2 Chapter 2. Nash Equilibrium
while by bidding x
i
either he wins and pays the same price or loses.
An equilibrium in which player j obtains the good is that in which b
1
< v
j
, b
j
> v
1
, and
b
i
= 0 for all players i /∈ {1, j}.
18.5 (War of attrition) The set of actions of each player i is A
i
= [0, ∞) and his payoff function
is
u
i
(t
1
, t
2
) =
−t
i
if t
i
< t
j
v
i
/2 − t
i
if t
i
= t
j
v
i
− t
j
if t
i
> t
j
where j ∈ {1, 2} \ {i}. Let (t
1
, t
2
) be a pair of actions. If t
1
= t
2
then by conceding slightly
later than t
1
player 1 can obtain the object in its entirety instead of getting just half of it,
so this is not an equilibrium. If 0 < t
1
< t
2
then player 1 can increase her payoff to zero by
deviating to t
1
= 0. Finally, if 0 = t
1
< t
2
then player 1 can increase her payoff by deviating
to a time slightly after t
2
unless v
1
− t
2
≤ 0. Similarly for 0 = t
2
< t
1
to constitute an
equilibrium we need v
2
− t
1
≤ 0. Hence (t
1
, t
2
) is a Nash equilibrium if and only if either
0 = t
1
< t
2
and t
2
≥ v
1
or 0 = t
2
< t
1
and t
1
≥ v
2
.
Comment An interesting feature of this result is that the equilibrium outcome is inde-
pendent of the players’ valuations of the object.
19.1 (Location game)
1
There are n players, each of whose set of actions is {Out } ∪ [0, 1]. (Note
that the model differs from Hotelling’s in that players choose whether or not to become
candidates.) Each player prefers an action profile in which he obtains more votes than any
other player to one in which he ties for the largest number of votes; he prefers an outcome
in which he ties for first place (regardless of the number of candidates with whom he ties)
to one in which he stays out of the competition; and he prefers to stay out than to enter
and lose.
Let F be the distribution function of the citizens’ favorite positions and let m = F
−1
(
1
2
)
be its median (which is unique, since the density f is everywhere positive).
It is easy to check that for n = 2 the game has a unique Nash equilibrium, in which both
players choose m.
The argument that for n = 3 the game has no Nash equilibrium is as follows.
• There is no equilibrium in which some player becomes a candidate and loses, since
that player could instead stay out of the competition. Thus in any equilibrium all
candidates must tie for first place.
• There is no equilibrium in which a single player becomes a candidate, since by choosing
the same position any of the remaining players ties for first place.
• There is no equilibrium in which two players become candidates, since by the argument
for n = 2 in any such equilibrium they must both choose the median position m, in
which case the third player can enter close to that position and win outright.
• There is no equilibrium in which all three players become candidates:
1
Correction to first printing of book : The first sentence on page 19 of the book should be
amended to read “There is a continuum of citizens, each of whom has a favorite position;
the distribution of favorite positions is given by a density function f on [0, 1] with f (x) > 0
for all x ∈ [0, 1].”
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