Before we prove Theorem 14.1, we present the following lemma.
Proof of Theorem 14.1 First, we have \(\hat{u}_i(x^0)=U_i(\sigma ^*)\) by CS. Consider \((\sigma _{i}^*,(g_i,\delta _i))\). This gives an active experience \(\phi _i=(g_i,\delta _i,\mathcal {E}_i(\sigma _{i}^*,(g_i,\delta _i)))\) and \(h_i=U_i(\sigma _{i}^*,(g_i,\delta _i))\). By CA, there is a state \(x'=((g_{i}^0,g_i),(\delta _{i}',\delta _i), y') \in X\) such that \(\hat{o}_i(x')=\phi _i\) and \(\hat{u}_i(x') =h_i\). Now, we take an arbitrary \(x=((g_{i}^0,g_i),(\delta _{i},\delta _i), y) \in X\). Lemma 14.7 implies \(h_i=\hat{u}_i(x')=\hat{u}_i(x)\). Since \(\sigma ^*\) is inductively stable, we have \(U_i(\sigma ^*) \ge h_i\) by Proposition 13.1. Hence, \(\hat{u}_i(x^0) = U_i(\sigma ^*)\ge h_i=\hat{u}_i(x')=\hat{u}_i(x)\). \(\square \)
Proof of Theorem 14.2 The onlyif part is Theorem 14.1. Thus, it suffices to show that if the model\(\mathcal {M} = \langle \hat{N}, \hat{Z}, \hat{o}_i, \hat{u}_i; x^0, X \rangle \) of player i is coherent with \(A_i(\sigma ^*)\) and satisfies inequality (14.1), then his stationary actions maximize her objective payoff function. Consider an arbitrary \((f_i,\delta _i)\). This induces an experience \([\phi _i;h_i]=[f_i,\delta _i,\mathcal {E}_i(f_{i}^*, f_i);U_i(\sigma _{i}^*,(f_i,\delta _i))] \in A_i(\sigma ^*)\). Since \(\mathcal {M}\) is coherent with active experiences \(A_i(\sigma ^*)\), there is a state \(x=(g,\delta , y) \in X\) such that \(g_{i}=g_{i}^0\), \(\hat{o}_i(x)=(f_i,\delta _i,\mathcal {E}_i(f_{i}^*, f_i))\), and \(\hat{u}_i(x)=U_i(\sigma _{i}^*,(f_i,\delta _i))\). Also, \(\hat{u}_i(x^0)=U_i(\sigma ^*)\) by CS. Then \(U_i(\sigma ^*)=\hat{u}_i(x^0)\ge \hat{u}_i(x)=U_i(\sigma _{i}^*,(f_i,\delta _i))\). \(\square \)
Proof of Theorem 14.4 For Part (i), let player
j come to the festival
k of player
i with the friendly action. Suppose that player
i takes the unfriendly action to
j’s presence. Then
\(U_i(\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}})))=0<U_i(\sigma ^*)\). By CP,
\(\hat{u}_i(g,\delta , y)=\hat{u}_i(g^0, ({\mathop {\frown }\limits ^{. \ .}}), y)=0\) holds for some
\((g,\delta , y)\). However, since
\(\hat{u}_i(g_i^0,({\mathop {\smile }\limits ^{. \ .}}), y^0)=U_i(\sigma ^*)>0\) by CS, we have
$$ \hat{u}_i(g,\delta , y)=\hat{u}_i(g_i^0, ({\mathop {\frown }\limits ^{. \ .}}), y)=0 <\hat{u}_i(g_i^0,({\mathop {\smile }\limits ^{. \ .}}), y^0) =\hat{u}_i(g,(\delta _{i},({\mathop {\smile }\limits ^{. \ .}})), y^0). $$
This violates rationalization.
For Part (ii), let player
j be a player in the smallest festival. Then his payoff
\(U_j(\sigma ^*)\) is the lowest. When
j comes to
\(f_j=f_i^*\) and take the friendly action,
\(U_j(\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}}))) \le U_j(\sigma ^*)\) holds; otherwise,
j would stay at
\(f_j\). By CP, there is
\((g,\delta , y)\) such that
\(g_i=g_i^0\),
\(\delta _i=r_i^*(f_{j}^*, f_j)=({\mathop {\smile }\limits ^{. \ .}})=\delta _i^0\), and
$$ \hat{u}_i(g_i^0,({\mathop {\smile }\limits ^{. \ .}}), y)=U_i(\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}})))=U_j(\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}}))) $$
hold. Thus,
\(\hat{u}_i(g_i^0,\delta _i^0,y) \le U_j(\sigma ^*) = \min _{k\in N} U_k(\sigma ^*)\). By rationalization, we have
$$\hat{u}_i(g_i^0,({\mathop {\frown }\limits ^{. \ .}}), y^0) \le \hat{u}_i(g_i^0,\delta _i^0,y) \le U_j(\sigma ^*) = \min _{k\in N} U_k(\sigma ^*).$$
\(\square \)Proof of Theorem 14.5 Let
\(\sigma ^*=(f^*, r^*)\) be an inductively stable stationary state satisfying FA. Let
\(f_j=f_i^*\) for player
j below. We define the constituents of a sophisticated hedonistic model other than utility function
\(\hat{u}_i\) as follows:
 SH1

\(\hat{N}\) is an arbitrary imaginary player set partitioned into nonempty disjoint ethnic groups \(\hat{N}_1,\ldots ,\hat{N}_{e_0}\) with \(i\in \hat{N}_{e(i)}\) and \(\hat{N}_{e(i)} \ge 2\);
 SH2

\(\hat{Z} =\{1,\ldots ,\ell \}^{\hat{N}} \times \{({\mathop {\smile }\limits ^{. \ .}}), ({\mathop {\frown }\limits ^{. \ .}}) \} ^{\hat{N}} \times \{1,0\}\);
 SH3

\(\hat{o}_i(x) = (g_i,\delta _i,\hat{E}_i(g))\) for all \(x=(g,\delta , y) \in \hat{Z}\);
 SH5

\(x^0 = (g^0,\delta ^0,0)\), where \(\delta ^0=(({\mathop {\smile }\limits ^{. \ .}}),\ldots ,({\mathop {\smile }\limits ^{. \ .}}))\) and \(g^0=(g_j^0)_{j \in \hat{N}}\) is defined by: for each \(j\in \hat{N}_e\) (\(e=1,\ldots , e_0\)), \(g_j^0 =f_k^*\) for some \(k\in N_e\);
 SH6

\(X=\{x^0 \} \cup X_A \cup X_{P}\), where
$$\begin{aligned} X_A= & {} \left\{ ((g_{i}^0,g_i),(\delta _{i}^0,\delta _i), 0) \right\} _{(g_i,\delta _i)} \\ X_{P}= & {} \{ ( (g_{j}^0,f_j),(\delta _{i}^0, r_i^*(f_j,\hat{E}_i(g_{j}^0,f_j))), 0) ) \} \cup \{(g^0,\delta ^0,1)\}. \end{aligned}$$
We define a utility function
\(\hat{u}_i:\{1,\ldots ,\ell \} \times \{({\mathop {\smile }\limits ^{. \ .}}), ({\mathop {\frown }\limits ^{. \ .}}) \} \times 2^{\{1, \ldots , e_0 \} } \times \{1, 0 \} \rightarrow \mathbb {R} \) by
$$ \hat{u}_i(g_i,\delta _i,E,y) = {\left\{ \begin{array}{ll} \delta _i(\mu _i(\sigma _{i}^*,(g_i,\delta _i))+y) &{} \mathrm{if \ } \mathcal {E}=\mathcal {E}_i(f_{i}^*,g_i) \mathrm{\ for \ some \ } g_i, \\ \delta _i(\mu _i(\sigma _{j}^*,(f_j,\delta _i))+y) &{} \mathrm{if \ } \mathcal {E}=\mathcal {E}_i(f_{j}^*,f_j) \mathrm{\ for \ some \ } j, \delta _i=r_i^*(f_{j}^*,f_j), \\ h^(g_i,\delta _i,E,y) &{} \mathrm{if \ } \mathcal {E}=\mathcal {E}_i(f_{j}^*,f_j) \mathrm{\ for \ some \ } j, \delta _i \ne r_i^*(f_{j}^*, f_j), \\ \mathrm{arbitrary} &{} \mathrm{otherwise,} \end{array}\right. } $$
where
\(h^(g_i,\delta _i,E, y)\) is a real number not greater than
\(U_i(\sigma _{j}^*,(f_j, 0))\). FA and Theorem
13.2 guarantee that the function is welldefined. The first case gives utility values to
i’s own deviations, which together with the observation function
\(\hat{o}_i\) implies coherence with active experiences. This covers also the case where an insider takes the unfriendly action or moves out of festival
\(f_i^*\). The second case gives utility values to the situations where outsiders come to festival
\(f_i^*\), which together with
\(\hat{o}_i\) implies coherence with passive experiences. The third case gives utility values to the unexperienced cases where an outsider come to festival
\(f_i^*\) but he took the action not prescribed by
\(r_i^*\). In fact, we set
\(h^(g_i,\delta _i,E, y)\) so that the sophisticated hedonistic model is rationalizable.
\(\square \) Proof of Theorem 14.6 The if part is straightforward. We show the onlyif part. By FA, \(\sigma ^*\) is a fully active equilibrium. If the players in each festival are all nondiscriminators or all discriminators toward each ethnicity of an outsider, then \(\sigma ^*\) enjoys sequential rationality over \(\cup _i P_i(\sigma ^*)\). Now we show the contrapositive of the onlyif part. Suppose that \(\sigma ^*\) does not enjoy sequential rationality over \(\cup _i P_i(\sigma ^*)\). Then some festival has nondiscriminators as well as discriminators toward some ethnicity.
Let k be a festival where some are discriminators and some are nondiscriminators when an outsider j comes to \(f_j\). Let i and \(i'\) be a discriminator and a nondiscriminator, respectively, in k toward the ethnicity of j. There are two cases to consider: (a) \(\mu _i(\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}}))) \ge m_0\); and (b) \(\mu _i(\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}}))) < m_0\).
In case (a), let x be the path determined by \((\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}})))\). Then \(\hat{u}_i(x)=U_i(\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}})))=0\). Let \(\sigma _i'=\sigma _j'=(f_j,({\mathop {\smile }\limits ^{. \ .}}))\), \(\sigma _{i,j}'=\sigma _{i, j}^*\), and \(x'\) be the path determined by \(\sigma '\). Then, we have \(0< U_i(\sigma ')=\hat{u}_i(x')\). Hence, \(\mathcal {TG}\) of player i is not rationalizable.
Consider case (b) in which \(\mu _{i'}(\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}})))<\mu _i(\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}})))<m_0\) holds. Define \(\sigma '\), the strategy profile given by \(\sigma _{i'}' =(f_j,({\mathop {\frown }\limits ^{. \ .}}))\), \(\sigma _{j}' =(f_j,({\mathop {\smile }\limits ^{. \ .}}))\), \(\sigma _{i',j}'=\sigma _{i', j}^*\), and let \(x'\) be the path determined by \(\sigma '\). Then \(\hat{u}_{i'}(x) = U_{i'}(\sigma _{j}^*,(f_j,({\mathop {\smile }\limits ^{. \ .}})))<0=U_{i'}(\sigma ') = \hat{u}_{i'}(x')\). Hence, \(\mathcal {TG}\) of player \(i'\) is not rationalizable. \(\square \)