# Spatial Parrondo games
and an interacting particle system

Sung Chan Choi Department of Mathematics, University of Utah, 155 S. 1400 E.,
Salt Lake City, UT 84112, USA. e-mail<EMAIL_ADDRESS>
###### Abstract

Parrondo games with spatial dependence were introduced by Toral (2001) and
have been studied extensively. In Toral’s model $N$ players are arranged in a
circle. The players play either game $A$ or game $B$. In game $A$, a randomly
chosen player wins or loses one unit according to the toss of a fair coin. In
game $B$, which depends on parameters $p_{0},p_{1},p_{2}\in[0,1]$, a randomly
chosen player, player $x$ say, wins or loses one unit according to the toss of
a $p_{m}$-coin, where $m\in\\{0,1,2\\}$ is the number of nearest neighbors of
player $x$ who won their most recent game. In this paper, we replace game $A$
by a spatially dependent game, which we call game $A^{\prime}$, introduced by
Xie et al. (2011). In game $A^{\prime}$, two nearest neighbors are chosen at
random, and one pays one unit to the other based on the toss of a fair coin.
Game $A^{\prime}$ is fair, so we say that the Parrondo effect occurs if game
$B$ is losing or fair and the game $C^{\prime}$, determined by a random or
periodic sequence of games $A^{\prime}$ and $B$, is winning. Here we give
sufficient conditions for convergence as $N\to\infty$ of the mean profit per
game played from game $C^{\prime}$. This requires ergodicity of an associated
interacting particle system (not necessarily a spin system), for which
sufficient conditions are found using the basic inequality.

##  1 Introduction

Toral (2001) introduced what he called cooperative Parrondo games with spatial
dependence. (We prefer the term spatially dependent Parrondo games so as to
avoid conflict with the field of cooperative game theory.) The games depend on
an integer parameter $N\geq 3$, the number of players, and four probability
parameters, $p_{0},p_{1},p_{2},p_{3}$. (This is a slight generalization of the
model described in the abstract.) The players are arranged in a circle and
labeled from 1 to $N$, so that players 1 and $N$ are adjacent. At each turn, a
player is chosen at random to play. Suppose player $x$ is chosen. In game $A$,
he tosses a fair coin. In game $B$, he tosses a $p_{m}$-coin (i.e., a coin
whose probability of heads is $p_{m}$), where $m\in\\{0,1,2,3\\}$ depends on
the winning or losing status of his two nearest neighbors. A player’s status
as winner (1) or loser (0) is decided by the result of his most recent game.
Specifically,

$m=\begin{cases}0&\text{if $x-1$ and $x+1$ are both losers,}\\\ 1&\text{if
$x-1$ is a loser and $x+1$ is a winner,}\\\ 2&\text{if $x-1$ is a winner and
$x+1$ is a loser,}\\\ 3&\text{if $x-1$ and $x+1$ are both
winners,}\end{cases}$

where $N+1:=1$ and $0:=N$ because of the circular arrangement of players.
Player $x$ wins one unit with heads and loses one unit with tails.

Game $A$ is fair, so we say that the Parrondo effect occurs if game $B$ is
losing or fair and the game $C$, determined by a random or periodic sequence
of games $A$ and $B$, is winning.

These games have been studied by Mihailović and Rajković (2003) and Ethier and
Lee (2012a,b, 2013a,b), and a strong law of large numbers and a central limit
theorem were obtained. In particular, the asymptotic cumulative profits per
turn exist and are the means in the SLLN. Further, it seems clear that these
means converges as $N\to\infty$. This has been proved under certain conditions
(see Ethier and Lee, 2013a).

Notice that Toral’s (2001) game $A$ is not spatially dependent (the rules of
the game do not depend on the spatial structure of the players). Xie et al.
(2011) proposed a modification of game $A$ that is spatially dependent as well
as being a fair game. To distinguish, we call that game $A^{\prime}$. As
before, the games depend on an integer parameter $N\geq 3$, the number of
players, and four probability parameters, $p_{0},p_{1},p_{2},p_{3}$. The
players are arranged in a circle and labeled from $1$ to $N$, so that players
$1$ and $N$ are adjacent. At each turn, a player is chosen at random to play.
Suppose player $x$ is chosen. In game $A^{\prime}$, he chooses one of his two
nearest neighbors at random and competes with that neighbor by tossing a fair
coin. The result is a transfer of one unit from one of the players to the
other, hence the wealth of the set of $N$ players is unchanged. Game $B$ is as
before. Player $x$ wins one unit with heads and loses one unit with tails.

Game $A^{\prime}$ is fair, so we say that the Parrondo effect occurs if game
$B$ is losing or fair and the game $C^{\prime}$, determined by a random or
periodic sequence of games $A^{\prime}$ and $B$, is winning.

These games were studied by Xie et al. (2011), Li et al. (2014), and Ethier
and Lee (2015). Only the case of random sequences was treated, and convergence
of the means has not yet been addressed. Our aim in this paper, together with
Choi (2020), is to fill in these gaps in the literature.

The Markov chain formalized by Mihailović and Rajković (2003) keeps track of
the status (loser or winner, 0 or 1) of each of the $N\geq 3$ players of game
$B$. Its state space is the product space

$\\{\eta=(\eta(1),\eta(2),\ldots,\eta(N)):\eta(x)\in\\{0,1\\}{\rm\ for\
}x=1,\ldots,N\\}=\\{0,1\\}^{N}$

with $2^{N}$ states. Let $m_{x}(\eta):=2\eta(x-1)+\eta(x+1)\in\\{0,1,2,3\\}$.
Of course $\eta(0):=\eta(N)$ and $\eta(N+1):=\eta(1)$ because of the circular
arrangement of players. Also, let $\eta_{x}$ be the element of $\\{0,1\\}^{N}$
equal to $\eta$ except at the $x$th coordinate. For example,
$\eta_{1}:=(1-\eta(1),\eta(2),\eta(3),\ldots,\eta(N))$.

The one-step transition matrix $\bm{P}_{B}$ for this Markov chain depends not
only on $N$ but on four parameters, $p_{0},p_{1},p_{2},p_{3}\in[0,1]$. It has
the form

$P_{B}(\eta,\eta_{x}):=\begin{cases}N^{-1}p_{m_{x}(\eta)}&\text{if
$\eta(x)=0$,}\\\ N^{-1}q_{m_{x}(\eta)}&\text{if $\eta(x)=1$,}\end{cases}\qquad
x=1,\ldots,N,\;\eta\in\\{0,1\\}^{N},$

and

$P_{B}(\eta,\eta):=N^{-1}\bigg{(}\sum_{x:\eta(x)=0}q_{m_{x}(\eta)}+\sum_{x:\eta(x)=1}p_{m_{x}(\eta)}\bigg{)},\qquad\eta\in\\{0,1\\}^{N},$

where $q_{m}:=1-p_{m}$ for $m=0,1,2,3$ and empty sums are 0. The Markov chain
is irreducible and aperiodic if $0<p_{m}<1$ for $m=0,1,2,3$. Under slightly
weaker assumptions (see Ethier and Lee 2013a), the Markov chain is ergodic,
which suffices. For example, if $p_{0}$ is arbitrary and $0<p_{m}<1$ for
$m=1,2,3$, or if $0<p_{m}<1$ for $m=0,1,2$ and $p_{3}$ is arbitrary, then
ergodicity holds.

A Markov chain in the same state space corresponds to game $A^{\prime}$. As
previously mentioned, if game $A^{\prime}$ is played, the profit to the set of
$N$ players is 0, since game $A^{\prime}$ simply redistributes capital among
the players. The transition probabilities require some new notation. Starting
from state $\eta$, let $\eta^{x,x\pm 1,\pm 1}\in\\{0,1\\}^{N}$ represent the
players’ status after player $x$ plays player $x\pm 1$ and wins (1) or loses
($-1$). Of course player 0 is player $N$ and player $N+1$ is player 1. For
example, $\eta^{1,2,-1}=(0,1,\eta(3),\ldots,\eta(N))$ (player 1 competes
against player 2 and loses, leaving player 1 a loser and player 2 a winner,
regardless of their previous status). The transition probabilities have the
form

$\displaystyle P_{A^{\prime}}(\eta,\zeta)$
$\displaystyle:=\frac{1}{4N}\sum_{x=1}^{N}[\delta(\eta^{x,x-1,-1},\zeta)+\delta(\eta^{x,x-1,1},\zeta)+\delta(\eta^{x,x+1,-1},\zeta)+\delta(\eta^{x,x+1,1},\zeta)],$

where $\delta(\eta,\zeta)=1$ if $\eta=\zeta$ and $=0$ otherwise.

Next, we want to regard the players, originally labeled from 1 to $N$, as
labeled from $l_{N}$ to $r_{N}$, where

$l_{N}:=\begin{cases}-(N-1)/2&\text{if $N$ is odd,}\\\ -N/2&\text{if $N$ is
even,}\end{cases}\quad r_{N}:=\begin{cases}(N-1)/2&\text{if $N$ is odd,}\\\
N/2-1&\text{if $N$ is even.}\end{cases}$ (1)

Then we can speed up time, playing $N$ games per unit of time, and our process
is described in the limit as $N\to\infty$ by an interacting particle system in
the state space $\Sigma:=\\{0,1\\}^{\bf Z}$. The details of this limit
operation are postponed to Section 3. Initially, our concern is with the
ergodicity of the limiting interacting particle system.

The generator $\Omega_{A^{\prime}}$ of the interacting particle system
corresponding to game $A^{\prime}$ can be described as follows. For
$\eta\in\Sigma:=\\{0,1\\}^{\bf Z}$ and $x\in{\bf Z}$, define $\eta_{x}$ and
${}_{x}\eta_{x+1}$ in $\Sigma$ by

$\eta_{x}(y):=\begin{cases}1-\eta(x)&\text{if $y=x$,}\\\
\eta(y)&\text{otherwise,}\end{cases}\quad\text{and}\quad_{x}\eta_{x+1}(y):=\begin{cases}\eta(x+1)&\text{if
$y=x$,}\\\ \eta(x)&\text{if $y=x+1$,}\\\ \eta(y)&\text{otherwise.}\end{cases}$

Then

$(\Omega_{A^{\prime}}f)(\eta):=\sum_{x}c^{\prime}(x,\eta)[f(\eta_{x})-f(\eta)]+\frac{1}{2}\sum_{x}[f(_{x}\eta_{x+1})-f(\eta)]$

for $f\in C(\Sigma)$ depending on only finitely many coordinates, where

$c^{\prime}(x,\eta):=\frac{1}{2}[\bm{1}_{\\{\eta(x)=\eta(x+1)\\}}+\bm{1}_{\\{\eta(x)=\eta(x-1)\\}}].$
(2)

Note that the generator $\Omega_{A^{\prime}}$ is the sum of a spin system
generator and an exclusion process generator.

The generator $\Omega_{B}$ of the interacting particle system corresponding to
game $B$ can be described as follows. For each $x\in{\bf Z}$, define
$\eta_{x}$ as above. Then, given parameters $p_{0},p_{1},p_{2},p_{3}\in[0,1]$,

$(\Omega_{B}f)(\eta):=\sum_{x}c(x,\eta)[f(\eta_{x})-f(\eta)]$

for $f\in C(\Sigma)$ depending on only finitely many coordinates, where

$c(x,\eta):=\begin{cases}p_{m_{x}(\eta)}&\text{if $\eta(x)=0$,}\\\
q_{m_{x}(\eta)}&\text{if $\eta(x)=1,$}\end{cases}$ (3)

$q_{m}:=1-p_{m}$ for $m=0,1,2,3$, and
$m_{x}(\eta):=2\eta(x-1)+\eta(x+1)\in\\{0,1,2,3\\}$.

Finally, the generator $\Omega_{C^{\prime}}$ of the interacting particle
system corresponding to game $C^{\prime}:=\gamma A^{\prime}+(1-\gamma)B$
(denoting the random mixture of games $A^{\prime}$ and $B$, i.e., the game in
which a $\gamma$-coin determines whether game $A^{\prime}$ or game $B$ is
played) is

$\Omega_{C^{\prime}}:=\gamma\Omega_{A^{\prime}}+(1-\gamma)\Omega_{B}.$

The generator $\Omega_{C^{\prime}}$, just like $\Omega_{A^{\prime}}$, is the
sum of a spin system generator and an exclusion process generator.

##  2 Ergodicity

First, $\Omega_{B}$ is a spin system generator, so the basic inequality yields
a sufficient condition for ergodicity (Liggett, 1985, Eq. (III.0.6)):

$\sup_{x\in{\bf Z}}\sum_{u:u\neq
x}\sup_{\eta\in\Sigma}|c(x,\eta)-c(x,\eta_{u})|<\inf_{x\in{\bf
Z},\;\eta\in\Sigma}[c(x,\eta)+c(x,\eta_{x})],$ (4)

where $c(x,\eta)$ is as in (3).

Next, $\Omega_{C^{\prime}}$, given by

$\displaystyle(\Omega_{C^{\prime}}f)(\eta)$
$\displaystyle:=\gamma(\Omega_{A^{\prime}}f)(\eta)+(1-\gamma)(\Omega_{B}f)(\eta)$
$\displaystyle\;=\gamma\sum_{x}c^{\prime}(x,\eta)[f(\eta_{x})-f(\eta)]+\frac{\gamma}{2}\sum_{x}[f(_{x}\eta_{x+1})-f(\eta)]$
$\displaystyle\qquad\qquad+(1-\gamma)\sum_{x}c(x,\eta)[f(\eta_{x})-f(\eta)],$
(5)

is not a spin system generator, so (4) does not apply. Instead, we apply a
more general form of the basic inequality. It assumes that the generator has
the form

$(\Omega f)(\eta)=\sum_{T\subset{\bf Z}\text{
finite}}\int_{\\{0,1\\}^{T}}c_{T}(\eta,d\zeta)[f(\eta^{\zeta})-f(\eta)],$ (6)

where $c_{T}(\eta,d\zeta)$ is a finite positive measure on $\\{0,1\\}^{T}$,
and

$\eta^{\zeta}(x):=\begin{cases}\zeta(x)&\text{if $x\in T$}\\\ \eta(x)&\text{if
$x\notin T$}\end{cases}$

for $\zeta$ $\in\\{0,1\\}^{T}$. The interpretation is that $\eta$ is the
current configuration, $c_{T}(\eta,\\{0,1\\}^{T})$ is the rate at which a
transition occurs involving coordinates in $T$, and
$c_{T}(\eta,d\zeta)/c_{T}(\eta,\\{0,1\\}^{T})$ is the distribution of the
restriction to $T$ of the new configuration after a transition.

We conclude from (2) and (6) that

$\displaystyle c_{\\{x\\}}(\eta,G)$
$\displaystyle=\delta_{1-\eta(x)}(G)[\gamma
c^{\prime}(x,\eta)+(1-\gamma)c(x,\eta)],$ $\displaystyle
c_{\\{x,x+1\\}}(\eta,H)$
$\displaystyle=\delta_{(\eta(x+1),\eta(x))}(H)\frac{\gamma}{2}\bm{1}_{\\{\eta(x)\neq\eta(x+1)\\}},$

where $\delta_{u}$ is the unit mass concentrated at $u$. Here
$G\subset\\{0,1\\}$ and $H\subset\\{(0,0),(0,1),(1,0),(1,1)\\}$.

The sufficient condition for ergodicity is that $M<\varepsilon$ (Liggett,
1985, Theorem I.4.1), where $M$ and $\varepsilon$ are constants, defined
below, that remain to be evaluated.

For $u\in{\bf Z}$ and finite $T\subset{\bf Z}$, let
$c_{T}(u):=\sup\\{\|c_{T}(\eta,d\zeta)-c_{T}(\eta^{\prime},d\zeta)\|_{\text{TV}}:\eta,\eta^{\prime}\in\Sigma,\eta(y)=\eta^{\prime}(y)\,\forall\,y\neq
u\\}$, where $\|\cdot\|_{\text{TV}}$ denotes the total variation norm of a
measure on $\\{0,1\\}^{T}$. Then

$\displaystyle M$ $\displaystyle:=\sup_{x\in{\bf Z}}\sum_{T\ni x}\sum_{u:u\neq
x}c_{T}(u)$ $\displaystyle=\sup_{x\in{\bf Z}}\sum_{T\ni x}\sum_{u:u\neq
x}\sup_{\eta,\eta^{\prime}\in\Sigma,\eta(y)=\eta^{\prime}(y)\,\forall\,y\neq
u}\|c_{T}(\eta,d\zeta)-c_{T}(\eta^{\prime},d\zeta)\|_{\text{TV}}$
$\displaystyle=\sup_{x\in{\bf Z}}\bigg{[}\sum_{u:u\neq
x}\sup_{\eta\in\Sigma}\|c_{\\{x\\}}(\eta,d\zeta)-c_{\\{x\\}}(\eta_{u},d\zeta)\|_{\text{TV}}$
$\displaystyle\qquad\;\;{}+\sum_{v:v\neq
x}\sup_{\eta\in\Sigma}\|c_{\\{x,x+1\\}}(\eta,d\zeta)-c_{\\{x,x+1\\}}(\eta_{v},d\zeta)\|_{\text{TV}}$
$\displaystyle\qquad\;\;{}+\sum_{w:w\neq
x}\sup_{\eta\in\Sigma}\|c_{\\{x-1,x\\}}(\eta,d\zeta)-c_{\\{x-1,x\\}}(\eta_{w},d\zeta)\|_{\text{TV}}\bigg{]}$
$\displaystyle=\sup_{x\in{\bf Z}}\bigg{[}\sum_{u:u\neq
x}\sup_{\eta\in\Sigma}\sup_{G\subset\\{0,1\\}}|c_{\\{x\\}}(\eta,G)-c_{\\{x\\}}(\eta_{u},G)|$
$\displaystyle\qquad\;\;{}+\sum_{v:v\neq
x}\sup_{\eta\in\Sigma}\sup_{H\subset\\{(0,0),(0,1),(1,0),(1,1)\\}}|c_{\\{x,x+1\\}}(\eta,H)-c_{\\{x,x+1\\}}(\eta_{v},H)|$
$\displaystyle\qquad\;\;{}+\sum_{w:w\neq
x}\sup_{\eta\in\Sigma}\sup_{H\subset\\{(0,0),(0,1),(1,0),(1,1)\\}}|c_{\\{x-1,x\\}}(\eta,H)-c_{\\{x-1,x\\}}(\eta_{w},H)|\bigg{]}$
$\displaystyle=\sup_{x\in{\bf Z}}\bigg{[}\sum_{u:u\neq
x}\sup_{\eta\in\Sigma}|\gamma c^{\prime}(x,\eta)+(1-\gamma)c(x,\eta)-[\gamma
c^{\prime}(x,\eta_{u})+(1-\gamma)c(x,\eta_{u})]|$
$\displaystyle\qquad\;\;{}+\sup_{\eta\in\Sigma}\sup_{H\subset\\{(0,0),(0,1),(1,0),(1,1)\\}}|c_{\\{x,x+1\\}}(\eta,H)-c_{\\{x,x+1\\}}(\eta_{x+1},H)|$
$\displaystyle\qquad\;\;{}+\sup_{\eta\in\Sigma}\sup_{H\subset\\{(0,0),(0,1),(1,0),(1,1)\\}}|c_{\\{x-1,x\\}}(\eta,H)-c_{\\{x-1,x\\}}(\eta_{x-1},H)|\bigg{]}$
$\displaystyle=\sup_{x\in{\bf Z}}\sum_{u=x\pm 1}\sup_{\eta\in\Sigma}|\gamma
c^{\prime}(x,\eta)+(1-\gamma)c(x,\eta)-[\gamma
c^{\prime}(x,\eta_{u})+(1-\gamma)c(x,\eta_{u})]|$
$\displaystyle\qquad\qquad\qquad{}+2\cdot\frac{\gamma}{2}$
$\displaystyle=\sup_{x\in{\bf
Z}}\bigg{[}\sup_{\eta\in\Sigma}|\gamma[c^{\prime}(x,\eta)-c^{\prime}(x,\eta_{x+1})]+(1-\gamma)[c(x,\eta)-c(x,\eta_{x+1})]|$
$\displaystyle\qquad\qquad{}+\sup_{\eta\in\Sigma}|\gamma[c^{\prime}(x,\eta)-c^{\prime}(x,\eta_{x-1})]+(1-\gamma)[c(x,\eta)-c(x,\eta_{x-1})]|\bigg{]}+\gamma$
$\displaystyle=\max\bigg{[}\bigg{|}\frac{\gamma}{2}+(1-\gamma)(p_{0}-p_{1})\bigg{|},\bigg{|}\frac{\gamma}{2}+(1-\gamma)(p_{2}-p_{3})\bigg{|}\bigg{]}$
$\displaystyle\qquad{}+\max\bigg{[}\bigg{|}\frac{\gamma}{2}+(1-\gamma)(p_{0}-p_{2})\bigg{|},\bigg{|}\frac{\gamma}{2}+(1-\gamma)(p_{1}-p_{3})\bigg{|}\bigg{]}+\gamma,$
(7)

where the last step requires clarification. Notice first that

$c^{\prime}(x,\eta)=\frac{1}{2}[\bm{1}_{\\{\eta(x)=\eta(x+1)\\}}+\bm{1}_{\\{\eta(x)=\eta(x-1)\\}}],$
$\displaystyle c^{\prime}(x,\eta_{x+1})$
$\displaystyle=\frac{1}{2}[\bm{1}_{\\{\eta_{x+1}(x)=\eta_{x+1}(x+1)\\}}+\bm{1}_{\\{\eta_{x+1}(x)=\eta_{x+1}(x-1)\\}}]$
$\displaystyle=\frac{1}{2}[\bm{1}_{\\{\eta(x)=1-\eta(x+1)\\}}+\bm{1}_{\\{\eta(x)=\eta(x-1)\\}}],$
$\displaystyle c^{\prime}(x,\eta_{x-1})$
$\displaystyle=\frac{1}{2}[\bm{1}_{\\{\eta_{x-1}(x)=\eta_{x-1}(x+1)\\}}+\bm{1}_{\\{\eta_{x-1}(x)=\eta_{x-1}(x-1)\\}}]$
$\displaystyle=\frac{1}{2}[\bm{1}_{\\{\eta(x)=\eta(x+1)\\}}+\bm{1}_{\\{\eta(x)=1-\eta(x-1)\\}}],$
$c(x,\eta)=\begin{cases}p_{m_{x}(\eta)}&\text{if $\eta(x)=0$,}\\\
q_{m_{x}(\eta)}&\text{if $\eta(x)=1$,}\end{cases}$
$c(x,\eta_{x+1})=\begin{cases}p_{m_{x}(\eta_{x+1})}&\text{if $\eta(x)=0$,}\\\
q_{m_{x}(\eta_{x+1})}&\text{if $\eta(x)=1$,}\end{cases}\quad
c(x,\eta_{x-1})=\begin{cases}p_{m_{x}(\eta_{x-1})}&\text{if $\eta(x)=0$,}\\\
q_{m_{x}(\eta_{x-1})}&\text{if $\eta(x)=1$,}\end{cases}$

where $q_{m}:=1-p_{m}$ for $m=0,1,2,3$ and
$m_{x}(\eta):=2\eta(x-1)+\eta(x+1)\in\\{0,1,2,3\\}$. The last line of (2) is
by direct calculation.

Next,

$\displaystyle\varepsilon$ $\displaystyle:=\inf_{u\in{\bf
Z}}\;\inf_{\eta=\eta^{\prime}\text{ off
}u,\;\eta(u)\neq\eta^{\prime}(u)}\sum_{T\ni
u}[c_{T}(\eta,\\{\zeta\in\\{0,1\\}^{T}:\zeta(u)=\eta^{\prime}(u)\\})$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad{}+c_{T}(\eta^{\prime},\\{\zeta\in\\{0,1\\}^{T}:\zeta(u)=\eta(u)\\})]$
$\displaystyle=\inf_{u\in{\bf
Z}}\inf_{\eta\in\Sigma}\big{[}c_{\\{u\\}}(\eta,\\{\eta_{u}(u)\\})+c_{\\{u\\}}(\eta_{u},\\{\eta(u)\\})$
$\displaystyle\qquad\qquad\quad{}+c_{\\{u,u+1\\}}(\eta,\\{\zeta\in\\{0,1\\}^{\\{u,u+1\\}}:\zeta(u)=\eta_{u}(u)\\})$
$\displaystyle\qquad\qquad\quad{}+c_{\\{u,u+1\\}}(\eta_{u},\\{\zeta\in\\{0,1\\}^{\\{u,u+1\\}}:\zeta(u)=\eta(u)\\})$
$\displaystyle\qquad\qquad\quad{}+c_{\\{u-1,u\\}}(\eta,\\{\zeta\in\\{0,1\\}^{\\{u-1,u\\}}:\zeta(u)=\eta_{u}(u)\\})$
$\displaystyle\qquad\qquad\quad{}+c_{\\{u-1,u\\}}(\eta_{u},\\{\zeta\in\\{0,1\\}^{\\{u-1,u\\}}:\zeta(u)=\eta(u)\\})\big{]}$
$\displaystyle=\inf_{u\in{\bf
Z}}\inf_{\eta\in\Sigma}\big{[}c_{\\{u\\}}(\eta,\\{1-\eta(u)\\})+c_{\\{u\\}}(\eta_{u},\\{1-\eta_{u}(u)\\})$
$\displaystyle\qquad\qquad\quad{}+c_{\\{u,u+1\\}}(\eta,\\{\zeta\in\\{0,1\\}^{\\{u,u+1\\}}:\zeta(u)=1-\eta(u)\\})$
$\displaystyle\qquad\qquad\quad{}+c_{\\{u,u+1\\}}(\eta_{u},\\{\zeta\in\\{0,1\\}^{\\{u,u+1\\}}:\zeta(u)=1-\eta_{u}(u)\\})$
$\displaystyle\qquad\qquad\quad{}+c_{\\{u-1,u\\}}(\eta,\\{\zeta\in\\{0,1\\}^{\\{u-1,u\\}}:\zeta(u)=1-\eta(u)\\})$
$\displaystyle\qquad\qquad\quad{}+c_{\\{u-1,u\\}}(\eta_{u},\\{\zeta\in\\{0,1\\}^{\\{u-1,u\\}}:\zeta(u)=1-\eta_{u}(u)\\})\big{]}$
$\displaystyle=\inf_{u\in{\bf
Z}}\inf_{\eta\in\Sigma}\big{[}\gamma[c^{\prime}(u,\eta)+c^{\prime}(u,\eta_{u})]+(1-\gamma)[c(u,\eta)+c(u,\eta_{u})]\big{]}+2\cdot\frac{\gamma}{2}$
$\displaystyle=1+\gamma,$ (8)

where the last line of (2) follows from

$\displaystyle c^{\prime}(u,\eta)+c^{\prime}(u,\eta_{u})$
$\displaystyle=\frac{1}{2}[\bm{1}_{\\{\eta(u)=\eta(u+1)\\}}+\bm{1}_{\\{\eta(u)=\eta(u-1)\\}}]$
$\displaystyle\quad{}+\frac{1}{2}[\bm{1}_{\\{\eta_{u}(u)=\eta_{u}(u+1)\\}}+\bm{1}_{\\{\eta_{u}(u)=\eta_{u}(u-1)\\}}]$
$\displaystyle=\frac{1}{2}[\bm{1}_{\\{\eta(u)=\eta(u+1)\\}}+\bm{1}_{\\{\eta(u)=\eta(u-1)\\}}]$
$\displaystyle\quad{}+\frac{1}{2}[\bm{1}_{\\{\eta(u)=1-\eta(u+1)\\}}+\bm{1}_{\\{\eta(u)=1-\eta(u-1)\\}}]$
$\displaystyle=1$

and $c(u,\eta)+c(u,\eta_{u})=1$. We have proved the following theorem.

######  Theorem 1.

The interacting particle system in $\Sigma:=\\{0,1\\}^{\bf Z}$ with generator
$\Omega_{C^{\prime}}:=\gamma\Omega_{A^{\prime}}+(1-\gamma)\Omega_{B}$, where
$0<\gamma<1$, is ergodic if

$\displaystyle\max\bigg{[}\bigg{|}\frac{\gamma}{2}+(1-\gamma)(p_{0}-p_{1})\bigg{|},\bigg{|}\frac{\gamma}{2}+(1-\gamma)(p_{2}-p_{3})\bigg{|}\bigg{]}$
$\displaystyle\qquad+\max\bigg{[}\bigg{|}\frac{\gamma}{2}+(1-\gamma)(p_{0}-p_{2})\bigg{|},\bigg{|}\frac{\gamma}{2}+(1-\gamma)(p_{1}-p_{3})\bigg{|}\bigg{]}<1.$
(9)

The volume of the subset of the parameter space $[0,1]^{4}$ for which (1)
holds with $\gamma=1/2$ is 5/6. If we assume that $p_{1}=p_{2}$, then the
volume of the subset of the parameter space $[0,1]^{3}$ for which (1) holds
with $\gamma=1/2$ is 3/4. In fact, the volume is $3/4$ if and only if
$\gamma\geq 1/3$.

##  3 Convergence of means

We would like to prove that
$\lim_{N\rightarrow\infty}\mu^{N}_{(\gamma,1-\gamma)^{\prime}}$ and
$\lim_{N\rightarrow\infty}\mu^{N}_{[r,s]^{\prime}}$ exist under certain
conditions, where $\mu^{N}_{(\gamma,1-\gamma)^{\prime}}$ denotes the mean
profit per turn at equilibrium to the $N$ players playing the
$(\gamma,1-\gamma)$ random mixture of games $A^{\prime}$ and $B$ (the Parrondo
games of Xie et al., 2011), and $\mu^{N}_{[r,s]^{\prime}}$ denotes the mean
profit per turn at equilibrium to the $N$ players playing games $A^{\prime}$
and $B$ in the nonrandom periodic pattern
$A^{\prime},A^{\prime},\dots,A^{\prime}$ ($r$ times), $B,B,\dots,B$ ($s$
times), $A^{\prime},A^{\prime},\dots,A^{\prime}$ ($r$ times), $B,B,\dots,B$
($s$ times), and so on. The first result is relatively straightforward, while
the second requires more work. The key step for the second result is to prove
that the sequence of discrete generators converges to the generator of an
interacting particle system.

We want to show that our sequence of discrete-time Markov chains, suitably
rescaled, converges in distribution to an interacting particle system on ${\bf
Z}$. The limiting process is characterized in terms of its generator. First,
we need to define generators corresponding to game $A^{\prime}$, game $B$, and
game $C^{\prime}$. The state space is

$\displaystyle\Sigma$ $\displaystyle:=\\{0,1\\}^{{\bf Z}}$
$\displaystyle\phantom{:}=\\{\eta=(\ldots,\eta(-2),\eta(-1),\eta(0),\eta(1),\eta(2),\ldots):\eta(x)\in\\{0,1\\}\text{
for all }x\in{\bf Z}\\}.$

In Section 1 we defined $\eta^{x,x\pm 1,\pm 1}$ for $\eta\in\\{0,1\\}^{N}$,
and that definition is easily extended to $\eta\in\Sigma$. For $\eta\in\Sigma$
and $x\in{\bf Z}$ define $\eta^{x,-1}$ and $\eta^{x,1}$ to be the elements of
$\Sigma$ given by

$\eta^{x,-1}(y):=\begin{cases}1&\text{if $y=x-1$,}\\\ 0&\text{if $y=x$,}\\\
\eta(y)&\text{otherwise,}\end{cases}\qquad\eta^{x,1}(y):=\begin{cases}0&\text{if
$y=x$,}\\\ 1&\text{if $y=x+1$,}\\\ \eta(y)&\text{otherwise.}\end{cases}$

For example,
$\eta^{0,1}:=(\ldots,\eta(-2),\eta(-1),0,1,\eta(2),\eta(3),\ldots)$. And let
$\eta_{x}$ be the element of $\Sigma$ equal to $\eta$ except at the $x$th
coordinate. Then the generators are

$\displaystyle(\Omega_{A^{\prime}}f)(\eta)$ $\displaystyle:=\sum_{x\in{\bf
Z}}\bigg{[}\frac{1}{4}f(\eta^{x,x-1,-1})+\frac{1}{4}f(\eta^{x,x-1,1})+\frac{1}{4}f(\eta^{x,x+1,-1})+\frac{1}{4}f(\eta^{x,x+1,1})-f(\eta)\bigg{]}$
$\displaystyle=\frac{1}{2}\sum_{x\in{\bf
Z}}[f(\eta^{x-1,x,1})-f(\eta)]+\frac{1}{2}\sum_{x\in{\bf
Z}}[f(\eta^{x,x+1,-1})-f(\eta)]$
$\displaystyle=\frac{1}{2}\bigg{[}\sum_{x:(\eta(x-1),\eta(x))=(0,0)}[f(\eta_{x-1})-f(\eta)]+\sum_{x:(\eta(x-1),\eta(x))=(0,1)}[f(_{x-1}\eta_{x})-f(\eta)]$
$\displaystyle\quad{}+\sum_{x:(\eta(x-1),\eta(x))=(1,1)}[f(\eta_{x})-f(\eta)]+\sum_{x:(\eta(x),\eta(x+1))=(0,0)}[f(\eta_{x+1})-f(\eta)]$
$\displaystyle\quad{}+\sum_{x:(\eta(x),\eta(x+1))=(1,0)}[f(_{x}\eta_{x+1})-f(\eta)]+\sum_{x:(\eta(x),\eta(x+1))=(1,1)}[f(\eta_{x})-f(\eta)]\bigg{]}$
$\displaystyle=\frac{1}{2}\bigg{[}\sum_{x:(\eta(x),\eta(x+1))=(0,0)}[f(\eta_{x})-f(\eta)]+\sum_{x:(\eta(x),\eta(x+1))=(0,1)}[f(_{x}\eta_{x+1})-f(\eta)]$
$\displaystyle\quad{}+\sum_{x:(\eta(x),\eta(x+1))=(1,1)}[f(\eta_{x+1})-f(\eta)]+\sum_{x:(\eta(x),\eta(x+1))=(0,0)}[f(\eta_{x+1})-f(\eta)]$
$\displaystyle\quad{}+\sum_{x:(\eta(x),\eta(x+1))=(1,0)}[f(_{x}\eta_{x+1})-f(\eta)]+\sum_{x:(\eta(x),\eta(x+1))=(1,1)}[f(\eta_{x})-f(\eta)]\bigg{]}$
$\displaystyle=\frac{1}{2}\bigg{[}\sum_{x:\eta(x)=\eta(x+1)}[f(\eta_{x})-f(\eta)]+\sum_{x:\eta(x)=\eta(x+1)}[f(\eta_{x+1})-f(\eta)]$
$\displaystyle\qquad\quad{}+\sum_{x\in{\bf
Z}}[f(_{x}\eta_{x+1})-f(\eta)]\bigg{]}$
$\displaystyle=\frac{1}{2}\bigg{[}\sum_{x:\eta(x)=\eta(x+1)}[f(\eta_{x})-f(\eta)]+\sum_{x:\eta(x-1)=\eta(x)}[f(\eta_{x})-f(\eta)]$
$\displaystyle\qquad\quad{}+\sum_{x\in{\bf
Z}}[f(_{x}\eta_{x+1})-f(\eta)]\bigg{]}$ $\displaystyle=\sum_{x\in{\bf
Z}}c^{\prime}(x,\eta)[f(\eta_{x})-f(\eta)]+\frac{1}{2}\sum_{x\in{\bf
Z}}[f(_{x}\eta_{x+1})-f(\eta)],$

where $c^{\prime}(x,\eta)$ is as in (2),

$(\Omega_{B}f)(\eta):=\sum_{x\in{\bf
Z}}c(x,\eta)\big{[}f(\eta_{x})-f(\eta)\big{]},$

where $c(x,\eta)$ is as in (3), and

$(\Omega_{C^{\prime}}f)(\eta):=\big{[}\gamma\Omega_{A^{\prime}}f+(1-\gamma)\Omega_{B}f\big{]}(\eta)$

for functions $f\in C(\Sigma)$ depending on only finitely many coordinates.

Next, it is necessary to show that this interacting particle system is the
limit in distribution of the $N$-player model as $N\to\infty$. Furthermore, we
need to adjust the state space by relabeling the players. Specifically, we let

$\Sigma_{N}:=\\{\eta=(\eta(l_{N}),\ldots,\eta(r_{N})):\eta(x)\in\\{0,1\\}\text{
for }x=l_{N},\ldots,r_{N}\\},$

where $l_{N}$ and $r_{N}$ are as in (1). It should be noted that players
$l_{N}$ and $r_{N}$ are nearest neighbors. We denote the Markov chain in
$\Sigma_{N}$ by $\\{X_{k}^{\cdot,N},\;k=0,1,2,\ldots\\}$, where $A^{\prime}$,
$B$, or $C^{\prime}$ appears in place of the dot.

First, let us analyze game $A^{\prime}$. The one-step transition matrix
$\bm{P}_{A^{\prime}}$ of the Markov chain in the state space $\Sigma_{N}$ has
the form

$\displaystyle P_{A^{\prime}}(\eta,\xi)$
$\displaystyle:=\frac{1}{4N}\sum_{l_{N}\leq x\leq
r_{N}}[\delta(\eta^{x,x-1,-1},\xi)+\delta(\eta^{x,x-1,1},\xi)$
$\displaystyle\qquad\qquad\qquad\qquad{}+\delta(\eta^{x,x+1,-1},\xi)+\delta(\eta^{x,x+1,1},\xi)]$
$\displaystyle\;=\frac{1}{2N}\sum_{l_{N}\leq x\leq
r_{N}}\big{[}\delta(\eta^{x,-1},\xi)+\delta(\eta^{x,1},\xi)\big{]},$

where $\delta(\eta,\xi)$ is the Kronecker delta, which is $1$ if $\eta=\xi$
and is $0$ otherwise; the sum over $x$ ranges over $\\{l_{N},\dots,r_{N}\\}$,
and $l_{N}-1:=r_{N}$ and $r_{N}+1:=l_{N}$. Next, we have the one-step
transition matrix $\bm{P}_{B}$ of the form

$P_{B}(\xi,\zeta):=\frac{1}{N}\sum_{y}[1-c(y,\xi)]\delta(\xi,\zeta)+\frac{1}{N}\sum_{y}c(y,\xi)\delta(\xi_{y},\zeta),$

where the sum over $y$ also ranges over $\\{l_{N},\dots,r_{N}\\}$; $c(y,\xi)$
is as in (3), except that $l_{N}-1:=r_{N}$ and $r_{N}+1:=l_{N}$.

We speed up time in the $N$-player model so that $N$ one-step transitions
occur per unit of time. Then the discrete generator corresponding to game
$A^{\prime}$ is

$\displaystyle(\Omega^{N}_{A^{\prime}}f)(\eta)$
$\displaystyle=N\text{E}\big{[}f(X_{1}^{A^{\prime},N})-f(\eta)\mid
X_{0}^{A^{\prime},N}=\eta\big{]}$
$\displaystyle=N\sum_{\xi\in\Sigma_{N}}P_{A^{\prime}}(\eta,\xi)\big{[}f(\xi)-f(\eta)\big{]}$
$\displaystyle=N\sum_{\xi\in\Sigma_{N}}\frac{1}{2N}\sum_{l_{N}\leq x\leq
r_{N}}\big{[}\delta(\eta^{x,-1},\xi)+\delta(\eta^{x,1},\xi)\big{]}\big{[}f(\xi)-f(\eta)\big{]}$
$\displaystyle=\frac{1}{2}\sum_{l_{N}\leq x\leq
r_{N}}\big{[}f(\eta^{x,-1})-f(\eta)+f(\eta^{x,1})-f(\eta)\big{]}$
$\displaystyle=\sum_{l_{N}\leq x\leq
r_{N}}\bigg{[}\frac{1}{2}f(\eta^{x,-1})+\frac{1}{2}f(\eta^{x,1})-f(\eta)\bigg{]}.$

The discrete generator corresponding to game $B$ is

$\displaystyle(\Omega^{N}_{B}f)(\eta)$
$\displaystyle=N\text{E}\big{[}f(X_{1}^{B,N})-f(\eta)\mid
X_{0}^{B,N}=\eta\big{]}$
$\displaystyle=N\sum_{\xi\in\Sigma_{N}}P_{B}(\eta,\xi)\big{[}f(\xi)-f(\eta)\big{]}$
$\displaystyle=\sum_{l_{N}\leq x\leq
r_{N}:\eta(x)=0}p_{m_{x}(\eta)}\big{[}f(\eta_{x})-f(\eta)\big{]}$
$\displaystyle\qquad\quad+\sum_{l_{N}\leq x\leq
r_{N}:\eta(x)=1}q_{m_{x}(\eta)}\big{[}f(\eta_{x})-f(\eta)\big{]}.$

Hence the discrete generator corresponding to game $C^{\prime}$ is

$\displaystyle(\Omega^{N}_{C^{\prime}}f)(\eta)$
$\displaystyle=N\text{E}\big{[}f(X_{1}^{C^{\prime},N})-f(\eta)\mid
X_{0}^{C^{\prime},N}=\eta\big{]}$
$\displaystyle=\big{[}\gamma\Omega_{A^{\prime}}^{N}f+(1-\gamma)\Omega_{B}^{N}f\big{]}(\eta)$
$\displaystyle=\gamma\sum_{l_{N}\leq x\leq
r_{N}}\bigg{[}\frac{1}{2}f(\eta^{x,-1})+\frac{1}{2}f(\eta^{x,1})-f(\eta)\bigg{]}$
$\displaystyle\qquad+(1-\gamma)\bigg{[}\sum_{l_{N}\leq x\leq
r_{N}:\eta(x)=0}p_{m_{x}(\eta)}\big{[}f(\eta_{x})-f(\eta)\big{]}$
$\displaystyle\qquad\qquad\qquad\qquad+\sum_{l_{N}\leq x\leq
r_{N}:\eta(x)=1}q_{m_{x}(\eta)}\big{[}f(\eta_{x})-f(\eta)\big{]}\bigg{]}.$

We define $\psi_{N}:B(\Sigma)\mapsto B(\Sigma_{N})$ by

$(\psi_{N}f)(\eta(l_{N}),\ldots,\eta(r_{N})):=f(\ldots,1,1,\eta(l_{N}),\ldots,\eta(r_{N}),1,1,\ldots).$

######  Lemma 2.

If $f\in C(\Sigma)$ depends on $\eta$ only through the $2K+1$ components
$\eta(x)$ for $-K\leq x\leq K$, then

$(\Omega^{N}_{A^{\prime}}\psi_{N}f)(\eta)=\psi_{N}(\Omega_{A^{\prime}}f)(\eta),$
$(\Omega^{N}_{B}\psi_{N}f)(\eta)=\psi_{N}(\Omega_{B}f)(\eta),$

and

$(\Omega^{N}_{C^{\prime}}\psi_{N}f)(\eta)=\psi_{N}(\Omega_{C^{\prime}}f)(\eta)$

for all $\eta\in\Sigma_{N}$ and $N\geq 2K+4$.

###### Proof.

The proof is straightforward. ∎

Lemma 2 implies that the process $\\{X_{\lfloor Nt\rfloor}^{C^{\prime},N}\\}$
converges in distribution to the interacting particle system $\\{X_{t}\\}$ by
Theorem 1.6.5 and 4.2.6 of Ethier and Kurtz (1986). More importantly, it
implies that, if the interacting particle system has a unique stationary
distribution, then the unique stationary distribution of the $N$-player Markov
chain converges to it in the topology of weak convergence, essentially by
Proposition I.2.14 of Liggett (1985). Let us assume that the interacting
particle system with generator $\Omega_{C^{\prime}}$ has a unique stationary
distribution $\pi$, and let us denote the unique stationary distribution of
the $N$-player Markov chain for the $(\gamma,1-\gamma)$ random mixture of
games $A^{\prime}$ and $B$ by $\pi^{N}$. Let us denote their $-1,1$ two-
dimensional marginals by $\pi_{-1,1}^{N}$ and $\pi_{-1,1}$. Then we have

$\displaystyle\mu_{(\gamma,1-\gamma)^{\prime}}^{N}$
$\displaystyle=(1-\gamma)[\pi^{N}_{-1,1}(0,0)(2p_{0}-1)+\pi^{N}_{-1,1}(0,1)(2p_{1}-1)$
$\displaystyle\qquad{}+\pi^{N}_{-1,1}(1,0)(2p_{2}-1)+\pi^{N}_{-1,1}(1,1)(2p_{3}-1)]$
$\displaystyle\to(1-\gamma)[\pi_{-1,1}(0,0)(2p_{0}-1)+\pi_{-1,1}(0,1)(2p_{1}-1)$
$\displaystyle\qquad{}+\pi_{-1,1}(1,0)(2p_{2}-1)+\pi_{-1,1}(1,1)(2p_{3}-1)]$
$\displaystyle=:\mu_{(\gamma,1-\gamma)^{\prime}},$ (10)

where the first equality is based on the following idea: Suppose player 0 is
chosen to play. If he plays game $A^{\prime}$ (probability $\gamma$) he wins
or loses 1 with probability 1/2 each, so his mean profit is 0; if he plays
game $B$ (probability $1-\gamma$), he tosses a $p_{m}$ coins with $m$
determined by the status of his nearest neighbors, hence his expected profit
is $2p_{m}-1$. We conclude that $\mu_{(\gamma,1-\gamma)^{\prime}}^{N}$, the
mean profit per turn at equilibrium to the $N$ players playing the
$(\gamma,1-\gamma)$ random mixture of games $A^{\prime}$ and $B$, converges as
$N\to\infty$ to a limit that can be expressed in terms of an interacting
particle system. We have proved the following.

######  Theorem 3.

Fix $\gamma\in(0,1)$. Assume that the interacting particle system on ${\bf Z}$
with generator
$\Omega_{C^{\prime}}:=\gamma\Omega_{A^{\prime}}+(1-\gamma)\Omega_{B}$ is
ergodic with unique stationary distribution $\pi$. Then
$\lim_{N\to\infty}\mu_{(\gamma,1-\gamma)^{\prime}}^{N}=\mu_{(\gamma,1-\gamma)^{\prime}}$,
where $\mu_{(\gamma,1-\gamma)^{\prime}}$ is as in (3).

The discrete generator for the nonrandom periodic pattern
$(A^{\prime})^{r}B^{s}$ has the form, for $f\in B(\Sigma_{N})$,

$(\Omega^{N}_{[r,s]^{\prime}}f)(\eta^{0})=\frac{N}{r+s}\sum_{\eta^{r+s}}[f(\eta^{r+s})-f(\eta^{0})](\bm{P}^{r}_{A^{\prime}}\bm{P}^{s}_{B})(\eta^{0},\eta^{r+s}).$

We begin by evaluating

$\displaystyle\\!\\!\\!\\!\\!(\bm{P}^{r}_{A^{\prime}}\bm{P}^{s}_{B})(\eta^{0},\eta^{r+s})$
$\displaystyle=\sum_{\eta^{1},\eta^{2},\ldots,\eta^{r+s-1}}P_{A^{\prime}}(\eta^{0},\eta^{1})P_{A^{\prime}}(\eta^{1},\eta^{2})\cdots
P_{A^{\prime}}(\eta^{r-1},\eta^{r})$ $\displaystyle\quad\cdot
P_{B}(\eta^{r},\eta^{r+1})P_{B}(\eta^{r+1},\eta^{r+2})\cdots
P_{B}(\eta^{r+s-1},\eta^{r+s})$
$\displaystyle=\sum_{\eta^{1},\eta^{2},\ldots,\eta^{r+s-1}}\prod_{i=1}^{r}P_{A^{\prime}}(\eta^{i-1},\eta^{i})\prod_{i=r+1}^{r+s}P_{B}(\eta^{i-1},\eta^{i})$
$\displaystyle=\sum_{\eta^{1},\eta^{2},\ldots,\eta^{r+s-1}}\prod_{i=1}^{r}\bigg{[}\frac{1}{2N}\sum_{x_{i}}[\delta((\eta^{i-1})^{x_{i},-1},\eta^{i})+\delta((\eta^{i-1})^{x_{i},1},\eta^{i})]\bigg{]}$
$\displaystyle\quad\cdot\prod_{i=r+1}^{r+s}\bigg{[}\frac{1}{N}\sum_{x_{i}}(1-c(x_{i},\eta^{i-1}))\delta(\eta^{i-1},\eta^{i})+\frac{1}{N}\sum_{x_{i}}c(x_{i},\eta^{i-1})\delta((\eta^{i-1})_{x_{i}},\eta^{i})\bigg{]}$
$\displaystyle=\frac{1}{2^{r}N^{r+s}}\sum_{A\subset\\{1,\ldots,r\\}}\sum_{B\subset\\{r+1,\ldots,r+s\\}}\sum_{\eta^{1},\eta^{2},\ldots,\eta^{r+s-1}}\prod_{i\in
A^{c}}\bigg{[}\sum_{x_{i}}\delta((\eta^{i-1})^{x_{i},-1},\eta^{i})\bigg{]}$
$\displaystyle\quad\cdot\prod_{i\in
A}\bigg{[}\sum_{x_{i}}\delta((\eta^{i-1})^{x_{i},1},\eta^{i})\bigg{]}\prod_{i\in
B^{c}}\bigg{[}\sum_{x_{i}}(1-c(x_{i},\eta^{i-1}))\delta(\eta^{i-1},\eta^{i})\bigg{]}$
$\displaystyle\quad\cdot\prod_{i\in
B}\bigg{[}\sum_{x_{i}}c(x_{i},\eta^{i-1})\delta((\eta^{i-1})_{x_{i}},\eta^{i})\bigg{]}$
$\displaystyle=\frac{1}{2^{r}N^{r+s}}\sum_{A\subset\\{1,\ldots,r\\}}\sum_{B\subset\\{r+1,\ldots,r+s\\}}\sum_{\eta^{1},\eta^{2},\ldots,\eta^{r+s-1}}\sum_{x_{i}:i\in
A^{c}}\sum_{x_{i}:i\in A}$ $\displaystyle\quad\cdot\prod_{j\in
A^{c}}\big{[}\delta((\eta^{j-1})^{x_{j},-1},\eta^{j})\big{]}\prod_{j\in
A}\big{[}\delta((\eta^{j-1})^{x_{j},1},\eta^{j})\big{]}\sum_{x_{i}:i\in
B^{c}}\sum_{x_{i}:i\in B}$ $\displaystyle\quad\cdot\prod_{j\in
B^{c}}\big{[}(1-c(x_{j},\eta^{j-1}))\delta(\eta^{j-1},\eta^{j})\big{]}\prod_{j\in
B}\big{[}c(x_{j},\eta^{j-1})\delta((\eta^{j-1})_{x_{j}},\eta^{j})\big{]}$
$\displaystyle=\frac{1}{2^{r}N^{r+s}}\sum_{A\subset\\{1,\ldots,r\\}}\sum_{B\subset\\{r+1,\ldots,r+s\\}}\sum_{x_{i}:i\in\\{1,2,\ldots,r+s\\}}$
$\displaystyle\quad\cdot\prod_{j\in
B^{c}}\big{[}1-c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})\big{]}$ $\displaystyle\quad\cdot\prod_{j\in
B}c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})$
$\displaystyle\qquad\times\delta(((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B\\}},\eta^{r+s}),$ (11)

where $A^{c}:=\\{1,2,\ldots,r\\}-A$ and $B^{c}:=\\{r+1,r+2,\ldots,r+s\\}-B$;
also $p\in\\{1,2,\ldots,r\\}$ and

$a_{p}=\begin{cases}-1&\text{if $p\in A^{c}$},\\\ \phantom{-}1&\text{if $p\in
A$}.\end{cases}$

Here, for example, $\eta_{\\{x_{l}:l\in B\\}}$ denotes $\eta$ with the spins
flipped at each site $x_{l}$ with $l\in B$. These site labels need not be
distinct, so if there are multiple flips at a single site, only the parity of
the number of flips is relevant.

Next, assume that $f\in B(\Sigma)$ depends only on
$\eta(-(K-2)),\ldots,\eta(K-2)$ for some integer $K\geq 2$, and put
$f_{N}:=\psi_{N}f\in B(\Sigma_{N})$. Then the discrete generator for the
pattern $(A^{\prime})^{r}B^{s}$, acting on $f_{N}$, reduces to

$\displaystyle(\Omega^{N}_{[r,s]^{\prime}}f_{N})(\eta^{0})$
$\displaystyle=\frac{N}{r+s}\sum_{\eta^{r+s}}[f_{N}(\eta^{r+s})-f_{N}(\eta^{0})](\bm{P}^{r}_{A^{\prime}}\bm{P}^{s}_{B})(\eta^{0},\eta^{r+s})$
$\displaystyle=\frac{1}{r+s}\
\frac{1}{2^{r}N^{r+s-1}}\sum_{A\subset\\{1,\ldots,r\\}}\sum_{B\subset\\{r+1,\ldots,r+s\\}}\sum_{x_{i}:i\in\\{1,2,\ldots,r+s\\}}$
$\displaystyle\quad\cdot\prod_{j\in
B^{c}}\big{[}1-c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})\big{]}$ $\displaystyle\quad\cdot\prod_{j\in
B}c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})$
$\displaystyle\quad\cdot[f_{N}(((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B\\}})-f_{N}(\eta^{0})]$ (12)

We replace $\sum_{x_{i}:i\in\\{1,2,\ldots,r+s\\}}$ in (3) by

$\displaystyle\bigg{(}\sum_{i\in A^{c}}+\sum_{i\in A}\bigg{)}\sum_{|x_{i}|\leq
K}\sum_{|x_{m}|>K:m\in\\{1,2,\ldots,r+s\\},m\neq i}$
$\displaystyle\quad{}+\sum_{i\in B}\sum_{|x_{i}|\leq
K}\sum_{|x_{m}|>K:m\in\\{1,2,\ldots,r+s\\},m\neq i}$ (13)

since each sum $\sum_{x_{i}}$ can be written as $\sum_{|x_{i}|\leq
K}+\sum_{|x_{i}|>K}$ resulting in $2^{r+s}$ multiple sums in which each of
those multiple sums with two or more sums of the form $\sum_{|x_{i}|\leq K}$
contributes at most $O(N^{-1})$ and those without the form $\sum_{|x_{i}|\leq
K},$ where $i\in\\{1,2,\ldots,r\\}\cup B$ are $0$. So it is enough to analyze
the cases in which only one of the $|x_{i}|$’s is less than or equal to $K$.

We consider first the first term in (3). It contributes

$\displaystyle\frac{1}{r+s}\;\frac{1}{2^{r}N^{r+s-1}}\sum_{A\subset\\{1,\ldots,r\\}}\sum_{B\subset\\{r+1,\ldots,r+s\\}}\bigg{[}\sum_{i\in
A^{c}}\sum_{|x_{i}|\leq K}[f_{N}((\eta^{0})^{x_{i},-1})-f_{N}(\eta^{0})]$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\sum_{i\in
A}\sum_{|x_{i}|\leq K}[f_{N}((\eta^{0})^{x_{i},1})-f_{N}(\eta^{0})]\bigg{]}$
$\displaystyle\quad\cdot\sum_{|x_{m}|>K:m\in\\{1,2,\ldots,r+s\\},m\neq
i}\prod_{j\in B^{c}}\big{[}1$
$\displaystyle\qquad\qquad\quad{}-c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})\big{]}$ $\displaystyle\quad\cdot\prod_{j\in
B}c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})$
$\displaystyle=\frac{1}{r+s}\,\frac{1}{2^{r}N^{r+s-1}}\\!\\!\\!\sum_{A\subset\\{1,\ldots,r\\}}\sum_{B\subset\\{r+1,\ldots,r+s\\}}\bigg{[}\sum_{i\in
A^{c}}\sum_{|x_{i}|\leq K}[f_{N}((\eta^{0})^{x_{i},-1})-f_{N}(\eta^{0})]$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;{}+\sum_{i\in
A}\sum_{|x_{i}|\leq K}[f_{N}((\eta^{0})^{x_{i},1})-f_{N}(\eta^{0})]\bigg{]}$
$\displaystyle\quad\cdot\sum_{|x_{m}|>K:m\in\\{1,2,\ldots,r+s\\},m\neq
i}\prod_{j\in B^{c}}\big{[}1$
$\displaystyle\qquad\qquad\qquad\qquad\qquad{}-c(x_{j},(\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})\big{]}$
$\displaystyle\quad\cdot\prod_{j\in
B}c(x_{j},(\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})$
$\displaystyle=\frac{1}{r+s}\,\frac{1}{2^{r}N^{r+s-1}}\big{[}N-(2K+1)\big{]}^{r+s-1}$
$\displaystyle\qquad\qquad\cdot{}\sum_{A\subset\\{1,\ldots,r\\}}\bigg{[}|A^{c}|\sum_{|x|\leq
K}[f_{N}((\eta^{0})^{x,-1})-f_{N}(\eta^{0})]$
$\displaystyle\qquad\qquad\qquad\qquad\qquad{}+|A|\sum_{|x|\leq
K}[f_{N}((\eta^{0})^{x,1})-f_{N}(\eta^{0})]\bigg{]}$
$\displaystyle=\frac{1}{r+s}\
\frac{1}{2^{r}N^{r+s-1}}\big{[}N-(2K+1)\big{]}^{r+s-1}\sum_{A\subset\\{1,\ldots,r\\}}\bigg{[}|A^{c}|\sum_{|x|\leq
K}f_{N}((\eta^{0})^{x,-1})$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad{}+|A|\sum_{|x|\leq
K}f_{N}((\eta^{0})^{x,1})-r\sum_{|x|\leq K}f_{N}(\eta^{0})\bigg{]}$
$\displaystyle=\frac{1}{r+s}\
\frac{1}{N^{r+s-1}}\big{[}N-(2K+1)\big{]}^{r+s-1}\bigg{[}\frac{r}{2}\sum_{|x|\leq
K}f_{N}((\eta^{0})^{x,-1})$
$\displaystyle\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\frac{r}{2}\sum_{|x|\leq
K}f_{N}((\eta^{0})^{x,1})-r\sum_{|x|\leq K}f_{N}(\eta^{0})\bigg{]}$
$\displaystyle=\frac{r}{r+s}\sum_{|x|\leq
K}\bigg{[}\frac{1}{2}f_{N}((\eta^{0})^{x,-1})+\frac{1}{2}f_{N}((\eta^{0})^{x,1})-f_{N}(\eta^{0})\bigg{]}+O(N^{-1})$
$\displaystyle=\frac{r}{r+s}\psi_{N}(\Omega_{A^{\prime}}f)(\eta^{0})+O(N^{-1}),$

where, in the second equality,

$\displaystyle\sum_{|x_{m}|>K:m\in\\{1,2,\ldots,r+s\\},m\neq
i}\sum_{B\subset\\{r+1,\ldots,r+s\\}}\prod_{j\in B^{c}}\big{[}1$
$\displaystyle\qquad\qquad\qquad\qquad\qquad{}-c(x_{j},(\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})\big{]}$
$\displaystyle\qquad\qquad\qquad\qquad\cdot\prod_{j\in
B}c(x_{j},(\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})$
$\displaystyle=\sum_{|x_{m}|>K:m\in\\{1,2,\ldots,r+s\\},m\neq
i}\prod_{j=r+1}^{r+s}\big{[}1$
$\displaystyle\qquad\qquad\qquad\qquad{}-c(x_{j},(\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})$
$\displaystyle\qquad\qquad\qquad\qquad{}+c(x_{j},(\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})\big{]}$
$\displaystyle=[N-(2K+1)]^{r+s-1},$

and in the fourth equality,

$\displaystyle\frac{1}{2^{r}}\sum_{A\subset\\{1,\ldots,r\\}}|A^{c}|=\frac{1}{2^{r}}\sum_{A\subset\\{1,\ldots,r\\}}|A|=\frac{1}{2^{r}}\sum_{i=0}^{r}i\
{r\choose i}=\frac{r}{2}.$

Next, we consider the second term in (3). It contributes

$\displaystyle\\!\\!\\!\\!\\!\frac{1}{r+s}\
\frac{1}{2^{r}N^{r+s-1}}\sum_{A\subset\\{1,\ldots,r\\}}\sum_{B\subset\\{r+1,\ldots,r+s\\}}\sum_{i\in
B}\sum_{|x_{i}|\leq K}[f_{N}((\eta^{0})_{x_{i}})-f_{N}(\eta^{0})]$
$\displaystyle\quad\cdot\sum_{|x_{m}|>K:m\in\\{1,2,\ldots,r+s\\},m\neq
i}\prod_{j\in B^{c}}\big{[}1$
$\displaystyle\qquad\qquad{}-c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})\big{]}$ $\displaystyle\quad\cdot\prod_{j\in
B}c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})$ $\displaystyle=\frac{1}{r+s}\
\frac{1}{2^{r}N^{r+s-1}}\sum_{A\subset\\{1,\ldots,r\\}}\sum_{i=r+1}^{r+s}\
\sum_{|x_{i}|\leq
K}c(x_{i},\eta^{0})[f_{N}((\eta^{0})_{x_{i}})-f_{N}(\eta^{0})]$
$\displaystyle\quad\cdot\sum_{|x_{m}|>K:m\in\\{1,2,\ldots,r+s\\},m\neq
i}\sum_{B\subset\\{r+1,\ldots,r+s\\}:i\in B}\prod_{j\in
B^{c}}\big{[}1-c(x_{j},\eta^{0})\big{]}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad{}\cdot\prod_{j\in
B-\\{i\\}}c(x_{j},\eta^{0})+O(N^{-1})$ $\displaystyle=\frac{1}{r+s}\
\frac{1}{2^{r}N^{r+s-1}}\big{[}N-(2K+1)\big{]}^{r+s-1}\sum_{A\subset\\{1,\ldots,r\\}}\sum_{i=r+1}^{r+s}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad{}\cdot\sum_{|x_{i}|\leq
K}c(x_{i},\eta^{0})[f_{N}((\eta^{0})_{x_{i}})-f_{N}(\eta^{0})]+O(N^{-1})$
$\displaystyle=\frac{1}{r+s}\
\frac{1}{2^{r}N^{r+s-1}}\big{[}N-(2K+1)\big{]}^{r+s-1}2^{r}\sum_{i=r+1}^{r+s}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad{}\cdot\sum_{|x_{i}|\leq
K}c(x_{i},\eta^{0})[f_{N}((\eta^{0})_{x_{i}})-f_{N}(\eta^{0})]+O(N^{-1})$
$\displaystyle=\frac{1}{r+s}\
\frac{1}{N^{r+s-1}}\big{[}N-(2K+1)\big{]}^{r+s-1}s\sum_{|x|\leq
K}c(x,\eta^{0})[f_{N}((\eta^{0})_{x})-f_{N}(\eta^{0})]$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad{}+O(N^{-1})$
$\displaystyle=\frac{s}{r+s}\ \sum_{|x|\leq
K}c(x,\eta^{0})[f_{N}((\eta^{0})_{x})-f_{N}(\eta^{0})]+O(N^{-1})$
$\displaystyle=\frac{s}{r+s}\psi_{N}(\Omega_{B}f)(\eta^{0})+O(N^{-1}),$

where the first and second equalities require clarification.

In the first equality we used

$\sum_{B\subset\\{r+1,\ldots,r+s\\}}\sum_{i\in
B}=\sum_{i=r+1}^{r+s}\sum_{B\subset\\{r+1,\ldots,r+s\\}:i\in B}$

and

$c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})=c(x_{j},\eta^{0})$

with possible exceptions if

$\\{x_{j}-1,x_{j},x_{j}+1\\}\cap\bigg{[}\bigcup_{p\in
A^{c}}\\{x_{p},x_{p}-1\\}\cup\bigcup_{p\in
A}\\{x_{p},x_{p}+1\\}\cup\bigcup_{p\in B}\\{x_{p}\\}\bigg{]}\neq\varnothing.$

That excludes at most $4r+3s$ of the $N$ possible values of $x_{j}$, hence
involves an error of at most $O(N^{-1})$. In the second equality,

$\displaystyle\sum_{|x_{m}|>K:m\in\\{1,2,\ldots,r+s\\},m\neq
i}\sum_{B\subset\\{r+1,\ldots,r+s\\}:i\in B}\prod_{j\in
B^{c}}\big{[}1-c(x_{j},\eta^{0})\big{]}\prod_{j\in
B-\\{i\\}}c(x_{j},\eta^{0})$
$\displaystyle=\sum_{|x_{m}|>K:m\in\\{1,2,\ldots,r+s\\},m\neq
i}\prod_{j\in\\{r+1,\ldots,r+s\\}-\\{i\\}}\big{[}1-c(x_{j},\eta^{0})+c(x_{j},\eta^{0})\big{]}$
$\displaystyle=[N-(2K+1)]^{r+s-1}.$

Therefore, we conclude that

$(\Omega^{N}_{[r,s]^{\prime}}\psi_{N}f)(\eta^{0})=\psi_{N}\bigg{(}\frac{r}{r+s}\Omega_{A^{\prime}}f+\frac{s}{r+s}\Omega_{B}f\bigg{)}(\eta^{0})+O(N^{-1}),$
(14)

as desired.

Since (14) holds, uniformly over $\Sigma_{N},$ the unique stationary
distribution $\pi^{N}$ of $\bm{P}_{A^{\prime}}^{r}\bm{P}_{B}^{s}$ converges
weakly to the unique stationary distribution $\pi^{r/(r+s)}$ of the
interacting particle system with generator $\Omega_{C^{\prime}},$ provided
that ergodicity holds for the limiting interacting particle system. Here,
$\Omega_{C^{\prime}}=\gamma\Omega_{A^{\prime}}+(1-\gamma)\Omega_{B}$ with
$\gamma:=r/(r+s)$, specifically

$\displaystyle(\Omega_{C^{\prime}}f)(\eta)$
$\displaystyle=\gamma\sum_{x}c^{\prime}(x,\eta)[f(\eta_{x})-f(\eta)]$
$\displaystyle\quad+\gamma\sum_{x}[f(_{x}\eta_{x+1})-f(\eta)]+(1-\gamma)\sum_{x}c(x,\eta)[f(\eta_{x})-f(\eta)].$
(15)

Our aim is to prove the following theorem.

######  Theorem 4.

Fix integers $r,s\geq 1$ and put $\gamma:=r/(r+s)$. Assume that the
interacting particle system on ${\bf Z}$ with generator $\Omega_{C^{\prime}}$
as in (3) is ergodic with unique stationary distribution $\pi^{\gamma}$. Then
$\lim_{N\to\infty}\mu_{[r,s]^{\prime}}^{N}=\mu_{(\gamma,1-\gamma)^{\prime}}$,
where $\mu_{(\gamma,1-\gamma)^{\prime}}$ is as in (3).

###### Proof.

The mean profit per turn to the ensemble of $N$ players playing the nonrandom
periodic pattern $(A^{\prime})^{r}B^{s}$ is

$\mu_{[r,s]^{\prime}}^{N}=\frac{1}{r+s}\sum_{v=0}^{s-1}\sum_{\eta\in\Sigma_{N}}(\bm{\pi}^{N}\bm{P}^{r}_{A^{\prime}}\bm{P}^{v}_{B})(\eta)\frac{1}{N}\sum_{z}[2p_{m_{z}(\eta)}-1].$
(16)

The sum over $\eta$ in (16) can be expressed, using (3), as

$\displaystyle\\!\\!\\!\\!\\!\sum_{\eta^{0},\eta}\pi^{N}(\eta^{0})(\bm{P}^{r}_{A^{\prime}}\bm{P}^{v}_{B})(\eta^{0},\eta)\frac{1}{N}\sum_{z}[2p_{m_{z}(\eta)}-1]$
$\displaystyle=\frac{1}{2^{r}N^{r+v}}\
\sum_{\eta^{0}}\pi^{N}(\eta^{0})\sum_{A\subset\\{1,\ldots,r\\}}\sum_{B\subset\\{r+1,\ldots,r+v\\}}\sum_{x_{i}:i\in\\{1,2,\ldots,r+v\\}}$
$\displaystyle\quad\cdot\prod_{j\in
B^{c}}\big{[}1-c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})\big{]}$ $\displaystyle\quad\cdot\prod_{j\in
B}c(x_{j},((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B,l<j\\}})$
$\displaystyle\quad\cdot\frac{1}{N}\sum_{z}[2p_{m_{z}(((\cdots((\cdots(((\eta^{0})^{x_{1},a_{1}})^{x_{2},a_{2}})\cdots)^{x_{p},a_{p}})\cdots)^{x_{r},a_{r}})_{\\{x_{l}:l\in
B\\}})}-1]$ $\displaystyle=\frac{1}{2^{r}N^{r+v}}\
\sum_{\eta^{0}}\pi^{N}(\eta^{0})\sum_{A\subset\\{1,\ldots,r\\}}\sum_{B\subset\\{r+1,\ldots,r+v\\}}\sum_{x_{i}:i\in\\{1,2,\ldots,r+v\\}}$
$\displaystyle\quad\cdot\prod_{j\in
B^{c}}\big{[}1-c(x_{j},\eta^{0})\big{]}\prod_{j\in
B}c(x_{j},\eta^{0})\frac{1}{N}\sum_{z}[2p_{m_{z}(\eta^{0})}-1]+O(N^{-1})$
$\displaystyle=\frac{1}{2^{r}N^{r+v}}\sum_{\eta^{0}}\pi^{N}(\eta^{0})\
(2^{r})(1)[N-(2k+1)]^{r+v}\frac{1}{N}\sum_{z}[2p_{m_{z}(\eta^{0})}-1]+O(N^{-1})$
$\displaystyle=\frac{1}{N}\sum_{z=1}^{N}\sum_{\eta^{0}}\pi^{N}(\eta^{0})[p_{m_{z}(\eta^{0})}-q_{m_{z}(\eta^{0})}]+O(N^{-1})$
$\displaystyle=\sum_{k=0}^{1}\sum_{l=0}^{1}(\pi^{N})_{-1,1}(k,l)[2p_{2k+l}-1]+O(N^{-1})$
$\displaystyle=\sum_{k=0}^{1}\sum_{l=0}^{1}(\pi^{r/(r+s)})_{-1,1}(k,l)[2p_{2k+l}-1]+o(1).$

So we have, with $\gamma:=r/(r+s)$,

$\mu^{N}_{[r,s]^{\prime}}\rightarrow(1-\gamma)\sum_{k=0}^{1}\sum_{l=0}^{1}(\pi^{\gamma})_{-1,1}(k,l)[2p_{2k+l}-1]=\mu_{(\gamma,1-\gamma)^{\prime}},$

as required. ∎

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