The refined best-response correspondence in normal form games

Abstract

This paper provides an in-depth study of the (most) refined best-response correspondence introduced by Balkenborg et al. (Theor Econ 8:165–192, 2013). An example demonstrates that this correspondence can be very different from the standard best-response correspondence. In two-player games, however, the refined best-response correspondence of a given game is the same as the best-response correspondence of a slightly modified game. The modified game is derived from the original game by reducing the payoff by a small amount for all pure strategies that are weakly inferior. Weakly inferior strategies, for two-player games, are pure strategies that are either weakly dominated or are equivalent to a proper mixture of pure strategies. Fixed points of the refined best-response correspondence are not equivalent to any known Nash equilibrium refinement. A class of simple communication games demonstrates the usefulness and intuitive appeal of the refined best-response correspondence.

Appendices

Appendix

On the generic equivalence of best responses and refined best responses

This appendix provides a proof of Theorem 4, which is organized in a number of steps: We will first fix some notations for the mappings and various submanifolds to be considered. Step 1 argues that the embedding of the uncorrelated strategy combinations into the set of beliefs has nice differentiability properties. Step 2 invokes the transversality theorem (see Guillemin and Pollack 1974) to show that for generic payoff functions we obtain the needed transversality conditions.¹⁵ Step 3 argues that we can restrict attention to completely mixed strategy combinations of the opponents. If the player is indifferent between several of his strategies against a given completely mixed strategy combination, step 4 shows how we can construct an arbitrarily nearby strategy combination, against which the player strictly prefers a given one among these strategies. Step 5 completes the argument.

For any finite set $M$ let $\mathbb {R}^{M}$ be the vector space of all mappings $y:M \rightarrow \mathbb {R}$. The dimension of $\mathbb {R}^{M}$ is the number of elements in $M$.

Let $q_{i}:\prod _{j\ne i}\mathbb {R}^{S_{j}} \rightarrow \mathbb {R}^{S_{-i}}$ be the mapping defined by

$$\begin{aligned} \left( q_{i}\left( x_{-i}\right) \right) \left( s_{-i}\right) :=\prod _{j\ne i}x_{j}\left( s_{j}\right) . \end{aligned}$$

The multilinear function $q_{i}$ describes the first step in the computation mentioned above. While $x_{-i}\in \prod _{j\ne i}\mathbb {R}^{S_{j}}$ denotes the usual strategy combinations of the opponents, we use $\chi _{-i}\in \mathbb {R}^{S_{-i}}$ to describe a “correlated strategy of the opponents”, i.e., a belief over the set $S_{-i}$ of pure strategy combinations of the opponents. $q_{i}$ maps mixed strategy combinations to such beliefs.

Let $L_{i}$ be the vector space of all linear mappings

$$\begin{aligned} v_{i}:\mathbb {R}^{S_{-i}}\rightarrow \mathbb {R}^{S_{i}}. \end{aligned}$$

If $\chi _{-i}\in \mathbb {R}^{S_{-i}}$ is a belief of player $i$ and $s_{i}\in S_{i}$ a pure strategy player $i$ chooses, then $\left( v_{i}\left( \chi _{-i}\right) \right) \left( s_{i}\right)$ is the payoff player $i$ expects with his strategy choice. In this context a vector $z\in \mathbb {R}^{S_{i}}$ represents the various gains a player could make, not probabilities. The linear function $v_{i}$ describes for every $s_{i}$ the second step in the computation of the expected payoff. Any $v_{i}\in L_{i}$ corresponds via $u_{i}\left( s_{-i} ,s_{i}\right) =\left( v_{i}\left( s_{-i}\right) \right) \left( s_{i}\right)$ uniquely to a payoff function

$$\begin{aligned} u_{i}:S\rightarrow \mathbb {R} \end{aligned}$$

in the standard notation (and this relation is a homeomorphism).

For $T_{i}\subseteq S_{i}$ set $Z_{i}\left( T_{i}\right) =\{z\in \mathbb {R}^{S_{i}}\mid \forall s_{i},t_{i}\in T_{i}:z\left( s_{i}\right) =z\left( t_{i}\right) \}$. $v_{i}\left( \chi _{-i}\right) \in Z_{i}\left( T_{i}\right)$ means that player $i$ is indifferent between all his strategies in $T_{i}$ when his belief is $\chi _{-i}$. Let $X_{j} :=\{x_{j}\in \mathbb {R}^{S_{j}}\mid \sum _{s_{j}\in S_{j}}x_{j}\left( s_{j}\right) =1\}$ be the affine space generated by player $j^{\prime }s$ strategy simplex for $j\ne i$ and let $X_{-i}:=\prod _{j\ne i}X_{j}$.

For $T_{j}\subseteq S_{j}$ $\left( j\ne i\right)$ and $T_{-i} :=\prod _{j\ne i}T_{j}$ set

$$\begin{aligned} X_{j}\left( T_{j}\right) :=\{x_{j}\in X_{j}\mid \forall s_{j}\notin T_{j}:x_{j}\left( s_{j}\right) =0\} \end{aligned}$$

and

$$\begin{aligned} X_{-i}\left( T_{-i}\right) :=\prod _{j\ne i}X_{j}\left( T_{j}\right) . \end{aligned}$$

The sets $\Theta _{-i}\cap X_{-i}\left( T_{-i}\right)$ describe the various faces of the polyhedron $\Theta _{-i}$. The strategies of player $j$ with support in $T_{j}$ have $X_{j}\left( T_{j}\right)$ as their affine hull.

Step 1: For all $T_{-i}$ the mapping $q_{i} :X_{-i}\left( T_{-i}\right) \rightarrow \mathbb {R}^{S_{-i}} \setminus \{0\}$ is a diffeomorphism onto its image (in particular $q_{i}\left( X_{-i}\left( T_{-i}\right) \right)$ is a closed submanifold of $\mathbb {R}^{S_{-i}} \setminus \{0\}$).

Proof

$X_{-i}\left( T_{-i}\right)$ is a closed affine submanifold in $\prod _{j\ne i}\left( \mathbb {R}^{S_{j}}\setminus \{0\}\right)$. It is straightforward to check that

$$\begin{aligned} q_{i|X_{-i}\left( T_{-i}\right) }:X_{-i}\left( T_{-i}\right) \rightarrow \mathbb {R}^{S_{-i}}\setminus \{0\} \end{aligned}$$

is well defined, is one-to-one, maps $X_{-i}\left( T_{-i}\right)$ onto a closed set, and has a derivative $dq_{i}|_{x_{-i}}$ with maximal rank everywhere.¹⁶ $\square$

Step 2: Let $Z\subseteq \mathbb {R}^{S_{i}}$ and $X\subseteq \mathbb {R}^{S_{-i} }\setminus \{0\}$ be submanifolds. Then for almost every $v_{i}\in L_{i}$ the mapping $v_{i}|_{X}:X\rightarrow \mathbb {R}^{S_{-i} }\setminus \{0\}$ is transversal to $Z$.

Proof

The family of linear maps $L_{i}$ defines a mapping

$$\begin{aligned} V_{i}:L_{i}\times \mathbb {R}^{S_{-i}}&\rightarrow \mathbb {R}^{S_{i}}\end{aligned}$$

(1)

$$\begin{aligned} \left( v_{i},\chi _{-i}\right)&\mapsto v_{i}\left( \chi _{-i}\right) . \end{aligned}$$

(2)

The derivative of $V_{i}$ at $\left( v_{i},\chi _{-i}\right)$ can be computed as

$$\begin{aligned} dV_{i}|_{\left( v_{i},\chi _{-i}\right) }:T_{v_{i}}L_{i}\times T_{\chi _{-i}} \mathbb {R}^{S_{-i}} \cong L_{i} \times \mathbb {R}^{S_{-i}}&\rightarrow \mathbb {R}^{S_{i}}\end{aligned}$$

(3)

$$\begin{aligned} \left( \nu _{i},\xi _{-i}\right)&\mapsto \nu _{i} \left( \chi _{-i}\right) + v_{i}\left( \xi _{-i}\right) . \end{aligned}$$

(4)

If $\chi _{-i} \ne 0$ we can find for every $\zeta _{i} \in \mathbb {R}^{S_{i}}$ some $\nu _{i}\in L_{i}$ with $\nu _{i}\left( \chi _{-i}\right) =\zeta _{i}$. Then $dV_{i}|_{\left( v_{i},\chi _{-i}\right) }\left( \nu _{i},0\right) =\zeta _{i}$.

Because for $\chi _{-i} \in X$ the tangent space $T_{\chi _{-i}} X \subseteq \mathbb {R}^{S_{-i}}$ contains the $0$-vector, $dV_{i}|_{\left( v_{i},\chi _{-i}\right) }: T_{v_{i}}L_{i}\times T_{\chi _{-i} }X\rightarrow \mathbb {R}^{S_{i}}$ is surjective. Thus $V_{i}:L_{i}\times X\rightarrow \mathbb {R}^{S_{i}}$ is transversal to $Z$ and our claim follows from the transversality theorem.

By step 1 and step 2 almost every $v_{i} \in L_{i}$ satisfies: $\square$

$\otimes$ :

For all subsets $T_j \subseteq S_j$ $\left( 1 \le j \le n\right)$ the mapping $\left( v_i \circ q_i\right) |_{X_{-i}\left( T_{-i}\right) }$ is transversal to $Z_i\left( T_i\right)$.

For given $v_{i}$ define $Y\left( T_{i}\right) =\{x_{-i}\in X_{-i}\mid \left( v_{i}\circ q_{i}\right) \left( x_{-i}\right) \in Z\left( T_{i}\right) \}$. $Y\left( T_{i}\right) \cap \Theta _{-i}$ is the set of strategy combinations of the opponents such that player $i$ is indifferent between the strategies in $T_{i}$ (i.e., they give the same payoff). If $T_{i}$ is a set of best replies against $x_{-i}$, then $x_{-i}\in Y\left( T_{i}\right) \cap \Theta _{-i}$.

Step 3: Suppose $v_{i}$ satisfies $\otimes$. For $T_{i}\subseteq S_{i}$ let $x_{-i}\in Y\left( T_{i}\right) \cap \Theta _{-i}$ and let $O_{-i}$ be a neighborhood of $x_{-i}$. Then $O_{-i}\cap Y\left( T_{i}\right)$ contains a point in the interior of $\Theta _{-i}$.

Proof

Suppose $x_{-i}$ is in the boundary of $\Theta _{-i}$. For each $j\ne i$ define $T_{j}:=\{s_{j}\in S_{j}\mid x_{j}\left( s_{j}\right) \ne 0\}$. Thus $x_{-i}\in X\left( T_{-i}\right) \cap \Theta _{-i}$. If $T_{j}=S_{j}$, $x_{j}$ is in the relative interior of $\Theta _{j}$. By assumption $T_{j}\ne S_{j}$ for at least one opponent $j$. Fix $j^{*}\ne i$ with $T_{j*}\ne S_{j^{*}}$ and $t_{j*}\notin T_{j*}$. Set $\tilde{T}_{j}:=T_{j}$ for $i\ne j\ne j^{*}$ and $\tilde{T}_{j*}:=T_{j*}\cup \{t_{j*}\}$. We show that $O_{-i}\cap Y\left( T_{i}\right)$ contains some $y_{-i}\in \Theta _{-i}\cap X\left( \tilde{T}_{-i}\right)$ such that $\tilde{T}_{j}=\{s_{j}\in S_{j}\mid y_{j}\left( s_{j}\right) \ne 0\}$ for all $j\ne i$. In other words: $y_{-i}$ is in the relative interior of the face $\Theta _{-i}\cap X\left( \tilde{T}_{-i}\right)$. The claim then follows by induction.

The transversality conditions imply that the submanifolds $X_{-i}\left( T_{-i}\right)$ and $Y\left( T_{i}\right) \left( \tilde{T} _{-i}\right)$ meet transversally in $X_{-i}\left( \tilde{T}_{-i}\right)$ (see Guillemin and Pollack 1974 , Exercise 1.6.7). Since $X_{-i}\left( T_{-i}\right)$ has codimension 1 in $X_{-i}\left( \tilde{T}_{-i}\right)$ it follows with arguments as in the next step that $X_{-i}\left( \tilde{T}_{-i}\right) \cap Y\left( T_{i}\right) \cap \{y_{-i}\mid y_{j*}\left( t_{j*}\right) >0\}\cap O_{-i}$ intersects the relative interior of $X_{-i}\left( \tilde{T}_{-i}\right) \cap \Theta _{-i}$. $\square$

Step 4: Suppose $v_{i}$ satisfies $\otimes$. For $T_{i}\subseteq S_{i}$ with $\#T_{i}\ge 2$ let $x_{-i}\in Y\left( T_{i}\right)$ be in the interior of $\Theta _{-i}$ and let $O_{-i}$ be a neighborhood of $x_{-i}$. Then we can find for every $s_{i}\in T_{i}$ some $y_{-i}\in O_{-i}\cap \Theta _{-i}$ such that

$$\begin{aligned} \left( v_{i}\circ q_{i}\right) \left( y_{-i}\right) \left( s_{i}\right) >\left( v_{i}\circ q_{i}\right) \left( y_{-i}\right) \left( t_{i}\right) \quad \text{ for } \text{ all } \ t_{i}\in T_{i}\setminus \{s_{i}\}. \end{aligned}$$

Proof

Because $v_{i}\circ q_{i}:X_{-i} \rightarrow \mathbb {R}^{S_{i}}$ is transversal to both $Z\left( T_{i}\right)$ and $Z\left( T_{i}\setminus \{s_{i}\}\right)$ it follows that $v_{i}\circ q_{i}:Y\left( T_{i}\setminus \{s_{i}\}\right) \rightarrow Z\left( T_{i}\setminus \{s_{i}\}\right)$ is transversal to $Z\left( T_{i}\right)$. From this we can deduce the existence of a tangent vector $\xi \in T_{x_{-i}}\left( Y\left( T_{i}\setminus \{s_{i}\}\right) \right)$ with $d\lambda _{|x_{-i}}\left( \xi \right) =1$, where $\lambda$ is the function

$$\begin{aligned} \lambda :Y_{i}\left( T_{i}\setminus \{s_{i}\}\right) \cap X_{-i}&\rightarrow \mathbb {R} \end{aligned}$$

(5)

$$\begin{aligned} y_{-i}&\rightarrow \left( v_{i}\circ q_{i}\right) \left( y_{-i}\right) \left( s_{i}\right) -\left( v_{i}\circ q_{i}\right) \left( y_{-i}\right) \left( t_{i}\right) \end{aligned}$$

(6)

defined for arbitrary but fixed $t_{i}\in T_{i}\setminus \{s_{i}\}$. We can therefore select a differentiable curve

$$\begin{aligned} c:\left( -\epsilon ,\epsilon \right) \rightarrow Y_{i}\left( T_{i} \setminus \{s_{i}\}\right) \end{aligned}$$

with $c\left( 0\right) =x_{-i}$ and $\left( \lambda \circ c\right) ^{\prime }\left( 0\right) =1$. For sufficiently small $0<\gamma <\epsilon \quad$the vector $y_{-i}:=c\left( \gamma \right)$ has the required properties.

Step 5: Suppose $s_{i}$ is a pure best response against $x_{-i}$. For every neighborhood $O_{-i}$ of $x_{-i}$ the continuity of the payoff function and the two steps above can be used to find $y_{-i}\in O_{-i}$ such that $s_{i}\in T_{i}$ is the unique best response against $y_{-i}$. Shrinking the open sets we can find a sequence of such $y_{-i}$’s converging to $x_{-i}$. Continuity yields an open set around each element in the sequence, where $s_{i}$ is the unique best response. $s_{i}$ is the unique best response on the union of these sets, which is again open. Thus $s_{i}$ is a refined best response against $x_{-i}$. $\square$

Refined best responses in two-player games

This appendix provides proofs of Theorems 1 and 2. In the case of two player games the payoff function is linear in the mixed strategy choice of the opponent. This allows the use of convex analysis (see Rockafellar 1970) to study the best-response correspondence of a player. The most direct consequence is the convexity of the region where a strategy is a best response. From this Theorem 1 follows immediately, the arguments are given after Lemma 1 below. More work is needed to obtain Theorem 2. We use conjugate functions and provide the proof after Lemma 2.

We will restrict attention to the best responses of player 1. Suppose player 2 has $K\ge 2$ strategies $s_{2}^{1},\cdots ,s_{2}^{K}$. It will be convenient to identify the mixed strategies $x_{2}\in \Theta _{2}$ with the vectors

$$\begin{aligned} x_{2}=\left( x_{2}^{1},x_{2}^{2},\cdots ,x_{2}^{K-1}\right) \in \mathbb {R}^{K-1} \end{aligned}$$

(7)

for which $x_{2}^{k}\ge 0$ for all $1\le k\le K-1$ and $x_{2}^{K} :=1-\sum _{k=1}^{K-1}x_{2}^{k}\le 0$. Notice that the zero vector corresponds to pure strategy $s_{2}^{K}.$

We define the function $f:\mathbb {R}^{K-1}\rightarrow \mathbb {R}$ by

$$\begin{aligned} f\left( x_{2}\right) =\left\{ \begin{array} [c]{lcc} {\max }_{s_{1}\in S_{1}}u_{1}\left( s_{1},x_{2}\right) &{} \text{ for } &{} x_{2} \in \Theta _{2}\\ +\infty &{} \text{ else } &{} \end{array} \right. \end{aligned}$$

(8)

Because $u_{1}$ is linear in $x_{2}$, $f$ is, in the terminology of Rockafellar (1970) a proper convex polyhedral function. A key idea explored in the following is that the strategies of player 1 that are refined best responses, correspond to the maximal compact faces of the epigraph

$$\begin{aligned} F=\left\{ \left( x_{2},\alpha \right) \in \mathbb {R}^{K-1}\times \mathbb {R}\, |\, f\left( x_{2}\right) \le \alpha \right\} \end{aligned}$$

of $f$, which is a convex (but not compact) polyhedron. Duality theory allows us to identify these faces with the extreme points of the epigraph $F^{*}$ of the conjugate function $f^{*}$ of $f$. This will be used in the proof of theorem Theorem 2.

Each strategy $x_{1}\in \Theta _{1}$ defines an affine function $a:\mathbb {R}^{K-1}\rightarrow \mathbb {R}$ by $a\left( x_{2}\right) =u_{1}\left( x_{1},x_{2}\right)$, which, for all $x_{2}\in \Theta _{2}$, satisfies the inequality $a\left( x_{2}\right) \le f\left( x_{2}\right)$ and $a\left( x_{2}\right) =f\left( x_{2}\right)$ holds if and only if $x_{1}\in \beta _{1}\left( x_{2}\right)$.

For a strategy $x_{1}\in \Theta _{1}$ we define the set

$$\begin{aligned} G\left( x_{1}\right) =\left\{ \left( x_{2},\alpha \right) \in \Theta _{2}\times \mathbb {R}\,|\,x_{1} \in \beta \left( x_{2}\right) \text{ and } \alpha =u_{1}\left( x_{1}, x_{2}\right) \right\} \end{aligned}$$

(9)

and the set $H\left( x_{1}\right) =\left\{ x_{2}\in \Theta _{2} \,|\, x_{1}\in \beta \left( x_{2}\right) \right\}$, the projection of $G\left( x_{1}\right)$ onto $\Theta _{2}$. $H\left( x_{1}\right)$ is the region where $x_{1}$ is a best response.

Lemma 1

The region $H\left( x_{1}\right)$ is a convex polyhedron.

Proof

$G\left( x_{1}\right)$ is a face of the epigraph $\left\{ \left( x_{2},\alpha \right) \in \Theta _{2}\times \mathbb {R}\,|\,f\left( x_{2}\right) \le \alpha \right\}$ of the function $f$, which is a convex polyhedron. $G\left( x_{1}\right)$ is hence a convex polyhedron. $H\left( x_{1}\right)$ is the image of the convex polyhedron $G\left( x_{1}\right)$ under the linear projection mapping and hence also a convex polyhedron. $\square$

Clearly $x_{1}$ is not weakly inferior if and only if the convex polyhedron $H\left( x_{1}\right)$ has non-empty interior $H^{\circ }\left( x_{1}\right)$. Moreover, $H\left( x_{1}\right)$ is the closure of $H^{\circ }\left( x_{1}\right)$ if $H^{\circ }\left( x_{1}\right)$ is not empty. Therefore, if $x_{1}$ is not weakly inferior and a best response against $x_{2},$ then $x_{2}$ is in the closure of the open set $H^{\circ }\left( x_{1}\right)$ and so $x_{1}$ is a refined best response against $x_{2}$. Given Definition 1 this implies immediately Theorem 1.

The remainder of this section aims at proving Theorem 2. We consider again the epigraph $F$ of the map $f$ defined above. We notice that $F$ is a polyhedral convex set whose compact faces are precisely the sets $G\left( x_{1}\right)$ with $x_{1}\in \Theta _{1}$. The non-compact faces are of the form $F\cap \left( \Theta _{1}^{\prime }\times \mathbb {R}\right)$, where $\Theta _{1}^{\prime }$ is a face of $\Theta _{1}$.

The conjugate function $f^{*}:\mathbb {R}^{K-1} \rightarrow \mathbb {R}$ of $f$ is defined by

$$\begin{aligned} f^{*}\left( x_{2}^{*}\right) =\sup _{x_{2}\in \mathbb {R}^{K-1}}\left\{ x_{2}^{*}\bullet x_{2}-f\left( x_{2}\right) \right\} =\max _{x_{2}\in \Theta _{2}}\left\{ x_{2}^{*}\bullet x_{2}-f\left( x_{2}\right) \right\} <\infty , \end{aligned}$$

(10)

where $x_{2}^{*}\bullet x_{2}$ denotes the usual scalar product $\sum _{_{k=1}}^{K-1}x_{2}^{*k}x_{2}^{k}$. As shown for any convex polyhedral function in Rockafellar (1970), the conjugate is again a convex polyhedral function and one has $f^{**}\left( x_{2}\right) =f\left( x_{2}\right)$.

Any two strategies $x_{1},x_{1}^{\prime }\in \Theta _{1}$ define the same affine function if and only if the two strategies are own-payoff equivalent. Without loss of generality we can thus identify $\Theta _{1}$ up to own-payoff equivalence with a subset of the affine functions on $\mathbb {R}^{K-1}$.

Any vector $\left( x_{2}^{*},\alpha \right)$ with $x_{2}^{*} \in \mathbb {R}^{K-1}$ and $\alpha \in \mathbb {R}$ defines one and only one affine function on $\mathbb {R}^{K-1}$ by

$$\begin{aligned} a\left( x_{2}\right) =-\alpha +\sum _{k=1}^{K-1}x_{2}^{*k}x_{2}^{k} \end{aligned}$$

(11)

We will identify affine functions with such vectors. For instance, $e=\left( 1,\ldots ,1\right)$ corresponds to the function $-x_{2}^{K+}=-1+\sum _{k=1}^{K-1}x_{2}^{k}$ that assigns $0$ to the first $K-1$ pure strategies and $-1$ to the last pure strategy of player 2.

Let $F^{*}$ be the epigraph of $f^{*}$.

Lemma 2

$F^{*}$ is a polyhedral convex set generated by extreme points $x_{1}$ that are refined best responses in $\Theta _{1}$ and the directions

$$\begin{aligned} -e_{k}=\left( -e_{k}^{1},\ldots ,-e_{k}^{K}\right) \in \mathbb {R}^{K} \text{ with } e_{k}^{l}=\left\{ \begin{array} [c]{ccc} -1 &{} \text{ for } &{} k=l\\ 0 &{} \text{ else } &{} \end{array} \right. \end{aligned}$$

(12)

for $k=1,\ldots ,K-1$ and

$$\begin{aligned} e=\left( 1,\ldots ,1\right) \in \mathbb {R}^{K} \end{aligned}$$

(13)

Proof

By definition $\left( x_{2}^{*},\alpha ^{*}\right) \in F^{*}$ if and only if $\alpha ^{*}\ge x_{2}^{*}\bullet x_{2}-f^{*}\left( x_{2}\right)$ for all $x_{2}\in \Theta _{2}$. $v\in \mathbb {R}^{K}$ is a direction in $F^{*}$ if and only if there exists $\left( x_{2}^{*},\alpha ^{*}\right) \in F^{*}$ such that all vectors $\left( x_{2}^{*},\alpha ^{*}\right) +\lambda v$ with $\lambda \ge 0$ are in $F^{*}$. We can write $v=-\sum _{k=1}^{K-1}\rho _{k}e_{k}+\rho _{K}e$ with $\rho _{1},\ldots ,\rho _{K}\in \mathbb {R}$ since $-e_{1},\ldots ,-e_{K-1},e$ form a vector basis of $\mathbb {R}^{K}$. We must show that $v$ is a direction in $F^{*}$ if and only if all $\rho _{i}$ are non-negative. Suppose that $v$ is a direction in $F^{*}$. Let $x_{2}=\left( 0,\ldots ,0\right) \in \Theta _{2}$. The condition that $\left( x_{2}^{*},\alpha ^{*}\right) +\lambda v\in F^{*}$ for all $\lambda \ge 0$ yields for this $x_{2}$ that $\alpha ^{*}+\lambda \rho _{K}\ge -f\left( x_{2}\right)$ holds for all $\lambda \ge 0$. This can be true only if $\rho _{K}\ge 0$. For $e_{k}\in \Theta _{2}$ ($1\le k\le K-1$) we obtain similarly $\alpha ^{*}+\lambda \rho _{K}\ge x_{2}^{*k}-\lambda \rho _{k}+\lambda \rho _{K}-f\left( e_{k}\right)$ for all $\lambda \ge 0$, which can hold only if $\rho _{k}\ge 0$. Thus only positive combinations of $-e_{1},\ldots ,-e_{K-1},e$ can be directions in $F^{*}$. For every combination $v=-\sum _{k=1}^{K-1}\rho _{k}e_{k}+\rho _{K}e$ with $\rho _{1} ,\ldots ,\rho _{K}\ge 0$, every $\lambda \ge 0,$ every $\left( x_{2}^{*},\alpha ^{*}\right) \in F^{*}$ and every $x_{2}\in \Theta _{2}$ we have conversely

$$\begin{aligned} \alpha ^{*}+\lambda \rho _{K}\ge x_{2}^{*\prime }x_{2}-\sum _{k=1} ^{K-1}\lambda \rho _{k}x_{2}^{k}+\lambda \rho _{K}-f\left( x_{2}\right) \end{aligned}$$

(14)

which proves that $v$ is a direction in $F^{*}$.

We have characterized the directions of $F^{*}$ and must now determine the extremal points of $F^{*}$. Suppose $\left( \hat{x}_{2}^{*},\hat{\alpha }^{*}\right)$ is an extremal point. Because $F^{*}$ has only finitely many extremal points, these are exposed points by Straszewick’s theorem (Theorem 18.6 in Rockafellar 1970). Therefore we can find $x_{2}\in \Theta _{2}$ such that the hyperplane $\left\{ x_{2}^{*}\bullet x_{2}=f\left( x_{2}\right) \right\}$ is a supporting hyperplane that meets $F^{*}$ only in $\left( \hat{x}_{2}^{*},f^{*}\left( \hat{x} _{2}^{*}\right) \right)$. Because $F^{*}$ has only finitely many extreme points and directions there exists an open neighborhood $U$ of $x_{2}$ in $\Theta _{2}$ for which the hyperplanes $\left\{ x_{2}^{*}\bullet y_{2}=f\left( y_{2}\right) \right\}$ are for all $y_{2}\in U$ supporting hyperplanes that intersect $F^{*}$ only in $\left( \hat{x}_{2}^{*},f^{*}\left( \hat{x}_{2}^{*}\right) \right)$. This implies that the graph of the affine function $\left( \hat{x}_{2}^{*},f^{*}\left( \hat{x}_{2}^{*}\right) \right)$ intersects $F$ in a $K-1$ dimensional face. It is therefore identical to an affine function defined by a strategy $x_{1}$ in $\Theta _{1}$ for which $H\left( x_{1}\right)$ is full dimensional. Given our identification, $\left( \hat{x}_{2}^{*},f^{*}\left( \hat{x}_{2}^{*}\right) \right)$ is consequently a not weakly inferior strategy in $\Theta _{1}$, which was to be shown. $\square$

Proof of Theorem 2

The lemma implies that all extreme points and hence all the points in the compact faces of $F^{*}$ are in $\Theta _{1}$.

However, no points on the compact faces of $F^{*}$ apart from the extreme points are not weakly inferior strategies. To see this, notice that a proper mixture $x_{1}=\sum _{l=1}^{L}\rho _{l}x_{1k}$ ($L>2,\rho _{l}>0$, $\sum _{l=1}^{L}\rho _{l}=1$) of non-equivalent not weakly inferior strategies in $\Theta _{1}$ is weakly inferior. Otherwise there would be an open set in $\Theta _{2}$ on which $x_{1}$ and hence all strategies $x_{1k}$ were best responses. They would yield identical payoffs on an open set and were hence (by Kalai and Samet Kalai and Samet 1984, Lemma 4) all own-payoff equivalent, contradicting the assumption. Per construction such a mixture is own-payoff equivalent to a proper mixture of strategies that are pairwise not own-payoff equivalent.

It remains to consider strategies in $\Theta _{1}$ that are not on a compact face of $F^{*}$. Such a strategy can be written as $x_{1}^{\prime } =x_{1}-\sum _{k=1}^{K}\rho _{k}e_{k}+\rho _{K}e$ where $x_{1}$ is on one of the compact faces of $F^{*}$ and, hence, in $\Theta _{1}$, and the $\rho _{k}$ are all non-negative and at least some of them are strictly positive. We obtain

$$\begin{aligned} u_{1}\left( x_{1}^{\prime },x_{2}\right) =u_{1}\left( x_{1},x_{2}\right) -\sum _{k=1}^{K-1}\rho _{k}x_{2}^{k}-\rho _{K}\left( 1-\sum _{k=1}^{K-1}x_{2} ^{k}\right) \le u_{1}\left( x_{1},x_{2}\right) , \end{aligned}$$

(15)

where this inequality holds as a strict one for the $k$-th pure strategy of player 2 whenever $\rho _{k}>0$. Thus $x_{1}^{\prime }$ is weakly dominated. It is a weakly inferior strategy because it is a best response only on a proper face of $\Theta _{1}$ (see Pearce 1984).

In summary, the only robust strategies in $\Theta _{1}$ are the extreme points of $F^{*}$. All other strategies are proper mixtures of not own-payoff equivalent not weakly inferior strategies or are weakly dominated and therefore weakly inferior. $\square$