Algorithms for cautious reasoning in games

Abstract

We provide comparable algorithms for the Dekel–Fudenberg procedure, iterated admissibility, proper rationalizability and full permissibility by means of the notions of likelihood orderings and preference restrictions. The algorithms model reasoning processes whereby each player’s preferences over his own strategies are completed by eliminating likelihood orderings. We apply the algorithms for comparing iterated admissibility, proper rationalizability and full permissibility, and provide a sufficient condition under which iterated admissibility does not rule out properly rationalizable strategies. We also use the algorithms to examine an economically relevant strategic situation, namely a bilateral commitment bargaining game. Finally, we discuss the relevance of our algorithms for epistemic analysis.

Appendix A: Proofs

Proof of Lemma 1

Let ${\mathcal {L}}_i$ be player i’s non-empty set of likelihood orderings.

Reflexivity of $\sim _i^{{\mathcal {L}}_i}$ and $\succsim _i^{{\mathcal {L}}_i}$. Consider any $\mu _i \in \Delta (S_i)$. Then, trivially, $u_i(\mu _i,s_{-i})$$= u_i(\mu _i,s_{-i})$ for every $s_{-i}$, so that $\mu _i \sim _i^{{\mathcal {L}}_i} \mu _i$, and thus, $\mu _i \succsim _i^{{\mathcal {L}}_i} \mu _i$.

Irreflexivity of $\succ _i^{{\mathcal {L}}_i}$. Consider any $\mu _i \in \Delta (S_i)$. Then, trivially, there exists no subset of opponent strategy profiles $S^{\prime }_{-i}\subseteq S_{-i}$ such that $\mu$ weakly dominates $\mu$ on $S^{\prime }_{-i}$. Hence, $\mu _i \nsucc _i^{{\mathcal {L}}_i} \mu _i$.

Transitivity of $\sim _i^{{\mathcal {L}}_i}$. If $\mu _i \sim _i^{{\mathcal {L}}_i} \mu '_i$ and $\mu '_i \sim _i^{{\mathcal {L}}_i} \mu ''_i$, then $u_i(\mu _i,s_{-i}) = u_i(\mu '_i,s_{-i}) = u_i(\mu ''_i,s_{-i})$ for every $s_{-i}$, so that $\mu _i \sim _i^{{\mathcal {L}}_i} \mu ''_i$.

Transitivity of $\succ _i^{{\mathcal {L}}_i}$. If $\mu _i \succ _i^{{\mathcal {L}}_i} \mu '_i$ and $\mu '_i \succ _i^{{\mathcal {L}}_i} \mu ''_i$, then, for all $L_i = (L_i^1, L_i^2, \ldots ,$$L_i^K) \in {\mathcal {L}}_i$, there exists $k' \in \{1,\ldots ,K\}$ such that $\mu _i$ weakly dominates $\mu '_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^{k'}$ and $k'' \in \{1,\ldots ,K\}$ such that $\mu '_i$ weakly dominates $\mu ''_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^{k''}$. For each $L_i \in {\mathcal {L}}_i$, choose $k = \min \{k',k''\}$. Then $\mu _i$ weakly dominates $\mu ''_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^{k}$ since $u_{i}(\mu _{i},s_{-i}) \ge u_{i}(\mu '_{i},s_{-i})$ and $u_{i}(\mu '_{i},s_{-i}) \ge u_{i}(\mu ''_{i},s_{-i})$ for every $s_{-i}\in L_{i}^{1}\cup \cdots \cup L_{i}^k$, with $u_{i}(\mu _{i},s'_{-i}) > u_{i}(\mu '_{i},s'_{-i})$ or $u_{i}(\mu '_{i},s'_{-i}) > u_{i}(\mu ''_{i},s'_{-i})$ for some $s'_{-i}\in L_{i}^{1}\cup \cdots \cup L_{i}^k$. Hence, for all $L_i \in {\mathcal {L}}_i$, there exists $k \in \{1,\ldots ,K\}$ such that $\mu _i$ weakly dominates $\mu ''_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^{k}$, so that so that $\mu _i \succ _i^{{\mathcal {L}}_i} \mu ''_i$.

Transitivity of $\succsim _i^{{\mathcal {L}}_i}$. Consider any $\mu _i$, $\mu '_i$, $\mu ''_i \in \Delta (S_i)$ such that $\mu _i \succsim _i^{{\mathcal {L}}_i} \mu '_i$ and $\mu '_i \succsim _i^{{\mathcal {L}}_i} \mu ''_i$. We must show that $\mu _i \succsim _i^{{\mathcal {L}}_i} \mu ''_i$.

Case 1: If $\mu _i \sim _i^{{\mathcal {L}}_i} \mu '_i$ and $\mu '_i \sim _i^{{\mathcal {L}}_i} \mu ''_i$, then it follows from the transitivity of $\sim _i^{{\mathcal {L}}_i}$ that $\mu _i \sim _i^{{\mathcal {L}}_i} \mu ''_i$, and thus, $\mu _i \succsim _i^{{\mathcal {L}}_i} \mu ''_i$.

Case 2a: If $\mu _i \succ _i^{{\mathcal {L}}_i} \mu '_i$ and $\mu '_i \sim _i^{{\mathcal {L}}_i} \mu ''_i$, then since $u_{i}(\mu '_{i},s_{-i}) = u_{i}(\mu '' _{i},s_{-i})$ for every $s_{-i}$, for all $L_i = (L_i^1, L_i^2, \ldots , L_i^K) \in {\mathcal {L}}_i$, there exists $k \in \{1,\ldots ,K\}$ such that $\mu _i$ weakly dominates $\mu ''_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^{k}$, so that $\mu _i \succ _i^{{\mathcal {L}}_i} \mu ''_i$, and thus, $\mu _i \succsim _i^{{\mathcal {L}}_i} \mu ''_i$.

Case 2b: Likewise if $\mu _i \sim _i^{{\mathcal {L}}_i} \mu '_i$ and $\mu '_i \succ _i^{{\mathcal {L}}_i} \mu ''_i$.

Case 3: If $\mu _i \succ _i^{{\mathcal {L}}_i} \mu '_i$ and $\mu '_i \succ _i^{{\mathcal {L}}_i} \mu ''_i$, then it follows from the transitivity of $\succ _i^{{\mathcal {L}}_i}$ that $\mu _i \succ _i^{{\mathcal {L}}_i} \mu ''_i$, and thus, $\mu _i \succsim _i^{{\mathcal {L}}_i} \mu ''_i$.

Objective independence of $\sim _i^{{\mathcal {L}}_i}$ Consider any $\mu _i$, $\mu '_i$, $\mu ''_i \in \Delta (S_i)$ and all $\gamma \in (0,1)$. Objective independence follows since $u_{i}(\mu '_{i},s_{-i}) = u_{i}(\mu '' _{i},s_{-i})$ for every $s_{-i} \in S_{-i}$ if and only if $u_{i}(\gamma \mu '_{i} + (1-\gamma ) \mu '''_{i},s_{-i}) = u_{i}(\gamma \mu ''_{i} + (1-\gamma ) \mu '''_{i},s_{-i})$ for every $s_{-i} \in S_{-i}$

Objective independence of $\succ _i^{{\mathcal {L}}_i}$. Consider any $\mu _i$, $\mu '_i$, $\mu ''_i \in \Delta (S_i)$ and all $\gamma \in (0,1)$. Objective independence follows since, for any $S'_{-i} \subseteq S_{-i}$, $\mu '_i$ weakly dominates $\mu ''_i$ on $S'_{-i}$ if and only if $\gamma \mu '_{i} + (1-\gamma ) \mu '''_{i}$ weakly dominates $\gamma \mu ''_{i} + (1-\gamma ) \mu '''_{i}$ on $S'_{-i}$. $\square$

Proof of Lemma 2

Let ${\mathcal {L}}_i$ be player i’s non-empty set of likelihood orderings, and consider two mixed strategies $\mu '_i$, $\mu ''_i \in \Delta (S_{i})$.

Assume $\mu '_i \succ _i^{{\mathcal {L}}_i} \mu ''_i$. Hence, for all $L_i = (L_i^1, L_i^2, \ldots , L_i^K) \in {\mathcal {L}}_i$, there exists $k \in \{1,\ldots ,K\}$ such that $\mu '_i$ weakly dominates $\mu ''_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^{k}$. Fix $L_i \in {\mathcal {L}}_i$. We need to show that every LPS $\lambda _{i}$ consistent with $L_i$ ranks $\mu '_i$ above $\mu ''_i$. This follows since $u_{i}(\mu '_i,\lambda _{i}^\ell ) \ge u_{i}(\mu ''_{i},\lambda _{i}^\ell )$ for all $\ell \in \{1, \ldots , k\}$ and $u_{i}(\mu '_i,\lambda _{i}^{\ell '}) > u_{i}(\mu ''_{i},\lambda _{i}^{\ell '})$ for some $\ell ' \in \{1, \ldots , k\}$, as $\mu '_i$ weakly dominates $\mu ''_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^{k} = \text {supp}\lambda _1 \cup \cdots \cup \text {supp}\lambda _k$.

Assume that, for all $L_i \in {\mathcal {L}}_i$, every LPS $\lambda _{i}$ consistent with $L_i$ ranks $\mu '_i$ above $\mu ''_i$. Fix $L_i = (L_i^1, L_i^2, \ldots , L_i^K) \in {\mathcal {L}}_i$. We need to show that there exists $k \in \{1,\ldots ,K\}$ such that $\mu '_i$ weakly dominates $\mu ''_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^{k}$. Suppose, by way of contradiction, that, for all $k \in \{1,\ldots ,K\}$, $\mu '_i$ does not weakly dominate $\mu ''_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^{k} = \text {supp}\lambda _1 \cup \cdots \cup \text {supp}\lambda _k$. Case 1:$u_{i}(\mu '_{i},s_{-i}) \le u_{i}(\mu '' _{i},s_{-i})$ for every $s_{-i} \in S_{-i}$. Then no LPS $\lambda _{i}$ consistent with $L_i$ ranks $\mu '_i$ above $\mu ''_i$. Case 2:$u_{i}(\mu '_{i},s'_{-i}) > u_{i}(\mu '' _{i},s'_{-i})$ for some $s'_{-i} \in S_{-i}$. W.l.o.g., choose $s'_{-i}$ and $k \in \{1, \ldots , K\}$ with the properties that $s'_{-i} \in L_i^k$ and $u_{i}(\mu '_{i},s_{-i}) \le u_{i}(\mu '' _{i},s_{-i})$ for every $s_{-i} \in L_{i}^{1}\cup \cdots \cup L_{i}^{k-1}$. Since $\mu '_i$ does not weakly dominate $\mu ''_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^{k}$, there exists $s''_{-i} \in L_{i}^{1}\cup \cdots \cup L_{i}^{k} = \text {supp}\lambda _1 \cup \cdots \cup \text {supp}\lambda _k$ such that $u_{i}(\mu '_{i},s''_{-i}) < u_{i}(\mu '' _{i},s''_{-i})$. Then it is possible to construct ${\tilde{\lambda }}_i$ consistent with $L_i$ such that $u_{i}(\mu '_i,{\tilde{\lambda }} _{i}^k) < u_{i}(\mu ''_{i},{\tilde{\lambda }} _{i}^k)$ and $u_{i}(\mu '_i,{\tilde{\lambda }} _{i}^\ell ) \le u_{i}(\mu ''_{i},{\tilde{\lambda }} _{i}^\ell )$ for all $\ell \in \{1, \cdots , k-1\}$, implying that ${\tilde{\lambda }}_i$ consistent with $L_i$ ranks $\mu ''_i$ above $\mu '_i$. In both cases, we obtain a contradiction to the claim that every LPS $\lambda _{i}$ consistent with $L_i$ ranks $\mu '_i$ above $\mu ''_i$.

Assume $\mu '_i \sim _i^{{\mathcal {L}}_i} \mu ''_i$. Hence, $u_{i}(\mu '_{i},s_{-i}) = u_{i}(\mu '' _{i},s_{-i})$ for every $s_{-i}$. Then, clearly, every LPS $\lambda _{i}$ deem $\mu '_i$ indifferent to $\mu ''_i$.

Assume that, for all $L_i \in {\mathcal {L}}_i$, every LPS $\lambda _{i}$ consistent with $L_i$, deem $\mu '_i$ indifferent to $\mu ''_i$. We need to show that $u_{i}(\mu '_{i},s_{-i}) = u_{i}(\mu '' _{i},s_{-i})$ for every $s_{-i}$. Suppose, by way of contradiction, that $u_{i}(\mu '_{i},s'_{-i}) \ne u_{i}(\mu '' _{i},s'_{-i})$ for some $s'_{-i} \in S_{-i}$. W.l.o.g., choose $s'_{-i}$, $L_i = (L_i^1, L_i^2, \ldots , L_i^K) \in {\mathcal {L}}_i$, and $k \in \{1, \ldots , K\}$ with the properties that $s'_{-i} \in L_i^k$ and $u_{i}(\mu '_{i},s_{-i}) = u_{i}(\mu '' _{i},s_{-i})$ for every $s_{-i} \in L_{i}^{1}\cup \cdots \cup L_{i}^{k-1}$. Then it is possible to construct ${\tilde{\lambda }}_i$ consistent with $L_i$ such that $u_{i}(\mu '_i,{\tilde{\lambda }} _{i}^{k}) \ne u_{i}(\mu ''_{i},{\tilde{\lambda }} _{i}^{k})$ and $u_{i}(\mu '_i,{\tilde{\lambda }} _{i}^{\ell }) = u_{i}(\mu ''_i,{\tilde{\lambda }} _{i}^{\ell })$ for all $\ell \in \{1,\ldots ,k-1\}$, implying that ${\tilde{\lambda }}_i$ consistent with $L_i$ does not deem $\mu '_i$ indifferent to $\mu ''_i$. This contradicts that every LPS $\lambda _{i}$ consistent with $L_i$, deem $\mu '_i$ indifferent to $\mu ''_i$. $\square$

Proof of Remark 2

Let ${\mathcal {L}}_i$ be player i’s non-empty set of likelihood orderings, and consider two pure opponent strategy profiles $s'_{-i}$, $s''_{-i} \in S_{-i}$. Assume $s_{-i}^{\prime }\gg _{i}^{{\mathcal {L}}_i}s_{-i}^{\prime \prime }$. Hence, for all $L_i = (L_i^1, L_i^2, \ldots , L_i^K) \in {\mathcal {L}}_i$, there exists $k \in \{1, \ldots , K\}$ such that $s'_{-i} \in L_i^1 \cup \cdots \cup L_i^k$ and $s''_{-i} \notin L_i^1 \cup \cdots \cup L_i^k$. If the mixed strategies $\mu '_i$, $\mu ''_i \in \Delta (S_i)$ satisfy $u_i(\mu '_i,s'_{-i}) > u_i(\mu ''_i,s'_{-i})$ and, for all $s_{-i} \in S_{-i} \backslash \{s'_{-i}, s''_{-i}\}$, $u_i(\mu '_i,s_{-i}) = u_i(\mu ''_i,s_{-i})$, then, for all $L_i = (L_i^1, L_i^2, \ldots , L_i^K) \in {\mathcal {L}}_i$, $\mu '_i$ weakly dominates $\mu ''_i$ on $L_{i}^{1}\cup \cdots \cup L_{i}^k$ by choosing $k \in \{1, \ldots , K\}$ such that $s'_{-i} \in L_i^1 \cup \cdots \cup L_i^k$ and $s''_{-i} \notin L_i^1 \cup \cdots \cup L_i^k$. Thus, $\mu '_i \succ _i^{{\mathcal {L}}_i} \mu ''_i$ whenever the mixed strategies $\mu '_i$, $\mu ''_i$ have these properties. $\square$

Proof of Lemma 3

Let $(s_i,A_i) \in R_{i}({\mathcal {L}}_{i})$, implying that there exists $\mu _{i}\in \Delta (A_{i})$ such that $\mu _i \succ _i^{{\mathcal {L}}_{i}} s_i$. Clearly, $\mu _i(s_i) < 1$ since $\succ _i^{{\mathcal {L}}_i}$ is irreflexive (cf. Lemma 1). If $\mu _i(s_i) = 0$, so that $s_i \notin \text {supp}\mu _i$, then $\mu _{i}\in \Delta (A_{i}\backslash \{s_i\})$ and $\mu _i \succ _i^{{\mathcal {L}}_{i}} s_i$, implying that $(s_i,A'_i) \in R_{i}({\mathcal {L}}_{i})$ where $A'_i = A_i \backslash \{s_i\}$. If $\mu _i(s_i) = (0,1)$, rewrite $\mu _i$ as $\mu _i(s_i)s_i + (1-\mu _i(s_i))\mu '_i$, where $\mu '_i$ is defined by $\mu '_i(s'_i) = \mu _i(s'_i)/(1-\mu _i(s_i))$ for all $s'_i \ne s_i$. Then

$$\begin{aligned} \mu _i(s_i)s_i + (1-\mu _i(s_i))\mu '_i = \mu _i \succ _i^{{\mathcal {L}}_{i}} s_i = \mu _i(s_i)s_i + (1-\mu _i(s_i))s_i . \end{aligned}$$

Hence, by the objective independence of $\succ _i^{{\mathcal {L}}_{i}}$ (cf. Lemma 1), $\mu '_i \succ _i^{{\mathcal {L}}_{i}} s_i$, where $\mu '_i \in \Delta (A_{i}\backslash \{s_i\})$. Thus, also in this case, $(s_i,A'_i) \in R_{i}({\mathcal {L}}_{i})$ where $A'_i = A_i \backslash \{s_i\}$. $\square$

Proof of Lemma 4

Assume there exists $\mu _i \in \Delta (S_i)$ such that $\mu _i \succ _i^{{\mathcal {L}}_{i}} s_i$. Then $(s_i, S_i) \in R_i({\mathcal {L}}_i)$ and $s_i \in S_i \backslash C_i(R_i({\mathcal {L}}_i))$. Hence,

$$\begin{aligned} C_i(R_i({\mathcal {L}}_i)) \subseteq \left\{ s_{i}\in S_{i}\mid \not \exists \mu _i \in \Delta (S_i) \text { such that } \mu _i \succ _i^{{\mathcal {L}}_{i}} s_i \right\} . \end{aligned}$$

(A1)

Assume there does not exist $\mu _i \in \Delta (S_i)$ such that $\mu _i \succ _i^{{\mathcal {L}}_{i}} s_i$. Then there does not exist $A_i \subseteq S_i$ with $(s_i,A_i)\in R_i({\mathcal {L}}_i)$, and $s_i \in C_i(R_i({\mathcal {L}}_i))$. Hence,

$$\begin{aligned} C_i(R_i({\mathcal {L}}_i)) \supseteq \left\{ s_{i}\in S_{i}\mid \not \exists \mu _i \in \Delta (S_i) \text { such that } \mu _i \succ _i^{{\mathcal {L}}_{i}} s_i \right\} . \end{aligned}$$

(A2)

The first part of the lemma follows from (A1) and (A2).

To show $C_i(R_i({\mathcal {L}}_i)) \ne \emptyset$, consider an LPS $\lambda _i$ that is consistent with some $L_i \in {\mathcal {L}}_i$. Since the ranking-above relation for given LPS $\lambda _i$ is transitive and irreflexive and $S_i$ is finite, there exists $s_i \in S_i$ such that $\lambda _i$ ranks no $s'_i \in S_i$ above $s_i$. Moreover, by the definition of the ranking-above relation it now follows that $\lambda _i$ ranks no $\mu _i \in \Delta (S_i)$ above $s_i$. Hence, by Lemma 2, there does not exist $\mu _i \in \Delta (S_i)$ such that $\mu _i \succ _i^{{\mathcal {L}}_i} s_i$, implying by the first part of the lemma that $s_i \in C_i(R_i({\mathcal {L}}_i))$. $\square$

Proof of Lemma 5

Assume that player i with non-empty set ${\mathcal {L}}_{i}$ of likelihoods orderings respects $R_{-i}$. Suppose $s'_{j}\notin C_{j}(R_{j})$ for some $j\ne i$, implying that there exists $A_{j}$ such that $(s_{j},A_{j})\in R_{j}$. Since i respects $R_{-i}$, $(L_{i}^1 \cap \{s_{j}\}) \times S_{-i,j}=\emptyset$ for all $L_i = (L_i^1, L_i^2, \ldots , L_i^K) \in {\mathcal {L}}_i$. Therefore, if $s_{-i}\in L_{i}^{1}$ for some $L_i = (L_i^1, L_i^2, \ldots , L_i^K) \in {\mathcal {L}}_i$, then $s_{-i}\in \prod _{j\ne i}C_{j}(R_{j})=C_{-i}(R_{-i})$. This implies that $L_i \subseteq C_{-i}(R_{-i})$ for all $L_i = (L_i^1, L_i^2, \ldots , L_i^K) \in {\mathcal {L}}_i$ and establishes the lemma. $\square$

Proof of Lemma 6

Only if. If there exists $\mu _{i}\in \Delta (A_{i})$ such that $\mu _{i}$ strictly dominates $s_{i}$ on $S^{\prime }_{-i}$, then, for every $(\emptyset \ne )\,S^{\prime \prime }_{-i}\subseteq S^{\prime }_{-i}$, $\mu _{i}\in \Delta (A_{i})$ weakly dominates $s_{i}$ on $S^{\prime \prime }_{-i}$ .

If. Suppose there does not exist $\mu _{i}\in \Delta (A_{i})$ such that $\mu _{i}$ strictly dominates $s_{i}$ on $S^{\prime }_{-i}$. Hence, by Pearce (1984, Lemma 3), there exists $\lambda _{i}\in \Delta (S^{\prime }_{-i})$ such that $u(s_{i},\lambda _{i})\ge u(s_{i}^{\prime },\lambda _{i})$ for all $s_{i}^{\prime }\in A_{i}$. Then, by Pearce (1984, Lemma 4), there does not exist $\mu ^{\prime }_{i}\in \Delta (A_{i})$ such that $\mu ^{\prime }_{i}$ weakly dominates $s_{i}$ on $S^{\prime \prime }_{-i}:= \text {supp}\lambda _{i}\subseteq S^{\prime }_{-i}$. $\square$

Proof of Proposition 1

Consider, for all players i, the sequence $\langle S_{i}^{n}\rangle _{n=0}^{\infty }$ defined in Definition 6. We show, by induction on n, that $C_{i}(R_{i}({\mathcal {L}}_{i}^{n}))=S_{i}^{n+1}$ for all players i and every $n\ge 0$.

Part (i). For $n=0,$ we have that ${\mathcal {L}}_{i}^{0}={\mathcal {L}} _{i}^{*}$ and hence,

$$\begin{aligned} R_{i}({\mathcal {L}}_{i}^{0})=\left\{ (s_{i},A_{i}) \mid \exists \mu _{i}\in \Delta (A_{i})\text { such that } s_i \text { is weakly dominated by }\mu _{i} \text { on }S_{-i}\right\} . \end{aligned}$$

Therefore, $C_{i}(R_{i}({\mathcal {L}}_{i}^{0})) = a_i(S_{-i}) = b_i(S_{-i}^0) \cap a_i(S_{-i}) = S_{i}^{1}$ for all players i, since $S_{-i}^0 = S_{-i}$ and $a_i(S_{-i}) \subseteq b_i(S_{-i})$.

Part (ii). Now, let $n\ge 1$, and assume that for all players i, $C_{i}(R_{i}({\mathcal {L}}_{i}^{n-1})) = S_{i}^{n}$. We show that, for all players i, $C_{i}(R_{i}({\mathcal {L}}_{i}^{n})) = S_{i}^{n+1}$.

Fix a player i. By definition, ${\mathcal {L}}_{i}^{n}={\mathcal {L}} _{i}^{b}(R_{-i}({\mathcal {L}}_{-i}^{n-1}))$. We have that

$$\begin{aligned} {\mathcal {L}}_{i}^{b}(R_{-i}({\mathcal {L}}_{-i}^{n-1}))= & {} \left\{ L_{i}\in {\mathcal {L}} _{i}^{*}\mid L_{i}\text { believes }C_{-i}(R_{-i}({\mathcal {L}}_{-i}^{n-1}))\right\} \\= & {} \{L_{i}\in {\mathcal {L}}_{i}^{*}\mid L_{i}\text { believes }S_{-i}^n \}\\= & {} \{L_{i}\in {\mathcal {L}}_{i}^{*}\mid L_{i}^{1}\subseteq S_{-i}^{n}\}, \end{aligned}$$

by our induction assumption. But then,

$$\begin{aligned} R_{i}({\mathcal {L}}_{i}^{n})= & {} \Bigg \{(s_{i},A_{i})\mid \text {for every }L_{i}^{1}\subseteq S_{-i}^{n}\text { there is }\mu _{i}\in \Delta (A_{i})\text { such that } \\&\quad s_{i} \text { is weakly dominated by } \mu _i \text { on } L_{i}^{1}\text { or on }S_{-i}\Bigg \} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} C_i(R_{i}({\mathcal {L}}_{i}^{n}))&= \{ s_i \in S_i \mid \exists (\emptyset \ne ) \, L_i^1 \subseteq S_{-i}^n \text { s.t. } s_i \in a_i(L_i^1) \cap a_i(S_{-i}) \} \\&= b_i(S_{-i}^n) \cap a_i(S_{-i}) = S_i^{n+1} \end{aligned} \end{aligned}$$

by (1) and Definition 6, thus concluding the proof.

The proof above would also apply to the case where ${\mathcal {L}}_{i}^{0}=\mathcal {{\tilde{L}}}_{i}^{*}$, considering only likelihood orderings that consist of one or two levels only. The reason is that the restrictions on the sets ${\mathcal {L}}_{i}^{n}$ of likelihood orderings only apply to the top level of the likelihood orderings. $\square$

Proof of Proposition 2

Consider, for all players i, the sequence $\langle S_{i}^{n}\rangle _{n=0}^{\infty }$ defined in Definition 7. We show, by induction on n, that $C_{i}(R_{i}({\mathcal {L}}_{i}^{n}))=S_{i}^{n+1}$ for all players i and every $n\ge 0$.

Part (i). For $n=0$, it follows by part (i) of the proof of Proposition 1, that $C_{i}(R_{i}({\mathcal {L}}_{i}^{0})) = a_i(S_{-i}) = a_i(S_{-i}^0) \cap S_i = S_{i}^{1}$ for all players i.

Part (ii). Let $n\ge 1$, and assume that, for all players i, $C_{i}(R_{i}({\mathcal {L}}_{i}^{m})) = S_{i}^{m+1}$ for every $m \in \{0, \ldots , \smash {n-1} \}$. We show that, for all players i, $C_{i}(R_{i}(\mathcal {L }_{i}^{n})) = S_{i}^{n+1}$.

Fix a player i. By definition, we have that

$$\begin{aligned} {\mathcal {L}}_{i}^{n}={\mathcal {L}}_{i}^{a}(R_{-i}(\,{\mathcal {L}}_{-i}^{0}))\cap {\mathcal {L}}_{i}^{a}(R_{-i}(\,{\mathcal {L}}_{-i}^{1}))\cap \cdots \cap {\mathcal {L}} _{i}^{a}(R_{-i}(\,{\mathcal {L}}_{-i}^{n-1})). \end{aligned}$$

By the induction assumption, we know that $C_{-i}(R_{-i}(\,{\mathcal {L}} _{-i}^{m})) = S_{i}^{m+1}$ for every $m \in \{0, \ldots , \smash {n-1} \}$, and hence

$$\begin{aligned} {\mathcal {L}}_{i}^{a}(R_{-i}(\,{\mathcal {L}}_{-i}^{m}))= & {} \left\{ L_{i}\in {\mathcal {L}} _{i}^{*}\mid L_{i}\text { assumes }C_{-i}(R_{-i}(\,{\mathcal {L}}_{-i}^{m}))\right\} \\= & {} \{L_{i}\in {\mathcal {L}}_{i}^{*}\mid L_{i}\text { assumes } S_{-i}^{m+1}\} \\= & {} \left\{ L_{i}\in {\mathcal {L}}_{i}^{*}\mid \exists k\in \{1,\ldots ,K\} \text { such that } L_{i}^{1}\cup \cdots \cup L_{i}^{k} = S_{-i}^{m+1}\right\} \end{aligned}$$

for every $m \in \{0, \ldots , \smash {n-1} \}$. This implies that

$$\begin{aligned} {\mathcal {L}}_{i}^{n}=\left\{ L_{i}\in {\mathcal {L}}_{i}^{*}\mid \forall m \in \{1, \ldots , n\}, \exists k\in \{1,\ldots ,K\} \text { such that } L_{i}^{1}\cup \cdots \cup L_{i}^{k} = S_{-i}^{m}\right\} . \end{aligned}$$

Therefore, $R_{i}({\mathcal {L}}_{i}^{n})$ contains exactly those preference restrictions $(s_{i},A_{i})$ such that $s_{i}$ is weakly dominated by some $\mu _{i}\in \Delta (A_{i})$ on some $S_{-i}^{m}$ with $m\le n$:

$$\begin{aligned} \begin{aligned} R_{i}({\mathcal {L}}_{i}^{n}) = \Bigg \{(s_{i},A_{i}) \mid&\text { there are } m \in \{0, \ldots , n\} \text { and }\mu _{i}\in \Delta (A_{i}) \\&\text { such that } s_i \text { is weakly dominated by } \mu _{i} \text { on } S_{-i}^m \Bigg \} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} C_{i}(R_{i}({\mathcal {L}}_{i}^{n})) = a_i(S_{-i}^0) \cap a_i(S_{-i}^1) \cap \cdots \cap a_i(S_{-i}^n) = S_{i}^{n+1}, \end{aligned}$$

which completes the proof. $\square$

Proof of Proposition 4

Consider, for both players i, the sequence $\langle \Sigma _{i}^{n}\rangle _{n=0}^{\infty }$ defined in Definition 8. Consider also, for both players i, the sequence $\langle \tilde{{\mathcal {L}}}_{i}^{n}\rangle _{n=0}^{\infty }$ defined by

Ini* :

For both players i, let $\tilde{{\mathcal {L}}}_i^0 = \tilde{{\mathcal {L}}}_i^*$.

and FP. Note that ${\mathcal {L}}_i^1 \subseteq \tilde{{\mathcal {L}}} _i^0 \subseteq {\mathcal {L}}_i^0$, so by induction, for every $n \ge 1$, ${\mathcal {L}}_i^{n+1} \subseteq \tilde{{\mathcal {L}}}_i^n \subseteq {\mathcal {L}} _i^n$. Since also the algorithm defined by Ini* and FP converges after a finite number of rounds, as the set of likelihood orderings is finite, we have that $\tilde{{\mathcal {L}}}_i^\infty := \bigcap _{n=1}^\infty \tilde{{\mathcal {L}}}_i^n$ equals ${\mathcal {L}}_i^\infty$. Thus, it is sufficient to show that there exists $L_i \in \tilde{{\mathcal {L}}} _i^n$ such that $A_i = C_{i}(R_{i}(L_i))$ if and only if $A_i \in \Sigma _{i}^{n+1}$, for both players i and every $n\ge 0$. We show this by induction on n.

Part (i). For $n=0,$ we have that $\tilde{{\mathcal {L}}}_i^0 = \tilde{{\mathcal {L}}}_i^*$ and thus, $L_i \in \tilde{{\mathcal {L}}}_i^0$ if and only if $L_i = (L_i^1) = S_j$ or $L_i = (L_i^1, L_i^2) = (S'_j, S_j\backslash S'_j)$ for some non-empty proper subset $S'_j$ of $S_j$. Hence, there is $L_i \in \tilde{{\mathcal {L}}}_i^0$ such that $(s_i,A_i) \in R_i(L_i)$ if and only if there exist $(\emptyset \ne ) \, S'_j \subseteq S_j$ and $\mu _{i}\in \Delta (A_i)$ such that $s_i$ is weakly dominated by $\mu _i$ on $S'_j$ or $S_j$. Therefore, there is $L_i \in \tilde{{\mathcal {L}}}_i^0$ such that $A_i \in C_i(R_i(L_i))$ if and only if $A_i = a_i(S'_j) \cap a_i(S_j)$ for some $(\emptyset \ne ) \, S'_j \subseteq S_j$. It now follows from the definition of the operator $\alpha _i(\Sigma '_j)$ that there is $L_i \in \tilde{{\mathcal {L}}}_i^0$ such that $A_i \in C_i(R_i(L_i))$ if and only if $A_i \in \alpha _i(\Sigma _j) = \alpha _i(\Sigma _j^0) = \Sigma _i^1$, since $\Sigma _j^0 = \Sigma _j$.

Part (ii). Now, let $n\ge 1$, and assume that for both players i , there exists $L_i \in \tilde{{\mathcal {L}}}_i^{n-1}$ such that $A_i = C_{i}(R_{i}(L_i))$ if and only if $A_i \in \Sigma _{i}^n$.

Fix a player i. By FP, $L_{i}\in \tilde{{\mathcal {L}}}_{i}^{n}$ is equivalent to there existing $(\emptyset \ne )\,{\mathcal {L}}_{j}\subseteq \tilde{{\mathcal {L}}}_j^{n-1}$ such that $L_i = (S'_j,S_j\backslash S'_j)$ if $S'_j \ne S_j$ and $L_i = (S_j)$ otherwise, where $S'_j = {\cup }_{L_j \in {\mathcal {L}}_j}C_j(R_j(L_j))$. By the induction assumption this is equivalent to there existing $(\emptyset \ne )\,\Sigma ''_j \subseteq \Sigma _j^n$ such that $L_i = (S'_j,S_j\backslash S'_j)$ if $S'_j \ne S_j$ and $L_i = (S_j)$ otherwise, where $S'_j = {\cup }_{A_j \in \Sigma _j^{n-1}}A_j$. Therefore, there is $L_i \in \tilde{{\mathcal {L}}}_i^n$ such that $A_i = C_i(R_i(L_i))$ if and only if $A_{i}=a_{i}({\cup }_{A_j \in \Sigma ''_j}A_{j})\cap a_{i}(S_{j})$ for some $(\emptyset \ne )\,\Sigma ''_j \subseteq \Sigma _j^n$. It now follows from the definition of the operator $\alpha _i(\Sigma '_j)$ that there is $L_i \in \tilde{{\mathcal {L}}}_i^n$ such that $A_i = C_i(R_i(L_i))$ if and only if $A_i \in \alpha _i(\Sigma _j^n) = \Sigma _i^{n+1}$, which completes the proof. $\square$

Proof of Proposition 5

Let $\langle {\mathcal {L}}_{1}^{n}, {\mathcal {L}}_{2}^{n}\rangle _{n=1}^{\infty }$ be the sequence of likelihood orderings according to the algorithm for proper rationalizability (cf. Sect. 3.3). It is sufficient to show, under the assumptions of the proposition, that for every $n \ge 0$ and both players i, it holds that, for every $s_{i}\in S_{i}\backslash S_{i}^{n+1}$ , $(s_{i},\{a_{i}\})\in R_{i}({\mathcal {L}}_{i}^{n})$ for every $a_{i}\in A_{i}^{n+1}$. In this case, namely, every properly rationalizable strategy is in $\bigcap _{n=1}^{\infty }S_{i}^{n}$. We show by induction that the statement above is true.

Part (i). Let $n=0$. If $S_{i}^{1}=S_{i}$, so that there is no $s_{i}\in S_{i}\backslash S_{i}^{1}$, then the statement is trivially true. If $S_{i}^{1} \ne S_{i}$, then, by the premise of the proposition, for every $s_i \in S_i \backslash S_i^1$, $s_i$ is weakly dominated by every $a_i \in A_i^1$ on $S_j$. Hence, by the full support assumption, $(s_{i},\{a_{i}\}) \in R_{i}({\mathcal {L}}_{i}^*) = R_{i}({\mathcal {L}} _{i}^{0})$, implying that the statement is true also in this case.

Part (ii). Let $n\ge 1$, and assume that, for every $m \in \{0, \ldots , \smash {n-1} \}$ and both players i, it holds that, for every $s_i \in S_i \backslash S_i^{m+1}$, $(s_i, \{a_i\}) \in R_i({\mathcal {L}}_{i}^{m})$ for every $a_i \in A_i^{m+1}$.

Fix a player i. We first make the observation that, for every $m \in \{1, \ldots , n \}$, every $L_i = (L_i^1, \ldots , L_i^K) \in {\mathcal {L}}_i^m$ satisfies that there exists $k \in \{1, \ldots , K \}$ such that $A_j^{m} \subseteq L_i^1 \cup \cdots \cup L_i^k \subseteq S_j^{m}$. This is true by the full support assumption if $S_j^{m} = S_j$ (and thus $A_j^{m} = S_j$, by the last bullet point of Proposition 5 and fact that $A_j^0 = S_j$). Assume now $S_j^{m} \ne S_j$. By the algorithm for proper rationalizability, every $L_i \in {\mathcal {L}}_i^m$ respects $R_j({\mathcal {L}} _{j}^{m-1})$, implying that there exists $k \in \{1, \ldots , K \}$ such that $(L_i^1 \cup \cdots \cup L_i^k) \cap \{a_j\} \ne \emptyset$ for every $a_j \in A_j^{m}$ and $(L_i^1 \cup \cdots \cup L_i^k) \cap \{s_j\} = \emptyset$ for every $s_j \in S_j \backslash S_j^m$, and the observation follows also in this case.

If $S_i^{n+1} = S_i$, then the statement is trivially true also for $n \ge 1$.

If $S_i^{n+1} \ne S_i$, let ($0 \le$) $m \le n$ satisfy $S_i^{n+1} = S_i^{m+1} \ne S_i^m$. By a premise of the proposition, for every $s_i \in S_i \backslash S_i^{m+1}$, $s_i$ is weakly dominated by every $a_i \in A_i^{m+1}$ on either ($A_j^{m}$ and $S_j^{m}$) or $S_j$. If $s_i$ is weakly dominated by $a_i$ on $A_j^{m}$ and $S_j^{m}$, then $s_i$ is weakly dominated by $a_i$ on each strategy set $S^{\prime }_j$ satisfying $A_j^{m} \subseteq S^{\prime }_j \subseteq S_j^{m}$. By the observation that every $L_i = (L_i^1, \ldots , L_i^K) \in {\mathcal {L}}_i^m$ satisfies that there exists $k \in \{1, \ldots , K \}$ such that $A_j^{m} \subseteq L_i^1 \cup \cdots \cup L_i^k \subseteq S_j^{m}$ it follows that $(s_i, \{a_i\}) \in R_i( {\mathcal {L}}_i^m)$. If $s_i$ is weakly dominated by $a_i$ on $S_j$, then by the full support assumption, $(s_i, \{a_i\}) \in R_i({\mathcal {L}}_i^{*}) = R_{i}({\mathcal {L}}_{i}^{0})$. Hence, since the sequence of sets of likelihood orderings is non-increasing, so that ${\mathcal {L}}_i^n \subseteq {\mathcal {L}} _i^m \subseteq {\mathcal {L}}_i^0$ and thus, $R_i({\mathcal {L}}_i^n) \supseteq R_i({\mathcal {L}}_i^m) \supseteq R_i({\mathcal {L}}_i^0)$, for every $s_i \in S_i \backslash S_i^{n+1}$, $(s_i, \{a_i\}) \in R_i({\mathcal {L}}_i^n)$ for every $a_i \in A_i^{n+1}$. $\square$

Proof of Proposition 6

The proof is divided into two parts. In part (i) we show that the strategies in $S_i{\backslash }\left( \{k\} \cup \{w\} \right)$ are not properly rationalizable. In part (ii) we show that k and ware properly rationalizable.

Part (i). Let $\langle {\mathcal {L}}_{1}^{n},{\mathcal {L}} _{2}^{n}\rangle _{n=1}^{\infty }$ be the sequence of sets of likelihood orderings for the finite version of Ellingsen and Miettinen’s (2008, Section I) bilateral commitment bargaining game with zero commitment cost, according to the algorithm for proper rationalizability (cf. Sect. 3.3). In order to show that the strategies in $S_{i}{\backslash } \left( \{k\}\cup \{w\}\right) =\{0,1,\ldots ,k-1\}$ are not properly rationalizable, it is sufficient to show that for each player i, it holds that (a) for every $s_{i}\in \{0,1,\ldots ,\beta _{i}\}$, $(s_{i},\{w\})\in R_{i}(\mathcal { L}_{i}^{0})$, and (b) for every $s_{i}\in \{\beta _{i}+1,\beta _{i}+2,\ldots ,k-1\}$, $(s_{i},\{k\})\in R_{i}({\mathcal {L}}_{i}^{1})$, keeping in mind that the sequence of sets of likelihood orderings is non-increasing, so that ${\mathcal {L}}_{i}^{n}\subseteq {\mathcal {L}}_{i}^{1}\subseteq {\mathcal {L}}_{i}^{0}$ and thus, $R_{i}({\mathcal {L}}_{i}^{n})\supseteq R_{i}({\mathcal {L}} _{i}^{1})\supseteq R_{i}({\mathcal {L}}_{i}^{0})$ for every $n\ge 1$.

Result (a) follows from the fact that, for each player i and for every $s_{i}\in \{0,1,\ldots ,\beta _{i}\}$, w weakly dominates $s_{i}$ on $S_{j}$. (To see this, note that if the opponent chooses w, then player i’s payoff by choosing w is $\beta _i$, while it is $\{0,1,\ldots ,\beta _{i}\}$ if player i commits to one of these demands, and if the opponent chooses $s_{j}\in \{0,1,\ldots ,k\}$, then player i’s payoff by choosing w is $1-s_j$, while it is no more than $1-s_j$ and sometimes 0 if $s_i \in \{0,1,\ldots ,\beta _{i}\}$.) Hence, for each player i and for every $s_{i}\in \{0,1,\ldots ,\beta _{i}\}$, $(s_{i},\{w\})\in R_{i}({\mathcal {L}}_{i}^{*}) = R_i({\mathcal {L}} _{i}^0)$. This result implies that, for each player i, every $L_{i}=(L_{i}^{1},\ldots ,L_{i}^{K})\in {\mathcal {L}}_{i}^1 = {\mathcal {L}} _{i}^r(R_j({\mathcal {L}}_{j}^0))$ satisfies that there exists $k\in \{1,\ldots ,K\}$ such that $\{w\}\subseteq L_{i}^{1}\cup \cdots \cup L_{i}^{k}\subseteq \{\beta _{j}+1,\beta _{j}+2,\ldots ,k\}\cup \{w\}$. Result (b) follows from the fact that, for each player i and for every $s_{i}\in \{\beta _{i}+1,\beta _{i}+2,\ldots ,k-1\}$, k weakly dominates $s_{i}$ on each strategy set $S^{\prime }_{j}$ satisfying $\{w\}\subseteq S^{\prime }_{j}\subseteq \{\beta _{j}+1,\beta _{j}+2,\ldots ,k\}\cup \{w\}$. Hence, for each player i and for every $s_{i}\in \{\beta _{i}+1,\beta _{i}+2,\ldots ,k-1\}$, $(s_{i},\{k\})\in R_{i}({\mathcal {L}}_{i}^1)$.

Part (ii). We establish that k and ware properly rationalizable in the finite version of Ellingsen and Miettinen’s (2008, Section I) bilateral commitment bargaining game with zero commitment cost, by showing that both k and w can be used with positive probability in a proper equilibrium; thus, they are properly rationalizable (Asheim 2001, Proposition 2). To prove this claim, consider the likelihood orderings

$$\begin{aligned} L_{1}&=\{\{w\},\{1\},\{2 \},\ldots ,\{\beta _{2}-1\},\{k\},\{k-1\},\ldots ,\{\beta _{2}+1\},\{\beta _{2}\},\{0\}\} \, , \\ L_{2}&=\{\{k\},\{k-1\},\ldots ,\{\beta _{1}+1\},\{w\},\{\beta _{1}\},\{\beta _{1}-1\},\ldots ,\{1\},\{0\}\} . \end{aligned}$$

Since each element in either of these partitions contains only one strategy, they determine a pair of LPSs. It is straightforward to check that this pair of LPSs determines a proper equilibrium, according to Blume et al.’s (1991b, Proposition 5) characterization, where player 1 chooses k with probability 1 and player 2 chooses w with probability 1. $\square$

Claim

Consider the finite version of Ellingsen and Miettinen’s (2008, Section I) bilateral commitment bargaining game with zero commitment cost. Assume that $x_1(s_1,s_2)=s_1$ and $x_2(s_2,s_1)=s_2$ if $s_1 + s_2 \le k$.

• (i) There exists a proper equilibrium where both players assign probability 1 to k.

• (ii) For both players i and any strategy $\ell \in \{\beta _i + 1, \beta _i + 2, \ldots , k-1\}$, there exists a perfect equilibrium where player i assigns positive probability to both w and $\ell$ and player j assigns probability 1 to k.

Proof

Part (i). Consider the LPSs

$$\begin{aligned} \lambda _1&= \left\{ \lambda _1^1, \ldots , \lambda _1^{k+1}\right\} \\ \lambda _2&= \left\{ \lambda _2^1, \ldots , \lambda _2^{k+1}\right\} , \end{aligned}$$

where for both players i and each $\ell \in \{1, \ldots , k+1\}$, the support of $\lambda _i^{\ell }$ is included in $\{w, k+1-\ell \}$ for $\ell \in \{1, \ldots , \beta _j + 1\}$, $\{w, 1\}$ for $\ell = \beta _j + 2$, $\{w, k+2-\ell \}$ for $\ell \in \{\beta _j +3, \ldots , k\}$, and $\{w, 0\}$ for $\ell = k+1$. Let, for each $\ell \in \{1, \ldots , k+1\}$, $\lambda _i^{\ell }$ be determined by $u_i(w,\lambda _i^{\ell }) = u_i(k-1,\lambda _i^{\ell })$. This means that

$$\begin{aligned} \begin{array}{lll} \lambda _i^{1}(w) = 0 \quad &{} \qquad &{}\quad \lambda _i^{1}(k) = 1 \\ \lambda _i^{2}(w) = \tfrac{1}{\beta _j} \quad &{} \qquad &{}\quad \lambda _i^{2}(k-1) = \tfrac{\beta _j - 1}{\beta _j} \\ \lambda _i^{3}(w) = \tfrac{2}{\beta _j + 1} \quad &{} \qquad &{}\quad \lambda _i^{3}(k-2) =\tfrac{\beta _j - 1}{\beta _j + 1} \\ \cdots \quad &{} \qquad &{}\quad \cdots \\ \lambda _i^{\beta _j + 1}(w) = \tfrac{\beta _j}{2\beta _j - 1} \quad &{}\quad &{}\quad \lambda _i^{\beta _j + 1}(\beta _i) = \tfrac{\beta _j - 1}{2\beta _j - 1} \\ \lambda _i^{\beta _j + 2}(w) = 0 \quad &{} \qquad &{}\quad \lambda _i^{\beta _j + 2}(1) = 1 \\ \lambda _i^{\beta _j + 3}(w) = \tfrac{\beta _j + 1}{2\beta _j} \quad &{}\quad &{}\quad \lambda _i^{\beta _j + 3}(\beta _i - 1) = \tfrac{\beta _j - 1}{2\beta _j} \\ \cdots \quad &{} \qquad &{}\quad \cdots \\ \lambda _i^{k}(w) = \tfrac{k - 2}{\beta _j + k - 3} \quad &{}\quad &{}\quad \lambda _i^{k}(2) = \tfrac{\beta _j - 1}{\beta _j + k - 3} \\ \lambda _i^{k+1}(w) = \tfrac{1}{\beta _j} \quad &{} \qquad &{}\quad \lambda _i^{k+1}(0) = \tfrac{\beta _j - 1}{\beta _j} \end{array} \end{aligned}$$

The LPSs $\lambda _1$ and $\lambda _2$ determine the following likelihood orderings:

$$\begin{aligned} L_{1}&=\left\{ \{k\},\{w, k-1\},\{k-2\},\ldots ,\{\beta _{1}+1\},\{\beta _{1}\},\{1\},\{\beta _{1}-1\},\ldots ,\{2\},\{0\}\right\} , \\ L_{2}&=\left\{ \{k\},\{w, k-1\},\{k-2\},\ldots ,\{\beta _{2}+1\},\{\beta _{2}\},\{1\},\{\beta _{2}-1\},\ldots ,\{2\},\{0\}\right\} . \end{aligned}$$

It can be checked that $L_1$ respects the preference restrictions that $u_2$ and $\lambda _2$ give rise to, and $L_2$ respects the preference restrictions that $u_1$ and $\lambda _1$ give rise to. To see this in the case of $L_1$ (the demonstration for $L_2$ is symmetric), note:

• (a) Player 2 ranks the commitment strategies 0, 2, $3, \ldots , k$ according to size since $u_2(s_2, \lambda _2^1) = 0$ and $u_2(s_2, \lambda _2^2) = s_2/\beta _1$ for $s_2 \in 0$, 2, $3, \ldots , k$.

• (b) Player 2 is indifferent between the commitment strategy $k-1$ and waiting w since, by construction, $u_2(w,\lambda _2^{\ell }) = u_2(k-1,\lambda _2^{\ell })$ for all $\ell \in \{1, \ldots , k+1\}$.

• (c) Player 2 ranks the commitment strategy 1 between the commitment strategies $\beta _1$ and $\beta _1 - 1$ since

  $$\begin{aligned} u_2(\beta _1, \lambda _2^1)&= u_2(1, \lambda _2^1) = u_2(\beta _1 - 1, \lambda _2^1) = 0 \, , \\ u_2(\beta _1, \lambda _2^2)&= u_2(1, \lambda _2^2) = 1> u_2(\beta _1 - 1, \lambda _2^2) = \tfrac{\beta _1 - 1}{\beta _1} \, , \\ u_2(\beta _1, \lambda _2^3)&= \tfrac{2\beta _1}{\beta _1 + 1} > u_2(1, \lambda _2^3) = 1, \end{aligned}$$

  since $\beta _1 > 1$ and $x_2(1, k-2) = 1$.

It follows from Blume et al.’s (1991b, Proposition 5) characterization that $(\lambda _1^{1},\lambda _2^{1})$, where $\lambda _2^1$ is the mixed strategy of player 1 and $\lambda _1^1$ is the mixed strategy of player 2, is a proper equilibrium. Note that, for both players i, $\lambda _i^{1}(k) = 1$.

Part (ii). Let $\ell$ be any player 1 strategy in $\{\beta _1 + 1, \beta _1 + 2, \ldots , k-1\}$. Consider the LPSs $\lambda _1 = \{\lambda _1^1, \ldots , \lambda _1^{k+1}\}$ and $\lambda _2 = \{\lambda _2^1, \lambda _2^2\}$ defined by

$$\begin{aligned} \begin{array}{lllll} \lambda _1^{1}(w) = 0 &{}\qquad &{}\lambda _1^{1}(k) = 1 &{}\qquad &{} \lambda _2^{1}(w) = \tfrac{\beta _2}{k} \qquad \lambda _2^{1}(\ell ) = 1 - \tfrac{\beta _2}{k}\\ \lambda _1^{2}(w) = 0 &{}\qquad &{}\lambda _1^{2}(k-\ell ) = 1 &{}\qquad &{} \lambda _2^{2}(s_1) = \tfrac{1}{k} \text { for all } s_1 \in S_1 \backslash \{w, \ell \} \\ \lambda _1^{3}(w) = \tfrac{1}{\ell - \beta _1 + 1} &{}\qquad &{} \lambda _1^{3}(k-1) = \tfrac{\ell -\beta _1}{\ell - \beta _1 + 1} &{}\qquad &{} \\ \cdots &{}\qquad &{} \cdots &{}\qquad &{} \\ \lambda _1^{\ell + 1}(w) = \tfrac{\ell - 1}{2 \ell - \beta _1 - 1} &{}\qquad &{}\lambda _1^{\ell + 1}(k-\ell +1) = \tfrac{\ell -\beta _1}{2 \ell - \beta _1 - 1} &{}\qquad &{} \\ \lambda _1^{\ell + 2}(w) = \tfrac{1}{\ell - \beta _1 + 1} &{}\qquad &{}\lambda _1^{\ell + 2}(k-\ell -1) = \tfrac{\ell -\beta _1}{\ell - \beta _1 + 1} &{}\qquad &{} \\ \cdots &{}\qquad &{} \cdots &{}\qquad &{} \\ \lambda _1^{k}(w) = \tfrac{k - \ell - 1}{k - \beta _1 - 1} &{}\qquad &{}\lambda _1^{k}(1) = \tfrac{\ell -\beta _1}{k - \beta _1 - 1} &{}\qquad &{} \\ \lambda _1^{k+1}(w) = \tfrac{k - \ell }{k - \beta _1} &{}\qquad &{}\lambda _1^{k + 1}(0) = \tfrac{\ell -\beta _1}{k - \beta _1} \, , &{}\qquad &{}\\ \end{array} \end{aligned}$$

with, for each level of these LPSs, zero probability assigned to other strategies.

These LPSs imply that player 1 is indifferent between w and $\ell$ and that player 1 prefers each of these strategies to any strategy in $S_1 \backslash \{w, \ell \}$, and that player 2 prefers k to any strategy in $S_2 \backslash \{k\}$. To see this, note:

• (a) It follows that player 1 strictly prefers each of w and $\ell$ to any strategy in $S_1 \backslash \{w, \ell \}$ since $u_1(s_1, \lambda _1^1) = 0$ for all $s_1 \in S_1$ and $u_1(w, \lambda _1^2) = u_1(\ell , \lambda _1^2) = \ell$, while $u_1(s_1, \lambda _1^2) = s_1 < \ell$ if $s_1$ is a commitment strategy in $\{1, 2, \ldots , \ell - 1 \}$ and $u_1(s_1, \lambda _1^2) = 0 < \ell$ if $s_1$ is a commitment strategy in $\{\ell + 1, \ell + 2, \ldots , k\}$. It follows that player 1 is indifferent between w and $\ell$ since $\lambda _1^3, \lambda _1^4, \ldots , \lambda _1^{k+1}$ have been constructed so that $u_1(w,\lambda _1^{m}) = u_1(\ell ,\lambda _1^{m})$ for each $m \in \{3, 4, \ldots , k+1\}$.

• (b) It follows that player 2 strictly prefers k to any strategy in $S_2 \backslash \{k\}$ since $u_2(k, \lambda _2^1) = \beta _2$ and $u_2(s_2, \lambda _2^1) < \beta _2$ for all $s_2 \in S_2 \backslash \{k\}$.

Since both $\lambda _1$ and $\lambda _2$ have full support on the set of opponent strategies, it follows from Blume et al.’s (1991b, Proposition 4) characterization that $(\lambda _1^1, \lambda _2^1)$, where $\lambda _2^1$ is the mixed strategy of player 1 and $\lambda _1^1$ is the mixed strategy of player 2, is a perfect equilibrium where player 1 assigns positive probability to both w and $\ell$ and player 2 assigns probability 1 to k.

In a similar fashion we can show that, for any player 2 strategy $\ell \in \{\beta _2 + 1, \beta _2 + 2, \ldots , k-1\}$, there exists a perfect equilibrium where player 1 assigns probability 1 to k and player 2 assigns positive probability to both w and $\ell$. $\square$