Nash equilibrium in games with ordered outcomes Текст научной статьи по специальности «Математика»

Victor V. Rozen

Saratov State University, Astrakhanskaya St. 83, Saratov, 410012, Russia E-mail: Rozenvv@info.sgu.ru

Abstract. We study Nash equilibrium in games with ordered outcomes.

Given game with ordered outcomes, we can construct its mixed extension.

For it the preference relations of players are to be extended to the set of probability measures. In this work we use the canonical extension of an order to the set of probability measures.

It is shown that a finding of Nash equilibrium points in mixed extension of a game with ordered outcomes can be reduced to search so called balanced matrices, which was introduced by the author. The necessary condition for existence of Nash equilibrium points in mixed extension of a game with ordered outcomes is a presence of balanced submatrices for the matrix of its realization function. We construct a certain method for searching of all balanced submatrices of given matrix using the concept of extreme balanced matrix. Necessary and sufficient conditions for Nash equilibrium point in mixed extension of a game with ordered outcomes are given also.

Keywords: Game with ordered outcomes, Nash equilibrium, Mixed extension of a game with ordered outcomes, Balanced matrix, Extreme balanced matrix, Balanced collection.

1. Introduction

A game G with ordered outcomes is a game in which preferences of players by partial ordered relations are given (all preliminaries see in Rozen (2010)).

Definition 1. A situation x0 is called Nash equilibrium point in game G if for any player i and its strategy xi the condition

holds, where ^ is a preference relation of player i.

We construct a mixed extension of a finite game G with ordered outcomes in the following way. Mixed strategies of players in game G is defined as usually that is probability measures on the sets of their pure strategies. A realization function F of a game G can be extended to a mapping F of mixed situations of game G into the set A consisting of probability measures on the set outcomes A. Now we need to extend the preference relations of players on A. We define the canonical extension of order w C A2 to the set A by formula

where /x, v G A, C(u>) is the set of all isotonic functions from A to R under order u> and /(/x) = (/, /x) is the standard scalar product.

F(x0 || xi) 1 F(x0)

(1)

(2)

Remark 1. In evident form the condition ^ S v means that /i(B) ^ v(B) for any subset B C A which is majorant stable under order w (see Rozen (1976)). Recall

that a subset B C A is called majorant stable one if the conditions a G B and

W

a' ^ a imply a' G B.

W

Corollary 1 (convexity of the canonical extension). Conditions /ik S vk, where ak > 0 (k = 1,... ,p), ak = 1, imply

22ak^h ^/2 ak Vk. (3)

k=1 k=l

Moreover, if at least one of these conditions is strict then condition (3) is strict also.

In this article we consider games of two players with ordered outcomes in the form

G = (X,Y,A,wi,w2,F)

where X is a set of strategies of player 1, Y is a set of strategies of player 2, A is a set of outcomes, wk is an order relation on A which represents preferences of player k = 1, 2 and F: X x Y ^ A is a realization function.

According to a general definition a situation (x0, y0) G X xY is Nash equilibrium point in game G if for any x G X and y G Y conditions

Wi ^ W2 ___

F(x,y0) S F(x0,y0) ,F(x0,y) S F(x0,y0)

hold.

In section 3 it is shown that, for game G, the existence of Nash equilibrium points in mixed strategies is connected with so called balanced submatrices its realization function. We study some basic properties of balanced matrices in section 2. Theorem 1 shows a simple connection between balanced matrices and well known in game theory balanced collections. Note that up to now, a concept of balanced collection was used for systems of coalitions in cooperative games only (see, for example, Bondareva (1968), Shapley (1965), Peleg (1965)). The main result of our work is Theorem 4 which states a necessary and sufficient conditions for Nash equilibrium point in mixed extension of game G with ordered outcomes. To illustrate these results we consider some examples.

2. Balanced matrices and balanced collections of subsets

We consider an arbitrary matrix M of size m x n over a set A as a mapping F: I x J ^ A where I = {1,..., m} ,J = {1,..., n}. Put M = ||F (*, j)||.

We denote by Sm the standard simplex consisting of m-components vectors:

m ^

x = (xi, ..., Xm) : Xi > 0,£ = 1 > .

i=1 )

For any x G Sm, y G Sn and a G A put

F(x,y) (a) = xi ' yj . (4)

F (i,j) = a (i,j) G 1 x J

In particular for any i = 1,...,m and j = 1,...,n we have

F(x,j)(a) = x^ F(i,y)(a) = yj. (5)

F (i,j)=a F (i,j)=a

Remark 2. We identify any index i G I with unit vector ei = (0,..., 1,..., 0) and also for any index j G J.

Lemma 1 (linearity property). The function F is linear under all arguments.

Proof (of lemma 1). Verify a linearity under first argument. Put xk G Sm,y G Sn, ak > 0 (k = 1,...,p)^Y!k=1 ak = 1. We have

(P \ P

Y,akxk) yj = E I>kxk yj =

k=1 ' F (i,j)= a k=1

(i,j) G Ix J (i,j) G Ix J

p p p kk

■ — " = ' a F(x«

ak xiyj =Y1 ak xkyj =Y1 ak F(xk,y) (a)

k=1 F (i,j)= a k=1 F (i,j)= a k=1

(i,j) Є I x J (i,j) Є I x J

which was to be proved. □

Definition 2. A matrix IF (i, j)|| is called balanced one if there exists such a pair of vectors (x0, y0) G Sm x Sn with positive components that for any x G Sm ,y G Sn and a G A the equalities

F(x, y0) (a) = F(x0, y0) (a) , F(x0, y) (a) = F(x0, y0) (a) (6)

hold. In this case the pair (x0, y°) is called a balance pair of vectors.

Note that (4) means the equality of corresponding measures:

F(x, y0) F(x0 , y0), F(x0 , y) F(x0 , y0) .

Definition 3. A vector x0 G Sm with positive components is called a row balance vector for matrix IF (i,j)|| (i G I,j G J) if for any j1 ,j2 G J the equality F(xo,j1) = F(x0,j2) holds. In other words, in this case, the measure F(x0,j) does not depend on j G J .A column balance vector is defined dually that is a vector y0 G Sn with positive components such that the measure F(it y0) does not depend on i G I.

Lemma 2. A pair of vectors (x°,y°) is a balance pair for matrix ||F (i, j)|| if and only if x0 is a row balance vector and y° is a column balance vector.

Indeed let (x0, y0) be a balance pair of vectors. Setting in (4) y = j1 and then y = j2 we have F(x0, j1) = F(xo, y0), F(x0, j2) = F(xo, y0) hence x0 is a row balance vector. Dually, y0 is a column balance vector. Conversely let x0 be a row balance vector and y0 a column balance vector. For arbitrary a G A put p (a) = F(x0,j) (a) (j G J). Fix a vector y = (y1,..., yn) G Sn. By using a linearity for F (see Lemma 1), we get

F(x0, y) (a) = F(x0,£jEJ yj ej) (a) =U yj F(x0,j) (a) =

j jeJ

= 53 yjp(a)=p (a)Y^ yj=p(a),

jeJ jeJ

i.e. the measure F(x0,y) does not depend on y G Sn. Then, for any y G Sn we have

F(x0,y) = F{x0, y0) and dually F(x, y0) = F(x0, y0), hence (x0, y0) is a balance pair of vectors.

Corollary 2. A balanced matrix is homogeneous (that is, any its row and any column contains a same set of elements).

For the proof it is sufficiently to remark that the set of elements consisting i-th row of matrix ||F (i,j)|| coincides with spectrum of the measure F(iy0) which, in our case, does not depend on i G I.

Recall the concept of balanced collection.

Definition 4. Let E be an arbitrary set and (E1,..., Ep) be a covering system of its subsets. The collection (E1,...,Ep) is called balanced one if there exists a representative vector for it, i.e. a vector A = (A1,..., Xp) with positive components such that for any e G E the equality

^ Ak = 1

Ek ^e

holds.

Definition 5. Two balanced collections (E^,..., E~1) and (E1,..., Ep) are said to be collinearly balanced if for them there exist collinear representative vectors A1 and A2 (in notation A1!A2).

Given a matrix F: I x J ^ A, we define for arbitrary a G A two collections of characteristic subsets (U°’)ieI and (VJa)jeJ by setting

j G Ua & F (ij) = a, i G Va ^ F (ij) = a. (7)

Theorem 1. A matrix F: I xJ ^ A is balanced if and only if for any a,1, a2 G pr2F

1. Collections (Ua1) j and (Uf2 )i£j are collinearly balanced ones;

2. Collections (V/1)jeJ and (V/2)je J are collinearly balanced ones.

Lemma 3. A matrix F: I x J ^ A has a row balanced vector if and only if

a) for any a G pr2F, the collection of its characteristic subsets (Uia)ieI is balanced one;

b) for any a1,a2 G pr2F, collections (U^1 )iej and (U“2)i£j are collinearly balanced ones.

Proof (of lemma 3). Necessity. Let x = (x1,... ,xm) be a row balance vector. For arbitrary a G pr2F put Aa = xi/p (a) where p (a) = F(x,j) (a) (the right part of this equality does not depend on j G J, see Definition 3). Check that the vector

Aa = (Aa)i j is a representative one for the collection (U“)ieI. Fix j G J. By using

(5), we have

£*? = £ *? = £ ,-£) = ,T75 £ x‘ = 7i~f{a} = 1'

UaBj F(i,j) = a F(i,j)=a p ( ) p ( ) F(i,j)=a p ( )

Furthermore, for any a1,a2 G pr2F, we get

a1 a2 xi xi p (a2)

A,- : X: = —;—- : —;—- = —;—- = const under * G 1,

i i p (a1) p a) p (a1)

i.e. representative vectors Aa1 and Aa2 are collinear ones.

Sufficiency. Let Aa = (Aa)ieI be a representative vector for balanced collection (U“)i£l and moreover, for any a1,a2 G pr2F corresponding vectors are collinear, Aai || Aa2. Fix a G pr2F and put

Aa

Xi = V * \a’X=(Xi)ieI-l^i’el Ai'

Because representative vectors are collinear, right parts of these equalities do not depend on element a G A. It is evident that xi > 0 and J2ieI xi = 1. By using that vector Aa is a representative one, we have for any j G J

F t \ Aa "I2ua3j Aa 1

F(x,j)(a)= 2^ -----^=const'

F (i,j) = a Uf3j i'eI ^ 2—'i'eI i' Z—*1' eI i'

Thus, the measure F(x, j) does not depend on j G J, i.e. x is a row balance vector for matrix | F (i, j)| .

Now, using Lemma 2, we obtain the proof of Theorem 1 from Lemma 3 and its dual. □

For matrix M over a set A and a G A we denote by x (Ma) a matrix with elements ma4, where

maj = 1 if F (i,j) = a maj = 0 if F (i,j) = a.

Theorem 2. Suppose M = | F (i, j)| is a balanced matrix. Then the following conditions are equivalent:

1. Matrix M has an unique pair of balance vectors;

2. Matrix M is a square one and Det x (Ma) = 0 for any a G A;

3. Matrix M is a square one and Det x (Ma) = 0 for some a G A.

The proof of Theorem 2 is based on the following lemmas.

Lemma 4. Let E be an arbitrary finite set, \E| = m and (E1, ..., En) be a balanced collections of its subsets. Then the collection (E1,..., En) has an unique representative vector if and only if the collection of characteristic vectors (x (E1) ,...,x (En)) is linearly independent.

Proof (of lemma 4). A vector A = (A1,..., An) is a representative one for the collection (E1,..., En) if and only if it is a positive solution for the following system of linear equations:

(A1,...,An) xm = (1,...,1) , (8)

where by xM denoted n x m matrix whose i-th row x (Ei) is the characteristic function of the subset Ei C E (i = 1,...,n) and (1,..., 1) is m-component vector consisting of 1. Because the collection (E1,..., En) is balanced, (8) is a compatible system consisting of m linear equations under A1 ,...,An. It is well known that such a system has an unique solution if and only if r = n where r = rank xM, hence, in this case, the system of its rows is linearly independent.

Conversely, assume r < n. In this case, the solution of (8) can be represented in the form A* + L where A* is a partial solution of (8) and L is a general solution of a corresponding homogeneous system (note that L is a linear space with dimension n — r). Let A* be a representative vector for collection (E1, ...,En) . Then all its components are positive. Fix some l* G L with l* = 0. Then there exists such positive e > 0 that all components of vector A* + el* remain be positive. Hence (A* + el*) is some positive solution of (8) with A* + el* = A*, then a representative vector for the collection (E1,...,En) is not unique. □

Lemma 5. Let M be a matrix over set A. Then the following conditions are equivalent:

1) Matrix M has an unique row balance vector;

2) For any a G pr2M, the collection of characteristic vectors (x (U“) ,...,x (U m)) is linearly independent;

3) For some a G pr2M, the collection of characteristic vectors (x (U“) ,...,x (Ua)) is linearly independent.

Proof (of lemma 5). 1) ^ 2). Fix a G pr2M. According to Lemma 3 a), the system of characteristic subsets (U“,..., U) forms a balanced collection on the set J.

Let A = (A -\_,..., Am) and A' = (A1,..., Am) be two its representative vectors. Then vectors x = (x -\_,..., xm) and x' = (x1,..., x'm) defined by

Ai A'

J2i'el Ai' 1 2i'el Ai'

are row balance vectors for matrix M (see the proof of Lemma 3). According to assumption 1) we have x = x', i.e. for any i G I holds the equality

Ai A'

J2i'e_I Ai' 2i'el Ai'

hence Ai = kAi where k = const. Fix j0 G J. Since A and A' are representative

vectors, we get

1 = E Ai = E kAi = kY.Ai = k,

j0 eu a j0eu a j0 e Ua

hence k =1 and A' = A. Thus, the balanced collection (U“,..., Um) has an unique representative vector and, according with Lemma 4, the collection (x (U“) ,...,x (Um)) is linearly independent.

It is evident that 2) ^ 3). Prove 3) ^ 1). Assume that the collection of characteristic vectors (x (U“) ,...,x (Um)) is linearly independent for some a G pr2M. Let x' = (x1,..., x'm) and x'' = (x'/,..., x'm) be two row balance vectors for matrix M. Then vectors A' and A'' defined by A/ = xi/p (a), A'/ = xi'/p (a) (i G I) are representative vectors for balanced collection (U“,..., Um) (see the proof of Lemma 3). Because the collection (x (U“) ,...,x (Um)) is linearly independent it follows from Lemma 4 that A' = A'' hence x' = x'' which was to be proved. □

Let us prove Theorem 2. Assume that matrix M has an unique pair of balance vectors. Then, according with Lemma 2, matrix M has an unique row balance vector. Fix an element a G pr2M and consider the matrix x (Ma) defined above. Since rows of matrix x (Ma) are vectors (x (U“), ...,x(Um)) and according to Lemma 5, we get that the collection of these vectors is linearly independent, then m = r where r = rank x (Ma) . Dually the equality n = r holds, hence we obtain m = n and Det x (Ma) = 0 which completes the proof of Theorem 2.

Now we consider a problem for finding of all balanced submatrix for given matrix. Given a matrix ||F (i, j)|| (i G I,j G J), fix two subsets I0 C I,J0 C J. We denote by B1 a convex polyhedron which consists of all vectors x G Sm satisfying the condition

(Vj 1,j2 G J0) F(x,j1) = F(x,j2). (9)

Dually is defined a convex polyhedron Bjo which consists of all vectors y G Sn satisfying the condition

(Vi 1,i2 G I0) F(iuy) = F(i2,y). (10)

Definition 6. A submatrix F (I0 x J0) = HF (i, j)|| (i G I0,j G J0) is called an extreme balanced submatrix if it is balanced one and then there exists a balance pair of vectors for it which is an extreme point in the convex polyhedron B1 x Bjo.

For arbitrary polyhedron P, we denote by Exf P the set of extreme point in P. It is easy to show that Ext {BlJo x Bj0) = Ext (Bj0) x Ext (Bjo) .

Lemma 6. A vector x0 G B^ is an extreme point in B^ if and only if it is minimal under spectrum, i.e. for any x G B10 the condition

Spx C Spx° =^ x = x° (11)

holds (we denote by Spx the spectrum of vector x : Spx = {i G I : xi =0}).

Proof (of lemma 6). Let x0 be an extreme point in B 1o. Consider a vector x G B 1o satisfying the condition Spx C Spx0. Assume that x = x0. Because

E xi = E x0 = 1

ieSp x ieSp x0

then between non-zero components of vector x there exists such xi' that 0 < x0' < xi' hence

x0

0 < e = min — < 1. (12)

ieSp x xi

It follows from (13) that \exi\ < x0 for all i G Sp x. Hence x0 ± exi > 0 i.e. all components of both vectors x° ± ex are non-negative. Put

1 ^0 | ex 2 ex

x " ; i x " •

1 + e 1 — e

Check that x1,x2 G Bj0. We have

1 xi +£Xi ^ 0 , e ^ , e 1

/ xi = / ~^---------------------- / xi + i------- / xi = ^-----------------------------H i-= 1

i 1 + e 1 + e^ i 1 + e^ 1 + e 1 + e

iel iel iel iel

E2 V- x'i — exi 1 0 e 1 e 1

xf = > —------- = - > x,u----> Xi =------------------------= 1.

* ^ 1 — £ l-£^ 1 l-£^ 1 — £ 1 — £

i£l i£l i£l i£l

Because x0, x G Bj0 and by using the linearity of F (see Lemma 1), we get for any jl,j2 G J0 ■

~ ~ 1 ~ e ~

Fix\h) = = —£f(x°^) + YT^~F{x’jl) =

1 ~ e ~ ~ ~

= Y^F<-x°’^ + ~\TeF{-x'j2) = F(wix0+Tfix’i2) = F(xl'j2')

We obtain the equality F^xi,j1) = ~xij2)\ the equality F(x2,j1) = ~{x2,j2) can be shown analogously. Thus, x1, x2 G BjQ. Since -^rx1 + ^-=^x2 = x° and + = 1

it follows from the definition of extreme point that x1 = x2. Then we get

o _ 1 + £ i 1 — e 2 _ 1 £ l , 1 — e l _ l

x - 2 X 2 X ““ 2 X 2 X ““ X ’

hence x^rE* = x° and x = x°. The last equality is in contradiction with our assumption which completes the proof of necessary condition.

Let us verify the sufficient condition. Assume that for vector x0 the condition (12) holds. Let x1 and x2 be two vectors such that x0 = a1 x1 + a2x2 where x1, x2 G

Bj0, a1,a2 > 0, a1 + a2 = 1. Then Sp x1 C Sp x0,Sp x2 C Sp x0, and according

with (12) we obtain x1 = x0 and x2 = x0 hence x1 = x2. Thus x0 is an extreme

point in B 1Jo which was to be proved. □

Corollary 3. Let M be an extreme balanced submatrix. Then

1. Submatrix M has an unique row balance vector;

2. Submatrix M has an unique collumn balance vector;

3. Submatrix M is a square one;

4. Det x (Ma) = 0 for arbitrary a which belongs to M.

Proof (of corollary 3). Let (x°, y°) be a balance pair of vectors for submatrix M = F (I0 x J0) provided x0 G ExtB1o and y° G ExtB2Io. Consider a row balance vector

x for F (I0 x J0), i.e. x G Bjo, Sp x = I0. Because Sp x = Sp x0, it follows from

Lemma (6) that x = x0. Thus, the assertion 1) is proved. Dually, the assertion 2) is true. Both assertions 3) and 4) follow from Theorem 2. □

The main result of this section states

Theorem 3. Let ||F (i, j) H (i G I,j G J) be an arbitrary matrix over A, I0 C I, J0 C J and M = F (I0 x J0) be its submatrix. A submatrix M is an extreme balance submatrix if and only if for any a, a1, a2 G pr2M the following conditions hold:

1) Submatrix M is a square one;

2) Det x (Ma) = 0;

3) All components of vector Aa = (1,..., 1) [x (Ma)] 1 are positive;

4) Vectors Aai and Xa2 are collinear;

5) All components of vector Sa = [x (Ma)] 1 (1,..., 1)T are positive;

6) Vectors Sai and Sa2 are collinear.

Proof (of Theorem 3). Necessity of conditions 1) and 2) is shown in Corollary 3. Put \I0\ = \J0\ = k. According to Theorem 1 for balanced submatrix M, the collection of characteristic subsets (U°')ieIo is balanced hence the following system of linear equations

(Aa,,...,Aak) x (Ma) = (1,...,1) (13)

has a positive solution for any a G pr2M. Setting Aa = (Aa,. ..,Ak) we obtain from (14) Aa = (1,..., 1) [x (Ma)] 1 and 3) is shown. Assertion 4) follows from Theorem 1. Dually we get 5) and 6).

Sufficiency. Assume the conditions 1)-6) hold. From 3) we obtain the equality

Aa x (M a ) = (1,...,1)

i.e. Aa is a representative vector for collection of characteristic subsets (Ua)ieI . It follows from 4) that collections (Ua1 )ieI and (U®2 )ieIo are collineary balanced ones. According with Lemma 3, a submatrix M has a row balance vector x0 and according to Lemma 5 this vector is an unique. Dually, submatrix M has an unique column balance vector y0. Because vector x0 is minimal under spectrum, from Lemma 6 we obtain that x° is an extreme point in B1o and dually y° is an extreme point in B^ . Thus, (x°,ya) is an extreme point in B 1o x B2o, i.e. submatrix M is an extreme balanced submatrix which completes the prove of Theorem 3. □

Combining Theorem 3 with Krein-Millman Theorem, we get a certain method for searching of all balanced submatrices of given matrix IF (i, j)|| (i G I,j G J).

3. A finding of Nash equilibrium points in mixed extension of games with ordered outcomes

Our method for finding of Nash equilibrium points is based on searching of balanced submatrices of realization function. Consider a game G of two players with ordered outcomes in the form

G = {I,J,A,^1,^2) , (14)

where I = {1,..., m} is a set of strategies of player 1, J = {1,...,n} is a set of strategies of player 2, A is a set of outcomes, is an order relation which represents preferences of player k =1, 2,F: I x J ^ A a realization function.

Lemma 7. A situation (x0,y0) in mixed strategies is Nash equilibrium point in mixed extension of game G if and only if for any i G I and j G J the following conditions hold:

~ wi ~ ~ ^2 ~

F(i,yo) = F(xo,yo) ,F(xo,j) S F(xo,yo).

Proof (of lemma 7). The necessity is obvious. Conversely, put x = (x1,..., xm) G Sm.

~ Wi _

Summarizing the conditions Ip(i.yo) S F(xo.yo) with weights xi and using the

~ Wi _

linearity property (see Lemma 1) we obtain F(x.yo) S F(xo.yo). The condition

W2 „

i'1(xo,y) S F(xo. yo) is proved analogously. □

Lemma 8. Let (x0,y0) be Nash equilibrium point in mixed extension of game G. Then the matrix which is a restriction of matrix of outcomes ||F (i, j)|| under the pair of spectra (Spx0,Spy0) is balanced one.

~ Wi _

Proof (of lemma 8). For any i G Sp x0 we have F(i^yo) S F(xo.yo). Assume that at least one of these conditions is strict. Then by using the convexity property (see Corollary 1) we get

£ xh.?) < £ xiF(x° .yo). (15)

iESp xo iESp xo

Using the linearity property (Lemma 1) we obtain from (16) inequality

~ Wi ~ ~ ~

F(xo.yo) < F(xo.yo) which is false. Thus, the equality Ip(i.yo) = F(xo.yo) holds for

any i G Spx0, i.e. y° is a column balance vector for submatrix F (Sp x0 x Spy0') . Analogously we can prove that x0 is a row balance vector for this submatrix. According to Lemma 2, (x0, y°) is a balance pair for submatrix F (Sp x0 x Sp y°) . □

Corollary 4. Suppose that there exists a Nash equilibrium point in mixed extension of game G. Then the matrix of outcomes of game G contains a balanced submatrix.

Now we state the main result of our work.

Theorem 4. A situation in mixed strategies (x°,y°) is Nash equilibrium point in mixed extension of game G if and only if the following conditions hoid:

1. The restriction of matrix of outcomes under the pair of spectra (Sp x0, Sp y°) is balanced matrix;

~ Wi „

2. F(i. yo) S F(xo. yo) for all i G Spx0;

~ W2 „

3. F(xo.j) S F(xo.yo) for all j G Spy0.

The proof of Theorem 4 follows directly from Lemma 7 and Lemma 8.

Corollary 5. A game G have a quite Nash equilibrium point in mixed strategies if and only if its matrix of outcomes is balanced. In this case, any pair of balance vector (x°, y°) is Nash equilibrium point with Sp x0 = I and Sp y0 = J.

Let us note that the last condition is defined only by realization function of game G and does not depend on order relations , k =1, 2.

Remark one special case of Theorem 4.

Corollary 6. Suppose I0 C I, J0 C J are two subsets and corresponding submatrix F (I0 x J0) is balanced. Assume for any i' G I0,j' G J0 there exist i0 G I0,j0 G J0 such that for all i G I0 and j G J0 hold the conditions:

F (i',j) S F ii0,j) ,

W (16)

F (i, j') S F(i,f)

hold. Then any balance pair for submatrix (xP,y°) , complemented by zero components for i G I \ I0 and j G J \ J0, is Nash equilibrium point in mixed extension of game G.

Appendix

1. Balanced matrices

Examples of balanced matrices together with their balance vectors by Table 1-3 are given.

Table 1.

I J 1 1 1

3 3 3

1 3 a b c

1 3 b c a

1 3 c a b

Table 2.

1—1 1 3 1 6 1 3 1 6

1 6 a a b b

1 3 a b b a

1 6 b b a a

1 3 b a a b

Remark that any Latin square forms a balanced matrix and moreover its balance pair is a pair of vectors whose components coincide (see Table 1).

To verify that a given pair of vectors (x0,y°) is a balance one for matrix

IF (i,j)II (i G I,j G J) we can check equalities F(ii^yo) = F(i2.yo) and F'(xo.ji) = F(xo.j2) for any i1, i2 G I and j1,j2 G J.

2. A finding of extreme balanced matrices

Consider a matrix of size 4 x 4 over set A = {a,b,c, d, e} given by Table 4

(a a b aba baa

Table 3.

1—1 1 6 i 6 i 6 1 6 1 6 1 6

1 6 a b c b a c

1 6 c b c a b a

1 3 b a a c c b

1 6 c c b a b a

1 6 a c b b a c

Table 4.

1—1 1 2 3 4

1 a a b a

2 a b a b

3 b a a e

4 c a d e

Let us verify that its submatrix M defined by the pair (I0, Jo), where I0 = {1, 2, 3} and J0 = {1, 2, 3} is an extreme balanced submatrix. We have

/1 1 0\

X (M“)= I 10 1 I \0 1 1/

and

We compute Det (x (M“)) = —2 and Det (x (M6)) = —1. Then we find inverse matrices:

(X (M“))-

( \

1 1 2 2

1 1 2 2

and

(x(m 6))-

'00 r 0 1 0 100,

1

1

We now compute

Aa = (1,1,1)

( \

1 1 1 2’ 2’ 2

0 0 1

A6 =(1,1,1) I 0 10 I =(1,1,1).

V100/

Thus, all components of vectors Aa and A6 are positive and Aa||A6.

Also

/II _I\

' 2 2 2 '

£a =

1 I

2 2

1 I

2 2

(m Л)т=[\\\

56 =

'00 1 G 1 G 1 G 0,

(1,1, lf = (1,1, lf .

We see that all components of vectors 5a and 56 are positive and 5a ||56. It is shown that all assumptions of Theorem 3 are satisfied, consequently a submatrix M is an extreme balance one. To find a balance pair of vectors for this submatrix, we use Theorem 1. We have xi = ^,x2 = ^,x% = \,yi = = 5,2/3 = Thus

the balance pair for submatrix M is x° = (^, , y° = (|, .

3. A finding of Nash equilibrium points in mixed strategies

Consider an antagonistic game with ordered outcomes given by realization function F (see Table 5) and order relation w for player 1 (see Diagram 1); an order relation for player 2 is defined as inverse order relation w-1.

Table 5.

F 1 2 з 4

1 a a b a

2 a b a b

3 b a a e

4 c a d e

It is shown above that submatrix M of matrix given by Table 5 is balanced and its balance pair is x° = (^, and y° = (^, (see Appendix 2).

Because hold the following conditions:

WWW

c ^ b, a ^ a, d ^ a;

WWW

a ^ a, b ^ b, e ^ a,

e

Fig. 1. Diagram 1

then by using Corollary 6, we get that the pair of vectors ((j, j, 0) , (j, j, g, 0))

is Nash equilibrium point in mixed extension of considered game.

References

Rozen, V. V. (2010). Equilibrium points in Games with ordered outcomes. Contributions to game theory and management. Vol.III. Collected papers on the Third International Conference Game Theory and Management /Editors Leon A. Petrosyan, Nikolay A. Zenkevich.- SPb.: Graduate School of Management SPbU, pp. 368-386.

Rozen, V. V. (1976). Mixed extensions of games with ordered outcomes (in Russian). Journal Vych. Matem. i Matem. Phis., 6, pp. 1436-1450.

Rozen, V. V. and Pankratova, J. N. (2000). Equilibrium points and balanced collections in games with ordered outcomes (in Russian). / Mathematic. Mechanic, 2, Saratov State univ., pp. 105-108.

Bondareva, O. N. (1968). Several applications of linear programming methods to the theory of cooperative games. Selected Russian papers in game theory. Princeton: Princeton univ. press, pp. 79-114.

Shapley, L. S. (1965). On Balanced Sets and Cores. RM-4601-PR. The Rand Corporation, 24 p.

Peleg, B. (1965). An inductive method for constructing minimal balanced collections of finite sets. Naval Res. Logist. Quart. 12, pp.155-162.