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MAST30022 Decision MakingProof of Nashs Theorem (solution)

Definition 1. A subset A ofRk is called

boundedif there exists r >0 such that x r for any x A.

closed ifRk\Ais open (i.e. for any x0 Rk\A, there exists >0 such that

x Rk :x x0<

is contained in Rk \ A). Intuitively A is closed if it contains all its boundary points.

compact if it is both bounded and closed.

convex if for any two points x1, x2 A, the line segment between them, namely {x1+ (1 )x2: 0 1},is contained in A.

Theorem 1 (John Nash 1951). Any non-cooperative 2-person game (zero-sum or non-zer-sum) with afinite number of pure strategies for each player has at least one equilibrium pair.

A proof of Nashs theorem requires the following well-known result.

Theorem 2. (Brouwers Fixed Point Theorem) LetC= be a compact convex set ofRk. Letf :C C

be a continuous function. Then there exists a point x

C such thatf(x

) = x

.

Such a pointx is called a fixed point.

Proof of Nashs Theorem. For any x X and y Y, define

ci= max{Ai yT xAyT, 0}, 1 i m,

dj = max{xBj xByT, 0}, 1 j n,

xi= xi+ci

1 +m

k=1ck, 1 i m,

yj = yj +dj

1 +n

k=1dk, 1 j n.

Let x = (x1, . . . , x

m) and y = (y1, . . . , yn). Then x

X and y Y.

Remark 1. ci can be interpreted as the regret function of player I in playing x instead of his purestrategyai. Similarly, dj can be interpreted as the regret function of player II in playing y instead ofher pure strategy Aj .

The function f(x, y) = (x, y) is a continuous function from X Y Rm+n to X Y.

Claim: (x, y) = (x, y) (i.e. (x, y) is a fixed point off) (x, y) is in equilibrium.

(x, y) is in equilibrium x X, xAyT xAyT. In particular, taking x = (0, . . . ,(i)

1 , . . . , 0), wehave xA = Ai, and so Ai y

T xAyT, implying ci = 0 for all i. Similarly, dj = 0 for all j. Hence,xi= xi and y

j =yj, i.e. (x, y) = (x, y).

Let (x, y) = (x, y) and suppose (x, y) is not an equilibrium pair. Then, either

xX s.t. xAyT >xAyT (1)

ory Y s.t. xByT >xByT. (2)

Consider case (1). Since

x= x1(1, 0, . . . , 0) +. . .+ xi(0, . . . ,(i)

1 , . . . , 0) +. . .+ xm(0, . . . , 1),

we havexAyT = x

1A1 y

T +. . .+ xmAm yT.

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So

xAyT >xAyT Ai yT >xAyT for somei

ci > 0 for somei

m

k=1

ck >0.

Now,

xAyT =x1A1 yT +. . .+xmAm y

T

i s.t. Ai yT xAyT and xi >0

ci = 0

xi =xi/(1 +

k

ck)< xi

x =x

(x, y)= (x, y),

which leads to a contradiction. Therefore (x, y) is an equilibrium pair. The reasoning is similar in case(2). The claim is then proved.

Note that X Y is a compact convex set ofRm+n. Hence by Brouwers Theorem and the claim, thereexists (x, y) X Y s.t. (x, y) = (x, y), i.e. (x, y) is an equilibrium.

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