Lecture 3 Nash’s Theorem

Nash’s Theorem

The Fundamental Theorem of Game Theory

Theorem(Nash 1950) Every finite $n$-person strategic game has a mixed Nash Equilibrium.

The Brouwer Fixed Point Theorem

Theorem(Brouwer, 1909) Every continuous function $f: D \to D$ mapping a compact and convex, nonempty subset $D \subseteq \mathbb{R}^m$ to itself has a “fixed point”, i.e., there is $x^\ast \in D$ such that $f(x^\ast) = x^\ast$.

Fact: The set of profiles $X = X_1 \times \cdots \times X_n$ is a compact and convex subset of $\mathbb{R}^m$, where $m = \sum_{i = 1}^n m_i$, with $m_i = \lvert S_i \rvert$.

Proof of Nash’s Theorem

Proof: We will define a continuous function $f: X \to X$, where $X = X_1 \times \cdots \times X_n$, and we will show that if $f(x^\ast) = x^\ast$ then $x^\ast = (x_1^\ast, \cdots, x_n^\ast)$ must be a Nash Equilibrium.

Claim: A profile $x^\ast = (x_1^\ast, \cdots, x_n^\ast) \in X$ is a Nash Equilibrium if and only if, for every Player $i$, and every pure strategy $\pi _{i, j}, j \in S_i$:

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

Proof of claim: If $x^\ast$ is a NE then, it is obvious by definition that $U_i(x^\ast) \geq U_i(x_{-i}^\ast; \pi_{i, j})$.

For the other direction: by calculation it is easy to see that for any mixed strategy $x_i \in X_i$,

\[U_i(x_{-i}^\ast; x_i) = \sum _{j= 1}^{m_i}x_i(j) \cdot U_i(x^\ast_{-i};\pi_{i, j}) \leq U_i(x^\ast)\]

The inequality is true, since we assume that $U_i(x^\ast) \geq U_i(x_{-i}^\ast;\pi_{i,j})$. Therefore, each $x_i^\ast$ is a best response to strategy $x_{-i}^\ast$. In other words, $x^\ast$ is a Nash Equilibrium.

With this Claim, we can rephrase our goal of proving $x^\ast$ is a NE to finding $x^\ast$ such that

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

for all Player $i \in N$, and all $j \in \lbrace 1, \cdots, m_i\rbrace$.

For a mixed profile $x = (x_1, \cdots, x_n) \in X$, let

\[\varphi_{i, j}(x) = \max\lbrace 0, U_i(x_{-i};\pi_{i, j}) - U_i(x)\rbrace\]

Define $f: X \to X$ as follows: For $x = (x_1, \cdots, x_n) \in X$, let

\[f(x) = (x_i^\prime, \cdots, x_n^\prime)\]

where for all $i$, and $j = 1, \cdots, m_i$,

\[x_i^\prime(j) = \frac{x_i(j) + \varphi_{i,j}(x)}{1 + \sum_{k=1}^{m_i}\varphi_{i, k}(x)}\]

Thus, by Brouwer, there exists $x^\ast$ such that $f(x^\ast) = x^\ast$.

Now we have to show $x^\ast$ is a NE.

For each $i$, and for $j = 1, \cdots, m_i$,

\[x_i^\ast(j) = \frac{x_i^\ast(j) + \varphi_{i, j}(x^\ast)}{1 + \sum_{k = 1}^{m_i}\varphi_{i, k}(x^\ast)}\]

hence,

\[x_i^*(j) \sum_{k = 1}^{m_i} \varphi_{i, k}(x^\ast) = \varphi_{i, j}(x^\ast)\]

We will show that in fact this implies $\varphi_{i, j}(x^\ast)$ must be equal to $0$ for all $j$.

Claim: For any mixed profile $x$, for each Player $i$, there is some $j$ such that $x_i(j) > 0$ and $\varphi_{i, j}(x) = 0$.

Proof of claim: Since $U_i(x)$ is the weighted average of $U_i(x_{-i}; \pi_{i, j})$’s, based on the weights in $x_i$, there must be some $j$ used in $x_i$, i.e., with $x_i(j) > 0$, such that $U_i(x_{-i};\pi_{i, j})$ is no more than the weighted average. i.e.,

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

Therefore,

\[\varphi_{i, j}(x) = \max\lbrace 0, U_i(x_{-i};\pi_{i, j}) - U_i(x)\rbrace = 0\]

Thus, for such a $j$, $x^\ast_i(j) > 0$ and

\[x^\ast_i(j) \sum_{k = 1}^{m_i} \varphi_{i, k}(x^\ast) = \varphi_{i, j}(x^\ast) = 0\]

But, since $\varphi_{i, k}(x^\ast)$’s are all $\geq 0$ and $x^\ast_i > 0$, this means $\varphi_{i, k}(x^\ast) = 0$ for all $k = 1, \cdots, m_i$. Thus, for all Player $i$, and for $j = 1, \cdots, m_i$,

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

Q.D.E.

Useful Corollary for Nash Equilibria

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

whenever $x^\ast_i(j) > 0$. i.e., each such $\pi_{i, j}$ is itself a best response to $x_{-i}^*$.

Proof: If $x^\ast$ is a NE, thus $U_i(x^\ast)$ must $\geq U_i(x^\ast_{-i}; \pi_{i, j})$, for all $j = 1, \cdots, m_i$.

Since $U_i(x^\ast) = \sum_{j = 1}^{m_i}x^\ast_i(j) \cdot U_i(x^\ast_{-i};\pi_{i, j})$ , consider the following two cases:

1. If $U_i(x^\ast) > U(x^\ast_{-i};\pi_{i, j})$ for all $j = 1, \cdots, m_i$, then $U_i(x^\ast) > \sum_{j = 1}^{m_i}x^\ast_i(j) \cdot U_i(x^\ast_{-i};\pi_{i, j})$ which is not true;
2. If $U_i(x^\ast) > U_i(x^\ast_{-i};\pi_{i, j})$ for some $j = 1, \cdots, m_i$, then there must exist $U_i(x^\ast) < U_i(x^\ast_{-i};\pi_{i, j^\prime})$ for some $j^\prime = 1, \cdots, m_i$ and $j^\prime \neq j$ when $U_i(x^\ast) = \sum_{j = 1}^{m_i}x^\ast_i(j) \cdot U_i(x^\ast_{-i};\pi_{i, j})$ holds, which is not true according to the property of NE.

In summary, $U_i(x^\ast)$ cannot be greater than $U_i(x_{-i}^\ast; \pi_{i,j})$ if $x_i^\ast(j) > 0$. Therefore, $U_i(x^\ast) = U_i(x^\ast_i; \pi_{i, j})$.

NE need not to be Pareto optimal

Given a profile $x \in X$ in an $n$-player game, the (purely utilitarian) social welfare is:

\[U_1(x) + U_2(x) + \cdots + U_n(x)\]

A profile $x \in X$ is Pareto efficient (a.k.a.,Pareto Optimal) if there is no other profile $x^\prime \in X$ such that $U_i(x^\prime) \geq U_i(x)$ for all $i$ and $U_j(x^\prime) > U_j(x)$ for some $j$.

Evolutionarily Stable Strategies

Definition: A 2-player game is symmetric if $S_1 = S_2$, and for all $s_1, s_2 \in S_1$, $u_1(s_1, s_2) = u_2(s_2, s_1)$.

Definition: In a 2p-sym-game, mixed strategy $x_1^\ast$ is an evolutionarily stable strategy (ESS) if:

1. $x_1^\ast$ is a best response to itself, i.e., $x^\ast = (x_1^\ast, x_1^\ast)$ is a symmetric NE &
2. If $x_1^\prime \neq x_1^\ast$ is another best response to $x_1^\ast$, then $U_1(x^\prime_1, x^\prime_1) < U_1(x_1^\ast, x^\prime_1)$.

Nash also proves that every symmetric game has a symmetric NE, $(x_1^\ast, x_1^\ast)$. However, not every symmetric game has a ESS.