location s^01 2 (s 1 ; s 2 ). To see this, note that
u 1 (s^01 ; s 2 ) =
s^01 + s 2 2
s 1 + s 2 2
= u 1 (s 1 ; s 2 ) :
Thus, in a pure-strategy Nash equilibrium both players choose the same location. Consider now the proÖle (s 1 = s; s 2 = s) where s 6 = 1= 2 : In this case, both players get 1 = 2 : However, if a player deviates and chooses 1 = 2 ; her payo§ is strictly greater than 1 = 2 : So, we are left with (s 1 = 1= 2 ; s 2 = 1=2) : In this case, no player has an incentive to deviate:
ui
si =
; sj =
s^0 i + (^12) 2
= ui
s^0 i; sj =
; where s^0 i <
ui
si =
; sj =
s^0 i + (^12) 2
= ui
s^0 i; sj =
; where s^0 i >
Thus, we conclude that the unique pure-strategy NE is (s 1 = 1= 2 ; s 2 = 1=2) :
1.2 Three vendors
We consider all cases of pure strategy proÖles and show that in each case at least one player has an incentive to deviate. First, suppose the players choose three di§erent locations, say s 1 < s 2 < s 3 : It is easy to check that each player has a proÖtable deviation. For example, player 1 has an incentive to choose s^01 2 (s 1 ; s 2 ). Now, suppose that two players, say 1 and 2 choose the same location s and player 3 chooses s 3 6 = s: If s 3 > s player 3 prefers to choose any s^03 2 (s; s 3 ), while if s 3 < s player 3 prefers to choose any s^03 2 (s 3 ; s). Finally, suppose all players choose the same location s: The payo§ of every player is 1 = 3 : Suppose that s 6 = 1= 2 : Then, player 1 can choose s 1 = 1= 2 and assure herself of a payo§ greater than 1 = 2 , which is greater than 1 = 3. Hence,
player 1 has an incentive to deviate. Suppose instead that s = 1= 2. Then, there exists an " > 0 su¢ ciently small such that
u 1 (s 1 = 1= 2 "; s 2 = 1= 2 ; s 3 = 1=2) >
2 Air strike
We have the following normal-form game. The set of players is fA; Bg : The sets of actions (pure strategies) are SA = SB = f 1 ; 2 ; 3 g : The playersí payo§s are described in the following matrix:
1 2 3 1 0 ; 0 v 1 ; v 1 v 1 ; v 1 2 v 2 ; v 2 0 ; 0 v 2 ; v 2 3 v 3 ; v 3 v 3 ; v 3 0 ; 0 where the total (non-destroyed) value of the three targets for player B is normalized to zero. Clearly, this game does not admit any pure-strategy NE. Player B would like to choose the same target as player A; while player A is better o§ when they choose di§erent targets. Hence, we have to look for mixed-strategy equilibria. There are four possible cases to consider: (i) A randomizes between target 1 and target 3 , but does not assign positive probability to target 2; (ii) A randomizes between target 2 and target 3; (iii) A randomizes between targets target 1 and target 2; (iv) A randomizes between all three targets. Some of these cases can be eliminated without any computational work. First note that in an NE, if player A assigns zero probability to one of the targets, then player B has to assign zero probability to the same target. (Think about why.) This, in turn, implies that there can be no NE in which player A assigns positive probability to target 3 and zero probability to one of the other targets. To see this, suppose that there is a NE in which player A randomizes between target 2 and target 3 with probabilities p 2 and p 3 = (1 p 2 ). Then, it must be that player B assigns zero probability to defending target 1. However, in that case, assigning probability p 3 to target 1 is a strictly proÖtable deviation for player A, which is a contradiction. Hence, there cannot be a NE in the form of case (ii). In a similar way, we can show that there cannot be a NE in the form of case (i). Therefore, we are left with cases (iii) and (iv). (Case iii) Let the strategy of player A be A = ( ; 1 ; 0) where 2 (0; 1). As argued above, player B will not defend target 3. That is, player B will use a strategy on the form B = ( ; 1 ; 0) where 2 [0; 1] : If ( ; 1 ; 0) is an equilibrium strategy for player A; it must be the case that:
uB (( ; ; 1 ) ; 1) = v 2 (1 ) v 3 = v 1 (1 ) v 3 = uB (( ; ; 1 ) ; 2) ; uB (( ; ; 1 ) ; 1) = v 2 (1 ) v 3 = v 1 v 2 = uB (( ; ; 1 ) ; 3) ; which implies (^) = v 2 v 3 v^1 v^2 +v^1 v^3 +v^2 v^3 = (^) v 1 v 2 +vv^11 vv^33 +v 2 v 3 : We conclude that if v 3 > (^) vv 11 +vv^22 ; then the Nash equilibrium of the game is A = ( ; ; 1 ^ ) ; B = ( ; ; 1 ^ ) : If v 3 < (^) vv 11 +vv^22 ; then player B assigns positive probability only to target 1 and target 2 : Thus, we have:
uB (( ; ; 1 ) ; 1) = v 2 (1 ) v 3 = v 1 (1 ) v 3 = uB (( ; ; 1 ) ; 2) ; uB (( ; ; 1 ) ; 1) = v 2 (1 ) v 3 > v 1 v 2 = uB (( ; ; 1 ) ; 3) ; which, in turn, implies:
(^) v 1 v 2 +vv^21 vv^33 +v 2 v 3 = (^) 2(v 1 v+^2 v 2 ) ; = v v^12 : To sum up, we have the following cases.
If v 3 < (^) vv 11 +vv^22 ; there exists a unique Nash equilibrium
A =
v 2 v 1 +v 2 ;^
v 1 v 1 +v 2 ;^0
B =
v 1 v 1 +v 2 ;^
v 2 v 1 +v 2 ;^0
If v 3 > (^) vv 11 +vv^22 ; there exists a unique Nash equilibrium
A =
v 2 v 3 v 1 v 2 +v 1 v 3 +v 2 v 3 ;^
v 1 v 3 v 1 v 2 +v 1 v 3 +v 2 v 3 ;^
v 1 v 2 v 1 v 2 +v 1 v 3 +v 2 v 3
B =
v 1 v 2 +v 1 v 3 v 2 v 3 v 1 v 2 +v 1 v 3 +v 2 v 3 ;^
v 1 v 2 v 1 v 3 +v 2 v 3 v 1 v 2 +v 1 v 3 +v 2 v 3 ;^
v 1 v 2 +v 1 v 3 +v 2 v 3 v 1 v 2 +v 1 v 3 +v 2 v 3
If v 3 = (^) vv 11 +vv^22 : there exist a continuum of Nash equilibria:
A =
; v v^12 ; 1
v 1 +v 2 v 2
B =
v 1 v 1 +v 2 ;^
v 2 v 1 +v 2 ;^0
where 2
h v 2 2(v 1 +v 2 ) ;^
v 2 (v 1 +v 2 )
i :
Clearly, as v 3 goes to zero we have only the Nash equilibrium in which both player A and player B randomize between target 1 and target 2. This is very intuitive. When the value of target 3 is negligible, player A does not have an incentive to attack it. Since player A does not attack target 3 ; player B does not defend it.
3 First-Price auction with di§erent valuations
First note that, in equilibrium, each player i = 1; : : : ; n can not get the object by bidding bi > vi: In fact, when bi > vi and player i wins the auction, her payo§ is negative (vi bi). But then, player i would have an incentive to deviate and bid, for example, zero. Now, suppose that player 1 does not get the object. Player 1 ís payo§ is zero. Moreover, we just argued that the highest bid then has to be smaller than or equal to v 2 : But since v 1 > v 2 ; a proÖtable deviation for player 1 is to choose a bid in the interval (v 2 ; v 1 ) and get a positive payo§. Notice that this game admits many pure-strategy Nash equilibria. Any combination of bids on the following form
b 1 = b; b 2 6 b; : : : ; bn 6 b with b 2 [v 2 ; v 1 ] and bj = b for at least one player j 2 f 2 ; : : : ; ng ; is an equilibrium:
4 A simple Bayesian game
t 1 = a t 1 = b L R U 2 ; 2 2 ; 0 D 0 ; 2 0 ; 0
L R
U 0 ; 2 1 ; 0
D 1 ; 2 2 ; 0
First of all notice that in any BNE, player 1 chooses D when her type is b. In fact, for any action of player 2 ; the payo§ of type b (of player 1 ) from action D is strictly greater than the payo§ from action U: Suppose that player 2 chooses L: Then type a (of player 1 ) chooses U and type b chooses D: The expected payo§ of player 2 is :9 (2) + :1 ( 2) = 1: 6 : Notice that if player 2 plays R her (expected) payo§ is 0. So the strategy proÖle ((U; D) ; L) constitutes a BNE. Suppose that player 2 chooses R: Then both type a and type b choose D: The (expected) payo§ of player 2 is 0 : If player 2 plays L; her (expected) payo§ is 2 : Thus we have another BNE: ((D; D) ; R) : Finally, suppose that player 2 randomizes between L and R. Let denote the probability that player 2 chooses L: Let denote the probability that type
