Social Capital, Communication Channels and Opinion Formation

Abstract

We study how different forms of social capital lead to different distributions of multidimensional opinions by affecting the channels through which individuals communicate. We develop a model to compare and contrast the evolution of opinions between societies whose members communicate through bonding associations (i.e., which bond similar people together) and societies where communication is through bridging associations (i.e., which bridge the gap among different people). Both processes converge towards opinion distributions where there are groups within which there is consensus in all issues. Bridging processes are more likely to lead to society-wide consensus and converge to distributions that have, on average, fewer opinion groups. The latter result holds even when the confidence bound that allows successful communication in the bridging process is much smaller than the respective bound in the bonding process.

Appendix

1.1 Proofs of general results

Proof of Theorem 1

For either of the two processes, given two vectors of initial opinions $\mathbf{x}(0)$ and $\mathbf{y}(0)$ and a sequence of interactions $\{z_t\}_{t=1}^{\infty }$, the evolution of opinions can be described by the vectors of initial opinions and a sequence of transition matrices for each issue, $\{A_x(t)\}_{t=1}^{\infty }$ and $\{A_y(t)\}_{t=1}^{\infty }$ respectively, where $A_x(t)=A_x(\mathbf{x}(t-1),\mathbf{y}(t-1), z_t)$ and $A_y(t)=A_y(\mathbf{x}(t-1),\mathbf{y}(t-1), z_t)$, for which $\mathbf{x}(t)=A_x(t)\mathbf{x}(t-1)$ and $\mathbf{y}(t)=A_y(t)\mathbf{y}(t-1)$.

At each t, by $z_t=((i_t,j_t),s_t,c_t)$ we know the pair $(i_t,j_t)$ that communicates, the issue $s_t$ discussed and whether communication was successful, $c_t \in \{0,1\}$. If $c_t=0$, then $A_x(t)=A_y(t)=I$, where I is the identity matrix, as no opinion is updated in this period. If $c_t=1$ and $s_t=x$, then $A_y(t)=I$ and for $A_x(t)$ we have that for all $k\ne i,j$ it holds that $a^t_{x,kk}=a^t_{y,kk}=1$ and $a^t_{x,kl}=a^t_{y,kl}=1$ for all $l \ne k$. For citizens i, j though we get that $a^t_{x,ik}=a^t_{y,ik}=a^t_{x,jk}=a^t_{y,jk}=0$ for all $k \ne i,j$, whereas $a^t_{y,ij}=a^t_{y,ji}=\mu$ and $a^t_{y,ii}=a^t_{y,jj}=1-\mu$. The same argument holds if $c_t=1$ and $s_t=y$, with x and y reversed.

Therefore, for each sequence $\{z_t\}_{t=1}^{\infty }$, we can consider the evolution of opinions in each of the issues separately and for convergence to hold it is sufficient to show that the sufficient conditions provided by Lorenz (2005) hold for both issue–specific sequences of transition matrices. We show that indeed this is the case:

(i) Self–confidence: At each period t, each citizen $i \in N$ puts positive weight to her own opinion, i.e. it holds that $a^t_{\cdot ,ii}>0$, which is clearly true in our case.

(ii) Mutual confidence: Zero entries in the transition matrix are symmetric. For every pair $i,j \in N$ it holds that $a^t_{\cdot ,ij}>0 \Leftrightarrow a^t_{\cdot ,ji} >0$. This is again clearly true in both our processes, given that after successful communication both citizens revise their opinions.

(iii) Positive weights do not converge to zero: There is a $\delta >0$ such that the lowest positive entry of the transition matrix is greater than $\delta$. This is also true in our case, considering $\delta =\min \{\mu ,1-\mu \}$ which is strictly positive.

These three conditions ensure for each issue the emergence of pairwise disjoint groups of citizens ${\mathcal {J}}_{x,1} \cup \dots \cup {\mathcal {J}}_{x,r}$ and ${\mathcal {J}}_{y,1} \cup \dots \cup {\mathcal {J}}_{y,s}$ who reach consensus in this issue. To complete the argument, let ${\mathcal {I}}_{kl}= {\mathcal {J}}_{x,k} \cap {\mathcal {J}}_{y,l}$. There is a total of $p=r \times s$ such groups and within each of them citizens reach consensus in both issues.²²

Note that the structure of the transition matrices does not depend on whether we consider a bonding or a bridging process, as these affect only the frequency of updating in each sequence $z_t$. $\square$

Proof of Proposition 1

For a bonding process, $({\overline{x}}_m,{\overline{y}}_m)\in Int\left( {\mathcal {N}}_{({\overline{x}}_l,{\overline{y}}_l)}\right) \Rightarrow |{\overline{x}}_m-{\overline{x}}_l|=\chi< d \text { and } |{\overline{y}}_m-{\overline{y}}_l|=\psi < d$ and without loss of generality let $\chi >\psi$. Consider some ${\widetilde{\epsilon }}>0$ that satisfies ${\widetilde{\epsilon }}<\frac{\delta -\chi }{2}$ and ${\widetilde{\epsilon }}<\frac{\mu \chi }{2(1+\mu )}$. If the probability that opinions converge to form groups ${\mathcal {I}}_1 \cup \dots \cup {\mathcal {I}}_p$ such that citizens belonging to group ${\mathcal {I}}_g$ reach consensus at $(\overline{x_g},\overline{y_g})$ for all $g\in \{1,\dots ,p\}$ is strictly positive, then for ${\widetilde{\epsilon }}$ (and in fact for all $\epsilon >0$) there must exist some ${\widehat{t}}_{{\widetilde{\epsilon }}}$ such that the probability that for all $t>{\widehat{t}}_{{\widetilde{\epsilon }}}$ holds that $|x_i^t-{\overline{x}}_g|<{\widetilde{\epsilon }}$ and $|y_i^t-{\overline{y}}_g|<{\widetilde{\epsilon }}$, for all $i \in {\mathcal {I}}_g$ and for all $g\in \{1,\dots ,p\}$ is strictly positive.

Now focus on issue x and notice that ${\widetilde{\epsilon }}<\frac{\delta -\chi }{2}$ guarantees that as long as citizens of groups ${\mathcal {I}}_l$ and ${\mathcal {I}}_m$ remain at a distance less than ${\widetilde{\epsilon }}$ from ${\overline{x}}_l$ and $\overline{x_m}$ respectively (and the same on issue y), they face a strictly positive probability of communicating successfully. This is because the maximum distance between a citizen of ${\mathcal {I}}_m$ and another citizen of ${\mathcal {I}}_l$ is at most $\chi +2{\widetilde{\epsilon }}<\delta$. Such a pair of citizens always exists because groups ${\mathcal {I}}_g$ are non–empty. On top of that, the condition ${\widetilde{\epsilon }}<\frac{\mu \chi }{2(1+\mu )}$ guarantees that a successful communication on issue x between some citizen in ${\mathcal {I}}_l$ and some citizen in ${\mathcal {I}}_m$ is sufficient to bring both of them at a distance larger than ${\widetilde{\epsilon }}$ from the respective consensus opinion. This is because any successful communication will incur an update of opinion between $\mu (\chi -2{\widetilde{\epsilon }})$ and $\mu (\chi +2{\widetilde{\epsilon }})$, which is always larger than the $2{\widetilde{\epsilon }}$. Therefore, as long as all citizens of groups ${\mathcal {I}}_l$ and ${\mathcal {I}}_g$ remain close to their consensus opinions, there is at least one pair of citizens whose successful communication can lead the citizens more than ${\widetilde{\epsilon }}$ far from that opinion.

Hence, let $H_t$ denote the event that all opinions in both issues are within ${\widetilde{\epsilon }}$ distance from the respective consensus opinion of the group at period t. Then, the event that for all $t>{\widehat{t}}_{{\widetilde{\epsilon }}}$ holds that $|x_i^t-{\overline{x}}_g|<{\widetilde{\epsilon }}$ and $|y_i^t-{\overline{y}}_g|<{\widetilde{\epsilon }}$, for all $i \in {\mathcal {I}}_g$ and for all $g\in \{1,\dots ,p\}$ is the infinite intersection of events $H_t$, i.e. $\bigcap \nolimits _{t={\widehat{t}}+1}^{\infty }H_t$. And conditional on $H_{{\widehat{t}}}$ being true, ${\mathbb {P}}\left[ \bigcap \nolimits _{t={\widehat{t}}+1}^{\infty }H_t|H_{{\widehat{t}}}\right] =\lim \nolimits _{\tau \rightarrow \infty }\prod \nolimits _{t={\widehat{t}}+1}^{\tau }{\mathbb {P}}\left[ H_t|H_{t-1}\right]$. Yet, for each t, ${\mathbb {P}}\left[ H_t|H_{t-1}\right] \le 1-{\underline{b}}$, where ${\underline{b}}$ is the strictly positive and independent of t (hence, bounded away from zero) lower bound of the probability of a pair that can communicate successfully to be actually drawn and indeed communicate successfully in one issue. We have already shown that such a pair exists. Therefore, ${\mathbb {P}}\left[ \bigcap \nolimits _{t={\widehat{t}}+1}^{\infty }H_t|H_{{\widehat{t}}}\right] =\lim \nolimits _{\tau \rightarrow \infty }\prod \nolimits _{t={\widehat{t}}+1}^{\tau }{\mathbb {P}}\left[ H_t|H_{t-1}\right] \le \lim \nolimits _{\tau \rightarrow \infty }(1-{\underline{b}})^{\tau -{\widehat{t}}}=0$.

The proof for the bridging process is identical, except of the difference in the definition of the neighboring area (which is inconsequential) and the value of confidence bound being k instead of d.

Explanatory note If $({\overline{x}}_m,{\overline{y}}_m)$ is at the boundary of ${\mathcal {N}}_{({\overline{x}}_l,{\overline{y}}_l)}$ then we can no longer find ${\widetilde{\epsilon }}<\frac{\delta -\chi }{2}$ to guarantee that all citizens of each group may communicate. If in particular it holds that $|{\overline{x}}_m-{\overline{x}}_l|=d$ and $|{\overline{y}}_m-{\overline{y}}_l|=d$ the result may not necessarily hold. Consider the following example with three citizens: Group ${\mathcal {I}}_1$ consists of a single citizen located at (0, 0) and group ${\mathcal {I}}_2$ consists of two citizens located at $(d+\frac{\epsilon }{2},d-\frac{\epsilon }{2})$ and $(d-\frac{\epsilon }{2},d+\frac{\epsilon }{2})$ for some small $\epsilon >0$, with consensus opinions (d, d). For any $\mu <1/2$, the two citizens of group ${\mathcal {I}}_2$ can keep communicating successfully, converging towards (d, d), but they can never communicate successfully with the citizen of group ${\mathcal {I}}_1$. For the remaining parts of the boundaries in bonding process and for all boundaries in a bridging process the result can be shown to still hold. $\square$

Proof of Proposition 2

The following two conditions are jointly equivalent to neighborhood stabilization: For each citizen $i \in N$ (i) i does not lose any of her current neighbors at a subsequent period (ii) no additional citizen enters i’s neighborhood at a subsequent period.

Remark 1 ensures that the existence of disjoint groups of citizens with common neighborhoods are necessary for conditions (i) and (ii) to be satisfied. Absent of such groups, there would be a positive probability that some citizen’s neighborhood will change. Given the existence of these disjoint groups of citizens, Remark 2 guarantees that a member of such a group will not face a change in her neighborhood, as long as the members of that group may communicate successfully only among themselves.

The second condition guarantees that the members of each of the disjoint groups may communicate successfully only among themselves. This is because at period t each citizen may only communicate successfully with another member of her own group. But successful communication between citizens $i,j \in {\mathcal {I}}_g$ weakly shrinks both the minimum bounding rectangle of the group and as a result also the neighboring area of the minimum bounding rectangle, i.e. $MBR_{{\mathcal {I}}_{g}}^{t+1} \subseteq MBR_{{\mathcal {I}}_{g}}^t$, which then implies that ${\mathcal {N}}_{MBR_{{\mathcal {I}}_g}^{t+1}} \subseteq {\mathcal {N}}_{MBR_{{\mathcal {I}}_g}^t}$. Therefore, if $MBR_{{\mathcal {I}}_{g'}}^t\cap {\mathcal {N}}_{MBR_{{\mathcal {I}}_g}^t}=\emptyset$, for all $g'\ne g$, it must also hold that $MBR_{{\mathcal {I}}_{g'}}^{t+1}\cap {\mathcal {N}}_{MBR_{{\mathcal {I}}_g}^{t+1}}=\emptyset$, for all $g'\ne g$. This completes the argument, as it guarantees that in all subsequent periods no additional citizen enters the neighborhood of any $i \in {\mathcal {I}}_g$. $\square$

Proof of Proposition 3

As explained in the proof of Proposition 2, the following two conditions are jointly equivalent with neighborhood stabilization: For each citizen $i \in N$ (i) i does not lose any of his current neighbors at a subsequent period (ii) no additional citizen enters i’s neighborhood at a subsequent period. Moreover, the first condition of Proposition 3 is necessary for neighborhood stabilization because of Remark 1 and the third condition of Proposition 3 guarantees that the members of each group may communicate successfully only among themselves, as in Proposition 2.

Combining the first with the second condition guarantees that no citizen will face a change in her neighborhood, as long as the members of each group may communicate successfully only among themselves. This is because the second condition is sufficient to guarantee that $MBR_{{\mathcal {I}}}^t \subseteq {\mathcal {N}}_{(x_i^t,y_i^t)}$ for all $i \in {\mathcal {I}}$, which we show below.

Without loss of generality, we prove the result considering $R_g$ to be of size $k \times 1$. $MBR_{{\mathcal {I}}_{g}}^t$ is a rectangle with sizes $|\max \nolimits _{i \in {\mathcal {I}}} x_i^t - \min \nolimits _{i \in {\mathcal {I}}} x_i^t| \times |\max \nolimits _{i \in {\mathcal {I}}} y_i^t - \min \nolimits _{i \in {\mathcal {I}}} y_i^t|$. Given that all $i \in {\mathcal {I}}$ are such that $(x_i^t,y_i^t) \in R_g$, it must hold that $|\max \nolimits _{i \in {\mathcal {I}}} x_i^t - \min \nolimits _{i \in {\mathcal {I}}} x_i^t|=k-\epsilon$ for some $\epsilon \in [0,k]$ and also the left bound of $R_g$, denote it ${\underline{x}}$, must satisfy ${\underline{x}} \in [\min \nolimits _{i \in {\mathcal {I}}} x_i^t-\epsilon ,\min \nolimits _{i \in {\mathcal {I}}} x_i^t]$. These last two conditions guarantee that the right bound of $R_g$, ${\overline{x}}$, will have to satisfy ${\overline{x}} \in [\max \nolimits _{i \in {\mathcal {I}}} x_i^t,\min \nolimits _{i \in {\mathcal {I}}} x_i^t+\epsilon ]$. Therefore, recalling that $R_g$ extends in the whole length of the vertical axis, it follows that $MBR_{{\mathcal {I}}}^t \subseteq R_g$. Furthermore, the neighboring area of the citizen $j=\mathop {\mathrm{argmin}}\nolimits _{i \in {\mathcal {I}}} x_i^t$ extends for all y at least until $\min \nolimits _{i \in {\mathcal {I}}} x_i^t -k\le {\underline{x}}$. Analogously, the neighboring area of the citizen $l=\mathop {\mathrm{argmin}}\nolimits _{i \in {\mathcal {I}}} x_i^t$ extends for all y at least until $\max \nolimits _{i \in {\mathcal {I}}} x_i^t -k\le {\overline{x}}$. Hence, it follows immediately that $R \subseteq {\mathcal {N}}_{(x_i^t,y_i^t)}$ for all $i \in {\mathcal {I}}$. Thus, also $MBR_{{\mathcal {I}}}^t \subseteq {\mathcal {N}}_{(x_i^t,y_i^t)}$ for all $i \in {\mathcal {I}}$. $\square$

1.2 Proofs of results on the restricted setup

Explanatory note For the results on the Restricted Setup we follow a constructive and rather primitive approach, which comes at a considerable computational cost. In particular, we derive analytically the exact probabilities (for the bonding process), or rather strict bounds of the probabilities (for the bridging process) with which each process converges to form one, two or three islands respectively. We provide the results for $d \in (0,1/2]$ and $k\in (0,1/4]$, nevertheless with similar calculations they could be extended to higher bounds, in which cases there could only exist either one or two islands. The reason we restrict k to be below 1/4 is that for higher values the closed form expressions of the probabilities we have calculated change. This would require to essentially repeat the same analysis, without adding much to our understanding of the problem. Finally, in order not to make the proofs excessively long, we present a concise version of the calculations. The complete and longer version of the calculations of all sequences of interactions is available upon request.

Lemma 1

In the Restricted Setup, consider a bonding process with $d \in (0,1/2]$. Opinions converge to form

• Three islands, with probability $1-12d^2+12d^3+36d^4-68d^5+\frac{88}{3}d^6$

• Two islands, with probability $\frac{1}{48}d^2(576-576d-2700d^2+4884d^3-2083d^4)$

• One island, with probability $\frac{9}{16}d^4(6-5d)^2$

Proof of Lemma 1

It is useful to observe that the random draws of the initial opinions can be decomposed into three independent draws. Namely, first we draw independently three points, each from a uniform distribution in [0, 1], which represent the opinions on x, call them $x_1\le x_2 \le x_3$. Without loss of generality, let us identify citizens according to their opinions on x, i.e. $\widehat{x_a}=x_1$, $\widehat{x_b}=x_2$ and $\widehat{x_c}=x_3$. Then, we draw independently three points, each from a uniform distribution in [0, 1], which represent the opinions on y, call them $y_1\le y_2\le y_3$. Finally, the three opinions on y are matched randomly with opinions on x, i.e. we draw one of the six permutations of (a, b, c) each with probability 1/6. If permutation (i, j, k) is selected, then $\widehat{y_i}=y_1$, $\widehat{y_j}=y_2$ and $\widehat{y_k}=y_3$. Importantly, the three sets of draws are independent, therefore we can perform the analysis separately for each issue.

There are five essentially different types of draws in each issue, depending on the relative position of the three points with respect to the confidence bound. They are presented graphically in Figure 7 and their probabilities of appearing are the following:

• All far: $6\int _0^{1-2d}\frac{(1-x_1-2d)^2}{2}dx_1=(1-2d)^3$

• All close: $6\left( \int _0^{1-d}\frac{d^2}{2}dx_1+\int _{1-d}^1\frac{(1-x_1)^2}{2}dx_1\right) =3d^2-2d^3$

• 12, 23 close, 13 far: $6\left( \int _0^{1-2d}\frac{d^2}{2}dx_1+\int _{1-2d}^{1-d}\frac{d^2}{2}-\frac{(x_1+2d-1)^2}{2}dx_1\right) =3d^2-4d^3$

• 12 close, 23 far: $6\left( \int _0^{1-2d}\frac{d^2}{2}+d(1-x_1-2d)dx_1+\int _{1-2d}^{1-d}\frac{(1-d-x_1)^2}{2}dx_1\right) =3d-9d^2+7d^3$

• 12 far, 23 close: $6\left( \int _0^{1-2d}\frac{d^2}{2}+d(1-x_1-2d)dx_1+\int _{1-2d}^{1-d}\frac{(1-d-x_1)^2}{2}dx_1\right) =3d-9d^2+7d^3$

Fig. 7

Cases of essentially different draws of three opinions on issue x. The sign (> or <) above the segment that connects $x_1$ with $x_2$ ($x_2$ with $x_3$) refers to the distance between the opinions $x_1$ and $x_2$ ($x_2$ and $x_3$), whereas the sign below $x_2$ refers to the distance between $x_1$ and $x_3$. The sign > (<) denotes that the respective distance is larger (smaller) than the confidence bound (d in bonding, k in bridging)

Let us focus first on the cases that lead to three islands. For this to happen, the initial opinions must be sufficiently far that no interaction can lead to an opinion updating.²³ We can calculate this probability by calculating and adding the probabilities of some mutually exclusive cases. To do so, we will use the fact that we have split the initial draws into three independent parts (draws for x, y and order), which allows to calculate the probabilities for each issue separately and then just multiply them. The (mutually exclusive) cases that lead to three islands in the bonding process are as follows:

• $(1-2d)^3$: All far in x.

• $(3d^2-2d^3)(1-2d)^3$: All close in x, all far in y.

• $(3d-9d^2+7d^3)(1-d)^2$: 12 close, 23 far in x, a and b far in y,

• $(3d-9d^2+7d^3)(1-d)^2$: 12 far, 23 close in x, b and c far in y,

• $(3d^2-4d^3)\left[ (1-2d)^3+\frac{2}{3}(3d-9d^2+7d^3)\right]$

  • 12, 23 close, 13 far in x, all far in y,

  • 12, 23 close, 13 far in x, 12 close, 23 far in y, permutations (a,c,b) and (c,a,b),

  • 12, 23 close, 13 far in x, 12 far, 23 close in y, permutations (b,a,c) and (b,c,a).

For “12 close, 23 far in x” (resp., “12 far, 23 close in x”) we simply need to consider that a and b (resp., b and c) are far in y, which occurs with probability $(1-d)^2$.

Overall, by adding those probabilities we get:

$$\begin{aligned} {\mathbb {P}}(\text {3 islands})&=(1-2d)^3+(3d^2-2d^3)(1-2d)^3+\dots \\&\quad \dots +2(3d-9d^2+7d^3)(1-d)^2+(3d^2-4d^3)\\&\quad \left[ (1-2d)^3+\frac{2}{3}(3d-9d^2+7d^3)\right] \\&=1-12d^2+12d^3+36d^4-68d^5+\frac{88}{3}d^6 \end{aligned}$$

Let us now turn our attention to the case of one island, i.e. the case in which the three citizens reach consensus in both issues. For this to happen, at some period all citizens must have opinions with distances at most d in both issues. This can happen under three circumstances: First, if they start with opinions that are at distance at most d in both issues. Second, if in one issue opinions are all close and in the other issue there is a distribution as in Figure 7(c) –$x_1$ close to $x_2$, $x_2$ close to $x_3$ but $x_1$ far from $x_3$, or analogously for y–. Third, if initial opinions are as in Figure 7(c) in both issues.

In the latter two cases, society-wide consensus is not certain, but it is possible. For instance, consider three citizens who start with all opinions close to each other in issue y and in issue x have opinions as in Figure 8. In this case, $x_1$ and $x_3$ are initially far (i.e., $x_3-x_1>d$), but get close to each other after $x_1$ and $x_2$ have been updated to $\frac{x_1+x_2}{2}$ (i.e., $x_3-\frac{x_1+x_2}{2}\le d$). However, this does not occur necessarily, as the initial opinions could be such that not only $x_1$ remains far from $x_3$, but also brings $x_2$ far from $x_3$, in which case society-wide consensus becomes impossible. In general, these indirect interactions may happen in a number of complex ways, yet in the specific problem with three agents and $\mu =1/2$ these are only possible for initial conditions that at least in one issue are of the form presented in Figure 8(a). Hence, two additional conditions are relevant here: (i) $x_3-\frac{x_1+x_2}{2}\lessgtr d$ and (ii) $\frac{x_3+x_2}{2}-x_1\lessgtr d$, as well as the two respective conditions for issue y.

Fig. 8

Example of opinions after successful communication between citizens with $x_1$ and $x_2$

The two additional conditions divide the set of triplets $(x_1,x_2,x_3)$ of Figure 7(c) to four subsets, which are depicted in Figure 9. We calculate the probability that a randomly drawn triplet will belong to each of these sets.

• 1. $x_2-x_1\le d$, $x_3-x_2\le d$, $x_3-x_1>d$ and $x_3-\frac{x_1+x_2}{2}\le d$, $\frac{x_3+x_2}{2}-x_1\le d$:

  • $6\left( \int _0^{1-4d/3}\frac{d^2}{6}dx_1+\int _{1-4d/3}^{1-d}\frac{d^2}{6}-\frac{3(x_1+\frac{4d}{3}-1)^2}{2}dx_1\right) =\frac{1}{9}(9d^2-10d^3)$

• 2. $x_2-x_1\le d$, $x_3-x_2\le d$, $x_3-x_1>d$ and $x_3-\frac{x_1+x_2}{2}\le d$, $\frac{x_3+x_2}{2}-x_1> d$:

  • $6\left( \int _0^{1-3d/2}\frac{d^2}{12}dx_1+\int _{1-3d/2}^{1-4d/3}\frac{d^2}{12}-(x_1+\frac{3d}{2}-1)^2dx_1+\int _{1-4d/3}^{1-d}\frac{(1-d-x_1)^2}{2}dx_1\right) =\frac{1}{36}(18d^2-23d^3)$

• 3. $x_2-x_1\le d$, $x_3-x_2\le d$, $x_3-x_1>d$ and $x_3-\frac{x_1+x_2}{2}> d$, $\frac{x_3+x_2}{2}-x_1\le d$:

  • $6\left( \int _0^{1-3d/2}\frac{d^2}{12}dx_1+\int _{1-3d/2}^{1-4d/3}\frac{d^2}{12}-(x_1+\frac{3d}{2}-1)^2dx_1+\int _{1-4d/3}^{1-d}\frac{(1-d-x_1)^2}{2}dx_1\right) =\frac{1}{36}(18d^2-23d^3)$

• 4. $x_2-x_1\le d$, $x_3-x_2\le d$, $x_3-x_1>d$ and $x_3-\frac{x_1+x_2}{2}> d$, $\frac{x_3+x_2}{2}-x_1> d$:

  • $6\left( \int _0^{1-2d}\frac{d^2}{6}dx_1+\int _{1-2d}^{1-3d/2}\frac{d^2}{6}-\frac{(x_1+2d-1)^2}{2}dx_1+\int _{1-3d/2}^{1-4d/3}\frac{3(1-\frac{4d}{3}-x_1)^2}{2}dx_1\right) =\frac{1}{18}(18d^2-29d^3)$

Fig. 9

Two–dimensional projection of the area in which $x_2-x_1\le d$, $x_3-x_2\le d$ and $x_3-x_1>d$, for some $x_1 \in [0,1-d]$. Areas 1–4 correspond to the four different areas with respect to the conditions $x_3-\frac{x_1+x_2}{2}\lessgtr d$ and $\frac{x_3+x_2}{2}-x_1\lessgtr d$. The origin of the axes is $(x_1,x_1)$ and the location of the line where $x_3=1$ depends on $x_1$

For these cases, the emergence or not of society-wide consensus does not depend only on the initial opinions, but also on the sequence of interactions. In fact, for the same profile of initial opinions, one sequence might lead to consensus, while another might not. Below, we present the probabilities with which each class of initial opinions leads to consensus. When we mention “case i” we refer to area $i=1,\dots ,4$ in Figure 9 for one issue and when we mention “case (i, j)” we refer to area i for issue x and to area j for issue y.

• $(3d^2-2d^3)^2$: All close in both x and y.

• $(3d^2-2d^3)\left[ \frac{1}{9}(9d^2-10d^3)+\frac{1}{36}(18d^2-23d^3)\right]$: All close in x, 12, 23 close, 13 far in y.

  • Case 1, with probability 1.

  • Case 2, with probability 1/2.

  • Case 3, with probability 1/2.

  • same argument and probabilities hold for:

    • 12, 23 close, 13 far in x, all close in y.

• $\left[ \frac{1}{9}(9d^2-10d^3)+\frac{1}{36}(18d^2-23d^3)\right] ^2$: 12, 23 close, 13 far in both x and y.

  • Case (1, 1), with probability 1.

  • Cases (1, 2), (2, 1), (1, 3), (3, 1) with probability 1/2.

  • Cases (2, 2), (2, 3), (3, 2), (3, 3) with probability 1/4.

Given that these cases are mutually exclusive, we can simply add the probabilities and get:

$$\begin{aligned} {\mathbb {P}}(\text {1 island})=\left[ (3d^2-2d^3)+\frac{1}{9}(9d^2-10d^3)+\frac{1}{36}(18d^2-23d^3)\right] ^2=\frac{9}{16}d^4(6-5d)^2 \end{aligned}$$

Finally, two islands arise with the remaining probability. $\square$

Lemma 2

In the Restricted Setup, consider a bridging process with $k \in (0,1/4]$. Opinions converge to form

• Three islands, with probability $(1-2k)^6$

• Two islands, with probability at most $\left( 12-93k+337k^2-725k^3+946k^4-535k^5\right) k$

• One island, with probability at least $\left( 33-177k+485k^2-754k^3+471k^4\right) k^2$

Proof of Lemma 2

In a bridging process the only way to end up with three islands is if all citizens’ initial opinions are located further than the confidence bound in both issues. The probability that the three opinions are all far from each other is $(1-2k)^3$ in each issue, draws are independent across issues and the permutations that match opinions on x with opinions on y do not affect the result. Thus ${\mathbb {P}}(\text {3 islands})=(1-2k)^6$.

In all the other cases the society ends with either one or two islands and, contrary to the bonding process, it is possible to reach society-wide consensus in each of them. In particular, if all citizens are close (see Figure 7b) in at least one of the two issues, then society-wide consensus is reached with probability one, as this guarantees that any two of them can communicate successfully at any period. The probability of all citizens being close in at least one of the two issues is equal to $2(3k^2-2k^3)-(3-2k)^2k^4$.

For the remaining cases one island occurs with probability lower than one. For some of them, we provide their exact values, whereas for others we provide a lower bound.

Fig. 10

Two-dimensional projections of the following areas: (Left) $x_2-x_1\le k$, $x_3-x_2\le k$ and $x_3-x_1>k$, for some $x_1 \in [0,1-k]$. Areas 1–4 correspond to the four different areas with respect to the conditions $x_3-\frac{x_1+x_2}{2}\lessgtr k$ and $\frac{x_3+x_2}{2}-x_1\lessgtr k$. (Right) $x_2-x_1\le k$ and $x_3-x_2>k$, for some $x_1 \in [0,1-k]$. Areas 5–8 correspond to the four different areas with respect to the conditions $\frac{x_3+x_2}{2}-x_1\lessgtr k$, $\frac{1}{2}\left( x_3-\frac{x_1+x_2}{2}\right) \lessgtr k$ and $\left| \frac{x_1+x_3}{2}-x_2\right| \lessgtr k$. For the last condition only $\frac{x_1+x_3}{2}-x_2\lessgtr k$ is relevant. The origins of the axes are $(x_1,x_1)$ in both subfigures and the location of the line where $x_3=1$ depends on $x_1$

Fig. 11

Two-dimensional projections of areas that satisfy $x_2-x_1>k$ and $x_3-x_2>k$. Areas 9–13 are obtained by taking into account the additional conditions $\frac{1}{2}\left( x_3-\frac{x_1+x_2}{2}\right) \lessgtr k$, $\frac{1}{2}\left( \frac{x_2+x_3}{2}-x_1\right) \lessgtr k$ and $x_3\lessgtr x_1+4k$

We first present the results for the cases that care as in Figure 7(d), that is $x_2-x_1\le k$ and $x_3-x_2>k$. Areas 5–8 on Figure 10 split these triplets into four areas within which the probabilities of reaching society-wide consensus are altered. Areas 7 and 8 are split into two parts each (7i, 7ii and 8i, 8ii respectively), where the additional splits are relevant only in some cases. Below we calculate explicitly the probability that three initial opinions in an issue belong to each of these areas. Areas 1–4 on Figure 10 correspond to draws as in Figure 7(c) and coincide with the areas presented in Figure 9, if one substitutes d with k, for which we have already calculated the probabilities of occurrence. Finally areas 9–13 on Figure 11 correspond to the case Figure 7(a) in which all opinions are far from each other and each of these areas is relevant in some cases. We also calculate explicitly their probabilities.

• 5. $x_2-x_1\le k$, $x_3-x_2>k$, $x_3-x_1>k$ and $\frac{x_3+x_2}{2}-x_1\le k$:

  • $6\left( \int _0^{1-2k}\frac{k^2}{4}dx_1+\int _{1-2k}^{1-\frac{3k}{2}}\frac{k^2}{4}-\frac{\left[ x_1-(1-2k)\right] ^2}{2}dx_1+\int _{1-\frac{3k}{2}}^{1-k}\frac{[(1-k)-x_1]^2}{2}dx_1\right) =\frac{3}{4}(2-3k)k^2$

• 6. $x_2-x_1\le k$, $x_3-x_2>k$, $x_3-x_1>k$, $\frac{x_3+x_2}{2}-x_1> k$ and $\frac{1}{2}\left( x_3-\frac{x_1+x_2}{2}\right) \le k$:

  • $6\left( \int _0^{1-\frac{5k}{2}}\frac{k^2}{2}dx_1+\int _{1-\frac{5k}{2}}^{1-2k}\frac{k^2}{2}-\left[ x_1-\left( 1-\frac{5k}{2}\right) \right] ^2dx_1+\int _{1-2k}^{1-\frac{3k}{2}}\left[ \left( 1-\frac{3k}{2}\right) -x_1\right] ^2dx_1\right) =3(1-2k)k^2$

• 7. $x_2-x_1\le k$, $x_3-x_2>k$, $x_3-x_1>k$, $\frac{1}{2}\left( x_3-\frac{x_1+x_2}{2}\right) >k$ and $\frac{x_1+x_3}{2}-x_2\le k$:

  • (7) $6\left( \int _0^{1-4k}\frac{3k^2}{4}dx_1+\int _{1-4k}^{1-\frac{5k}{2}}\frac{3k^2}{4}-\frac{\left[ x_1-\left( 1-4k\right) \right] ^2}{4}dx_1+\int _{1-\frac{5k}{2}}^{1-2k}\frac{3}{4}\left[ \left( 1-2k\right) -x_1\right] ^2dx_1\right) =\frac{3}{4}(6-17k)k^2$

  • (7ii) $6\left( \int _0^{1-4k}\frac{k^2}{6}dx_1+\int _{1-4k}^{1-\frac{10k}{3}}\frac{k^2}{6}-\frac{\left[ x_1-\left( 1-4k\right) \right] ^2}{4}dx_1+\int _{1-\frac{10k}{3}}^{1-3k}\frac{1}{2}\left[ \left( 1-3k\right) -x_1\right] ^2dx_1\right) =\frac{1}{9}(9-31k)k^2$

  • (7i) $\frac{3}{4}(6-17k)k^2-\frac{1}{9}(9-31k)k^2=\frac{1}{36}(126-335k)k^2$

• 8. $x_2-x_1\le k$, $x_3-x_2>k$, $x_3-x_1>k$ and $\frac{x_1+x_3}{2}-x_2> k$:

  • (8) $6\left( \int _0^{1-4k}k^2+k\left[ (1-4k)-x_1\right] dx_1+\int _{1-4k}^{1-2k}\frac{\left[ \left( 1-2k\right) -x_1\right] ^2}{4}dx_1\right) =(3-18k+28k^2)k$

  • (8ii) $6\left( \int _0^{1-4k}\frac{k^2}{3}+k\left[ (1-4k)-x_1\right] dx_1+\int _{1-4k}^{1-\frac{10k}{3}}\frac{3\left[ \left( 1-\frac{10k}{3}\right) -x_1\right] ^2}{4}dx_1\right) =(3-22k+\frac{364}{9}k^2)k$

  • (8i) $(3-18k+28k^2)k-(3-22k+\frac{364}{9}k^2)k=\frac{4}{9}(9-28k)k^2$

• 9. $x_2-x_1> k$, $x_3-x_2>k$, $\frac{1}{2}\left( \frac{x_3+x_2}{2}-x_1\right) \le k$ and $\frac{1}{2}\left( x_3-\frac{x_1+x_2}{2}\right) \le k$:

  • $6\left( \int _0^{1-\frac{8k}{3}}\frac{k^2}{6}dx_1+\int _{1-\frac{8k}{3}}^{1-\frac{5k}{2}}\frac{k^2}{6}-\frac{3\left( x_1-(1-\frac{8k}{3})\right] ^2}{2}dx_1+\int _{1-\frac{5k}{2}}^{1-2k}\frac{[(1-2k)-x_1]^2}{2}dx_1\right) =\frac{1}{18}(18-43k)k^2$

• 10. $x_2-x_1> k$, $x_3-x_2>k$, $\frac{1}{2}\left( \frac{x_3+x_2}{2}-x_1\right) \le k$ and $\frac{1}{2}\left( x_3-\frac{x_1+x_2}{2}\right) > k$:

  • $6\left( \int _0^{1-3k}\frac{k^2}{12}dx_1+\int _{1-3k}^{1-\frac{8k}{3}}\frac{k^2}{12}-\frac{\left[ \left( 1-3k\right) -x_1\right] ^2}{2}dx_1+\int _{1-\frac{8k}{3}}^{1-\frac{5k}{2}}\left[ \left( 1-\frac{5k}{2}\right) -x_1\right] ^2dx_1\right) =\frac{1}{36}(18-49k)k^2$

• 11. $x_2-x_1> k$, $x_3-x_2>k$, $\frac{1}{2}\left( \frac{x_3+x_2}{2}-x_1\right) > k$ and $\frac{1}{2}\left( x_3-\frac{x_1+x_2}{2}\right) <k$:

  • $6\left( \int _0^{1-3k}\frac{k^2}{12}dx_1+\int _{1-3k}^{1-\frac{8k}{3}}\frac{k^2}{12}-\frac{\left[ \left( 1-3k\right) -x_1\right] ^2}{2}dx_1+\int _{1-\frac{8k}{3}}^{1-\frac{5k}{2}}\left[ \left( 1-\frac{5k}{2}\right) -x_1\right] ^2dx_1\right) =\frac{1}{36}(18-49k)k^2$

• 12. $x_2-x_1> k$, $x_3-x_2>k$, $\frac{1}{2}\left( \frac{x_3+x_2}{2}-x_1\right) > k$, $\frac{1}{2}\left( x_3-\frac{x_1+x_2}{2}\right) >k$ and $x_3\le x_1+4k:$

  • $6\left( \int _0^{1-4k}\frac{5k^2}{3}dx_1+\int _{1-4k}^{1-3k}\frac{k^2}{6}+\frac{(1-k-x_1)(1-3k-x_1)}{2}dx_1+\int _{1-3k}^{1-\frac{8k}{3}}\frac{3\left( 1-\frac{8k}{3}-x_1\right) ^2}{2}\right) =\frac{2}{9}(45-157k)k^2$

• 13. $x_2-x_1> k$, $x_3-x_2>k$ and $x_3>x_1-4k$:

  • $6\int _0^{1-4k}\frac{(1-x_1)(1-x_1-4k)}{2}dx_1=1-6k+32k^3$

We can now calculate the probabilities of society-wide consensus for each of the cases.

• $\left( 6-54k+178k^2-252k^3+\frac{392}{3}k^4\right) k^2$: 12 close, 23 far in x and all far in y

  • if $\left| \frac{y_1+y_3}{2}-y_2\right| \le k$, with probability 1/3 –permutations (a,c,b) or (b,c,a)–.

  • same argument and probabilities hold for:

    • 12 far, 23 close in x and all far in y,

    • all far in x and 12 close, 23 far in y and

    • all far in x and 12 far, 23 close in y.

• $(10-51k+\frac{259}{3}k^2-49k^3)k^3$: 12 close, 23 far in x and 12 close, 23 far in y

  • Case (5, 5), with probability 2/3.

  • Cases (5, 6), (6, 5) with probability 5/9.

  • Cases (5, 7), (7, 5), (6, 6) with probability 4/9.

  • Cases (5, 8), (8, 5), (6, 7), (7, 6) with probability 1/3.

  • Cases (6, 8), (8, 6), (7, 7) with probability 2/9.

  • Cases (7, 8), (8, 7) with probability 1/9.

  • same argument and probabilities hold for:

    • 12 close, 23 far in x and 12 far, 23 close in y

    • 12 far, 23 close in x and 12 close, 23 far in y

    • 12 far, 23 close in x and 12 far, 23 close in y

• $>\left( \frac{58067}{8424}-\frac{3769369}{151362}k+\frac{22810741}{909792}k^2+\frac{6168103}{1259712}k^3\right) k^3$: 12 close, 23 far in x and 12, 23 close, 13 far in y

  • For permutation (a, b, c) - Cases: $(\cdot ,1)$ w.p. 1, (5, 2) w.p. 1, (6, 2) w.p. 5/6, (7i, 2) w.p. 5/6, (8i, 2) w.p. 5/6, (7ii, 2) w.p. 39/54, (8ii, 2) w.p. $>55/81$, (5, 3) w.p. 2/3, (6, 3) w.p. 2/3, (7i, 3) w.p. 2/3, (8i, 3) w.p. 2/3, (7ii, 3) w.p. 43/72, (8ii, 3) w.p. $>359/648$, (5, 4) w.p. 2/3, (6, 4) w.p. 1/2, (7i, 4) w.p. 7/18, (8i, 4) w.p. 7/18, (7ii, 4) w.p. 23/72, (8ii, 4) w.p. $>161/468$.

  • For permutation (b, a, c) - Cases: $(\cdot ,1)$ w.p. 1, (5, 2) w.p. 1, (6, 2) w.p. 11/12, (7i, 2) w.p. 29/36, (8i, 2) w.p. 13/18, (7ii, 2) w.p. 29/36, (8ii, 2) w.p. $>55/81$, (5, 3) w.p. 2/3, (6, 3) w.p. 2/3, (7i, 3) w.p. 2/3, (8i, 3) w.p. 2/3, (7ii, 3) w.p. 2/3, (8ii, 3) w.p. $>359/648$, (5, 4) w.p. 2/3, (6, 4) w.p. 7/12, (7i, 4) w.p. 17/36, (8i, 4) w.p. 7/18, (7ii, 4) w.p. 17/36, (8ii, 4) w.p. $>161/468$.

  • For permutation (a, c, b) - Cases: $(\cdot ,1)$ w.p. 1, (5, 2) w.p. 1, (6, 2) w.p. 8/9, (7i, 2) w.p. 91/108, (8i, 2) w.p. 91/108, (7ii, 2) w.p. 91/108, (8ii, 2) w.p. $>733/972$, (5, 3) w.p. 1, (6, 3) w.p. 1, (7i, 3) w.p. 35/36, (8i, 3) w.p. 31/36, (7ii, 3) w.p. 25/27, (8ii, 3) w.p. $>755/972$, (5, 4) w.p. 1, (6, 4) w.p. 8/9, (7i, 4) w.p. 85/108, (8i, 4) w.p. 73/108, (7ii, 4) w.p. 80/108, (8ii, 4) w.p. $>14/27$.

  • By symmetry, these give us also the total probabilities for the other three permutations. (c, b, a) is analogous to (a, b, c), (b, c, a) is analogous to (b, a, c) and (c, a, b) to (a, c, b).

  • same argument and probabilities hold for:

    • 12 far, 23 close in x and 12, 23 close, 13 far in y

    • 12, 23 close, 13 far in x and 12 close, 23 far in y

    • 12, 23 close, 13 far in x and 12 far, 23 close in y

• $>\left( \frac{545}{324}-\frac{70663}{5832}k+\frac{75203}{1944}k^2-\frac{811}{9}k^3+\frac{22145}{288}k^4\right) k^2$: 12 close, 23 far in x and all far in y

  • For permutation (a, b, c) - Cases: $(\cdot ,1)$ w.p. 1, (9, 2) w.p. 2/3, (10, 2) w.p. 43/72, (11, 2) w.p. 2/3, (12, 2) w.p. 43/72, (13, 2) w.p. $>181/324$, (9, 3) w.p. 2/3, (10, 3) w.p. 2/3, (11, 3) w.p. 43/72, (12, 3) w.p. 43/72, (13, 3) w.p. $>181/324$, (9, 4) w.p. 1/3, (10, 4) w.p. 19/72, (11, 4) w.p. 19/72, (12, 4) w.p. 7/36, (13, 4) w.p. $>10/81$.

  • For permutation (b, a, c) - Cases: $(\cdot ,1)$ w.p. 1, (9, 2) w.p. 1, (10, 2) w.p. 67/72, (11, 2) w.p. 1, (12, 2) w.p. 67/72, (13, 2) w.p. $>181/324$, (9, 3) w.p. 2/3, (10, 3) w.p. 2/3, (11, 3) w.p. 2/3, (12, 3) w.p. 2/3, (13, 3) w.p. $>181/324$, (9, 4) w.p. 2/3, (10, 4) w.p. 43/72, (11, 4) w.p. 2/3, (12, 4) w.p. 43/72, (13, 4) w.p. $>10/81$.

  • By symmetry, these give us also the total probabilities for the remaining permutations. (c, b, a) is analogous to (a, b, c), and all others are analogous (b, a, c) and (c, a, b).

  • same argument and probabilities hold for:

    • all far in x and 12, 23 close, 13 far in y

• $\frac{1}{48}\left( 412-1092k+723k^2\right) k^4$: 12, 23 close, 13 far in both x and y

  • Cases $(1,\cdot )$, $(\cdot ,1)$ with probability 1.

  • Cases (2, 2), (2, 3), (3, 2), (3, 3) with probability 103/108.

  • Cases (2, 4), (4, 2), (3, 4), (4, 3) with probability 98/108.

  • Case (4, 4) with probability 88/108.

Given that these cases are mutually exclusive, we can simply add the probabilities (recalling that some of the cases appear multiple times and we get the lower bound:

$$\begin{aligned} {\mathbb {P}}(\text {1 island})&>\left( \frac{5405}{162}-\frac{6696853}{37908}k+\frac{18405017}{37908}k^2-\frac{171347717}{227448}k^3+\frac{148623383}{314928}k^4\right) k^2>\\&>(33-177k+485k^2-754k^3+471k^4)k^2 \end{aligned}$$

$\square$

Proof of Proposition 4

The result is implied immediately by Lemma 1. ${\mathbb {P}}(\text {1 island})$ increases in d, whereas ${\mathbb {P}}(\text {3 islands})$ decreases in d, both for all $d\in (0,1/2]$. Thus, part (i) is obtained immediately and part (ii) is obtained by observing that ${\mathbb {E}}(\text {islands})={\mathbb {P}}(\text {1 island})+2{\mathbb {P}}(\text {2 islands})+3{\mathbb {P}}(\text {3 islands})=2-{\mathbb {P}}(\text {1 island})+{\mathbb {P}}(\text {3 islands})$. $\square$

Proof of Proposition 5

The result is implied immediately by Lemma 2. The lower bound of ${\mathbb {P}}(\text {1 island})$ increases in k, whereas ${\mathbb {P}}(\text {3 islands})$ decreases in k, both for all $k\in (0,1/4]$. Thus, part (i) is obtained immediately and part (ii) is obtained by recalling that ${\mathbb {E}}(\text {islands})=2-{\mathbb {P}}(\text {1 island})+{\mathbb {P}}(\text {3 islands})$, thus substituting the lower bound of ${\mathbb {P}}(\text {1 island})$ and the actual value of ${\mathbb {P}}(\text {3 islands})$ gives an upper bound of ${\mathbb {E}}(\text {islands})$, which then decreases in k. $\square$

Proof of Proposition 6

The proofs of both parts are almost identical and both make use of the Intermediate Value Theorem.

We start by proving the result on the probability of society-wide consensus. For $k=0$ the number of islands is equal to three with probability one.²⁴ Therefore, the probability of society-wide consensus in such a bridging process is zero, which is strictly lower than the respective probability for the bonding process with confidence bound d, for any $d\in (0,1/4]$, which is always strictly positive. For $k=d$, we know from Lemmas 1 and 2 that the probability of ending up with one island in a bonding process with bound d is lower than the lower bound of the probability of ending up with one island in a bridging process with bound $k=d$, hence it is obviously also lower than the actual probability. Moreover, given that the lower bound is continuous in k (as a polynomial) and it is also strictly increasing for all $k\in (0,1/4]$, we know from Intermediate Value Theorem that there will be some ${\overline{k}}\in (0,1/4)$ such that the lower bound of the probability in the bridging process becomes equal to the probability in the bonding process with confidence bound d and for all $k>{\overline{k}}$ the lower bound, thus also the actual probability, in the bridging process will be strictly larger than the respective probability in the bonding process.

The arguments for the expected number of islands is analogous. For $k=0$, the expected number of islands in the bridging process is equal to three, which is strictly higher than for that of the bonding process with confidence bound d, for any $d\in (0,1/2]$. For $k=d$ we already mentioned that the probability of having 1 island in the bonding process with d is lower even than the lower bound of the probability of having 1 island in the bridging process with $k=d$. We also know (again from Lemmas 1 and 2) that the probability of having 3 islands is larger in the bonding than in the bridging process. Thus, recalling that ${\mathbb {E}}(\text {islands})=2-{\mathbb {P}}(\text {1 island})+{\mathbb {P}}(\text {3 islands})$, the expected number of islands in the bonding process is larger than in the bridging process. Therefore, given that the upper bound of the expected number of islands in the bridging process is continuous, again from Intermediate Value Theorem, there will be some ${\tilde{k}}\in (0,1/4)$ such that the upper bound of the expected number of islands in the bridging process will be exactly equal to the expected number of islands in the bonding process with confidence bound d. Moreover, given that it is also strictly decreasing in k, for all $k>{\tilde{k}}$ the upper bound for the bridging process, thus also the actual expected number of islands, will be strictly smaller than the expected number of islands in the bonding process with confidence bound d. Both ${\overline{k}}$ and ${\tilde{k}}$ are in the interior of (0, 1/4], hence both results hold simultaneously for any $k>{\widehat{k}}=\max \{{\overline{k}},{\tilde{k}}\}\in (0,1/4)$. $\square$

Proof of Proposition 7

The result can be obtained immediately by simple calculations, using the values obtained in Lemmas 1 and 2. $\square$

Table 4 Quadratic probability of successful communication. Mean, median and standard deviation of number of islands and frequency of society-wide consensus for the two processes for different population sizes

Table 5 Quadratic probability of successful communication. Fractionalization Indices – $\mu _x$, ${\widetilde{x}}$ and $\sigma _{x}$ correspond to the mean, median and standard deviation of the two fractionalization indices respectively, for the two processes and for different populations sizes. The values denoted by 0.000 refer to quantities that are positive but lower than $10^{-3}$, whereas the values denoted by 0 refer to quantities with value exactly equal to zero