The paper proves that in two-player logit form symmetric contests with concave success function, commitment to a particular strategy does not increase a player's payoff, while in contests with more than two players it does. The paper also provides a contest-like game in which commitment does not increase a player's payoff for any number of players.contests, commitment

Alex Possajennikov

English

Research Papers in Economics

      
UNIVERSITY OF NOTTINGHAM 
 
 
 
Discussion Papers in Economics 
 
___________________________________________________ 
Discussion Paper 
No. 08/13 
 
 
Commitment in Symmetric Contests 
 
By Alex Possajennikov 
 
 
December 2008 
 
 
__________________________________________________________________ 
 2008               DP 08/13 Commitment in Symmetric Contests
Alex Possajennikov,y
University of Nottingham
December 2008
Abstract
The paper proves that in two-player logit form symmetric contests
with concave success function, commitment to a particular strategy
does not increase a player's payo, while in contests with more than
two players it does. The paper also provides a contest-like game in
which commitment does not increase a player's payo for any number
of players.
Keywords: contests, commitment
JEL Codes: C72, D72
1 Introduction
It has been noted that in two-player symmetric contest games, locally com-
mitment to an action does not increase a player's payo (Dixit, 1987), while
in contests with more than two players it does. However, a proof of the global
result for general contests has not been provided (Kr akel, 2002, has a proof
of a related result for contests with linear success function).
The local result is based on the fact that the slope of the opponent's
reaction function is zero at the symmetric equilibrium, and thus coincides
with the slope of the level curve of a player's payo function. This is a
necessary condition for the result but is not sucient. A comment on Dixit's
paper by Baye and Shin (1999) discusses an example of a contest game where
the result does not hold. They also provide a sucient condition for there not
to be an increase in a player's payo for local deviations from equilibrium.
School of Economics, University of Nottingham, University Park, Nottingham
NG7 2RD, United Kingdom. Email: Alex.Possajennikov@nottingham.ac.uk; URL:
http://www.nottingham.ac.uk/~lezap/.
yI thank Maria Montero for useful comments.
1This paper proves that when the contest probability of winning has the
logit form with concave success function, then the result holds globally, i.e.
a player does not increase payo from commitment to any strategy, not
necessarily close to equilibrium.
Two-player contests may appear special as the result does not extend
to contests with more than two players. I present a game in which the
result holds for any number of players. The game is a modication of the
usual contest game in which each player participates in a contest against
the average eort of other players, making the game similar to a two-player
setup.
2 Two-player logit contests
Consider two-player symmetric contests in which Players 1 and 2 simultane-
ously choose an eort or investment levels xi 2 [0;1), i = 1;2. The payos
are
u1(x1;x2) =
f(x1)
f(x1) + f(x2)
V   x1; (1)
where V > 0 is the value of the prize, f(x)  0 when x = 0, f0(x) > 0 for all
x, f00(x)  0 for all x, and u2(x1;x2) = u1(x2;x1).
The rst order conditions for Nash equilibrium are
@u1
@x1
=
f0(x1)f(x2)
(f(x1) + f(x2))2V   1 = 0; (2)
@u2
@x2
=
f0(x2)f(x1)
(f(x1) + f(x2))2V   1 = 0: (3)
The second order conditions @2ui=@x2
i = f(xj)(f00(xi)(f(xi) + f(xj))  
2(f0(xi))2)V=(f(xi)+f(xj))3 < 0 are satised for all interior x1;x2. Therefore
the rst order conditions dene the reaction functions of the players, provided
that x1;x2 satisfying them are positive.
At a symmetric equilibrium x1 = x2 = x the rst order conditions
become f0(x)V=(4f(x))   1 = 0. Since f0(x) is decreasing and f(x) is in-
creasing, if limx!0 f0(x)V=(4f(x)) > 1 and limx!1 f0(x)V=(4f(x)) < 1, then
there is unique interior symmetric equilibrium. The conditions are satised
for example by function f(x) = xr for r  1.
The second cross-derivative of the payo function is
@2ui
@xi@xj
=
f0(xi)f0(xj)(f(xi)   f(xj))
(f(xi) + f(xj))3 V: (4)
2From the rst order conditions, the slope of the reaction function ^ xi(xj) of
Player i is d^ xi=dxj =  (@2ui=@xi@xj)=(@2ui=@x2
i). Thus
d^ xi
dxj
=  
f0(^ xi)f0(xj)(f(^ xi)   f(xj))
f(xj)(f00(^ xi)(f(^ xi) + f(xj))   2(f0(^ xi))2)
: (5)
Then d^ xi=dxj > 0 when ^ xi > xj and d^ xi=dxj < 0 when ^ xi < xj. Therefore
^ xi(xj) is decreasing when xj > x. It hits zero when f0(0)V=f(xj)   1 = 0,
thus the reaction function is dened by the rst order condition when xj <
f 1(f0(0)V ).
Suppose that Player 1 can commit to an action x1, observable by Player
2 who then chooses x2. Player 1 then maximizes u1(x1; ^ x2(x1)). The rst
order condition for maximization is
du1
dx1
=
@u1
@x1
+
@u1
@x2
d^ x2
dx1
= 0:
At the symmetric equilibrium x the rst term @u1=@x1 = 0. From
equation (4), at this equilibrium d^ x2=dx1 = 0. Therefore du1=dx1 = 0 at x1 =
x. The necessary condition for maximization is satised at the simultaneous
move game equilibrium x. This is the result noted by Dixit (1987). However,
whether x1 = x is indeed a global maximum is left open, although Dixit
notes that this depends on the curvatures of the best response function and
of level contours.
The second order condition for maximization is
d2u1
dx2
1
=
@2u1
@x2
1
+
@2u1
@x2@x1
d^ x2
dx1
+

@2u1
@x1@x2
+
@2u1
@x2
2
d^ x2
dx1

d^ x2
dx1
+
@u1
@x2
d2^ x2
dx2
1
< 0:
At symmetric equilibrium d^ x2=dx1 = 0, thus @2u1=@x2
1+@u1=@x2d2^ x2=dx2
1 <
0. From (5), and because at equilibrium d(d^ x2=dx1)=dx1 = @(d^ x2=dx1)=@x1,
d2^ x2
dx2
1
(x
) =
(f0(x))3
2f(x)(f00(x)f(x)   (f0(x))2)
:
Since @u1=@x2 =  f(x1)f0(x2)=(f(x1) + f(x2))2, at symmetric equilibrium
d2u1
dx2
1
=
f00(x)f(x)   (f0(x))2
4(f(x))2 V  
f0(x)
4f(x)
V
(f0(x))3
2f(x)(f00(x)f(x)   (f0(x))2)
:
The rst term is negative while the second is positive, thus the sign
of the whole expression is unclear yet. From the rst order conditions, at
equilibrium f0(x)V = 4f(x). Then
d2u1
dx2
1
=
2(f00(x)V   16f(x))2   162f(x)2
8(f00(x)f(x)   (f0(x))2)
3Since f00  0, the minimum of the numerator is achieved when f00 = 0. Then
the expression is positive, thus the numerator is positive for all x. Since the
denominator is negative, d2u1=dx2
1 < 0 at equilibrium. Therefore locally the
second order condition for a maximum is satised. This can also be checked
by using the condition in Baye and Shin (1999) on the derivatives of the
contest winning probability function.
It was not possible to sign d2u1=dx2
1 for all x1. To prove that x1 = x is
global maximum, consider the following. The level curve of Player 1 passing
through the symmetric equilibrium x where f0(x)V = 4f(x) is
f(x1)
f(x1) + f(x2)
V   x1 =
1
2
V   x
: (6)
The reaction function of Player 2 is given by equation (3). If one can show
that the level curve and the reaction function have only x in common, then
the local second order condition proved above is sucient for a global max-
imum because the reaction function is always on the side of the level curve
that represents a lower payo for Player 1.
From the level curve, f(x1) + f(x2) = f(x1)V=(V=2   (x   x1)), or
f(x2) = f(x1)
V=2 + (x   x1)
V=2   (x   x1)
: (7)
Substituted into the reaction function, f(x1)f0(x2)V = (f(x1)V=(V=2 (x 
x1)))2, or
f
0(x2) = f(x1)
V
(V=2   (x   x1))2: (8)
A point on both the reaction function of Player 2 and the equilibrium payo
level curve of Player 1 satises the two equations (7) and (8).
The derivative of the right-hand side of (8) w.r.t. x1 is V (f0(x1)(V=2  
(x   x1))   2f(x1))=(V=2   (x   x1))3. Since V=2   (x   x1) > 0 (as
x  V=2, otherwise a player's payo would be negative in equilibrium), the
denominator is positive. Consider the numerator. Its derivative w.r.t. x1
is f00(x1)(V=2   (x   x1))   f0(x1) < 0. Since the numerator is zero when
x1 = x and is decreasing, f(x1)V (1=(V=2   (x   x1)))2 is decreasing when
x1 > x and increasing when x1 < x.
The derivative of the right-hand side of (7) w.r.t. x1 is (f0(x1)(V 2=4  
(x   x1)2)   V f(x1))=(V=2   (x   x1))2. The derivative of the numerator
is f00(x1)(V 2=4   (x   x1)2) + f0(x1)(2(x   x1)   V ). From equation (6),
x x1 = (1=2 f(x1)=(f(x1)+f(x2)))V . For positive x1;x2, jx x1j < V=2.
Therefore the derivative of the numerator is negative. Since the numerator is
zero when x1 = x, the same conclusion as in the previous paragraph follows:
4f(x1)(V=2 + (x   x1))=(V=2   (x   x1)) is decreasing when x1 > x and
increasing when x1 < x.
When x1 = x, then equation (7) becomes f(x2) = f(x). Since f is a
strictly increasing function, then x2 = x. Equation (8) becomes f0(x) =
4f(x)=V , which is satised. Thus x1 = x2 = x is one point satisfying the
two equations.
Consider x1 < x. Since the right-hand side of equation (7) is increasing,
f(x1)(V=2+(x  x1))=(V=2 (x  x1)) < f(x). Then the x2 that satises
equation (7) is less than x. Then f0(x2)  f0(x) = 4f(x)=V . Since the
right-hand side of equation (8) is also increasing, f(x1)V (1=(V=2   (x  
x1)))2 < 4f(x)=V . Thus the second equation is not satised. A similar
reasoning shows that the two equations cannot be satised for x1 > x.
Therefore x1 = x2 = x is the unique point that satises the two equations.
Since the local second order conditions are satised for x1 = x2 = x and
this is the only point that the level curve passing through it has in common
with the reaction function of Player 2, any other points on the reaction
function lie on a lower level curve of Player 1. Thus
Theorem 1 Suppose that in two-player symmetric contests with payo func-
tion (1), where the contest success function satises f(0)  0, f0(x) >
0;f00(x)  0 for all x > 0, there exists a simultaneous move interior sym-
metric equilibrium x. Then the subgame perfect equilibrium outcome when
Player 1 can commit to an observable action before Player 2 is x1 = x2 = x.
As the equality of slopes of the reaction function and the level curve at
the simultaneous move equilibrium is necessary but not sucient, Baye and
Shin (1999) discuss contest games where a player can gain by deviating from
equilibrium because the local second order condition do not hold. They show
that among such games are logit form contests with f(x) = xr for r 2 (
p
2;2].
3 Contests with more than two players
A symmetric n-player contest with the logit form probability of winning the
prize has payo functions
ui(x1;:::;xn) =
f(xi)
Pn
j=1 f(xj)
V   xi; (9)
with the same assumptions on f(x) as in the previous section, f(0)  0,
f0(x) > 0, f00(x)  0 for all x > 0.
5The derivatives of the payo function are:
@ui
@xi
=
f0(xi)
P
k6=i f(xk)
(
Pn
k=1 f(xk))
2 V   1,
@ui
@xj
=
 f(xi)f0(xj)
(
Pn
k=1 f(xk))
2V;
@2ui
@x2
i
=
(f00(xi)
Pn
k=1 f(xk)   2(f0(xi))2)
P
k6=i f(xk)
(
Pn
k=1 f(xk))
3 V;
@2ui
@xi@xj
=
f0(xi)f0(xj)(f(xi)  
P
k6=i f(xk))
(
Pn
k=1 f(xk))
3 V:
At an interior (x;:::;x) @ui=@xi = (n 1)f0(x)=n2f(x)V  1 and @2ui=@x2
i <
0. A solution to (n   1)f0(x)V=(n2f(x))   1 = 0 and thus an interior sym-
metric equilibrium x exists when limx!0(n   1)f0(x)V=(n2f(x)) > 1 and
limx!1(n   1)f0(x)V=(n2f(x)) < 1. These conditions are satised e.g. by
f(x) = xr for r  1.
Suppose that Player 1 can commit to an action x1, observable by all
other players who then choose their actions simultaneously. The rst order
condition for maximization of u1(x1; ^ x2(x1);:::; ^ xn(x1)) is
du1
dx1
=
@u1
@x1
+
@u1
@x2
d^ x2
dx1
+ ::: +
@u1
@xn
d^ xn
dx1
:
Equilibrium reaction functions ^ xi(x1) for i 6= 1 are given implicitly by the
rst order conditions @ui=@xi(x1;:::;xn) = 0, i = 2;:::;n. Dierentiating
each of the equations w.r.t. x1 gives
@2ui
@x1@xi
+
@2ui
@x2@xi
d^ x2
dx1
+ ::: +
@2ui
@xn@xi
d^ xn
dx1
= 0, i = 2;:::;n:
At symmetric equilibrium xi = x for all i = 1;:::;n. Then @2ui=@x2
i =
(n   1)(nf00(x)f(x)   2(f0(x))2)V=(n3f(x)2) for all i = 2;:::;n and
@2ui=@xj@xi = (2   n)(f0(x))2V=(n3f(x)2) for all i;j 6= i. Summing up
the n   1 equations in the above display gives
(n   1)
@2ui
@xj@xi
+

@2ui
@x2
i
+ (n   2)
@2ui
@xj@xi

d^ x2
dx1
+ ::: +
d^ xn
dx1

= 0: (10)
Since at equilibrium @u1=@xi =  f0(x)V=(n2f(x)) 6= 0 for all i 6= 1, and
@u1=@x1 = 0, the rst order condition for maximization becomes du1=dx1 =
@u1=@xi  (d^ x2=dx1 + ::: + d^ xn=dx1) = 0. However, from equation (10)
d^ x2=dx1+:::+d^ xn=dx1 6= 0 for n > 2, since @2ui=@xj@xi 6= 0 then. Therefore
the symmetric equilibrium x cannot be a part of subgame perfect equilib-
rium when Player 1 can commit to an action.
6Proposition 1 In symmetric n-player contests with payo function (9), the
subgame perfect equilibrium outcome when one player can commit to an ob-
servable action before other players is dierent from the simultaneous move
equilibrium x when n > 2.
The result is a particular case of a result in Dixit (1987) where asym-
metric contests are also allowed. Note that the result hinges on whether
@2ui=@xj@xi = 0 as this determines whether d^ xi=dx1 = 0.
The contests analyzed so far appear to have a dierence between cases
n = 2 and n > 2. However, the payo functions can be modied to construct
games in which commitment does not give an advantage for other values of
n or indeed for any value of n. Consider the payo function
ui(x1;:::;xn) =
kf(xi)
kf(xi) + m
P
j6=i f(xj)
V   xi:
The second cross-derivative of this payo function is
@2ui
@xi@xj
=
kmf0(xi)f0(xj)(kf(xi)   m
P
j6=i f(xj))

kf(xi) + m
P
j6=i f(xj)
3 V
and at symmetric equilibrium x
@2u1
@xi@xj
=
km(f0(x))2(k   m(n   1))
(k + m(n   1))3f(x)2 V:
This expression is zero when k = m(n   1). For various values of n, one
can construct games so that the necessary condition for maximization of
u1(x1; ^ x2(x1);:::; ^ xn(x1)) is satised. For example, when n = 3, then taking
m = 1=2 and k = 1 gives a game for which it is satised.
Taking m = 1 and k = n   1 gives a game in which the slope of the
reaction function is zero at symmetric equilibrium for all n. Rewriting this
game's payo function as
ui(x1;:::;xn) =
f(xi)
f(xi) + 1
n 1
P
j6=i f(xj)
V   x
i (11)
gives an interpretation that each player engages in a bilateral contest against
the average eort of all other players. Since the chances to win the prize do
not always add up to 1 when summed over all players, the game can be seen
as a contest for a variable total prize. For this game, independently of n,
commitment may have no advantage.
7To show that there is a game in which the possibility of commitment by
Player 1 gives no advantage for any n, consider the case f(x) = x. The
symmetric interior equilibrium for this game is found from the rst order
conditions
1
n 1
P
j6=i xj
(xi + 1
n 1
P
j6=i xj)2V   1 = 0:
When xi = xj = x, then x = V=4.
For any x1 > 0, in an interior equilibrium for other players ^ xi = ^ xj.
To show this, note that from the rst order condition of Player i, (x1 + P
j6=i ^ xj)V=(n   1) = (^ xi + (x1 +
P
j6=i ^ xj)=(n   1))2. Subtracting the rst
order condition for Player j from the one of Player i gives (^ xj   ^ xi)V=n =
(^ xj   ^ xi)(1=(n   1)   1)(1 + 1=(n   1)(^ xi + ^ xj) + 2=(n   1)(x1 +
P
k6=i;j ^ xk)).
If ^ xj 6= ^ xi, then ^ xj   ^ xi can be cancelled from the two sides. But then the
left-hand side is positive while the right-hand side is negative. Therefore
^ xj = ^ xi at an interior equilibrium.
When ^ xi = ^ x for all i 6= 1, the level curve of Player 1 through the
symmetric equilibrium is
x1
x1 + ^ x
  x1 =
V
4
(12)
and ^ x satises the rst order condition for Players i, i 6= 1
1
n 1(x1 + (n   2)^ x)
(^ x + 1
n 1(x1 + (n   2)^ x))2V   1 = 0: (13)
From the level curve ^ x = x1(3V=4 x1)=(V=4+x1). The rst order condition
can be rewritten as F(x1; ^ x) = (n 1)(x1+(n 2)^ x)V  ((2n 3)^ x+x1)2 = 0.
Substituting ^ x from the level curve, simplifying and factorizing the expression
gives (4x1 V )2(x1(4n2 4n+4)+V (8n 3n2 5)) = 0. The last parenthesis
is zero when x1 = V (3n2   8n + 5)=(4n2   4n + 4). The right-hand side is
larger than V=4 for n > 1 and then ^ x < 0 from the level curve. Thus the
only solution with both x1; ^ x positive is x1 = V=4 and ^ x = V=4.
Player 1 maximizes u1(x1; ^ x(x1);:::; ^ x(x1)). The local second order con-
dition for maximum is d2u1=dx2
1 < 0. At equilibrium d2u1=dx2
1 = @2u1=@x2
1+
(n   1)@u1=@^ x  d2^ x=dx2
1. From the reaction function of the other players,
d^ x=dx1 =  (@F=@x1)=(@F=@^ x) and at equilibrium @F=@x1 = 0. Then
d2^ x=dx2
1 =  (@2F=@x2
1)=(@F=@^ x). Evaluating the appropriate derivatives
at x1 = ^ x = V=4 gives d2u1=dx2
1 = ( 4(n 1)+2)=(V (n 1)) < 0 for n > 1.
Since the local second order condition is satised and the reaction function
of players i, i 6= 1 does not have common points with the level curve of Player
81 other than the symmetric equilibrium point x1 = x = V=4, choosing x1
dierent from V=4 cannot give Player 1 higher payo. Therefore
Proposition 2 In the game with payo function (11) with f(x) = x, the
outcome of the subgame perfect equilibrium when Player 1 can commit to an
observable action is the same as the outcome in the equilibrium of simulta-
neous move game x = V=4 for any n.
4 Conclusion
This paper has proven that in two-player logit form symmetric contests
with concave success functions, the possibility of commitment does not give
advantage to the player who can commit. In strategic situations (when
@ui=@xj 6= 0), the necessary condition for this is that at equilibrium the
reaction function of the other player has the same slope as the player's pay-
o level curve. This condition is not sucient in general but in the contests
analyzed, commitment indeed does not lead to a higher payo.
In n-player symmetric contests with n > 2 the possibility of commitment
is always advantageous. Modifying the payo function to represent the game
as bilateral contest against the average eort of other players leads to a game
where commitment does not work for any n.
Commitment may have several interpretations apart from the direct com-
mitment to actions. For example, delegation (e.g. Vickers, 1985) or indirect
evolution of preferences (G uth and Yaari, 1992) can be seen as using commit-
ment. In those cases the committed player has a dierent reaction function
and therefore the outcome shifts along the reaction function of the opponent.
The results of this paper and of Possajennikov (2008) imply that in games
where commitment to an action does not increase a player's payo, prefer-
ences coinciding with material payos are stable, or delegates are provided
with incentives to maximize principal's payo. Two-player contests and bi-
lateral contests against the average eort of other players represent examples
of such games.
References
Baye, M.R., Shin, O. (1999) \Strategic Behavior in Contests: Comment",
American Economic Review 89(3), 691{693.
Dixit, A. (1987) \Strategic Behavior in Contests", American Economic Re-
view 77(5), 891{898.
9G uth, W., Yaari, M. (1992) \Explaining Reciprocal Behavior in Simple
Strategic Games: An Evolutionary Approach", in Witt, U. (ed.) Explaining
Forces and Changes: Approaches to Evolutionary Economics, University
of Michigan Press.
Kr akel, M. (2002) \Delegation and Strategic Incentives for Managers in Con-
tests", Managerial and Decision Economics 23, 461{470.
Possajennikov, A. (2008) \On the Survival of Payo Maximizing Behavior
and Delegation in Contests", University of Nottingham, working paper,
available on http://www.nottingham.ac.uk/~lezap/.
Vickers, J. (1985) \Delegation and the Theory of the Firm", Economic Jour-
nal 95, 138{147.
10

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