2.5 The Fundamental Theorem of Game Theory For any 2-person zero-sum game there exists a pair (x*,y*) in S  T such that min {x*V. j : j=1,...,n} = max{min{xV.j : j=1,...,n}: x in S} = v 1, max {V i. y* : i=1,...,m} = min{max{V i. y: i=1,...,m}: y in T} = v 2, and v 1 =v 2.

Remember the following example? What can you tell about the values of v 1 and v 2 ?

Proof. It is convenient to assume that both v 1 and v 2 are strictly positive, namely that v 1 > 0 and v 2 > 0. This is a mere technicality, because we can always add to V a large constant without changing the nature of the game. (See later) The plan is to show that the problems faced by the players can be expressed as LP problems and one is the dual of the other.

By definition of v 1, v 1 := max {s(x): x in S} = max {min {xV. j : j=1,...,n}: x in S} (Theorem 1.4.1) This is equivalent to: v 1 = max u subject to xV.j ≥ u, j=1,...,n x in S. Observe that this is an LP problem, namely

v 1  max u,x u s.t. xV. 1  u ≥ 0. 2  u ≥ n  u0 x 1 ...  x m  1 x 1,...,x m 0 Observe that u is a decision variable! ≥ ≥

We can complete the proof using this form of the LP problem, but it will not be “elegant”. Let us then beautify the formulation. It is clear that if V is strictly positive, so is the optimal value of u. Thus, with no loss of generality, we can restrict the analysis to positive values of u, and divide the constraints by u. This yields:

Observe that maximizing u is equivalent to minimizing 1/u, thus the problem under consideration is equivalent to:

If we then set x’:=x/u, we obtain

Substituting the equality constraint for 1/u in the objective function, we obtain the following equivalent problem:

If we thus let b = (1, 1,..., 1) and c = (1, 1,..., 1), we can rewrite the problem as follows:

If we repeat the process for Player II, we discover that her problem is equivalent to

By “inspection” we conclude that both problems are feasible and have optimal solutions. Thus duality theory tells us that v 1 = v 2. To obtain the optimal strategies from the solutions to these LP problems we have to multiply them by the optimal value of the objective function, that is, v 1 or v 2.

Recipe Player IPlayer II

1.5.1 Example Check: no saddle Check whether v > 0 - IMPORTANT! The two linear programming problems in this case are as follows:

We prefer Player II’s formulation. Why? Player I Player II

Final tableau: y’* = (1/7, 1/7), v’ 2 = 2/7; v 2 = 1/v’ 2 = 7/2 y* = v 2 y’* = x’* = (5/28, 9/84) (why?) x* = v 2 x’* = Check the results for consistency.

a1a1 A1A1 a2a2 a3a3 A2A2

We now know how to satisfy Principle I. How about equilibrium? Do we need to test equilibrium every time?

Are the strategies yielding the optimal security levels in equilibrium? –YES!! –YIPEEE!!! –THANKS to COROLLARY !!!! Corollary Let (x*,y*) be any element of S  T such that v 1 = s(x*) =  (y*) = v 2. Then, this pair is in equilibrium.

Proof: Need to show that x*Vy ≥ x*Vy* ≥ xVy* for all x in S and y in T. For (x*,y*) we have that s(x*) = v 1 and  (y*) = v 2. Since v 1 = v 2, if follows that v 1 = s(x*) = min{x*Vy: y in T} ≤ x*Vy* ≤ max{xVy*: x in S} =  (y*) = v 2 = v 1 So the ≤ must be = and min{x*Vy: y in T} = x*Vy* = max{xVy*: x in S} x*Vy ≥ min{x*Vy: y in T} = x*Vy* = max{xVy*: x in S} ≥ xVy* hence (x*,y*) is in equilibrium.

The converse is also true Theorem. Suppose that the strategy pair (x*, y*) is in equilibrium. Then this pair is optimal. Proof: If (x*, y*) is in equilibrium, then xVy* ≤ x*Vy* ≤ x*Vy for all (x,y) in S  T (definition). Now, v 1 := max{{min xVy: y in T}: x in S} so by definition of max implies v 1 ≥ min {x*Vy: y in T} = x*Vy* (by definition of equilibrium) Similarly for v 2 we obtain, v 2 ≤ x*Vy*. So v 2 ≤ x*Vy* ≤ v 1 implies xVy* ≤ x*Vy* ≤ x*Vy for all (x,y) in S  T

So v 2 ≤ x*Vy* ≤ v 1. But by The Fundamental Theorem v 1 =v 2 thus v 2 = x*Vy* = v 1 i.e. (x*,y*) is an optimal pair. So for zero sum 2-person games, we have: an optimal pair is also an equilibrium pair AND an equilibrium pair is an optimal pair.

Summary For ANY 2-person zero-sum game, there exists a strategy pair (x*, y*) such that x* is optimal for Player I, y* is optimal for Player II, their respective security levels are equal and the pair is in equilibrium. More good news: such a pair can be computed by the simplex method.