Lecture 4 Zero-sum games, and the Minimax Theorem

Min-maximizing strategies

Suppose Player 1 chooses a mixed strategy $x_1^\ast \in X_1$, by trying to maximum the “worst that can happen would be for Player 2 to choose $x_2$ which minimizes $(x_1^\ast)^\top Ax_2$”, where $A$ is the payoff matrix $[u(i,j)]_{m_1\times m_2}$.

Definition: $x_1^\ast \in X_1$ is a minmaximizer for Player 1 if

\[\min_{x_2\in X_2}(x_1^\ast)^\top Ax_2 = \max_{x_1\in X_1}\min_{x_2\in X_2}(x_1)^\top Ax_2\]

Similarly, $x_2^\ast \in X_2$ is a maxminimizer for Player 2 if

\[\max_{x_1\in X_1}(x_1)^\top Ax_2^\ast = \min_{x_2\in X_2}\max_{x_1 \in X_1}(x_1)^\top A{x_2}\]

Note that (obviously)

\[\min_{x_2 \in X_2}(x_1^\ast)^\top Ax_2 \leq (x_1^\ast)^\top Ax_2^\ast \leq \max_{x_1\in X_1}(x_1)^\top Ax_2^\ast\]

The Minimax Theorem

Theorem(von Neumann) Let a 2p-zs game $\Gamma$ be given by an $(m_1 \times m_2)$-matrix $A$ of real numbers. There exists a unique value $v^\ast \in \mathbb{R}$, such that there exists $x^\ast = (x_1^\ast, x_2^\ast)\in X$ such that

1. $((x_1^\ast)^\top A)_j \geq v^\ast$, for $j = 1, \cdots, m_2$.
2. $(Ax_2^\ast)_j \leq v^\ast$, for $j = 1, \cdots, m_1$.

3. And (thus) $v^\ast = (x_1^\ast)^\top Ax_2^\ast$ and

   \[\min_{x_2 \in X_2}(x_1^\ast)^\top Ax_2 = (x_1^\ast)^\top Ax_2^\ast = \max_{x_1\in X_1}(x_1)^\top Ax_2^\ast\]
4. In fact, the above conditions all hold precisely when $x^\ast = (x_1^\ast, x_2^\ast)$ is any Nash Equilibrium. Equivalently, they hold precisely when $x_1^\ast$ is any minmaximizer and $x_2^\ast$ is any maxminimizer.

Note:

1. (1.) says $x_1^\ast$ guarantees Player 1 at least expected profit $v^\ast$ and (2.) says $x_2^\ast$ guarantees Player 2 at most expected loss $v^\ast$.
2. We call the unique $v^\ast$ the minimax value of game $\Gamma$, and call any such $x^\ast = (x_1^\ast, x_2^\ast)$ a minimax profile.

3. It is obvious that the maximum profit that Player 1 can guarantee for itself should be $\leq$ the minimum loss that Player 2 can guarantee for itself, i.e., that

   \[\max_{x_1\in X_1}\min_{x_2\in X_2}(x_1)^\top Ax_2 \leq \min_{x_2\in X_2}\max_{x_1\in X_1}(x_1)^\top Ax_2\]

   (Look back the whole definition.)

Proof: Let $x^\ast = (x_1^\ast, x_2^\ast) \in X$ be a NE of the 2-player zero-sum game $\Gamma$, with matrix $A$.

Let $v^\ast := (x_1^\ast)^\top Ax_2^\ast = U_1(x^\ast) = - U_2(x^\ast)$.

Since $x_1^\ast$ and $x_2^\ast$ are “best responses” to each other, we know that for $i \in \lbrace 1, 2\rbrace$

\[U_i(x_{-i}^\ast; \pi_{i,j}) \leq U_i(x^\ast)\]

Then we have

1. $U_1(x_{-1}^\ast; \pi_{1, j}) = (Ax_2^\ast)_j$, thus,

   \[(Ax_2^\ast)_j \leq U_1(x^\ast) = v^\ast\]

   The (2.) Q.E.D.

2. $U_2(x_{-2}^\ast; \pi_{2, j}) = -((x_1^\ast)^\top A)_j$, thus

   \[- ((x_1^\ast)^\top A)_j \leq U_2(x^\ast) = - v^\ast \\ s.t.\ ((x_1^\ast)^\top A)_j \geq v^\ast\]

   The (1.) Q.E.D.

3. Since 1. and $(x_1)^\top Ax_2^\ast$ is a weighted average of $(Ax_2^\ast)_j$, thus

   \[\max_{x_1\in X_1}(x_1)^\top Ax_2^\ast \leq v^\ast\]

   Similarly, since 2. and $(x_1^\ast)^\top Ax_2$ is a weighted average of $((x_1^\ast)^\top A)_j$, thus

   \[\min_{x_2\in X_2}(x_1^\ast)^\top Ax_2 \geq v^\ast\]

   So we finally have

   \[\max_{x_1\in X_1}(x_1)^\top Ax_2^\ast \leq v^\ast = (x_1^\ast)^\top Ax_2^\ast \leq \min_{x_2\in X_2}(x_1^\ast)^\top Ax_2\]

   which is entirely the opposite inequality we earlier noted

   \[\min_{x_2 \in X_2}(x_1^\ast)^\top Ax_2 \leq (x_1^\ast)^\top Ax_2^\ast \leq \max_{x_1\in X_1}(x_1)^\top Ax_2^\ast\]

   Therefore,

   \[\min_{x_2 \in X_2}(x_1^\ast)^\top Ax_2 = (x_1^\ast)^\top Ax_2^\ast = \max_{x_1\in X_1}(x_1)^\top Ax_2^\ast\]

4. We didn’t assume anything about the particular NE we chose. So, for every NE, $x^\ast$, letting $v^\prime = (x^\ast)^\top Ax_2^\ast$,

   \[\max_{x_1\in X_1}\min_{x_2 \in X_2}(x_1)^\top Ax_2 = v^\prime = v^\ast = (x_1^\ast)^\top Ax_2^\ast = \min_{x_2\in X_2}\max_{x_1\in X_1}(x_1)^\top Ax_2\]

   Moreover, if $x^\ast = (x_1^\ast, x_2^\ast)$ satisfies conditions (1.) and (2.) for some $v^\ast$, then $x^\ast$ must be a NE.

Useful Corollary for Minimax

In a minimax profile $x^\ast = (x_1^\ast, x_2^\ast)$,

1. if $x_2^\ast(j) > 0$, then $((x^\ast_1)^\top A)_j = (x^\ast_1)^\top A x_2^\ast = v^\ast$.
2. if $x_1^\ast(j) > 0$, then $(Ax^\ast_2)_j = (x_1^\ast)^\top A x_2^\ast = v^\ast$.

Minimax as an optimization problem

Maximize $v$

Subject to:

\[\begin{align*}&(x_1^\top A)_j \geq v \quad \text{for }j =1, \cdots, m_2, \\&x_1(1) + \cdots + x_1(m_1) = 1, \\&x_1(j) \geq 0 \quad \text{for }j = 1, \cdots, m_1 \end{align*}\]