# Probability of Consensus of Hegselmann-Krause Dynamics

###### Abstract.

The original Hegselmann-Krause (HK) model comprises a set of $n$ agents
characterized by their opinion, a number in $[0,1]$. Agent $i$ updates its
opinion $x_{i}$ via taking the average opinion of its neighbors whose opinion
differs by at most $\epsilon$ from $x_{i}$. In the article, the opinion space
is extended to $\mathbf{R^{d}}.$ The main result is to derive bounds for the
probability of consensus. In general, we have a positive lower bound for the
probability of consensus and demonstrate a lower bound for the probability of
consensus on a unit cube. In particular for one dimensional case, we derive an
upper bound and a better lower bound for the probability of consensus and
demonstrate them on a unit interval.

###### Key words and phrases:

synchronous Hegselmann-Krause model, probability of consensus, connected-
preserving

###### 1991 Mathematics Subject Classification:

28A05, 28A12, 28A20, 28A25, 05C07, 05C12, 05C40

Hsin-Lun Li

School of Mathematical and Statistical Sciences,

Arizona State University, Tempe, AZ 85287, USA

##  1\. Introduction

The original Hegselmann-Krause (HK) model consists of a set of $n$ agents
characterized by their opinion, a number in $[0,1]$. Agent $i$ updates its
opinion $x_{i}$ via taking the average opinion of its neighbors whose opinion
differs by at most $\epsilon$ from $x_{i}$ for a confidence bound
$\epsilon>0.$ In this essay, the opinion space is extended to
$\mathbf{R^{d}}$. The aim is to derive a lower bound for the probability of
consensus for the synchronous HK model as follows:

$x(t+1)=A(t)x(t)\hbox{ for }t\in\mathbf{N},$ (1)
$\begin{array}[]{c}\displaystyle A_{ij}(t)=\mathbbm{1}\\{j\in
N_{i}(t)\\}/|N_{i}(t)|\hbox{ and }\displaystyle
x(t)=(x_{1}(t),\ldots,x_{n}(t))^{\prime}=\hbox{transpose of
}(x_{1}(t),\ldots,x_{n}(t))\end{array}$

for $[n]:=\\{1,2,\ldots,n\\}$,
$N_{i}(t)=\\{j\in[n]:\|x_{j}(t)-x_{i}(t)\|\leq\epsilon\\}$ the collection of
agent $i^{\prime}s$ neighbors at time $t$ and $\|\ \,\|$ the Euclidean norm.
[5] gives an overview of HK models. [3, 4] elaborate that (1) has finite-time
convergence property. [1] further illustrates that the _termination time_

$T_{n}=\inf\\{t\geq 0:x(t)=x(s)\text{ for all }s\geq t\\}$

is bounded from above. Finite-time convergence property is enough to imply

$\lim_{t\rightarrow\infty}\max_{i,j\in[n]}\|x_{i}(t)-x_{j}(t)\|\hbox{
exists}.$

Let the initial opinions $x_{i}(0)$ be independent and identically distributed
random variables with a convex support $S\subset\mathbf{R^{d}}$ of positive
Lebesgue measure and a probability density function $f$, where $P(x_{i}\in
B)=\int_{B}f(x_{i})dm(x_{i})$ for all $i\in[n]$, $B$ a Borel set and $m$ the
Lebesgue measure. Here, we say a function or a set is measurable if it is
Lebesgue measurable. A _profile_ at time $t$ is an undirected graph
$\mathscr{G}(t)=(\mathscr{V}(t),\mathscr{E}(t))$ with the vertex set and edge
set

$\mathscr{V}(t)=[n]\hbox{ and }\mathscr{E}(t)=\\{(i,j):i\neq j\hbox{ and
}\|x_{i}(t)-x_{j}(t)\|\leq\epsilon\\}.$

A profile $\mathscr{G}(t)$ is _$\delta$ -trivial_ if any two vertices are at a
distance of at most $\delta$ apart. Observe that a consensus is reached at
time $t+1$ if $\mathscr{G}(t)$ is $\epsilon$-trivial.

##  2\. Main results

Define

$\mathscr{C}=\\{\lim_{t\rightarrow\infty}\max_{i,j\in[n]}\|x_{i}(t)-x_{j}(t)\|=0\\},$

the collection of all sample points that lead to a consensus.

######  Theorem 1.

$P(\mathscr{C})\geq P(\mathscr{G}(0)\hbox{ is connected})\hbox{ for }1\leq
n\leq 4.$

In general,

$\displaystyle P(\mathscr{C})$ $\displaystyle\geq P(\mathscr{G}(0)\text{ is
}\epsilon\text{-trivial})$ $\displaystyle\geq P(x_{i}(0)\in
B(x_{1}(0),\epsilon/2)\text{ for all }i\in[n])$
$\displaystyle=\int_{\mathbf{R^{d}}}f(x_{1})\left(\int_{B(x_{1},\epsilon/2)}f(x)dm(x)\right)^{n-1}dm(x_{1})>0\text{
for }n\geq 1.$

In particular, the probability of consensus is positive.

######  Corollary 1.

Assume that $S=[0,1]^{d}$ and $x_{i}(0)=\operatorname{Uniform\,}([0,1]^{d})$.
Then,

$P(\mathscr{C})\geq\left((\frac{\epsilon}{2})^{d}m(B(0,1))\right)^{n-1}(1-\epsilon)^{d}=\left((\frac{\epsilon}{2})^{d}\frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2}+1)}\right)^{n-1}(1-\epsilon)^{d}$

for all $i\in[n]$ and $\epsilon\in(0,1).$

Define

$(1)=\arg\min_{k\in[n]}x_{k}\text{ and
}(i)=\arg\min_{k\in[n]-\\{(j)\\}_{j=1}^{i-1}}x_{k}\text{ for }i\geq 2.$

Namely $x_{(i)}$ is the $i$-th smallest number among $(x_{k})_{k=1}^{n}.$ For
$n\geq 4$, let $m=\lfloor\frac{n-4}{3}\rfloor$ and $k=n-m-1.$ Say
$\mathscr{G}(t)$ satisfies $(\ast)$ if

$((m+2),(k))\in\mathscr{E}(t)\text{ and
}(x_{(n)}-x_{(k)}+x_{(m+2)}-x_{(1)})(t)\leq\epsilon.$

Say $\mathscr{G}(t)$ satisfies $(\ast\ast)$ if

$\max\left((x_{(n)}-x_{(n-i-1)})(t),\ (x_{(n-i-1)}-x_{(i+2)})(t),\
(x_{(i+2)}-x_{(1)})(t)\right)\leq\frac{\epsilon}{2}\text{ for some }0\leq
i\leq m.$

######  Theorem 2 ($d=1$).

$P(\mathscr{C})=P(\mathscr{G}(0)\text{ is connected})\text{ for }1\leq n\leq
4.$ $P(\mathscr{C})\geq P(\mathscr{G}(0)\text{ satisfies }(\ast))\hbox{ for
}5\leq n\leq 7.$

In general,

$\displaystyle P(\mathscr{G}(0)\text{ is connected})\geq P(\mathscr{C})\geq
P(\mathscr{G}(0)\text{ is }\epsilon\text{-trivial or satisfies
}(\ast\ast))\text{ for }n\geq 1.$

######  Corollary 2.

Let $S=[0,1]$, $d=1$, $\epsilon\in(0,1)$ and
$x_{i}(0)=\operatorname{Uniform\,}([0,1])$ for all $i\in[n]$. Then,
for $n=2,$

$P(\mathscr{C})=\epsilon(2-\epsilon)$

for $n=3,$

$P(\mathscr{C})=\left\\{\begin{array}[]{cc}6\epsilon^{2}(1-\epsilon)&\epsilon\in(0,\frac{1}{2})\\\
1-2(1-\epsilon)^{3}&\epsilon\in[\frac{1}{2},1)\end{array}\right.$

for $n=4,$

$P(\mathscr{C})=\left\\{\begin{array}[]{cc}24\epsilon^{3}(1-3\epsilon)+36\epsilon^{4}&\epsilon\in(0,\frac{1}{3})\\\
\\\
19\epsilon^{4}-4\epsilon^{3}(1-2\epsilon)+(1-2\epsilon)^{4}-6\epsilon^{2}(3\epsilon-1)^{2}\\\
-4\epsilon(1-2\epsilon)^{3}+12\epsilon^{3}(1-2\epsilon)+12\epsilon^{2}(1-2\epsilon)^{2}&\epsilon\in[\frac{1}{3},\frac{1}{2})\\\
\\\
\epsilon^{4}+4\epsilon^{3}(1-\epsilon)+6\epsilon^{2}(1-\epsilon)^{2}+4\epsilon(1-\epsilon)^{3}-2(1-\epsilon)^{4}&\epsilon\in[\frac{1}{2},1)\end{array}\right.$

for $n\geq 1,$

$\displaystyle P(\mathscr{G}(0)\text{ is
}\epsilon\text{-trivial})=\epsilon^{n-1}[n-(n-1)\epsilon]$ $\displaystyle
P(x_{i}(0)\in B(x_{1}(0),\epsilon/2)\text{ for all
}i\in[n])=\frac{2}{n}\epsilon^{n}(1-\frac{1}{2^{n}})+\epsilon^{n-1}(1-\epsilon).$

In general,

$P(\mathscr{C})\geq P(\mathscr{G}(0)\text{ is
}\epsilon\text{-trivial})=\epsilon^{n-1}[n-(n-1)\epsilon]\text{ for }n\geq 1.$

Figure 1. Bounds for $P(\mathscr{C})$

From Figure 1, labels $\epsilon/2,$ $\epsilon$ and $c$ denote respectively the
lines of

$P(x_{i}(0)\in B(x_{1}(0),\epsilon/2)\text{ for all }i\in[n]),\
P(\mathscr{G}(0)\text{ is }\epsilon\text{-trivial})\hbox{ and
}P(\mathscr{G}(0)\text{ is connected}).$

The black points are simulations for the probability of consensus. For $n=2,$

$P(\mathscr{C})=P(\mathscr{G}(0)\text{ is connected})=P(\mathscr{G}(0)\text{
is }\epsilon\text{-trivial})$

so $\epsilon$ and $c$-lines overlap. Observe that the points for the
probability of consensus are around $c$-line for $2\leq n\leq 4$, which meets
the theory. For $5\leq n\leq 7,$ the gray points and the solid line are
respectively simulations and numerical integrals of $P(\mathscr{G}(0)\text{
satisfies }(\ast))$, suggesting that theoretically $P(\mathscr{G}(0)\text{
satisfies }(\ast))$ is a better lower bound for $P(\mathscr{C})$ than
$P(\mathscr{G}(0)\text{ is }\epsilon\text{-trivial})$. For $n=10,$ the dark
gray points are simulations of $P(\mathscr{G}(0)\text{ is
}\epsilon\text{-trivial or satisfies }(\ast\ast)),$ and the points and solid
line are respectively simulations and numerical integrals of
$P(\mathscr{G}(0)\text{ satisfies }(\ast\ast)\text{ and }i=(m)).$ Suggest that

$P(\mathscr{G}(0)\text{ is }\epsilon\text{-trivial})\vee
P(\mathscr{G}(0)\hbox{ satisfies }(\ast\ast)\text{ and }i=(m))$

is a better lower bound than each of the two for the probability of consensus.

##  3\. Probability of consensus

To derive a better lower bound for the probability of consensus, we study
properties other than $\epsilon$-triviality that leads to a consensus. If a
profile is connected-preserving, then a consensus can be achieved in finite
time. We illustrate that any profile $\mathscr{G}$ is connected-preserving for
$1\leq n\leq 4$ and some profile $\mathscr{G}$ of some configuration $x$ fails
to remain connected for $n>4.$ Thus $P(\mathscr{C})\geq P(\mathscr{G}(0)\text{
is connected})$ for $1\leq n\leq 4.$ It is straightforward in general,
$P(\mathscr{G}(0)\text{ is }\epsilon\text{-trivial})$ is a lower bound for
$P(\mathscr{C})$ but it is uneasy to calculate in high dimensions. Therefore
we provide an easier calculated lower bound for the probability of consensus
and also depict that the probability of consensus is positive.

Lemma 1 is the key to depict that any profile $\mathscr{G}$ is connected-
preserving for $1\leq n\leq 4.$

######  Lemma 1 ([2]).

Given $\lambda_{1},...,\lambda_{n}$ in $\mathbf{R}$ with
$\sum_{i=1}^{n}\lambda_{i}=0$ and $x_{1},...,x_{n}$ in $\mathbf{R^{d}}$. Then
for $\lambda_{1}x_{1}+\lambda_{2}x_{2}+...+\lambda_{n}x_{n},$ the terms with
positive coefficients can be matched with the terms with negative coefficients
in the sense that

$\displaystyle\sum_{i=1}^{n}\lambda_{i}x_{i}=\sum_{i,c_{i}\geq
0,j,k\in[n]}c_{i}(x_{j}-x_{k})\text{ and
}\sum_{i}c_{i}=\sum_{j,\lambda_{j}\geq 0}\lambda_{j}.$

From Lemma 1, we derive a good upper bound for $\|x_{i}(t+1)-x_{j}(t+1)\|$ for
any $(i,j)\in\mathscr{E}(t).$

######  Lemma 2.

Assume that $(i,j)\in\mathscr{E}(t)$ and that $|N_{i}(t)|\leq|N_{j}(t)|$.
Then,

$\displaystyle\|x_{i}(t+1)-x_{j}(t+1)\|\leq\epsilon\left(3-|N_{i}(t)\cap
N_{j}(t)|(\frac{2}{|N_{j}(t)|}+\frac{1}{|N_{i}(t)|})\right).$

###### Proof.

Let $x=x(t)$, $x^{\prime}=x(t+1)$ and $N_{i}=N_{i}(t)$ for any $i\in[n]$. Via
Lemma 1, for any $i,j\in[n]$,

$\displaystyle x_{i}^{\prime}-x_{j}^{\prime}=\frac{1}{|N_{i}|}\sum_{k\in
N_{i}}x_{k}-\frac{1}{|N_{j}|}\sum_{k\in N_{j}}x_{k}$
$\displaystyle=(\frac{1}{|N_{i}|}-\frac{1}{|N_{j}|})\sum_{k\in N_{i}\cap
N_{j}}x_{k}+\frac{1}{|N_{i}|}\sum_{k\in
N_{i}-N_{j}}x_{k}-\frac{1}{|N_{j}|}\sum_{k\in N_{j}-N_{i}}x_{k}$
$\displaystyle=\sum_{p\in N_{i}\cap N_{j},q\in
N_{j}-N_{i}}a_{r}(x_{p}-x_{q})+\sum_{p\in N_{i}-N_{j},q\in
N_{j}-N_{i}}b_{r}(x_{p}-x_{q})$

where $a_{r}$, $b_{r}\geq 0$,
$\sum_{r}a_{r}=(\frac{1}{|N_{i}|}-\frac{1}{|N_{j}|})|N_{i}\cap N_{j}|$ and
$\sum_{r}b_{r}=|N_{i}-N_{j}|/|N_{i}|.$ Thus by the triangle inequality,

$\displaystyle\|x_{i}^{\prime}-x_{j}^{\prime}\|\leq\sum_{p\in N_{i}\cap
N_{j},q\in N_{j}-N_{i}}a_{r}(\|x_{p}-x_{j}\|+\|x_{j}-x_{q}\|)$
$\displaystyle+\sum_{p\in N_{i}-N_{j},q\in
N_{j}-N_{i}}b_{r}(\|x_{p}-x_{i}\|+\|x_{i}-x_{j}\|+\|x_{j}-x_{q}\|)$
$\displaystyle\leq(\frac{1}{|N_{i}|}-\frac{1}{|N_{j}|})|N_{i}\cap
N_{j}|(\epsilon+\epsilon)+\frac{|N_{i}-N_{j}|}{|N_{i}|}(\epsilon+\epsilon+\epsilon)$
$\displaystyle=(\frac{1}{|N_{i}|}-\frac{1}{|N_{j}|})|N_{i}\cap
N_{j}|2\epsilon+(1-\frac{|N_{i}\cap N_{j}|}{|N_{i}|})3\epsilon$
$\displaystyle=\epsilon\left(3-|N_{i}\cap
N_{j}|(\frac{2}{|N_{j}|}+\frac{1}{|N_{i}|})\right).$

∎

Thus $\epsilon\left(3-|N_{i}(t)\cap
N_{j}(t)|(\frac{2}{|N_{j}(t)|}+\frac{1}{|N_{i}(t)|})\right)\leq\epsilon$
implies $(i,j)\in\mathscr{E}(t+1).$ For the following lemmas, assume $x=x(t)$,
$x^{\prime}=x(t+1)$ and $N_{i}=N_{i}(t)$ for any $i\in[n]$ without specifying.

######  Lemma 3.

Assume that $(i,j)\in\mathscr{E}(t)$ with $|N_{i}(t)|\leq|N_{j}(t)|$ and that

$|N_{i}(t)\cap N_{j}(t)|(\frac{2}{|N_{j}(t)|}+\frac{1}{|N_{i}(t)|})\geq 2.$

Then, $(i,j)\in\mathscr{E}(t+1).$

###### Proof.

By Lemma 2,

$\displaystyle\|x_{i}^{\prime}-x_{j}^{\prime}\|\leq\epsilon\left(3-|N_{i}\cap
N_{j}|(\frac{2}{|N_{j}|}+\frac{1}{|N_{i}|})\right)\leq\epsilon(3-2)=\epsilon.$

Thus $(i,j)\in\mathscr{E}(t+1).$ ∎

Observe that $|N_{i}(t)\cap N_{j}(t)|\geq 2$ for $(i,j)\in\mathscr{E}(t)$ and
that the inequality $\frac{2}{|N_{j}(t)|}+\frac{1}{|N_{i}(t)|}\geq 1$
automatically holds for $1\leq n\leq 3.$ It is not straight forward to see
Lemma 3 works for $n=4.$ However, categorizing the degrees of the pair
$(i,j)\in\mathscr{E}(t)$, a profile $\mathscr{G}$ remains connected for $n=4.$

######  Lemma 4 (connected-preserving).

For $1\leq n\leq 4,$ a profile $\mathscr{G}$ is connected-preserving.

###### Proof.

Since $i,j\in N_{i}\cap N_{j}$ for $(i,j)\in\mathscr{E}(t)$ and $|N_{i}|\leq
n$ for all $i\in[n],$

$|N_{i}\cap N_{j}|(\frac{2}{|N_{j}|}+\frac{1}{|N_{i}|})\geq
2(\frac{2}{n}+\frac{1}{n})\geq 2(\frac{2}{3}+\frac{1}{3})=2\text{ for }1\leq
n\leq 3.$

From Lemma 3, $(i,j)\in\mathscr{E}(t+1).$ Thus any edge in $\mathscr{E}(t)$
remains in $\mathscr{E}(t+1).$ Hence a profile is connected-preserving for
$1\leq n\leq 3$.
For $n=4,$ let $d_{i}=d_{i}(t)$=the degree of vertex $i$ at time $t$. For
$(i,j)\in\mathscr{E}(t)$ and $d_{i}\leq d_{j},$ $i$ is either a leaf or a non-
leaf, and $i$ and $j$ can not be both leaves. So the cases of $(d_{i},d_{j})$
are as follows:

$\begin{bmatrix}d_{i}&1&1&2&2&3\\\ d_{j}&2&3&2&3&3\end{bmatrix}.$

Thus the cases of corresponding $(|N_{i}|,|N_{j}|)$ are

$\begin{bmatrix}|N_{i}|&2&2&3&3&4\\\ |N_{j}|&3&4&3&4&4\end{bmatrix}.$

From Lemma 3, if $\frac{2}{|N_{j}|}+\frac{1}{|N_{i}|}\geq 1$ or $|N_{i}\cap
N_{j}|(\frac{2}{|N_{j}|}+\frac{1}{|N_{i}|})\geq 2$, then
$(i,j)\in\mathscr{E}(t+1).$ We check if each case meets one of the two
conditions:

$\displaystyle(2,3):\ $
$\displaystyle\frac{2}{3}+\frac{1}{2}>\frac{1}{2}+\frac{1}{2}=1$
$\displaystyle(2,4):\ $
$\displaystyle\frac{2}{4}+\frac{1}{2}=\frac{1}{2}+\frac{1}{2}=1$
$\displaystyle(3,3):\ $ $\displaystyle\frac{2}{3}+\frac{1}{3}=1$
$\displaystyle(3,4):\ $ $\displaystyle
3(\frac{2}{4}+\frac{1}{3})=\frac{3}{2}+1>1+1=2$ $\displaystyle(4,4):\ $
$\displaystyle 4(\frac{2}{4}+\frac{1}{4})=2+1=3>2.$

Since each case satisfies one of the two conditions,
$(i,j)\in\mathscr{E}(t+1)$ for each case above, so a profile is connected-
preserving for $n=4$. ∎

Lemma 3 does not work for $n=5$ even by categorizing the degrees of the pair
$(i,j)\in\mathscr{E}(t)$. But indeed some profile $\mathscr{G}$ of some
configuration $x$ fails to remain connected.

Figure 2.

######  Lemma 5.

For $n\geq 5,$ some profile $\mathscr{G}$ of some configuration $x$ is not
connected-preserving.

###### Proof.

Need only show that there is a configuration $x$ with the profile
$\mathscr{G}(0)$ connected but $\mathscr{G}(1)$ disconnected for $n=5$ and
$d=1$. Consider

$\epsilon=1,\ x_{1}(0)=-1,\ x_{2}(0)=0,\ x_{3}(0)=1\ \hbox{and}\
x_{4}(0)=x_{5}(0)=2.$

Then, $\mathscr{G}(0)$ as Figure 3 is connected,

$x_{1}(1)=-0.5,\ x_{2}(1)=0,\ x_{3}(1)=1.25\ \hbox{and}\
x_{4}(1)=x_{5}(1)=\frac{5}{3}.$

So $\mathscr{G}(1)$ as Figure 3 is disconnected. For $n>5,$ let the new added
vertices whose opinion be -1 or 2. Then, at the next time step, opinion 0 goes
much closer to -1 or opinion 1 goes much closer to 2, and so $\mathscr{G}(0)$
connected but $\mathscr{G}(1)$ disconnected. This completes the proof. ∎

Hence a profile is connected-preserving for $n\leq 4$, and some profile
$\mathscr{G}$ of some configuration $x$ fails to remain connected for $n>4.$
We can estimate the probability of consensus via the initial profiles
$\mathscr{G}(0).$ Lemmas 6-9 indicate the probability of consensus is
positive.

######  Lemma 6.

Let $g>0$ be a measurable function on a measurable set $A$ with $m(A)>0$.
Then, $\int_{A}gdm>0.$

###### Proof.

Let $E_{k}=\\{g>\frac{1}{k}\\}$. Then, $A=\cup_{k\geq 1}E_{k}.$ Suppose by
contradiction that $\int_{A}gdm=0$. Then,

$\frac{1}{k}m(E_{k})\leq\int_{E_{k}}gdm\leq\int_{A}gdm=0\text{ for all }k\geq
1.$

Thus via the subadditivity of a measure,

$m(A)\leq\sum_{k\geq 1}m(E_{k})=0,\text{ a contradiction}.$

∎

######  Lemma 7.

(i) The intersection of convex sets is convex. (ii) The closure of a convex
set is convex.

###### Proof.

(i) Let $Q_{\alpha}^{\prime}s$ be convex sets. If
$\cap_{\alpha}Q_{\alpha}=\emptyset$, then clearly it is convex. Else, for
$a,b\in\cap_{\alpha}Q_{\alpha},a,b\in Q_{\alpha}$ for all $\alpha.$ So by
convexity of convex sets, any point on the segment $\overline{\rm ab}$ is in
$Q_{\alpha}$ for all $\alpha.$ Thus any point on the segment $\overline{\rm
ab}$ is in $\cap_{\alpha}Q_{\alpha}.$
(ii) Let $V$ be a convex set. For $v\in\overline{V}$, there exists
$(v_{n})_{n\geq 1}\subset V$ with $v_{n}\rightarrow v$ as
$n\rightarrow\infty.$ For $u,v\in\overline{V},$

$tu+(1-t)v=\lim_{n,m\rightarrow\infty}[tu_{n}+(1-t)v_{m}]$

where $u_{n},v_{m}\in V$ for all $n,m\geq 1$ and $t\in(0,1)$. By convexity of
$V,\ tu_{n}+(1-t)v_{m}\in V$, so $tu+(1-t)v\in\overline{V}.$ ∎

A _convex hull_ generated by $v_{1},\ldots,v_{k}\in\mathbf{R^{d}}$, denoted by
$C(\\{v_{1},\ldots,v_{k}\\})$, is the smallest convex set containing
$v_{1},\ldots,v_{k}$, $i.e.,\
C(\\{v_{1},\ldots,v_{k}\\})=\\{v:v=\sum_{i=1}^{k}a_{i}v_{i},(a_{i})_{i=1}^{k}\text{
is stochastic}\\}$.

######  Lemma 8.

A convex set in $\mathbf{R^{d}}$ is measurable.

###### Proof.

Let $\mathscr{L}$ be the collection of all Lebesgue sets in $\mathbf{R^{d}}$
and $V\subset\mathbf{R^{d}}$ be a convex set.

$\text{Claim: }m(\partial{V})=0.$

For $V^{\circ}=\emptyset$, if $m(\overline{V})>0$, then $\overline{V}$ is
uncountable, and there exist $d+1$ distinct points, $v_{1},v_{2},...,v_{d+1}$,
in $\overline{V}$ not in any hyperplane in $\mathbf{R^{d}}$. By convexity of
$\overline{V},$ $C(\\{v_{1},...,v_{d+1}\\})\subseteq\overline{V}$ with its
interior nonempty, a contradiction.

For $V^{\circ}\neq\emptyset,$ since measurability is shift-invariant, may
assume zero vector $\vec{0}\in V^{\circ}.$ Then $B(0,r)\subset V$ for some
$0<r<1$. For $n\in\mathbf{Z^{+}},$ let $A_{n}=B(0,n)\cap V$ then by Lemma 7,
$A_{n}$ is bounded and convex. For $q\in\partial A_{n},$ by convexity of
$\overline{V}$ and $A_{n}\supset B(0,r),$

$p=s\cdot q+(1-s)\cdot\vec{0}\in A_{n}^{\circ}\text{ for all }s\in(0,1).$

Thus $q\in\frac{1}{s}A_{n}^{\circ}.$ Since $\frac{1}{s}A_{n}^{\circ}\supset
A_{n}^{\circ},$

$\displaystyle m(\partial A_{n})\leq
m(\frac{1}{s}A_{n}^{\circ}-A_{n}^{\circ})=m(\frac{1}{s}A_{n}^{\circ})-m(A_{n}^{\circ})$
$\displaystyle=(\frac{1}{s})^{d}m(A_{n}^{\circ})-m(A_{n}^{\circ})\rightarrow
0\text{ as }s\rightarrow 1.$

Since $\cup_{n\geq 1}\partial A_{n}\supset\partial V,$

$m(\partial V)\leq m(\cup_{n\geq 1}\partial A_{n})\leq\sum_{n\geq 1}m(\partial
A_{n})=0.$

Thus $\partial V$ is a null set. By the completion of Lebesgue measure
$\partial V\cap V\in\mathscr{L}.$ Hence

$V=V^{\circ}\cap(\partial V\cap V)\in\mathscr{L}.$

∎

######  Lemma 9.

Let $V\subset\mathbf{R^{d}}$ be a convex set with $m(V)>0$. Then,

$m(V\cap B(x,r))>0\text{ for any }x\in\overline{V}\text{ and }r>0.$

###### Proof.

From the proof of Lemma 8, $m(\partial V)=0$ so

$m(V^{\circ})=m(V)-m(\partial V\cap V)=m(V)>0.$

Thus $V^{\circ}$ is uncountable and $V^{\circ}=\cup_{u\in
V^{\circ}}B(u,r_{u})$ for some $r_{u}>0$. For $x\in\overline{V}$ and $r>0$, by
the convexity of $\overline{V}$, there exists $y\in V^{\circ}$ with
$\|y-x\|<\frac{r}{2},$ so by the triangle inequality, $B(y,\frac{r}{2})\subset
B(x,r)$. Hence

$V\cap B(x,r)\supset B(y,r_{y})\cap
B(y,\frac{r}{2})=B(y,r_{y}\wedge\frac{r}{2}),\text{ so }m(V\cap B(x,r))\geq
m(B(y,r_{y}\wedge\frac{r}{2}))>0.$

This completes the proof. ∎

###### Proof of Theorem 1.

From Lemma 4, $\\{\mathscr{G}(0)\hbox{ is connected}\\}\subset\mathscr{C}$ for
$1\leq n\leq 4$ so

$P(\mathscr{C})\geq P(\mathscr{G}(0)\hbox{ is connected})\hbox{ for }1\leq
n\leq 4.$

Observe that $\mathscr{C}\supset\\{\mathscr{G}(0)\text{ is
}\epsilon\text{-trivial}\\}\supset\\{x_{i}(0)\in B(x_{1}(0),\epsilon/2)\text{
for all }i\in[n]\\}$ so

$\displaystyle P(\mathscr{C})$ $\displaystyle\geq P(\mathscr{G}(0)\text{ is
}\epsilon\text{-trivial})$ $\displaystyle\geq P(x_{i}(0)\in
B(x_{1}(0),\epsilon/2)\text{ for all }i\in[n])$
$\displaystyle=\int_{\mathbf{R^{d}}}\int_{B(x_{1},\epsilon/2)}...\int_{B(x_{1},\epsilon/2)}\prod_{i=1}^{n}f(x_{i})dm(x_{n})...dm(x_{1})$
$\displaystyle:=\int_{\mathbf{R^{d}}}(\int_{B(x_{1},\epsilon/2)})^{n-1}\prod_{i=1}^{n}f(x_{i})dm(x_{n})...dm(x_{1})$
$\displaystyle=\int_{\mathbf{R^{d}}}f(x_{1})\left(\int_{B(x_{1},\epsilon/2)}f(x)dm(x)\right)^{n-1}dm(x_{1}).$

Observe that $f>0$ on the convex set $S$ and $m(B(x_{1},\epsilon/2)\cap S)>0$
for all $x_{1}\in S$ from Lemma 9. Hence via Lemma 6,

$\displaystyle P(\mathscr{C})$
$\displaystyle\geq\int_{\mathbf{R^{d}}}f(x_{1})\left(\int_{B(x_{1},\epsilon/2)}f(x)dm(x)\right)^{n-1}dm(x_{1})$
$\displaystyle=\int_{S}f(x_{1})\left(\int_{B(x_{1},\epsilon/2)\cap
S}f(x)dm(x)\right)^{n-1}dm(x_{1})>0.$

∎

###### Proof of Corollary 1.

From theorem 1,

$\displaystyle P(\mathscr{C})$ $\displaystyle\geq P(x_{i}(0)\in
B(x_{1}(0),\epsilon/2)\text{ for all }i\in[n])$
$\displaystyle\geq\int_{[\epsilon/2,1-\epsilon/2]^{d}}dm(x_{1})\left(\int_{B(x_{1},\epsilon/2)}1dm(x)\right)^{n-1}$
$\displaystyle=\int_{[\epsilon/2,1-\epsilon/2]^{d}}m(B(x_{1},\frac{\epsilon}{2}))^{n-1}dm(x_{1})$
$\displaystyle=\left((\frac{\epsilon}{2})^{d}m(B(0,1))\right)^{n-1}(1-\epsilon)^{d}=\left((\frac{\epsilon}{2})^{d}\frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2}+1)}\right)^{n-1}(1-\epsilon)^{d}.$

∎

##  4\. One dimensional probability of consensus

In this section, we focus on the one dimensional HK model. Apart from higher
dimensions, opinions in one dimension are ordered by $\leq$. We demonstrate
that opinions are order-preserving and profiles are disconnected-preserving.
Hence $P(\mathscr{C})=P(\mathscr{G}(0)\text{ is connected})$ for $1\leq n\leq
4$ and in general $P(\mathscr{G}(0)\text{ is connected})$ is an upper bound
for the probability of consensus. Furthermore, we demonstrate the probability
of consensus on $[0,1].$

######  Lemma 10 (order-preserving).

For d=1, if $x_{i}(t)\leq x_{j}(t)$ then $x_{i}(t+1)\leq x_{j}(t+1).$

###### Proof.

Let $x=x(t),x^{\prime}=x(t+1),$ and $N_{i}=N_{i}(t)$ for all $i\in[n].$ From
Lemma 1,

$\displaystyle
x_{j}^{\prime}-x_{i}^{\prime}=(\frac{1}{|N_{j}|}-\frac{1}{|N_{i}|})\sum_{k\in
N_{i}\cap N_{j}}x_{k}+\frac{1}{|N_{j}|}\sum_{k\in
N_{j}-N_{i}}x_{k}-\frac{1}{|N_{i}|}\sum_{k\in N_{i}-N_{j}}x_{k}$
$\displaystyle=\left\\{\begin{array}[]{lr}\displaystyle\sum_{k,p\in
N_{j}-N_{i},q\in N_{i}}a_{k}(x_{p}-x_{q})&\hbox{if}\ |N_{j}|\geq|N_{i}|\\\
\displaystyle\sum_{k,p\in N_{j},q\in
N_{i}-N_{j}}a_{k}(x_{p}-x_{q})&\hbox{else,}\end{array}\right.$

where $a_{k}\geq 0$ for all $k.$ We claim that

  1. (1)

If $a\in N_{i}$ and $b\in N_{j}-N_{i}$, then $x_{a}<x_{b}$.

  2. (2)

If $a\in N_{i}-N_{j}$ and $b\in N_{j}$, then $x_{a}<x_{b}$.

###### Proof of Claim 1.

Assume by contradiction that there exist $a\in N_{i}$ and $b\in N_{j}-N_{i}$
such that $x_{a}\geq x_{b}$. Then, $x_{j}-\epsilon\leq x_{b}<x_{i}-\epsilon$,
a contradiction. ∎

###### Proof of Claim 2.

Assume by contradiction that there exist $a\in N_{i}-N_{j}$ and $b\in N_{j}$
such that $x_{a}\geq x_{b}$. Then, $x_{i}+\epsilon\geq x_{a}>x_{j}+\epsilon$,
a contradiction. ∎

Either way, $x_{j}^{\prime}-x_{i}^{\prime}\geq 0.$ This completes the proof. ∎

######  Lemma 11 (disconnected-preserving).

For $d=1,$ if $\mathscr{G}(t)$ is disconnected then $\mathscr{G}(t+1)$ is
disconnected.

###### Proof.

Assume $x_{1}(t)\leq x_{2}(t)\leq...\leq x_{n}(t).$ Since $\mathscr{G}(t)$ is
disconnected,

$x_{i+1}(t)-x_{i}(t)>\epsilon\text{ for some }i\in[n-1].$

Since vertices $i$ and $i+1$ have respectively no neighbors on its right and
left at time $t$,

$x_{i}(t+1)\leq x_{i}(t)\text{ and }x_{i+1}(t)\leq x_{i+1}(t+1).$

Hence $x_{i+1}(t+1)-x_{i}(t+1)>\epsilon.$ From Lemma 10,

$x_{1}(t+1)\leq x_{2}(t+1)\leq...\leq x_{n}(t+1).$

Thus $\mathscr{G}(t+1)$ is disconnected. ∎

Next, we consider several circumstances under which a profile is connected at
the next time step. Let $M\subset\mathbf{R}$ be a finite nonempty set and
$\overline{M}=\frac{\sum_{x\in M}x}{|M|}$ be the average on $M.$ It is clear
that

$\overline{a+M}>\overline{M}\iff a>\overline{M}.$ (2)

######  Lemma 12.

For $4\leq n\leq 7$, if $\mathscr{G}(t)$ satisfies $(\ast)$, then so does
$\mathscr{G}(t+1).$

###### Proof.

Let $x=x(t)$, $x^{\prime}=x(t+1),\ \epsilon_{1}=x_{(m+2)}-x_{(1)},\
\epsilon_{2}=x_{(n)}-x_{(k)},\ y_{1}=x_{(1)},\ y_{2}=x_{(m+2)},\
y_{3}=x_{(k)},\ y_{4}=x_{(n)}$. Since $x_{i}^{\prime}s$ are order-preserving,
need only consider $y_{3}^{\prime}-y_{2}^{\prime},\
y_{2}^{\prime}-y_{1}^{\prime},\text{ and }y_{4}^{\prime}-y_{3}^{\prime}.$ By
(2),

$\displaystyle\max_{(x_{i})_{i=1}^{n}-\\{y_{2},y_{3}\\}}(y_{3}^{\prime}-y_{2}^{\prime})=\frac{y_{2}+y_{3}+(y_{3}+\epsilon_{2})+[sy_{2}+(k-m-3-s)y_{3}]+m(y_{3}+\epsilon_{2})}{k}$
$\displaystyle-\frac{(y_{2}-\epsilon_{1})+y_{2}+y_{3}+m(y_{2}-\epsilon_{1})+[sy_{2}+(k-m-3-s)y_{3}]}{k}\text{
for some }0\leq s\leq k-m-3,$
$\displaystyle=\frac{(y_{3}-y_{2})+\epsilon_{2}+\epsilon_{1}+m(y_{3}-y_{2})+m(\epsilon_{2}+\epsilon_{1})}{k}=\frac{(m+1)(y_{3}-y_{2})+(m+1)(\epsilon_{1}+\epsilon_{2})}{k}$
$\displaystyle\leq\frac{2(m+1)}{k}\epsilon<\epsilon\text{ so
}((m+2),(k))\in\mathscr{E}(t+1).$
$\displaystyle\max_{(x_{i})_{i=1}^{n}-\\{y_{1},y_{2}\\}}(y_{2}^{\prime}-y_{1}^{\prime})=\frac{y_{1}+y_{2}+(n-m-2)(y_{2}+\epsilon)+[sy_{1}+(m-s)y_{2}]}{n}$
$\displaystyle-\frac{y_{1}+y_{2}+[sy_{1}+(m-s)y_{2}]}{m+2}\text{ for some
}0\leq s\leq m.$

Since

$\partial_{s}\max_{(x_{i})_{i=1}^{n}-\\{y_{1},y_{2}\\}}(y_{2}^{\prime}-y_{1}^{\prime})=\frac{y_{1}-y_{2}}{n}-\frac{y_{1}-y_{2}}{m+2}=(y_{1}-y_{2})(\frac{1}{n}-\frac{1}{m+2})\geq
0,\text{ set }s=m,$
$\displaystyle\max_{(x_{i})_{i=1}^{n}-\\{y_{1},y_{2}\\}}(y_{2}^{\prime}-y_{1}^{\prime})$
$\displaystyle=\frac{(m+2)[(m+1)y_{1}+(n-m-1)y_{2}+(n-m-2)\epsilon]-n[(m+1)y_{1}+y_{2}]}{n(m+2)}$
$\displaystyle=\frac{(m+1)(n-m-2)(y_{2}-y_{1})+(m+2)(n-m-2)\epsilon}{n(m+2)}.$

By symmetry,

$\max_{(x_{i})_{i=1}^{n}-\\{y_{3},y_{4}\\}}(y_{4}^{\prime}-y_{3}^{\prime})=\frac{(m+1)(n-m-2)(y_{4}-y_{3})+(m+2)(n-m-2)\epsilon}{n(m+2)}.$

Hence

$\displaystyle\max_{(x_{i})_{i=1}^{n}-\\{y_{1},y_{2}\\}}(y_{2}^{\prime}-y_{1}^{\prime})+\max_{(x_{i})_{i=1}^{n}-\\{y_{3},y_{4}\\}}(y_{4}^{\prime}-y_{3}^{\prime})$
$\displaystyle=\frac{(m+1)(n-m-2)(y_{2}-y_{1}+y_{4}-y_{3})+(m+2)(n-m-2)\epsilon}{n(m+2)}$
$\displaystyle\leq\frac{(n-m-2)(2m+3)}{n(m+2)}\epsilon\leq\epsilon\text{ for
}4\leq n\leq 7.$

So $y_{4}^{\prime}-y_{3}^{\prime}+y_{2}^{\prime}-y_{1}^{\prime}\leq\epsilon$
for $4\leq n\leq 7$. This completes the proof. ∎

Observe that a profile is connected-preserving if it satisfies $(\ast)$. It is
clear that an $\epsilon$-trivial profile satisfies $(\ast)$ and there exists
an $\epsilon$-nontrivial profile satisfies $(\ast).$ Thus
$\\{\mathscr{G}(0)\text{ satisfies }(\ast)\\}\supsetneq\\{\mathscr{G}(0)\text{
is }\epsilon\text{-trivial}\\}$.

######  Lemma 13.

For any $0\leq i\leq m$ and $n\geq 4$, assume that

$\max\left((x_{(n)}-x_{(n-i-1)})(t),\ (x_{(n-i-1)}-x_{(i+2)})(t),\
(x_{(i+2)}-x_{(1)})(t)\right)\leq\frac{\epsilon}{2}.$

Then,

$\max\left((x_{(n)}-x_{(n-i-1)})(t+1),\ (x_{(n-i-1)}-x_{(i+2)})(t+1),\
(x_{(i+2)}-x_{(1)})(t+1)\right)<\frac{\epsilon}{2}.$

###### Proof.

Let $x=x(t),\ x^{\prime}=x(t+1),\ y_{1}=x_{(1)},\ y_{2}=x_{(i+2)},\
y_{3}=x_{(n-i-1)},\ y_{4}=x_{(n)}.$ By the assumption, the neighborhood of
$(i+2)$ is the same as that of $(n-i-1),$ so
$y_{3}^{\prime}-y_{2}^{\prime}=0.$ Via (2),

$\displaystyle\max_{(x_{i})_{i=1}^{n}-\\{y_{1},y_{2}\\}}(y_{2}^{\prime}-y_{1}^{\prime})$
$\displaystyle=\frac{y_{1}+y_{2}+[s_{1}y_{1}+(i-s_{1})y_{2}]+[s_{2}y_{2}+(n-2i-3-s_{2})(y_{2}+\frac{\epsilon}{2})]+(i+1)(y_{2}+\epsilon)}{n}$
$\displaystyle-\frac{y_{1}+y_{2}+[s_{1}y_{1}+(i-s_{1})y_{2}]+[s_{2}y_{2}+(n-2i-3-s_{2})(y_{2}+\frac{\epsilon}{2})]}{n-i-1}$

for some $0\leq s_{1}\leq i$ and $0\leq s_{2}\leq n-2i-3$. Since $3m+4\leq
n\leq 3m+6$, $0\leq i\leq m$,
$\partial_{s_{1}}\max_{(x_{i})_{i=1}^{n}-\\{y_{1},y_{2}\\}}(y_{2}^{\prime}-y_{1}^{\prime})=(y_{1}-y_{2})(\frac{1}{n}-\frac{1}{n-i-1})\geq
0$, and
$\partial_{s_{2}}\max_{(x_{i})_{i=1}^{n}-\\{y_{1},y_{2}\\}}(y_{2}^{\prime}-y_{1}^{\prime})=[y_{2}-(y_{2}+\frac{\epsilon}{2})](\frac{1}{n}-\frac{1}{n-i-1})\geq
0$, set $s_{1}=i$ and $s_{2}=n-2i-3,$

$\displaystyle\max_{(x_{i})_{i=1}^{n}-\\{y_{1},y_{2}\\}}(y_{2}^{\prime}-y_{1}^{\prime})$
$\displaystyle=\frac{(i+1)y_{1}+(n-i-1)y_{2}+(i+1)\epsilon}{n}-\frac{(i+1)y_{1}+(n-2i-2)y_{2}}{n-i-1}$
$\displaystyle=\frac{(n-i-1)[(i+1)y_{1}+(n-i-1)y_{2}+(i+1)\epsilon]-n[(i+1)y_{1}+(n-2i-2)y_{2}]}{n(n-i-1)}$
$\displaystyle=\frac{(i+1)^{2}(y_{2}-y_{1})+(n-i-1)(i+1)\epsilon}{n(n-i-1)}\leq\frac{\frac{(i+1)^{2}}{2}+(n-i-1)(i+1)}{n(n-i-1)}\epsilon=\frac{1}{2}\frac{(i+1)(2n-i-1)}{n(n-i-1)}\epsilon$
$\displaystyle\leq\frac{1}{2}\frac{(m+1)(2n-1)}{n(n-m-1)}\epsilon\leq\frac{1}{2}\frac{(m+1)(6m+11)}{(3m+4)(2m+3)}\epsilon<\frac{\epsilon}{2}.$

By symmetry,

$\max_{(x_{i})_{i=1}^{n}-\\{y_{3},y_{4}\\}}(y_{4}^{\prime}-y_{3}^{\prime})=\frac{(i+1)^{2}(y_{4}-y_{3})+(n-i-1)(i+1)\epsilon}{n(n-i-1)}<\frac{\epsilon}{2}.$

This completes the proof. ∎

Observe that an $\epsilon$-trivial profile may not satisfies the assumption of
Lemma 13. Consider $n>1,\ x_{1}(t)=0,\ x_{i}(t)=\epsilon$ for $i>1$ then
$\mathscr{G}(t)$ is $\epsilon$-trivial but does not satisfy the assumption of
Lemma 13. Observe that $\mathscr{G}(s)$ satisfies $(\ast\ast)$ for all $s\geq
t.$

###### Proof of Theorem 2.

From Lemmas 4 and 11, $\mathscr{G}(t)$ is connected-preserving and
disconnected-preserving for $1\leq n\leq 4$ so
$\mathscr{C}=\\{\mathscr{G}(0)\text{ is connected}\\}.$ Thus

$P(\mathscr{C})=P(\mathscr{G}(0)\text{ is connected})\text{ for }1\leq n\leq
4.$

Since $\mathscr{G}(t)$ is disconnected-preserving for all $n\geq 1$ and
$\mathscr{G}(t)$ is connected-preserving if it satisfies $(\ast\ast)$ for all
$n\geq 4,$

$\\{\mathscr{G}(0)\text{ is
connected}\\}\supset\mathscr{C}\supset\\{\mathscr{G}(0)\text{ is
}\epsilon\text{-trivial}\\}\cup\\{\mathscr{G}(0)\text{ satisfies
}(\ast\ast)\\}.$

Hence

$P(\mathscr{G}(0)\text{ is connected})\geq P(\mathscr{C})\geq
P(\mathscr{G}(0)\text{ is }\epsilon\text{-trivial or satisfies }(\ast\ast)).$

∎

###### Proof of Corollary 2.

For $n=2,$

$P(\mathscr{C})=2!\int_{[0,1]}dx_{1}\int_{[x_{1},x_{1}+\epsilon]\cap[0,1]}dx_{2}.$
$\displaystyle\int_{[0,1]}dx_{1}\int_{[x_{1},x_{1}+\epsilon]\cap[0,1]}dx_{2}=\int_{[0,1-\epsilon]+[1-\epsilon,1]}\left[(x_{1}+\epsilon)\wedge
1-x_{1}\right]dx_{1}$ $\displaystyle=\int_{[0,1-\epsilon]}\epsilon
dx_{1}+\int_{[1-\epsilon,1]}1-x_{1}dx_{1}$
$\displaystyle=\epsilon(1-\epsilon)-\left[\frac{(1-x_{1})^{2}}{2}\right]_{1-\epsilon}^{1}=\epsilon(1-\epsilon)+\frac{1}{2}\epsilon^{2}=\epsilon(1-\frac{\epsilon}{2}).$

Thus

$P(\mathscr{C})=\epsilon(2-\epsilon).$

For $n=3,$

$\displaystyle
P(\mathscr{C})=3!\int_{[0,1]}dx_{1}\int_{[x_{1},x_{1}+\epsilon]\cap[0,1]}dx_{2}\int_{[x_{2},x_{2}+\epsilon]\cap[0,1]}dx_{3}.$

(i) $\epsilon\in[\frac{1}{2},1)$

$\displaystyle\int_{[0,1]}dx_{1}\int_{[x_{1},x_{1}+\epsilon]\cap[0,1]}dx_{2}\int_{[x_{2},x_{2}+\epsilon]\cap[0,1]}dx_{3}$
$\displaystyle=\int_{[0,1]}dx_{1}\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]}dx_{2}\int_{[x_{2},(x_{2}+\epsilon)\wedge 1]}dx_{3}$
$\displaystyle=\int_{[0,1]}dx_{1}\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]}(x_{2}+\epsilon)\wedge 1-x_{2}dx_{2}$
$\displaystyle=\int_{[0,1]}dx_{1}\left(\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[0,1-\epsilon]}\epsilon dx_{2}+\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[1-\epsilon,1]}1-x_{2}dx_{2}\right)$
$\displaystyle=\int_{[0,1-\epsilon]}dx_{1}\left(\int_{[x_{1},1-\epsilon]}\epsilon
dx_{2}+\int_{[1-\epsilon,x_{1}+\epsilon]}1-x_{2}dx_{2}\right)+\int_{[1-\epsilon,1]}dx_{1}\int_{[x_{1},1]}1-x_{2}dx_{2}$
$\displaystyle=\int_{[0,1-\epsilon]}\epsilon(1-\epsilon-
x_{1})-\left[\frac{(1-x_{2})^{2}}{2}\right]_{x_{2}=1-\epsilon}^{x_{1}+\epsilon}dx_{1}+\int_{[1-\epsilon,1]}-\left[\frac{(1-x_{2})^{2}}{2}\right]_{x_{2}=x_{1}}^{1}dx_{1}$
$\displaystyle\frac{-\epsilon(1-\epsilon-
x_{1})^{2}}{2}|_{0}^{1-\epsilon}+\frac{1}{2}\int_{[0,1-\epsilon]}\epsilon^{2}-(1-\epsilon-
x_{1})^{2}dx_{1}+\frac{1}{2}\int_{[1-\epsilon,1]}(1-x_{1})^{2}dx_{1}$
$\displaystyle=\frac{1}{2}\left\\{\epsilon(1-\epsilon)^{2}+\epsilon^{2}(1-\epsilon)+\left[\frac{(1-\epsilon-
x_{1})^{3}}{3}\right]_{0}^{1-\epsilon}-\left[\frac{(1-x_{1})^{3}}{3}\right]_{1-\epsilon}^{1}\right\\}$
$\displaystyle=\frac{1}{2}\left\\{\epsilon(1-\epsilon)-\frac{(1-\epsilon)^{3}}{3}+\frac{\epsilon^{3}}{3}\right\\}\text{
so }$ $\displaystyle
P(\mathscr{C})=3\epsilon(1-\epsilon)-(1-\epsilon)^{3}+\epsilon^{3}=\epsilon^{3}+(1-\epsilon)^{3}+3\epsilon(1-\epsilon)-2(1-\epsilon)^{3}$
$\displaystyle=\epsilon^{2}-\epsilon(1-\epsilon)+(1-\epsilon)^{2}+3\epsilon(1-\epsilon)-2(1-\epsilon)^{2}$
$\displaystyle=\epsilon^{2}+2\epsilon(1-\epsilon)+(1-\epsilon)^{2}-2(1-\epsilon)^{3}=1-2(1-\epsilon)^{3}.$

(ii) $\epsilon\in(0,\frac{1}{2})$

$\displaystyle\int_{[0,1]}dx_{1}\int_{[x_{1},x_{1}+\epsilon]\cap[0,1]}dx_{2}\int_{[x_{2},x_{2}+\epsilon]\cap[0,1]}dx_{3}$
$\displaystyle=\int_{[0,1]}dx_{1}\left(\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[0,1-\epsilon]}\epsilon dx_{2}+\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[1-\epsilon,1]}1-x_{2}dx_{2}\right)$
$\displaystyle=\int_{[0,1-2\epsilon]}dx_{1}\left(\int_{[x_{1},x_{1}+\epsilon]\cap[0,1-\epsilon]}\epsilon
dx_{2}+\int_{[x_{1},x_{1}+\epsilon]\cap[1-\epsilon,1]}1-x_{2}dx_{2}\right)$
$\displaystyle+\int_{[1-2\epsilon,1-\epsilon]}dx_{1}\left(\int_{[x_{1},x_{1}+\epsilon]\cap[0,1-\epsilon]}\epsilon
dx_{2}+\int_{[x_{1},x_{1}+\epsilon]\cap[1-\epsilon,1]}1-x_{2}dx_{2}\right)$
$\displaystyle+\int_{[1-\epsilon,1]}dx_{1}\left(\int_{[x_{1},1]\cap[0,1-\epsilon]}\epsilon
dx_{2}+\int_{[x_{1},1]\cap[1-\epsilon,1]}1-x_{2}dx_{2}\right)$
$\displaystyle=\int_{[0,1-2\epsilon]}\epsilon^{2}dx_{1}+\left(\int_{[1-2\epsilon,1-\epsilon]}\epsilon(1-\epsilon-
x_{1})-\left[\frac{(1-x_{2})^{2}}{2}\right]_{x_{2}=1-\epsilon}^{x_{1}+\epsilon}dx_{1}\right)$
$\displaystyle+\int_{[1-\epsilon,1]}-\left[\frac{(1-x_{2})^{2}}{2}\right]_{x_{2}=x_{1}}^{1}dx_{1}$
$\displaystyle=\epsilon^{2}(1-2\epsilon)+\int_{[1-2\epsilon,1-\epsilon]}\epsilon(1-\epsilon-
x_{1})$ $\displaystyle+\frac{1}{2}\left[\epsilon^{2}-(1-\epsilon-
x_{1})^{2}\right]dx_{1}+\frac{1}{2}\int_{[1-\epsilon,1]}(1-x_{1})^{2}dx_{1}$
$\displaystyle=\epsilon^{2}(1-2\epsilon)-\left[\frac{\epsilon(1-\epsilon-
x_{1})^{2}}{2}\right]_{1-2\epsilon}^{1-\epsilon}+\frac{1}{2}\left\\{\epsilon^{3}+\left[\frac{(1-\epsilon-
x_{1})^{3}}{3}\right]_{x_{1}=1-2\epsilon}^{1-\epsilon}\right\\}$
$\displaystyle-\frac{1}{2}\left[\frac{(1-x_{1})^{3}}{3}\right]_{1-\epsilon}^{1}$
$\displaystyle=\epsilon^{2}(1-2\epsilon)+\frac{1}{2}(\epsilon^{3}+\epsilon^{3}-\frac{\epsilon^{3}}{3}+\frac{\epsilon^{3}}{3})=\epsilon^{2}(1-2\epsilon)+\epsilon^{3}=\epsilon^{2}(1-\epsilon)\text{
so}$ $\displaystyle P(\mathscr{C})=6\epsilon^{2}(1-\epsilon).$

Thus

$P(\mathscr{C})=\left\\{\begin{array}[]{cc}6\epsilon^{2}(1-\epsilon)&\epsilon\in(0,\frac{1}{2})\\\
1-2(1-\epsilon)^{3}&\epsilon\in[\frac{1}{2},1)\end{array}\right.$

For $n=4,$

$P(\mathscr{C})=4!\int_{[0,1]}dx_{1}\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]}dx_{2}\int_{[x_{2},(x_{2}+\epsilon)\wedge
1]}dx_{3}\int_{[x_{3},(x_{3}+\epsilon)\wedge 1]}dx_{4}.$

(i) $\epsilon\in[\frac{1}{2},1)$

$\displaystyle\int_{[0,1]}dx_{1}\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]}dx_{2}\int_{[x_{2},(x_{2}+\epsilon)\wedge
1]}dx_{3}\int_{[x_{3},(x_{3}+\epsilon)\wedge 1]}dx_{4}$
$\displaystyle=\int_{[0,1]}dx_{1}\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]}dx_{2}\int_{[x_{2},(x_{2}+\epsilon)\wedge 1]}(x_{3}+\epsilon)\wedge
1-x_{3}dx_{3}$
$\displaystyle=\int_{[0,1]}dx_{1}\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]}dx_{2}\left(\int_{[x_{2},(x_{2}+\epsilon)\wedge
1]\cap[0,1-\epsilon]}\epsilon dx_{3}+\int_{[x_{2},(x_{2}+\epsilon)\wedge
1]\cap[1-\epsilon,1]}1-x_{3}dx_{3}\right)$
$\displaystyle=\int_{[0,1]}dx_{1}\left[\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[0,1-\epsilon]}\left(\int_{[x_{2},1-\epsilon]}\epsilon
dx_{3}+\int_{[1-\epsilon,x_{2}+\epsilon]}1-x_{3}dx_{3}\right)\right.$
$\displaystyle\left.+\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[1-\epsilon,1]}dx_{2}\left(\int_{[x_{2},1]}1-x_{3}dx_{3}\right)\right]$
$\displaystyle=\int_{[0,1]}dx_{1}\left(\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[0,1-\epsilon]}\epsilon(1-\epsilon-
x_{2})-\left[\frac{(1-x_{3})^{2}}{2}\right]_{1-\epsilon}^{x_{2}+\epsilon}dx_{2}\right.$
$\displaystyle\left.+\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[1-\epsilon,1]}-\left[\frac{(1-x_{3})^{2}}{2}\right]_{x_{2}}^{1}dx_{2}\right)$
$\displaystyle=\int_{[0,1-\epsilon]}dx_{1}\left(\int_{[x_{1},1-\epsilon]}\epsilon(1-\epsilon-
x_{2})+\frac{1}{2}[\epsilon^{2}-(1-\epsilon-x_{2})^{2}]dx_{2}\right.$
$\displaystyle\left.+\frac{1}{2}\int_{[1-\epsilon,x_{1}+\epsilon]}(1-x_{2})^{2}dx_{2}\right)+\int_{[1-\epsilon,1]}dx_{1}\left(\frac{1}{2}\int_{[x_{1},1]}(1-x_{2})^{2}dx_{2}\right)$
$\displaystyle=\int_{[0,1-\epsilon]}-\left[\frac{\epsilon(1-\epsilon-
x_{2})^{2}}{2}\right]_{x_{1}}^{1-\epsilon}+\frac{1}{2}\left(\epsilon^{2}(1-\epsilon-
x_{1})+\left[\frac{(1-\epsilon-
x_{2})^{3}}{3}\right]_{x_{1}}^{1-\epsilon}\right)$
$\displaystyle-\frac{1}{2}\left[\frac{(1-x_{2})^{3}}{3}\right]_{1-\epsilon}^{x_{1}+\epsilon}dx_{1}+\int_{[1-\epsilon,1]}-\frac{1}{2}\left[\frac{(1-x_{2})^{3}}{3}\right]_{x_{1}}^{1}dx_{1}$
$\displaystyle=\frac{1}{2}\left(\int_{[0,1-\epsilon]}\epsilon(1-\epsilon-
x_{1})^{2}+\epsilon^{2}(1-\epsilon-x_{1})-\frac{(1-\epsilon-
x_{1})^{3}}{3}\right.$ $\displaystyle\left.+\frac{[\epsilon^{3}-(1-\epsilon-
x_{1})^{3}]}{3}dx_{1}+\int_{[1-\epsilon,1]}\frac{(1-x_{1})^{3}}{3}dx_{1}\right)$
$\displaystyle=\frac{1}{2}\left\\{-\left[\frac{\epsilon(1-\epsilon-
x_{1})^{3}}{3}\right]_{0}^{1-\epsilon}-\left[\frac{\epsilon^{2}(1-\epsilon-
x_{1})^{2}}{2}\right]_{0}^{1-\epsilon}\right.$
$\displaystyle\left.+\frac{1}{3}\left(\epsilon^{3}(1-\epsilon)+\left[\frac{2(1-\epsilon-
x_{1})^{4}}{4}\right]_{0}^{1-\epsilon}-\frac{1}{3}\left[\frac{(1-x_{1})^{4}}{4}\right]_{1-\epsilon}^{1}\right)\right\\}$
$\displaystyle=\frac{1}{2}\left\\{\frac{\epsilon(1-\epsilon)^{3}}{3}+\frac{\epsilon^{2}(1-\epsilon)^{2}}{2}+\frac{1}{3}\left[\epsilon^{3}(1-\epsilon)-\frac{2(1-\epsilon)^{4}}{4}\right]+\frac{\epsilon^{4}}{12}\right\\}$
$\displaystyle=\frac{1}{2}\left(\frac{\epsilon(1-\epsilon)^{3}}{3}+\frac{\epsilon^{2}(1-\epsilon)^{2}}{2}+\frac{\epsilon^{3}(1-\epsilon)}{3}-\frac{(1-\epsilon)^{4}}{6}+\frac{\epsilon^{4}}{12}\right)$
$\displaystyle=\frac{1}{24}\left(\epsilon^{4}+4\epsilon^{3}(1-\epsilon)+6\epsilon^{2}(1-\epsilon)^{2}+4\epsilon(1-\epsilon)^{3}-2(1-\epsilon)^{4}\right)\text{
so }$ $\displaystyle
P(\mathscr{C})=\epsilon^{4}+4\epsilon^{3}(1-\epsilon)+6\epsilon^{2}(1-\epsilon)^{2}+4\epsilon(1-\epsilon)^{3}-2(1-\epsilon)^{4}.$

(ii) $\epsilon\in[\frac{1}{3},\frac{1}{2})$

$\displaystyle\int_{[0,1]}dx_{1}\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]}dx_{2}\int_{[x_{2},(x_{2}+\epsilon)\wedge
1]}dx_{3}\int_{[x_{3},(x_{3}+\epsilon)\wedge 1]}dx_{4}$
$\displaystyle=\int_{[0,1]}dx_{1}\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]}dx_{2}\left(\int_{[x_{2},(x_{2}+\epsilon)\wedge
1]\cap[0,1-\epsilon]}\epsilon dx_{3}+\int_{[x_{2},(x_{2}+\epsilon)\wedge
1]}1-x_{3}dx_{3}\right)$
$\displaystyle=\int_{[0,1]}dx_{1}\left[\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[0,1-2\epsilon]}dx_{2}\left(\int_{[x_{2},x_{2}+\epsilon]}\epsilon
dx_{3}\right)\right.$ $\displaystyle+\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[1-2\epsilon,1-\epsilon]}dx_{2}\left(\int_{[x_{2},1-\epsilon]}\epsilon
dx_{3}+\int_{[1-\epsilon,x_{2}+\epsilon]}1-x_{3}dx_{3}\right)$
$\displaystyle\left.+\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[1-\epsilon,1]}dx_{2}\int_{[x_{2},1]}1-x_{3}dx_{3}\right]$
$\displaystyle=\int_{[0,1]}dx_{1}\left(\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[0,1-2\epsilon]}\epsilon^{2}dx_{2}\right.$
$\displaystyle+\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[1-2\epsilon,1-\epsilon]}\epsilon(1-\epsilon-
x_{2})-\left[\frac{(1-x_{3})^{2}}{2}\right]_{1-\epsilon}^{x_{2}+\epsilon}dx_{2}$
$\displaystyle\left.+\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[1-\epsilon,1]}-\left[\frac{(1-x_{3})^{2}}{2}\right]_{x_{2}}^{1}dx_{2}\right)$
$\displaystyle=\int_{[0,1-2\epsilon]}dx_{1}\left(\int_{[x_{1},1-2\epsilon]}\epsilon^{2}dx_{2}+\int_{[1-2\epsilon,x_{1}+\epsilon]}\epsilon(1-\epsilon-
x_{2})\right.$ $\displaystyle\left.+\frac{1}{2}[\epsilon^{2}-(1-\epsilon-
x_{2})^{2}]dx_{2}\right)$
$\displaystyle+\int_{[1-2\epsilon,1-\epsilon]}dx_{1}\left(\int_{[x_{1},1-\epsilon]}\epsilon(1-\epsilon-
x_{2})+\frac{1}{2}[\epsilon^{2}-(1-\epsilon-x_{2})^{2}]dx_{2}\right.$
$\displaystyle\left.+\frac{1}{2}\int_{[1-\epsilon,x_{1}+\epsilon]}(1-x_{2})^{2}dx_{2}\right)+\frac{1}{2}\int_{[1-\epsilon,1]}dx_{1}\int_{[x_{1},1]}(1-x_{2})^{2}dx_{2}$
$\displaystyle=\int_{[0,1-2\epsilon]}\epsilon^{2}(1-2\epsilon-
x_{1})-\left[\frac{\epsilon(1-\epsilon-
x_{2})^{2}}{2}\right]_{1-2\epsilon}^{x_{1}+\epsilon}$
$\displaystyle+\frac{1}{2}\left(\epsilon^{2}(x_{1}+3\epsilon-1)+\left[\frac{(1-\epsilon-
x_{2})^{3}}{3}\right]_{1-2\epsilon}^{x_{1}+\epsilon}\right)dx_{1}$
$\displaystyle+\int_{[1-2\epsilon,1-\epsilon]}-\left[\frac{\epsilon(1-\epsilon-
x_{2})^{2}}{2}\right]_{x_{1}}^{1-\epsilon}+\frac{1}{2}\left(\epsilon^{2}(1-\epsilon-
x_{1})+\left[\frac{(1-\epsilon-
x_{2})^{3}}{3}\right]_{x_{1}}^{1-\epsilon}\right)$
$\displaystyle-\frac{1}{2}\left[\frac{(1-x_{2})^{3}}{3}\right]_{1-\epsilon}^{x_{1}+\epsilon}dx_{1}-\frac{1}{2}\int_{[1-\epsilon,1]}\left[\frac{(1-x_{2})^{3}}{3}\right]_{x_{1}}^{1}dx_{1}$
$\displaystyle=\int_{[0,1-2\epsilon]}\epsilon^{2}(1-2\epsilon-
x_{1})+\frac{1}{2}[\epsilon^{3}-\epsilon(1-2\epsilon-x_{1})^{2}]$
$\displaystyle+\frac{1}{2}\left(\epsilon^{2}(x_{1}+3\epsilon-1)+\frac{1}{3}[(1-2\epsilon-
x_{1})^{3}-\epsilon^{3}]\right)dx_{1}+\int_{[1-2\epsilon,1-\epsilon]}\frac{\epsilon(1-\epsilon-
x_{1})^{2}}{2}$ $\displaystyle+\frac{1}{2}\left(\epsilon^{2}(1-\epsilon-
x_{1})-\frac{(1-\epsilon-
x_{1})^{3}}{3}\right)+\frac{1}{6}[\epsilon^{3}-(1-\epsilon-x_{1})^{3}]dx_{1}$
$\displaystyle+\frac{1}{2}\int_{[1-\epsilon,1]}\frac{1}{3}(1-x_{1})^{3}dx_{1}$
$\displaystyle=-\left[\frac{\epsilon^{2}(1-2\epsilon-
x_{1})^{2}}{2}\right]_{0}^{1-2\epsilon}+\frac{1}{2}\left(\epsilon^{3}(1-2\epsilon)+\left[\frac{\epsilon(1-2\epsilon-
x_{1})^{3}}{3}\right]_{0}^{1-2\epsilon}\right)$
$\displaystyle+\frac{1}{2}\left[\left[\frac{\epsilon^{2}(x_{1}+3\epsilon-1)^{2}}{2}\right]_{0}^{1-2\epsilon}+\frac{1}{3}\left(-\left[\frac{(1-2\epsilon-
x_{1})^{4}}{4}\right]_{0}^{1-2\epsilon}-\epsilon^{3}(1-2\epsilon)\right)\right]$
$\displaystyle+\frac{1}{2}\left\\{-\left[\frac{\epsilon(1-\epsilon-
x_{1})^{3}}{3}\right]_{1-2\epsilon}^{1-\epsilon}-\left[\frac{\epsilon^{2}(1-\epsilon-
x_{1})^{2}}{2}\right]_{1-2\epsilon}^{1-\epsilon}+\left[\frac{(1-\epsilon-
x_{1})^{4}}{12}\right]_{1-2\epsilon}^{1-\epsilon}\right.$
$\displaystyle\left.+\frac{1}{3}\left(\epsilon^{4}+\left[\frac{(1-\epsilon-
x_{1})^{4}}{4}\right]_{1-2\epsilon}^{1-\epsilon}\right)\right\\}-\frac{1}{6}\left[\frac{(1-x_{1})^{4}}{4}\right]_{1-\epsilon}^{1}$
$\displaystyle=\frac{\epsilon^{2}(1-2\epsilon)^{2}}{2}+\frac{1}{2}[\epsilon^{3}(1-2\epsilon)-\frac{\epsilon(1-2\epsilon)^{3}}{3}]+\frac{1}{4}[\epsilon^{4}-\epsilon^{2}(3\epsilon-1)^{2}]+\frac{1}{24}(1-2\epsilon)^{4}$
$\displaystyle-\frac{1}{6}\epsilon^{3}(1-2\epsilon)+\frac{1}{6}\epsilon^{4}+\frac{1}{4}\epsilon^{4}-\frac{1}{24}\epsilon^{4}+\frac{1}{6}\epsilon^{4}-\frac{1}{24}\epsilon^{4}+\frac{1}{24}\epsilon^{4}$
$\displaystyle=\frac{19}{24}\epsilon^{4}-\frac{1}{6}\epsilon^{3}(1-2\epsilon)+\frac{1}{24}(1-2\epsilon)^{4}-\frac{1}{4}\epsilon^{2}(3\epsilon-1)^{2}-\frac{1}{6}\epsilon(1-2\epsilon)^{3}$
$\displaystyle+\frac{1}{2}\epsilon^{3}(1-2\epsilon)+\frac{1}{2}\epsilon^{2}(1-2\epsilon)^{2}\text{
so}$ $\displaystyle P(\mathscr{C})$
$\displaystyle=19\epsilon^{4}-4\epsilon^{3}(1-2\epsilon)+(1-2\epsilon)^{4}-6\epsilon^{2}(3\epsilon-1)^{2}-4\epsilon(1-2\epsilon)^{3}$
$\displaystyle+12\epsilon^{3}(1-2\epsilon)+12\epsilon^{2}(1-2\epsilon)^{2}.$

(iii) $\epsilon\in(0,\frac{1}{3})$

$\displaystyle\int_{[0,1]}dx_{1}\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]}dx_{2}\int_{[x_{2},(x_{2}+\epsilon)\wedge
1]}dx_{3}\int_{[x_{3},(x_{3}+\epsilon)\wedge 1]}dx_{4}$
$\displaystyle=\int_{[0,1]}dx_{1}\left(\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[0,1-2\epsilon]}\epsilon^{2}dx_{2}+\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[1-2\epsilon,1-\epsilon]}\epsilon(1-\epsilon-x_{2})\right.$
$\displaystyle\left.-\left[\frac{(1-x_{3})^{2}}{2}\right]_{1-\epsilon}^{x_{2}+\epsilon}dx_{2}+\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]\cap[1-\epsilon,1]}-\left[\frac{(1-x_{3})^{2}}{2}\right]_{x_{2}}^{1}dx_{2}\right)$
$\displaystyle=\int_{[0,1-3\epsilon]}dx_{1}\int_{[x_{1},x_{1}+\epsilon]}\epsilon^{2}dx_{2}+\int_{[1-3\epsilon,1-2\epsilon]}dx_{1}\left(\int_{[x_{1},1-2\epsilon]}\epsilon^{2}dx_{2}\right.$
$\displaystyle\left.+\int_{[1-2\epsilon,x_{1}+\epsilon]}\epsilon(1-\epsilon-
x_{2})+\frac{1}{2}[\epsilon^{2}-(1-\epsilon-x_{2})^{2}]dx_{2}\right)$
$\displaystyle+\int_{[1-2\epsilon,1-\epsilon]}dx_{1}\left(\int_{[x_{1},1-\epsilon]}\epsilon(1-\epsilon-
x_{2})+\frac{1}{2}[\epsilon^{2}-(1-\epsilon-x_{2})^{2}]dx_{2}\right.$
$\displaystyle\left.+\frac{1}{2}\int_{[1-\epsilon,x_{1}+\epsilon]}(1-x_{2})^{2}dx_{2}\right)+\frac{1}{2}\int_{[1-\epsilon,1]}dx_{1}\int_{[x_{1},1]}(1-x_{2})^{2}dx_{2}$
$\displaystyle=\int_{[0,1-3\epsilon]}\epsilon^{3}dx_{1}+\int_{[1-3\epsilon,1-2\epsilon]}\epsilon^{2}(1-2\epsilon-
x_{1})-\left[\frac{\epsilon(1-\epsilon-
x_{2})^{2}}{2}\right]_{1-2\epsilon}^{x_{1}+\epsilon}$
$\displaystyle+\frac{1}{2}\left(\epsilon^{2}(x_{1}+3\epsilon-1)+\left[\frac{(1-\epsilon-
x_{2})^{3}}{3}\right]_{1-2\epsilon}^{x_{1}+\epsilon}\right)dx_{1}$
$\displaystyle+\int_{[1-2\epsilon,1-\epsilon]}-\left[\frac{\epsilon(1-\epsilon-
x_{2})^{2}}{2}\right]_{x_{1}}^{1-\epsilon}+\frac{1}{2}\left(\epsilon^{2}(1-\epsilon-
x_{1})+\left[\frac{(1-\epsilon-
x_{2})^{3}}{3}\right]_{x_{1}}^{1-\epsilon}\right)$
$\displaystyle-\frac{1}{2}\left[\frac{(1-x_{2})^{3}}{3}\right]_{1-\epsilon}^{x_{1}+\epsilon}dx_{1}+\frac{1}{2}\int_{[1-\epsilon,1]}-\left[\frac{(1-x_{2})^{3}}{3}\right]_{x_{1}}^{1}dx_{1}$
$\displaystyle=\epsilon^{3}(1-3\epsilon)+\int_{[1-3\epsilon,1-2\epsilon]}\epsilon^{2}(1-2\epsilon-
x_{1})+\frac{1}{2}[\epsilon^{3}-\epsilon(1-2\epsilon-x_{1})^{2}]$
$\displaystyle+\frac{1}{2}\epsilon^{2}(x_{1}+3\epsilon-1)-\frac{1}{6}[\epsilon^{3}-(1-2\epsilon-
x_{1})^{3}]dx_{1}$
$\displaystyle+\frac{1}{2}\int_{[1-2\epsilon,1-\epsilon]}\epsilon(1-\epsilon-
x_{1})^{2}+\epsilon^{2}(1-\epsilon-x_{1})-\frac{(1-\epsilon-x_{1})^{3}}{3}$
$\displaystyle+\frac{1}{3}[\epsilon^{3}-(1-\epsilon-
x_{1})^{3}]dx_{1}+\frac{1}{6}\int_{[1-\epsilon,1]}(1-x_{1})^{3}dx_{1}$
$\displaystyle=\epsilon^{3}(1-3\epsilon)-\left[\frac{\epsilon^{2}(1-2\epsilon-
x_{1})^{2}}{2}\right]_{1-3\epsilon}^{1-2\epsilon}+\frac{1}{2}\epsilon^{4}+\frac{1}{2}\left[\frac{\epsilon(1-2\epsilon-
x_{1})^{3}}{3}\right]_{1-3\epsilon}^{1-2\epsilon}$
$\displaystyle+\frac{1}{2}\left[\frac{\epsilon^{2}(x_{1}+3\epsilon-1)^{2}}{2}\right]_{1-3\epsilon}^{1-2\epsilon}-\frac{1}{6}\epsilon^{4}-\frac{1}{6}\left[\frac{(1-2\epsilon-
x_{1})^{4}}{4}\right]_{1-3\epsilon}^{1-2\epsilon}$
$\displaystyle\frac{1}{2}\left(-\left[\frac{\epsilon(1-\epsilon-
x_{1})^{3}}{3}\right]_{1-2\epsilon}^{1-\epsilon}-\left[\frac{\epsilon^{2}(1-\epsilon-
x_{2})^{2}}{2}\right]_{1-2\epsilon}^{1-\epsilon}+\frac{1}{3}\left[\frac{(1-\epsilon-
x_{1})^{4}}{4}\right]_{1-2\epsilon}^{1-\epsilon}\right.$
$\displaystyle\left.+\frac{1}{3}\epsilon^{4}+\frac{1}{3}\left[\frac{(1-\epsilon-
x_{1})^{4}}{4}\right]_{1-2\epsilon}^{1-\epsilon}\right)-\frac{1}{6}\left[\frac{(1-x_{1})^{4}}{4}\right]_{1-\epsilon}^{1}$
$\displaystyle=\epsilon^{3}(1-3\epsilon)+\frac{1}{2}\epsilon^{4}+\frac{1}{2}\epsilon^{4}-\frac{1}{6}\epsilon^{4}+\frac{1}{4}\epsilon^{4}-\frac{1}{6}\epsilon^{4}+\frac{1}{24}\epsilon^{4}+\frac{1}{6}\epsilon^{4}+\frac{1}{4}\epsilon^{4}-\frac{1}{24}\epsilon^{4}$
$\displaystyle+\frac{1}{6}\epsilon^{4}-\frac{1}{24}\epsilon^{4}+\frac{1}{24}\epsilon^{4}$
$\displaystyle=\epsilon^{3}(1-3\epsilon)+\frac{3}{2}\epsilon^{4}\text{ so}$
$\displaystyle P(\mathscr{C})=24\epsilon^{3}(1-3\epsilon)+36\epsilon^{4}.$

For $n\geq 2,$

$\displaystyle\int_{[0,1]}dx_{1}\int_{[x_{1},x_{1}+\epsilon]\cap[0,1]}dx_{n}\left(\int_{[x_{1},x_{n}]}dx_{n}\right)^{n-2}=\int_{[0,1]}dx_{1}\int_{[x_{1},(x_{1}+\epsilon)\wedge
1]}(x_{n}-x_{1})^{n-2}dx_{n}$
$\displaystyle=\int_{[0,1]}\frac{(x_{n}-x_{1})^{n-1}}{n-1}|_{x_{n}=x_{1}}^{(x_{1}+\epsilon)\wedge
1}dx_{1}$ $\displaystyle=\frac{1}{n-1}\int_{[0,1]}\left[(x_{1}+\epsilon)\wedge
1-x_{1}\right]^{n-1}dx_{1}$
$\displaystyle=\frac{1}{n-1}\left\\{\int_{[0,1-\epsilon]}\epsilon^{n-1}dx_{1}+\int_{[1-\epsilon,1]}(1-x_{1})^{n-1}dx_{1}\right\\}$
$\displaystyle=\frac{1}{n-1}\left\\{\epsilon^{n-1}(1-\epsilon)-\left[\frac{(1-x_{1})^{n}}{n}\right]_{1-\epsilon}^{1}\right\\}=\frac{1}{n-1}\left\\{\epsilon^{n-1}(1-\epsilon)+\frac{1}{n}\epsilon^{n}\right\\}$
$\displaystyle=\frac{\epsilon^{n-1}}{n-1}(1-\epsilon+\frac{\epsilon}{n})=\frac{\epsilon^{n-1}}{n-1}\left[1-(1-\frac{1}{n})\epsilon\right]\text{
so}$ $\displaystyle P(\mathscr{G}(0)\text{ is
}\epsilon\text{-trivial})=\epsilon^{n-1}\left[n-(n-1)\epsilon\right].$

Observe that $\epsilon^{n-1}\left[n-(n-1)\epsilon\right]=1$ for $n=1.$ So

$P(\mathscr{G}(0)\text{ is
}\epsilon\text{-trivial})=\epsilon^{n-1}\left[n-(n-1)\epsilon\right]\text{ for
}n\geq 1.$ $\displaystyle P(x_{i}(0)\in B(x_{1}(0),\epsilon/2)\text{ for all
}i\in[n])=\int_{[0,1]}dx_{1}\left(\int_{[x_{1}-\frac{\epsilon}{2},x_{1}+\frac{\epsilon}{2}]\cap[0,1]}dx\right)^{n-1}$
$\displaystyle=\int_{[0,1]}\left[(x_{1}+\frac{\epsilon}{2})\wedge
1-(x_{1}-\frac{\epsilon}{2})\vee 0\right]^{n-1}dx_{1}$
$\displaystyle=\int_{[0,\frac{\epsilon}{2}]}(x_{1}+\frac{\epsilon}{2})^{n-1}dx_{1}+\int_{[\frac{\epsilon}{2},1-\frac{\epsilon}{2}]}[(x_{1}+\frac{\epsilon}{2})-(x_{1}-\frac{\epsilon}{2})]^{n-1}dx_{1}$
$\displaystyle+\int_{[1-\frac{\epsilon}{2},1]}[1-(x_{1}-\frac{\epsilon}{2})]^{n-1}$
$\displaystyle=\frac{(x_{1}+\frac{\epsilon}{2})^{n}}{n}|_{0}^{\frac{\epsilon}{2}}+\epsilon^{n-1}(1-\epsilon)-\left[\frac{(1+\frac{\epsilon}{2}-x_{1})^{n}}{n}\right]_{1-\frac{\epsilon}{2}}^{1}$
$\displaystyle=\frac{1}{n}[\epsilon^{n}-(\frac{\epsilon}{2})^{n}]+\epsilon^{n-1}(1-\epsilon)+\frac{1}{n}[\epsilon^{n}-(\frac{\epsilon}{2})^{n}]$
$\displaystyle=\frac{2}{n}\epsilon^{n}(1-\frac{1}{2^{n}})+\epsilon^{n-1}(1-\epsilon)$

∎

In conclusion, the probability of consensus is positive and is bounded from
below by the one that an initial profile is connected for $1\leq n\leq 4$. In
particular for one dimension, the disconnected-preserving property for a
profile engenders an upper bound $P(\mathscr{G}(0)\hbox{ is connected})$ for
the probability of consensus, and therefore
$P(\mathscr{C})=P(\mathscr{G}(0)\hbox{ is connected})$ for $1\leq n\leq 4$.

## References

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