%! TEX root = TA.tex
% vim: tw=50
% 01/02/2024 12PM

Two people two outcomes:
\begin{align*}
  A(\bf{p}, \bf{q})
  &= \sum p_i a_{ij} q_j \\
  B(\bf{p}, \bf{q})
  &= \sum_i p_i a_{ij} q_j
\end{align*}
Von Neumann zero sum games: $a_{ij} = -b_{ij}$.
Then can show there exists $\bf{p}^*, \bf{q}^*$
such that
\begin{align*}
  A(\bf{p}^*, \bf{q}^*)
  &= \sup_f \inf_{\bf{q}} A(\bf{p}^*, \bf{q}) \\
  B(\bf{p}^*, \bf{q}^*)
  &= \sup_f \inf_{\bf{p}^*} B(\bf{p}^*, \bf{q}^*)
\end{align*}

\begin{center}
  \begin{tabular}{c|c|c}
    & you admit & you don't admit \\
    \hline
    he admits & $2$ & $8$ \\
    \hline
    he does not admit & $\half$ & $1$
  \end{tabular}
\end{center}

Nash showed that there is at least one $(\bf{p}^*,
\bf{q}^*)$ such that
\begin{align*}
  A(\bf{p}^*, \bf{q}^*)
  &\ge A(\bf{p}, \bf{q}^*) \qquad \forall \bf{p} \\
  B(\bf{p}^*, \bf{q}^*)
  &\ge B(\bf{p}^*, \bf{q}) \qquad \forall \bf{q} \\
\end{align*}
That is to say there is no reason unilaterally to
change your choice.

\begin{proof}
  Use \nameref{brouwer}. Let
  \[ \Gamma = \{(p, 1 - p, q, 1 - q) : 1 \ge p \ge
  0, 1 \ge q \ge 0\} \]
  2 dimensional square. We will define $T : \Gamma
  \to \Gamma$. First define $u_1$:
  \[ u_1 = \max(A(1, 0, q_1, q_2) - A(p_1, p_2,
  q_1, q_2), 0) \]
  $u_1 > 0$ if worth moving towards $(1, 0)$ for
  $\bf{p}$. $u_1$ is continuous. Define
  \[ u_2 = \max(A(0, 1, \bf{q}) - A(\bf{p},
  \bf{q}), 0) \]
  and similarly define $v_1, v_2$ for $q$.

  Then we define
  \[ T(\bf{p}, \bf{q}) = \left( \frac{p_1 + u_1}{1
  + u_1 + u_2}, \frac{p_2 + u_2}{1 + u_1 + u_2},
  \frac{q_1 + v_1}{1 + v_1 + v_2}, \frac{q_2 +
  v_2}{1 + v_1 + v_2} \right) \]
  $u_1, u_2$ continuous, and $u_1, u_2 \ge 0$
  gives that
  \[ \frac{p_1 + u_1}{1 + u_1 + u_2} \]
  is a continuous function of $(\bf{p}, \bf{q})$.
  Also, $\frac{u_1 + p_1}{1 + u_1 + u_2} \ge 0$
  and
  \[ \frac{u_1 + p_1}{1 + u_1 + u_2} + \frac{u_2 +
  p_2}{1 + u_1 + u_2} = \frac{u_1 + u_2 + p_1 +
  p_2}{1 + u_1 + u_2} = 1 .\]
  Similarly for $u_2, v_1, v_2$. So $T$ is a
  continuous map from $\Gamma$ to $\Gamma$. Thus
  there is at least one fixed point. Since at most
  one of $u_1, u_2$ is non-zero, we may suppose
  $u_2 = 0$. But $T(\bf{p}^*, \bf{q}^*) =
  (\bf{p}^*, \bf{q}^*)$, so $u_1 = u_2 = 0$ so
  either $1 > p_1 > 0$ and $A(1, 0, \bf{q}^*) =
  A(0, 1, \bf{q}^*) = A(\bf{p}, \bf{q})$ so no
  reasn for $A$ to change (note $A(p, q)$ is affine
  in $p_1$). Otherwise, $p_1 = 0$ or $p_1 = 1$.
  Without loss of generality, $A(1, 0, \bf{q}) \ge
  A(1 - p', p', \bf{q})$, but we cannot.
\end{proof}

\begin{example*}[Game of chicken]
  $A$ and $B$ drive cars towards each other. If
  both swerve, they both lose $1$ prestige point.
  If one swerves and the other does not, then the
  swerver loses $5$ points and the swerver gains
  $10$ points. If neither swerves, then both lose
  $100$ points.
  \[ A(p, q) = -pq - 5p(1 - q) + 10(1 - p)q -
  100(1 - p)(1 - q) \]
  Easiest to differentiate:
  \begin{align*}
    \pfrac{A}{p}
    &= -q - 5(1 - q) - 10q + 100(1 - q) \\
    &= 95 - 106q
  \end{align*}
  so if $\pfrac{A}{p} = $, then $106q = 95$, so $q
  = \frac{95}{106}$. This tells us that $\left(
  \frac{95}{106}, \frac{95}{106} \right)$ is a
  Nash point.

  However, we have only examined interior points
  (because a local extrema is only found by
  $\pfrac{A}{p}$ if it is not on the boundary).
  We must check $p = 1$ for example. Upon doing
  this, we also find that $(1, 0)$ and $(0, 1)$
  are Nash points.
\end{example*}

\begin{remark*}
  Recall that \nameref{brouwer} has
  a multidimensional extension. It is quite easy
  to extend from $2$ people, $2$ choices to $2$
  people, $n$ choices. It can also be extended to
  more than $2$ players (which Von Neumann can't).
  However, what Nash then says is that there
  exist points $(p^*, q^*, v^*)$ where it is to no
  player's advantage to change unilaterally. This
  is much weaker than it sounds. Winkn, Blykn and
  Nod who must divide $\pounds 90$ by majority vote.
\end{remark*}
