The ESS and replicator equation in matrix games under time constraints

Abstract

Recently, we introduced the class of matrix games under time constraints and characterized the concept of (monomorphic) evolutionarily stable strategy (ESS) in them. We are now interested in how the ESS is related to the existence and stability of equilibria for polymorphic populations. We point out that, although the ESS may no longer be a polymorphic equilibrium, there is a connection between them. Specifically, the polymorphic state at which the average strategy of the active individuals in the population is equal to the ESS is an equilibrium of the polymorphic model. Moreover, in the case when there are only two pure strategies, a polymorphic equilibrium is locally asymptotically stable under the replicator equation for the pure-strategy polymorphic model if and only if it corresponds to an ESS. Finally, we prove that a strict Nash equilibrium is a pure-strategy ESS that is a locally asymptotically stable equilibrium of the replicator equation in n-strategy time-constrained matrix games.

Appendix

1.1 A.1. Preliminaries

We first state some technical lemmas necessary later, although sometimes they are used only tacitly.

Lemma 6.1

(Garay et al. 2017, Lemma 2, p. 7) The following system of nonlinear equations in n variables,

$$\begin{aligned} u_i=\frac{1}{1+\sum _{j=1}^n c_{ij}u_j},\quad 1\le i\le n, \end{aligned}$$

(24)

where the coefficients $c_{ij}$ are positive numbers, has a unique solution in the unit hypercube $[0,1]^{n}$.

We also claim the following.

Lemma 6.2

The solution $\mathbf {u}=(u_1,u_2,\ldots ,u_n)\in [0,1]^{n}$ of (24)

• (i) is a continuous function in

  $$\begin{aligned}\mathbf {c}:=(c_{11},\ldots , c_{1n},c_{21},\ldots ,c_{2n},\ldots , c_{n1},\ldots ,c_{nn})\in \mathbb {R}_{\ge 0}^{n^2}, \end{aligned}$$

• (ii) has positive coordinates uniformly separated from zero, namely

  $$\begin{aligned} \frac{1}{1+\mathop {\sum }\nolimits _{ij}c_{ij}}\le u_l\le 1\quad (1\le l \le n). \end{aligned}$$

Proof

To the proof of (i), assume that the sequence

$$\begin{aligned} \mathbf {c}_k= (c_{11}^{(k)},\ldots ,c_{1n}^{(k)},c_{21}^{(k)}, \ldots ,c_{2n}^{(k)},\ldots , c_{n1}^{(k)},\ldots ,c_{nn}^{(k)})\rightarrow \mathbf {c}. \end{aligned}$$

For each positive integer k, let $\mathbf {u}_k= (u_{1}^{(k)},u_{2}^{(k)},\ldots ,u_n^{(k)})$ be the unique solution of the system of equations

$$\begin{aligned} u_i=\frac{1}{1+\sum _{j=1}^nc_{ij}^{(k)}u_j}, \quad 1\le i\le n, \end{aligned}$$

in the unit hypercube $[0,1]^{n}$. As $\mathbf {u}_k\in [0,1]^{n}$ it is a bounded sequence. To see that $\mathbf {u}_k$ tends to $\mathbf {u}$ it, therefore, suffices to prove that any $\mathbf {u}_{k_l}$ convergent subsequence of $\mathbf {u}_k$ tends to the same $\mathbf {u}$. Denote by $\hat{\mathbf {u}}$ the limit of $\mathbf {u}_{k_l}$. Since

$$\begin{aligned} \frac{1}{1+\sum _{j=1}^nb_{ij}x_j}-x_i \end{aligned}$$

is a continuous function in $(\mathbf {b},\mathbf {x})$ (consider it as a function of $n^2+n$ variables on $\mathbb {R}_{\ge 0}^{n^2}\times [0,1]^{n}$), it follows that

$$\begin{aligned} 0=\frac{1}{1+\sum _{j=1}^nc_{ij}^{(k_l)}u_j^{(k_l)}}-u_i^{(k_l)}\rightarrow \frac{1}{1+\sum _{j=1}^nc_{ij}\hat{u}_j}-\hat{u}_i \end{aligned}$$

which implies that

$$\begin{aligned} 0= \frac{1}{1+\sum _{j=1}^nc_{ij}\hat{u}_j}-\hat{u}_i\quad \text {or, equivalently,}\quad \hat{u}_i=\frac{1}{1+\sum _{j=1}^nc_{ij}\hat{u}_j}. \end{aligned}$$

Because of the uniqueness of the solution, the limit $\hat{\mathbf {u}}$ must be the same for any convergent subsequent of $\mathbf {u}^{(k)}$ and $\hat{\mathbf u}$ must be equal to $\mathbf u$. Since $\mathbf {u}^{(k)}$ tends to this unique solution in $[0,1]^n$, the solution is continuous in $\mathbf {c}$.

The lower estimate in (ii) is apparent from (24) because $u_i\le 1$ for every $1\le i \le n$.$\square$

Corollary 6.3

Consider a population of types $\mathbf p_0,\mathbf p_1,\ldots ,\mathbf p_m\in S_n$ with frequencies $x_0,x_1,\ldots ,x_m$.

• (i) The active part $\varrho _i$ of the different types given by the solution of the system:

  $$\begin{aligned} \varrho _i=\frac{1}{1+\mathbf p_iT[\sum _{j=0}^mx_j\varrho _j\mathbf p_j]} \end{aligned}$$

  continuously depends on $\mathbf {x}=(x_0,x_1,\ldots ,x_m)$. We use the notation $\varrho _i(\mathbf {x})=\varrho _i(\mathbf {x},\mathbf p_0,\ldots ,\mathbf p_m)$ in this sense.

• (ii) Let

  $$\begin{aligned} \bar{\varrho }(\mathbf {x})=\bar{\varrho }(\mathbf {x},\mathbf p_0, \ldots ,\mathbf p_m): =\sum _{i=0}^{m}x_i \varrho _i(\mathbf {x},\mathbf p_0,\ldots ,\mathbf p_m) \end{aligned}$$

  and

  $$\begin{aligned} \bar{\mathbf h}(\mathbf {x})=\bar{\mathbf h}(\mathbf {x},\mathbf p_0,\ldots ,\mathbf p_m): =\sum _{i=0}^{m}\frac{x_i \varrho _i(\mathbf {x},\mathbf p_0,\ldots ,\mathbf p_m)}{\bar{\varrho }(\mathbf {x},\mathbf p_0,\ldots ,\mathbf p_m)}\mathbf p_i. \end{aligned}$$

  If $\mathbf {y}$ is another frequency distribution such that $\bar{\mathbf h}(\mathbf {y}) =\bar{\mathbf h}(\mathbf {x})=:\bar{\mathbf h}$, then both $\bar{\varrho }(\mathbf {x})=\bar{\varrho }(\mathbf {y})$ and $\varrho _i(\mathbf {x})=\varrho _i(\mathbf {y})$ ($0\le i\le m$).

• (iii) If $\bar{\mathbf h}$ can be uniquely represented as a convex combination of $\mathbf p_0,\ldots ,\mathbf p_m$¹⁴ then $x_i$ must be equal to $y_i$ for every i.

• (iv) If $\mathbf p$ is a convex combination of $\mathbf p_0,\ldots ,\mathbf p_m$ then there is an $\mathbf x=(x_0,x_1,\ldots ,x_m)\in S_{m+1}$ such that $\bar{\mathbf h}(\mathbf x)=\mathbf p$. Namely, if $\mathbf p=\sum _{i=0}^m\alpha _i \mathbf p_i$ then

  $$\begin{aligned} x_i=\frac{\rho _{\mathbf p}}{\rho _i(\mathbf p)}\alpha _i \end{aligned}$$

  where $\rho _{\mathbf p}$ is the unique solution in [0, 1] to the equation

  $$\begin{aligned} \rho =\frac{1}{1+\mathbf {p}T\rho \mathbf {p}} \end{aligned}$$

  (25)

  and $\rho _i(\mathbf p)$ denotes the expression

  $$\begin{aligned} \frac{1}{1+\mathbf {p}_{i}T\rho _{\mathbf p}\mathbf {p}} \quad 0\le i\le m. \end{aligned}$$

  (26)

Proof

• (i) The continuity of $\varrho _i$ in $\mathbf {x}$ immediately follows from Lemma 6.2. (Set $c_{ij}$ equal to $\mathbf p_iT x_j\mathbf p_j$.)

• (ii) Since $\bar{\varrho }(\mathbf x)=(1-x)\varrho _0(\mathbf x)+ x \tilde{\varrho }(\mathbf x)$, Proposition 3.1 shows that

  $$\begin{aligned} \bar{\varrho }(\mathbf {x})=\frac{1}{1+\bar{\mathbf h} T\bar{\varrho }(\mathbf {x})\bar{\mathbf h}} \end{aligned}$$

  and

  $$\begin{aligned} \bar{\varrho }(\mathbf {y})=\frac{1}{1+\bar{\mathbf h}T\bar{\varrho } (\mathbf {y})\bar{\mathbf h}} \end{aligned}$$

  hold (as if the population consisted of only $\bar{\mathbf h}$-strategists). Lemma 6.1 says that the equation

  $$\begin{aligned} \bar{\varrho }=\frac{1}{1+\bar{\mathbf h}T\bar{\varrho }\bar{\mathbf h}} \end{aligned}$$

  (with $\bar{\varrho }$ as the unknown) has a unique solution in [0, 1] which implies at once that $\bar{\varrho }(\mathbf {x})=\bar{\varrho }(\mathbf {y})$. Also,

  $$\begin{aligned} \varrho _i=\frac{1}{1+\mathbf p_iT\sum _jx_j\varrho _j\mathbf p_j} =\frac{1}{1+\mathbf p_iT\bar{\varrho }\bar{\mathbf h}} \end{aligned}$$

  has a unique solution (in the unknowns $\varrho _i$) in $[0,1]^{n+1}$ for every i from which $\varrho _i(\mathbf {x})=\varrho _i(\mathbf {y})$ follows for every $0\le i \le m$.

• (iii) This is an immediate consequence of (ii) by comparing the coefficients in the representations $\bar{\mathbf h}(\mathbf {x})$ and $\bar{\mathbf h}(\mathbf {y})$.

• (iv) Take the frequencies

  $$\begin{aligned} x_i=\frac{\rho _{\mathbf p}}{\rho _{i}(\mathbf p)}\alpha _{i}. \end{aligned}$$

  Then, by (25) and (26), we have that

  $$\begin{aligned} \sum _{i=0}^mx_i&=\sum _{i=0}^m \frac{\rho _{\mathbf p}}{\rho _{i}(\mathbf p)}\alpha _{i}= \sum _{i=0}^m\alpha _i \frac{1+\mathbf {p}_iT\rho _{\mathbf p}\mathbf {p}}{1+\mathbf {p}T\rho _{\mathbf p}\mathbf {p}}=1, \end{aligned}$$

  that is $\mathbf x=(x_0,x_1,\ldots ,x_m)$ is a frequency distribution. Consider the polymorphic population of strategies $\mathbf p_0,\mathbf p_1,\ldots ,\mathbf p_m$ with this frequency distribution $\mathbf x$. Then, $\rho _0(\mathbf p), \rho _1(\mathbf p),\ldots ,\rho _m(\mathbf p)$ satisfy the following system of equations corresponding the system (5) with $\mathbf p^*=\mathbf p_0$:

  $$\begin{aligned} \varrho _{i} =\frac{1}{1+\mathbf {p}_{i}T\big [\sum _{j=0}^{m}x_{j}\varrho _{j}\mathbf {p}_j\big ]}. \end{aligned}$$

  (27)

  Indeed, by a simple replacement, we get that

  $$\begin{aligned} \frac{1}{1+\mathbf {p}_{i}T\big [\sum _{j=0}^{m}x_{j} \rho _{j}(\mathbf p)\mathbf {p}_{j}\big ]}&= \frac{1}{1+\mathbf {p}_{i}T\big [\sum _{j=0}^{m} \alpha _j\frac{\rho _{\mathbf p}}{\rho _j(\mathbf p)} \rho _{j}(\mathbf p)\mathbf {p}_{j}\big ]}\\&= \frac{1}{1+\mathbf {p}_{i}T\big [\sum _{j=0}^{m} \rho _{\mathbf p}\alpha _j\mathbf {p}_{j}\big ]}= \frac{1}{1+\mathbf {p}_{i}T \rho _{\mathbf p}\mathbf p}=\rho _i(\mathbf p) \end{aligned}$$

  On the other hand, $\varrho _0(\mathbf x), \varrho _1(\mathbf x),\ldots ,\varrho _n(\mathbf x)$ are also the solutions of the system of equations (27). By the uniqueness (see Lemma 6.1 in A.1.) we have that $\varrho _i(\mathbf x)=\rho _i(\mathbf p)$. Consequently, we have that

  $$\begin{aligned} \bar{\varrho }(\mathbf x)=\sum _{i=0}^m x_i \varrho _i(\mathbf x)= \sum _{i=0}^m x_i \rho _i(\mathbf p)= \sum _{i=0}^m \left( \frac{\rho _{\mathbf p}}{\rho _i(\mathbf p)}\alpha _i\right) \rho _i(\mathbf p)= \rho _{\mathbf p}\sum _{i=0}^m\alpha _i=\rho _{\mathbf p} \end{aligned}$$

  and

  $$\begin{aligned} \bar{\mathbf h}(\mathbf x)=\frac{1}{\bar{\varrho }(\mathbf x)}\sum _{i=0}^m x_i \varrho _i(\mathbf x)\mathbf p_i= \frac{1}{\rho _{\mathbf p}}\sum _{i=0}^m \left( \frac{\rho _{\mathbf p}}{\rho _i(\mathbf p)}\alpha _i\right) \rho _i(\mathbf p) \mathbf p_i= \sum _{i=0}^m\alpha _i\mathbf p_i=\mathbf p. \end{aligned}$$

  $\square$

Lemma 6.4

Let $\mathbf p,\mathbf r\in S_2$. Denote by $\varrho _{\mathbf p}(\varepsilon ),\varrho _{\mathbf r}(\varepsilon )$ the unique solution in $[0,1]^2$ of the system:

$$\begin{aligned} \varrho _{\mathbf p}=\frac{1}{1+\mathbf p T [(1-\varepsilon )\varrho _{\mathbf p} \mathbf p+\varepsilon \varrho _{\mathbf r} \mathbf r]},\\ \varrho _{\mathbf r}=\frac{1}{1+\mathbf r T [(1-\varepsilon )\varrho _{\mathbf p} \mathbf p+\varepsilon \varrho _{\mathbf r} \mathbf r]}. \end{aligned}$$

Furthermore, $\bar{\varrho }(\varepsilon ):=(1-\varepsilon )\varrho _{\mathbf p}(\varepsilon )+ \varepsilon \varrho _{\mathbf r}(\varepsilon )$ and

$$\begin{aligned} \mathbf q(\varepsilon ):=\frac{1}{\bar{\varrho }(\varepsilon )} [(1-\varepsilon )\varrho _{\mathbf p}(\varepsilon )\mathbf p+ \varepsilon \varrho _{\mathbf r}(\varepsilon )\mathbf r]. \end{aligned}$$

Then $\mathbf q(0)=\mathbf p$, $\mathbf q(1)=\mathbf r$ and $\mathbf q(\varepsilon )$ uniquely runs through the line segment between $\mathbf p$ and $\mathbf r$ as $\varepsilon$ runs from 0 to 1 in such a way that $0\le \varepsilon _1<\varepsilon _2\le 1$ implies that $||\mathbf q(\varepsilon _1)-\mathbf p||<||\mathbf q(\varepsilon _2)-\mathbf p||$.

Proof

Corollary 6.3 (i) with the choice $\mathbf p_0=\mathbf p$ and $\mathbf p_1=\mathbf r$ shows that $\varrho _{\mathbf p}(\varepsilon )$ and $\varrho _{\mathbf r}(\varepsilon )$ are continuous in $\varepsilon \in [0,1]$. Therefore, both $\bar{\varrho }(\varepsilon )$ and $\mathbf q(\varepsilon )$ are continuous in $\varepsilon$. Since $\mathbf q(0)=\mathbf p$ and $\mathbf q(1)=\mathbf q$ the Bolzano-Darboux property of continuous function (intermediate value theorem) ensures that $\mathbf q(\varepsilon )$ runs through the line segment between $\mathbf p$ and $\mathbf r$ as $\varepsilon$ runs from 0 to 1. Furthermore, by Corollary 6.3 (iii), $\mathbf q(\varepsilon _1)\not = \mathbf q(\varepsilon _2)$ also holds. Since $\mathbf q(0)=\mathbf p$, it follows that $||\mathbf q(\varepsilon _1)-\mathbf p||<||\mathbf q(\varepsilon _2)-\mathbf p||$.$\square$

1.2 A.2.

Lemma 6.5

Let $\mathbf p^*\in S_2$ and $\hat{\mathbf x}=(\hat{x}_1,\hat{x}_2)\in S_2$ be the unique solution of $\bar{\mathbf h}(\mathbf x)=\mathbf p^*$. Then $\mathbf p^*$ is an ESS if and only if there is a $\delta >0$ such that

$$\begin{aligned} \hat{x}_1W_1(\mathbf x)+\hat{x}_2 W_2(\mathbf x)> x_1W_1(\mathbf x)+x_2 W_2(\mathbf x)=\bar{W}(\mathbf x) \end{aligned}$$

whenever $|x_1-\hat{x}_1|<\delta .$

Proof

Consider a $\mathbf p^*\in S_2$. Without loss of generality, it can be assumed that $\mathbf p^*\not = \mathbf e_2$. We use the following notations:

$$\begin{aligned} \bar{\varrho }(\mathbf x):=x_1\varrho _1(\mathbf x) + x_2\varrho _2(\mathbf x)\quad \text {and}\quad \bar{\mathbf h}(\mathbf x):= x_1\frac{\varrho _1(\mathbf x)}{\bar{\varrho }(\mathbf x)} \mathbf e_1+ x_2\frac{\varrho _2(\mathbf x)}{\bar{\varrho }(\mathbf x)} \mathbf e_2. \end{aligned}$$

(28)

Similarly, we introduce the notations $\hat{\varrho }(\mathbf x)$ and $\hat{\mathbf h}(\mathbf x)$, respectively, as follows:

$$\begin{aligned} \hat{\varrho }(\mathbf x):=\hat{x}_1\varrho _1(\mathbf x) + \hat{x}_2\varrho _2(\mathbf x)\quad \text {and}\quad \hat{\mathbf h}(\mathbf x):= \hat{x}_1\frac{\varrho _1(\mathbf x)}{\hat{\varrho }(\mathbf x)} \mathbf e_1+ \hat{x}_2\frac{\varrho _2(\mathbf x)}{\hat{\varrho }(\mathbf x)} \mathbf e_2. \end{aligned}$$

(29)

Assume that $\hat{x}_2<x_2$ (the case $x_2<\hat{x}_2$ can be handled in the same way). It is easy to check this is equivalent to the inequality

$$\begin{aligned} \hat{x}_2\frac{\varrho _2(\mathbf x)}{\hat{\varrho }(\mathbf x)} <x_2\frac{\varrho _2(\mathbf x)}{\bar{\varrho }(\mathbf x)} \end{aligned}$$

which implies that $\bar{\mathbf h}(\mathbf x)$ lies on the line segment between $\hat{\mathbf h}(\mathbf x)$ and $\mathbf e_2$. By Lemma 6.4, it is also true that $\bar{\mathbf h}(\mathbf x)$ is located on the line segment between $\mathbf p^*$ and $\mathbf e_2$. Thus, there is an $\hat{\eta }= \hat{\eta }(\mathbf x)$ and an $\eta =\eta (\mathbf x)$, respectively, between 0 and 1, such that

$$\begin{aligned} \bar{\mathbf h}(\mathbf x)=(1-\hat{\eta })\frac{\rho _{\hat{\mathbf h}(\mathbf x)} (\hat{\mathbf h}(\mathbf x),\mathbf e_2,\hat{\eta })}{\bar{\rho }(\hat{\mathbf h}(\mathbf x),\mathbf e_2,\hat{\eta })}\hat{\mathbf h}(\mathbf x)+ \hat{\eta }\frac{\rho _{\mathbf e_2} (\hat{\mathbf h}(\mathbf x),\mathbf e_2,\hat{\eta })}{\bar{\rho }(\hat{\mathbf h}(\mathbf x),\mathbf e_2,\hat{\eta })}\mathbf e_2 \end{aligned}$$

(30)

and

$$\begin{aligned} \bar{\mathbf h}(\mathbf x)=(1-\eta )\frac{\rho _{\mathbf p^*} (\mathbf p^*,\mathbf e_2,\eta )}{\bar{\rho }(\mathbf p^*,\mathbf e_2, \eta )}\mathbf p^*+ \eta \frac{\rho _{\mathbf e_2} (\mathbf p^*,\mathbf e_2, \eta )}{\bar{\rho }(\mathbf p^*,\mathbf e_2, \eta )}\mathbf e_2, \end{aligned}$$

(31)

respectively. Observe that $\bar{\rho }(\hat{\mathbf h}(\mathbf x),\mathbf e_2,\hat{\eta })=\bar{\rho }(\mathbf p^*,\mathbf e_2, \eta )=\bar{\varrho }(\mathbf x)$ and $\rho _{\mathbf e_2} (\hat{\mathbf h}(\mathbf x),\mathbf e_2,\hat{\eta })= \rho _{\mathbf e_2} (\mathbf p^*,\mathbf e_2, \eta )=\varrho _2(\mathbf x).$¹⁵ Note that $\eta$ and $x_2$ mutually determine each other so we can write $\eta =\eta (x_2)$ or $x_2=x_2(\eta )$ depending on what is given. A similar observation is true for the relationship between $\hat{\eta }$ and $x_2$.

Suppose, for some $x_2$, we have

$$\begin{aligned} \rho _{\mathbf p^*} (\mathbf p^*,\mathbf e_2,\eta )\mathbf p^*A&\underbrace{\bar{\rho }(\mathbf p^*,\mathbf e_2, \eta ) \left[ (1-\eta )\frac{\rho _{\mathbf p^*} (\mathbf p^*,\mathbf e_2,\eta )}{\bar{\rho }(\mathbf p^*,\mathbf e_2, \eta )}\mathbf p^*+ \eta \frac{\rho _{\mathbf e_2} (\mathbf p^*,\mathbf e_2, \eta )}{\bar{\rho }(\mathbf p^*,\mathbf e_2, \eta )}\mathbf e_2\right] }_ {\bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x)}\nonumber \\&> \rho _{\mathbf e_2} (\mathbf p^*,\mathbf e_2,\eta )\mathbf e_2A\bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x)= \varrho _2(\mathbf x) \mathbf e_2 A \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x). \end{aligned}$$

(32)

Multiply both sides of inequality (32) by $(1-\eta )$ then add $\eta \varrho _2(\mathbf x) \mathbf e_2 A \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x)$ to get that

$$\begin{aligned} \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x) A \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x) > \varrho _2(\mathbf x) \mathbf e_2 A \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x) \end{aligned}$$

(33)

By subtracting $\hat{\eta }\varrho _2(\mathbf x) \mathbf e_2 A \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x)$ from both sides of (33), simplifying by $(1-\hat{\eta })$ and using (30) we obtain inequality:

$$\begin{aligned} \rho _{\hat{\mathbf h}(\mathbf x)} (\hat{\mathbf h}(\mathbf x),\mathbf e_2,\hat{\eta }) \hat{\mathbf h}(\mathbf x) A \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x) > \varrho _2(\mathbf x) \mathbf e_2 A \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x). \end{aligned}$$

(34)

Multiplying (34) by $\hat{\eta }$ and adding $(1-\hat{\eta }) \rho _{\hat{\mathbf h}(\mathbf x)} (\hat{\mathbf h}(\mathbf x),\mathbf e_2,\hat{\eta }) \hat{\mathbf h}(\mathbf x) A \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x)$ to both sides results in

$$\begin{aligned} \rho _{\hat{\mathbf h}(\mathbf x)} (\hat{\mathbf h}(\mathbf x),\mathbf e_2,\hat{\eta }) \hat{\mathbf h}(\mathbf x) A \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x) >\bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x)A \bar{\varrho }(\mathbf x)\bar{\mathbf h}(\mathbf x) \end{aligned}$$

(35)

where the right-hand side is just equal to $x_1W_1(\mathbf x)+x_2 W_2(\mathbf x)$. As regards the left-hand side, observe that, by Lemma 6.1, $\rho _{\hat{\mathbf h}(\mathbf x)}(\hat{\mathbf h}(\mathbf x),\mathbf e_2,\hat{\eta })= \hat{\varrho }(\mathbf x)$.¹⁶ Therefore the left-hand side just equals $\hat{x}_1W_1(\mathbf x)+\hat{x}_2 W_2(\mathbf x)$. The above reasoning shows the equivalence of inequalities (32), (33), (34) and (35) which means that

$$\begin{aligned} \hat{x}_1 W_1(\mathbf x)+\hat{x}_2 W_2(\mathbf x)> x_1 W_1(\mathbf x)+x_2 W_2(\mathbf x) \end{aligned}$$

(36)

holds if and only if inequality (32) does.

If $\mathbf p^*$ is an ESS then consider $\varepsilon _0$ from Definition 2.1. By continuity (see Corollary 6.3), there exists a $\delta >0$ such that if $0<|x_2-\hat{x}_2|<\delta$ then $0<\eta (x_2)<\varepsilon _0$. Therefore, (32) holds which, as we have just seen, is equivalent to (36).

Conversely, assume the existence of a $\delta >0$ such that (36) holds whenever $0<|x_2-\hat{x}_2|<\delta$. By Corollary 6.3 and Lemma 6.4, there is an $\varepsilon _0(\mathbf e_2)$ such that $0<|x_2(\eta )-\hat{x}_2|<\delta$ whenever $0<\eta <\varepsilon _0$.¹⁷ Consider the strategy $\mathbf r\in S_2$ with $r_2>p_2^*$ (the case $r_2<p_2^*$ can be treated in a similar way). Then $\mathbf r$ is on the segment between $\mathbf p^*$ and

$$\begin{aligned} \mathbf h(\mathbf e_2,\varepsilon _0(\mathbf e_2)):=(1-\varepsilon _0(\mathbf e_2)) \frac{\rho _{\mathbf p^*}(\mathbf p^*,\mathbf e_2,\varepsilon _0(\mathbf e_2))}{\bar{\rho }(\mathbf p^*,\mathbf e_2,\varepsilon _0(\mathbf e_2))}\mathbf p^*+ \varepsilon _0(\mathbf e_2) \frac{\rho _{\mathbf e_2}(\mathbf p^*,\mathbf e_2,\varepsilon _0 (\mathbf e_2))}{\bar{\rho }(\mathbf p^*,\mathbf e_2,\varepsilon _0(\mathbf e_2))}\mathbf e_2 \end{aligned}$$

or on the segment between $\mathbf h(\mathbf e_2,\varepsilon _0(\mathbf e_2))$ and $\mathbf e_2$. In the former case set $\varepsilon _0(\mathbf r)$ to be 1, in the latter one set $\varepsilon _0(\mathbf r)$ to be the unique $\zeta _0$ for which $\mathbf h(\mathbf e_2,\varepsilon _0(\mathbf e_2))=\mathbf h(\mathbf r,\zeta _0)$ (such $\zeta _0$ exists by Lemma 6.4). Then, again by Lemma 6.4, for any $0<\zeta <\varepsilon _0(\mathbf r)$ there is an $0<\eta <\varepsilon _0(\mathbf e_2)$ with $\mathbf h(\mathbf e_2,\eta )=\mathbf h(\mathbf r,\zeta )$. Following the reasoning in the proof of Proposition 3.1 we see that $\bar{\rho }(\mathbf p^*,\mathbf e_2,\eta )=\bar{\rho }(\mathbf p^*,\mathbf r,\zeta )$ and, hence, $\rho _{\mathbf p^*}(\mathbf p^*,\mathbf e_2,\eta )= \rho _{\mathbf p^*}(\mathbf p^*,\mathbf r,\zeta )$. From these observations and the argument above that inequalities (32), (33), (34) and (35) are equivalent, we conclude that the ensuing inequalities are equivalent with each other:

$$\begin{aligned} \rho _{\mathbf p^*}(\mathbf p^*,\mathbf r,\zeta )\mathbf p^*A\underbrace{\left[ (1-\zeta )\frac{\rho _{\mathbf p^*}(\mathbf p^*,\mathbf r,\zeta )}{\bar{\rho }(\mathbf p^*,\mathbf r,\zeta )}\mathbf p^*+ \zeta \frac{\rho _{\mathbf p^*}(\mathbf p^*,\mathbf r,\zeta )}{\bar{\rho }(\mathbf p^*,\mathbf r,\zeta )}\mathbf r\right] } _{\bar{\rho }(\mathbf p^*,\mathbf r,\zeta )\mathbf h(\mathbf r,\zeta )} \end{aligned}$$

(37)

$$\begin{aligned}&> \rho _{\mathbf r}(\mathbf p^*,\mathbf r,\zeta )\mathbf rA \bar{\rho }(\mathbf p^*,\mathbf r,\zeta )\mathbf h(\mathbf r,\zeta )\\ \rho _{\mathbf p^*}(\mathbf p^*,\mathbf r,\zeta )\mathbf p^* A\bar{\rho }(\mathbf p^*,\mathbf r,\zeta )\mathbf h(\mathbf r,\zeta )&> \bar{\rho }(\mathbf p^*,\mathbf r,\zeta )\mathbf h(\mathbf r,\zeta )A \bar{\rho }(\mathbf p^*,\mathbf r,\zeta )\mathbf h(\mathbf r,\zeta )\\ \rho _{\mathbf p^*}(\mathbf p^*,\mathbf e_2,\eta )\mathbf p^* A\bar{\rho }(\mathbf p^*,\mathbf e_2,\eta )\mathbf h(\mathbf e_2,\eta )&>\bar{\rho }(\mathbf p^*,\mathbf e_2,\eta )\mathbf h(\mathbf e_2,\eta )A \bar{\rho }(\mathbf p^*,\mathbf e_2,\eta )\mathbf h(\mathbf e_2,\eta )\\ \rho _{\mathbf p^*}(\mathbf p^*,\mathbf e_2,\eta )\mathbf p^* A \bar{\rho }(\mathbf p^*,\mathbf e_2,\eta )\mathbf h(\mathbf e_2,\eta )&>\rho _{\mathbf e_2}(\mathbf p^*,\mathbf e_2,\eta )\mathbf e_2 A \bar{\rho }(\mathbf p^*,\mathbf e_2,\eta )\mathbf h(\mathbf e_2,\eta ). \end{aligned}$$

We have seen that the last inequality is true for any $0<\eta <\varepsilon _0(\mathbf e_2)$ whenever (36) holds for $(\hat{x}_1,\hat{x}_2)$. This proves that $\mathbf p^*$ is an ESS.$\square$