Interaction and imitation in a world of Quixotes and Sanchos

Abstract

This paper studies a two-population evolutionary game in a new setting in between a symmetric and an asymmetric evolutionary model. It distinguishes two types of agents: Sanchos, whose payoffs are defined by a prisoner’s dilemma game, and Quixotes, whose payoffs are defined by a snowdrift game. Considering an imitative revision protocol, a revising agent is paired with someone from his own population or the other population. When matched, they observe payoffs, but not identities. Thus, agents in one population interact and imitate agents from their own population and from the other population. In this setting we prove that a unique mixed-strategy asymptotically stable fixed point of the evolutionary dynamics exists. Taking as an example the compliance with social norms, and depending on the parameters, two type of equilibrium are possible, one with full compliance among Quixotes and partial compliance among Sanchos, or another with partial compliance among Quixotes and defection among Sanchos. In the former type, Sanchos comply above their Nash equilibrium (as they imitate compliant Quixotes). In the latter type, Quixotes comply below their Nash equilibrium (as they imitate defecting Sanchos).

Appendices

Appendix A: Proof of Proposition 2

• Equilibrium under condition (18):

  Since $r^{\text{\tiny S}\text{\tiny S}}_{\text{\tiny C}\text{\tiny D}}$ = 0, the dynamics of the compliance share among Sanchos for q = 1 reads:

  $$ \dot{p}=(1-p)r^{\text{\tiny S}\text{\tiny Q}}_{\text{\tiny D}\text{\tiny C}}-p(1-p)r^{\text{\tiny S}\text{\tiny S}}_{\text{\tiny C}\text{\tiny D}}=(1-p)\sigma\left[{\Delta}_{1}-y-p\left( y+\frac{d-\phi}{\sigma}\right)\right]. $$

  Hence, taking into account that y = (1 + p)/2, then $\dot {p}/((1-p)\sigma )$ is given by the second order polynomial:

  $$ -p^{2}-2p\frac{c-d}{\sigma}+2\frac{\alpha\varepsilon-(d-\phi)}{\sigma}-1. $$

  This polynomial has one stable and one unstable real root. The stable solution defines $p^{*}_{\text {\tiny S}\text {\tiny Q}}$ in Eq. 19.

  Since an equilibrium in this regions requires that all Quixotes comply, the share of compliance in the global population can be obtained as: $y^{*}_{\text {\tiny S}\text {\tiny Q}}=(1+p^{*}_{\text {\tiny S}\text {\tiny Q}})/2$, that is:

  $$ y^{*}_{\text{\tiny S}\text{\tiny Q}}=\frac{-(d-\phi)+\sqrt{(d-\phi)^{2}+2\alpha\varepsilon\sigma}}{2\sigma }, $$

  (23)

  which is always positive and lower than 1. Furthermore, it lies within region IV or III_α> 1/2 if and only if $y^{*}_{\text {\tiny S}\text {\tiny Q}}\le {\Delta }_{1}$, which is satisfied under condition (18).

  It is easy to show that condition (18), which guarantees the existence of this type of equilibrium, is also a necessary and sufficient condition for the share of compliant Sanchos in Eq. 19 to be positive. Moreover, p SQ∗ is strictly lower than 1 as long as Quixotes play a snowdrift game, which occurs under condition (11).

• Equilibrium under condition (20):Whenever y < Δ, compliant Quixotes do not abandon compliance when paired with defecting Quixotes, r CDQQ = 0. Furthermore, since p = 0, the dynamics for the compliance share among Quixotes reads:

  $$ \dot{q}=(1-q)qr^{\text{\tiny Q}\text{\tiny Q}}_{\text{\tiny D}\text{\tiny C}}-qr^{\text{\tiny Q}\text{\tiny S}}_{\text{\tiny C}\text{\tiny D}}=q\sigma\left[(1-q)[{\Delta}-y]-(y-{\Delta}_{1})\right]. $$

  Taking into account that now y = q/2, then $\dot {q}/(q\sigma )$ is given by the second order polynomial:

  $$ q^{2}-2q\left[1+{\Delta}\right]+4{\Delta}-2(1-\alpha)\frac{\varepsilon}{\sigma}. $$

  The stable root for this polynomial is given in Eq. 21.

  For this solution to be feasible, it must remain within regions III_α< 1/2 or II, i.e. $y^{*}_{\text {\tiny Q}}>{\Delta }_{1}$. After some algebra, it is easy to conclude that this inequality is equivalent to condition (20). It is also easy to prove that this is also equivalent to $q^{*}_{\text {\tiny Q}}<1$.

Appendix B: Proof of Proposition 3

First, we compute the Jacobian matrix, J(p, q), of the system of differential equations within the region where the equilibrium is located. Then we compute the two eigenvalues of J(p^∗, q^∗).

The equilibrium $(p^{*}_{\text {\tiny S}\text {\tiny Q}},1)$, which lies within region¹⁵III_α> 1/2, has associated the two eigenvalues:

$$ \begin{array}{@{}rcl@{}} &&\lambda^{1}_{\text{\tiny S}\text{\tiny Q}}=(d-\phi)-2\varepsilon+\sqrt{(d-\phi)^{2}+2\alpha\varepsilon\sigma},\\ &&\lambda^{2}_{\text{\tiny S}\text{\tiny Q}}=(d-\phi)^{2}+2\alpha\varepsilon\sigma-(\sigma+c-d)\sqrt{(d-\phi)^{2}+2\alpha\varepsilon\sigma}. \end{array} $$

After some calculus, it is easy to see that $\lambda ^{1}_{\text {\tiny S}\text {\tiny Q}}<0$ is equivalent to:

$$ \varepsilon>\alpha\frac{c-\phi}{2}+(1-\alpha)(d-\phi). $$

But since α ∈ [0, 1], then from condition (18) it is immediately obvious that ε > αε > (c − ϕ)/2 > α(c − ϕ)/2. Hence condition above follows, and $\lambda ^{1}_{\text {\tiny S}\text {\tiny Q}}<0$.

Likewise, $\lambda ^{2}_{\text {\tiny S}\text {\tiny Q}}<0$ is equivalent to:

$$ \alpha\varepsilon\sigma[2\alpha\varepsilon-4(c-d)]<(d-\phi)^{2}[(c-d)-\alpha\varepsilon]. $$

This condition can be proved based on the fact that $p^{*}_{\text {\tiny S}\text {\tiny Q}}\le 1$, which, after some calculus, is equivalent to 2αε − 4(c − d) ≤ 0. Moreover, since αε ≤ ε < c − d, then it is immediately obvious that the LHS of previous inequality is non-positive, while the RHS is positive, which concludes the proof of $\lambda ^{2}_{\text {\tiny S}\text {\tiny Q}}<0$.

The equilibrium $(0,q^{*}_{\text {\tiny Q}})$ which lies within region¹⁶III_α< 1/2, has associated the two eigenvalues:

$$ \begin{array}{@{}rcl@{}} &&\lambda^{1}_{\text{\tiny Q}}=\frac{1}{4}\left\{-(c-d)-\varepsilon+\sqrt{Dis}\right\},\quad Dis=(c-d)^{2} - 2\alpha\varepsilon\sigma-2(d-\phi)\varepsilon+\varepsilon^{2},\\ &&\lambda^{2}_{\text{\tiny Q}}=\frac{1}{2\sigma}\left\{Dis-[\sigma+\varepsilon-(d-\phi)]\sqrt{Dis}\right\}. \end{array} $$

It is easy to see that $\lambda ^{1}_{\text {\tiny Q}}<0$ is equivalent to:

$$ -2\alpha\varepsilon\sigma-2(d-\phi)\varepsilon<2\varepsilon(c-d). $$

The LHS in this inequality is negative and the RHS positive. Therefore, it always holds.

Finally, $\lambda ^{2}_{\text {\tiny Q}}<0$ can be equivalently written as αε − 2(d − ϕ) > −(c − d). But this condition is true as long as $q^{*}_{\text {\tiny Q}}$ in Eq. 21 takes a positive value.

Appendix C: Proof of Proposition 4

Under condition (20), the stable equilibrium is $(0,q^{*}_{\text {\tiny Q}})$, and the Nash equilibrium is (0,q^NE). Obviously $p^{\text {\tiny NE}}=p^{*}_{\text {\tiny Q}}=0$. Moreover, taking into account Eq. 21 and Eq. 13, one gets that $q^{\text {\tiny NE}}>q^{*}_{\text {\tiny Q}}$ if and only if:

$$ \sqrt{(1-{\Delta})^{2}+2(1-\alpha)\varepsilon/\sigma}>(1+{\Delta})-2{\Delta}=1-{\Delta}, $$

which immediately holds.

Conversely, under condition (18) the stable equilibrium is $(p^{*}_{\text {\tiny S}\text {\tiny Q}},1)$, and the Nash equilibrium is (0,q^NE). From Eq. 18 it immediately follows that $p^{*}_{\text {\tiny S}\text {\tiny Q}}>p^{\text {\tiny NE}}=0$. However, from condition (18) and the fact that α ∈ [0, 1] one gets:

$$ \varepsilon\ge\alpha\varepsilon>\frac{c-\phi}{2}. $$

This inequalities imply 2Δ > 1 and, taking into account Eq. 13, q^NE = 1.

Appendix D: Proof of Remark 3

It is easy to see that:

$$q^{\text{\tiny NE}}<1 \text{if} \varepsilon<\frac{c-\phi}{2}, \text{and} q^{\text{\tiny NE}}=1 \text{if} \varepsilon\in\left( \frac{c-\phi}{2},\frac{c-\phi}{2\alpha}\right). $$

Thus, under condition (18), ε > αε > (c − ϕ)/2 and q^NE = 1. The gap in the average payoff for Quixotes between the asymptotically stable equilibrium and the Nash equilibrium reads:

$$ \bar{\pi}^{\text{\tiny Q}}(p^{*}_{\text{\tiny S}\text{\tiny Q}},1)-\bar{\pi}^{\text{\tiny Q}}(0,1)=\frac{dp^{*}_{\text{\tiny S}\text{\tiny Q}}}{2}>0. $$

Under the alternative condition (20), although p^NE = 0, q^NE can be equal to or less than 1.

• 1. αε < ε < (c − ϕ)/2 and hence q^NE < 1:

  The gap in the average payoff for Sanchos between the asymptotically stable equilibrium and the Nash equilibrium reads:

  $$ \bar{\pi}^{\text{\tiny S}}(0,q^{*}_{\text{\tiny Q}})-\bar{\pi}^{\text{\tiny S}}(0,q^{\text{\tiny NE}})=\frac{1}{2}(c-d+\phi)(q^{*}_{\text{\tiny Q}}-q^{\text{\tiny NE}})<0. $$

• 2. αε < (c − ϕ)/2 < ε and hence q^NE = 1:

  The gap in the average payoff for Sanchos between the asymptotically stable equilibrium and the Nash equilibrium reads:

  $$ \bar{\pi}^{\text{\tiny S}}(0,q^{*}_{\text{\tiny Q}})-\bar{\pi}^{\text{\tiny S}}(0,1)=-\frac{1}{2}\sigma(1-q^{*}_{\text{\tiny Q}})<0. $$