2. Model

In this section, we describe the model studied in this paper. The model consists of a network of n agents described by an undirected graph $G_n=(V_n,E_n)$, where the nodes in V _n (with $\lvert V_n \rvert=n$) represent the agents and the edges in E _n represent the connections between the agents. For each agent (node) $u \in V_n$, we denote by $N_u = \left\{v:(u,v)\in E_n\right\}$, the set of neighbors of u.

Time is assumed to be discrete and at each discrete time instant $t \in \mathbb{Z}_+$, each agent is assumed to have an opinion in the set $\{0,1\}$. Without loss of generality, we assume that 1 is the superior opinion. Let $X_u(t) \in \{0,1\}$ denote the opinion of agent u at time t. At t = 0, the opinions of the agents are initialized such that $\vert \{u\in V_n: X_u(0)=1\}\vert = \lceil pn \rceil$ for some $p \in [0,1)$. Hence, p fraction of agents initially have the superior opinion 1.

At each instant $t \geq 0$, an agent, sampled uniformly at random, updates its opinion: with probability α, it performs a biased update in which it adopts the superior opinion irrespective of its current opinion and the opinions of all other agents; with probability $1-\alpha$, it performs a regular update following the 2-choices rule in which the updating agent samples two neighbors uniformly at random (with replacementFootnote ¹) and adopts the majority opinion among the sampled agents and the agent itself. Therefore, if U(t) denotes the randomly sampled agent at time t, then at time t + 1 the opinion of the agent is given by

(1)\begin{equation} X_{U(t)}(t+1)=\begin{cases} 1 & \text{w.p. } \alpha, \\ M(t) & \text{w.p. } 1-\alpha, \end{cases} \end{equation}

where $M(t)={1}{\left(X_{U(t)}(t)+X_{N_1(t)}(t)+X_{N_2(t)}(t)\geq 2\right)}$ denotes the majority opinion among two randomly sampled neighbors $N_1(t)$ and $N_2(t)$ of U(t) and the agent U(t) itself. The parameter $\alpha \in (0,1]$ represents the bias toward the superior opinion and is referred to as the bias parameter of the model.

The state of the network at any time $t \geq 0$ can be represented by the vector $\mathbf{X}(t)=(X_u(t), u \in V_n)$ of opinions of all the agents. The process $\mathbf{X}(\cdot)$ is Markov on the state space $\left\{0,1\right\}^n$ with a single absorbing state 1 where all agents have opinion 1. We refer to this absorbing state as the consensus state. Since it is possible to reach the consensus state from any other state in a finite number of steps, with probability one, the chain is absorbed in the consensus state in a finite time. We refer to this time as the consensus time. The objective of the rest of the paper is to analyze the mean consensus time for different values of the parameters $n,\alpha,p$ and for different classes of graphs.

3. Analysis for complete graphs

In this section, we assume that G _n is a complete graph on n nodes. For complete graphs, the process $\bar X^n=(\bar X^n(t)=\vert \{u\in V_n: X_u(t)=1\} \vert,\ t\geq 0)$ counting the number of agents with opinion 1 is a Markov chain on $\mathbb{Z}_+$ with absorbing state n. For the chain $\bar X^n$, the transition probability $\tilde{p}_{i,j}$ from state i to j is given by $\tilde p_{i,j}={p}_{i,j}+o(1/n)$, where

(2)\begin{equation} p_{i,j}=\begin{cases} \left(1-\frac{i}{n}\right)\left(\alpha+(1-\alpha)\left(\frac{i}{n}\right)^2\right), &\text{if } j=i+1,\\ \frac{i}{n}(1-\alpha)\left(1-\frac{i}{n}\right)^2, &\text{if } j=i-1,\\ 1-p_{i,i+1}-p_{i,i-1}, &\text{if } j=i,\\ 0, &\text{otherwise.} \end{cases} \end{equation}

Note that $p_{i,j}$ denotes the transition probability of a slightly modified chain $\bar Y^n$ (with the same absorbing state n), where an agent during a regular update can sample itself in addition to its neighbors. Further we analyze this modified chain as it is simpler to do so and the asymptotic (in n) results we obtain for this modified chain also hold for the original chain (see Remark 3.7 for more explanation).

Below we characterize the scaling law of $\bar T_n(p)=\mathbb{E}_{\lceil pn \rceil}[T_n]$, where $\mathbb{E}_{x}\left[\cdot\right]$ denotes expectation conditioned on $\bar Y^n(0)=x$ and $T_k=\inf\{t\geq 0: \bar Y^n(t)=k\}$ denotes the first time the network reaches state k. Our main result is described in Theorem 3.1.

The above theorem implies that when the bias parameter is sufficiently high ( $\alpha \gt1/9$), the network quickly (in $\Theta(n\, \log n)$ time) reaches consensus on the superior opinion irrespective of its initial state. This is in sharp contrast to the result of [Reference Anagnostopoulos, Becchetti, Cruciani, Pasquale and Rizzo2], where the consensus time is exponential in n for all values of the bias parameter α. The theorem further implies that even with low value of the bias parameter α (for $\alpha \lt 1/9$), fast consensus can be achieved as long as the initial proportion p of agents with the superior opinion is above a threshold value denoted by $p_c(\alpha)$. We explicitly characterize this threshold $p_c(\alpha)$ required to ensure fast consensus when the bias is low. If the bias parameter is low ( $\alpha \lt 1/9$) and the initial proportion of agents with the superior opinion is below the threshold, that is, $p \lt p_c(\alpha)$, then the mean consensus time is exponential in n, which corresponds to slow speed of convergence. Thus, our model exhibits rich behavior in terms of the parameters α and p.

We breakdown the proof of the above theorem into several simpler steps. The first step is to make the following simple observation, which expresses the mean consensus time as a function of the number of visits of the chain $\bar Y^n$ to different states in its state-space $\{0,1,\ldots,n\}$.

Lemma 3.2 Let Z _k denote the number of visits of the chain $\bar Y^n$ to the state $k \in \{0,1,\ldots,n\}$ before absorption. Then, for any starting state $x =\bar{X}^n(0)\in \{0,1,\ldots,n\}$, the average consensus time is given by

(4)\begin{equation} \mathbb{E}_{x}\left[T_n\right] =\sum_{k=0}^{n-1} \frac{\mathbb{E}_{x}\left[Z_k\right]}{(1-p_{k,k})}, \end{equation}

where $p_{k,k}$ is the transition probability from state k to itself given by Eq. (2).

Proof. Observe that the consensus time T _n can be written as

(5)\begin{equation} T_n=\sum_{k=0}^{n-1}\sum_{j=1}^{Z_k} M_{k,j}, \end{equation}

where Z _k denotes the number of visits to state k before absorption and $M_{k,j}$ denotes the time spent in state k in the $j{\textrm{th}}$ visit. Clearly, the random variables Z _k and $(M_{k,j})_{j\geq 1}$ are independent of each other. Furthermore, $(M_{k,j})_{j\geq 1}$ is a sequence of i.i.d. random variables with geometric distribution given by $\mathbb{P}_{x}(M_{k,j}=i)=p_{k,k}^{i-1}(1-p_{k,k})$ for all j. Hence, applying Wald’s identity to Eq. (5), we have

(6)\begin{align} \bar T_{n}(p) &=\mathbb{E}_{x}\left[T_n\right] \nonumber\\ &=\sum_{k=0}^{n-1}\mathbb{E}_{x}\left[\sum_{j=1}^{Z_k}M_{k,j}\right]\nonumber\\ &=\sum_{k=0}^{n-1} \mathbb{E}_{x}\left[Z_k\right] \mathbb{E}_{x}\left[M_{k,j}\right] \nonumber\\ &=\sum_{k=0}^{n-1} \frac{\mathbb{E}_{x}\left[Z_k\right]}{(1-p_{k,k})}, \end{align}

where the last step follows from the fact that $\mathbb{E}_{x}[M_{k,j}]=1/(1-p_{k,k})$ for each $j \in [Z_k]$.

From the above lemma, it is evident that in order to obtain bounds on $\mathbb{E}_{x}[T_n]$, we need to obtain bounds of the expected number of visits $\mathbb{E}_{x}[Z_k]$ to different states and the transition probabilities $p_{k,k}$. Using this approach, we first obtain a lower bound on $\bar T_n(p)$.

Proof. From Eq. (2), we obtain

\begin{align*} 1-p_{k,k} &= \left(1-\frac{k}{n}\right)\left(\alpha+\left(1-\alpha\right)\frac{k}{n}\right)\\ &\leq \max\left(\alpha,1-\alpha\right) \left(1-\frac{k}{n}\right)\left(1+\frac{k}{n}\right).r_\alpha \end{align*}

Furthermore, we have $\mathbb{E}_{x}[Z_k] \geq {1}{(k \geq x)}$ since states $k \geq x$ are visited at least once. Hence, using the above two inequalities in Eq. (4), we obtain

\begin{align*} \bar T_n(p) & \geq \sum_{k=0}^{n-1} \frac{{1}{\left(k \geq \lceil np \rceil\right)}}{\max\left(\alpha,1-\alpha\right) \left(1-\frac{k}{n}\right)\left(1+\frac{k}{n}\right)}\\ & = \frac{n}{2\max\left(\alpha,1-\alpha\right)}\sum_{k=\lceil np \rceil}^{n-1}\left(\frac{1}{n-k}+\frac{1}{n+k}\right)\\ & \gt \frac{n}{2\max\left(\alpha,1-\alpha\right)}\sum_{k=\lceil np \rceil}^{n-1}\left(\frac{1}{n-k}\right)\\ & \geq \frac{n}{2\max\left(\alpha,1-\alpha\right)}\log\left(n-\lceil np \rceil+1\right), \end{align*}

which completes the proof.

To obtain an upper bound on the mean consensus time $\bar T_n(p)$ similarly, we need an upper bound on the expected number of visits $\mathbb{E}_{x}\left[Z_k\right]$ to state k for each $k \in \left\{0,1,2,\ldots,n\right\}$. We obtain such an upper bound using a technique developed in [Reference Mukhopadhyay, Mazumdar and Roy20], where the number of visits to each state is expressed as a function of a branching process. Specifically, we define ζ _k to be the number of jumps from state k to state k − 1 for any $k\in \{1,2,\ldots,n\}$. Clearly, $\zeta_n=0$, and if the starting state is $x=\bar Y^n(0) \in [1,n)$, then ζ _k satisfies the following recursion:

(7)\begin{equation} \zeta_k=\begin{cases} \sum_{l=0}^{\zeta_{k+1}}\xi_{l,k}, &\text{if } k \in \{x,x+1,\ldots,n-1\},\\ \sum_{l=1}^{\zeta_{k+1}}\xi_{l,k}, &\text{if } k \in \{0,1,\ldots,x-1\}, \end{cases} \end{equation}

where $\xi_{l,k}$ denotes the number of jumps from state k to state k − 1 between the $l{\textrm{th}}$ and $(l+1){\textrm{th}}$ visit to state k + 1. Note that the first sum in Eq. (7) (for $k \geq x$) starts from l = 0, whereas the second sum starts from l = 1. This is because for $k\geq x$, jumps from k to k − 1 can occur even before the first jump from k + 1 to k. This is not the case for states k < x. Furthermore, $\{\xi_{l,k}\}_{l\geq 0}$ is a sequence of i.i.d. random variables independent of $\zeta_{k+1}$. Hence, the sequence $\{\xi_{n-k}\}_k$ defines a branching process. Moreover, the number of visits to state k can be written as

(8)\begin{equation} {Z_k} = \begin{cases} 1+\zeta_k+\zeta_{k+1}, &\text{if } k \in \{x,x+1,\ldots,n-1\},\\ \zeta_k+\zeta_{k+1}, &\text{if } k \in \{0,1,\ldots,x-1\}, \end{cases} \end{equation}

where the additional one appears in the first case (for $k \geq x$) because the states $k \geq x$ are visited at least once before the chain is absorbed in state n. It is the above characterization of the Z _k that we shall use to obtain upper bounds on $\mathbb{E}_{x}[Z_k]$. Precisely, we obtain bounds on $\mathbb{E}_{x}[\zeta_k]$ to bound $\mathbb{E}_{x}[Z_k]$. In the lemma below, we express $\mathbb{E}_{x}[\zeta_k]$ in terms of the transition probabilities of the chain $\bar{Y}^n$.

Lemma 3.4 For any $k\in \{1,2,\ldots,n\}$, let ζ _k denote the number of jumps of the chain $\bar Y^n$ from state k to state k − 1. We have $\zeta_n=0$ and

(9)\begin{equation} \mathbb{E}_{x}\left[\zeta_{k}\right]=\begin{cases} \sum_{t=k}^{n-1} \prod_{i=k}^{t} \frac{p_{i,i-1}}{p_{i,i+1}}, \quad \text{for } x \leq k \leq n-1,\\ \left(\prod_{i=k}^{x-1} \frac{p_{i,i-1}}{p_{i,i+1}}\right)\mathbb{E}_{x}\left[\zeta_{x}\right], \quad \text{for } 0 \leq k \lt x, \end{cases} \end{equation}

where for each $i, j \in \{0,1,\ldots,n\}$, $p_{i,j}$ denotes the transition probability of the Markov chain $\bar Y^n$ from state i to state j and is given by Eq. (2).

Proof. We observe that $\zeta_{k+1}$ and $\left\{\xi_{l,k}\right\}_{l\geq 0}$ are independent of each other and $\left\{\xi_{l,k}\right\}_{l\geq 0}$ is a sequence of i.i.d. random variables with

\begin{equation*}\mathbb{P}_{x}\left(\xi_{l,k}=i\right)=\left(\frac{p_{k,k-1}}{1-p_{k,k}}\right)^i\left(\frac{p_{k,k+1}}{1-p_{k,k}}\right)\end{equation*}

for all $i \in \mathbb{Z}_+$ and all $l \geq 0$. Hence, we have

\begin{equation*}\mathbb{E}_{x}\left[\xi_{l,k}\right]=\frac{p_{k,k-1}}{p_{k,k+1}},\end{equation*}

for all $l \geq 0$. Applying Wald’s identity to Eq. (7), we obtain the following recursions

(10)\begin{equation} \mathbb{E}_{x}\left[\zeta_k\right]=\begin{cases} \left(1+\mathbb{E}_{x}\left[\zeta_{k+1}\right]\right)\frac{p_{k,k-1}}{p_{k,k+1}}, &\text{if } k \geq x,\\ \mathbb{E}_{x}\left[\zeta_{k+1}\right]\frac{p_{k,k-1}}{p_{k,k+1}}, &\text{if } k \lt x, \end{cases} \end{equation}

Upon solving the above recursions with boundary condition ${\zeta_n}=0$, we obtain desired result.

From Eq. (9), we observe that the ratio $p_{i,i-1}/p_{i,i+1}$ plays a crucial role in the expression of $\mathbb{E}_{x}\left[\zeta_k\right]$. Hence, by characterizing this ratio, we can characterize $\mathbb{E}_{x}\left[\zeta_k\right]$. We note from Eq. (2) that

(11)\begin{equation} \frac{p_{i,i-1}}{p_{i,i+1}}=f_\alpha(i/n), \end{equation}

where $f_\alpha:[0,1]\to \mathbb{R}_+$, for each $\alpha \in (0,1)$, is as defined in Theorem 3.1. In the lemma below, we obtain a list of some important properties of the function f _α.

Lemma 3.5 For $\alpha \in (0,1)$, define $f_\alpha:[0,1]\to \mathbb{R}_+$

(12)\begin{equation} f_\alpha(x)\triangleq\frac{(1-\alpha)x(1-x)}{\alpha+(1-\alpha)x^2}. \end{equation}

Then f _α satisfies the following properties

1. 1. For all $\alpha \in (0,1)$, f _α is strictly increasing in $[0,x_\alpha)$, strictly decreasing in $(x_\alpha, 1]$, and, in the domain $[0,1]$, attains its maximum value at x _α, where

   (13)\begin{equation} x_\alpha\triangleq \sqrt{\left(\frac{\alpha}{1-\alpha}\right)^2+\frac{\alpha}{1-\alpha}}-\frac{\alpha}{1-\alpha} \in [0,1). \end{equation}

2. 2. Let $r_\alpha\triangleq f_\alpha(x_\alpha)=\max_{x \in [0,1]}f_\alpha(x)$. Then, for $\alpha \in (1/9,1)$, we have $r_\alpha \lt 1$, and for $\alpha \in (0,1/9)$, we have $r_\alpha \gt 1$.

3. 3. For $\alpha \in (0,1/9]$, we have $f_\alpha(x) \geq 1$ iff $x \in [\underline{x}_{\alpha},\bar{x}_\alpha]$, where

   (14)\begin{align} \underline{x}_\alpha &= \frac{1}{4}\left(1-\sqrt{1-\frac{8\alpha}{1-\alpha}}\right), \end{align}

   (15)\begin{align} \bar x_\alpha &=\frac{1}{4}\left(1+\sqrt{1-\frac{8\alpha}{1-\alpha}}\right). \end{align}

   Furthermore, $f_\alpha(\underline{x}_\alpha)=f_\alpha(\bar{x}_\alpha)=1$ and $x_\alpha \in [\underline{x}_{\alpha},\bar{x}_\alpha]$.

4. 4. For $\alpha \in (0,1/9)$, define $g_\alpha(x)\triangleq\int_{\underline{x}_\alpha}^x \log(\ f_\alpha(x))\,{\rm d}x$. Then g _α has a unique root $p_c(\alpha) \in (\bar x_\alpha,1)$. Furthermore, $g_\alpha(p) \gt 0$ if $p \in [\bar x_\alpha, p_c(\alpha))$ and $g_\alpha(p) \lt 0$ if $p \in (p_c(\alpha),1).$

Proof. Taking the derivative of Eq. (12) with respect x, we obtain

\begin{equation*} f_\alpha^\prime(x)=\frac{-(1-\alpha)^2 x^2-2\alpha(1-\alpha)x+\alpha(1-\alpha)}{\left(\alpha+(1-\alpha)x^2\right)^2}. \end{equation*}

We note that the numerator of the above expression is zero when $x=x_\alpha$, positive when $x \in [0, x_\alpha)$, and negative when $x \in (x_\alpha, 1]$, where x _α is as defined in Eq. (13). This proves the first statement of the lemma.

Next, we note from Eq. (12) that the condition $f_\alpha(x) \lesseqgtr 1$ is equivalent to $2x^2-x+\frac{\alpha}{1-\alpha} \gtreqless 0$. Hence, $f_\alpha(x) \lt 1$ for all $x \in [0,1]$ if and only if $1-8\frac{\alpha}{1-\alpha} \lt0$ or equivalently iff $\alpha \gt 1/9$. Furthermore, for $\alpha \in (0,1/9]$, we have $2x^2-x+\frac{\alpha}{1-\alpha}=(x-\underline{x}_\alpha)(x-\bar x_\alpha)$, where $\underline{x}_\alpha$ and $\bar{x}_\alpha$ are as defined in Eqs. (14) and (15), respectively, and for $\alpha \lt 1/9$, we have $\underline{x}_\alpha \lt \bar{x}_\alpha$ and $x_\alpha \in (\underline{x}_\alpha,\bar{x}_\alpha)$. Combining the above facts, we have the second and the third statements of the lemma.

We now turn to the last statement of the lemma. Note that for $\alpha \in (0,1/9)$, we have

\begin{equation*}g_\alpha(\bar x_\alpha)=\int_{\underline{x}_\alpha}^{\bar x_\alpha}\, \log(f_\alpha(x))\,{\rm dx} \geq \log(f_\alpha(x_\alpha)) =\log(r_\alpha) \gt 0,\end{equation*}

where the last inequality follows from the fact that $r_\alpha=f_\alpha(x_\alpha) \gt 1$ for $\alpha \in (0,1/9)$ as established in the previous paragraph. Using Eq. (12), we can compute g _α in closed form. This is given as follows:

(16)\begin{align} g_\alpha(x) &= x\,\log\left(\frac{\left(1-\alpha\right)x}{\alpha+\left(1-\alpha\right)x^2}\right)- \left(1-x\right)\,\log\left(1-x\right)+\log\left(1-\underline{x}_\alpha\right) \nonumber\\ & -2\sqrt{\frac{\alpha}{1-\alpha}}\left(\arctan\left(\sqrt{\frac{1-\alpha}{\alpha}}x\right)-\arctan\left(\sqrt{\frac{1-\alpha}{\alpha}}\underline{x}_\alpha\right)\right). \end{align}

From the above, it follows that $\lim_{x \to 1^{-}} g(x) \lt 0$, since $\underline{x}_\alpha \lt 1$ and $\arctan(\cdot)$ is an increasing function. Furthermore, we have $g^{\prime}_\alpha(x)=\log(f_\alpha(x)) \lt 0$ for $x\in (\bar x_\alpha,1)$. Hence, there must exist a unique root $p_c(\alpha)$ of g _α in $(\bar x_\alpha, 1)$, and $g_\alpha(p)$ must be strictly negative for $p \in (p_c(\alpha),1)$ and strictly positive for $p \in [\bar{x}_\alpha,p_c(\alpha))$.

We are now in a position to complete the proof of Theorem 3.1. We use Lemma 3.4 and the properties of f _α proved in Lemma 3.5 to obtain upper bounds of $\mathbb{E}_{x}\left[\zeta_k\right]$. These upper bounds, in turn, provide upper bounds on $\mathbb{E}_{x}\left[Z_k\right]$ using Eq. (8). Finally, we use Lemma 3.2 and the upper bounds on $\mathbb{E}_{x}[Z_k]$ to obtain an upper bound on the mean consensus time $\bar T_n(p)$. The complete proof is given below.

Proof of Theorem 3.1

From Lemma 3.2, it follows that

(17)\begin{align} \bar T_n(p) &=\sum_{k=0}^{n-1} \frac{\mathbb{E}_{\lceil pn \rceil}\left[Z_k\right]}{(1-p_{k,k})} \nonumber\\ &= \sum_{k=0}^{n-1} \frac{\mathbb{E}_{\lceil pn \rceil}\left[Z_k\right]}{\left(1-\frac{k}{n}\right)\left(\alpha+\left(1-\alpha\right)\frac{k}{n}\right)} \nonumber\\ &=n\sum_{k=0}^{n-1} \mathbb{E}_{\lceil pn \rceil}\left[Z_k\right]\left(\frac{1}{n-k}+\frac{1}{\frac{\alpha}{1-\alpha}n+k}\right). \end{align}

Hence, to prove $T_n(p)=O(n\, \log n)$, it is sufficient to establish that $\mathbb{E}_{\lceil p n \rceil}\left[Z_k\right] \leq C$ for all $k \in \{0,1,\ldots,n\}$, where C > 0 is a constant independent of k. Indeed, if $\mathbb{E}_{\lceil p n \rceil}\left[Z_k\right] \leq C$ for each state $k \in \{0,1,\ldots,n\}$, then for $n \geq \frac{1-\alpha}{\alpha}$, we have

(18)\begin{align} \bar T_n(p) & \leq n {C}\sum_{k=0}^{n-1} \left(\frac{1}{n-k}+\frac{1}{k+1}\right) \nonumber \\ &=2 n {C}\sum_{k=1}^{n-1}\frac{1}{k} \nonumber\\ &=O\left(n\,\log n\right). \end{align}

Hence, to prove the first two statements of the theorem, it is sufficient to show that, under the conditions stated in these statements, $\mathbb{E}_{\lceil pn \rceil}\left[Z_k\right]$ is uniformly bounded by some constant for all $k \in \left\{0,1,\ldots,n-1\right\}$. This is what we prove next.

Using the first two properties of f _α proved in Lemma 3.5 and Eq. (9), we see that for $\alpha \in (1/9,1)$ we have

(19)\begin{equation} \mathbb{E}_{x}\left[\zeta_{k}\right]\leq \begin{cases} \sum_{t=k}^{n-1} r_\alpha^{t-k+1} \lt \frac{r_\alpha}{1-r_\alpha}, &\quad \text{for } x \leq k \leq n-1,\\ \mathbb{E}_{x}\left[\zeta_{x}\right] \lt \frac{r_\alpha}{1-r_\alpha}, &\quad \text{for } k \lt x, \end{cases} \end{equation}

where $r_\alpha \lt 1$ is as defined in Lemma 3.5. Hence, from Eq. (8), we have that for all $\alpha \in (1/9,1)$ and all $k \in \left\{0,1,\ldots,n-1\right\}$,

\begin{equation*}\mathbb{E}_{x}\left[Z_k\right]\leq \frac{1+r_\alpha}{1-r_\alpha}.\end{equation*}

This establishes the first statement of the theorem.

We now prove the second statement of the theorem. For $\alpha \in (0,1/9)$, the third statement of Lemma 3.5 implies $p_c(\alpha) \gt \bar x_\alpha$. Furthermore, by the first and third statements of Lemma 3.5, it follows that f _α is strictly decreasing in the range $[\bar x_\alpha,1)$. Hence, for $p \geq p_c(\alpha) \gt \bar x_\alpha$, we have $f_\alpha(p) \lt f_\alpha(\bar x_\alpha) =1$. Furthermore, for all $k \geq x \geq pn$, we have

\begin{equation*}\frac{p_{k,k-1}}{p_{k,k+1}}=f_\alpha(k/n)\leq f_\alpha(p) \lt 1.\end{equation*}

Let $r_p\triangleq f_\alpha(p) \lt 1$. Then, from Eq. (9), we have that for $k \geq x$

\begin{equation*}\mathbb{E}_{x}\left[\zeta_k\right]\leq \sum_{t=k}^{n-1} r_p^{t-k+1} \lt \frac{r_p}{1-r_p}.\end{equation*}

Hence, from Eq. (8), we have that

\begin{equation*}\mathbb{E}_{x}\left[Z_k\right] \leq \frac{1+r_p}{1-r_p}\end{equation*}

for $k \geq x$. Moreover, for $k\lt n\bar{x}_\alpha \leq x= \lceil pn \rceil$, we have

\begin{align*} \prod_{i=k}^{x-1} \frac{p_{i,i-1}}{p_{i,i+1}} &=\prod_{i=k}^{\lceil pn \rceil-1} f_\alpha(i/n)\\ & \overset{(a)}{\leq} \prod_{i=\lceil n\underline{x}_\alpha \rceil}^{\lceil pn \rceil-1} f(i/n)\\ &=\exp\left(\sum_{i=\lceil n\underline{x}_\alpha \rceil}^{\lceil pn \rceil-1}\log\left(\ f_\alpha(i/n)\right)\right)\\ &{\overset{(b)}{=}\exp\left(n\left( \int_{\underline{x}_\alpha}^{p} \log (f_\alpha(x))\,{\rm d}x +O(1/n)\right)\right)}\\ & \overset{(c)}{=}\exp\left(n {g_\alpha(p)} +O(1)\right)\\ &\overset{(d)}{=}O(1), \end{align*}

where (a) follows from the facts that $f_\alpha(i/n) \leq 1$ for $i \leq n\underline{x}_\alpha$ and $f_\alpha(i/n) \geq 1$ for $n\underline{x}_\alpha \leq i \leq n\bar{x}_\alpha$; (b) follows from the fact that the Riemannian sum $(1/n)\sum_{i=\lceil n\underline{x}_\alpha \rceil}^{\lceil pn \rceil-1}\log\left(f_\alpha(i/n)\right)$ converges to the integral $\int_{\underline{x}_\alpha}^{p} \log (f_\alpha(x))\,{\rm d}x$ as $n \to \infty$ with an error of $O(1/n)$; (c) follows from the definition of g _α in Lemma 3.5; (d) follows from the fact that for $p \geq p_c(\alpha) \gt \bar x_\alpha$, we have $g_\alpha(p) \leq 0$ by the last statement of Lemma 3.5. For $x \gt k \geq n \bar x_\alpha$, we have

\begin{align*} \prod_{i=k}^{x-1} \frac{p_{i,i-1}}{p_{i,i+1}} &=\prod_{i=k}^{\lceil pn \rceil-1} f(i/n) \leq 1, \end{align*}

because for $i \geq n \bar{x}_\alpha$, we have $f_\alpha(i/n) \leq 1$. Hence, for all k < x, it follows from Eq. (9) that $\mathbb{E}_{x}\left[\zeta_k\right]=O(1) \mathbb{E}_{x}\left[\zeta_x\right]=O(1)$. Finally, from Eq. (8), we obtain that $\mathbb{E}_{x}\left[Z_k\right] =O(1)$ for all k < x. This establishes the second statement of the theorem.

We now turn to the third statement of the theorem. Note from Eq. (17) that to prove this statement, it suffices to show that $\mathbb{E}_{\lceil pn \rceil}\left[Z_k\right]=\Omega(\exp(\Theta(n)))$ for $k =\lceil n\underline{x}_\alpha \rceil$ when $\alpha \in (0,1/9)$ and $p \in (0, p_c(\alpha))$. First, note that for $\alpha \in (0,1/9)$, $p \in (\underline{x}_\alpha,p_c(\alpha))$, and $k=\lceil n\underline{x}_\alpha \rceil$, we have

\begin{align*} \prod_{i=\lceil n\underline{x}_\alpha \rceil}^{x-1} \frac{p_{i,i-1}}{p_{i,i+1}} &=\prod_{i=\lceil n\underline{x}_\alpha \rceil}^{\lceil np \rceil-1} f_\alpha(i/n)\\ &=\exp\left(n \int_{\underline{x}_\alpha}^{p}\log(f_\alpha(x))\,{\rm d}x +O(1)\right)\\ &=\Omega(\exp(\Theta(n))), \end{align*}

where the last line follows from the fact that $\int_{\underline{x}_\alpha}^{p}\log(f_\alpha(x))\,{\rm d}x=g_\alpha(p) \gt 0$ for $p\in (\underline{x}_\alpha,p_c(\alpha))$. Hence, $\mathbb{E}_{x}\left[\zeta_{\lceil \underline{x}_\alpha n \rceil}\right]=\Omega(\exp(\Theta(n)))\mathbb{E}_{x}\left[\zeta_x\right] =f_\alpha(p)\Omega(\exp(\Theta(n)))$, since $\mathbb{E}_{x}\left[\zeta_x\right] \gt f_\alpha(x/n)=f_\alpha(p)+o(1)$. Hence, $\mathbb{E}_{x}\left[Z_{\lceil n\underline{x}_\alpha \rceil}\right]=\Omega(\exp(\Theta(n)))$. For $p \in [0,\underline{x}_\alpha]$, we have

\begin{align*} \mathbb{E}_{x}\left[\zeta_{\lceil \underline{x}_\alpha n \rceil}\right]\gt \prod_{i=\lceil n\underline{x}_\alpha \rceil}^{\lceil n\bar{x}_\alpha \rceil-1} \frac{p_{i,i-1}}{p_{i,i+1}} &=\prod_{i=\lceil n\underline{x}_\alpha \rceil}^{\lceil n\bar{x}_\alpha \rceil-1} f_\alpha(i/n)\\ &=\exp\left(n \int_{\underline{x}_\alpha}^{\bar{x}_\alpha}\log(f_\alpha(x))\,{\rm d}x +O(1)\right)\\ &=\Omega(\exp(\Theta(n))). \end{align*}

This proves the third part of the theorem.

Remark 3.7. Although we have proved the theorem for the modified chain $\bar{Y}^n$, the proof can be extended to the original chain $\bar X^n$. This is because the results of Lemma 3.2 and Lemma 3.4 are also applicable to the original chain if the transition probabilities are replaced by those of the original chain. Furthermore, the transition probabilities of the two chains satisfy the following relations:

\begin{equation*}1-\tilde{p}_{k,k}=1-p_{k,k}+\left(1-\alpha\right)\frac{k}{n}\left(1-\frac{k}{n}\right)\left(\frac{n^2}{\left(n-1\right)^2}-1\right)\leq 1-p_{k,k},\end{equation*}

and $\tilde{p}_{k,k-1}/\tilde{p}_{k,k+1}={p_{k,k-1}}/{p_{k,k+1}}+o(1)$, where we recall that $\tilde{p}_{k,k}$ denotes the transition probability from state k to itself for the original chain $\bar X^n$. We note that the same lower bound as in Lemma 3.3 can be obtained for the chain $\bar X^n$ since we have $1-\tilde{p}_{k,k} \leq (\max(\alpha,1-\alpha)+o(1))(1-k/n)(1+k/n)$. Similarly, the upper and lower bounds on the expected number of visits to each state derived for the modified chain also hold for the original chain for large enough n. Combining the above, we obtain the same asymptotic bounds for the original chain $\bar X^n$ as we have for the modified chain $\bar Y^n$.

4. Numerical Results

In this section, we present numerical results for the model presented in this paper. We first present results to support our theoretical findings on complete graphs. We then present simulation results for other classes of graphs. Error bars in the plots represent $95\%$ confidence intervals. Also, to understand the growth rates better, we overlay the plots obtained from simulations on theoretical growth rates obtained by plotting appropriately scaled functions.

4.1. Complete graphs

For complete graphs, the mean consensus time $\bar T_n(p)$ can be computed numerically using the first step analysis of the Markov chain $\bar Y^n$. This method is more exact and computationally less expensive than simulating the chain a large number of times to obtain the average absorption time. Hence, we adopt this method for complete graphs. With a slight abuse of notation, let $\bar T_n(k)$ denote the average absorption time of the chain starting from state $k \in \left\{0,1,\ldots,n\right\}$. Then, from first step analysis, we have for each $k \in \left\{0,1,\ldots,n\right\}$,

(20)\begin{equation} \bar T_n(k)=1+p_{k,k-1}\bar{T}_n(k-1)+p_{k,k+1}\bar{T}_n(k+1)+p_{k,k}\bar{T}_n(k), \end{equation}

where $p_{i,j}$ is as defined in Eq. (2). Simplifying the above and using the boundary condition $\bar T_n(n)=0$, we obtain

(21)\begin{equation} \bar{T}_n(p):= \bar T_n\left(\lceil np \rceil\right)=\sum_{k=\lceil np \rceil}^{n-1}S_n(k), \end{equation}

where $S_n(k)$ satisfies the following recursion:

(22)\begin{equation} S_n(k)=\frac{1}{p_{k,k+1}}+\frac{p_{k,k-1}}{p_{k,k+1}}S_n(k-1), \end{equation}

with $S_n(0)=1/\alpha$. We find the expected consensus time numerically by solving the above recursion for different values of n, $\alpha,$ and p. First, we choose $\alpha=0.1 \lt 1/9$. For this value of α, we can numerically compute the value of $p_c(\alpha)$ by solving Eq. (3). This value turns out to be 0.431 (accurate to the third decimal place). In Figure 1(a) and (b), we plot the normalized (by the network size) average consensus time as a function of the network size for $p=0.4 \lt p_c(\alpha)=0.431$ and $p=0.5 \gt p_c(\alpha)=0.431$, respectively. As expected from Theorem 3.1, we observe that the mean consensus time grows exponentially for $p \lt p_c(\alpha)$ and as $\Theta(n \log n)$ for $p \gt p_c(\alpha)$.

Figure 1. Mean consensus time per node $\bar{T}_n(p)/n$ as a function of the network size for $\alpha=0.1 \lt 1/9$ for complete graphs. (a) $\alpha=0.1 \lt 1/9, p=0.4 \lt p_c(\alpha)=0.431$. (b) $\alpha=0.1 \lt 1/9, p=0.5 \gt p_c(\alpha)=0.431$.

Next, we choose $\alpha =0.125 \gt 1/9$. For this choice of α, we expect (from Theorem 3.1) the mean consensus time to grow as $\Theta(n \,\log n)$ for all values of p. This is verified in Figure 2(a) and (b), where we choose p = 0 and p = 0.5, respectively. We observe that in both cases $\bar{T}_n(p)=\Theta(n\, \log n)$.

Figure 2. Mean consensus time per node $\bar{T}_n(p)/n$ as a function of the network size for $\alpha=0.125 \gt 1/9$ for complete graphs. (a) $\alpha=0.125 \gt 1/9, p=0$. (b) $\alpha=0.125 \gt 1/9, p=0.5$.

4.2. Random d-regular graphs with $d=\Theta(\log n)$

We now present simulation results for random d-regular graphs where the degree d for each node is chosen to be $d=\lceil \log n \rceil$. We use the $\texttt{networkx}$ package to generate random d-regular graphs. For each randomly generated graph, we run the protocol until the network reaches consensus. The above procedure is repeated 500 times, and the mean consensus time is computed over these 500 runs. We further repeat this for different values of n, α, and p. In Figure 3(a) and (b), we fix α to be 0.05 and plot the normalized mean consensus time as a function of the network size for p = 0.05 and p = 0.8, respectively. Similar to complete graphs, we observe that the mean consensus time grows exponentially with n when both α and p are low. In contrast, when the initial proportion of agents having the superior opinion is sufficiently large, the mean consensus time grows as $\Theta(n\,\log n)$ even when the bias parameter α is small.

Figure 3. Mean consensus time per node $\bar{T}_n(p)/n$ as a function of the network size for α = 0.05 for random d-regular graphs with $d=\lceil \log n \rceil$. (a) $\alpha=0.05, p=0.05$. (b) $\alpha=0.05, p=0.8$.

In Figure 4(a) and (b), we choose α = 0.8 and plot the normalized mean consensus time as a function of the network size for p = 0.005 and p = 0.8, respectively. We observe that in both cases, the mean consensus time grows as $\Theta(n\, \log n)$. This indicates that if the bias parameter α is sufficiently large, then the network reaches consensus to the superior opinion in $\Theta(n \,\log n)$ time irrespective of the initial proportion of agents having the superior opinion.

Figure 4. Mean consensus time per node $\bar{T}_n(p)/n$ as a function of the network size for α = 0.8 for random d-regular graphs with $d=\lceil \log n \rceil$. (a) $\alpha=0.8, p=0.05$. (b) $\alpha=0.8, p=0.8$.

4.3. Random d-regular graphs with $d=O(1)$

We now investigate the biased opinion dynamics on random d-regular with constant degree, that is, $d=O(1)$. Specifically, we set d = 5 and simulate the network for different values of n, $\alpha,$ and p. As before, we generate a random 5-regular graph using the networkx package and run the protocol on this randomly generated instance until consensus is reached. The above procedure is repeated multiple times (each with a newly generated random graph) to obtain the mean consensus time within the desired confidence bounds. In Figure 5(a) and (b), we fix α = 0.05 and choose p = 0.05 and p = 0.8, respectively. We observe that the normalized mean consensus time grows approximately linearly in n (i.e., $\bar T_n(p)= O(n^2)$) when both α and p are small. This is unlike the previous results for dense graphs, where, for small values of α and p, the mean consensus time is exponential in n. This reduction in absorption time can be intuitively explained by the fact that a smaller neighborhood size reduces the chance of an agent updating to the worse opinion when a fixed proportion of all the agents has the worse opinion. Based on this intuition, we conjecture that for small values of p and α, the mean consensus time grows polynomially in n when the neighborhood size d is a constant higher than 2. The case of cycles (where d = 2) has already been considered in [Reference Anagnostopoulos, Becchetti, Cruciani, Pasquale and Rizzo2]. But in the case of cycles, there is no phase transition; indeed, it has been shown in [Reference Anagnostopoulos, Becchetti, Cruciani, Pasquale and Rizzo2] that for cycles, consensus is achieved on the superior opinion in $O(n\,\log n)$ time for all values of α even when p = 0.

Figure 5. Mean consensus time per node $\bar{T}_n(p)/n$ as a function of the network size for α = 0.05 for random d-regular graphs with d = 5. (a) $\alpha=0.05, p=0.05$. (b) $\alpha=0.05, p=0.8$.

Figure 6. Mean consensus time per node $\bar{T}_n(p)/n$ as a function of the network size for α = 0.8 for random d-regular graphs with d = 5. (a) $\alpha=0.8, p=0.05$. (b) $\alpha=0.8, p=0.8$.

In Figure 6(a) and (b), we plot the mean consensus time as a function of the network size for p = 0.05 and p = 0.8, respectively, keeping α = 0.8. In both cases, we observe that the mean consensus time grows as $\Theta(n\, \log n)$. Thus, in these regimes, the behavior of random d-regular graphs with $d=O(1)$ is similar to that of random d-regular graphs with $d=\Theta(\log n)$.

4.4. Erdős–Rényi graphs with edge probability $\log n/n$

Finally, we present results for Erdős–Rényi graphs where the edge probability is fixed at the connectivity threshold $\log n/n$. The experiments are designed in the same way as described for other classes of graphs. The normalized mean consensus time as a function of the network size n is plotted for different values of α and p: in Figure 7(a) and (b) for α = 0.05 and in Figure 8(a) and (b) for α = 0.8. We observe similar phase transitions as in the case of complete graphs, that is, the mean consensus time grows as $\Theta(n\log n)$ in all cases except when both α and p are low.

Figure 7. Mean consensus time per node $\bar{T}_n(p)/n$ as a function of the network size for α = 0.05 for Erdős–Rényi graphs with edge probability $\log n/n$. (a) $\alpha=0.05, p=0.05$. (b) $\alpha=0.05, p=0.8$.

Figure 8. Mean consensus time per node $\bar{T}_n(p)/n$ as a function of the network size for α = 0.8 for Erdős–Rényi graphs with edge probability $\log n/n$. (a) $\alpha=0.8, p=0.05$. (b) $\alpha=0.8, p=0.8$.

5. Conclusion and future work

In this paper, we have studied a model of binary opinion dynamics where the agents show a strong form of bias toward one of the opinions, called the superior opinion. We showed that for complete graphs, the model exhibits rich phase transitions based on the values of the bias parameter and the initial proportion of agents with the superior opinion. Specifically, we proved that fast consensus can be achieved on the superior opinion irrespective of the initial configuration of the network when bias is sufficiently high. When bias is low, we show that fast consensus can only be achieved when the initial proportion of agents with the superior opinion is above a certain threshold. If this is not the case, then we show that consensus takes exponentially long time. Through simulations, we observed similar behavior for several classes of dense graphs where the average degree scales at least logarithmically with the network size. For sparse graphs, where the average degree is constant, we observed that phase transitions do occur but the behavior below criticality is different to that of dense graphs. Specifically, we observed that when both bias and the initial proportion of agents with the superior opinion are low, the average consensus time is still polynomial in the network size.

Several directions remain open for theoretical investigation. One immediate problem is to theoretically establish the observed phase transitions for dense graphs. Here, it will be interesting to find out how “dense” a graph must be in order for it to exhibit the same phase transitions as in complete graphs. Similar questions remain open for sparse graphs such as d-regular expanders with constant d. Here, it will be of interest to find out how the threshold valued of the parameters depend on the average degree d or the spectral properties of the expander. It is also worth analyzing the exact behavior below criticality. The numerical experiments conducted in this paper suggests that the consensus time is still polynomial below criticality (unlike in complete graphs where it is exponential), but it would be challenging to obtain exact bounds in this case.