\section{Monotone Stopping Game.}           \label{sec:MonoStop}

A monotone rule is introduced to sum up individual declarirations in 
a multi-variate stopping problem\cite{YNK82}. The rule is defined 
by a monotone logical function and is equivalent to the winning class 
of Kadane\cite{Kad78}.
There given $p$-dimensional random process $\{X_n; n=1,2,\cdots \}$ 
and a stopping rule $\pi$ by which the group decision determined
from the declararion of $p$ players at each stage.
The stopping rule is $p$-variate $\{0,1\}$-valued monotone 
logical function.  We consider two cases of rules with $p = 3$ 
as follows:
\begin{equation}                                 \label{rule(a)}
  \pi(x_1,x_2,x_3) = x_1 + x_2
\end{equation}
and
\begin{equation}                                 \label{rule(b)}
  \pi(x_1,x_2,x_3) = x_1 x_2 + x_1 x_3.
\end{equation}
  That is, in case of the case (\ref{rule(a)}), 
if either of player 1 and 2 declares
stop, then the system stops neglecting of $x_3$. 
In case of (\ref{rule(a)}), we have the system stops when 
either of player 1 and 2 declares stop accompanying with player 1.
%
\begin{figure}
%  \begin{center}
  \begin{minipage}{.5\textwidth}
  \begin{center}
  \begin{tabular}[t]{|ll|c|} \hline               \label{fig:1}
  $x_1$ &  $x_2$ &  $\pi(x_1,x_2,x_3)$ \\ \hline
    0   &    0   &   0 \\
    0   &    1   &   1 \\
    1   &    0   &   1 \\
    1   &    1   &   1 \\ \hline
  \end{tabular}
  \caption{$\pi(x_1,x_2,x_3) = x_1+x_2$ for any $ x_3$}
  \end{center}
  \end{minipage}
%  �󔒍s�Ȃ�
  \begin{minipage}{.5\textwidth}
  \begin{center}
  \begin{tabular}[t]{|lll|c|} \hline               \label{fig:2}
  $x_1$ &  $x_2$ & $x_3$ &  $\pi(x_1,x_2,x_3)$ \\ \hline
    0   &    0   &   0   &    0 \\
    0   &    0   &   1   &    0 \\
    0   &    1   &   0   &    0 \\
    0   &    1   &   1   &    0 \\
    1   &    0   &   0   &    0 \\
    1   &    0   &   1   &    1 \\
    1   &    1   &   0   &    1 \\
    1   &    1   &   1   &    1 \\ \hline
  \end{tabular}
  \caption{$\pi(x_1,x_2,x_3) = x_1x_2 + x_1x_3$}
  \end{center}
  \end{minipage}
%  \end{center}
\end{figure}
%
Without loss of generality, we can assume that each $X_n$ takes
the uniformly distribution on $[0,1]$. 
Then equilibrium expected values for each player is given as
\begin{center}
 {\doublerulesep=\arrayrulewidth  % �r���̑�����2�{�� \hline*2
 \begin{tabular}{l|ccc} \hline\hline \hline \\[-2ex]    
      & Player 1 & Player 2 & Player 3 \\[1.5ex] \hline \hline\hline \\[-1.5ex]
   $\pi(x_1,x_2,x_3) = x_1 + x_2$ & $\dfrac{\sqrt{5}-1}{2}$ & 
       $\dfrac{\sqrt{5}-1}{2}$ & $0.5$     \\[2ex] \hline \\[-1.5ex]
   $\pi(x_1,x_2,x_3) = x_1x_2 + x_1x_3$ &  $1$   &  $\dfrac{\sqrt{5}-1}{2}$    
       & $\dfrac{\sqrt{5}-1}{2}$    \\[2ex] \hline\hline 
 \end{tabular}
 } %doublerule
 \\[1.5ex] 
 {Table 1. The equilibrium expected value for each players.}
\end{center}

\vspace{2ex}
In order to derive the value $ \phi^{-1} = \dfrac{\sqrt{5}-1}{2}$ , 
we consider an equilibrium stopping strategy of threshold type 
in the form $\{X_n > a\}$ for some $a$.
Bellman type equation for this game version will be given as 
\cite{YNK82}.
That is, each player declares  ^^ ^^ stop" or ^^ ^^ continue" 
if the observed value exceeds some $a$ or not. 
The event of the occurence is denoted by 
$D_n^{i} = \{ \mbox {Player $i$ declares stop}\}$.
Two trivial cases are the whole event $\Omega $ and the empty event 
$\emptyset$. 

In generally a logical function is assumed ^^ ^^ monotone" so
its function can be written as
$$
   \pi(x^1,\cdots, x^p ) = x^i \cdot \pi(x^1,\cdots , \stackrel{i}{1} , 
       \cdots x^p) + \overline{ x^i} \cdot \pi(x^1,\cdots , \stackrel{i}{0} , 
       \cdots x^p)  \quad x^i \in \{ 0, 1\} \,\, \,  \forall i
$$ 
where $\overline{ x^i} = 1-  x^i$.  Corresponding to this expression,
$$
   \Pi(D^1,\cdots, D^p ) = D^i \cdot \Pi(D^1,\cdots , \stackrel{i}{\Omega } , 
       \cdots D^p) + \overline{ D^i} \cdot 
       \Pi(D^1,\cdots , \stackrel{i}{\emptyset } , \cdots D^p) 
$$ 
where $\overline{ D^i} $ is the complement of the event $D^i$.
The general equation for the expected for player $i$ equals
\begin{equation}                                    \label{eq:101}
   E\left[(X_n^i - v^i)^{+} 
      {\bf 1}_{\Pi(D_n^1,\cdots , \stackrel{i}{\Omega}, \cdots D_n^p)}\right]
   + E\left[(X_n^i - v^i)^{-} 
      {\bf 1}_{\Pi(D_n^1,\cdots , \stackrel{i}{\emptyset}, 
      \cdots D_n^p)}\right]
\end{equation}
where $D_n^i = \{ X_n^i \geq v^i\}$ and 
$(x)^+ = \max\{x,0\}, (x)^- = \min\{x,0\}$. 
If we assume an independence case 
between player's random variable $X_n^i$ for each $i$. The equation
(\ref{eq:101}) becomes as
\begin{equation}                                    \label{eq:102}
   \beta_n^{\Pi(i) } E\left[(X_n^i - v^i)^{+}\right]
   - \alpha_n^{\Pi(i) } E\left[(X_n^i - v^i)^{-} \right]
\end{equation}
where
{rule(a)}Our objection is to find an equilibrium strategy and values of playes 
for a given monotone rule as the rule (\ref{rule(a)}) and (\ref{rule(b)}).
A sequence of expected value (a net gain) under the situation formulated 
in the section is obtained as 
\begin{equation}                                    \label{eq:103}
   v_{n+1}^i = v_n^i + \beta_n^{\Pi(i) } E\left[(X_n^i - v_n^i)^{+}\right]
   - \alpha_n^{\Pi(i) } E\left[(X_n^i - v_n^i)^{-} \right]
\end{equation}
for player $ i = 1, \cdots, p$ and $n$  denotes a time-to-go.
The details refer to Theorem 2.1 in YKN\cite{YNK82}.
%
Under these derivation, now we are able to calculate the optimal (equilibrium) 
value $ v^i = \lim_{n } v_n^i$ for player $i= 1,2,3$ for the rule (\ref{rule(a)}) 
and (\ref{rule(b)}) in the table .
