Evolution of vulnerability to pain in interpersonal relations as a strategic trait aiding cooperation

Abstract

Why are humans so vulnerable to pain in interpersonal relations and can so easily hurt others physically and emotionally? We theoretically examine whether being offensively strong but defensively weak can evolve as a strategic trait that fosters cooperation. We study a population comprised of “thick-skinned” and “thin-skinned” agents by using an indirect evolution model that combines rational choice in strategic interactions with evolutionary selection across generations. We find that (a) the relatively vulnerable and cooperative thin-skins cannot evolve under purely random matching, (b) with some assortment thin-skins evolve and can take over the entire population, (c) vulnerability to greater pain makes it easier for thin-skins to evolve, and (d) proximate pain which merely feels bad but does not lower fitness helps thin-skins evolve even more than pain which accurately reflects fitness consequences. We draw contrast with the Hawk-Dove model and identify several ways in which rationality hinders the evolution of the relatively vulnerable and peaceful type of agent.

Appendix

1.1 Proof of Proposition 7

For a thick-skinned agent, HALF is a dominated strategy regardless of the type of opponent in its dyad. Thus, agent of type A demands ALL. A thin-skinned agent prefers to demand HALF against A and mix against B, but he can only estimate which type he is facing in his dyad according to (7). Suppose each B agent plays ALL with probability θ ∈ [0,1]. Then, the expected payoffs to a B agent from playing ALL and HALF are

$$ \displaystyle \label{eq37} \pi _B ( \mathit{ALL} )=( {1-p^B} )F+p^B( {\theta F+( {1-\theta } )} ) $$

(30)

$$ \displaystyle \label{eq38} \pi _B ( \mathit{HALF} )=( {1-p^B} )0+p^B( {\theta 0+\frac{1}{2}( {1-\theta } )} )=\frac{1}{2}p^B( {1-\theta } ) $$

(31)

Setting π _B (HALF) = π _B (ALL) and solving for p assuming θ = 0 gives $\bar {p}( a )\equiv 1-\frac{1}{( {1-a} )( {1-2F} )}$. This curve partitions the space of all populations and assortment degrees into two regions, as follows.

• Case I If $0\le p<\bar {p}(a)$, then π _B (HALF) > π _B (ALL) for any θ ∈ [0,1] and therefore B always demands HALF. This is the case in which there are few other thin-skins and/or assortment is weak, so that it’s best for a thin-skin to play it safe by demanding HALF. The average fitness of the two types are:

  $$ \displaystyle \label{eq39} V(A)=( {1-p^A} )F_{A\!A} +p^A=F_{A\!A} +p( {1-a} )( {1-F_{A\!A} } ) $$

  (32)
  $$ \displaystyle \label{eq40} V( B )=\frac{1}{2}p^B=\frac{1}{2}( {a+p( {1-a} )} ) $$

  (33)

  If a stable polymorphic equilibrium exists at $\hat {p}$, then V(A) = V(B), which implies (15). Setting $\hat {p}=0$ and solving for a gives (19). Setting $\hat {p}=\bar {p}$ and solving for a gives (20).

• Case II If $\bar {p}( a )\le p\le 1$, then there exists a unique θ ^∗ ∈ [0,1] such that π _B (ALL) = π _B (HALF). Solving this equation for θ gives (18). The average fitness of the two types when thin skins mix with probability θ ∗  is:

  $$ \displaystyle \label{eq41} V(A)= ( {1-p^A})F_{A\!A} +p^A ({\theta ^\ast F_{A\!B} + ({1-\theta ^\ast})}) $$

  (34)
  $$ \displaystyle \label{eq42} V(B)=\frac{1}{2}p^B({1-\theta ^\ast } )=\frac{F}{2F-1} $$

  (35)

  If a stable polymorphic population exists at $\hat {p}$, then V(A) = V(B), which yields (16)–(17). Setting $\hat {p}=\bar {p}$ and solving for a gives (20) and establishes the continuity of polymorphic equilibria on the boundary between Cases I and II. Setting $\hat {p}=1$ and solving for a gives (21). It is straightforward to confirm that if p = 1, then V(B) > V(A) for all $a\in ( {a_{\max }^{pop} ,1} )$, which implies stability. □

1.2 Proof of Proposition 8

Subtracting (19) from (10) establishes that $a_{\min }^{pop} <a_{\min }$, which implies that on $[ {0,a_{\min}^{pop}}]$ there are no thin-skins under either information regime and on $( {a_{\min }^{pop} }, {a_{\min}}]$ there are thin-skins under the population-level information but not under perfect information. Subtracting (21) from (11) establishes that $a_{\max}^{pop} <a_{\max}$, which implies that on [a _max ,1] thin-skins have 100% population share under either information regime and on $[{a_{\max }^{pop} ,a_{\max }}]$ thin-skins have 100% population share under population-level information but not under perfect information. On $a\in ( {a_{\min } ,\bar {a}} )$, comparing (8) with (32) and (9) with (33) shows that thick-skins have the same average fitness under both regimes but thin-skins have strictly higher fitness under population-level information. By continuity, this implies that thin-skins hold higher population share for any $a\in ( {a_{\min } ,\bar {a}} )$. On $a\in ( {\bar {a},a_{\max }^{pop} } )$, comparing (8) with (34) shows that thick-skins have lower average fitness under population-level information than under perfect information; comparing and (9) with (35) shows that thin-skins have higher average fitness under population-level information than under perfect information. By continuity, this implies that thin-skins hold higher population share for any $a\in ( {\bar {a},a_{\max }^{pop} } )$. □

1.3 Proof of Proposition 11

A thick-skinned agent maximizes its payoff by demanding ALL under all circumstances. A thin-skinned agent moving second maximizes its payoff by demanding HALF in response to ALL and vice versa. A first-moving thin-skinned agent who demands HALF earns $\pi _B^{f\/irst} ( \mathit{HALF} )=0$; if it instead demands ALL, the expected payoff is $\pi _B^{f\/irst} ( \mathit{ALL} )= ( {1-p^B} )F+p^B$. Setting $\pi _B^{f\/irst} ( \mathit{HALF})=\pi _B^{f\/irst} ( \mathit{ALL} )$ and solving for proportion of thin-skins gives $\bar {p}( a )\equiv 1-\frac{1}{( {1-a} )( {1-F} )}$. This curve partitions the space of all populations and assortment degrees into two regions, as follows. For all $( {p,a} )\in \{ {( {p,a} )\vert p<\bar {p}( a )} \}$ first-moving thin-skins demand HALF. For all $( {p,a} )\in \{ {( {p,a} )\vert p>\bar {p}( a )} \}$ first-moving thin-skins demand ALL. On the boundary, $( {p,a} )\in \{ {( {p,a} )\vert p=\bar {p}( a )} \}$ first-moving thin-skins are indifferent between demanding ALL and HALF. The proof proceeds by considering the two regions and the boundary as separate cases:

• Case I $( {p,a} )\in \{ {( {p,a} )\vert p<\bar {p}( a )} \}$. This is the case in which there are few other thin-skins and/or the assortment is weak, so that it’s best for the thin-skinned agent to play it safe by demanding HALF. The average fitness of the two types are the same as in the full-information case of Section 6.1:

  $$ \displaystyle \label{eq43} V( A )=( {1-p^A} )F_{A\!A} +p^A $$

  (36)
  $$ \displaystyle \label{eq44} V( B )=\frac{1}{2}p^B $$

  (37)

  If a stable polymorphic equilibrium exists at $\hat{p}$, then V(A) = V(B), which implies (12″). Differentiating (12″) shows that $\hat {p}$ is monotonically increasing in a. Setting $\hat {p}=0$ and solving for a gives (25). Setting $\hat {p}=\bar {p}$ and solving for a gives (26). The population $\hat {p}=0$ is stable on [0, a _min ) since V(A) > V(B) for all a ∈ [0, a _min ) such that $p<\bar {p}( a )$. The population given by (12″) is stable on $[ {a_{\min } } , {\bar {a}} )$ since V(A) = V(B) for $p=\hat {p}( a )$, V(A) < V(B) for $p<\hat {p}( a )$, and V(A) > V(B) for $p>\hat {p}(a)$.

• Case II $( {p,a} )\in \{ {( {p,a} )\vert p>\bar {p}( a )} \}$ This is the case in which there are many other thin-skins and/or the assortment is strong, so that the payoff-maximizing strategy for the thin-skinned agent is to assume the opponent in his dyad is another a thin-skin and try to exploit its vulnerability to pain by demanding ALL. The average fitness of the two types are:

  $$ \displaystyle \label{eq45} V( A )=( {1-p^A} )F_{A\!A} +\frac{1}{2}p^A( {F_{A\!B} +1} ) $$

  (38)
  $$ \displaystyle \label{eq46} V( B )=\frac{1}{2}( {1-p^B} )F+\frac{1}{2}p^B $$

  (39)

  Monomorphic thick-skin population is stable if V(A) > V(B) when p = 0. This holds if a < a ₂, where a ₂ is given by (29). Monomorphic thin-skin population is stable if V(A) < V(B) when p = 1. This holds if a > a ₁, where a ₁ is given by (28). If F > 2F _AA  − F _AB , then a ₁ > a ₂; solving V(A) = V(B) for p gives (24). The population given by (24) is stable since V(A) < V(B) for $p<\hat {p}(a)$, and V(A) > V(B) for $p>\hat {p}( a )$.

• Case III $( {p,a} )\in \{ {( {p,a} )\vert p=\bar {p}( a )} \}$ This is the case in which the first-moving thin-skin is indifferent between demanding ALL or HALF. Let x be the fraction of thin-skins that demand ALL when moving first in a dyad. The average fitness of thick- and thin-skins are, respectively:

  $$ \displaystyle \label{eq47} V( A )= ( {1-p^A} )F_{A\!A} +\frac{1}{2}p^A( {( {xF_{A\!B} + ( {1-x} )} )+1} ) $$

  (40)
  $$ \displaystyle \label{eq48} V( B )=\frac{1}{2}( {1-p^B} )xF+\frac{1}{2}p^B $$

  (41)

  Solving V(A) = V(B) for x gives (23), which is the unique proportion of “assertive” thin-skins necessary for evolutionary stability of a polymorphic population along the boundary between Cases I and II. Differentiating (23) shows that $\hat{x}$ is monotonically increasing in a, and is thus uniquely determined for each a or p. It is straightforward to confirm that $\hat {x}\in [ {0,1} ]$ for all $a\in [ {\bar {a},a_0 } ]$ and $\hat {x}<0$ for all $a\in [ 0 , {\bar {a}} )$. Solving $\bar {p}( a )=0$ gives (27) and establishes the upper bound on assortment in this Case III. This population is stable because for $p<\bar {p}( a )$ the average fitness of thin-skins (37) is higher than average fitness of thick-skins (36), and for $p>\bar {p}( a )$ the average fitness of thick-skins (38) is higher than average fitness of thin-skins (39). □