A signalling explanation for preferential attachment in the evolution of social networks

Abstract

The network theory developed by physicists has several critical drawbacks in characterizing the structure of social networks. First, they largely neglect considering the link cost and the link benefit that agents usually take into account in forming their links. Second, they view a social network as a consequence of unilateral decisions of agents, not of bilateral decisions of linking parties, although a link of an agent can be formed only after he obtains the consent of the other side. Third, there is no logical justification for the assumption of preferential attachment upon which their analysis heavily relies. In this paper, we provide several models that overcome the three drawbacks. By analyzing the models, we can explain preferential attachment as rational equilibrium behavior. The main idea is that people are not certain of the value that they can obtain from forming a link with someone. Based on this assumption, we will argue that a person has an incentive to form a link with another who has many links because the number of his links can convey some information about his value.

Appendix

Proof of Lemma 2

(i) Since player i has been once linked with some j, either v _i  = L or v _j  = L if d(i; g _{t − 1, 2}) = 0. Thus, $\alpha_t (i)=\frac{\alpha}{1+\alpha}<\alpha_t (t-1)=\alpha$. (ii) If (i, j) ∈ g _{t − 1, 2} for some j < t − 1, α _t (i) = 1 > α _t (t − 1) = α by Lemma 1. (iii) Let (i, j) ∈ g _{i, 1}. Clearly, j < i. If (i, t − 1) ∈ g _{t − 1, 2} and d(i; g _{t − 1, 2}) = 1, this implies that (i,j) has been deleted. This implies that either v _i  = L or v _j  = L. Thus, $\alpha_t (i{\kern1.5pt})=\frac{\alpha}{1+\alpha}<\alpha_t (t-1)=\alpha$. □

Proof of Lemma 4

It is clear that g _{2, 1} = g _{2, 2} satisfies [UNC], since it is connected. Suppose g _{t − 1, 2} satisfies [UNC]. From Lemma 2(i), it is clear that g _{t, 1} also satisfies [UNC]. It only remains to show that g _{t, 2} satisfies [UNC]. Two possible cases are that either (i) (i, t) ∈ g _{t, 1} for i ≠ t − 1 or (ii) (t − 1, t) ∈ g _{t, 1}. In the case (i), supposing (t − 1, k) ∈ g _{t, 1} for some k ∈ N _t − 2 and deleting (t − 1, k) decomposes g _{t, 1} − (t − 1, k) into more than one nonempty component. This means that d(t − 1; g _{t, 1}) ≥ 2, which is not possible. Consider the other case (ii). Since g _{t − 1, 2} satisfies [UNC], d(t − 1; g _{t − 1, 2}) = 1. Let the neighbor of player t − 1 be k ∈ N _t − 1, i.e., (t − 1, k) ∈ g _{t − 1, 2}. If d(k; g _{t − 1, 2}) = 1, g _{t, 2} − (t − 1, k) still satisfies [UNC]. If d(k; g _{t − 1, 2}) ≥ 2, it is a contradiction to (t, t − 1) ∈ g _{t, 1} due to Lemma 2(ii). □

Proof Proposition 1

For any t, define

$$ \begin{array}{rcl} N_{t-1}^{0}&=&\left\{i\in N_{t-2}\mid d(i; g_{t-1, 2})=0\right\},\\ N_{t-1}^{1}&=& \left\{i\in N_{t-2}\mid d(i; g_{t-1, 2})=1\right\},\\ N_{t-1}^{2}&=&\left\{i\in N_{t-2}\mid d(i; g_{t-1, 2})\geq 2\right\}. \end{array}$$

Then, $N_{t-1}=\{t-1\}\cup N_{t-1}^{0}\cup N_{t-1}^{1}\cup N_{t-1}^{2}$. Also, define

$$ \begin{array}{rcl} N_{t-1}^{1, 0}&=&\left\{i\in N_{t-1}^{1}\mid (i, j{\kern1.5pt})\in g_{t-1, 2}\; \mbox{for $j=t-1$}\right\},\\ N_{t-1}^{1, 2}&=&\left\{i\in N_{t-1}^{1}\mid (i, j{\kern1.5pt})\in g_{t-1, 2}\; \mbox{for $j\neq t-1$}\right\}.\end{array}$$

From Lemma 2, it is clear that (i) t − 1 ≻  _t i for $i\in N_{t-1}^{0}\cup N_{t-1}^{1, 0}$ and (ii) i ≻  _t t − 1 for $i\in N_{t-1}^{2}\cup N_{t-1}^{1, 2} \equiv \bar{N}_{t-1}$. If $\bar{N}_{t-1}\neq \emptyset$, it is optimal for player t to choose to link with player $i\in \bar{N}_{t-1}$. If $\bar{N}_{t-1}=\emptyset$, it is optimal for him to choose to link with player t − 1. Thus, the decision of agent t to choose to link with a player $i\in N_{t-1}^{2}$ if $\bar{N}_{t-1}\neq \emptyset$ and to link with a player t − 1 otherwise satisfies preferential attachment. □

Proof of Lemma 5

Suppose v _i  = L. If i = t − 1, d(i; g _{t − 1, 2}) = 1( < 2), so it must be that i < t − 1. Then, the links of player i except the link with t − 1 would have been deleted at latest by period t − 1. Thus, d(i; g _{t − 1, 2}) could be at most one. Contradiction. □

Proof of Lemma 6

(i) It is clear, because α _t (i) = α _t (j) = 1 by Lemma 5. (ii) This is also clear because α _t (i) = 1 > α _t (j). (iii) We have $\alpha_t (j{\kern1.5pt})=\frac{\alpha}{1+\alpha}$ for j such that d(j; g _{t − 1, 2}) = 0. On the other hand, consider player i such that d(i; g _{t − 1, 2}) = 0. Either i = t − 1 or i ≠ t − 1. If i = t − 1, $\mbox{Prob}(v_i =H)=\alpha$. If i ≠ t − 1, $\mbox{Prob}(v_i =H \mid i\neq t-1)=\frac{\alpha}{1+\alpha}$. Thus, $\alpha_t (i)=\theta \alpha+(1-\theta)\frac{\alpha}{1+\alpha}$ for some θ ∈ (0, 1). Since $\alpha >\frac{\alpha}{1+\alpha}$, $\alpha_t (i{\kern1.5pt})>\frac{\alpha}{1+\alpha}=\alpha_t (j{\kern1.5pt})$. □

Proof of Proposition 3

(i) $(\Longrightarrow )$ Under (A2), the myopic equilibrium is not viable, as argued in the text, so this is clear. $(\Longleftarrow)$ Note that $\frac{\delta}{2}(M-c)$ is the maximum gain attainable by postponing severing a link for one or more than one period, while c − L is the minimum loss from postponing severing a link. (Postponing severance for more than one period incurs c − L + δ(c − L) + ⋯.) Thus, if $c-L>\frac{\delta}{2}(M-c)$, no other deviation will be profitable, implying that the myopic equilibrium is sustainable. (ii) The proof is similar. □

Proof of Proposition 4

Let x(t) be the number of H types in N _t , that is, $x(t)=\sum_{i=1}^{t} x_i$ where x _i  = 1 if v _i  = H and x _i  = 0 if v _i  = L. Since E(x _i ) = α and $\mbox{Var}(x_i )=\alpha (1-\alpha)$, we have μ ≡ E(x(t)) = αt and $\sigma^2 \equiv \mbox{Var}(x(t))=\alpha (1-\alpha) t$.

Suppose that for some t ^* x(t ^* ) = n for some large n such that $L-c+\delta \frac{M-c}{n}<0$, or equivalently, $n>\delta \frac{M-c}{c-L}$. Then, for i, j ≤ t ^* − 1, link (i, j) with v _i  = L or v _j  = L will not survive the second stage of period t ^*. Knowing this, it is optimal for agent i (i ≥ t ^* + 1) to choose the equilibrium strategy of a myopic agent.

It remains to show that with probability one, there exists t ^* such that x(t ^* ) = n, in other words, that x(t) < n for all t with probability zero. By Chebyshev’s inequality, we have

$$ P(|x(t)-\mu| >k\sigma )\leq 1/k^2 , \; \mbox{for all $k>0$}.\label{eq2} $$

(2)

Let 1/k ² = ε, i.e., $k=1/\sqrt{\epsilon}$. Then, inequality (2) can be written as

$$ P\left(|x(t)-\mu|>\sigma /\sqrt{\epsilon}\,\right)\leq \epsilon, \; \forall \epsilon. $$

Note that inequality (2) holds for any t. Thus, we have $P\big(x(t)<\mu-\sigma/\sqrt{\epsilon}\big)<\epsilon$. Since μ = αt and σ ² = α(1 − α)t, we have

$$ P(x(t)<\bar{x}(t))<\epsilon, $$

where $\bar{x}(t)=\alpha t -\sqrt{\frac{\alpha (1-\alpha)t}{\epsilon}}$. Note that $d\bar{x}(t)/dt>0$ if t ≥ t ₁ for some large t ₁ and that $\lim_{t\rightarrow \infty} \bar{x}(t)=\infty$. Since $\lim_{t\rightarrow \infty} \bar{x}(t)=\infty$, we can take t ₂ such that $\bar{x}(t_2 )\equiv \alpha t_2 -\sqrt{\frac{\alpha (1-\alpha)t_2}{\epsilon}}>n$. Take $\bar{t}=\max\{t_1, t_2\}$. Then, $P(x(t)<n)<P\big(x(t)<\alpha t -\sqrt{\frac{\alpha (1-\alpha)t}{\epsilon}}\big)<\epsilon$ for all $t\geq \bar{t}$. This implies that there exists t ^* such that x (t ^* ) = n with probability one. □