Possibilistic beliefs in strategic games

Abstract

We introduce possibilistic beliefs into strategic games, describing a player’s belief about his opponents’ strategies as the set of their strategies he regards as possible. We formulate possibilistic strategic games where each player has preferences over his own strategies conditional on his possibilistic belief about his opponents’ strategies. We define several solution concepts for possibilistic strategic games such as (strict) equilibria, rationalizable sets, iterated elimination of never-best responses, and iterated elimination of strictly dominated strategies, and we study their properties and relationships. We develop a class of possibilistic strategic games called possibilistic supermodular games to relate supermodular games to possibilistic strategic games. Lastly, we discuss a direction of extending our possibilistic framework to games with incomplete information.

Appendix: Proofs

Proof of Proposition 1

Since $R^*$ includes any rationalizable set, it suffices to show that $R^*$ is a rationalizable set to prove that $R^*$ is the largest rationalizable set. Since the empty set is a rationalizable set, we assume that $R^*$ is nonempty. Choose any $s \in R^*$. Since $R^* = \cup _{R \in {\mathcal {R}}} R$, there exists $R \in {\mathcal {R}}$ such that $s \in R$. Then we have $s_i \in {\bar{b}}_i( {\tilde{R}}_{-i}(s_i) )$ for any $i \in N$. Since $R \subseteq R^*$, we have ${\tilde{R}}_{-i}(s_i) \subseteq {\tilde{R}}_{-i}^*(s_i)$. Since ${\bar{b}}_i$ is nondecreasing, we have $s_i \in {\bar{b}}_i( {\tilde{R}}_{-i}^*(s_i) )$ for any $i \in N$. Hence, $R^*$ is a rationalizable set.

Next, we show that $R^*$ is in product form. Choose any $s \in \times _{j \in N} {\tilde{R}}_{j}^*$ and any $i \in N$. Then there exists $s'_{-i} \in S_{-i}$ such that $(s_i, s'_{-i}) \in R^*$. Since $R^*$ is a rationalizable set, we have $s_i \in {\bar{b}}_i( {\tilde{R}}_{-i}^*(s_i) )$. Since ${\tilde{R}}_{-i}^*(s_i) \subseteq \times _{j \ne i} {\tilde{R}}_{j}^*$, we have $s_i \in {\bar{b}}_i( \times _{j \ne i} {\tilde{R}}_{j}^* )$. Hence, $\times _{j \in N} {\tilde{R}}_{j}^*$ is also a rationalizable set. Since $R^* \subseteq \times _{j \in N} {\tilde{R}}_{j}^*$ and $R^*$ is the largest rationalizable set, it must be that $R^* = \times _{j \in N} {\tilde{R}}_{j}^*$.$\square$

Proof of Proposition 2

Since r is nondecreasing and ${\mathcal {P}}(S)$ is a complete lattice, there exists the largest fixed point of r, which is equal to the largest post fixed point, by Tarski’s fixed point theorem. By Proposition 1, $R^*$ is the largest post fixed point of r. Since the game is finite, the set ${\mathcal {P}}(S)$ is finite. Note that S is the largest element of ${\mathcal {P}}(S)$. Then the largest fixed point of r can be obtained as $r^k(S)$ for some finite k (see, for example, Echenique, 2005, Sect. 4). If $B = \times _{j \in N} B_j$ is a product belief, we have ${\tilde{B}}_{-i}(s_i) = B_{-i}:= \times _{j \ne i} B_j$ for any $s_i \in B_i$ and ${\tilde{B}}_{-i}(s_i) = \varnothing$ for any $s_i \notin B_i$, for any $i \in N$. Hence, we have $r(B) = \times _{i \in N} {\bar{b}}_i (B_{-i})$ for any product belief $B \in \times _{i \in N} {\mathcal {P}}(S_i)$. Suppose that $r(B) \ne \varnothing$ for every $B \in \times _{i \in N} {\mathcal {P}}_0(S_i)$. Then $r^k(S)$ is a nonempty product belief for all $k = 1,2,\ldots$, and thus $R^*$ is nonempty. Suppose that $b_i$ is nondecreasing on $\times _{j \ne i} {\mathcal {P}}(S_j)$ for every $i \in N$. Then $r(B) = e(B)$ for all $B \in \times _{i \in N} {\mathcal {P}}(S_i)$. It follows that $R^*$ is given by $e^k(S)$ for some finite k, and it is a (post) fixed point of e. Hence, $R^*$ is a (strict) equilibrium. Since any equilibrium is contained in $R^*$, it is the largest (strict) equilibrium. $\square$

Proof of Proposition 3

The results that there exists an IENBR procedure and that $R^{\infty } = R^*$ for any IENBR procedure $\{ R^{\lambda } \}_{\lambda \in \Lambda }$ can be proven as in the proof of Proposition 2 of Chen et al. (2016). The results that there exists an IESDS procedure and that all IESDS procedures yield the same $D^{\infty }$ can be proven as in the proof of Theorem 1 of Chen et al. (2007). As shown in Proposition 1, $R^*$ is a rationalizable set. Choose any $i \in N$ and any $s \in R^*$. Then $s_i$ is not a never-best response given $R^*$. This implies that $s_i$ is not strictly dominated given $R^*$, which in turn implies that $s_i$ is not strictly dominated given any $S' \subseteq S$ such that $R^* \subseteq S'$. Hence, for any IESDS procedure $\{ D^{\lambda } \}_{\lambda \in \Lambda }$, no $s \in R^*$ is deleted in any round of the procedure, and thus $R^* \subseteq D^{\infty }$ holds. $\square$

Proof of Proposition 4

Since $R^{\infty }$ is nonempty and $R^{\infty } \subseteq D^{\infty }$, $D^{\infty }$ is nonempty as well. Moreover, both $R^{\infty }$ and $D^{\infty }$ are product beliefs.

(i) Let $B^* \subseteq S$ be an equilibrium of G. Since the empty belief is an equilibrium of $G |_{R^{\infty }}$, we assume that $B^*$ is nonempty. Choose any $s \in B^*$ and any $i \in N$. Then we have $s_i \in b_i({\tilde{B}}^*_{-i}(s_i))$. Since $B^* \subseteq R^* = R^{\infty }$, we have $s \in R^{\infty }$ and ${\tilde{B}}^*_{-i}(s_i) \subseteq R^{\infty }_{-i}$. Since there is no $s'_i \in S_i$ such that $s'_i \succ _{i, {\tilde{B}}^*_{-i}(s_i)} s_i$, there is no $s'_i \in R^{\infty }_i$ such that $s'_i \succ _{i, {\tilde{B}}^*_{-i}(s_i)} s_i$. Hence, $s_i$ is a best response to ${\tilde{B}}^*_{-i}(s_i)$ in the reduced game $G |_{R^{\infty }}$. This implies that $B^*$ be an equilibrium of $G |_{R^{\infty }}$.

(ii) For any player $i \in N$, any belief $B_{-i} \subseteq S_{-i}$ of his, and any nonempty subset $S'_i \subseteq S_i$, let us define

$$\begin{aligned} b_i(B_{-i}; S'_i) = \{ s_i \in S'_i : \not \exists s'_i \in S'_i \text { s.t. } s'_i \succ _{i, B_{-i}} s_i \}. \end{aligned}$$

That is, $b_i(B_{-i}; S'_i)$ is the set of best responses to player i’s belief $B_{-i}$ when his strategy set is restricted to $S'_i$. Note that $b_i(B_{-i}) = b_i(B_{-i}; S_i)$ for any $i \in N$ and any $B_{-i} \subseteq S_{-i}$.

Let $B^* \subseteq R^{\infty }$ be a nonempty equilibrium of $G |_{R^{\infty }}$. Choose any $s \in B^*$ and any $i \in N$. Then we have $s_i \in b_i({\tilde{B}}^*_{-i}(s_i); R^{\infty }_{i})$. To show that $s_i \in b_i({\tilde{B}}^*_{-i}(s_i))$, suppose to the contrary that $s_i \notin b_i({\tilde{B}}^*_{-i}(s_i))$. Since the game G has the well-defined best-response property on $B^*$, there exists $s'_i \in S_i$ such that $s'_i \in b_i({\tilde{B}}^*_{-i}(s_i))$ and $s'_i \succ _{i, {\tilde{B}}^*_{-i}(s_i)} s_i$. Since ${\tilde{B}}^*_{-i}(s_i) \subseteq R^{\infty }_{-i}$, we have $s'_i \in R^{\infty }_{i}$. Then it contradicts that $s_i \in b_i({\tilde{B}}^*_{-i}(s_i); R^{\infty }_{i})$. Thus, $B^*$ is an equilibrium of G.

Let $B^* \subseteq R^{\infty }$ be a nonempty strict equilibrium of $G |_{R^{\infty }}$. Using the well-defined best-response property of G on $B^*$, we have already shown that $B^*$ is an equilibrium of G. To show that $B^*$ is a strict equilibrium of G, suppose to the contrary that there exists $s \notin B^*$ such that $s_i \in b_i({\tilde{B}}^*_{-i}(s_i))$ for all $i \in N$. Since $B^* \subseteq R^{\infty }$, we have $s \in R^{\infty }$. Since $B^*$ is a strict equilibrium of $G |_{R^{\infty }}$ and $s \in R^{\infty } {\setminus } B^*$, there exists $i \in N$ such that $s_i \notin b_i({\tilde{B}}^*_{-i}(s_i); R^{\infty }_{i})$. This implies that $s_i \notin b_i({\tilde{B}}^*_{-i}(s_i))$, which is a contradiction. Hence, $B^*$ is a strict equilibrium of G.

Let $B^* \subseteq S$ be a nonempty strict equilibrium of G. We know that $B^* \subseteq R^{\infty }$ and have already shown in (i) that $B^*$ is an equilibrium of $G |_{R^{\infty }}$. To show that $B^*$ is a strict equilibrium of $G |_{R^{\infty }}$, suppose to the contrary that there exists $s \in R^{\infty } \setminus B^*$ such that $s_i \in b_i({\tilde{B}}^*_{-i}(s_i); R^{\infty }_{i})$ for all $i \in N$. Since $B^*$ is a strict equilibrium of G and $s \notin B^*$, there exists $i \in N$ such that $s_i \notin b_i({\tilde{B}}^*_{-i}(s_i))$. Using the well-defined best-response property of G on $B^*$, we can derive a contradiction to $s_i \in b_i({\tilde{B}}^*_{-i}(s_i); R^{\infty }_{i})$ as in the above argument. Thus, $B^*$ is a strict equilibrium of $G |_{R^{\infty }}$.

The results for $G |_{D^{\infty }}$ can be proven analogously. $\square$

Proof of Proposition 5

Let ${\underline{\rho }}(s) = ({\underline{\rho }}_i(s_{-i}))_{i \in N}$ and ${\overline{\rho }}(s) = ({\overline{\rho }}_i(s_{-i}))_{i \in N}$ for all $s \in S$. Let ${\underline{s}}^0 = \inf (S)$ and ${\overline{s}}^0 = \sup (S)$, and let ${\underline{s}}^k = {\underline{\rho }}({\underline{s}}^{k-1})$ and ${\overline{s}}^k = {\overline{\rho }}({\overline{s}}^{k-1})$ for all $k = 1,2,\ldots$. Then $\{ {\underline{s}}^k \}$ is nondecreasing and $\{ {\overline{s}}^k \}$ is nonincreasing, and the sequences have order limits ${\underline{s}}^* = \inf {\underline{s}}^k$ and ${\overline{s}}^* = \sup {\underline{s}}^k$, which are the smallest and largest pure strategy Nash equilibria, respectively, of the supermodular game $\langle N, (S_i), (u_i) \rangle$ (see Theorem 5 of Milgrom and Roberts, 1990). Note also that ${\underline{s}}^*$ and ${\overline{s}}^*$ are the unique fixed points of ${\underline{\rho }}$ and ${\overline{\rho }}$, respectively (see Sobel, 2019). We have $r^k(S) = [{\underline{s}}^k, {\overline{s}}^k]$ for all $k = 1,2,\ldots$, and thus $r^{\omega }(S) = [{\underline{s}}^*, {\overline{s}}^*]$. Since ${\underline{s}}^*$ and ${\overline{s}}^*$ are the fixed points of ${\underline{\rho }}$ and ${\overline{\rho }}$, respectively, $r^{\omega }(S)$ is a fixed point of r. Then by Proposition 3, we have $R^* = r^{\omega }(S) = [{\underline{s}}^*, {\overline{s}}^*]$. $\square$