On Uniform Conditions for the Existence of Mixed Strategy
Equilibria
Pavlo Prokopovych1 and Nicholas C. Yannelis2
Abstract
Embarking from the concept of uniform payo¤ security (Monteiro
P.K., Page F.H, J Econ Theory 134: 566-575, 2007), we introduce two other uniform
conditions and then study the existence of mixed strategy Nash equilibria in games
where the sum of the payo¤ functions is not necessarily upper semicontinuous.
Keywords
Discontinuous game; Diagonally transfer continuous game; Payo¤
secure game; Mixed strategy equilibrium; Transfer lower semicontinuity
JEL classi…cation Numbers
Date
1
C65; C72
March 12, 2012
Pavlo Prokopovych, Kyiv School of Economics, 13 Yakira, 3d ‡oor, Kyiv 04119, Ukraine, Email:
[email protected]
2
Nicholas C. Yannelis, Economics - School of Social Sciences, The University of Manchester,
Oxford Road, Manchester M13 9PL, UK; Department of Economics, Tippie College of Business,
University of Iowa, Iowa City, IA 52242-1994, USA, E-mail: [email protected]
1
Introduction
We study the existence of mixed strategy equilibria in compact Borel games in which
the sum of the payo¤ functions is not necessarily upper semicontinuous. The existence of a mixed strategy Nash equilibrium in a game is often shown by invoking
Reny’s equilibrium existence theorem (Reny, 1999), according to which a compact
Borel game has a mixed strategy Nash equilibrium if its mixed extension is betterreply secure.3 In its turn, better-reply security is implied by two conditions, namely
reciprocal (or weak reciprocal) upper semicontinuity and payo¤ security. As was
shown by Simon (1987, Example 5), not every reciprocally upper semicontinuous
game has a reciprocally upper semicontinuous mixed extension. Consequently, imposing the assumption that the mixed extension of a game is reciprocally upper
semicontinuous is pretty close to assuming explicitly that the sum of the payo¤ functions is upper semicontinuous. Dasgupta and Maskin (1986, Theorem 5b) provided
a set of su¢ cient conditions for the existence of a mixed strategy Nash equilibrium
in reciprocally upper semicontinuous games.
Establishing the payo¤ security of the mixed extension of a game is also often
quite challenging. The concept of uniform payo¤ security, introduced by Monteiro
and Page (2007), makes the problem considerably more tractable in games where it
is applicable, including catalog games (Page and Monteiro, 2003) and voting models
(Carbonell-Nicolau and Ok, 2007). In this paper we propose two modi…cations of the
uniform payo¤ security condition in order to broaden the class of games in which the
existence of mixed strategy Nash equilibria follows from a uniform, easily tractable
condition
First we consider games with diagonally transfer continuous mixed extensions.
Baye, Tian, and Zhou (1993) showed that the existence of a pure strategy Nash
3
A number of results generalizing Reny’s equilibrium existence theorem have been obtained
recently (see, e.g., Barelli and Soza 2009; Bich 2009; Carmona 2011; de Castro 2011; McLennan,
Monteiro, and Tourky 2011)
2
equilibrium in diagonally transfer continuous games follows from a generalization of
the Knuster–Kuratowski–Mazurkiewicz (KKM) lemma. In Section 2 of this paper,
the Ky Fan minimax inequality, in a slightly generalized form, is used to prove that
every compact Borel game whose mixed extension is diagonally transfer continuous
has a mixed strategy Nash equilibrium. In particular, the mixed extension of a
game is diagonally transfer continuous if the conventional assumptions hold: the
extension is payo¤ secure and the game is upper semicontinuous-sum. Then, by
introducing uniform diagonal security in Section 3, we extend the concept of uniform
payo¤ security to diagonally transfer continuous games. In upper semicontinuoussum games, these two uniform conditions coincide. However, if a compact Borel game
is uniformly diagonally secure, it has a mixed strategy Nash equilibrium, which makes
it possible to avoid verifying the reciprocal upper semicontinuity of the game’s mixed
extension.
Example 1 is a slight modi…cation of the Tullock rent-seeking game in which
it is additionally assumed that the favor the players vie for may be granted to a
third party with probability one-half if at least one player exerts no e¤ort at all.
Notwithstanding the fact that the game is not better-reply secure, it is not only
diagonally transfer continuous, it is also uniformly diagonally secure; that is, the
game has a mixed strategy Nash equilibrium.
In Section 4 we introduce another uniform condition, called uniform hospitality.
The notion of a hospitable game is based on the notion of a hospitable strategy, introduced by Duggan (2007). Though uniform hospitality, in a certain sense, is stronger
that uniform payo¤ security (see, e.g., Lemma 6), it might be useful in applications.
For instance, all-pay auctions are both uniformly payo¤ secure and uniformly hospitable. At the same time, if, in a proof, some payo¤s are to be modi…ed, uniform
hospitality tends to be used side by side with uniform payo¤ security.
In Section 5 we study the existence of mixed strategy equilibria in two player
games on the unit square. The focus is on Theorem 5b of Dasgupta and Maskin
(1986). Bagh (2010) showed with the aid of examples that the hypotheses of Theorem 5b are not in full harmony with its proof and put forward a modi…cation of
the theorem covering mostly Bertrand-Edgeworth games. The games covered by the
3
modi…cation of Theorem 5b presented in Section 5 (Theorem 3) possess the following two properties: (a) they are uniformly payo¤ secure; (b) if a player’s payo¤ is
to be modi…ed at a discontinuity point (z; z), the player’s payo¤ function is strongly
uniformly hospitable at z. As an application of Theorem 3, a probabilistic spatial
voting model from Ball (1999) is considered.
2
The Model and Some Facts
We consider a game G = (Xi ; ui )i2I , where I = f1; : : : ; ng, each player i’s pure
strategy set Xi is a nonempty, compact subset of a metrizable topological vector
space, and each payo¤ function ui is a bounded Borel measurable function from the
Cartesian product X = i2I Xi , equipped with the product topology, to R. Under
these conditions, G = (Xi ; ui )i2I is called a compact Borel game. In this paper, by
a game we mean a compact Borel game. The following de…nition of a payo¤ secure
game was given by Reny (1999).
De…nition 1 In G = (Xi ; ui )i2I , player i can secure a payo¤ of 2 R at x 2 X if
for all x0 i in some open neighborhood
there exists di 2 Xi such that ui (di ; x0 i )
of x i . The game G is payo¤ secure if, for every x 2 X and every " > 0, each player
i can secure a payo¤ of ui (x) " at x.
Payo¤ security can be reformulated in terms of transfer lower semicontinuity, due
to Tian (1992).
De…nition 2 Let Z and Y be two topological spaces. A function f : Z Y 7! R is
-transfer lower semicontinuous in y if, for every (z; y) 2 Z Y , f (z; y) > implies
that there exists some point z 0 2 Z and some neighborhood NY (y) of y in Y such
that f (z 0 ; w) > for all w 2 NY (y): A function f : Z Y 7! R is transfer lower
semicontinuous in y if f is -transfer lower semicontinuous in y for every 2 R.
A game is payo¤ secure if and only if each player’s payo¤ function is transfer
lower semicontinuous in the other players’strategies (Prokopovych, 2011).
4
The graph of G is de…ned by GrG = f(x; u) 2 X Rn j ui (x) = ui for all i 2 N g.
For a subset B of a topological vector space X, we denote the interior of B in X by
intX B, the boundary of B by @B, the closure of B by clB, and the convex hull of B
by coB. In a metric space Y , we denote by BY (y; r) the open ball centered at y and
with radius r > 0. Denote by EG the set of pure strategy Nash equilibria of G in X.
De…nition 3 A game G = (Xi ; ui )i2I is better-reply secure if, whenever (x ; u ) 2
clGrG and x 2 XnEG , some player i can secure a payo¤ strictly above ui at x .
Reny’s (1999) equilibrium existence theorem states that every compact, quasiconcave, better-reply secure game has a Nash equilibrium in pure strategies. Better-reply
security is equivalent to transfer reciprocal upper semicontinuity in payo¤ secure
games.4
Another approach to studying the equilibrium existence problem in discontinuous
games was proposed by Baye, Tian, and Zhou (1993).
For G = (Xi ; ui )i2I , de…ne the following aggregator functions:
AG : X
where, as usual, the
X ! R by AG (d; x) =
X
ui (di ; x i );
i2I
i subscript on x stands for "all players except i,"
A0G : X ! R by A0G (x) =
X
ui (x),
i2I
and
FG : X
X ! R by FG (d; x) = AG (d; x)
A0G (x).
A strategy pro…le x 2 X is a Nash equilibrium of G i¤ FG (d; x)
0 for all d 2 X.
De…nition 4 A game G = (Xi ; ui )i2I is diagonally transfer continuous if, for every
x 2 XnEG , there exist some d 2 X and some neighborhood NX (x) of x in X such
that FG (d; z) > 0 for all z 2 NX (x).
4
We refer the reader to Reny (1999), Bagh and Jofre (2006), and Prokopovych (2011) for further
details regarding reciprocal upper semicontinuity and its generalizations.
5
It is important to notice that G is diagonally transfer continuous i¤ FG is 0transfer lower semicontinuos in x.
Every payo¤ secure game with an upper semicontinuous A0G is diagonally transfer
continuous.
Lemma 1 If, in a game G = (Xi ; ui )i2I , each ui : X ! R is transfer lower semicontinuous in x i and the aggregator function A0G : X ! R is upper semicontinuous,
then G is diagonally transfer continuous.
Proof. We shall …rst show that AG (d; x) is transfer lower semicontinuous in x. Let
(d; x) 2 X X and 2 R be such that AG (d; x) > . Then there are 1 ; : : : ; n 2 R
such that = 1 + : : : + n and ui (di ; x i ) > i for all i 2 I. Since each ui is transfer
lower semicontinuous in x i , there exist di 2 Xi and an open neighborhood NX i (x i )
of x i in X i such that ui (di ; z i ) > i for all z i 2 NX i (x i ). Consequently,
AG (d; z) > for every z 2 \i2I fXi NX i (x i )g.
Since A0G is upper semicontinuos on X, the transfer lower semicontinuity of AG
in x implies the transfer lower semicontinuity of FG in x. Therefore FG is 0-transfer
lower semicontinuous in x.
Now we de…ne the mixed extension of a game G = (Xi ; ui )i2I . Denote by 4(Xi )
the set of Borel probability measures on Xi and by ca(Xi ) the set of Borel signed
measures with …nite total variation on Xi . A basic open neighborhood of i 2 ca(Xi )
in the weak topology on ca(Xi ) is a set of the form
i
2 ca(Xi ) :
Z
fj (d
d i ) < "; j = 1; : : : ; m
i
for some f1 ; : : : ; fm 2 C(Xi ) and " > 0. The set ca(Xi ) is a Hausdor¤ topological
vector space equipped with the weak topology. The topology induced on 4(Xi )
by the weak topology is compact.5 Let each of the Cartesian products ca(X) =
5
In order to make 4(Xi ) a subset of a linear space, we embed it in the space ca(Xi ) of signed
measures with …nite total variation on Xi . Sometimes it is possible to proceed without the embedding (see, e.g., the proof of the compactness of the set of probability measures given by Glycopantis
and Muir 2004).
6
ca(X1 ) : : : ca(Xn ) and 4(X) = 4(X1 ) : : : 4(Xn ) be equipped with the
product topology. The set ca(X) is a Hausdor¤ topological vector space in which
the operations of addition and scalar multiplication are de…ned as follows: for =
( 1 ; : : : ; n ) 2 ca(X) and 2 R the scalar multiplication of by is the element
given by
= ( 1 ; : : : ; n ). The addition of = ( 1 ; : : : ; n ) 2 ca(X) and
v = ( 1 ; : : : ; n ) 2 ca(X) gives + v = ( 1 + 1 ; : : : ; n + n ).
The mixed extension of the game G is the n-player normal form game
=
(4(Xi ); Ui )i2I , where 4(Xi ) is player i’s strategy set and player i’s payo¤ function
Ui : 4(X) ! R is de…ned by
Ui ( ) =
Z
X1
Z
X2
:::
Z
ui (x1 ; : : : ; xn )d
1
:::d
n:
Xn
Since Fubini’s theorem holds in this context, the integral in the de…nition of player
i’s expected payo¤ is properly de…ned.
For the game , we also de…ne the aggregator functions A : 4(X) 4(X) ! R,
0
A : 4(X) ! R, and F : 4(X) 4(X) ! R (see the corresponding de…nitions for
G).
Theorem 1 If the mixed extension of a game G = (Xi ; ui )i2I is diagonally transfer
continuous, then G possesses a mixed strategy Nash equilibrium.
Proof. The set 4(X) is a compact, convex subset of ca(X). Consider the aggregator
function F ( ; ) : 4(X) 4(X) ! R. Since F is linear in and 0-transfer lower
semicontinuous in , the mixed extension of G has a Nash equilibrium in pure
strategies by the Ky Fan minimax inequality (see Lemma 9 in the Appendix).
Another proof of Theorem 1 can be obtained by using the fact that every diagonally transfer continuous mixed extension has the single deviation property (see
Reny, 2009, footnote 4; Reny, 2011, p. 19; Prokopovych, 2012 for some details).
Since the upper semicontinuity of A0G implies the upper semicontinuity of A0 ,
verifying whether a game has a mixed strategy Nash equilibirum usually means
verifying the following two properties: (a) the upper semicontinuity of the sum of
the payo¤ functions; and (b) the payo¤ security of its mixed extension. If these
7
properties hold, the mixed extension of the game is not only better-reply secure but,
by Lemma 1, diagonally transfer continuous.
The diagonal transfer continuity of a game does not imply that its mixed extension
is diagonally transfer continuous. For instance, Sion and Wolfe’s zero-sum game (Sion
and Wolfe, 1957) is payo¤ secure (see Carmona, 2005, Example 4) and its aggregator
function A0G is constant. Thus, the game is diagonally transfer continuous by Lemma
1. However, since the game has no mixed strategy Nash equilibrium, its mixed
extension is not diagonally transfer continuous.
3
Uniform Security
An easily veri…able condition for the mixed extension of a game to be payo¤ secure
is that of uniform payo¤ security, due to Monteiro and Page (2007).
De…nition 5 A game G = (Xi ; ui )i2I is uniformly payo¤ secure if, for every xi 2 Xi
and every " > 0, there is di 2 Xi such that, for every x i 2 X i , ui (di ; w i )
ui (xi ; x i ) " for all w i in some open neighborhood NX i (x i ) of x i in X i .
It is useful to notice that: (a) replacing the pure deviation strategy di in De…nition 5 with a mixed deviation strategy i 2 4(Xi ) would not a¤ect the validity
of the proof of Theorem 1 of Monteiro and Page (2007); (b) uniform payo¤ security
can be introduced in a pointwise manner. We will need the following de…nition. In
G = (Xi ; ui )i2I , player i’s payo¤ function ui : Xi X i ! R is uniformly transfer
lower semicontinuous in x i at xi 2 Xi if, for every " > 0, there is di 2 Xi such that,
for every x i 2 X i , there exists a neighborhood NX i (x i ) of x i in X i such that
ui (di ; w i ) > ui (xi ; x i ) " for all w i 2 NX i (x i ).
Corollary 1 If G = (Xi ; ui )i2I is uniformly payo¤ secure and its aggregator function A0G is upper semicontinuous, then the mixed extension is diagonally transfer
continuous, and, therefore, G possesses a mixed strategy Nash equilibrium.
The notion of a uniformly payo¤ secure game can be extended to diagonally
transfer continuous games.
8
De…nition 6 A game G = (Xi ; ui )i2I is uniformly diagonally secure if, for every
d 2 X and every " > 0, there is d 2 X such that, for every x 2 X, FG (d; w) >
FG (d; x) " for all w in some open neighborhood NX (x) of x in X.
An upper semicontinuous-sum game G is uniformly diagonally secure if it is
uniformly payo¤ secure.
Lemma 2 If a game G = (Xi ; ui )i2I is uniformly payo¤ secure and the aggregator
function A0G : X ! R is upper semicontinuous, then G is uniformly diagonally
secure.
Proof. Fix some d 2 X. By the uniform payo¤ security of G, for every " > 0 and
each i 2 I, there is a deviation strategy di 2 Xi such that, for every x i 2 X i ,
"
ui (di ; w i )
ui (di ; x i ) 2n
for all w i in some open neighborhood NX i (x i ) of
x i in X i . Consider some x 2 X and denote NX1 (x) = \i2I fXi NX i (x i )g.
Then AG (d; w) AG (d; x) 2" for all w 2 NX1 (x). Since the function A0G is lower
semicontinuous on X, there exists a neighborhood NX2 (x) such that A0G (w) >
A0G (x) 2" for all w 2 NX2 (x). Then FG (d; w) > FG (d; x) " for all w 2 NX1 (x) \
NX2 (x).
Theorem 2 If a game G = (Xi ; ui )i2I is uniformly diagonally secure, then its mixed
extension is diagonally transfer continuous, and, therefore, G possesses a mixed
strategy Nash equilibrium.
The proof of Theorem 2 follows the lines of the proof of Theorem 1 of Monteiro
and Page (2007) and is given in the Appendix.
The concept of uniform diagonal security may be of help in studying equilibirum
existence in games whose aggregator function A0G is not upper semicontinuous.
Example 1 Consider a slight modi…cation of the rent-seeking game due to Tullock (1980). Two players simultaneously bid for a political favor commonly known
worth V dollars. Their bids, denoted by x1 and x2 , in‡uence the probability of receiving the favor. Player i’s strategy set is the segment [0; V ]. Let i (x1 ; x2 ) denote
9
the probability player i wins. The function
function, is speci…ed as follows:
8
>
>
<
i,
often called player i’s contest success
1
4
1
2
if x1 = x2 = 0,
if xi > x i = 0,
i (x1 ; x2 ) =
>
r
>
: r xi r otherwise,
x +x
i
i
where r > 0. Player i’s payo¤ function ui is
ui (x1 ; x2 ) =
i (x1 ; x2 )V
xi :
The only di¤erence of the model from the Tullock rent-seeking game is the assumption that if the lowest bid submitted is equal to zero (or, in other words, at least
one player exerts no e¤ort at all), the favor may be granted to a third party with
probability one-half. Consequently, the aggregator function A0G is not upper semicontinuous.
Let, for speci…city, V = 2 and r = 3. In this case, the game has no pure strategy
Nash equilibria (see Baye, Kovenock, and de Vries 1994 for a related discussion). For
instance, one can check that the only candidate point for being an interior solution
is (1:5; 1:5), a strategy pro…le where both players get negative expected payo¤s.
However, each of them can avoid getting a negative payo¤ by bidding zero.
To verify that the game is not better-reply secure, consider the sequence fxk g
with xk = ( k1 ; k1 ) for k = 1; 2; : : : :Then the corresponding sequence of payo¤ vectors
f(u1 (xk ); u2 (xk ))g converges to (1; 1). It is clear that no player can secure a payo¤
strictly above 1 at (0; 0).
On the other hand, the game is not only diagonally transfer continuous, it is
uniformly diagonally secure (see the Appendix for details). Therefore, the game
possesses a mixed strategy Nash equilibrium.
10
4
Uniform Hospitality
In this section we extend the notion of a hospitable strategy, initially developed by
Duggan (2007) for zero-sum games, to our framework.
De…nition 7 A strategy i 2 4(Xi ) of player i is called hospitable if Ui ( i ; ) :
R
X i ! R de…ned by Ui ( i ; x i ) = Xi ui (xi ; x i )d i is continuous.
De…nition 8 A game G = (Xi ; ui )i2I is called hospitable if, for each i 2 I, every
2 4(X), and every " > 0, there is a hospitable strategy i such that Ui ( i ; i ) >
Ui ( ) ".
There is no ambiguity associated with the notation used in De…nition 7 since
R
Ui ( i ; i ) = X i Ui ( i ; x i )d i by Fubini’s theorem.
Lemma 3 If G = (Xi ; ui )i2I is a hospitable game, then its mixed extension
payo¤ secure.
is
Proof. Fix some = ( 1 ; : : : ; n ) 2 4(X), " > 0, and i 2 I. We have to show
that there are a strategy i 2 4(Xi ) and a neighborhood N4(X i ) ( i ) such that
Ui ( i ; 0 i ) Ui ( ) " for all 0 i 2 N4(X i ) ( i ).
Since G is a hospitable game, Ui ( i ; i ) > Ui ( ) " for some hospitable i 2
4(Xi ). Since Ui ( i ; ) : X i ! R is continuous, Ui ( i ; ) : 4(X i ) ! R is continuous as well (see Aliprantis and Border, 2006, Theorem 15.5). Therefore, there
exists a neighborhood N4(X i ) ( i ) such that Ui ( i ; 0 i ) > Ui ( ) " for all 0 i 2
N4(X i ) ( i ).
Corollary 2 If G = (Xi ; ui )i2I is a hospitable game with an upper semicontinuous
aggregator function A0G , then it has a mixed strategy Nash equilibrium.
We now give some su¢ cient conditions for a game to be hospitable.
De…nition 9 A game G = (Xi ; ui )i2I is uniformly hospitable if, for each i 2 I,
every xi 2 Xi , and every " > 0, there is a hospitable strategy i 2 4(Xi ) such that
Ui ( i ; x i ) > ui (xi ; x i ) " for every x i 2 X i .
11
Lemma 4 Every uniformly hospitable game G = (Xi ; ui )i2I is hospitable.
Proof. Fix some i 2 f1; : : : ; ng, 2 4(X), and " > 0. We have to show that there
is a hospitable strategy i such that Ui ( i ; i ) > Ui ( ) ".
First, pick a strategy xi 2 Xi such that Ui (xi ; i ) Ui ( ). Since G is uniformly
hospitable, there exists a hospitable strategy i 2 4(Xi ) such that Ui ( i ; i ) >
Ui (xi ; i ) ". Therefore Ui ( i ; i ) > Ui ( ) ".
From now on, we impose the additional assumption that each Xi is a subset of a
…nite-dimensional Euclidean space. Denote by i the Lebesgue measure on Xi . The
proof of the next lemma is based on some ideas from Duggan (2007, Propositions 2
and 3) and is relegated to the Appendix.
Lemma 5 Let each Xi , i 2 I, be a compact subset of a …nite-dimensional Euclidean
space. A game G = (Xi ; ui )i2I is uniformly hospitable if
(i) for each i 2 N and every x i 2 X i ,
i (fxi
2 Xi : ui (xi ; ) is discontinuous at x i g) = 0,
(ii) for each i 2 N , every xi 2 Xi , and every " > 0, there is an open ball Bi+ (xi ; ") in
Xi such that ui (x0i ; x i ) > ui (xi ; x i ) " for every x0i 2 Bi+ (xi ; ") and every x i 2 X i .
It is worth noticing that Bi+ (xi ; ") is not necessarily centered at xi .
Assumption (i) of Lemma 5 is akin to the one-to-one assumption used by Dasgupta and Maskin (1986, de…nitions on p. 7 and Example 4 on p. 21) and is not
super‡uous.
Example 2 Consider G = (Xi ; ui )i2f1;2g where X1 = X2 = [0; 1] and player i’s
payo¤ function is de…ned as follows:
u1 (x1 ; x2 ) =
u2 (x1 ; x2 ) =
(
(
x1 if (x1 ; x2 ) 2 [0; 1] [0; 1),
2 x1 if (x1 ; x2 ) 2 [0; 1] f1g;
x2 if (x1 ; x2 ) 2 (0; 1] [0; 1] and (x1 ; x2 ) = (0; 1),
3 x2 if (x1 ; x2 ) 2 f0g [0; 1).
12
It is easy to see that the sum of the payo¤ functions is upper semicontinuous on
X. The game is compact, quasiconcave, and satis…es assumption (ii) of Lemma 5.
However it has no mixed strategy Nash equilibria. To show this, assume, by way of
contradiction, that the game has a mixed strategy Nash equilibrium, = ( 1 ; 2 ).
For each i 2 f1; 2g de…ne BRi : 4(X i )
4(Xi ) by
BRi (
i)
=f
i
2 4(Xi ) : Ui ( i ;
i)
Ui ( i ;
i)
for every
i
2 4(Xi )g:
Since is a Nash equilibrium of , i 2 BRi ( i ) for i = 1; 2. Denote by x the
Dirac measure concentrated at x 2 [0; 1].
If 1 (f0g) = 0, then BR2 ( 1 ) = 1 . However, if 2 = 1 , then BR1 ( 2 ) = 0 ,
and, therefore, 1 (f0g) = 1, a contradiction.
If 2 (f1g) = 0, then BR1 ( 2 ) = 1 , which, in turn, implies that BR2 ( 1 ) = 1 ,
another impossibility.
Therefore, both 1 (f0g) > 0 and 2 (f1g) > 0. Then the fact that 2 2 BR2 ( 1 )
implies that U2 ( 1 ; 1 ) U2 ( 1 ; 2 ) for every 2 2 4(X2 ). By de…nition, U2 ( 1 ; 1 ) =
R
0
1 (f0g)u2 (0; 1) + (0;1] u2 (x1 ; 1)d 1 . Fix some " 2 (0; 1 (f0g)) and pick some x2 close
enough to one such that u2 (x1 ; x02 ) > u2 (x1 ; 1) " for every x1 2 [0; 1]. Then
U2 ( 1 ; x02 ) U2 ( 1 ; 1 ) > 1 (f0g)(u2 (0; x01 ) u2 (0; 1)) " > 0, which contradicts the
initial premise that is a Nash equilibrium of .
In …nite-dimensional applications, it is sometimes possible to apply the notion
of a uniformly hospitable game in place of the notion of a uniformly payo¤ secure
game.
Example 3 Consider the following uniformly payo¤ secure game G = (Xi ; ui )i2I
(see Monteiro and Page, 2007, Example 1; Baye, Kovenock, and de Vries, 1996).
There are n bidders, competing for an object with a known value of unity. Each
player i submits a sealed bid bi 2 [0; 1]. Let b = maxi2I bi , H = fi 2 I : bi = b g,
and jHj be the cardinality of H. Bidder i’s payo¤ is as follows:
ui (b) =
(
1
jHj
bi if bi = b ,
bi if bi < b .
13
Let " 2 (0; 1) and bi 2 Xi = [0; 1]. To show that the auction is uniformly
hospitable, we have to consider two cases.
Case 1. If bi 2 [0; 1), pick bi > bi and > 0 such that bi < bi
and bi + <
+
minf1; bi + "g. Put Bi (bi ; ") = BXi (bi ; ).
Case 2. If bi = 1, then player i’s payo¤ is either zero or negative. Put Bi+ (xi ; ") =
BXi (0; ").
It is clear that in each of the cases ui (bi ; b i ) > ui (bi ; b i ) " for every bi 2 Si+ (bi ; ")
and every b i 2 X i . By Lemma 5, the game is uniformly hospitable.
5
Uniform Conditions in Two Player Games on
the Unit Square
Theorem 5b of Dasgupta and Maskin (1986) (hereinafter called Theorem 5b) is an
important equilibrium existence result in which the assumption that the sum of the
players’ payo¤s functions is upper semicontinuous is relaxed. As shown by Bagh
(2010) with the aid of examples, the proof of Theorem 5b is not in full harmony with
its statement. The modi…cation of the theorem proposed by Bagh (2010, Theorem
5.1) is applicable mostly to Bertrand-Edgeworth games (see also Remark 1 below).
Making use of the concepts of uniform payo¤ security and uniform hospitality leads
us to another modi…cation of Theorem 5b.
We consider a two-player game G = (Xi ; ui )i2f1;2g on the unit square, X = X1
X2 = [0; 1] [0; 1], where the payo¤ functions’discontinuities lie on the main diagonal
of X. For such games, we will need the following strengthening of uniform hospitality.
Player i’s payo¤ function ui : [0; 1] [0; 1] ! R is called strongly uniformly hospitable
at xi 2 [0; 1] if, for every " > 0, there exists (") 2 Rnf0g such that xi + (") 2 [0; 1]
and, for every di 2 (xi ; xi + (")), ui (di ; x i ) > ui (xi ; x i ) " for all x i 2 X i .
The game G = (Xi ; ui )i2f1;2g is strongly uniformly hospitable if each ui is strongly
uniformly hospitable at every xi 2 Xi .
The set X consists of three subsets: S 1 = f(x1 ; x2 ) 2 [0; 1] [0; 1] : x1 2 [0; 1]
and x2 > x1 g, S 2 = f(x1 ; x2 ) 2 [0; 1] [0; 1] : x1 2 [0; 1] and x1 > x2 g, and
14
S = f(x1 ; x2 ) : [0; 1]
[0; 1] : x1 = x2 g.
Lemma 6 Consider a two-player game G = (Xi ; ui )i2f1;2g on the unit square X =
[0; 1] [0; 1]. Suppose that there are continuous functions lij : clS j ! R, i = 1; 2;
j = 1; 2 such that ui (x) = lij (x) for all x 2 S j and all i; j 2 f1; 2g. If player i’s payo¤
function ui is strongly uniformly hospitable at some xi 2 [0; 1], then it is uniformly
transfer lower semicontinuous in x i at xi 2 Xi .
Since each lij is de…ned on clS j , limk!1 ui (xk ) = lij (x) for every sequence fxk g
S j converging to x 2 S.
Proof. Since ui is strongly uniformly hospitable at xi , for every " > 0 there exists
(") 2 Rnf0g such that xi + (") 2 [0; 1] and, for every di 2 (xi ; xi + (")), ui (di ; x i ) >
ui (xi ; x i ) " for all x i 2 X i . Fix some arbitrary " > 0 and some di 2 (xi ; xi + ( 2" )).
We have to show that, for every x i 2 X i , there exists a neighborhood NX i (x i )
of x i in X i such that ui (di ; w i ) > ui (xi ; x i ) " for all w i 2 NX i (x i ).
If x i 6= di , then ui is continuous at (di ; x i ), and we can choose a suitable
neighborhood NX i (x i ). If x i = di , then (xi ; x i ) 2 S j for some j 2 f1; 2g. Pick
r > 0 such that (xi ; w i ) 2 S j and jui (xi ; w i ) ui (xi ; x i )j < 2" for all w i 2
BX i (x i ; r). Then ui (di ; w i ) > ui (xi ; w i ) 2" > ui (xi ; x i ) " for every w i 2
BX i (x i ; r).
The next lemma states that any game covered by Theorem 5b is uniformly payo¤
secure.
Lemma 7 Consider a two-player game G = (Xi ; ui )i2f1;2g on the unit square X =
[0; 1] [0; 1]. Assume that
(i) there are continuous functions lij : clS j ! R, i = 1; 2; j = 1; 2 such that ui (x) =
lij (x) for all x 2 S j and all i; j 2 f1; 2g;
(ii) for each i 2 f1; 2g and every x 2 S, there exists j 2 f1; 2g such that
lij (x)
ui (x)
Then G is uniformly payo¤ secure.
15
li j (x);
Proof. Fix some i 2 f1; 2g, xi 2 (0; 1), and " > 0. Then, by (ii), there exists
j 2 f1; 2g such that lij (xi ; xi )
ui (xi ; xi )
li j (xi ; xi ). Suppose that i = 1 and
j = 2. The rest of the cases can be handled similarly.
Since l11 and l12 are uniformly continuous on their respective compact domains,
there exists 2 (0; minfx1 ; 1 x1 g) such that, for each j 2 f1; 2g, l1j (x0 ) l1j (x00 ) < 2"
for all x0 and x00 in clS j with kx0 x00 k < , where kx0 x00 k denotes the Euclidean
distance between x0 and x00 .
We shall show that, for any …xed d1 2 (x1 ; x1 + ), u1 (d1 ; x2 ) > u1 (x1 ; x2 ) "
for all x2 2 X2 . If x2 2 X2 n[x1 ; d1 ], then (d1 ; x2 ) and (x1 ; x2 ) lie in the same S j
and, therefore, u1 (d1 ; x2 ) > u1 (x1 ; x2 ) ". If x2 = x1 , then u1 (x1 ; x1 ) l12 (x1 ; x1 ) <
l12 (d1 ; x1 ) + " = u1 (d1 ; x1 ) + ". If x2 2 (x1 ; d1 ], then u1 (x1 ; x2 ) < l11 (x1 ; x1 ) + 2"
l12 (x1 ; x1 ) + 2" < l12 (d1 ; x2 ) + ". Therefore, u1 (x1 ; x2 ) < u1 (d1 ; x2 ) + " for all x2 2
(x1 ; d1 ). If x2 = d1 , then u1 (x1 ; d1 ) = l11 (x1 ; d1 ) < l11 (d1 ; d1 ) + 2" . Then u1 (x1 ; d1 ) <
minfl11 (d1 ; d1 ) + "; l12 (d1 ; d1 ) + "g u1 (d1 ; d1 ) + ".
Thus we conclude that each ui is strongly uniformly hospitable at every xi 2 (0; 1),
which, by Lemma 6, implies that ui is uniformly transfer lower semicontinuous in
x i at every xi 2 (0; 1).
It is not di¢ cult to see that ui need not be strongly uniformly hospitable at xi = 0
or at xi = 1. However it is still uniformly transfer lower semicontinuous in x i at
those xi ’s. For instance, if xi = 0 and lii (0; 0) ui (0; 0) li i (0; 0), then put di = 0.
It is clear that for every " > 0 and every x i 2 X i , there exists a neighborhood
NX i (x i ) of x i in X i such that ui (0; w i ) ui (0; x i ) " for all w i 2 NX i (x i ).
ui (0; 0)
li i (0; 0), then it is not di¢ cult to show that ui is strongly
If lii (0; 0)
uniformly hospitable at xi = 0 by repeating the above argument.
In addition to all-pay auctions, Lemma 7 is applicable to …rst-price and secondprice sealed-bid auctions, a number of location games and voting models. If the hypotheses of Lemma 7 are satis…ed for a game and the game is upper semicontinuoussum, then it has a mixed strategy Nash equilibrium.
Another important corollary of Lemma 7 is that modi…cations of payo¤s similar
to those used by Dasgupta and Maskin (1986) in the proof of Theorem 5b do not
lead out of the class of uniformly payo¤ secure games.
16
We need one more auxiliary lemma. Denote SD = fx 2 X : limsupy!x A0G (y) >
A0G (x)g.
Lemma 8 Consider a two-player game G = (Xi ; ui )i2f1;2g on the unit square X =
[0; 1] [0; 1]. Suppose that the hypotheses of Lemma 7 hold and
(i) for every point x = (z; z) 2 SD , there exist i; j 2 f1; 2g and some sequence
f(xki ; z)g S j converging to x such that limn!1 lij (xki ; z) > ui (x);
S j converging to x =
(ii) if, for some i; j 2 f1; 2g and some sequence f(xki ; z)g
(z; z) 2 SD , limn!1 lij (xki ; z) > ui (x), then limn!1 lj i (z; xki ) < u i (x).
Then each ui is strongly uniformly hospitable at any z 2 Xi such that (z; z) 2 SD .
Proof. As we have shown in the proof of Lemma 7, each ui is strongly uniformly
hospitable at every xi 2 (0; 1). Let (0; 0) 2 SD (the argument is similar if (1; 1) 2
SD ). We shall show that both u1 and u2 are strongly uniformly hospitable at 0. Since
(0; 0) 2 SD , li i (0; 0) > ui (0; 0) for some i 2 f1; 2g by (i). Then l ii (0; 0) < u i (0; 0)
by (ii). Therefore, li i (0; 0) ui (0; 0) lii (0; 0) and li i (0; 0) u i (0; 0) l ii (0; 0),
which implies that each ui is strongly uniformly hospitable at 0.
The following result is a modi…cation of Theorem 5b.
Theorem 3 Consider a two-player game G = (Xi ; ui )i2f1;2g on the unit square X =
[0; 1] [0; 1] in which the restriction of A0G to S is a continuous function from S to
R. Assume that
(i) there are continuous functions lij : clS j ! R, i = 1; 2; j = 1; 2 such that ui (x) =
lij (x) for all x 2 S j and all i; j 2 f1; 2g;
(ii) for each i 2 f1; 2g and every x 2 S, there exists j 2 f1; 2g such that
lij (x)
ui (x)
li j (x);
(iii) for every point x = (z; z) 2 SD , there exist i; j 2 f1; 2g and some sequence
f(xki ; z)g S j converging to x such that limn!1 lij (xki ; z) > ui (x);
(iv) if, for some i; j 2 f1; 2g and some sequence f(xki ; z)g
S j converging to x =
(z; z) 2 SD , limn!1 lij (xki ; z) > ui (x), then limn!1 lj i (z; xki ) < u i (x).
Then G has a mixed strategy Nash equilibrium.
17
As shown by Example A.1 of Bagh (2010), Theorem 5b lacks an assumption
regarding the players’payo¤s along the main diagonal of the unit square. The extra
assumption that the restriction of A0G to S is a continuous function plays a crucial
role in Theorem 3. At the same time, some hypotheses of Theorem 3 are weaker
than their counterparts in Theorem 5b. For instance, (iii) and (iv) are to be held
only on SD , not at every point of discontinuity.
Proof. It is not di¢ cult to see that SD = fx 2 S : maxj2f1;2g fl1j (x) + l2j (x)g >
A0G (x)g, and, therefore, it is a relatively open subset of S. In particular, the latter
implies that SD is a Borel subset of X. Another useful fact is the following: if lij (x) >
ui (x) for some x = (z; z) 2 SD and some i; j 2 f1; 2g, then limn!1 lij (xni ; z) > ui (x)
for some sequence f(xni ; z)g S j converging to x, and, therefore, u i (x) > lj i (x) by
(iv) (see the proof of Lemma 8 for an explanation).
The plan of the proof is similar to that used by Dasgupta and Maskin (1986,
Theorem 5b). First we will modify payo¤s on SD so as to make the sum of the payo¤
functions upper semicontinuous on X.
De…ne the modi…ed payo¤ functions as follows: for i 2 f1; 2g, u
bi (x) = ui (x) for
all x 2 XnSD , and
u
bi (x) = maxfui (x); max flij (x)
j2f1;2g
maxf0; u i (x)
lj i (x)ggg for x 2 SD .
Since SD is Borel, each u
bi is Borel measurable on X. By Lemma 7, the game
b
G = (Xi ; u
bi )i2f1;2g is uniformly payo¤ secure.
In order to show that A0Gb (x) is upper semicontinuous at every x 2 clSD , consider a
sequence fxk g X converging to x 2 clSD such that limk!1 A0Gb (xk ) exists. We have
to show that limk!1 A0Gb (xk )
A0Gb (x) for any such sequence. We have to consider
two cases: (a) x 2 SD ; and (b) x 2 rbdSD , where rbdSD denotes the boundary
of SD in S. De…ne Ji : S
f1; 2g by Ji (x) = fj 2 f1; 2g : lij (x) > ui (x) and
lij (x) ui (x) > u i (x) lj i (x)g. Notice that values of Ji might be empty.
Case 1. Let x 2 SD . Consider a sequence fxk g lying in some S j . Since A0Gb (xk ) =
A0G (xk ) for each k, we have that limk!1 A0Gb (xk ) = l1j (x) + l2j (x). If l1j (x) + l2j (x)
u1 (x)+u2 (x), then limk!1 A0Gb (xk ) A0G (x) A0Gb (x). If l1j (x)+l2j (x) > u1 (x)+u2 (x),
18
then j 2 Ji (x) for some i 2 f1; 2g. By (iv), u i (x) lj i (x) > 0, and, therefore,
u
bi (x) = lij (x) + lj i (x) u i (x). Then u
bi (x) + u
b i (x) u
bi (x) + u i (x) = l1j (x) + l2j (x).
Now consider a sequence fxk g lying in SD . Since x 2 SD , Ji (x) 6= ? for at least
one i 2 f1; 2g. Let us show that each Ji has open lower sections; that is, if j 2 Ji (x),
then j 2 Ji (w) for all w in some open neighborhood NS (x) of x in S. Assume, by way
of contradiction, that, despite the fact that fxk g tends to x and j 2 Ji (x), j 2
= Ji (xk )
for each k. Since the restriction of A0G to S is continuous, there is no loss in generality
to assume that lij (xk ) ui (xk ) > u i (xk ) lj i (xk ) for each k. On the other hand,
since lij and li j are continuous on S and lij (x) > ui (x) li j (x), there is no loss of
generality to assume that lij (xk )
ui (xk )
li j (xk ) for all k. Then j 2
= Ji (xk ) for
each k only if lij (xk ) = ui (xk ) for each k. Therefore, 0 > u i (xk ) lj i (xk ) for each k.
However, by (iv), lij (xk ) < ui (xk ) for each k, a contradiction. Thus, we assume that
if j 2 Ji (x) for some i; j 2 f1; 2g, then j 2 Ji (xk ) for all k.
We have to consider two cases: (a) J1 (x) 6= ? and J2 (x) 6= ?; and (b) Ji (x) 6= f?g
for some i and J i (x) = f?g.
P
If J1 (x) 6= f?g and J2 (x) 6= f?g, then u
b1 (x) + u
b2 (x) = t2f1;2g (l1t (x) + l2t (x))
u1 (x) u2 (x). Since the restriction of A0G to S is continuous and each Ji has open
lower sections, limk!1 A0Gb (xk ) = A0Gb (x).
If j 2 Ji (x) and J i (x) = f?g for some i; j 2 f1; 2g, then u
bi (x) = l1j (x) +
l2j (x) u i (x), u
b i (x) = u i (x), and, therefore, A0Gb (x) = l1j (x) + l2j (x). If J i (xk ) =
f?g, then u
b i (xk ) = u i (xk ) and A0Gb (xk ) = l1j (xk ) + l2j (xk ). If j 2 J i (xk ), then
P
A0Gb (xk ) = t2f1;2g (l1t (xk ) + l2t (xk )) u1 (xk ) u2 (xk ). The set J i (x) is empty if
either u i (x)
l ij (x) or l ij (x) u i (x)
ui (x) li j (x). Since ui (x)
li j (x)
by (ii), we have, in both cases, that li j (x) + l ij (x)
ui (x) + u i (x). Therefore,
P
j
j
t
t
l1 (x)+l2 (x). Thus, limk!1 A0Gb (xk ) A0Gb (x).
t2f1;2g (l1 (x)+l2 (x)) u1 (x) u2 (x)
Case 2. Let x 2 rbdSD . First notice that, since x 2
= SD , we have that u
bi (x) =
j
j
ui (x) for each i and A0G (x)
maxj2f1;2g fl1 (x) + l2 (x)g. Obviously, if xk 2 XnSD
for each k, then limk!1 A0Gb (xk ) = limk!1 A0G (xk )
A0G (x) = A0Gb (x). Consider a
sequence fxk g lying in SD .
If there are i; j 2 f1; 2g such that j 2 Ji (xk ) and J i (xk ) = f?g for in…nitely
many k, then, for such k’s, A0Gb (xk ) = l1j (xk )+l2j (xk ), and, therefore, limk!1 A0Gb (xk )
19
A0Gb (x). If there are i; j 2 f1; 2g such that j 2 Ji (xk ) and j 2 J i (xk ) for in…nitely
P
many k’s, then, for such k’s, A0Gb (xk ) = t2f1;2g (l1t (xk ) + l2t (xk )) u1 (xk ) u2 (xk ).
Since u1 (x) + u2 (x)
maxt2f1;2g fl1t (x) + l2t (x)g and the restriction of A0G to S is
A0G (x) =
mint2f1;2g fl1t (x) + l2t (x)g
continuous, we have that limk!1 A0Gb (xk )
A0Gb (x). Thus, we have shown that A0Gb is upper semicontinuous on X.
b has a mixed strategy Nash equilibrium, b = (b1 ; b2 ).
By Lemma 7, G
Let us show that b(SD ) = 0. Assume, by way of contradiction, that b(SD ) > 0.
It is not di¢ cult to see that b(SD ) > 0 i¤ b(x) > 0 for some x = (z; z) 2 SD
(see the Appendix, Lemma 10). Then bi (z) > 0 for each i 2 f1; 2g. By (iii),
there exist i; j 2 f1; 2g and some sequence f(xki ; z)g S j converging to x such that
bi (x). Let, without loss of generality, j = i. By Lemma 8, u
bi
limn!1 lij (xki ; z) > u
is strongly uniformly hospitable at z. That is, for every " > 0, there exists (") > 0
such that z + (") 2 [0; 1] and, for every di 2 (z; z + (")), u
bi (di ; x i ) > u
bi (z; x i ) "
for all x i 2 X i . It is worth noticing that (") is positive since j = i.
bi ( z ; b i ) = max 24(X ) U
bi ( i ; b i ). On the other hand,
Since bi (z) > 0, U
i
i
R
bi ( z ; b i ) = b i (z)ui (z; z)+
U
u
b (z; x i )db i . Fix some " 2 (0; 12 b i (z)(li i (z; z)
[0;1]nfzg i
u
bi (z; z))). Pick some di 2 (z; z + (")) such that u
bi (di ; z) = li i (di ; z) > 21 (li i (z; z) +
bi ( d ; b i ) U
bi ( z ; b i ) > b i (z)(b
u
bi (z; z)). Then U
ui (di ; z) u
bi (z; z)) " > 0, a
i
contradiction.
bi (b) = Ui (b), i = 1; 2. Since b is a mixed strategy equilibrium of
Therefore, U
b and, by construction, U
bi ( ) Ui ( ) for each i 2 f1; 2g and every 2 4(X), we
G
conclude that b is also a mixed strategy Nash equilibrium of G.
P
Remark 1 Since u
b1 (x) + u
b2 (x) = t2f1;2g (l1t (x) + l2t (x)) u1 (x) u2 (x) in some
cases, assuming that the restriction of A0G to S is only upper semicontinuous would
not be su¢ cient for the purposes of Theorem 3 even if assumption (ii) of Theorem 3
were replaced with the stronger assumption (iv) of Theorem 5.1 of Bagh (2010). On
the other hand, it is useful to notice that the restrictions of u
b1 and u
b2 to S need not
be reciprocally upper semicontinuous (see Bagh, 2010, p. 1265, footnote 24, for the
de…nition) if the restrictions of u1 and u2 to S are reciprocally upper semicontinuous.
20
Example 4 demonstrates that Theorem 3 can be used in place of Theorem 5b to
replicate the equilibrium existence results obtained by Ball (1999).
Example 4 Consider the following spatial voting model (see Ball, 1999, Example
1). Two candidates are competing in an election for public o¢ ce. The electorate
is distributed uniformly along the ideological spectrum [0; 1]. During the electoral
campaign, each candidate i announces, simultaneously with the other candidate, a
platform, denoted by xi . The probability Pi (xi ; x i ) that candidate i candidate wins
the election is de…ned as follows:
8 x +x
i
i
>
for 0 xi < x i 1;
<
2
1
Pi (xi ; x i ) =
for 0 xi = x i 1;
2
>
:
i
1 xi +x
for 0 x i < xi 1:
2
As in Ball (1999, p. 541), candidates 1 and 2’s policy preferences on [0; 1] are
represented by h1 (z) = 21 (z 1)2 and h2 (z) = 21 z 2 . The candidates are assumed
to be o¢ ce-motivated. Let the candidates’o¢ ce motivation parameters be k1 = :05
and k2 = 3, respectively. Then candidate i’s payo¤ function is
ui (xi ; x i ) = Pi (xi ; x i )(hi (xi ) + ki ) + (1
Pi (xi ; x i ))hi (x i ):
The game has no pure strategy Nash equilibrium.
It is strongly uniformly hospitable since SD = Snf( 12 ; 12 )g and the payo¤ functions
are continuous at ( 21 ; 12 ). Obviously, the restriction of u1 + u2 to the main diagonal
of the unit square is a continuous function. By Theorem 3, the game has a mixed
strategy Nash equilibrium.
Appendix
The Ky Fan Minimax Inequality
In Theorem 1 we use the Ky Fan minimax inequality in the following, slightly generalized form. (see, e.g., Kalmoun, 2001, Theorem 3.6; Balaj and Muresan, 2005,
Theorem 9 for more general results).
21
Lemma 9 Let X be a compact convex set in a Hausdor¤ topological vector space,
and let f : X X
R satisfy:
(i) f (x; x) 0 for each x 2 X;
(i) f ( ; y) is quasiconcave for each y 2 X,
(ii) f is 0-transfer lower semicontinuous in y.
Then there exists y 2 X such that f (x; y) 0 for all x 2 X.
Lemma 9 can be shown in a number of ways. Its conventional proofs are based
either on the KKM lemma or on Browder’s …xed point theorem, which are two
equivalent results (see, for an in-depth discussion, Yannelis 1991). Let us give an
outline of the proof using Browder’s …xed point theorem. It proceeds by assuming, to
the contrary, that, for each y 2 X, there exists x 2 X such that f (x; y) > 0. Then the
correspondence M : X
X de…ned by M (y) = fx 2 X : f (x; y) > 0g has nonempty
values. The quasiconcavity of f in x implies that M has convex values. Since f is
0-transfer lower semicontinuous in y, M has a multivalued selection with open lower
sections (see, e.g., Prokopovych, 2011 for details), denoted by M0 : X
X. Then,
by Lemma 5.1 of Yannelis and Prabhakar (1983), the convex-valued correspondence
M0 : X
X de…ned by M 0 (x) = coM0 (x) also has open lower sections. Therefore,
by Browder’s …xed point theorem, the selection has a …xed point, which contradicts
(i).
Proof of Theorem 2
Suppose
= ( 1 ; : : : ; n ) 2 4(X)nE . Then there exists
2 4(X) such that
F ( ; ) > 0. Since is a vector of probability measures, there exists d = (d1 ; : : : ; dn ) 2
X such that F ( d ; ) > 0, where d = ( d1 ; : : : ; dn ) is the vector of Dirac measures
concentrated at d1 ; : : : ; dn . With some abuse of notation, we will write F (d; )
in place of F ( d ; ). Put " = F (d; ) and denote liminf w!x FG (d; w) by F G (d; x).
Since G is uniformly diagonally secure, there is d 2 X such that F G (d; x) > FG (d; x)
R
"
"
for
all
x
2
X.
Therefore
F
(d;
)
>
F
(d;
)
,
where
F
(d;
)
=
F G (d; x)d .
2
2
X
The lower semicontinuity of F G in x implies that F (d; ) : 4(X) ! R is lower
semicontinuous (see Aliprantis and Border, 2006, Theorem 15.5). Consequently,
22
F (d; 0 ) > F (d; ) "2 > 0 for all 0 in some open neighborhood N4(X) ( ) of .
Since FG (d; x) F G (d; x) for all x 2 X, we conclude that F (d; 0 ) F (d; 0 ) > 0
for all 0 2 N4(X) ( ), which means that is diagonally transfer continuous.
Proof of Lemma 5
We have to show that for each i 2 I, every xi 2 Xi , and every " > 0, there is
a hospitable strategy i 2 4(Xi ) such that Ui ( i ; x i ) > ui (xi ; x i ) " for every
x i 2 X i.
Fix some i 2 I, xi 2 Xi , and " > 0. Let i denote the uniform probability measure on Bi+ (xi ; "). By de…nition, i (C) = (B +i (C)
for every Borel subi
i (xi ;"))
set C of Bi+ (xi ; "), therefore, i is absolutely continuous with respect to i . Since
Ui ( i ; x i ) > ui (xi ; x i ) " for every x i 2 X i , the only fact left to be proven is
that i is a hospitable strategy.
Consider some x i 2 X i and a sequence of fxk i g, xk i 2 X i , converging to it.
Then ui (xi ; xk i ) tends to ui (xi ; x i ) for i -almost every xi 2 Xi . By the absolute
continuity of i , we conclude that ui (xi ; xk i ) converges to ui (xi ; x i ) for i -almost
every xi 2 Xi . Then, by Lebesgue’s dominated convergence theorem, Ui ( i ; xk i )
tends to Ui ( i ; x i ).
An Auxiliary Lemma
Lemma 10 Let X = [0; 1] [0; 1], and i 2 4([0; 1]), i = 1; 2, and let D be a Borel
subset of [0; 1]. Then the set D = f(z; z) 2 [0; 1] [0; 1] : z 2 Dg has measure 0 for
the product measure = 1
(x) = 0 for every x 2 D.
2 i¤
Proof. Assume that (x) = 0 for every x 2 D. We have to show that (D) = 0.
First consider the case where, for some i 2 f1; 2g, i (x) = 0 for all x 2 D. Fix
some arbitrary " > 0. Since i has no atoms on D, there exists a …nite disjoint
m
collection fDk gm
". Then
k=1 of subsets of D such that D = [k=1 Dk and i (Dk )
P
P
(D) = k (Dk Dk ) = k i (Dk ) i (Dk )
" i (D)
", which implies that
(D) = 0.
23
Let both 1 and 2 have atoms on D. Each i , as a …nite measure, can have
only a countable number of atoms. Denote by Ci the set of atoms of i on D. Since
(x) = 0 for every x 2 D, C1 \ C2 = ?.
Fix some arbitrary " > 0. Since C1 \ C2 = ?, it is possible to associate with
every x 2 C1 a Borel set D(x)
[0; 1] containing x such that 1 (D(x))
2 1 (x)
"
and 2 (D(x)) 2 . Let D1 = DnC1 . The set D1 is Borel, and 1 is nonatomic on it.
Denote by C 1 (D1 ) the set f(z; z) 2 [0; 1] [0; 1] : z 2 C1 (z 2 D1 )g. It is clear that C 1
is countable and (D1 ) = 0. Then, since D = C 1 [D1 and C 1 \D1 = ?, we have that
P
P
"
(D) = (C 1 ) + (D1 ) = (C 1 )
":
x2C1 2 1 (x) 2
x2C1 1 (D(x)) 2 (D(x))
Example 1: Details
Let us show that the game described in Example 1 with V = 2 and r = 3 is
uniformly diagonally secure. Let d 2 X and, without loss of generality, " 2 (0; 1).
De…ne d = (d1 ; d2 ) as follows:
di =
(
di if di > 0;
if di = 0;
"
4
for i = 1; 2:
r
For any x 2 (0; 1] denote by y(x) 2 (0; x) the root of the equation xrx+yr V =
"
V
. Such y(x) exists and is unique since the function f : [0; x] ! R de…ned by
4
r
f (y) = xrx+yr V is strictly decreasing and continuous, and, moreover, f (0) = V and
f (x) = V2 . We will often use the fact that y(x) < x for every x 2 (0; 1].
There are a number of cases to consider.
Case 1. Let d = (0; 0). Then d1 = d2 = 4" . We shall show that for every x 2 X,
there exists a neighborhood NX (x) of x in X such that FG (d; w) > FG (d; x) " for
all w 2 NX (x):
For x = (0; 0), put NX (x) = BX (x; 4" ). Since FG (d; x) = 0, we have to show that
FG (d; w) > " for all w 2 NX (x).
"
"
If w = (0; 0), then FG (d; w) = AG (d; w) A0G (w) = (V
) V2 = V2
> ":
2
2
24
If w = (w1 ; 0) with w1 > 0, then, since w1 < 4" , FG (d; w) > (V
".
"
If w = (w1 ; w2 ) with w1 w2 > 0, then FG (d; w) > (V
) (V
2
Let x = (x1 ; 0) with x1 2 (0; V ]. Then FG (d; x) = V4 ( V2
BX (x; 41 minfx1 ; y( 4" )g). Consider w 2 NX (x).
"
"
If w = (w1 ; 0), then FG (d; w) > ( V2
) ( V2 w1 ).
4
4
If w = (w1 ; w2 ) with w1 w2 > 0, then FG (d; w) > (V 4" d1
It is clear that FG (d; w) FG (d; x) > " for all w 2 NX (x).
"
)
2
w1
( V2
w1 ) >
w2 ) >
":
x1 ). Put NX (x) =
d2 ) (V
w1
w2 ).
For x = (x1 ; x2 ) 2 (0; V ] (0; V ], FG (d; x) = A0G (x) = (V
x1 x2 ). Put
"
1
NX (x) = BX (x; 4 minfx1 ; x2 ; y( 4 )g). For w 2 NX (x) we have FG (d; w) = AG (d; w)
(V
w1 w2 ) > ( d1 d2 ) (V
w1 w2 ). Therefore, FG (d; w) FG (d; x) >
d1 d2 + w1 + w2 x1 x2 > ".
Case 2. Let d = (d1 ; 0) with d1 2 (0; V ].
V
For x = (0; 0), FG (d; x) = ( V2
d1 + V4 )
= V4
d1 . Put NX (x) =
2
BX (x; 41 minf"; y(d1 ); y( 4" )g) and consider w 2 NX (x).
If w = (0; 0), then FG (d; w) = (V d1 4" ) V2 = V2 d1 4" .
"
"
If w = (w1 ; 0) and w1 > 0, then FG (d; w) > ( V2 d1 + V
) ( V2 w1 ) =
4
4
V d1 2" + w1 .
"
"
If w = (0; w2 ) and w2 > 0, then FG (d; w) > (V
d1 + V2
) ( V2 w2 ) =
4
4
V d1 2" + w2 .
If w = (w1 ; w2 ) with w1 w2 > 0, then FG (d; w) > (V 4" d1 + V 4" 4" ) (V
+ w1 + w2 .
w1 w2 ) = V d1 3"
4
Therefore, FG (d; w) FG (d; x) > " for all w 2 NX (x).
For x = (x1 ; 0) with x1 2 (0; V ], FG (d; x) = ( V2 d1 ) ( V2
BX (x; 41 minf"; x1 ; y(d1 )g) and consider w 2 NX (x).
If w = (w1 ; 0), then FG (d; w) > ( V2 d1 4" ) ( V2 w1 ).
If w = (w1 ; w2 ) with w1 w2 > 0, then FG (d; w) > (V 4" d1
Therefore, FG (d; w) FG (d; x) > " for all w 2 NX (x).
25
x1 ). Put NX (x) =
"
)
4
(V
w1
w2 ).
For x = (0; x2 ) with x2 2 (0; V ], FG (d; x) = (u1 (d1 ; x2 )+ V4 ) ( V2 x2 ). Since u1 is
continuous at (d1 ; x2 ), there exists 0 < < x22 such that u1 (d1 ; w2 ) u1 (d1 ; x2 ) > 4"
for any w2 2 (x2
; x2 + ) \ [0; V ]. Put NX (x) = BX (x; 14 minf ; x2 ; y( 4" )g) and
consider w 2 NX (x).
"
) ( V2 w2 ).
If w = (0; w2 ), then FG (d; w) = (u1 (d1 ; w2 ) + V2
4
If w = (w1 ; w2 ) with w1 w2 > 0, then FG (d; w) > (u1 (d1 ; w2 ) + V 4" 4" ) (V
w1 w2 ).
It is obvious that FG (d; w) FG (d; x) > " for all w 2 NX (x).
For x = (x1 ; x2 ) with x1 x2 > 0, FG (d; x) = u1 (d1 ; x2 ) (V x1 x2 ). There exists
0 < < x22 such that u1 (d1 ; w2 ) u1 (d1 ; x2 ) > 4" for any w2 2 (x2
; x2 + )\[0; V ].
1
Put NX (x) = BX (x; 4 minf"; ; x1 ; x2 g) and consider w 2 NX (x). Then FG (d; w) >
(u1 (d1 ; w2 ) 4" ) (V w1 w2 ), and, therefore, FG (d; w) FG (d; x) > ".
Case 3. Let d = (d1 ; d2 ) 2 (0; V ]
(0; V ].
For x = (0; 0), FG (d; x) = V2 d1 d2 . Put NX (x) = BX (x; 14 minf"; y(d1 ); y(d2 )g)
and consider w 2 NX (x).
If w = (0; 0), then FG (d; w) = FG (d; x).
If w = (w1 ; 0), then FG (d; w) > ( V2 d1 + V 4" d2 ) ( V2 w1 ), and, therefore,
FG (d; w) FG (d; x) > ".
If w = (w1 ; w2 ) with w1 w2 > 0, then FG (d; w) > (V 4" d1 + V 4" d2 ) (V
w1 w2 ).
Therefore, FG (d; w) FG (d; x) > " for all w 2 NX (x).
For x = (x1 ; 0) with x1 2 (0; V ], FG (d; x) = ( V2 d1 + u2 (x1 ; d2 )) ( V2 x1 ). Pick
0 < < x21 such that u2 (w1 ; d2 ) u2 (x1 ; d2 ) > 4" for any w1 2 (x1
; x1 + )\[0; V ].
Put NX (x) = BX (x; 41 minf"; ; x1 ; y(d1 )g) and consider w 2 NX (x).
If w = (w1 ; 0), then FG (d; w) > ( V2 d1 + u2 (w1 ; d2 )) ( V2 w1 ).
If w = (w1 ; w2 ) with w1 w2 > 0, then FG (d; w) > (V 4" d1 + u2 (w1 ; d2 )) (V
w1 w2 ).
Therefore, FG (d; w) FG (d; x) > " for all w 2 NX (x).
For x = (x1 ; x2 ) 2 (0; V ] (0; V ], FG (d; x) = u1 (d1 ; x2 )+u2 (x1 ; d2 ) (V x1 x2 ).
Pick 0 < < 21 minfx1 ; x2 g such that, for each i 2 f1; 2g, ui (di ; w i ) ui (di ; x i ) >
26
"
4
for all w i 2 (x i
; x i + ) \ [0; V ]. Put NX (x) = BX (x; 41 minf"; g). For
w = (w1 ; w2 ) 2 NX (x) we have FG (d; w) = u1 (d1 ; w2 ) + u2 (w1 ; d2 ) (V w1 w2 ).
Therefore, FG (d; w) FG (d; x) > ".
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