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Previously we dealt with untrustworthy
individuals. There are occasions when people do
act trustworthy.

For example, contracts enforcable by law, or
individuals whomust make repeated transactions
with each other.

With high symmetry, it is not too hard. For
example, if Wynken, Blykn and Noel need to divide
$\$90$, then they will just split it evenly, if
they are incentivised to behave fairly.

In our game of chicken with rules as set out
before, we could do ``players swerve
alternately''.

More different if there is no obvious symmetry.
``What do we then mean by fair''.

Nash has an interesting model. $n$ participants.
$E \subseteq \RR^n$. Think of $\bf{x} \in E$ as a
choice. If $x_j > y_j$, then $j$ preferse $\bf{x}$
to $\bf{y}$.

Mathematical condition: $E$ is closed and bounded
(natural if we want there to exist some kind of
`best').

More interesting:
\begin{enumerate}[(1)]
  \item Demand $E$ be convex, i.e. $\bf{x}, \bf{y}
    \in E$, $0 \le \lambda \le 1$, then $\lambda
    \bf{x} + (1 - \lambda) \bf{y} \in E$ (for
    mathmos, choose $\bf{x}$ with probability
    $\lambda$, $\bf{y}$ with probability $1 -
    \lambda$; for non-mathmos, exchange sums of
    cash or other horse trading).
  \item There is a status quo point $\bf{s} \in
    E$. If no agreement reached then the outcome
    is $\bf{s}$. Could be ``continue as before''
    or ``strike'' or ``lock out''. Could consider
    $E \cap \{\bf{x} : x_j \ge s_j ~\forall j\}$.
  \item Pareto optimality. It is a well kept
    secret amongst politicians and economists that
    function $f : X \to \RR^n$ do not have maxima
    for $n \ge 2$. Pareto says that we should aim
    for a ``Pareto optimum'' $x^*$ such that
    $\nexists j$ and $\bf{z} \in E$ such that
    $x_j^* > z_j$ and $x_k^* \ge z_k$ for all $k
    \neq j$.
  \item Independence of irrelevant conditions. If
    $(E, s)$, $(E', s)$ given and $E' \supseteq
    E$, if $x^*$ is chosen for $(E', s)$ and
    $\bf{x}^* \in E$ then $\bf{x}$ is chosen for
    $(E, s)$.
  \item Symmetry: Suppose $(x_1, \ldots, x_n) \in
    E$, $\sigma \in S_n$ implies $(x_{\sigma(1)},
    \ldots, x_{\sigma(n)} \in E$ and $\bf{s} = (s,
    \ldots, s)$, then our solution $\bf{x}^*$ will
    have $x_1^* = x_2^* = \cdots = x_n^*$.
  \item You can't improve things by exaggeration.
    If $T(x, x_n) = (a_1 x_1 + b_1, a_2 x_2 + b_2,
    \ldots, a_n x_n + b_n)$ then if $x^*$ is the
    outcome for $(E, \bf{s})$, then $T x^*$ is the
    outcome for $(TE, T\bf{s})$.
\end{enumerate}

\begin{lemma*}
  If $E$ closed and convex. % , $\bf{0} \in E$.
  Suppose $(1, 1, \ldots, 1) \in E$ and $\prod_{j
  = 1}^n x_j \le 1$ for all $\bf{x} \in E$. Then
  $x_1 + x_2 + \cdots + x_n \le n$ for all $\bf{x}
  \in E$.
\end{lemma*}

\begin{proof}
  Suppose $\bf{x} \in E$. Then since $\bf{1} \in E$,
  convexity gives
  \[ (1 - \delta) \bf{1} + \delta \bf{x} \in E \]
  so $\prod ((1 - \delta) + \delta x_j) \le 1$,
  i.e.
  \[ \prod (1 + \delta(x_j - 1)) \le 1 .\]
  So $1 + \sum \delta(x_j - 1) + A(\delta) \le 1$
  with $\frac{A(\delta)}{\delta} \to )$ as $\delta
  \to 0$. Then $\sum (x_j - 1) +
  \frac{A(\delta)}{\delta} \le 0$, so $\sum (x_j
  - 1) \le 1$.
\end{proof}

\begin{corollary*}
  If $\bf{s} = 0$, $(1, 1, \ldots, 1) \in E$ and
  $\prod x_j \le 1$ for all $x_j \in E$, then $(1,
  1, \ldots, 1)$ is the choice.
\end{corollary*}

\begin{proof}
  By our previous lemma, $E \subseteq \Gamma =
  \{x_j \le n\}$. Look at our problem for
  $(\Gamma, \bf{0})$. This a symmetric problem (in
  the sense of (5)) so solution must satisfy $x_1
  = x_2 = \cdots = x_n$. By Pareto optimality,
  $x_1 = x_2 = \cdots = x_n = 1$. So we have a
  unique solution. But $E \subseteq \Gamma$ and
  $(1, 1, \ldots, 1) \in E$ so by (4) independence
  of irrelevant conditions, we have that $(1, 1,
  \ldots, 1)$ is optimal for $(E, \bf{0})$
  problem.
\end{proof}

We now use condition (6) (problem invariant under
affine transformations $\bf{x} \mapsto (a_1 x_1 +
b_1, \ldots, a_n x_n + b_n)$, with $a_j > 0$).

Suppose $(E, \bf{s})$ given claim solution is
unique $x^*$ which maximises $\prod_{i = 1}^n (x_i
- s_i)$ subject to $\bf{x} \in E$. By compactness
of $E$, such a point exists.
\[ \prod_{i = 1}^n (x_i^* - s_i) \ge \prod_{i =
1}^n (x_i - s_i) \qquad \forall x \in X .\]
(We assume $\prod_{i = 1}^n (x_i^* - s_i) > 0$).
If we make the transformation $z_i = \frac{(x_i -
s_i)}{(x_i^* - s_i)}$.