 # Notes on the model of Smaldino & Epstein (2015)

•  Jeremy Van Cleve

# Background

In a recent paper, (Smaldino 2015) provide a model for conformist bias that allows individuals to deviate from the mean trait value, $$\bar{x}$$, via a “distinctiveness” trait, $$\delta_i$$ for individual $$i$$. Each individual updates its trait value $$x_i$$ via the following recursion $x_i(t+1) = (1-k) x_i(t) + k \left( \bar{x}(t) + \delta_i \sigma(t) \right)$ where $$\sigma(t)$$ is the standard deviation of $$x_i$$ at time $$t$$ and $$k$$ scales the rate of adjustment.

# Proof of “conformity” result

Figure 2 of the paper provides numerical evidence that the explosion of the variance in $$x$$, $$\sigma^2$$, occurs when the variance in $$\delta_i$$, which I call $$V_\delta$$, is greater than one. Equation D12 provides analytical evidence of this pattern for a special case where $$\delta_i$$ takes only two values in the population. This increase in the variance is interpreted as non-conformity.

However, this result holds for any distribution of $$\delta_i$$, namely that $$\sigma^2(t) > 0$$ as $$t \to \infty$$ when $$V_\delta > 1$$. To show this mathematically, I will show that the equilibrium point of zero variance is unstable. To begin proving this, I need the recursion for the variance in the trait, $\label{eq-sig2} \sigma^2(t+1) = (1-k)^2 \sigma^2(t) + 2 k (1-k) \sigma \text{Cov}(x,\delta)(t) + k^2 \sigma^2(t) V_\delta \:,$ which holds for any distribution on $$\delta$$. Unlike the case in (Smaldino 2015) where $$\delta$$ is a constant, the recursion for $$\sigma^2$$ contains the covariance between $$x$$ and $$\delta$$. Even though this covariance maybe small initially, it may build up over time. Thus, I need the recursion for the covariance as well: $\label{eq-cov} \text{Cov}(x,\delta)(t+1) = (1-k) \text{Cov}(x,\delta)(t) + k \sigma(t) V_\delta \: .$ The only equilibrium point of these two equations is $$(\sigma^2,\text{Cov}(x,\delta))=(0,0)$$, which is the conformity outcome. In order to show that this equilibrium point is unstable, I apply standard linear stability analysis for discrete-time systems. This involves calculating the Jacobian matrix, $$J$$, from equations \ref{eq-sig2} and \ref{eq-cov}, $\label{eq-J} J = \begin{pmatrix} (1-k)^2+\frac{k \text{Cov}(x,\delta ) (1-k)}{\sigma}+k^2 V_{\delta } & 2 (1-k) k \sigma \\ \frac{k V_{\delta }}{2 \sigma} & 1-k \\ \end{pmatrix} \: ,$ and determining the conditions under which the eigenvalues of $$J$$ have magnitude greater than one. A sufficient condition for the eigenvalues, $$\lambda$$, to be greater than one in magnitude is that the characteristic polynomial, $$P(\lambda)$$, evaluated at one is negative (this comes from the so-called “Jury condition”). Applied to the Jacobian in \ref{eq-J}, $P(1) = k^2 \left( (1-k) \left(1-\frac{\text{Cov}(x,\delta )}{\sigma }\right)+1-V_{\delta } \right) \:.$ At the equilibrium point with zero variance (zero standard deviation) and zero covariance, $P(1) = k^2 ( 1 - V_\delta) \: ,$ which is negative when $V_\delta > 1 \: .$