Decision Accuracy and the Role of Spatial Interaction in Opinion Dynamics

Abstract

The opinions and actions of individuals within interacting groups are frequently determined by both social and personal information. When sociality (or the pressure to conform) is strong and individual preferences are weak, groups will remain cohesive until a consensus decision is reached. When group decisions are subject to a bias, representing for example private information known by some members of the population or imperfect information known by all, then the accuracy achieved for a fixed level of bias will increase with population size. In this work we determine how the scaling between accuracy and group size can be related to the microscopic properties of the decision-making process. By simulating a spatial model of opinion dynamics we show that the relationship between the instantaneous fraction of leaders in the population (L), system size (N), and accuracy depends on the frequency of individual opinion switches and the level of population viscosity. When social mixing is slow, and individual opinion changes are frequent, accuracy is determined by the absolute number of informed individuals. As mixing rates increase, or the rate of opinion updates decrease, a transition occurs to a regime where accuracy is determined by the value of $L\sqrt{ N}$. We investigate the transition between different scaling regimes analytically by examining a well-mixed limit.

Appendices

Appendix A: Derivation of Absorption Probability

Absorption probabilities for the Markov process may be calculated exactly as shown in [24]. This derivation proceeds as follows. We firstly define the number of individuals in the correct state as a one step Markov process. Each state is therefore an integer value ranging from 0 (all individuals are incorrect) to N (all individuals are correct). As each boundary is an absorbing state, the probability of absorption U _i in state N (all correct) from state i satisfies the equation

(25)

where t ⁺ and t ⁻ are the probabilities of transitioning to the state one step above or below respectively. Rewriting the expression for the interior states gives,

$$ t^-_i (U_{i-1}-U_i) + t^+_i (U_{i+1}-U_i) = 0 $$

(26)

We now define

$$ V_i = U_{i}-U_{i-1} $$

(27)

so that

(28)

All values of V can therefore be calculated from V ₁,

$$ V_i = \Biggl(\,\prod_{k=1}^{i-1} \frac{t^-_k}{t^+_k} \Biggr) V_1. $$

(29)

To solve for V ₁ we use the relation

$$ \sum_{i=1}^{N}V_i = U_N - U_0 $$

(30)

that arises from the definition of V and the cancelling of interior terms of the series. From Eqs. (29) and (30),

$$ V_1 + \sum_{i=2}^{N} \Biggl(\, \prod_{k=1}^{i-1} \frac{t^-_k}{t^+_k} \Biggr) V_1 = 1 $$

(31)

therefore,

$$ V_1 = \Biggl(1 + \sum_{j=2}^{N} \Biggl(\,\prod_{k=1}^{j-1} \frac {t^-_k}{t^+_k} \Biggr) \Biggr)^{-1}. $$

(32)

Again by definition,

$$ U_i = \sum_{j=1}^{i}V_j + U_0 $$

(33)

which, in combination with the expression for V ₁ and V _i , and the value of U ₀=0, leads to

$$ U_i = \frac{1 + \sum_{j=2}^{i} \bigl(\prod_{k=1}^{j-1} \frac{t^-_k}{t^+_k} \bigr) }{1 + \sum_{j=2}^{N} \bigl(\prod_{k=1}^{j-1} \frac {t^-_k}{t^+_k} \bigr)}. $$

(34)

The quantity of interest is the absorption probability of the correct state when initially the system begins with equal numbers of individuals with each opinion, i.e. U _0.5N . This value is

$$ \frac{1+\sum_{j=2}^{0.5N} \bigl(\prod_{k=1}^{j-1} \frac {t^-_k}{t^+_k} \bigr) }{1+ \sum_{j=2}^{N} \bigl(\prod_{k=1}^{j-1} \frac {t^-_k}{t^+_k} \bigr)}. $$

(35)

Appendix B: Derivation of $\frac{\partial B}{\partial x}$

Given the binomial probability

$$ B(x,S) = \sum_{k=0.5(S+1)}^{S} \binom{S}{k} { (x )}^k { (1-x )}^{S-k} $$

(36)

we wish to find

$$ \frac{\partial B}{\partial x} \biggl|_{x=0.5} $$

(37)

We take derivatives, then rearrange to give

(38)

Since

$$ \sum_{k=0}^{n} \binom{n}{ k} = 2^n $$

(39)

and the binomial coefficients are symmetric, this reduces to

(40)

This is an exact result, however a simpler expression for large S may be obtained by introducing Stirling’s approximation for the binomial coefficients.

Define T=(S−1)/2, then

(41)

Therefore,

(42)