Collective Opinion as Tendency Towards Consensus

Group beliefs in social networks are often construed as arising from individual beliefs through processes of update and aggregation. In this paper, we explore an alternative ‘arational’ perspective. More specifically, we focus on group attitudes as neutral tendencies toward alignment of opinions driven by influence patterns among agents modeled in a Markov dynamics. In addition, we investigate logical patterns in the resulting potential group beliefs or, in more neutral arational terminology: collective opinion structures.

1. In such a graph representation, there is an edge from i to j iff \(\mathfrak {I}_{ij} > 0\).

This research is supported by the Major Program of the National Social Science Foundation of China (NO. 17ZDA026). The author is supported by the “Shuimu Scholars” programme of Tsinghua University. The author would like to thank Fenrong Liu, Sonja Smets and Alexandru Baltag for bringing to his attention the fascinating topic on social network and information flow and keeping giving him advice during the writing of this paper. Especially, he would like to thank Johan van Benthem for his substantial feedback and help, which put the results presented here into perspective. Furthermore, the author would like to thank the two referees for their invaluable comments and suggestions. At last, the author would like to thank the Tsinghua - Amsterdam Joint Research Centre for Logic for making the exchange visits possible to conduct the research in this paper.

Authors and Affiliations

1. Tsinghua-Amsterdam Joint Research Centre for Logic, Department of Philosophy, Tsinghua University, Beijing, China

   Chenwei Shi

Corresponding author

Correspondence to Chenwei Shi.

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Appendix

1.1 A1 Proof of Theorems 2 and 3

Theorem 2

When an influence matrix is strongly connected, 0 and 1 are the only absorbing states in the generated transition matrix (Theorem 2). This is implied by the following: (a) A group state b is absorbing in \(\mathbb {T}\) generated by \(\mathfrak {I}\) if and only if Ib = b, (b) For a strongly connected influence matrix \(\mathfrak {I}\), every column vector x with \(\mathfrak {I}\mathbf {x} = \mathbf {x}\) is a constant vector (Theorem 11.10 in [13]).

Theorem 3

Next we prove, using also the regularity, that all states except 1 and 0 can reach both absorbing states.

Lemma

Given a group with a strongly connected and aperiodic influence matrix \(\mathfrak {I}\), for any vector \(\mathbf {b}\in \mathfrak {b}\) with some i ∈ G such that b_i = 1, there exists a walk from b to 1. I.e., there is a sequence p⁰,p¹,…,pⁿ of vectors in \(\mathfrak {b}\) such that p⁰ = b, pⁿ = 1 and for all natural numbers k ∈ [0,n − 1], \(\mathbb {T}_{p^{k}, p^{k+1}} >0\). Similarly, there is a walk from b≠1 to 0.

Proof

First, it is well-known, [24], that when a stochastic matrix \(\mathfrak {I}\) is strongly connected and aperiodic, there is a natural number N such that for all n ≥ N, \(\mathfrak {I}^{n}_{ij} > 0\) for all i and j in the group (see also [30, Appendix D.3, Theorem D.3.1] for a concrete proof).

Now take any binary vector \(\mathbf {b}\in \mathfrak {b}\) such that b_i = 1 for some i ∈ G. We fix such an agent, ι, satisfying b_ι = 1. We try to find a natural number n and n + 1 binary vectors p⁰,…,pⁿ such that \(\mathbb {T}_{p^{k}, p^{k+1}} >0\) for all natural numbers k ∈ [0,n − 1] and p⁰ = b and pⁿ = 1.

We first show how to find the sequence of binary vectors. Let p⁰ = b. Suppose that p^k has been defined, then we set \(p^{k+1}_{i} := 1\), if there is y ∈ G such that \(\mathfrak {I}_{iy}>0\) and \({p^{k}_{y}} = 1\), and := 0, otherwise. □

Claim

This definition implies that for all positive integers k, \(\mathbb {T}_{p^{k}, p^{k+1}} >0\).

Proof

Take any i ∈ G. (a) If there is a y ∈ G such that \(\mathfrak {I}_{iy}>0\) and \({p^{k}_{y}} = 1\), according to the definition of \(P(Bv_{i}^{\prime }=1|Bv=p^{k})\) given by (2), \(\mathfrak {I}_{iy}>0\) and \({p^{k}_{y}} = 1\) imply that \(P(Bv_{i}^{\prime }=1|Bv=p^{k}) > 0\). Since in this case \(p^{k+1}_{i} = 1\), it clearly follows that \(P(Bv_{i}^{\prime }=p^{k+1}_{i}|Bv=p^{k}) > 0\). (b) If there is no y ∈ G such that \(\mathfrak {I}_{iy}>0\) and \({p^{k}_{y}} = 1\), then \(P(Bv_{i}^{\prime }=1|Bv=p^{k}) = 0\), which implies that \(P(Bv_{i}^{\prime }=0|Bv=p^{k}) =1 > 0\). In this case \(p^{k+1}_{i} = 0\). Therefore, we have \(P(Bv_{i}^{\prime }=p^{k+1}_{i}|Bv=p^{k}) > 0\).

It follows that, for all i ∈ G, \(P(Bv_{i}^{\prime }=p^{k+1}_{i}|Bv=p^{k}) > 0\), which implies that \(\mathbb {T}_{p^{k}, p^{k+1}} = {\prod }_{i\in G}P(Bv_{i}^{\prime }=p^{k+1}_{i}|Bv=p^{k}) > 0\).

Next, we show that by taking any number n satisfying \(\mathfrak {I}_{ij}^{n} > 0\) for all i,j ∈ G (the existence of such numbers n is ensured by the property of the influence matrix \(\mathfrak {I}\) observed at the beginning of this proof), pⁿ = 1.

Assume that n is the number satisfying \(\mathfrak {I}_{ij}^{n} > 0\) for all i,j ∈ G. Then for all x ∈ G, there must be a walk from x to ι in the graph representation of the matrix \(\mathfrak {I}\)^{Footnote 1} whose length is n. For each x ∈ G, fix such a walk w_x from x to ι. Let S_n = {w_x∣x ∈ G}. Keep in mind that for all walks in the associated graph, their length is n and they start from x and end at ι. □

Claim

pⁿ = 1.

Proof

For all x ∈ G, let \({w_{x}^{m}}\) be the agent m steps away from ι along the opposite direction of the path w_x ∈ S_n. For example, \({w_{x}^{0}} = \iota \). (a) We first show that for all x ∈ G and all m > 0, \(p^{m}_{{w^{m}_{x}}} = 1\) by induction. Obviously, \(p^{1}_{{w^{1}_{x}}} = 1\), because there is y ∈ G, which is \({w^{0}_{x}}\) (i.e., ι), satisfying \(\mathfrak {I}_{{w^{1}_{x}}y} > 0\) and \({p^{0}_{y}} = 1\). (b) Now suppose that \(p^{m}_{{w^{m}_{x}}} = 1\). By noticing that \({w^{m}_{x}}\) satisfies \(\mathfrak {I}_{w^{m+1}_{x}{w^{m}_{x}}} > 0\), we get \(p^{m+1}_{w^{m+1}_{x}} = 1\). (c) Next, we show that for the number n, \(G = \{{w^{n}_{x}}\mid x\in G\}\). Recall the definition of w_x. For all x ∈ G, it follows by the definition of w_x that \({w^{n}_{x}} = x\).

Therefore, for all x ∈ G, \({p^{n}_{x}} = p^{n}_{{w^{n}_{x}}}= 1\). That is pⁿ = 1.

By a similar argument there is a walk from any state b≠1 to 0. □

1.2 A2 Proof of Theorem 5

The proof is by induction. We first prove the base case for all states of a group \(\mathbf {b}\in \mathfrak {b}\) and all group members i ∈ G, \(\mathfrak {I}_{i\ast }\cdot \mathbf {b} = {\sum }_{\mathbf {s}_{i} = 1}\mathbb {T}_{\mathbf {bs}}\).

Claim

\({\sum }_{\{\mathbf {s}\in \mathfrak {b}\mid \mathbf {s}_{i} =1\}}{\prod }_{\{x\in G\mid x\neq i\}} P(Bv_{x}^{\prime } = \mathbf {s}_{x}|Bv = \mathbf {b}) = 1\)

Proof

Recall that \(P(Bv^{\prime } = \mathbf {s}| Bv = \mathbf {b}) = \mathbb {T}_{\mathbf {bs}}\). Since \(\mathbb {T}_{\mathbf {bs}} = {\prod }_{x\in G} P(Bv_{x}^{\prime } = \mathbf {s}_{x}|Bv = \mathbf {b})\), for all \(\mathbf {s}\in \mathfrak {b}\) such that s_i = 1: \( P(Bv^{\prime } = \mathbf {s}| Bv = \mathbf {b}, Bv_{i}^{\prime } = 1) = \frac {P(Bv ' = \mathbf {s}| Bv = \mathbf {b})}{P(Bv_{i}^{\prime } = 1|Bv =\mathbf {b})} = {\prod }_{\{x\in G\mid x\neq i\}} P(Bv_{x}^{\prime } = \mathbf {s}_{x}|Bv = \mathbf {b}) \). Obviously, \({\sum }_{\{\mathbf {s}\in \mathfrak {b}\mid \mathbf {s}_{i} =1\}}P(Bv^{\prime } = \mathbf {s}\mid Bv = \mathbf {b}, Bv_{i}^{\prime } = 1) = 1\).

Making use of the claim, we prove the base case:

Next we prove \(\mathfrak {I}^{n+1}_{i\ast }\cdot \mathbf {b} = {\sum }_{\mathbf {s}_{i} = 1}\mathbb {T}^{n+1}_{\mathbf {bs}}\) assuming that \(\mathfrak {I}^{n}_{i\ast }\cdot \mathbf {b} = {\sum }_{\mathbf {s}_{i} = 1}\mathbb {T}^{n}_{\mathbf {bs}} \enspace \).

This completes the proof.

Next, we use Theorem 5 to prove the direction from 3 to 4 in Theorem 6.

First, using Theorem 5, we prove that as n tends to \(\infty \), if the powers of the transition matrix \(\mathbb {T}^{n}\) converge, so do \(\mathfrak {I}^{n}\), the powers of the influence matrix . □

Proof

Let the powers \(\mathbb {T}^{n}\) converge as n tends to \(\infty \). By Theorem 5, \(\mathfrak {I}_{ij}^{n} = \mathfrak {I}_{i\ast }^{n}\mathbf {e_{j}} = {\sum }_{\mathbf {s}_{i} = 1}\mathbb {T}^{n}_{\mathbf {e}_{j}\mathbf {s}}\) for all \(n\in \mathbb {N}\), where e_j is the vector with i th entry 1 and all other entries 0. Because \(\mathbb {T}^{n}\) converges to \(\mathbb {T}^{\infty }\), \(\mathfrak {I}_{ij}^{n}\) converges to \({\sum }_{\mathbf {s}_{i} = 1}\mathbb {T}^{\infty }_{\mathbf {e}_{j}\mathbf {s}}\). Each entry of \(\mathfrak {I}\) thus converges to a number, and hence \(\mathfrak {I}\) converges. □

Second, we prove that if in the limiting matrix \(\mathbb {T}^{\infty }\), \(\mathbb {T}^{\infty }_{\mathbf {bd}} = 0\) for each state b and each state d≠1,0, no other entries are zero and \(\mathbb {T}^{\infty }_{\mathbf {11}} = \mathbb {T}^{\infty }_{\mathbf {00}} = 1\), then the rows of \(\mathfrak {I}^{\infty }\) are all the same strictly positive stochastic vector.

Proof

Given the assumption, for all i ∈ G, \(\mathfrak {I}_{ij}^{\infty } = \mathfrak {I}^{\infty }_{i\ast }\mathbf {e}_{j} = {\sum }_{\mathbf {s}_{i} = 1}\mathbb {T}^{\infty }_{\mathbf {e}_{j}\mathbf {s}} = \mathbb {T}^{\infty }_{\mathbf {e}_{j}\mathbf {1}}\). The second equality is what we have just proved. The third equality holds as the only non-zero entries in \(\mathbb {T}^{\infty }_{\mathbf {e}_{j}\ast }\) are \(\mathbb {T}^{\infty }_{\mathbf {e}_{j}\mathbf {1}}\), \(\mathbb {T}^{\infty }_{\mathbf {e}_{j}\mathbf {0}}\). □

Finally, the fact that the row vector in \(\mathfrak {I}^{\infty }\) sums to 1 follows from the fact that any powers of a probability matrix is a probability matrix (namely, each row vector sums to 1 and is non-negative).

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