In this section, we consider the majority rule model in the presence of ‘stubborn agents’. These are agents that never update their opinions. The other agents, referred to as the non-stubborn agents, are assumed to update their opinions at all points of the Poisson processes associated with themselves. We focus on the case where the updates occur according to the majority rule model. The voter model with stubborn agents was studied before in [34] using coalescing random walks. However, this technique does not apply to the majority rule model. We use mean field techniques to study the opinion dynamics under the majority rule model.
We denote by \(\gamma _i\), \(i \in \{0,1\}\), the fraction of agents in network who are stubborn and have opinion i at all times. Thus, \((1-\gamma _0-\gamma _1)\) is fraction of non-stubborn agents in the network. The presence of stubborn agents prevents the network from reaching a consensus state. This is because at all times there are at least \(N \gamma _0\) stubborn agents having opinion \(\{0\}\) and \(N \gamma _1\) stubborn agents having opinion \(\{1\}\). Furthermore, since each non-stubborn agent may interact with some stubborn agents at every update instant, it is always possible for the non-stubborn agent to change its opinion. Below we characterise the equilibrium fraction of non-stubborn agents having opinion \(\{1\}\) in the network for large N using mean field techniques. For analytical tractability, we consider the case \(K=1\), i.e., when an agent sample two agents at each update instant. However, similar results hold even for larger values of K.
Let \(x^{(N)}(t)\) denote the fraction of non-stubborn agents having opinion \(\{1\}\) at time \(t \ge 0\). Clearly, \(x^{(N)}(\cdot )\) is a Markov process with possible jumps at the points of a rate \(N(1-\gamma _0-\gamma _1)\) Poisson process. The process \(x^{(N)}(\cdot )\) jumps from the state x to the state \(x+1/N(1-\gamma _0-\gamma _1)\) when one of the non-stubborn agents having opinion \(\{0\}\) becomes active (which happens with rate \(N(1-\gamma _0-\gamma _1)(1-x)\)) and samples two agents with opinion \(\{1\}\). The probability of sampling an agent having opinion \(\{1\}\) from the entire network is \((1-\gamma _0-\gamma _1)x+\gamma _1\). Hence, the total rate at which the process transits from state x to the state \(x+1/N(1-\gamma _0-\gamma _1)\) is given by
$$\begin{aligned} q\left( {x \rightarrow x+\frac{1}{N(1-\gamma _0-\gamma _1)}}\right)&=N(1-\gamma _0-\gamma _1)(1-x) \nonumber \\&\quad \times [(1-\gamma _0-\gamma _1)x+\gamma _1]^2. \end{aligned}$$
(16)
Similarly, the rate of the other possible transition is given by
$$\begin{aligned} q\left( {x \rightarrow x-\frac{1}{N(1-\gamma _0-\gamma _1)}}\right)&=N(1-\gamma _0-\gamma _1)x \nonumber \\&\quad \times [(1-\gamma _0-\gamma _1)(1-x)+\gamma _0]^2. \end{aligned}$$
(17)
As in Theorem 2, it can be shown from the above transition rates that the process \(x^{(N)}(\cdot )\) converges weakly to the mean field limit \(x(\cdot )\) which satisfies the following differential equation
$$\begin{aligned} {\dot{x}}(t)= & {} (1-x(t))[(1-\gamma _0-\gamma _1)x(t)+\gamma _1]^2 \nonumber \\&-x(t) [(1-\gamma _0-\gamma _1)(1-x(t))+\gamma _0]^2. \end{aligned}$$
(18)
We now study the equilibrium distribution \(\pi _N\) of the process \(x^{(N)}(\cdot )\) for large N via the equilibrium points of the mean field \(x(\cdot )\).
From (18) we see that \({\dot{x}}(t)\) is a cubic polynomial in x(t). Hence, the process \(x(\cdot )\) can have at most three equilibrium points in [0, 1]. We first characterise the stability of these equilibrium points.
Proposition 1
The process \(x(\cdot )\) defined by (18) has at least one equilibrium point in (0, 1). Furthermore, the number of stable equilibrium points of \(x(\cdot )\) in (0, 1) is either two or one. If there exists only one equilibrium point of \(x(\cdot )\) in (0, 1), then the equilibrium point must be globally stable (attractive).
Proof
Define \(f(x)=(1-x)[(1-\gamma _0-\gamma _1)x+\gamma _1]^2-x[(1-\gamma _0-\gamma _1)(1-x)+\gamma _0]^2\). Clearly, \(f(0)=\gamma _1^2 > 0\) and \(f(1)=-\gamma _0^2 < 0\). Hence, there exists at least one root of \(f(x)=0\) in (0, 1). This proves the existence of an equilibrium point of \(x(\cdot )\) in (0, 1).
Since f(x) is a cubic polynomial and \(f(0)f(1) < 0\), either all three roots of \(f(x)=0\) lie in (0, 1) or exactly one root of \(f(x)=0\) lies in (0, 1). Let the three (possibly complex and non-distinct) roots of \(f(x)=0\) be denoted by \(r_1, r_2, r_3\), respectively. By expanding f(x) we see that the coefficient of the cubic term is \(-2(1-\gamma _0-\gamma _1)^2\). Hence, f(x) can be written as
$$\begin{aligned} f(x)=-2(1-\gamma _0-\gamma _1)^2(x-r_1)(x-r_2)(x-r_3). \end{aligned}$$
(19)
We first consider the case when \(0< r_1, r_2, r_3 <1\) and not all of them are equal. Let us suppose, without loss of generality, that the roots are arranged in the increasing order, i.e., \(0< r_1 \le r_2< r_3 < 1\) or \(0< r_1< r_2 \le r_3 < 1\). From (19) and (18), it is clear that, if \(x(t)> r_2\) and \(x(t) > r_3\), then \({\dot{x}}(t) < 0\). Similarly, if \(x(t)> r_2\) and \(x(t) < r_3\), then \({\dot{x}}(t) > 0\) . Hence, if \(x(0) > r_2\) then \(x(t) \rightarrow r_3\) as \(t \rightarrow \infty \). Using similar arguments we have that for \(x(0) < r_2\), \(x(t) \rightarrow r_1\) as \(t \rightarrow \infty \). Hence, \(r_1,r_3\) are the stable equilibrium points of \(x(\cdot )\). This proves that there exist at most two stable equilibrium points of the mean field \(x(\cdot )\).
Now suppose that there exists only one equilibrium point of \(x(\cdot )\) in (0, 1). This is possible either i) if there exists exactly one real root of \(f(x)=0\) in (0, 1), or ii) if all the roots of \(f(x)=0\) are equal and lie in (0, 1). Let \(r_1\) be a root of \(f(x)=0\) in (0, 1). Now by expanding f(x) from (19), we see that the product of the roots must be \(\gamma _1^2/2(1-\gamma _0-\gamma _1)^2 >0\). This implies that the other roots, \(r_2\) and \(r_3\), must satisfy one of the following conditions: 1) \(r_2, r_3 >1\), 2) \(r_2, r_3 < 0\), 3) \(r_2, r_3\) are complex conjugates, 4) \(r_2=r_3=r_1\).
In all the above cases, we have that \((x-r_2)(x-r_3) \ge 0\) for all \(x \in [0,1]\) with equality if and only if \(x=r_1=r_2=r_3\). Hence, from (19) and (18), it is easy to see that \({\dot{x}}(t) > 0\) when \(0 \le x(t) < r_1\) and \({\dot{x}}(t) < 0\) when \(1 \ge x(t) > r_1\). This implies that \(x(t) \rightarrow r_1\) for all \(x(0) \in [0,1]\). In other words, \(r_1\) is globally stable. \(\square \)
In the next proposition, we provide the conditions on \(\gamma _0\) and \(\gamma _1\) for which there exist multiple stable equilibrium points of the mean field \(x(\cdot )\).
Proposition 2
There exist two distinct stable equilibrium points of the mean field \(x(\cdot )\) in (0, 1) if and only if
-
1.
\(D(\gamma _0,\gamma _1)=(\gamma _0-\gamma _1)^2+3(1-2\gamma _0-2\gamma _1) > 0\)
-
2.
\(0< z_1, z_2 < 1\), where
$$\begin{aligned} z_1&= \frac{(3-\gamma _0-5\gamma _1)+ \sqrt{D(\gamma _0,\gamma _1)}}{6(1-\gamma _0-\gamma _1)}, \end{aligned}$$
(20)
$$\begin{aligned} z_2&= \frac{(3-\gamma _0-5\gamma _1)- \sqrt{D(\gamma _0,\gamma _1)}}{6(1-\gamma _0-\gamma _1)}. \end{aligned}$$
(21)
-
3.
\(f(z_1)f(z_2) \le 0\), where \(f(x)=(1-x)[(1-\gamma _0-\gamma _1)x+\gamma _1]^2-x[(1-\gamma _0-\gamma _1)(1-x)+\gamma _0]^2\).
If any one of the above conditions is not satisfied then \(x(\cdot )\) has a unique, globally stable equilibrium point in (0, 1).
Proof
From Proposition 1, we have seen that \(x(\cdot )\) has two stable equilibrium points in (0, 1) if and only if \(f(x)=0\) has three real roots in (0, 1) among which at least two are distinct. This happens if and only if \(f'(x)=0\) has two distinct real roots \(z_1, z_2\) in the interval (0, 1) and \(f(z_1) f(z_2) \le 0\). Since \(f'(x)\) is a quadratic polynomial in x, the above conditions are satisfied if and only if
-
1.
The discriminant of \(f'(x)=0\) is positive. This corresponds to the first condition of the proposition.
-
2.
The two roots \(z_1,z_2\) of \(f'(x) =0\) must lie in (0, 1). This corresponds to the second condition of the proposition.
-
3.
\(f(z_1) f(z_2) \le 0\). This is the third condition of the proposition.
Clearly, if any one of the above conditions is not satisfied, then \(x(\cdot )\) has a unique equilibrium point in (0, 1). According to Proposition 1 this equilibrium point must be globally stable. \(\square \)
Hence, depending on the values of \(\gamma _0\) and \(\gamma _1\) there may exist of multiple stable equilibrium points of the mean field \(x(\cdot )\). However, for every finite N, the process \(x^{(N)}(\cdot )\) has a unique stationary distribution \(\pi _N\) (since it is irreducible on a finite state space). In the next result, we establish that any limit point of the sequence of stationary probability distributions \((\pi _N)_N\) is a convex combination of the Dirac measures concentrated on the equilibrium points of the mean field \(x(\cdot )\) in [0, 1].
Theorem 6
Any limit point of the sequence of probability measures \((\pi _N)_N\) is a convex combination of the Dirac measures concentrated on the equilibrium points of \(x(\cdot )\) in [0, 1]. In particular, if there exists a unique equilibrium point r of \(x(\cdot )\) in [0, 1] then \(\pi _N \Rightarrow \delta _r\), where \(\delta _{r}\) denotes the Dirac measure concentrated at the point r.
Proof
We first note that since the sequence of probability measures \((\pi _N)_N\) is defined on the compact space [0, 1], it must be tight. Hence, Prokhorov’s theorem implies that \((\pi _N)_N\) is relatively compact. Let \(\pi \) be any limit point of the sequence \((\pi _N)_N\). Then by the mean field convergence result we know that \(\pi \) must be an invariant distribution of the maps \(\alpha \mapsto x(t,\alpha )\) for all \(t \ge 0\), i.e., \(\int \varphi (x(t,\alpha ))d\pi (\alpha )=\int \varphi (\alpha )d\pi (\alpha )\), for all \(t \ge 0\), and all continuous (and hence bounded) functions \(\varphi : [0,1] \mapsto \mathbb {R}\). In the above, \(x(t,\alpha )\) denotes the process \(x(\cdot )\) started at \(x(0)=\alpha \). Hence we have
$$\begin{aligned} \int \varphi (\alpha )d\pi (\alpha )&= \lim _{t \rightarrow \infty } \int \varphi (x(t,\alpha ))d\pi (\alpha ) \end{aligned}$$
(22)
$$\begin{aligned}&= \int \varphi \left( {\lim _{t \rightarrow \infty } x(t,\alpha )}\right) d\pi (\alpha ). \end{aligned}$$
(23)
The second equality follows from the first by the Dominated convergence theorem and the continuity of \(\varphi \). Now, let \(r_1, r_2\), and \(r_3\) denote the three equilibrium points of the mean field \(x(\cdot )\). Hence, by Proposition 1 we have that for each \(\alpha \in [0,1]\), \(\varphi (\lim _{t \rightarrow \infty }x(t,\alpha )) =\varphi (r_1) I_{N_{r_1}}(\alpha )+\varphi (r_2) I_{N_{r_2}}(\alpha )+\varphi (r_3) I_{N_{r_3}}(\alpha )\), where for \(i=1,2,3\), \(N_{r_i} \in [0,1]\) denotes the set for which if \(x(0) \in N_{r_i}\) then \(x(t) \rightarrow r_i\) as \(t \rightarrow \infty \), and I denotes the indicator function. Hence, by (23) we have that for all continuous functions \(\varphi : [0,1] \mapsto \mathbb {R}\)
$$\begin{aligned} \int \varphi (\alpha )d\pi (\alpha ) = \varphi (r_1) \pi (N_{r_1})+\varphi (r_2) \pi (N_{r_2})+\varphi (r_3) \pi (N_{r_3}). \end{aligned}$$
(24)
This proves that \(\pi \) must be of the form \(\pi =c_1 \delta _{r_1}+c_2 \delta _{r_2}+c_3 \delta _{r_3}\), where \(c_1, c_2, c_3 \in [0,1]\) are such that \(c_1+c_2+c_3=1\). This completes the proof. \(\square \)
Thus, according to the above theorem, if there exists a unique equilibrium point of the process \(x(\cdot )\) in [0,1], then the sequence of stationary distributions \((\pi _N)_N\) concentrates on that equilibrium point as \(N \rightarrow \infty \). In other words, for large N, the fraction of non-stubborn agents having opinion \(\{1\}\) (at equilibrium) will approximately be equal to the unique equilibrium point of the mean field.
Simulation Results
In Fig. 5, we plot the equilibrium point of \(x(\cdot )\) (when it is unique) as a function of the fraction \(\gamma _1\) of agents having opinion \(\{1\}\) who are stubborn keeping the fraction \(\gamma _0\) of stubborn agents having opinion \(\{0\}\) fixed. We choose the parameter values so that there exists a unique equilibrium point of \(x(\cdot )\) in [0, 1] (such parameter settings can be obtained using the conditions of Proposition 2). We see that as \(\gamma _1\) is increased in the range \((0,1-\gamma _0)\), the equilibrium point shifts closer to unity. This is expected since increasing the fraction of stubborn agents with opinion \(\{1\}\) increases the probability with which a non-stubborn agent samples an agent with opinion \(\{1\}\) at an update instant.
If there exist multiple equilibrium points of the process \(x(\cdot )\) then the convergence \(x^{(N)}(\cdot ) \Rightarrow x(\cdot )\) implies that at steady state the process \(x^{(N)}(\cdot )\) spends intervals near the region corresponding to one of the stable equilibrium points of \(x(\cdot )\). Then due to some rare events, it reaches, via the unstable equilibrium point, to a region corresponding to the other stable equilibrium point of \(x(\cdot )\). This fluctuation repeats giving the process \(x^{(N)}(\cdot )\) a unique stationary distribution. This behavior is formally known as metastability.
To demonstrate metastability, we simulate a network with \(N=100\) agents and \(\gamma _0=\gamma _1=0.2\). For the above parameters, the mean field \(x(\cdot )\) has two stable equilibrium points at 0.127322 and 0.872678. In Fig. 6, we show the sample path of the process \(x^{(N)}(\cdot )\). We see that at steady state the process switches back and forth between regions corresponding to the stable equilibrium points of \(x(\cdot )\). This provides numerical evidence of the metastable behavior of the finite system.