We adopt the standard evolutionary setting and construct a model with a single population of players randomly matched to play the mini ultimatum game. Since the mini ultimatum game is an asymmetric one, the roles in the game, i.e. proposer and responder, are assigned at random. Consequently, the position of all players ex-ante is exactly the same. We assume, though, the conditioning of behavior on the assigned role, that is, a strategy is a plan describing a behavior in both roles. We are interested in studying the evolution of a distribution of pure strategies (describing behavior in both roles).
Strategies and Payoffs
In a symmetric version of the ultimatum game, strategies of players must prescribe behavior in both proposer and responder roles. Therefore, a strategy of a player is a pair (α,β), where α∈[0,1] is an offer of a player in the role of the proposer and β∈[0,1] is an acceptance level, that is, a player in the role of the responder accepts only offers not lower than β. For simplicity we narrow down the possible offers to just three, α,∈{0,δ,1}, where 0<δ<1. We also assume that acceptance levels cannot be larger than offers, that is, β≤α. Consequently, we are left with six possible strategies:
The payoff matrices, one for each role (bold zeros correspond to rejections), read
$$P = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}1 & \mathbf{0} & 1 & \mathbf{0} & \mathbf{0} & 1 \\1-\delta & 1-\delta & 1-\delta & \mathbf{0} & 1-\delta & 1-\delta \\1-\delta & 1-\delta & 1-\delta & \mathbf{0} & 1-\delta & 1-\delta \\0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0\end{array} \right ] \quad\text{and}\quad R=\left [ \begin{array}{l@{\quad}l@{\quad}l@{\quad}l@{\quad}l@{\quad}l}0 & \mathbf{0} & 0 & \mathbf{0} & \mathbf{0} & 0 \\\delta & \delta & \delta & \mathbf{0} & \delta & \delta \\\delta & \delta & \delta & \mathbf{0} & \delta & \delta \\1 & 1 & 1 & 1 & 1 & 1 \\1 & 1 & 1 & 1 & 1 & 1 \\1 & 1 & 1 & 1 & 1 & 1\end{array} \right ] $$
for a proposer and a responder, respectively, where the first strategy is the egoistic one, then follow two medium strategies, and finally three altruistic ones.
Unperturbed Replicator Dynamics
We consider a single population of players. They are repeatedly matched into pairs in which their roles are attached at random. Fractions of the population using given strategies are denoted by x
1,…,x
6, respectively. The vector of fractions is denoted by \(\mathbf {x}\in \triangle \), where △ is a simplex,
$$\triangle = \Biggl\{ \mathbf {x}\in\mathbf{R}^{6}, \sum_{i=1}^{6} x_{i}=1, x_{i} \geq0,i=1 \ldots6 \Biggr\}.$$
Since we consider only a single population in which roles are attached randomly to players, average payoffs are given by the matrix \(A = (P+R^{\mathrm{T}})/2\), where \(R^{\mathrm{T}}\) is the transpose of R. For this matrix we consider the standard replicator dynamics of the form
$$ \dot{x}_i = x_i \bigl([A\mathbf {x}]_i -\mathbf {x}^{\mathrm{T}} A \mathbf {x}\bigr),\quad\text{for } i=1, \ldots, 6.$$
(1)
where time derivatives are denoted by \(\dot{x}_{i}\).
We first notice that all altruistic strategies disappear in the long run for any interior initial condition.Footnote 2 Namely, let \(\xi(\mathbf {x}^{0},t)\) be the solution of (1) with the initial condition \(\mathbf {x}^{0} \in \mathrm {int}(\triangle )\), then
$$\lim\limits _{t \rightarrow\infty} \xi_{i}\bigl(\mathbf {x}^{0},t\bigr) = 0\quad\text{for } i=4,5,6.$$
Therefore, we can consider only first three strategies. Such a game is reminiscent of the ultimatum mini game considered in [4] with the additional strategy (δ,0). The replicator dynamics for variables x
i
, i=1,2,3 takes the following form:
$$ \left \{ \begin{aligned}\dot{x}_1 &= f_1 ( \mathbf {x}) = \frac{1}{2}x_1 (x_1 - 1) (x_2 - \delta),\\\dot{x}_2 &= f_2 ( \mathbf {x}) = \frac{1}{2}x_1 x_2 ( x_2 - \delta),\\\dot{x}_3 &= f_3 ( \mathbf {x}) = \frac{1}{2}x_1 x_3 ( x_2 - \delta).\end{aligned}\right .$$
(2)
We write \(\dot{ \mathbf {x}} = \mathbf {f}( \mathbf {x})\) for short, where \(\mathbf {f}= (f_{1}, f_{2},f_{3})\), then (2) can be written in the following form:
$$ \dot{ \mathbf {x}} = \mathbf {f}( \mathbf {x}) = \frac{1}{2}x_1 ( x_2 - \delta) \begin{bmatrix}x_1 -1 \\x_2 \\x_3\end{bmatrix} =\frac{1}{2} x_1 ( x_2 - \delta) \left( \mathbf {x}- \begin{bmatrix}1\\0\\0\end{bmatrix} \right).$$
(3)
The next three propositions follow directly from (3).
Proposition 1
(Equilibria)
The set
E
of equilibria of dynamics (2) consists of four components, E=E
1∪E
2∪E
3∪E
4, where
E
1={(1,0,0)}, \(E_{2} = \{ \mathbf {x}\in \triangle : x_{2} = \delta\}\), \(E_{3} = \{ \mathbf {x}\in \triangle : x_{1} = 0, x_{2} < \delta\}\), and
\(E_{4} = \{\mathbf {x}\in \triangle : x_{1} = 0, x_{2} > \delta\}\).
Proposition 2
(Trajectories)
Let
\(\mathbf {x}^{0} \in \triangle \)
be an initial condition. The trajectory
\(\{\mathbf {x}\in \triangle : \mathbf {x}=\xi(\mathbf {x}^{0},t)\), t≥0} of (2) is contained in the straight line passing through points
\(\mathbf {x}^{0}\)
and (1,0,0).
Proposition 3
(Stability)
Equilibrium (1,0,0) is locally asymptotically stable. Each equilibrium in
E
4
is Lyapunov stable. Each equilibrium in
E
2∪E
3
is unstable.
Figure 1(a) shows the phase portrait of the unperturbed replicator dynamics (2).
Perturbed Replicator Dynamics
We would like to see how equilibria and their stabilities change under perturbations of the replicator dynamics. Following [4] we consider the situation where the original evolution takes place with a probability 1−ϵ and with a small probability ϵ strategies are chosen randomly, that is, each one with the probability 1/3. The random choice of strategies introduces the term ϵ(1/3−x
i
) into the replicator dynamics and we obtain the following perturbed version of the dynamics (2):
$$ \left \{ \begin{aligned}\dot{x}_1 &= f_{1}(\mathbf {x}, \epsilon ) = \frac{1}{2} (1-\epsilon )x_1 (x_1-1) (x_2 - \delta) + \epsilon \biggl(\frac{1}{3} - x_1 \biggr),\\\dot{x}_2 &= f_{2}(\mathbf {x}, \epsilon ) = \frac{1}{2} (1-\epsilon ) x_1 x_2 (x_2 - \delta) + \epsilon \biggl(\frac{1}{3} - x_2 \biggr),\\\dot{x}_3 &= f_{3}(\mathbf {x}, \epsilon ) = \frac{1}{2} (1-\epsilon ) x_1 x_3 (x_2 - \delta) + \epsilon \biggl(\frac{1}{3} - x_3 \biggr).\end{aligned} \right .$$
(4)
Let us notice that from the last two equations of (4) it follows that the segment x
2=x
3 is invariant under the flow (4). Moreover, x
2=x
3 at any equilibrium. Hence we may have at most three equilibria of (4).
Proposition 4
(Equilibria)
Let
ϵ>0 be small enough. If
δ≥1/2, then dynamics (4) has only one equilibrium
\(\hat{\mathbf {x}}^{1} \in \triangle \). If
δ<1/2, then dynamics (4) has three equilibria: \(\hat{\mathbf {x}}^{1}\), \(\hat{\mathbf {x}}^{2}\), \(\hat{\mathbf {x}}^{3} \in \triangle \). In the limit
ϵ↓0 we have
$$\hat{\mathbf {x}}^1 \rightarrow(1, 0, 0), \qquad \hat{\mathbf {x}}^2\rightarrow(1-2\delta, \delta, \delta) \quad\text{\textit{and}}\quad \hat{\mathbf {x}}^3 \rightarrow(0, 1/2, 1/2).$$
Proof
Let \(\hat{\mathbf {x}}\) be an equilibrium, i.e. \(\mathbf {f}( \hat{\mathbf {x}}, \epsilon )=0\). From \(f_{2}(\hat{\mathbf {x}},\epsilon )=f_{3}(\hat{\mathbf {x}},\epsilon )=0\) it follows that \(\hat {x}_{2} = \hat{x}_{3}\) and therefore \(\hat{x}_{1} = 1-2 \hat{x}_{2}\). We substitute this into \(f_{2}(\mathbf {x}, \epsilon )=0\) and get the following cubic equation:
$$f_2(x_2, \epsilon ) = \frac{1}{2}(1-\epsilon) (1-2x_2 ) x_2 (x_2-\delta ) + \epsilon \biggl(\frac{1}{3}- x_{2} \biggr)=0.$$
Depending on the value of δ we have two cases. For δ<1/2, cf. Fig. 2(a), there are three different roots of f
2(x
2,0)=0, namely \(\hat {x}_{2}^{1}=0\), \(\hat{x}_{2}^{2}=\delta\) and \(\hat{x}_{2}^{3}=1/2\). All these roots give rise to stationary points of the flow (4) that are contained in △. It is easy to see that for any small enough ϵ>0 all these stationary points still remain in the simplex, hence there are three equilibria for the perturbed flow \(\mathbf {f}(\mathbf {x}, \epsilon )\). It is clear that as ϵ↓0 we have the desired convergence.
For δ≥1/2 and ϵ>0, cf. Figure 2(b), there is only a single root \(\hat{x}_{2}^{1}(\epsilon )\) of f
2(x
2,ϵ)=0 that gives rise to a stationary point of \(\mathbf {f}(\mathbf {x}, \epsilon )\) that is contained in a simplex △. Also here, it is clear that as ϵ↓0 we have the desired convergence. □
As we have already mentioned, a line x
2=x
3 is invariant under the perturbed dynamics. This plays a crucial role in the following proposition.
Proposition 5
(Global Convergence)
Let
\(\mathbf {x}\in \triangle \)
be any initial condition and let
\(\xi(\mathbf {x}, t)\)
be the solution (4). Then
\(\xi(\mathbf {x}, t)\)
converges to one of the equilibria
\(\hat{x}^{i}\).
Proof
System (4) is a planar system and so by the Poincaré–Bendixson theorem we know that \(\xi(\mathbf {x}, t)\) converges either to a stationary point or a limit cycle. Therefore, in order to prove the proposition it is enough to show that there cannot be any limit cycles.
Suppose that \(\xi( \mathbf {x}, t)\) converges to a limit cycle defining a region γ that is invariant under \(\mathbf {f}(\mathbf {x},\epsilon )\). It is well knownFootnote 3 that there must be a stationary point \(\hat{x}\) of \(\mathbf {f}(\mathbf {x}, \epsilon )\) such that \(\hat{x} \in \mathrm {int}(\gamma)\). But then a limit cycle has to cross the invariant line x
2=x
3 which cannot happen. □
To complete the analysis of the perturbed system we focus on local phase portraits around equilibria of (4).
Proposition 6
(Local Phase Portraits)
Let
δ<1/2 and
ϵ>0 be small enough. Equilibria
\(\hat{\mathbf {x}}^{1}\)
and
\(\hat{\mathbf {x}}^{3}\)
are asymptotically stable and equilibrium
\(\hat{\mathbf {x}}^{2}\)
is a saddle.
Proof
We linearize \(\mathbf {f}(\mathbf {x}, \epsilon )\) and calculate eigenvalues to get
$$\begin{aligned}\lambda_1 &= \frac{1}{2} (\epsilon-1) x_1 (\delta-x_2 )-\epsilon,\\\lambda_2 &= \frac{1}{2} \bigl(\delta-(\delta+2) \epsilon+(\epsilon -1) x_1 (2 \delta-3 x_2 )+(\epsilon-1)x_2 \bigr).\end{aligned} $$
Firstly, we see that, at any equilibrium, λ
1<0 leads to
$$(1-\epsilon ) \hat{x}_1 (\hat{x}_2 - \delta) < 2 \epsilon \quad\Leftrightarrow\quad (1-\epsilon ) \biggl(\hat{x}_2 - \frac{1}{2}\biggr) (\hat{x}_2 - \delta) > -\epsilon . $$
(5)
Also, at any equilibrium we have \(f_{1}(\hat{x}_{2}, \epsilon )=0\) that leads to
$$(1-\epsilon ) \biggl(\hat{x}_2 - \frac{1}{2} \biggr) (\hat{x}_2 - \delta) = -\frac{2 \epsilon (\hat{x}_2 - \frac{1}{3} )}{\hat{x}_2}. $$
(6)
Inserting (6) into (5) we get \(\hat {x}_{2} < 2/3\), a condition that is satisfied at any equilibrium.
Secondly, we notice that after substituting \(\hat{x}_{1} = 1-2 \hat{x}_{2}\) into λ
2 we get
$$\lambda_2 = \frac{1}{2} \bigl(6 (\epsilon -1) \hat{x}_{2}^2-2(2 \delta+1) (\epsilon -1) \hat{x}_{2}+\delta (\epsilon -1)-2 \epsilon \bigr) =\frac{df_2(\hat{x}_2, \epsilon )}{dx_2}. $$
(7)
Direct inspection of f
2(x
2,ϵ) shows, cf. Fig. 2, that λ
2 is negative at both \(\hat{\mathbf {x}}^{1}\) and \(\hat{\mathbf {x}}^{3}\) and positive at \(\hat{\mathbf {x}}^{2}\). □
The case of δ≥1/2 is trivial. The proof of Proposition 6 works for the only equilibrium \(\hat{\mathbf {x}}^{1}\) in the case of δ≥1/2 as well. We could also argue differently. We know by Proposition 5 that any trajectory converges to one of the equilibria and all of them are on a line x
2=x
3. Therefore, trajectories come arbitrarily close to this line and by the continuity of (4) their behavior is qualitatively identical to the behavior of (4) on that line.