**Amit Agarwal - Vikas Bansal
**

*Date:* August 02 2002

**Contents**

- Strategic Games

- Examples of Nash equilibrium in some games

- Iterated Deletion to compute Nash Equilibrium.
- A general two player constant sum game

- Minimax Theorem
- Proof of Minimax Theorem

Nash Equilibrium:
In this lecture we consider several examples of strategic games and try to find
Nash Equlibria for them and answer the following questions regarding
Nash Equilibrium for a strategic game.

- Does Nash Equilibrium always exist for a game ?
- If it exists how to compute it and is it always possible to compute it ?

We also analyze a general constant sum game and Nash Equilibria for it.

- a set of players
- for each player, a set of actions/strategies
- for each player, preferences over the set of action profiles.

, where is the utility of player .

The above definition neither implies that a strategic game has a Nash equilibrium, nor that it has only one. Examples in the next section show that some games have a single Nash equilibrium, some possess no Nash equilibria and others have many Nash equilibria.

*Pure Strategies:*A pure strategy is one where players deterministically choose their moves.*Mixed Strategies:*A mixed strategy is one where players randomly choose one out of many different strategies. For example players can choose a probability distribution over the the set of possible strategies and randomly pick one before playing the game.

This situation may be modeled as a strategic game with :

We can represent the suspects' preference orderings with a payoff function and represent the game compactly with the payoff matrix:

C | D | |

C | 5,5 | 0,10 |

D | 10,0 | 1,1 |

C | M | |

C | 10,5 | 0,0 |

M | 0,0 | 5,10 |

H | T | |

H | 1,-1 | -1,1 |

T | 1,-1 | 1,-1 |

C | 5 | 0 | Dominates |

D | 10 | 1 |

Note: Here the payoffs are in the negative sense.

I II | A | B | C | D |

A | 5,2 | 2,6 | 1,4 | 0,4 |

B | 0,0 | 3,2 | 2,1 | 1,1 |

C | 7,0 | 2,2 | 1,5 | 5,1 |

D | 9,5 | 1,3 | 0,2 | 4,8 |

This strategy however does not always succeed in giving the Nash Equilibrium. We might reach a stage from which we cannot delete further rows/columns to obtain the Nash Equilibrium.

For example, in the battle of sexes game described previously, this method does not give the Nash Equilibria since no column or row is dominated by any other column or row respectively.

Since the game is a constant sum game,

A general two-player constant sum game can be represented as a pair of payoff matrices.
Consider a general 2 player game in which the row player has choices of strategy and the column player has choices. The payoff matrix for this game would be a x matrix.
where 0, 1

and

Similarly, the mixed strategy of the column player is represented as a -tuple,

Similarly, the payoff to the column player is

and for column player B,

( )If is a Nash Equilibrium, obviously is the maximum value in its column(it is the payoff to the column player). Similarly for the row player, it is the minimum in its row since (c - ) is the payoff to the row player. Hence, (,) is a saddle point.

.

: Since by definition, saddle points are minima in their rows and
maxima in their columns,
and
,
Hence,

Also, is a saddle point since is a minimum in its row i, and is a maximum in its column . Similarly, is a saddle point.
*
***Definition**: Let
be the guaranteed payoff to row player if he chooses row .
Let

** Lemma: **For any matrix A,
** Proof: **
We have,
.
Hence
, .

This implies , . Therefore

The above lemma also holds for the case of mixed strategies. The proof is given later

### Theorem: Matrix A has a saddle point .

* ***Proof:**
(
) Let be a saddle point of A.
By definition, .

Also, and . Hence,

Combining these two, . But from the previous lemma .

Hence .

( ) Choose s.t., .

Now choose s.t. .

Since is the minimum in row and column such that .

.

This implies , . Therefore

The above lemma also holds for the case of mixed strategies. The proof is given later

Also, and . Hence,

Combining these two, . But from the previous lemma .

Hence .

( ) Choose s.t., .

Now choose s.t. .

Since is the minimum in row and column such that .

Thus .
Since is a minimum in its row,
.

Thus, is also a minimum its row.

which proves that is a saddle point of A.##
Minimax Theorem

** Definition** : Row value for a mixed strategy is defined as

Thus, is also a minimum its row.

which proves that is a saddle point of A.

Similarly column value is defined as,

In other words, is the amount of payoff that the row player is guaranteed to win on the average, assuming that he plays rationally.

** Lemma: **For any matrix A, .

**Proof** :
We observe that
. Taking the maximum over all
on both sides,
. The RHS is , thus the previous equation can be re-written as
. Therefore
. This proves
.

: is a Nash Equilibrium iff .

**Proof** : (
) As
is a Nash Equilibrium of A,
. Also,
and
. Hence,
. Combining these two we get,
. But from previous lemma
. This proves
.

( ) Choose s.t., . Now choose s.t. . Since is the minimum over all and strategy s.t.

.

Thus . Since is the minimum over all , .

Thus is also minimum over all for the same . . Thus is a Nash Equilibrium.

For any two person zero-sum game specified by matrix , optimal mixed strategies exist for both players. Moreover the row and column values are equal. In other words,

or . Also if and denote the optimal strategies for the row and the column player respectively, then*publicly* committed to play the optimal strategy, it is possible for the other player to play the game with a pure strategy and still receive the optimal expected payoff.
##
Proof of Minimax Theorem

This proof requires the duality theorem, a well known result in linear programming.
A linear programming problem can be defined in terms of constraints
and
, and a cost vector
. The goal is to minimize the cost
subject to the constraints and given a cost vector. This is called the *primal* problem.
Associated with every primal problem is a *dual* problem stated as follows.
The constraints now become
and
, the new cost vector is
and the goal is to maximize
.

If either problem (primal or dual) has a best vector (called or ), then so does the other. The minimum equals the maximum

In terms of (the row player) and (column player), we want to minimize (primal) (called ) subject to the constraints and . We also want to maximize (dual) (called ) subject to the constraints and , where and are unit vectors. In light of this formulation and the duality theorem, we can state

Therefore, our probability distributions that correspond to our optimal vectors and are obtained by setting and .

**[Proof of Von Neumanns' Minimax Theorem]**
Since
,
. And since
, this implies that
. This gives a lower bound on how much is winning (
). Similarly,
implies that
and that
is an upper bound on 's loss.
Therefore,

: is a Nash Equilibrium iff .

( ) Choose s.t., . Now choose s.t. . Since is the minimum over all and strategy s.t.

.

Thus . Since is the minimum over all , .

Thus is also minimum over all for the same . . Thus is a Nash Equilibrium.

For any two person zero-sum game specified by matrix , optimal mixed strategies exist for both players. Moreover the row and column values are equal. In other words,

(1) |

or . Also if and denote the optimal strategies for the row and the column player respectively, then

- (,) is a Nash Equilibrium for this two player game.

If either problem (primal or dual) has a best vector (called or ), then so does the other. The minimum equals the maximum

(2) |

In terms of (the row player) and (column player), we want to minimize (primal) (called ) subject to the constraints and . We also want to maximize (dual) (called ) subject to the constraints and , where and are unit vectors. In light of this formulation and the duality theorem, we can state

(3) |

Therefore, our probability distributions that correspond to our optimal vectors and are obtained by setting and .