Convergence in the $p$-contest

TL;DR

This paper investigates the long-term behavior of a Markov system called the $p$-contest, analyzing how the configuration of points evolves and converges depending on the parameter $p$ and the distribution of new points.

Contribution

It extends previous results on the $p$-contest by establishing convergence criteria for different $p$ values and point distributions, especially for $p eq 1$, using novel supermartingale techniques.

Findings

For $p<1$, the configuration converges to zero.

For $p>1$, the configuration converges to zero or one, with examples of both.

When $p>1$, $N=3$, and $ ext{distribution} o U[0,1]$, convergence is almost sure to one.

Abstract

We study asymptotic properties of the following Markov system of $N \geq 3$ points in~$[0,1]$. At each time step, the point farthest from the current centre of mass, multiplied by a constant $p>0$, is removed and replaced by an independent $\zeta$-distributed point; the problem, inspired by variants of the Bak--Sneppen model of evolution and called a $p$-contest, was posed in [Grinfeld, M, Knight, P.A., and Wade, A.R. Rank-driven Markov processes, J. Stat. Phys. 146 (2012)]. We obtain various criteria for the convergences of the system, both for $p<1$ and $p>1$. In particular, when $p<1$ and $\zeta\sim U[0,1]$, we show that the limiting configuration converges to zero. When $p>1$, we show that the configuration must converge to either zero or one, and we present an example where both outcomes are possible. Finally, when $p>1$, $N=3$ and $\zeta$ satisfies certain conditions (e.g.~$\zeta\sim U[0,1]$), we prove that the configuration can only converge to one a.s. Our paper substantially extends the results of [Grinfeld, M., Volkov, S., and Wade, A.R. Convergence in a multidimensional randomized Keynesian beauty contest. Adv. in Appl. Probab. 47 (2015)] and [Kennerberg, P., and Volkov, S. Jante's law process. Adv. in Appl. Probab. 50 (2018)] where it was assumed that $p=1$. Unlike the previous models, one can no longer use the Lyapunov function based just on the radius of gyration; when $0<p<1$ one has to find a much finer tuned function which turns out to be a supermartingale; the proof of this fact constitutes an unwieldy, albeit necessary, part of the paper.

Equations

$${\mu }_N(x_1,\dots ,x_N)$$

$$\mathcal{X}(t+1)={\mathcal{X}}^{\prime }(t)\cup \{{\zeta }_{t+1}\},$$

$$\mathbb{P}(\zeta \in (a,b))\ge f(b-a)>0\text{for all }0\le a<b\le 1.$$

$$\{\exists \lim {\mathcal{X}}^{\prime }(t)\in (0,1]\}\subseteq \bigcup _{q\in Q\cap (0,1]}\bigcup _{T>0}\bigcap _{t\ge T}{A}_{q,t}$$

$$\mathbb{P}(A_{q,t+1}|{\mathcal{F}}_t,A_{q,t})<1-{\nu }_q\text{for all }t$$

$$\mathbb{P}\left(\bigcap _{t\ge T}{A}_{q,t}\right)=0$$

$$p{\mu }_N({\mathcal{X}}^{\prime }({\tau }_k)\cup \{\zeta \})$$

$$\mathbb{P}\left(A_{q,t+1}|{\mathcal{F}}_t,A_{q,t}\right)\le 1-\mathbb{P}\left(\zeta \in (q,rq)\right)\le 1-f(pq-q)=:1-{\nu }_q$$

$$\mu \le \frac{(N-1)x_{(1)}+(N-1)x_{(N)}}{N}=\frac{2p(N-1)}{N}\cdot \frac{x_{(1)}+x_{(N)}}{2p}<\frac{x_{(1)}+x_{(N)}}{2p}$$

$$x_{(N)}-p\mu >p\mu -x_{(1)}⟺x_{(N)}-p\mu >|p\mu -x_{(1)}|$$

$$x-p\mu >|p\mu -x_{(1)}|⟺x-p\mu >p\mu -x_{(1)}⟺x\left(1-\frac{2p}{N}\right)>2p\frac{x_1+\cdots +x_{N-1}}{N}-x_{(1)}$$

$$x\left(1-\frac{2p}{N}\right)>\frac{x}{3}>2x_{(N-1)}>2p{x}_{(N-1)}>2p\frac{x_1+\cdots +x_{N-1}}{N}$$

$$h(y)=F(y)+k\mu (y{)}^2,$$

$$F(y)=\sum _{i=1}^{N-1}(y_i-\mu (y){)}^2,\mu (y)=\frac{1}{N-1}\sum _{i=1}^{N-1}{y}_i,k=\frac{(N-1{)}^2(1-p)}{N-2}.$$

$${\mathbb{E}[h({\mathcal{X}}^{\prime }(t+1))-h({\mathcal{X}}^{\prime }(t))|{\mathcal{F}}_t]|}_{x(t)=y}\le 0$$

$$z=6U_t\text{ (the newly sampled point)},a=\min y<1\text{ (the leftmost point)}$$

$$p\ge \frac{N}{2(N-1)}=\frac{M+1}{2M}=\frac{1}{2}+\frac{1}{2M}.$$

$$\begin{array}[]{rlrl}F_{old}&=F(y),&F_{new}&=F\left((y\cup\{z\})^{\prime}\right)\\ \mu^{\prime}_{old}&=\mu(y),&\mu^{\prime}_{new}&=\mu\left((y\cup\{z\})^{\prime}\right),\\ h_{old}&=F_{old}+k\left(\mu_{old}^{\prime}\right)^{2},&h_{new}&=F_{new}+k\left(\mu_{new}^{\prime}\right)^{2}\end{array}$$

$$\mathbb{E}[h_{new}-h_{old}|{\mathcal{F}}_t]\le 0.$$

$$F(\tilde{y})-F(y)$$

$$\mu (\tilde{y}{)}^2-\mu (y{)}^2$$

$${\Delta }_a(z):=h_{new}-h_{old}=\frac{z-a}{M}\left[(M-1)z+(M+1)a-2M\mu (y)+\frac{k}{M}(z-a+2M\mu (y))\right]$$

$${\Delta }_1(z):=h_{new}-h_{old}=\frac{z-1}{M}\left[(M-1)z+(M+1)-2M\mu (y)+\frac{k}{M}(z-1+2M\mu (y))\right]$$

$$m(z)=p\cdot \frac{y_1+\cdots +y_M+z}{M+1}=p\cdot \frac{M\mu +z}{M+1},$$

$$z=t_{z1}:=\frac{M(2p\mu -1)-1}{M+1-2p}\in \{\begin{array}{ll}(-\infty ,0)& \text{ if }p<p_1:=\frac{M+1}{2M\mu }\\ (0,a)& \text{ if }{p}_1<p<p_2:=\frac{(M+1)(a+1)}{2M\mu +2a}\\ (a,+\infty )& \text{ if }p>p_2.\end{array}$$

$$z=t_{a1}:=\frac{(M+1)(a+1)-2M\mu p}{2p}\in \{\begin{array}{ll}(1,+\infty )& \text{ if }p<p_3:=\frac{(M+1)(a+1)}{2M\mu +2},\\ (a,1)& \text{ if }{p}_3<p<p_2,\\ (-\infty ,a)& \text{ if }p>p_2.\end{array}$$

$$z=t_{za}:=\frac{2M\mu p-(M+1)a}{M+1-2p}\in \{\begin{array}{ll}(-\infty ,1)& \text{ if }p<p_3,\\ (1,+\infty )& \text{ if }p>p_3.\end{array}$$

$$p_2-p_1$$

$$p_2-p_3$$

$$p_1<p_3⟺Ma\mu >1⟺f>\frac{1-a-a^2(M-1)}{a(M-2)(1-a)}\text{ (when M>2)}$$

$$6-t_{za}$$

Full text

In particular, when $p<1$ and $\zeta\sim U[0,1]$, we show that the limiting configuration converges to zero. When $p>1$, we show that the configuration must converge to either zero or one, and we present an example where both outcomes are possible. Finally, when $p>1$, $N=3$ and $\zeta$ satisfies certain mild conditions (e.g. $\zeta\sim U[0,1]$), we prove that the configuration converges to one a.s.

Our paper substantially extends the results of [3, 5] where it was assumed that $p=1$. Unlike the previous models, one can no longer use the Lyapunov function based just on the radius of gyration; when $0<p<1$ one has to find a more finely tuned function which turns out to be a supermartingale; the proof of this fact constitutes an unwieldy, albeit necessary, part of the paper.

Keywords: Keynesian beauty contest; Jante’s law, rank-driven process.

AMS 2010 Subject Classifications: 60J05 (Primary) 60D05, 60F15, 60K35, 82C22, 91A15 (Secondary)

1 Introduction

This paper extends the results of [3] and [5] on the so-called Keynesian beauty contest, or, as it was called in [5], Jante’s law process. Following [3], we recall that in the Keynesian beauty contest, we have $N$ players guessing a number, and the person who guesses closest to the mean of all the $N$ guesses wins; see [6, Ch. 12, §V]. The formal version, suggested by Moulin [7, p. 72], assumes that this game is played by choosing numbers on the interval $[0,1]$, the “$p$-beauty contest”, in which the target is the mean value, multiplied by a constant $p>0$. For the applications of the $p$-contest in the game theory, we refer the reader to e.g. [1]; see also [2] and [3] and references therein for further applications and other relevant papers.

The version of the $p$-contest with $p\equiv 1$ was studied in [3, 5]. In [3] it was shown that in the model where at each unit of time the point farthest from the center of mass is replaced by a point chosen uniformly on $[0,1]$, then eventually all (but one) points converge almost surely to some random limit the support of which is the whole interval $[0,1]$; many of the results were extended for the version of the model on ${\mathbb{R}}^{d}$, $d\geq 2$. The results of [3] were further generalized in [5], by removing the assumption that a new point is chosen uniformly on $[0,1]$, as well as by removing more than one point at once, these points being chosen in such a way that the moment of inertia of the resulting configuration is minimized. However, the case $p\neq 1$ was not addressed in either of these two papers.

Let us now formally define the model; the notation will be similar to those in [3, 5]. Let ${\cal X}=\{x_{1},x_{2},\ldots,x_{N}\}\in{\mathbb{R}}^{N}$ be an unordered $N$-tuple of points in ${\mathbb{R}}$, and $(x_{(1)},x_{(2)},\dots,x_{(N)})$ be these points put in non-decreasing order, that is, $x_{(1)}\leq x_{(2)}\leq\dots\leq x_{(N)}$. As in [3, 5] let us define the barycentre of the configuration as

[TABLE]

Fix some $p>0$ and also define the $p-$centre of mass as $p\mu_{N}(x_{1},\dots,x_{N})$.

The point, farthest from the $p-$centre of mass, is called the extreme point of ${\cal X}$, and it can be either $x_{(1)}$ or $x_{(N)}$ (with possibility of a tie), and the core of ${\cal X}$, denoted by ${\cal X}^{\prime}$, is constructed from ${\cal X}$ by removing the extreme point; in case of a tie between the left-most and the right-most point, we choose either of them with equal probability (same as in [3, 5]). Throughout the rest of the paper, $x_{(1)}(t),\dots,x_{(N-1)}(t)$ shall denote the points of the core222rather than of ${\cal X}(t)$ ${\cal X}^{\prime}(t)$ put into non-decreasing order.

Our process runs as follows. Let ${\cal X}(t)=\{X_{1}(t),\dots,X_{N}(t)\}$ be an unordered $N$-tuple of points in ${\mathbb{R}}$ at time $t=0,1,2,\dots$. Given ${\cal X}(t)$, let ${\cal X}^{\prime}(t)$ be the core of ${\cal X}(t)$ and replace ${\cal X}(t)\setminus{\cal X}^{\prime}(t)$ by a $\zeta$-distributed random variable so that

[TABLE]

where $\zeta_{t}$, $t=1,2,\dots$, are i.i.d. random variables with a common distribution $\zeta$.

Finally, to finish the specification of our process, we allow the initial configuration ${\cal X}(0)$ to be arbitrary or random, with the only requirement being that all the points of ${\cal X}(0)$ must lie in the support of $\zeta$.

Throughout the paper we will use the notation $A\underset{\text{a.s.}}{\Longrightarrow}B$ for two events $A$ and $B$, whenever ${\mathbb{P}}(A\cap B^{c})=0$, that is, when $A\subseteq B$ up to a set of measure [math]. We will also write, with some abuse of notations, that $\lim_{t\to\infty}{\cal X}^{\prime}(t)=a\in{\mathbb{R}}$ or equivalently ${\cal X}^{\prime}(t)\to a$ as $t\to\infty$ if ${\cal X}^{\prime}(t)\to(a,a,\dots,a)\in{\mathbb{R}}^{N-1}$, i.e. $\lim_{t\to\infty}x_{(i)}(t)=a$ for all $i=1,2,\dots,N-1$. Similarly, for an interval $(a,b)$ we will write ${\cal X}^{\prime}(t)\in(a,b)$ whenever all $x_{(1)}(t),\dots,x_{(N-1)}(t)\in(a,b)$. Finally, we will assume that $\inf\emptyset=+\infty$, and use the notation $y^{+}=\max(y,0)$ for $y\in{\mathbb{R}}$.

Also we require that $\zeta$ has a full support on $[0,1]$, that is, ${\mathbb{P}}(\zeta\in(a,b))>0$ for all $a,b$ such that $0\leq a<b\leq 1$.

2 The case $p<1$

Throughout this Section we assume that $0<p<1$ and that $\mathop{\mathrm{supp}}\zeta=[0,1]$. Because of the scaling invariance, our results may be trivially extended to the case when $\mathop{\mathrm{supp}}\zeta=[0,A]$, $A\in(0,\infty)$; some of them are even true when $A=\infty$; however, to simplify the presentation from now on we will deal only with the case $A=1$.

First, we present some general statements; more precise results will follow in case where $\zeta\sim U[0,1]$.

Proposition 1**.**

We have

• (a)

$\liminf_{t\to\infty}x_{(N-1)}(t)=0$;

• (b)

${\mathbb{P}}\left(\exists\lim_{t\to\infty}{\cal X}^{\prime}(t)\in(0,1]\right)=0$;

• (c)

if $p<\frac{1}{2}+\frac{1}{2(N-1)}$ then ${\mathbb{P}}\left(\lim_{t\to\infty}{\cal X}^{\prime}(t)=0\right)=1$;

• (d)

if $p<\frac{1}{2}+\frac{1}{N-2}$ then $\{x_{(1)}(t)\to 0\}\underset{\text{a.s.}}{\Longrightarrow}\{\lim_{t\to\infty}{\cal X}^{\prime}(t)=0\}$.

Proof.

(a) Since $\zeta$ has full support on $[0,1]$ it follows that (see [5], Proposition 1) there exists a function $f:{\mathbb{R}}^{+}\to{\mathbb{R}}^{+}$ such that

[TABLE]

Also, to simplify notations, we write $\mu=\mu_{N}({\cal X}(t))$ throughout the proof.

Fix a small positive $\varepsilon$ such that $p+2\varepsilon<1$. Suppose that for some $t$ we have $x_{(N-1)}(t)\leq b\leq 1$. We will show that $x_{(N-1)}(t+N)\leq b(1-\varepsilon)$ with a strictly positive probability which only depends on $p,b,\epsilon$ and $N$. Assume that we have $\zeta_{t+1},\dots,\zeta_{t+N-1}\in\left(pb,(p+\varepsilon)b\right)\subset(pb,b)$; this happens with probability no less than $\left[f(p\varepsilon b)\right]^{N-1}$. We claim that by the time $t+N$ we have $x_{(N-1)}(t+N-1)<(p+\varepsilon)b$. Indeed, $p\mu\leq pb$ always lies to the left of the newly sampled points, therefore either there are no more points to the right of $(p+\varepsilon)b$ at some time $s\in[t,t+N-1]$ (which implies that there will be no points there at time $t+N$ due to the sampling range of the new points), or one of the older points, i.e. present at time $t$, gets removed (it can be the one to the left of $pb$). Since we eventually have to replace all the $N-1$ old points, then $x_{(N-1)}(t+N)\leq b(1-\varepsilon)$.

Fix a $\delta>0$ and find $M$ so large that $(1-\varepsilon)^{M}<\delta$. Let the event $C(s)=\{x_{(N-1)}(s)<\delta\}$. By iterating the above argument, we get that ${\mathbb{P}}(C(t+NM)|{\cal{F}}_{t})\geq\prod_{i=1}^{M}\left[f(p\varepsilon(1-\varepsilon)^{i-1})\right]^{N-1}$, since at time $t$ we can set $b=1$. Therefore, $\sum_{m}{\mathbb{P}}(C(NM(m+1))|{\cal{F}}_{NMm})=\infty$ and by Lévy’s extension of the Borel-Cantelli lemma (see e.g. [8]) infinitely many $C(s)$ occur. Since $\delta>0$ is arbitrary, we get $\liminf_{t\to\infty}x_{(N-1)}(t)=0$.

(b) Let $r=\frac{1+p^{-1}}{2}>1$. Suppose that the core converges to some point $x\in(0,1]$; then there exist a rational $q\in(0,1]$ and a $T>0$ such that ${\cal X}^{\prime}(t)\in(q,rq)$ for all $t\geq T$, formally

[TABLE]

where $A_{q,t}=\{{\cal X}^{\prime}(t)\in(q,rq)\}$. We will show that

[TABLE]

for some $\nu_{q}>0$. This will imply, in turn, that

[TABLE]

and hence the RHS (and thus the LHS as well) of (2.2) has the probability [math].

Suppose $A_{q,t}$ has occurred and the newly sampled point $\zeta\in(pq,q)$. Then

[TABLE]

Consequently, $x_{(N-1)}$ lies further from the $p-$center of mass, and hence it should be removed. The new configuration will, however, contain the point $\zeta\notin(q,rq)$ and hence $A_{q,t+1}$ does not occur. Thus

[TABLE]

as required.

(c) First, we will show that it is the right-most point of the configuration which should be always removed; note that it suffices to check this only when $x_{(N)}>0$. Indeed, by the assumption on $p$ we have

[TABLE]

implying that

[TABLE]

Therefore, $x_{(N)}$ is the farthest point from the $p-$centre of mass. This implies that $x_{(N-1)}(t)$ is non-increasing and therefore result now easily follows from part (a) since $x_{(N-1)}(t)$ is an upper bound for all the core points.

(d) Apply Corollary 3 with $k=1$; it is possible because of Remark 3. ∎

We are ready to present the main result of this Section.

Theorem 1**.**

Suppose that $\zeta\sim U[0,1]$. Then ${\cal X}^{\prime}(t)\to 0$ a.s.

Proof.

Proposition 1 (c) implies that we now only need to consider the case $p\geq\frac{N}{2(N-1)}$, which we will assume from now on.

Let us introduce a modification of this process on $[0,+\infty)$ which we will call the borderless $p$-contest; it is essentially the same process as the one in Section 3.4 of [3]. In order to do this, we need the following statement.

Lemma 1**.**

Suppose that $x_{1},\dots,x_{N-1}>0$. Then there exists an $R=R(x_{(N-1)})\geq 0$ such that $x$ is the farthest point from $p\mu=\frac{p}{N}(x_{1}+\dots+x_{N-1}+x)$ whenever $x>R$.

Proof of Lemma 1..

Set $R=6x_{(N-1)}$. Then $x>x_{(1)}$ is farther from the centre of mass than $x_{(1)}$ if and only if

[TABLE]

This is true, due to the fact that $x>R$ and

[TABLE]

since $p<1$ and $N\geq 3$. ∎

The borderless process is constructed as follows. Our core configuration starts as before in $[0,1]$, and we use the same rejection/acceptance criteria for new points. However, we will now allow points to be generated to the right of $1$ as well. Let $R_{t}=R(x_{(N-1)}(t))$ where $R$ is taken from Lemma 1. Then a new point is sampled uniformly and independently of the past on the interval $[0,R_{t}]$; formally, it is given by $R_{t}U_{t}$ where $U_{t}$ are i.i.d. uniform $[0,1]$ random variables independent of everything. Observe that if we consider the embedded process only at the times when the core configuration changes, then the exact form of the function $R(\cdot)$ is irrelevant, due to the fact that the uniform distribution conditioned on a subinterval is also uniform on that subinterval.

Next, for $y=\{y_{1},\dots,y_{N-1}\}$ define the function

[TABLE]

where

[TABLE]

We continue with the following

Lemma 2**.**

For the borderless $p$-contest the sequence of random variables $h\left({\cal X}^{\prime}(t)\right)\geq 0$, $t=1,2,\dots$, is a supermartingale.

Remark 1**.**

Note that the function $F(\cdot)$ defined above is a Lyapunov function for the process in [3]; this is no longer the case as long as $p\neq 1$; that is why we have to use a carefully chosen “correction” factor which involves the barycentre of the configuration.

Proof of Lemma 2.

Assume that $x_{(N-1)}(t)>0$ (otherwise the process has already stopped, and the result is trivial). The inequality, which we want to obtain is

[TABLE]

for all $y=(y_{1},\dots,y_{N-1})$ with $y_{i}\in[0,1]$. Note that the function $h(y)$ is homogeneous of degree $2$ in $y$, therefore w.l.o.g. we can assume that $\max y\equiv 1$.

For simplicity let $M=N-1\geq 2$, and let

[TABLE]

Note also that

[TABLE]

Define

[TABLE]

Thus we need to establish

[TABLE]

First of all, observe that if $\tilde{y}=(y\setminus\{y_{i}\})\cup\{z\}$, that is, $\tilde{y}$ is obtained from $y$ by replacing $y_{i}$ with $y_{0}$, then

[TABLE]

In particular, if we replace point $a$ by the new point $z$, then

[TABLE]

and if we replace point $1$, then

[TABLE]

Note that both $\Delta_{a}$ and $\Delta_{1}$ depend only on four variables $(a,z,\mu,M)$ but not the whole configuration. Let us also define

[TABLE]

the $p-$centre of mass of the old core and the newly sampled point.

There are three different cases that can occur: either (a) the point $a$ is removed, (b) $1$, the rightmost point of the previous core, is removed, or (c) the newly sampled point $z$ is removed. In the third case the core remains unchanged, and the change in the value of the function $h$ is trivially zero. The point $a$ can only be removed if $z>a$; the point $1$ can only be removed if $z<1$; the point $z$ can be possibly removed only if $z\in(0,a)$ or $z\in(1,\infty)$. Let us compute the critical values for $z$, for which there is a tie between the farthest points.

Which point to remove?

$(i)$ Suppose $\boxed{z<a}$. Then there is a tie between $z$ and $1$ if and only if $m(z)=\frac{z+1}{2}$, that is if

[TABLE]

Thus, we have:

• •

when $p<p_{1}$, point $1$ is removed;

• •

when $p_{1}<p<p_{2}$, if $z<t_{z1}$ then $z$ is removed; if $z>t_{z1}$ point $1$ is removed;

• •

when $p>p_{2}$, point $z$ is removed.

$(ii)$ Suppose $\boxed{a<z<1}$. There is a tie between $a$ and $1$ if and only if $m(z)=\frac{a+1}{2}$, that is if

[TABLE]

Thus, we have:

• •

when $p<p_{3}$, point $1$ is removed;

• •

when $p_{3}<p<p_{2}$, if $z<t_{a1}$ then $1$ is removed; if $z>t_{a1}$ then point $a$ is removed;

• •

when $p>p_{2}$, point $a$ is removed.

$(iii)$ Suppose $\boxed{z>1}$. There is a tie between $z$ and $a$ if and only if $m(z)=\frac{z+a}{2}$, that is if

[TABLE]

Thus, we have:

• •

when $p<p_{3}$, point $z$ is removed;

• •

when $p>p_{3}$, if $z<t_{za}$ then $a$ is removed; if $z>t_{za}$ then point $z$ is removed.

We always have $p_{1}<p_{2}$, $p_{3}<p_{2}$ since

[TABLE]

while

[TABLE]

The final observation is that $t_{za}<6$, so there is indeed no need to sample the new point outside of the range $(0,6)$; this holds since $M\geq 2$ and

[TABLE]

The five cases for the removal:

• •

$p<\min\{p_{1},p_{3}\}$:

• –

when $z<1$, point $1$ is removed

• –

when $z>1$, point $z$ is removed

• •

$p>p_{2}$:

• –

when $z<a$ or $z>t_{za}\in(1,\infty)$ point $z$ is removed

• –

when $a<z<t_{za}$, point $a$ is removed

• •

$\max\{p_{1},p_{3}\}<p<p_{2}$

• –

when $z<t_{z1}\in(0,a)$ or $t>t_{za}\in(1,+\infty)$, point $z$ is removed

• –

when $t_{z1}<z<t_{a1}\in(a,1)$, point $1$ is removed

• –

when $t_{a1}<z<t_{za}$, point $a$ is removed

• •

$p_{1}<p<p_{3}\ (<p_{2})$:

• –

when $z<t_{z1}\in(0,a)$ or $z>1$, point $z$ is removed

• –

when $t_{z1}<z<1$, point $1$ is removed

• •

$p_{3}<p<p_{1}\ (<p_{2})$:

• –

when $z<t_{a1}\in(a,1)$, point $1$ is removed

• –

when $t_{a1}<z<t_{za}\in(1,+\infty)$, point $a$ is removed

• –

when $z>t_{za}$, point $z$ is removed

Let

[TABLE]

Define

[TABLE]

Since $\max y=1$, because of the comment on the restriction of the uniform distribution on a subinterval, we have $\tilde{\bf I}_{j}=c_{j}{\bf I}_{j}$, $j=1,2,3,4,5$, where $c_{j}$’s are some positive constants and

[TABLE]

Thus to establish (2.5), it suffices to show that ${\bf I}_{j}\leq 0$ for each $j=1,2,3,4,5$. This is done by very extensive and tedious calculations, which can be found in the Appendix. ∎

We now return to our original $p$-contest process ${\cal X}(t)$. For $L\geq 2$ define

[TABLE]

note that $\tau_{L}$ is a.s. finite for every $L$ by Proposition 1. Let $W(s)=\{w_{1}(s),\dots,w_{N}(s)\}$ be a borderless $p$-contest with $W(0)={\cal X}(\tau_{L})$; let $W^{\prime}(s)$ be its core. By Lemma 2 the quantity $\xi_{t}=h(W^{\prime}(t\wedge\eta_{L}))$ is a supermartingale, that converges to some $\xi_{\infty}$. Since $\xi_{t}$ is bounded,

[TABLE]

since on $\{\eta_{L}<\infty\}$ we have $\xi_{\infty}=W^{\prime}(\eta_{L})$ and the largest coordinate of $W^{\prime}(\eta_{L})$ is larger than $1/2$, implying that $\mu(W^{\prime}(\eta_{L}))\geq\frac{1}{2(N-1)}$ and thus $h(W^{\prime}(\eta_{L}))=F(W^{\prime}(\eta_{L}))+k\mu(W^{\prime}(\eta_{L}))^{2}\geq\frac{k}{(2(N-1))^{2}}$. We also have

[TABLE]

since ${\cal X}^{\prime}(\tau_{L})\subset[0,1/L]$ and so $\mu({\cal X}^{\prime}(\tau_{L}))\in[0,1/L]$.

Combining the above inequalities, we conclude that ${\mathbb{P}}(\eta_{L}<\infty)\to 0$ as $L\to\infty$. However, on $\eta_{L}=\infty$ the core of the regular $p$-contest process can be trivially coupled with the core of the borderless process $W^{\prime}(s)$ which converges to zero, so ${\cal X}^{\prime}(t)\to 0$ as well. Since ${\mathbb{P}}(\eta_{L}=\infty)$ can be made arbitrarily close to $1$ by choosing a large $L$, we conclude that ${\cal X}^{\prime}(t)\to 0$ a.s. ∎

3 The case $p>1$

Throughout this section we suppose that $\zeta$ has a full support on $[0,1]$, and, unless explicitly stated otherwise, that $p>1$.

Theorem 2**.**

• (a)

${\mathbb{P}}\left(\{{\cal X}^{\prime}(t)\to 0\}\cup\{{\cal X}^{\prime}(t)\to 1\}\right)=1;$**

• (b)

if $x_{(1)}(0)\geq 1/p$ then ${\mathbb{P}}({\cal X}^{\prime}(t)\to 1)=1$;

• (c)

if $x_{(k)}(0)>0$, where $k$ satisfies

[TABLE]

then ${\mathbb{P}}\left({\cal X}^{\prime}(t)\to 1\right)>0.$

Remark 2**.**

In general, both convergences can have a positive probability. Let $N=3$, $p\in(1,3/2)$, and

[TABLE]

where $U\in U[0,1]$ (so $\zeta$ has full support). Suppose we sample the points of ${\cal X}(0)$ from $\zeta$. If they all start off in [math], then $p\mu\leq p/3<1/2$, so they cannot escape from [math]. On the other hand, there is a positive probability they all start in $(1/p,1]$, and then Theorem 2(b) says that they converge to $1$.

The key idea behind the proof of Theorem 2 is that one can actually find the “ruling” order statistic of the core; namely, there exists some non-random $k=k(N,p)\in\{1,2,\dots,N-1\}$ such that $x_{(k)}(t)\to 0$ implies ${\cal X}^{\prime}(t)\overset{\text{a.s.}}{\longrightarrow}0$, while $x_{(k)}(t)\not\to 0$ implies that ${\cal X}^{\prime}(t)\overset{\text{a.s.}}{\longrightarrow}1$.

We start with the following two results, which tells us that there is an absorbing area $[\frac{1}{p},1]$ for the process, such that, once the core enters this area, it will never leave it, and moreover the core will keep moving to the right.

Claim 1**.**

Suppose that $x_{1}\leq x_{2}\leq x_{3}\leq\dots\leq x_{N}\leq 1$ and $x_{2}\geq p^{-1}$. Then $\{x_{1},\cdots,x_{N}\}^{\prime}=\{x_{2},\cdots,x_{N}\}$

Proof.

Let $\mu=\frac{x_{1}+\cdots x_{N}}{N}$. If $p\mu\geq x_{N}$ then the claim follows immediately; assume instead that $p\mu<x_{N}$. We need to check if $p\mu-x_{1}>x_{N}-p\mu$, that is, if

[TABLE]

However, since $x_{i}\geq x_{2}$ for $i=3,\dots,N-1$ we have

[TABLE]

while $(N-2p)(x_{1}+x_{N})\leq 2(N-2p)<2(N-2).$ Hence (3.7) follows. ∎

Lemma 3**.**

If $x_{(1)}(t_{0})\geq 1/p$ for some $t_{0}$, then ${\cal X}^{\prime}(t)\to 1$ a.s.

Proof.

If $x_{(1)}(t_{0})\geq 1/p$, then any point that lands in $[0,1/p)$ is extreme, so $x_{(2)}(t)\geq 1/p$ for all $t\geq t_{0}$. Choose any positive $\varepsilon<1-\frac{1}{p}$, and let $A_{t}=\left\{\zeta_{t+1},\dots,\zeta_{t+N-1}\in\left(1-\varepsilon,1\right]\right\}$. Then if $A_{t}$ happens for $s>t_{0}$, any point in $\left[0,1-\varepsilon\right]$ is removed in preference to any of the new points coming in, so $x_{(2)}(s+N-1)>1-\varepsilon$. As a result, by Claim 1 we get that ${\cal X}^{\prime}(t)\in\left[0,1-\varepsilon\right]$ for all $t\geq s$.

On the other hand, ${\mathbb{P}}(A_{t})\geq[f(\varepsilon)]^{N-1}>0$ (see (2.1)) for any $t$, and the events $A_{t},A_{t+N},A_{t+2N},\dots$ are independent. Hence, eventually with probability $1$, one of the $A_{t}$’s must happen for some $t>t_{0}$, so a.s. ${\cal X}^{\prime}(t)\in\left[0,1-\varepsilon\right]$ for all large $t$. Since $\varepsilon$ can be chosen arbitrary small, we get the result. ∎

The next two results show that if the is some $\varepsilon>0$ such that infinitely often the core does not have any points in $[0,\varepsilon)$, then it must, in fact, converge to $1$.

Lemma 4**.**

If $x_{(1)}(t_{0})\geq\varepsilon$ for some $t_{0}$ and $\varepsilon>0$, then ${\mathbb{P}}(x_{(1)}(t_{0}+\ell)\geq p^{-1}|{\cal{F}}_{t})\geq\delta$ for some $\ell=\ell(\varepsilon)$ and $\delta=\delta(\varepsilon)>0$.

Proof.

Suppose that for some $t$ we have $x_{(1)}(t)\geq\varepsilon$. We claim that it is possible to move $x_{(1)}$ to the right of $\frac{1+p}{2}\varepsilon$ in at most $N-1$ steps with positive probability, depending only on $p$ and $\varepsilon$. Indeed, if $x_{(1)}(t)>\frac{1+p}{2}\varepsilon$ then we are already done. Otherwise, if the new point $\zeta_{t+1}$ is sampled in $\left(\frac{1+p}{2}\varepsilon,p\varepsilon\right]\subset[0,1]$ it cannot be rejected. If at this stage $x_{(1)}(t+1)>\frac{1+p}{2}\varepsilon$, then we are done. If not, we proceed again by sampling $\zeta_{t+2}\in\left(\frac{1+p}{2}\varepsilon,p\varepsilon\right]$, etc. After at most $N-1$ steps of sampling new points in $\left(\frac{1+p}{2}\varepsilon,p\varepsilon\right]$, the leftmost point $x_{(1)}$ will have moved to the right of $\frac{1+p}{2}\varepsilon$.

Thus, in no more than $N-1$ steps, with probability no less than $\left[f\left(\frac{p-1}{2}\varepsilon\right)\right]^{N-1}>0$, $x_{(1)}$ is to the right of $\frac{1+p}{2}\varepsilon$. By iterating this argument at most $m$ times, where $m\in{\mathbb{N}}$ is chosen such that $\left[\frac{1+p}{2}\right]^{m}\,\varepsilon>1/p$, we achieve that $x_{(1)}$ is to the right of $1/p$ (for definiteness, one can chose $\ell=(N-1)m$ and $\delta=\left[f\left(\frac{p-1}{2}\varepsilon\right)\right]^{(N-1)m}$.) ∎

Lemma 5**.**

Let $\varepsilon\in(0,1)$, and define $B(\varepsilon):=\{x_{(1)}(t)\geq\varepsilon\text{ i.o.}\}$ Then $B(\varepsilon)\underset{\text{a.s.}}{\Longrightarrow}\{{\cal X}^{\prime}(t)\to 1\}.$

Corollary 1**.**

We have $\left\{\liminf_{t\to\infty}x_{(1)}(t)>0\right\}\underset{\text{a.s.}}{\Longrightarrow}\{{\cal X}^{\prime}(t)\to 1\}.$

Proof of Lemma 5.

Assume that $\varepsilon<\frac{1}{p}$ (otherwise the result immediately follows from Lemma 3). Also suppose that ${\mathbb{P}}\left(B(\varepsilon)\right)>0$, since otherwise the result is trivial. Let $\ell$ and $\delta$ be the quantities from Lemma 4.

Define

[TABLE]

with the convention that if $\tau_{k}=\infty$ then $\tau_{m}=\infty$ for all $m>k$. Notice that $B(\varepsilon)=\bigcap_{k=0}^{\infty}\{\tau_{k}<\infty\}$. On $B(\varepsilon)$ we can also define $D_{\tau_{k}}=\{x_{(1)}(\tau_{k}+\ell)\geq 1/p\}$. Since $\tau_{k}-\tau_{k-1}>\ell$ whenever both are finite, we have from Lemma 4 we have ${\mathbb{P}}(D_{\tau_{k+1}}|{\cal{F}}_{\tau_{k}})\geq\delta$. Therefore,

[TABLE]

hence by Lévy’s extension of the Borel-Cantelli lemma it follows that a.s. on $B(\varepsilon))$ infinitely many (and hence at least one) of $D_{\tau_{k}}$ occur, that is, $x_{(1)}(\tau_{k}+\ell)\geq 1/p$. Now the result follows from Lemma 3. ∎

Assume for now that $p<\frac{N}{2}$; in this case $N-\frac{N}{2p}+1<N$ (see (3.6)). The case $p\geq\frac{N}{2}$ will be dealt with separately.

The following statement shows that if all the points to the right of $x_{(k)}$ lie very near each other, while the left-most one lies near zero, then it is to be removed.

Claim 2**.**

Let $a\in(0,1]$ and suppose that $k\in\{2,\dots,N-1\}$ satisfies (3.6). Then there exist small $\delta,\Delta>0$, depending on $N,k,p,a$ such that if

[TABLE]

then $\{x_{1},\dots,x_{N}\}^{\prime}=\{x_{2},\dots,x_{N}\}$.

Proof.

The condition to remove the leftmost point is $p\mu-\frac{x_{1}+x_{N}}{2}>0$ where $\mu=(x_{1}+\dots+x_{N})/N$. However,

[TABLE]

The RHS is linear in $\delta$ and $\Delta$, and when $\delta=\Delta=0$ it is strictly positive by the assumption on $k$; hence it can also be made positive, by allowing $\delta>0$ and $\Delta>0$ to be sufficiently small. ∎

Corollary 2**.**

Suppose that ${\cal X}(t)=\{x_{1},\dots,x_{N}\}$ satisfies the conditions of Claim 2 for some $a$ and $k$. Let $\delta$ be the quantity from this claim. Then

[TABLE]

Proof.

The probability to sample a new point $\zeta\in(a(1-\Delta),a]$ is bounded below by $f(a\Delta)$ where $f$ is the same function as in (2.1). On the other hand, if the new point is sampled in $(a(1-\Delta),a]$ then ${\cal X}(t+1)$ continues to satisfy the conditions of Claim 2 as long as the leftmost point is in $[0,\delta]$. By repeating this argument at most $k$ times and using the induction, we get the result with $c=\left[f(a\Delta)\right]^{k}>0$. ∎

Lemma 6**.**

Let $k\in{\mathbb{N}}$ satisfy (3.6). Then

[TABLE]

Proof.

Note that by Lemma 5, it suffices to show that $\left\{x_{(k)}(t)\not\to 0\right\}\underset{\text{a.s.}}{\Longrightarrow}\left\{x_{(1)}(t)\not\to 0\right\}$.

If $x_{(k)}(t)\not\to 0$, there exists an $a>0$ such that $x_{(k)}(t)\geq a$ for infinitely many $t$’s. Let $s$ be such a time. Now suppose that $\zeta_{s+i}\in I:=(a(1-\Delta),a]$ for $i=0,1,\dots,N-1$ where $\Delta$ is defined in Claim 2; the probability of this event is strictly positive and depends only on $a$ and $\delta$ (see (2.1)). As long as there are points of ${\cal X}(s+i)$ on both sides of the interval $I$, none of the points inside $I$ can be removed; hence, for some $u\in\{s,s+1,\dots,s+N-1\}$ we have that either $\min{\cal X}(u)>a(1-\Delta)$ or $\max{\cal X}(u)\leq a$. In the first case, $x_{(1)}(u)>a(1-\Delta)$.

In the latter case, both $x_{(N)}(u)\in I$ and $x_{(k)}(u)\in I$, since every time we replaced a point, the number of points to the left of $I$ did not increase (and there were initially at most $k-1$ of them). As a result

[TABLE]

Together with Corollary 2, this yields

[TABLE]

which proves Lemma 6. ∎

Claim 3**.**

Let $A_{i}:=\left\{x_{(i)}(t)\to 0\right\}$ and suppose that for some $1\leq k\leq N-2$ we have

[TABLE]

Then $A_{k}\subseteq\{\exists\lim_{t\to\infty}x_{(k+1)}(t)\}.$

Proof.

Fix any $a>0$. Let $\delta>0$ be so small that

[TABLE]

In the event $A_{k}$ there exists a finite $\tau=\tau_{\delta}(\omega)$ such that

[TABLE]

From now on assume that $t\geq\tau$. We will show below that $x_{(k+1)}(t+1)\leq\max\{x_{(k+1)}(t),a\}$.

To begin, let us prove that $x_{(k+1)}(t+1)\leq x_{(k+1)}(t)$ as long as $x_{(k+1)}(t)>\delta$. Indeed, if the new point $\zeta$ is sampled to the left of $x_{(k+1)}(t)$, then regardless of which point is to be removed, $x_{(k+1)}(t+1)\leq x_{(k+1)}(t)$. If the new point $\zeta$ is sampled to the right, then the farthest point from the $p-$centre of mass must be the rightmost one (and hence $x_{(k+1)}(t+1)=x_{(k+1)}(t)$) since there are exactly $k$ points in $[0,\delta]$ and none of these can be removed by the definition of $\tau$.

On the other hand, if $x_{(k+1)}(t)\leq\delta$ then either $x_{(k+2)}(t)\leq a$ or $x_{(k+2)}(t)>a$. In the first case, $x_{(k+1)}(t+1)\leq x_{(k+2)}(t)\leq a$ even if $x_{(1)}$ is removed. In the other case, when $x_{(k+2)}(t)>a$, we have $x_{(N-1)}>a$ as well, and

[TABLE]

by (3.9), so $x_{(N)}=x_{(N)}(t)$ must be removed and thus $x_{(k+1)}(t+1)\leq x_{(k+1)}(t)$.

Consequently, we obtained

[TABLE]

since $a>0$ is arbitrary, we get

[TABLE]

∎

Lemma 7**.**

Suppose that (3.8) holds for some $1\leq k\leq N-2$. Then $A_{k}\underset{\text{a.s.}}{\Longrightarrow}A_{k+1}$.

Proof.

Let ${\tilde{A}}_{k+1}^{\geq a}:=\left\{\lim_{t\to\infty}x_{(k+1)}(t)\geq a\right\}$ (the existence of this limit on $A_{k}$ follows from Claim 3). It suffices to show that ${\mathbb{P}}\left(A_{k}\cap\tilde{A}_{k+1}^{\geq a}\right)=0$ for all $a>0$; then from the continuity of probability we get that ${\mathbb{P}}\left(A_{k}\cap\{\lim_{t\to\infty}x_{(k+1)}(t)>0\}\right)=0$ and hence $A_{k}\underset{\text{a.s.}}{\Longrightarrow}A_{k+1}$.

Fix an $a>0$. Let

[TABLE]

then

[TABLE]

If the probability of the LHS is positive, then, using the continuity of probability and the fact that $\bar{C}_{T}$ is an increasing sequence of events, we obtain that $\lim_{T\to\infty}{\mathbb{P}}(\bar{C}_{T})>0$. Consequently, there exists a non-random $T_{0}$ such that ${\mathbb{P}}(\bar{C}_{T_{0}})>0$.

This is, however, impossible, as at each time point $t$, with probability at least $f(a/3)$ (see (2.1)) the new point $\zeta_{t}$ is sampled in $B:=\left(\frac{a}{3},\frac{2a}{3}\right)$ and then either $x_{(k)}(t+1)\in B$ or $x_{(k+1)}(t+1)\in B$. Formally, this means that

[TABLE]

By induction, for all $k\geq 1$,

[TABLE]

Since $k$ is arbitrary, and $f(a/3)>0$, by taking the expectation, we conclude that ${\mathbb{P}}(\bar{C}_{T_{0}})=0$ yielding a contradiction.

Hence the probability of the event $A_{k}\cap\tilde{A}_{k+1}^{\geq a}$ is zero. ∎

Corollary 3**.**

Suppose that (3.8) holds for some $1\leq k\leq N-2$. Then

[TABLE]

Proof.

Observe that if $k$ satisfies (3.8) then $k+1$ satisfies (3.8) as well. Thus by iterating Lemma 7 we obtain that $A_{k}\underset{\text{a.s.}}{\Longrightarrow}A_{k+1}\underset{\text{a.s.}}{\Longrightarrow}A_{k+2}\underset{\text{a.s.}}{\Longrightarrow}\dots\underset{\text{a.s.}}{\Longrightarrow}A_{N-1}$, i.e. $x_{(N-1)}(t)\to 0$, which is equivalent to the statement of Corollary. ∎

Remark 3**.**

Note that the condition (3.8) does not assume $p>1$; hence the conclusion of Corollary 3 holds for the case $0<p\leq 1$ as well.

For the case $p\geq\frac{N}{2}$ we have

Lemma 8**.**

If $p\geq\frac{N}{2}$ then ${\cal X}^{\prime}(t)\to 1$ a.s.

Proof.

The case $p>\frac{N}{2}$ is easy: with a positive probability the newly sampled point $\zeta>0$ and then

[TABLE]

hence it is the left-most point which is always removed, implying $\liminf_{t\to\infty}x_{(1)}(t)>0$. Hence by Corollary 1, ${\cal X}^{\prime}(t)\to 1$ a.s.

For the case $p=\frac{N}{2}$ we notice that at each moment of time we either have a tie (between the left-most and right-most point) or remove the left-most point. However, we can only have a tie if $x_{(1)}(t)=...=x_{(N-1)}(t)=0$; in this case, eventually the right-most point will be kept and the left-most removed. After this moment of time, there will be more ties, and the left-most point will always be removed, leading to the same conclusion as in the case $p>N/2$. ∎

Proof of Theorem 2.

Part (b) follows from Lemma 3.

To prove part (c), note that unless $x_{(1)}(0)>0$ already, by repeating the arguments from the beginning of the proof of Lemma 6, with a positive probability we can “drag” the whole configuration in at most $N-1$ steps to the right of zero, that is, there is $0\leq t_{0}\leq N-1$ such that ${\mathbb{P}}(\min{\cal X}^{\prime}(t_{0})>0)>0$. Now we can apply Lemma 4 and then Lemma 3.

Let us now prove part (a). First, assume $p<\frac{N}{2}$. It is always possible to find an integer $k$ which satisfies both (3.6) and (3.8), so let $k$ be such that

[TABLE]

(if $N/(2p)\in{\mathbb{N}}$ this $k$ will be unique). Now the statement of the theorem follows from Corollary 3 and Lemma 6.

Finally, in case $p\geq\frac{N}{2}$ the theorem follows from Lemma 8. ∎

4 Non-convergence to zero for $p>1$ and $N=3$

In this section we prove the following

Theorem 3**.**

Suppose that $N=3$, $p>1$ and $\zeta$, restricted to some neighbourhood of zero, is a continuous random variable with a non-decreasing density (e.g. uniformly distributed). Then ${\cal X}^{\prime}(t)\to 1$ as $t\to\infty$ a.s.

Remark 4**.**

**

• •

In case $p\geq 3/2$ we already know that ${\cal X}^{\prime}(t)\to 1$ for any initial configuration and any distribution (see Lemma 8), so we have to prove the theorem only for $p\in(1,3/2)$.

• •

Simulations suggest that the statement of Theorem 3 holds, in fact, for a much more general class of distributions $\zeta$.

Let $\varepsilon\in(0,1/2)$ be such that $\zeta$ conditioned333note that the full support assumption ensures that the probability of this event is positive on $\{\zeta\leq 2\varepsilon\}$ has a non-decreasing density; according to the statement of the Theorem 3 such an $\varepsilon$ must exist. Let us fix this $\varepsilon$ from now on.

The idea of the proof will be based on finding a non-negative function $h:(0,1]^{2}\to{\mathbb{R}}_{+}$ which has the following three properties:

• (i)

$h(\cdot,\cdot)$ is non-increasing in each of its arguments;

• (ii)

$h\left(x_{(1)}(t),x_{(2)}(t)\right)$ is a supermartingale as long as $x_{(2)}(t)\leq\varepsilon$;

• (iii)

$h(\cdot,\cdot)$ goes to infinity when the first coordinate goes to zero.

From the supermartingale convergence theorem it would then follow that

[TABLE]

Let us formally prepare for the proof of Theorem 3. As before, denote by $x_{1},\dots,x_{N}$ $N$ distinct points on $[0,1]$, and let $x_{(1)},\dots,x_{(N)}$ be this unordered $N$-tuple sorted in the increasing order. Let

[TABLE]

be the unordered $N$-tuple $\{x_{1},\dots,x_{N}\}$ with the farthest point from $p$-centre of mass removed; w.l.o.g. assume that $y_{i}$ are already in the increasing order.

Lemma 9**.**

The operation $\{\dots\}^{\prime}_{p}$ is monotone in $p$, that is, if $\hat{p}\geq\tilde{p}$ and

[TABLE]

then $\hat{y}_{i}\geq\tilde{y}_{i}$, $i=1,\dots,N-1$.

Proof.

Assume w.l.o.g. $x_{1}\leq...\leq x_{N}$, and let $\mu=\mu\left(\{x_{1},\dots,x_{N}\}\right)$. Notice that, regardless of the value of $p$, the only points which can possibly be removed are $x_{1}$ or $x_{N}$ (since they are the two extreme points). Therefore, it suffices to show that $\{x_{1},\dots,x_{N}\}^{\prime}_{\tilde{p}}=\{x_{2},\dots,x_{N}\}$ implies $\{x_{1},\dots,x_{N}\}^{\prime}_{\hat{p}}=\{x_{2},\dots,x_{N}\}$. Note also that $|x_{1}-p\mu|=p\mu-x_{1}$ for all $p\geq 1$.

If $\tilde{p}\mu-x_{1}>|\tilde{p}\mu-x_{N}|$ and $\tilde{p}\mu-x_{N}>0$, that is, the $p-$centre of mass lies to the right of $x_{N}$, then $\hat{p}\mu>\tilde{p}\mu>x_{N}$ as well, and hence $x_{1}$ is discarded.

On the other hand, if $\tilde{p}\mu-x_{1}>|\tilde{p}\mu-x_{N}|$ and $\tilde{p}\mu<x_{N}$ then either $\hat{p}\mu<x_{N}$, or $\hat{p}\mu\geq x_{N}$. In the first case,

[TABLE]

so $x_{1}$ is discarded. In the second case, $p-$centre of mass lies to the right of $x_{N}$ and so $x_{1}$ is also discarded. ∎

Lemma 10**.**

Let $h$ be a real-valued function on the sets of $N$ real numbers. Suppose that $h$ is non-increasing in each of its arguments, namely

[TABLE]

whenever $x_{i}^{\prime}\geq x_{i}$. Let ${\cal E}_{t}$ be some ${\cal{F}}_{t}$-measurable event, and suppose that

[TABLE]

for $p=1$. Then (4.10) holds for $p>1$ as well.

Proof.

Let

[TABLE]

be the new core after the new point $\zeta_{t+1}$ is sampled and the farthest point from the $p-$centre of mass is removed; note that ${\cal X}^{\prime}(t+1)=G_{p}({\cal X}^{\prime}(t),\zeta_{t+1})$. Then on ${\cal E}_{t}$

[TABLE]

since the operation $\{\dots\}^{\prime}_{p}$ is monotone in $p$ by Lemma 9 and $h$ is decreasing in each argument. ∎

From now on assume $N=3$ and $p=1$. Denote $x_{(1)}(t)=a$, $x_{(2)}(t)=b$ and consider the events

[TABLE]

(we assume that $b$ is smaller than $1/2$, yielding $2b-a<1$.) If $x_{(2)}(t)\leq\varepsilon$ then ${\cal X}^{\prime}(t+1)\not={\cal X}^{\prime}(t)$ implies that one of the events $L_{b}$, $B_{b}$, $B_{a}$ or $R_{a}$ occurs (i.e. all points sampled outside of $\left((2a-b)^{+},2b-a\right)$ are rejected at time $t+1$). Let us study the core ${\cal X}^{\prime}(t+1)=\{\zeta_{t+1},a,b\}^{\prime}$ on these events: on $L_{b}$ and $B_{b}$ we have ${\cal X}^{\prime}(t+1)=\{x,a\}$, while on $B_{a}$ and $R_{a}$ we have ${\cal X}^{\prime}(t+1)=\{x,b\}$.

We have, assuming $x_{(1)}(t)=a$ and $x_{(2)}(t)=b$,

[TABLE]

When $0\leq a\leq b\leq\varepsilon$ we have $2b-a\leq 2\varepsilon$. Define

[TABLE]

which is positive in the first two cases, and negative in the next two. Let $\varphi(x)$ be the density of $\zeta$ conditioned on $\{\zeta\in[0,2\varepsilon]\}$. By the monotonicity of $\varphi$ and the positivity (negativity resp.) of $g$ on the first (second resp.) interval,

[TABLE]

where

[TABLE]

So if we can establish that $\Lambda\leq 0$ for a suitable function $h$, then indeed $\Delta(a,b)\leq 0$, and the supermartingale property follows.

Remark 5**.**

Notice that the method of proof, presented here, could possibly work for $N>3$ as well; that is, if one can find a function $h(x_{1},\dots,x_{N-1})$, which is positive and decreasing in each of its arguments, and $h({\cal X}^{\prime}(t))$ is a supermartingale provided $\max{\cal X}^{\prime}(t)<\varepsilon$ for some $\varepsilon>0$. Unfortunately, however, we were not able to find such a function.

Set

[TABLE]

it is easy to check $h$ is indeed monotone in each of its arguments as long as $x,y\in(0,1]$. Let us now compute the integrals in the expression for $\Lambda$. We have

[TABLE]

It turns out that $h({\cal X}^{\prime}(t))$ indeed has a non-positive drift, provided $0<a\leq b\leq\varepsilon$, as is shown by the following

Lemma 11**.**

$\Lambda\leq 0$* for $a,b\in(0,1/2]$.*

Proof.

Substitute $a=b\nu$ in the expression for $\Lambda$. Then for $\nu\leq 1/3$ we easily obtain $\Lambda=-b\left[3\nu(1-\ln 2)+\ln 8-2\right]\leq 0.$

For $1/3<\nu\leq 1/2$ we have $2\Lambda=-bC_{1}(\nu)\leq 0$ where

[TABLE]

since $\frac{\partial^{2}C_{1}(\nu)}{\partial^{2}\nu}=-\frac{1}{\nu^{2}(1+\nu)}<0$ and hence $\min_{1/3\leq\nu\leq 1/2}C_{1}(\nu)$ is achieved at one of the endpoints $\nu=1/3$ or $\nu=1/2$; the values there are $C_{1}(1/3)=\ln(4)-1>0$ and $C(1/2)=\frac{1}{2}\ln\left(\frac{512}{27}\right)-1>0$ respectively.

For $1/2<\nu\leq 2/3$ we have $\Lambda=-bC_{2}(\nu)\leq 0$ where

[TABLE]

since $\frac{\partial^{2}C_{2}(\nu)}{\partial^{2}\nu}=\frac{2\nu^{2}+6\nu+3}{\nu^{2}(1+\nu)}>0$ and $\left.\frac{\partial C_{2}(\nu)}{\partial\nu}\right|_{\nu=2/3}=\ln\left(\frac{256}{45}\right)-\frac{5}{2}<0$ implies that $\frac{\partial C_{2}(\nu)}{\partial\nu}<0$ for all $\nu\in[1/2,2/3]$ and hence $\min_{1/2\leq\nu\leq 2/3}C_{2}(\nu)=C_{2}(2/3)=\frac{1}{3}\ln\left(\frac{104976}{3125}\right)-1>0.$

Finally, for $2/3<\nu\leq 1$ we have $\Lambda=-bC_{3}(\nu)\leq 0$, where

[TABLE]

since

[TABLE]

changes its sign from $-$ to $+$ at $1/\sqrt{2}\in(2/3,1)$ and therefore $\frac{\partial C_{3}(\nu)}{\partial\nu}$ achieves its maximum at the endpoints of the interval; thus

[TABLE]

Therefore, $C_{3}(\nu)$ is decreasing and hence $\min_{2/3\leq\nu\leq 1}C_{3}(\nu)=C_{3}(1)=0$. ∎

Proof of Theorem 3.

We will show that ${\mathbb{P}}({\cal X}^{\prime}(t)\to 0)=0$, which will imply by Theorem 2(a) that ${\mathbb{P}}({\cal X}^{\prime}(t)\to 1)=1$; we shall do it by showing that

[TABLE]

Indeed, fix some $\varepsilon\in(0,1/2)$ and let $\tau_{0}=0$. For $\ell=1,2,\dots,$ define the sequence of stopping times

[TABLE]

so that $\tau_{0}<\eta_{1}<\tau_{1}<\eta_{2}<\tau_{2}<\dots$ with the conventions that if one of the stopping times is infinite, so is the rest of them. Define also $\ell^{*}=\inf\{\ell\geq 1:\ \tau_{\ell}=+\infty\}$.

If $\ell^{*}=\infty$, that is, $\tau_{\ell}<\infty$ for all $\ell$, we immediately have $\limsup_{t\to\infty}x_{(2)}(t)\geq\varepsilon$ and we are done; so assume that for some $\ell^{*}\geq 1$ we have $\tau_{\ell^{*}-1}<\infty=\tau_{\ell^{*}}$. If $\eta_{\ell^{*}}=\infty$, then $x_{(2)}(t)>\varepsilon$ for all $t\geq\tau_{\ell^{*}-1}$ and thus again $\limsup_{t\to\infty}x_{(2)}(t)\geq\varepsilon$; hence $\ell^{*}<\infty$ and $\eta_{\ell^{*}}<\infty$ on the event $\left\{\limsup_{t\to\infty}x_{(2)}(t)<\varepsilon\right\}$.

On the other hand, as long as $\eta_{\ell}<\infty$, we can define

[TABLE]

where $h$ is given by (4.11).

By Lemmas 10 and 11 we have

[TABLE]

hence $\tilde{\xi}^{(\ell)}_{s}$, the process $\xi^{(\ell)}$ stopped at the time when $x_{(2)}$ exceeds $\varepsilon$, is a non-negative supermartingale, hence it must converge to a finite value. In case $\tau_{\ell}=+\infty$ this means that $\liminf_{t\to\infty}x_{(1)}(t)>0$ since the function $h(a,b)$ goes to infinity when $a\downarrow 0$. Thus we have established (4.12). ∎

5 Appendix: The calculations for the proof of Lemma 2.

Observe that all expressions for ${\bf A}_{j}$ are fractions of the polynomials in $(a,f,p,M)$; moreover, their denominators

[TABLE]

are always positive. Throughout the rest of the proof let ${\sf n}(w)$ denote the numerator of such a fraction $w$.

Case 1: ${\bf I}_{1}\leq 0$

Observe that

[TABLE]

and the term in the square brackets is positive as $M\mu\geq 1$, so the maximum of ${\sf n}({\bf A}_{1})$ is achieved at the highest possible value of $p$. However, in this case we have $p\leq p_{1}$, hence

[TABLE]

where

[TABLE]

Hence $s_{1}\geq 0$ for $M=2,3,\dots$ and thus ${\bf I}_{1}\leq 0$.

Case 2: ${\bf I}_{2}\leq 0$

Here

[TABLE]

where

[TABLE]

and we need to show that $s_{2}\geq 0$.

Assume first $M=2$. Then (using the fact that $\mu=(1+a)/2$)

[TABLE]

which is impossible; so from now on $M\geq 3$.

To establish ${\bf I}_{2}\leq 0$, it will suffice to demonstrate that

[TABLE]

as ${\bf I}_{2}$ has a factor $1_{X_{2}\geq 0}$, and $s_{2}1_{X_{2}\geq 0}\geq\frac{s_{3}}{2M}1_{X_{2}\geq 0}$. Substituting

[TABLE]

where $w\in[0,1)$ corresponding to the condition (2.4), we get

[TABLE]

The expression in the square brakets is non-negative for $M\geq 3$, so the minimum of $s_{3}$ is achieved when $a=1$; i.e.

[TABLE]

But

[TABLE]

so

[TABLE]

Case 3: ${\bf I}_{3}\leq 0$

Here

[TABLE]

and it suffices to show that $s_{5}\geq 0$. If $M=2$, then $\mu=(a+1)/2$ and $p\geq 3/4$, so

[TABLE]

For $M\geq 3$, let $M=3+\delta$, $\delta=0,1,\dots$. Then $s_{5}=\sum_{i=0}^{5}e_{i+1}\delta^{i}$ where we will show that all $e_{i}\geq 0$. Indeed, we have

[TABLE]

[TABLE]

The fact that $e_{6}\geq 0$ is trivial; we will prove separately that $e_{1},\dots,e_{5}\geq 0$ below. In what follows, we substitute $p=\frac{1+\nu}{2}$, where $\nu\in(0,1)$.

Proof that $e_{1}\geq 0$

We have

[TABLE]

hence $e_{1}$ achieves its minimum at

[TABLE]

which solves $\frac{\partial e_{1}}{\partial a}=0$. Note that it is possible that $a_{cr}\not\in[0,1]$. However, in any case,

[TABLE]

so it will suffice to show that

[TABLE]

is positive. We have

[TABLE]

Similarly,

[TABLE]

where

[TABLE]

and

[TABLE]

So, $c_{1b1},c_{1b2}>0$ $\Longrightarrow$ $c_{1b}>0$ and since $c_{1a}>0$ we have $c_{1}\geq 0$ and thus $e_{1}\geq 0$.

Proof that $e_{2}\geq 0$

We have

[TABLE]

so, similarly to the previous case,

[TABLE]

where

[TABLE]

solves $\frac{\partial e_{2}}{\partial a}=0$ and

[TABLE]

Now,

[TABLE]

hence the minimum of $c_{2}$ w.r.t. $\mu\in[0,1]$ can be achieved either at $\mu=0$ or at $\mu=1$. At the same time

[TABLE]

so $c_{2}\geq 0$ and hence $e_{2}\geq 0$.

Proof that $e_{3}\geq 0$

We have

[TABLE]

so, similarly to the previous case,

[TABLE]

where

[TABLE]

solves $\frac{\partial e_{3}}{\partial a}=0$ and

[TABLE]

Hence $e_{3}\geq 0$.

Proof that $e_{4}\geq 0$

We have

[TABLE]

so, similarly to the previous case,

[TABLE]

where

[TABLE]

solves $\frac{\partial e_{4}}{\partial a}=0$ and

[TABLE]

Hence $e_{4}\geq 0$.

Proof that $e_{5}\geq 0$

We have

[TABLE]

so, similarly to the previous case,

[TABLE]

where

[TABLE]

solves $\frac{\partial e_{5}}{\partial a}=0$ and

[TABLE]

Hence $e_{5}\geq 0$.

As a result, $s_{5}\geq 0$ and thus ${\bf I}_{3}\leq 0$.

Case 4: ${\bf I}_{4}\leq 0$

Here

[TABLE]

Then, substituting $M=2+\delta$,

[TABLE]

and as a result for $\delta\geq 0$ we have

[TABLE]

Case 5: ${\bf I}_{5}\leq 0$

Here

[TABLE]

We need to show that $s_{7}\geq 0$ when $X_{1}\leq 0$ and $X_{3}\geq 0$.

Since $X_{1}\leq 0$, we have $2Mp\mu\leq M+1$. Together with $X_{3}\geq 0$ this implies

[TABLE]

whence

[TABLE]

Let us show that for this $a$ we have $s_{7}\geq 0$; substitute $a=b\cdot\frac{2p}{M+1}$, where $b\in[0,1]$.

First, let $M=2$ , then $\mu=\frac{1+a}{2}$, $p\in[3/4,1)$, and $s_{7}=\frac{3-2p}{27}s_{8}$ where

[TABLE]

Note that we can write $s_{8}=e_{1}+e_{2}(1-p)+e_{3}(1-p)^{2}$, where

[TABLE]

with $p=\frac{3+\nu}{4}$, $\nu\in[0,1]$. Consequently, since $(1-p)^{2}<1$ and $e_{1}>0$,

[TABLE]

and thus $s_{7}\geq 0$ as required.

For $\boxed{M\geq 3}$ , set $M=3+\delta$, $\delta\geq 0$. Then

[TABLE]

where

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and the expressions for $e_{9}$ and $e_{10}$ are given a little bit further.

First, we will show that $e_{i}\geq 0$, $i=1,\dots,8$.

Proof that $e_{1},\dots,e_{8}>0$

It turns out that it is easiest is to use a computer-assisted proof in this case; to this end we developed the method which we call a Box method; it may have been described by other authors, but since we do not have the reference to the right source, we give its description below.

First of all, we substitute

[TABLE]

Let ${\rm m}=\min_{a_{i}\leq x_{i}\leq b_{i},i=1,2,3}f(x_{1},x_{2},x_{3})$ where

[TABLE]

and $f_{+}$ and $f_{-}$ are polynomials with non-negative coefficients. We want to show that ${\rm m}>0$.

Let

[TABLE]

Since

[TABLE]

as $M\to\infty$, we conclude that ${\rm m}>0$ if and only if $G_{f,M}\geq 0$ for some $M\geq 1$. Checking that $G_{f,M}\geq 0$ can be quite tedious and time-consuming for large $M$, however, this could be easily accomplished with the help of a computer; please note, that the results are still completely rigorous, unlike e.g. simulations.

The results of application of this method to $e_{1},\dots,e_{8}$ are presented in the following table:

[TABLE]

Consequently, $e_{j}>0$ for all $j=1,\dots,8$.

Proof that $e_{9}\geq 0$ and $e_{10}\geq 0$

The Box method of the previous section would not work for $e_{9}$ and $e_{10}$, since these functions do touch zero in the required area, and hence the minimum is, in fact, [math]. Therefore, we have to handle these two cases analytically.

We have

[TABLE]

hence, the minimum is achieved either at $b=0$ or $b=1$.

For $\mu<1/(2p)$ we have $e_{9}\geq e_{9a}$, where

[TABLE]

where $s=2p\mu\in[0,1]$.

In case $\mu\geq 1/(2p)$ we have $e_{9}\geq e_{9b}$, where

[TABLE]

where $\mu=\frac{1}{2p}+s\left(1-\frac{1}{2p}\right)$, $s\in[0,1]$. Now,

[TABLE]

so the minimum of $e_{9b}$ w.r.t. $s$ is achieved where $\frac{\partial}{\partial s}e_{9b}=0$, i.e.

[TABLE]

and equals

[TABLE]

for $p\geq 1/2$.

Finally, trivially, we have $e_{10}=3p(2\mu p-1)^{2}\geq 0.$ Consequently, $s_{7}\geq 0$ and ${\bf I}_{5}\leq 0$.

Combining this with the previously established inequalities ${\bf I}_{j}\leq 0$, $j=1,2,3,4$, we complete the proof Lemma 2.

Acknowledgement

We would like to thank the anonymous referees for very careful reading our manuscript and for giving many useful suggestions and corrections.