Proof of Lemma 1. Since \(zu(a,H)+(1-z)u(a,L)\) is continuous with respect to \(z\), \(B(a^{*})\) is a closed set. Suppose \(z_1\in B(a^{*})\) and \(z_2\in B(a^{*})\). It follows that \(z_1u(a^{*},H)+(1-z_1)u(a^{*},L)\geq z_1u(a,H)+(1-z_1)u(a,L)\) and \(z_2u(a^{*},H)+(1-z_2)u(a^{*},L)\geq z_2u(a,H)+(1-z_2)u(a,L)\) for any \(a\in A\). Take any \(t\in [0,1]\), then we have \[\begin{align} &[tz_1+(1-t)z_2]u(a^{*},H)+[1-tz_1-(1-t)z_2]u(a^{*},L)\\ &= t[z_1u(a^{*},H)+(1-z_1)u(a^{*},L)]+(1-t)[z_2u(a^{*},H)+(1-z_2)u(a^{*},L)]\\ &\geq t[z_1u(a,H)+(1-z_1)u(a,L)]+(1-t)[z_2u(a,H)+(1-z_2)u(a,L)]\\ &=[tz_1+(1-t)z_2]u(a,H)+[1-tz_1-(1-t)z_2]u(a,L). \end{align}\] Hence, \(tz_1+(1-t)z_2\in B(a^{*})\). ◻

Proof of Lemma 2. Take any \(a_{0}\in B^{-1}(z_1)\cap B^{-1}(z_2)\). Then, \(z_1u(a_{0},H)+(1-z_1)u(a_{0},L)\geq z_1u(a,H)+(1-z_1)u(a,L)\) and \(z_2u(a_{0},H)+(1-z_2)u(a_{0},L)\geq z_2u(a,H)+(1-z_2)u(a,L)\) for all \(a\in A\). Note that at least one inequality holds strictly if \(a\notin B^{-1}(z_1)\cap B^{-1}(z_2)\). Hence, for any \(w\in (z_1,z_2)\), \[\begin{align} &wu(a_{0},H)+(1-w)u(a_{0},L)\\ & = \frac{w-z_2}{z_1-z_2}[ z_1u(a_{0},H)+(1-z_1)u(a_{0},L)] \\ & \quad \quad \quad \quad \quad \quad \quad \quad+\left(1-\frac{w-z_2}{z_1-z_2}\right) [ z_2u(a_{0},H)+(1-z_2)u(a_{0},L)]\\ &\geq \frac{w-z_2}{z_1-z_2}[z_1u(a,H)+(1-z_1)u(a,L)] \\ & \quad \quad \quad \quad \quad \quad \quad \quad+\left(1-\frac{w-z_2}{z_1-z_2}\right)[ z_2u(a,H)+(1-z_2)u(a,L)]\\ &= wu(a,H)+(1-w)u(a,L) \end{align}\] for all \(a\in A\) and strict inequality holds for all \(a\notin B^{-1}(z_1)\cap B^{-1}(z_2)\). Thus, \(B^{-1}(w)=B^{-1}(z_{1})\cap B^{-1}(z_{2})\). ◻

Proof of Lemma 3. If \(a_{0}\in B^{-1}(1)\), the statement holds because \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*}) &=\mu_{0}u(a_{0},H)+(1-\mu_{0})u(a_{0},L)\\ &=\mu_{0}[(1-p^{i})u(a_{0},H)+p^{i}u(a_{0},H)]+(1-\mu_{0})u(a_{0},L)\\ &=\mu_{0}[(1-p^{i})u(a_{1},H)+p^{i}u(a_{0},H)]+(1-\mu_{0})u(a_{0},L). \end{align}\] Suppose \(a_{0}\notin B^{-1}(1)\). Take any equilibrium under \(\pi\). Then, by Lemma 2, agent 1 chooses \(a_{0}\) if and only if he receives \(s\notin S(1)\). Agent 2 chooses an action from \(B^{-1}(1)\) if she receives \(s\in S(1)\) or agent 1 takes an action other than \(a_{0}\) because she knows that the state is \(H\). Notably, the public belief after observing \(a_{0}\) is less than \(\mu_{0}\) and Lemma 2 implies that \(B^{-1}(z)=\{a_{0}\}\) for all \(z\in[0,x]\). Hence, agent 2 must choose \(a_{0}\) if she receives \(s\notin S(1)\) and agent 1 chooses \(a_{0}\). Analogously, agent \(i\) takes action from \(B^{-1}(1)\) if and only if he receives \(s\in S(1)\) or at least one previous agent chooses an action other than \(a_{0}\). Otherwise, agent \(i\) takes \(a_{0}\). Therefore, we have \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=\mu_{0}[(1-p^{i})u(a_{1},H)+p^{i}u(a_{0},H)]+(1-\mu_{0})u(a_{0},L). \end{align}\] ◻

Proof of Lemma 4. Suppose \(\pi\succsim_{B} \pi'\) and \(\rho \succsim_{B} \rho'\). Then, there exist Markov kernels \(\gamma_{1}\) and \(\gamma_{2}\) such that \[\pi'(s'|\omega)=\sum_{s\in \text{supp}(\pi)}\gamma_{1}(s'|s)\pi(s|\omega) \quad \text{and} \quad \rho'(t'|\omega)=\sum_{t\in \text{supp}(\rho)}\gamma_{2}(t'|t)\rho(t|\omega)\] for all \(s'\in \text{supp}(\pi')\) and \(t'\in \text{supp}(\rho')\). Then, we have \[\begin{align} (\pi'\otimes \rho')((s',t')|\omega)&=\pi'(s'|\omega)\rho'(t'|\omega)\\ &=\sum_{s\in \text{supp}(\pi)}\gamma_{1}(s'|s)\pi(s|\omega)\sum_{t\in \text{supp}(\rho)}\gamma_{2}(t'|t)\rho(t|\omega)\\ &=\sum_{(s,t)\in \text{supp}(\pi\otimes \rho)}\gamma_{1}(s'|s)\gamma_{2}(t'|t)\pi(s|\omega)\rho(t|\omega)\\ &=\sum_{(s,t)\in \text{supp}(\pi\otimes \rho)}\gamma((s',t')|(s,t))(\pi\otimes\rho)((s,t)|\omega), \end{align}\] where \(\gamma((s',t')|(s,t))=\gamma_{1}(s'|s)\gamma_{2}(t'|t)\). Since \(\gamma\) is a Markov kernel, \(\pi'\otimes \rho'\) is a garbling of \(\pi \otimes \rho\). ◻

Proof of Lemma 5. Take any \(\mathcal{D},\pi\), and \(\boldsymbol{\sigma}\). Note that \(\overline{V}_{1}^{\mathcal{D}}(\pi)= V_{1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma})\). Fix \(i\geq 2\). For each \(s\in S^{i-1}\), define \(f_{i-1}(s)\in \Delta (A^{i-1})\) as \[\begin{align} f_{i-1}(\boldsymbol{a}|s)=\prod_{k=1}^{i-1}\sigma_{k}(a_{k}|a_{1},\dots,a_{k-1},s_{k}). \end{align}\] Hence, \(f_{i-1}(a|s)\) is the probability that agent \(1\) to agent \(i-1\) takes actions \(\boldsymbol{a}=(a_{1},\dots,a_{i-1})\) when agent \(1\) to agent \(i-1\) receives private signal \(\boldsymbol{s}=(s_{1},\dots,s_{i-1})\). Then, \[\begin{align} \alpha_{\leq i-1}^{\omega}(\boldsymbol{a}|\pi,\boldsymbol{\sigma})=&\sum_{\boldsymbol{s}\in S^{i-1}}\prod_{k=1}^{i-1}\sigma_{k}(a_{k}|a_{1},\dots,a_{k-1},s_{k})\pi(s_{k}|\omega)\\=&\sum_{\boldsymbol{s}\in S^{i-1}}f_{i-1}(\boldsymbol{a}|\boldsymbol{s})\pi^{\otimes i-1}(\boldsymbol{s}|\omega). \end{align}\] Thus, \(\alpha_{\leq i-1}(\cdot|\pi,\boldsymbol{\sigma})\) is a garbling of \(\pi^{\otimes i-1}\), where \(\alpha_{\leq i-1}\) is an ex-ante action distribution induced by \(\alpha_{\leq i-1}^{\omega}\).³³ By Lemma 4, we have \(\pi^{\otimes i-1}\otimes \pi \succsim_{B} \alpha_{\leq i-1}(\cdot|\pi,\boldsymbol{\sigma})\otimes\pi\). Hence, \(\overline{V}_{i}^{\mathcal{D}}(\pi)\geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma})\) holds for all \(i\), \(\mathcal{D}\), \(\pi\), and \(\boldsymbol{\sigma}\). ◻

Proof of Theorem 1. Since \(\overline{V}_{i}^{\mathcal{D}}(\pi')\geq V_{i}^{\mathcal{D}}(\pi',\boldsymbol{\sigma}')\) holds for every strategy profile \(\boldsymbol{\sigma}'\) by Lemma 5, \(\pi\succsim_{S} \pi'\) holds if \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})\geq \overline{V}_{i}^{\mathcal{D}}(\pi')\).

Conversely, suppose \(\pi\succsim_{S}\pi'\). Take any \(\mathcal{D}=(A,u)\), equilibrium \(\boldsymbol{\sigma}^{*}\) under \(\pi:\Omega\to \Delta(S)\), and equilibrium \(\boldsymbol{\sigma}^{**}\) under \(\pi':\Omega\to \Delta(S')\). Then, \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})\geq V_{i}^{\mathcal{D}}(\pi',\boldsymbol{\sigma}^{**})\) by \(\pi\succsim_{S}\pi'\). Consider the decision problem \(\overline{\mathcal{D}}=(\overline{A},\overline{u})\), where \(\overline{A}=\{(a,k)\mid a\in A,k\in S'\}\) and \(\overline{u}((a,k),\omega)=u(a,\omega)\) for all \(a\in A,\omega\in \Omega\). Fix \(s_{1}\in S'\) and define strategy profile \(\boldsymbol{\sigma}=(\sigma_{i})_{i\in \mathbb{N}}\) under \((\overline{\mathcal{D}},\pi)\) as following: \[\begin{align} \begin{cases} \sigma_{i}((a,s_{1})|(a_{1},k_{1}),(a_{2},k_{2}),\dots,(a_{i-1},k_{i-1}),s)=\sigma^{*}_{i}(a| a_{1},a_{2},\dots,a_{i-1},s)\\ \sigma_{i}((a,k)|(a_{1},k_{1}),(a_{2},k_{2}),\dots,(a_{i-1},k_{i-1}),s)=0, \end{cases} \end{align}\] for all \(a\in A\), \(s\in S\), \((a_{1},\dots,a_{i-1})\in A^{i-1}\), \(k_{1},k_{2},\dots,k_{i-1}\in S'\), and \(k\in S'\setminus \{s_{1}\}\). Note that \(\boldsymbol{\sigma}\) is an equilibrium under \((\overline{\mathcal{D}},\pi)\). Moreover, it follows that \(V_{i}^{\overline{\mathcal{D}}}(\pi,\boldsymbol{\sigma})=V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})\). Under \((\overline{\mathcal{D}},\pi')\), if we consider the following equilibrium \(\boldsymbol{\sigma}'\), the expected payoff of agent \(i\) at equilibrium (\(V_{i}^{\overline{\mathcal{D}}}(\pi',\boldsymbol{\sigma}')\)) coincides with \(\overline{V}_{i}^{\mathcal{D}}(\pi')\). Specifically, each agent \(i\) chooses an action that maximizes his expected payoff on the equilibrium path, but always chooses an action of the form \((a,k)\) (\(a\in A\)) when the received signal is \(k\in S'\). Since each agent can observe signals received by their predecessor on the equilibrium path, it follows that \(V_{i}^{\overline{\mathcal{D}}}(\pi',\boldsymbol{\sigma}')=\overline{V}_{i}^{\overline{\mathcal{D}}}(\pi')=\overline{V}_{i}^{\mathcal{D}}(\pi')\). Therefore, \[V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})= V_{i}^{\overline{\mathcal{D}}}(\pi,\boldsymbol{\sigma})\geq V_{i}^{\overline{\mathcal{D}}}(\pi',\boldsymbol{\sigma}')=\overline{V}_{i}^{\mathcal{D}}(\pi').\] ◻

Proof of Corollary 1. Prove by contradiction. Suppose co\((\text{supp}(\mu))\neq [0,1]\). Then, either \(1\notin \text{supp}(\mu)\) or \(0 \notin \text{supp}(\mu)\). By symmetry, it suffices to consider the case where \(1\notin \text{supp}(\mu)\). Since \(\text{supp}(\mu)\) is a closed set, there exists \(r\in[\mu_{0},1)\) such that \(\text{supp}(\mu)\subseteq [0,r]\). Consider the following decision problem \(\mathcal{D}=(A,u)\): \(A=\{a_{0},a_{1}\}\), \(u(a_{0},L)=u(a_{0},H)=0\), \(u(a_{1},H)=1-r\) and \(u(a_{1},L)=-r\). Then, the strategy profile \(\boldsymbol{\sigma}^{*}\) that all agents always choose \(a_{0}\) is an equilibrium under \((\mathcal{D},\pi)\). It follows that \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=0\). Since \(\pi'\) is not no information, repeated observations of \(\pi'\) allow agents to learn the state in the limit. Hence, \(\overline{V}_{i}^{\mathcal{D}}(\pi')>0\) holds for sufficiently large \(i\). By Theorem 1, \(\pi\) is not more socially valuable than \(\pi'\). ◻

Proof of Lemma 6. First, it is easily calculated that \[\begin{align} \overline{V}_{i}^{\mathcal{D}}(\pi)=\mu_{0}[(1-p^{i})U_{1}]+(1-\mu_{0})[(1-p^{i})U_{0}]+p^{i}U_{\mu_{0}}. \end{align}\] We now show that \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=\overline{V}_{i}^{\mathcal{D}}(\pi)\). First, this obviously holds for agent \(1\): \[\begin{align} V_{1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})= \overline{V}_{1}^{\mathcal{D}}(\pi) = \mu_{0} (1-p)U_{1}+(1-\mu_{0})(1-p)U_{0}+pU_{\mu_{0}}. \end{align}\] Then, for each \(i\geq 2\), we consider a strategy in which agent \(i\) chooses the optimal actions upon receiving conclusive signals about each state and mimics agent \(i-1\)’s action otherwise. By the optimality of the equilibrium strategy, for each \(i\geq 2\), we have \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})&\geq \mu_{0} (1-p)U_{1}+(1-\mu_{0})(1-p)U_{0} \\ & \quad +p\left[\mu_{0}\sum_{a}\alpha_{i-1}^{H}(a|\pi,\boldsymbol{\sigma}^{*})u(a,H) +(1-\mu_{0})\sum_{a}\alpha_{i-1}^{L}(a|\pi,\boldsymbol{\sigma}^{*})u(a,L)\right]\\ &= \mu_{0} (1-p)U_{1}+(1-\mu_{0})(1-p)U_{0}+pV_{i-1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*}). \end{align}\] Conversely, from Lemma 5, for each \(i\) \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*}) \leq \overline{V}_{i}^{\mathcal{D}}(\pi) = \mu_{0} (1-p^{i})U_{1}+(1-\mu_{0})(1-p^{i})U_{0}+p^{i}U_{\mu_{0}}. \end{align}\] Fix some \(j\geq 1\) and suppose \(V_{j-1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=\overline{V}_{j-1}^{\mathcal{D}}(\pi)\). Then, \[\begin{align} V_{j}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})&\geq \mu_{0} (1-p)U_{1}+(1-\mu_{0})(1-p)U_{0}+pV_{j-1}^{\mathcal{D}}(\pi,\sigma^{*})\\ &=\mu_{0} (1-p)U_{1}+(1-\mu_{0})(1-p)U_{0} \\ & \quad \quad \quad \quad +p\left[\mu_{0} (1-p^{j-1})U_{1}+(1-\mu_{0})(1-p^{j-1})U_{0}+p^{j-1}U_{\mu_{0}}\right]\\ &=\mu_{0} (1-p^j)U_{1}+(1-\mu_{0})(1-p^j)U_{0}+p^jU_{\mu_{0}}\\ &=\overline{V}_j^{\mathcal{D}}(\pi). \end{align}\] Hence, we have \(V_{j}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=\overline{V}_j^{\mathcal{D}}(\pi)\). By mathematical induction, it follows that \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=\overline{V}_{i}^{\mathcal{D}}(\pi)\) for all \(i\). ◻

Proof of Lemma 7. Take any \(\mathcal{D}=(A,u)\). Take arbitrary equilibrium \(\boldsymbol{\sigma}^{*}\) under \((\mathcal{D},\pi)\). From Lemma 6, we have \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=\overline{V}_{i}^{\mathcal{D}}(\pi)\). Hence, the expected payoff of agent \(i\) in any equilibrium is the same as the expected payoff of agent \(i\) when agent \(i\) can observe the signal realizations of past agents rather than the actions taken by them.

Next, take any equilibrium \(\boldsymbol{\sigma}^{**}\) under \((\mathcal{D},\pi')\). Note that \(\overline{V}_{i}^{\mathcal{D}}(\pi')\geq V_{i}^{\mathcal{D}}(\pi',\boldsymbol{\sigma}^{**})\) holds by Lemma 5. Since \(\pi^{\otimes i} \succsim_{B} \pi'^{\;\otimes i}\) by Lemma 4, we have \(\overline{V}_{i}^{\mathcal{D}}(\pi)\geq \overline{V}_{i}^{\mathcal{D}}(\pi')\). Hence, it follows that \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})= \overline{V}_{i}^{\mathcal{D}}(\pi)\geq \overline{V}_{i}^{\mathcal{D}}(\pi')\geq V_{i}^{\mathcal{D}}(\pi',\boldsymbol{\sigma}^{**}). \end{align}\] Therefore, we obtain \(\pi \succsim_{S} \pi'\). ◻

Proof of Lemma 8. Let \(q_{L}=\frac{\pi'(\mu'=0|L)}{\pi(\mu=0|L)}\) and \(q_{H}=\frac{\pi'(\mu'=1|H)}{\pi(\mu=1|H)}\). Take any \(\mathcal{D}\) and define \(\boldsymbol{\sigma}^{**}=(\sigma_{i}^{**})_{i\in \mathbb{N}}\) as the following strategy under \((\mathcal{D},\pi)\). Agent 1 chooses \(a_{0}\in B^{-1}(0)\) with probability \(q_{L}\) and chooses \(a_{2}\in B^{-1}(\mu_{0})\) with probability \(1-q_{L}\) if he receives conclusive signal about \(\omega=L\). Agent 1 chooses \(a_{1}\in B^{-1}(1)\) with probability \(q_{H}\) and chooses \(a_{2}\) with probability \(1-q_{H}\) if he receives conclusive signal about \(\omega=H\). Otherwise, agent 1 chooses \(a_{2}\). For \(i\geq 2\), agent \(i\) chooses \(a_{0}\) with probability \(q_{L}\) and chooses the same action as agent \(i-1\) with probability \(1-q_{L}\) if he receives a conclusive signal about \(\omega=L\). Agent \(i\) chooses \(a_{1}\) with probability \(q_{H}\) and chooses the same action as agent \(i-1\) with probability \(1-q_{H}\) if he receives a conclusive signal about \(\omega=H\). Otherwise, agent \(i\) chooses the same action as agent \(i-1\). First, note that \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{**}) &=\mu_{0}[(1-p^{i})U_{1}]+(1-\mu_{0})[(1-p^{i})U_{0}]+p^{i}U_{\mu_{0}}\\ &=\overline{V}_{i}^{\mathcal{D}}(\pi'), \end{align}\] where the last equality comes from Lemma 6.

Fix any equilibrium \(\boldsymbol{\sigma}^{*}\) under \((\mathcal{D},\pi)\) and define \(\boldsymbol{\sigma}(k)\) as \[\begin{align} \boldsymbol{\sigma}(k)=(\sigma^{*}_1,\sigma^{*}_2,\dots,\sigma^{*}_k,\sigma^{**}_{k+1},\sigma^{**}_{k+2},\dots). \end{align}\] We show that if \(i\geq k+1\), \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k))=\mu_{0}(1-p)U_{1}+(1-\mu_{0})(1-p)U_{0}+p V_{i-1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k)). \end{align}\] Note that \[\begin{align} V_{i-1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k))=&\mu_{0}\sum_{a\in A}\alpha_{i-1}^{H}(a|\pi,\boldsymbol{\sigma}(k))u(a,H)\\ & \quad \quad \quad \quad +(1-\mu_{0})\sum_{a\in A}\alpha_{i-1}^{L}(a|\pi,\boldsymbol{\sigma}(k))u(a,L). \end{align}\] \(i\geq k+1\) implies that \(\boldsymbol{\sigma}(k)_{i}=\sigma^{**}_{i}\). Hence, \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k)) &= \mu_{0}\left[\pi(\mu=1|H)q_{H}U_{1}+(1-\pi(\mu=1|H)q_{H})\sum_{a\in A}\alpha_{i-1}^{H}(a|\pi,\boldsymbol{\sigma}(k))u(a,H)\right]\\ & + (1-\mu_{0})\left[\pi(\mu=0|L)q_{L}U_{0}+(1-\pi(\mu=0|L)q_{L})\sum_{a\in A}\alpha_{i-1}^{L}(a|\pi,\boldsymbol{\sigma}(k))u(a,L)\right]\\ & = \mu_{0}\left[(1-p)U_{1}+p\sum_{a\in A}\alpha_{i-1}^{H}(a|\pi,\boldsymbol{\sigma}(k))u(a,H)\right]\\ &\quad \quad \quad \quad \quad \quad +(1-\mu_{0})\left[(1-p)U_{0}+p\sum_{a\in A}\alpha_{i-1}^{L}(a|\pi,\boldsymbol{\sigma}(k))u(a,L)\right]\\ & = \mu_{0} (1-p)U_{1}+(1-\mu_{0})(1-p)U_{0}+pV_{i-1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k)). \end{align}\] By the definition of \(\boldsymbol{\sigma}(k)\), \[\begin{align} \begin{cases} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k)) & \text{ if } i<k+1,\\ V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})\geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k)) & \text{ if } i=k+1. \end{cases} \end{align}\] The second inequality is held by the optimality of \(\sigma^{*}_{i}\). We now show that \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k+1))\geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k)) \end{align}\] for all \(i\) and \(k\). First, if \(k\geq i-1\), we have \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k+1))= V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*}) \geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k))\). Next, we have \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-1))\geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-2))\) for \(i\geq 2\) since \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-2))&=\mu_{0}(1-p)U_{1}+(1-\mu_{0})(1-p)U_{0}+pV_{i-1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-2))\\ &\leq \mu_{0}(1-p)U_{1}+(1-\mu_{0})(1-p)U_{0}+pV_{i-1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-1))\\ &=V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-1)). \end{align}\] Then, we have \(V_{i}(\pi,\boldsymbol{\sigma}(i-2))\geq V_{i}(\pi,\boldsymbol{\sigma}(i-3))\) for \(i\geq 3\) since \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-3))&=\mu_{0}(1-p)U_{1}+(1-\mu_{0})(1-p)U_{0}+pV_{i-1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-3))\\ &\leq \mu_{0}(1-p)U_{1}+(1-\mu_{0})(1-p)U_{0}+pV_{i-1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-2))\\ &=V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-2)). \end{align}\] Analogously, it follows that \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-m))\geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i-m-1))\) for all \(i,m\) that satisfies \(i-m-1\geq 0\). Hence, \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k+1))\geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(k))\) for all \(i,k\). Therefore, we have \[V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*}) = V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(i)) \geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}(0)) =V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{**})=\overline{V}_{i}^{\mathcal{D}}(\pi').\] ◻

Proof of Theorem 2. Suppose that \(\pi\succsim_{B} \pi'' \succsim_{B} \pi'\) and \(\text{supp}(\mu'')=\{0,\mu_{0},1\}\). From Lemma 7, we conclude that \(\pi''\succsim_{S} \pi'\) holds. Since \(\pi\succsim_{B} \pi''\) and \(\text{supp}(\mu'')=\{0,\mu_{0},1\}\), it follows that \(\min\{\pi(\mu=0|L),\pi(\mu=1|H)\}\geq \pi''(\mu''=\mu_{0}|L)=\pi''(\mu''=\mu_{0}|H)\). Thus, from Lemma 8, we also conclude that \(\pi\succsim_{S} \pi''\) holds. Therefore, we have \(\pi\succsim_{S} \pi'\). ◻

Proof of Proposition 2. Let \(S'=\text{supp}(\pi')\). Suppose \(\text{supp}(\mu'')=\{0,\mu_{0},1\}\). We now show that \(\pi \succsim_{B} \pi''\) is equivalent to \[\pi(\mu=0|L)\geq \pi''(\mu''=0|L) \quad \text{and} \quad \pi(\mu=1|H)\geq \pi''(\mu''=1|H).\] Note that \[\begin{align} &\pi''(\mu''=0|\omega)=\frac{\pi''(\mu''=0|L)}{\pi(\mu=0|L)}\pi(\mu=0|\omega)\\ &\pi''(\mu''=1|\omega)=\frac{\pi''(\mu''=1|H)}{\pi(\mu=1|H)}\pi(\mu=1|\omega)\\ &\pi''(\mu''=\mu_{0}|\omega)=\left[1-\frac{\pi''(\mu''=0|L)}{\pi(\mu=0|L)}\right] \pi(\mu=0|\omega)+ \left[1-\frac{\pi''(\mu''=1|H)}{\pi(\mu=1|H)}\right]\pi(\mu=1|\omega)\\ & +\sum_{x\in \text{supp}(\mu)\setminus\{0,1\}}\pi(\mu=x|\omega). \end{align}\] Thus, \(\pi''\) is a garbling of \(\pi\) if \(\pi(\mu=0|L)\geq \pi''(\mu''=0|L)\) and \(\pi(\mu=1|H)\geq \pi''(\mu''=1|H)\). Conversely, suppose \(\pi''\) is a garbling of \(\pi\). Then, we have \[\begin{align} &\pi''(\mu''=0|\omega)=\sum_{x\in \text{supp}(\mu)}\gamma(x)\pi(\mu=x|\omega) \end{align}\] for some \(\gamma\in \Delta (\text{supp}(\mu))\). Since \(\pi''(\mu''=0|H)=0\) and \(\pi(\mu=x|H)>0\) for all \(x\neq 0\), it follows that \(\gamma(x)=0\) for all \(x\neq 0\). Hence, \(\pi(\mu=0|L)\geq \pi''(\mu''=0|L)\). Similarly, we have \(\pi(\mu=1|H)\geq \pi''(\mu''=1|H).\)

Next, we show that \(\pi'' \succsim_{B} \pi'\) is equivalent to \[\begin{align} \pi''(\mu''=\mu_{0}|L)&=\pi''(\mu''=\mu_{0}|H)\leq \sum_{s\in S'}\min\{\pi'(s|L),\pi'(s|H)\}. \end{align}\] Suppose \(\pi''(\mu''=\mu_{0}|L)=\pi''(\mu''=\mu_{0}|H) \leq \sum_{s\in S'}\min\{\pi'(s|L),\pi'(s|H)\}\). Define \(\rho:\Omega\to \Delta\{s_{0},s_{1},s_{2}\}\) that satisfies \[\begin{align} \rho(s_{1}|L)=0, \quad \rho(s_{0}|H)=0, \quad \rho(s_{2}|H)=\rho(s_{2}|L)=\sum_{s\in S'}\min\{\pi'(s|L),\pi'(s|H)\}. \end{align}\] Then, we have \(\pi''\succsim_{B}\rho\) as \(\text{supp}(\mu'')=\{0,\mu_{0},1\}\).

If \(\rho(s_{2}|L)=\rho(s_{2}|H)=1\), \(\rho\succsim_{B} \pi'\) as \(\pi'\) is no information. If \(\rho(s_{2}|L)=\rho(s_{2}|H)=0\), \(\rho\succsim_{B} \pi'\) as both \(\rho\) and \(\pi'\) are full information. Otherwise, \[\begin{align} \pi'(s|\omega)= &\frac{\max\{\pi'(s|L)-\pi'(s|H),0\}}{\rho(s_{0}|L)}\rho(s_{0}|\omega)+\frac{\max\{\pi'(s|H)-\pi'(s|L),0\}}{\rho(s_{1}|H)}\rho(s_{1}|\omega)\\&\quad\quad\quad\quad\quad\quad+\frac{\min\{\pi'(s|L),\pi'(s|H)\}}{\rho(s_{2}|L)}\rho(s_{2}|\omega) \end{align}\] and \[\begin{align} &\sum_{s\in S'}\frac{\max\{\pi'(s|L)-\pi'(s|H),0\}}{\rho(s_{0}|L)} =\frac{\sum_{s\in S'}\max\{\pi'(s|L)-\pi'(s|H),0\}}{1-\sum_{s\in S'}\min\{\pi'(s|L),\pi'(s|H)\}}=1\\ &\sum_{s\in S'}\frac{\max\{\pi'(s|H)-\pi'(s|L),0\}}{\rho(s_{1}|H)} =\frac{\sum_{s\in S'}\max\{\pi'(s|H)-\pi'(s|L),0\}}{1-\sum_{s\in S'}\min\{\pi'(s|L),\pi'(s|H)\}}=1\\ &\sum_{s\in S'}\frac{\min\{\pi'(s|L),\pi'(s|H)\}}{\rho(s_{2}|L)}=1. \end{align}\] Hence, \(\pi'\) is a garbling of \(\rho\) and we have \(\rho\succsim_{B} \pi'\). Note that \(\pi''\succsim_{B} \rho\) and \(\rho\succsim_{B} \pi'\) implies \(\pi''\succsim_{B}\pi'\). Therefore, \(\pi''(\mu''=\mu_{0}|H)=\pi''(\mu''=\mu_{0}|L)\leq \sum_{s\in S'}\min\{\pi'(s|L),\pi'(s|H)\}\) is a sufficient condition for \(\pi'' \succsim_{B} \pi'\).

Conversely, suppose \(\pi'' \succsim_{B} \pi'\). Then, there exists probability distribution \(\gamma_{0}\), \(\gamma_{1}\), and \(\gamma_{\mu_{0}}\) over \(S'\) such that \[\pi'(s|\omega)=\gamma_{0}(s)\pi''(\mu''=0|\omega)+\gamma_{1}(s)\pi''(\mu''=1|\omega)+\gamma_{\mu_{0}}(s)\pi''(\mu''=\mu_{0}|\omega)\] for all \(s\in S'\) and \(\omega\in \Omega\). Then, for each \(\omega\in\Omega\), \[\begin{align} & \sum_{s\in S'}\min\{\pi'(s|L),\pi'(s|H)\}\\ &= \sum_{s\in S'} \min\left\{ \begin{aligned} & \gamma_{0}(s)\pi''(\mu''=0|L)+\gamma_{\mu_{0}}(s)\pi''(\mu''=\mu_{0}|L), \\ & \quad \quad \quad \gamma_{1}(s)\pi''(\mu''=1|H)+\gamma_{\mu_{0}}(s)\pi''(\mu''=\mu_{0}|H) \end{aligned} \right\}\\ &= \sum_{s\in S'}\left[\min\left\{\gamma_{0}(s)\pi''(\mu''=0|L),\gamma_{1}(s)\pi''(\mu''=1|H)\right\}+\gamma_{\mu_{0}}(s)\pi''(\mu''=\mu_{0}|L)\right]\\ & \geq \sum_{s\in S'}\gamma_{\mu_{0}}(s)\pi''(\mu''=\mu_{0}|L)\\ &= \pi''(\mu''=\mu_{0}|\omega). \end{align}\] Hence, \(\pi''(\mu''=\mu_{0}|L)=\pi''(\mu''=\mu_{0}|H)\leq \sum_{s\in S'}\min\{\pi'(s|L),\pi'(s|H)\}\) is a necessary condition for \(\pi'' \succsim_{B} \pi'\). Therefore, \(\pi'' \succsim_{B} \pi'\) is equivalent to \(\pi''(\mu''=\mu_{0}|L)=\pi''(\mu''=\mu_{0}|H)\leq \sum_{s\in S'}\min\{\pi'(s|L),\pi'(s|H)\}\), or \(\pi''(\mu''=0|L)=\pi''(\mu''=1|H)\geq 1- \sum_{s\in S'}\min\{\pi'(s|L),\pi'(s|H)\}\). By combining the first half and the second half, it can be seen that Proposition 2 holds. ◻

Derivation in Example 2. For the first step, we derive the expected payoff \(\overline{V}^{\overline{\mathcal{D}}}_{i}(\pi')\) for each sufficiently large \(i\). We first decompose the expected payoff into two parts. First, if an agent observes a conclusive signal about \(H\) (resp. \(L\)), then she chooses the optimal action \(a_{1}\) (resp. \(a_{0}\)). Hence, the expected payoff from the events in which at least one of the first \(i\) agents observes a conclusive signal is \(\frac{1}{2}\cdot [1-(1-\lambda')^{i}](1-r)\). We then focus on the event in which all agents from \(1\) to \(i\) observe either \(s_{l}\) or \(s_{h}\) and derive the corresponding interim payoff. Note that this interim payoff coincides with that obtained from \(i\) conditionally independent observations of the symmetric binary signal \(\pi'_{p'}\) where \(S_{p'}=\{s_{l},s_{h}\}\), \(\pi'_{p'}(s_{l}|L)=\pi'_{p'}(s_{h}|H)=1-p'\), and \(\pi'_{p'}(s_{h}|L)=\pi'_{p'}(s_{l}|H)=p'\) with \(p'\in(0,1/2)\).

Let \(l(s)\) denote the log-likelihood ratio after observing signal \(s\) from \(\pi'_{p'}\), that is, \(l(s)=\log \frac{\pi'_{p'}(s|H)}{\pi'_{p'}(s|L)}\). Then, we have \(\mathbb{E}[l(s)|L] < 0 < \mathbb{E}[l(s)|H]\). Let \(K^{L}:\mathbb{R}\to\mathbb{R}\) denote the cumulant generating function conditional on state \(L\), defined as follows: \[K^{L}(t) = \log \sum_{s\in S_{p'}} \pi'_{p'}(s|L)\left(\frac{\pi'_{p'}(s|H)}{\pi'_{p'}(s|L)} \right)^{t} =\log\left((1-p')\left(\frac{p'}{1-p'}\right)^{t} + p' \left(\frac{1-p'}{p'}\right)^{t}\right),\] for every \(t\in \mathbb{R}\).

By Cramér’s theorem [54], for each \(r\in(1-p,1)\) and \(p\in (0,1/2)\) (the parameter for \(\pi\)), we have \[\begin{align} \pi_{p'}'^{\; \otimes i}(\mu'\geq r|L) & = \mathbb{P}\left[l(s_{1})+\dots+l(s_{i})\geq \log \frac{r}{1-r}\mid L\right] \\ & = e^{i\min_{t}K^{L}(t) + o(i)}. \end{align}\] As \(\min_{t}K^{L}(t)=\log(2\sqrt{p'(1-p')})\) by the arithmetic–geometric mean inequality, we have \(\pi_{p'}'^{\; \otimes i}(\mu'\geq r|L)=(2\sqrt{p'(1-p')})^{i}e^{o(i)}\). By the symmetric argument, we obtain \(\pi_{p'}'^{\; \otimes i}(\mu'\leq r|H)=(2\sqrt{p'(1-p')})^{i}e^{o(i)}\).

Therefore, the interim payoff is \[\begin{align} & \frac{1}{2}\cdot \left((1-\pi_{p'}'^{\; \otimes i}(\mu'\leq r|H))(1-r)+\pi_{p'}'^{\; \otimes i}(\mu'\geq r|L)(-r)\right) \\ & = \frac{1}{2}\cdot \left(1-r-\left(2\sqrt{p'(1-p')}\right)^{i}e^{o(i)} \right). \end{align}\] Thus, for each sufficiently large \(i\), the expected payoff is \[\overline{V}^{\overline{\mathcal{D}}}_{i}(\pi')=\frac{1}{2}\left[ [1-(1-\lambda')^{i}](1-r)+(1-\lambda')^{i}\left(1-r-\left(2\sqrt{p'(1-p')}\right)^{i}e^{o(i)} \right)\right].\]

Then, a necessary condition for \(\pi \succsim_{S} \pi'\) is \(V^{\overline{\mathcal{D}}}_{i}(\pi,\boldsymbol{\sigma}) \geq \overline{V}^{\overline{\mathcal{D}}}_{i}(\pi')\) for all \(r\in(1-p,1)\) and sufficiently large \(i\). By the above derivation, the necessary condition is \[\begin{align} (1-\lambda')^{i}\left(2\sqrt{p'(1-p')}\right)^{i}e^{o(i)} \geq (1-\lambda)^{i}(1-r), \end{align}\] which is equivaent to \[\begin{align} r \geq 1- \left(\frac{1-\lambda'}{1-\lambda} \cdot 2\sqrt{p'(1-p')}\right)^{i}e^{o(i)}. \end{align}\] The left-hand-side is smallest when \(r\) is close to \(1-p\), and \(\pi \succsim_{S} \pi'\) does not hold if the inequality does not hold at \(r=1-p\). Thus, applying the \(1/i\)-th power to both sides at \(r=1-p\) yields the necessary condition as \[\frac{1-\lambda'}{1-\lambda} \cdot 2\sqrt{p'(1-p')}e^{\frac{o(i)}{i}} \geq p^{\frac{1}{i}}.\] Therefore, for sufficiently large \(i\), we can derive a necessary condition as \[\lambda \geq 1-2(1-\lambda')\sqrt{p'(1-p')}.\] ◻

Proof of Corollary 2. By Theorem 2, it is enough to show the necessity part. Take any \(\pi\) and \(\pi'\) such that \(\pi(\mu=1|H)=\pi'(\mu'=1|H)>0\) and \(\pi(\mu=0|L)=\pi'(\mu'=0|L)>0\). As a contraposition, suppose that there is no \(\pi''\) such that \(\text{supp}(\mu'')=\{0,\mu_{0},1\}\) and \(\pi \succsim_{B} \pi'' \succsim_{B} \pi'\). Then, under \(\pi'\), there exists a signal \(s'\) inducing a private belief in \((0,1)\setminus \{\mu_{0}\}\).³⁴ As in the construction behind the proof of Corollary 1, there exists some decision problem and its auxiliary problem, in which the observable signal setting of \(\pi'\) yields strictly higher payoff than \(\pi\) for some sufficiently later agent \(i\) because observing signal \(s'\) \(i\)-times induce the strictly better payoff by \(\pi(\mu=1|H)=\pi'(\mu'=1|H)\) and \(\pi(\mu=0|L)=\pi'(\mu'=0|L)\). ◻

Proof of Proposition 4. Suppose \(\pi \succsim_{ES} \pi'\). Take \(N\in \mathbb{N}\) such that \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})\geq \overline{V}_{i}^{\mathcal{D}}(\pi')\) for all \(\mathcal{D}\), \(i\geq N\), and equilibrium \(\boldsymbol{\sigma}^{*}\) under \((\mathcal{D},\pi)\). By Lemma 5, we have \(\overline{V}_{i}^{\mathcal{D}}(\pi)\geq \overline{V}_{i}^{\mathcal{D}}(\pi')\) for all \(\mathcal{D}\) and \(i\geq N\). Thus, \(\pi \succsim_{EB} \pi'\).

Consider the same example as Example 1. Then, we have \(V_{i}^{\mathcal{D}}(\pi',\boldsymbol{\sigma}')>V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma})\) for all \(i\geq 2\). Hence, \(\pi\) is not eventually more socially valuable than \(\pi'\), although \(\pi \succsim_{EB} \pi'\) holds as \(\pi \succsim_{B} \pi'\) and Lemma 4. ◻

Proof of Proposition 5. Since \(\succsim_{ES}\) satisfies transitivity, showing \(\pi \succsim_{ES} \pi''\) and \(\pi'' \succsim_{ES} \pi'\) is sufficient. First, we have \(\pi'' \succsim_{ES} \pi'\) because \[V_{i}^{\mathcal{D}}(\pi'',\boldsymbol{\sigma}^{*})=\overline{V}_{i}^{\mathcal{D}}(\pi'') \geq \overline{V}_{i}^{\mathcal{D}}(\pi')\geq V_{i}^{\mathcal{D}}(\pi',\boldsymbol{\sigma}^{**})\] for sufficiently large \(i\), where \(\boldsymbol{\sigma}^{*}\) is an arbitrary equilibrium under \((\mathcal{D},\pi'')\) and \(\boldsymbol{\sigma}^{**}\) is an arbitrary equilibrium under \((\mathcal{D},\pi')\). Each equality or inequality follows from Lemma 6, \(\pi'' \succsim_{EB} \pi'\), and Lemma 5 respectively. Thus, \(\pi'' \succsim_{ES} \pi'\).

To show \(\pi \succsim_{ES} \pi''\), note that if \(\text{supp}(\mu'')=\{0,\mu_{0},1\}\), the support of private belief distribution induced by \(\pi''^{\; \otimes i}\) is also \(\{0,\mu_{0},1\}\) for all \(i\). Then by the proof of Proposition 2, \(\pi\succsim_{EB}\pi''\) if and only if there exists \(N\in\mathbb{N}\) such that \(\pi^{\otimes i}(\mu=0|L) \geq \pi''^{\; \otimes i}(\mu''=0|L)\) and \(\pi^{\otimes i}(\mu=1|H) \geq \pi''^{\; \otimes i}(\mu''=1|H)\) for all \(i\geq N\). By the nature of conclusive signals, agents form the extreme beliefs \(\mu=1\) or \(0\) if they observe the conclusive signal at least once. Thus, these conditions are equivalent to \[1-[1-\pi(\mu=0|L)]^{i}\geq 1-[1-\pi''(\mu''=0|L)]^{i},\] and \[1-[1-\pi(\mu=1|H)]^{i}\geq 1-[1-\pi''(\mu''=1|H)]^{i},\] for all sufficiently large \(i\). Furthermore, these inequalities are equivalent to \[\pi(\mu=0|L)\geq \pi''(\mu''=0|L) \quad \text{and} \quad \pi(\mu=1|H)\geq \pi''(\mu''=1|H).\] Thus, under \(\text{supp}(\mu'')=\{0,\mu_{0},1\}\), \(\pi\succsim_{EB} \pi''\) if and only if \(\pi\succsim_{B} \pi''\). Therefore, we also have \(\pi \succsim_{S} \pi''\) by Lemma 8. ◻

Proof of Theorem 3. Suppose that \(\pi\) does not induce unbounded beliefs. By symmetry, we can assume \(1\notin \text{supp}(\mu)\) without loss of generality. Take \(r\in (0,1)\) such that \(\text{supp}(\mu) \subset [0,r]\). Let \(\mathcal{D}=(A,u)\) be the decision problem that is the same as in the proof of Corollary 1. Define \(\overline{\mathcal{D}}\) as the replicated decision problem in the same way as in the proof of Theorem 1. Then, we can construct the equilibrium \(\boldsymbol{\sigma}^{**}\) under \((\overline{\mathcal{D}},\pi')\) whose history completely reveals the signals received by each agent. Fix \(k_{1}\in \text{supp}(\pi').\) Note that it is always an equilibrium under \((\overline{\mathcal{D}}, \pi)\) for all agents to choose \((a_{0}, k_{1})\). Then, it follows that \(V_{i}^{\overline{\mathcal{D}}}(\pi,\boldsymbol{\sigma}^{*})=0\) for all \(i\). Since \(\pi'\) is not no information, for any \(\varepsilon>0\), \(V_{i}^{\overline{\mathcal{D}}}(\pi',\boldsymbol{\sigma}^{**})>\mu_{0}(1-r)-\varepsilon\) for sufficiently large \(i\). Thus, \(\lim_{i\to \infty}V_{i}^{\overline{\mathcal{D}}}(\pi',\boldsymbol{\sigma}^{**})>\lim_{i\to \infty}V_{i}^{\overline{\mathcal{D}}}(\pi,\boldsymbol{\sigma}^{*})\), and hence \(\pi \succsim_{LES} \pi'\) does not hold.

Conversely, suppose that \(\pi\) induces unbounded beliefs. Note that \(V_{i}^{\mathcal{D}}(\pi',\boldsymbol{\sigma}^{**})\leq \mu_{0} U_{1}+(1-\mu_{0})U_{0}\), where \(U_{0}=\max_{a\in A} u(a,L)\) and \(U_{1}=\max_{a\in A} u(a,H)\), because the right-hand side is an expected utility under full information. Hence, it is enough to show that, for any \(\mathcal{D}\), \(\lim_{i\to \infty}V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=\mu_{0} U_{1}+(1-\mu_{0})U_{0}\) for any equilibrium \(\boldsymbol{\sigma}^{*}\) under \((\mathcal{D},\pi)\). Take any decision problem \(\mathcal{D}=(A,u)\). Let \(q_{L}=\frac{\min\{\pi(\mu=1|H),\pi(\mu=0|L)\}}{\pi(\mu=0|L)}\) and \(q_{H}=\frac{\min\{\pi(\mu=1|H),\pi(\mu=0|L)\}}{\pi(\mu=1|H)}\). Note that \(q_{L},q_{H}>0\) holds since \(\pi\) induces unbounded beliefs. Define \(\boldsymbol{\sigma}^{**}\) in the same way as in the proof of Lemma 8. Then, by the same argument as the proof of Lemma 8, we have \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})\geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{**})\) for any equilibrium \(\boldsymbol{\sigma}^{*}\) under \((\mathcal{D},\pi)\). Since \[V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{**}) =\mu_{0}[(1-p^{i})U_{1}]+(1-\mu_{0})[(1-p^{i})U_{0}]+p^{i}U_{\mu_{0}},\] where \(p=1-\min\{\pi(\mu=1|H),\pi(\mu=0|L)\}\) and \(U_{\mu_{0}}=\mathop{\rm arg~max}\limits_{a\in A} [\mu_{0}u(a,H)+(1-\mu_{0})u(a,L)]\), it follows that \[\begin{align} \mu_{0} U_{1}+(1-\mu_{0})U_{0} & \geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})\\ & \geq \mu_{0}[(1-p^{i})U_{1}]+(1-\mu_{0})[(1-p^{i})U_{0}]+p^{i}U_{\mu_{0}}. \end{align}\] Since \(p<1\), we have \(\mu_{0}[(1-p^{i})U_{1}]+(1-\mu_{0})[(1-p^{i})U_{0}]+p^{i}U_{\mu_{0}}\to \mu_{0} U_{1}+(1-\mu_{0})U_{0}\) as \(i\to \infty\). Thus, by the squeeze theorem, it follows that \(\lim_{i\to \infty}V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=\mu_{0} U_{1}+(1-\mu_{0})U_{0}\) for any \(\mathcal{D}\) and any equilibrium \(\boldsymbol{\sigma}^{*}\) under \((\mathcal{D},\pi)\). Thus, \(\pi \succsim_{LES} \pi'\) for any \(\pi'\). ◻

Proof of Proposition 6. Suppose \(\text{supp}(\mu)\subseteq\{0,\mu_{0},1\}\). Then, we have \[\begin{align} 1-\sum_{s\in\text{supp}(\pi')}\min\{\pi'(s|L),\pi'(s|H)\} & = 1-\pi(\mu=\mu_{0}|L) \\ & = \pi(\mu=0|L)=\min \{\pi(\mu=0|L),\pi(\mu=1|H)\}. \end{align}\] Therefore, by Theorem 2 and Proposition 2, we have \(\pi\succsim_{S} \pi\).

Now, we show that \(\pi \succsim_{S} \pi\) does not hold if \(\text{supp}(\mu)\nsubseteq\{0,\mu_{0},1\}\). It is sufficient to show for the case where there exists some \(x\in(\mu_{0},1)\) such that \(x\in \text{supp}(\mu )\). Take \(r\in [0,1]\) that satisfies \(x<r<\frac{x^2}{x^2+\frac{\mu_{0}}{1-\mu_{0}}(1-x)^2}\). Consider the decision problem \(\mathcal{D}=(A,u)\): \(A=\{a_{0},a_{1}\}\) and the payoff function is defined as \(u(a_{0},H)=u(a_{0},L)=0\), \(u(a_{1},H)=1-r\), and \(u(a_{1},L)=-r\). Take any equilibrium \(\boldsymbol{\sigma}^{*}=(\sigma^{*}_{i})_{i\in\mathbb{N}}\) and \(s_{1},s_{2}\in S(x)\). Then, it follows that \(\sigma^{*}_1(a_{0}|s_{1})=1\) and \(\sigma^{*}_2(a_{0}|a_{0},s_{2})=1\). Thus, \(V_{2}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*}|(s_{1},s_{2}))=0\). Additionally, we have \(\overline{V}_2^{\mathcal{D}}(\pi|(s_{1},s_{2}))>0\) since \(r<\frac{x^2}{x^2+\frac{\mu_{0}}{1-\mu_{0}}(1-x)^2}\). Note that for all \(s_{1}',s_{2}'\in S\), \(\overline{V}_2^{\mathcal{D}}(\pi|(s_{1}',s_{2}'))\geq V_{2}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*}|(s_{1}',s_{2}'))\).³⁵ Therefore, \(\overline{V}_2^{\mathcal{D}}(\pi)>V_{2}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})\). By Theorem 1, it follows that \(\pi\succsim_{S}\pi\) does not hold. ◻

Proof of Proposition 7. Without loss of generality, assume that \(x>\mu_{0}\), \(\text{supp}(\pi)=\{s_{0},s_{1},s_{2}\}\) and \(\pi(s_{0}|H)=0\), \(\pi(s_{1}|H)=1-\varepsilon\), \(\pi(s_{2}|H)=\varepsilon\), \(\pi(s_{0}|L)=1-\delta\), \(\pi(s_{1}|L)=0\), and \(\pi(s_{2}|L)=\delta\), where \(\varepsilon\) and \(\delta\) satisfy the condition that \(x=\frac{\mu_{0}\varepsilon}{\mu_{0}\varepsilon+(1-\mu_{0})\delta}\). We divide decision problem \(\mathcal{D}\) into three cases and construct the following equilibrium \(\boldsymbol{\sigma}^{*}\) under \((\mathcal{D},\pi)\).

Case (i): \(B^{-1}(0)\cap B^{-1}(1)\neq \emptyset\). Fix \(a^{*}\in B^{-1}(0)\cap B^{-1}(1)\). In this case, all agents choose \(a^{*}\) regardless of private signal and action histories.

Case (ii): \(B^{-1}(1)\cap B^{-1}(x)= \emptyset\), \(B^{-1}(0)= B^{-1}(x)=\{a_{0}\}\) for some \(a_{0}\in A\). Fix any \(a_{1}\in B^{-1}(1)\). Agent 1 chooses \(a_{0}\) if he receives \(s_{0}\) or \(s_{2}\) and chooses \(a_{1}\) otherwise. For \(i\geq 2\), agent \(i\) chooses \(a_{0}\) if she receives \(s_{0}\), or receives \(s_{2}\) and all previous agents take \(a_{0}\). Otherwise, \(i\) chooses \(a_{1}\).

Case (iii): Otherwise. First, fix \(a_{0}\in B^{-1}(0)\) such that for all \(z\in [x,1]\), \(B^{-1}(z)\neq \{a_{0}\}\). (Such \(a_{0}\) must exist by Lemma 2.) In this case, agent \(1\) chooses action \(a_{0}\) if he receives \(s_{0}\), chooses action from \(B^{-1}(1)\) if he receives \(s_{1}\), and chooses action from \(B^{-1}(x)\) if he receives \(s_{2}\). For \(i\geq 2,\) agent \(i\) chooses action \(a_{0}\) if she receives \(s_{0}\) or at least one agent before \(i\) has taken \(a_{0}\), chooses action from \(B^{-1}(1)\) if she receives \(s_{1}\), and chooses action from \(B^{-1}(\frac{x^{i}}{x^{i}+(\frac{\mu_{0}}{1-\mu_{0}})^{i-1}(1-x)^{i}})\setminus\{a_{0}\}\) if she receives \(s_{2}\) and no one before \(i\) has taken action \(a_{0}\) or action from \(B^{-1}(1)\). Otherwise, she chooses the same action as agent \(i-1\).

In Case (i), it is always optimal to take \(a^{*}\) regardless of the posterior belief. Hence, this strategy \(\boldsymbol{\sigma}^{*}\) is an equilibrium and we have \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=\overline{V}_{i}^{\mathcal{D}}(\pi)\). In Case (iii), action \(a_{0}\) is taken if someone has received the signal \(s_{0}\) in the past, an action from \(B^{-1}(1)\) is taken if someone has received the signal \(s_{1}\) in the past, and an action from \(B^{-1}(\frac{x^{i}}{x^{i}+(\frac{\mu_{0}}{1-\mu_{0}})^{i-1}(1-x)^{i}})\) or an action yielding the same expected payoff is taken when all past agents have received \(s_{2}\). Therefore, we have \(V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})=\overline{V}_{i}^{\mathcal{D}}(\pi)\). Hence, \(\boldsymbol{\sigma}^{*}\) is an equilibrium. Then, in Case (i) and Case (iii), by the same argument as Lemma 7, \[\begin{align} V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*})= \overline{V}_{i}^{\mathcal{D}}(\pi)\geq \overline{V}_{i}^{\mathcal{D}}(\pi')\geq V_{i}^{\mathcal{D}}(\pi',\boldsymbol{\sigma}^{**}), \end{align}\] for any equilibrium \(\boldsymbol{\sigma}^{**}\) under \((\mathcal{D},\pi')\).

The only case left is Case (ii). In Case (ii), from Lemma 3, \[\begin{align} V_{i}^{\mathcal{D}}(\pi',\boldsymbol{\sigma}^{**})=& \mu_{0}[(1-(1-\pi'(s_{1}|H))^{i})u(a_{1},H)+\\&\quad\quad(1-\pi'(s_{1}|H))^{i}u(a_{0},H)]+(1-\mu_{0})u(a_{0},L), \end{align}\] for any equilibrium \(\boldsymbol{\sigma}^{**}\) under \((\mathcal{D},\pi')\). Since \(\pi'(s_{1}|H)\leq 1-\varepsilon\) (by \(\pi \succsim_{B} \pi'\)) and \(u(a_{1},H)>u(a_{0},H)\), it follows that \[\begin{align} &\mu_{0}[(1-(1-\pi'(s_{1}|H))^{i})u(a_{1},H)+(1-\pi'(s_{1}|H))^{i}u(a_{0},H)]+(1-\mu_{0})u(a_{0},L)\\ &=\mu_{0}u(a_{1},H)-\mu_{0}(1-\pi'(s_{1}|H))^{i}[u(a_{1},H)-u(a_{0},H)]+(1-\mu_{0})u(a_{0},L)\\ & \leq \mu_{0}u(a_{1},H)-\mu_{0}\varepsilon^{i}[u(a_{1},H)-u(a_{0},H)]+(1-\mu_{0})u(a_{0},L)\\ & =V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma}^{*}), \end{align}\] where \(\boldsymbol{\sigma}^{*}\) is an equilibrium described above. Therefore, \(\pi \succsim_{W} \pi'\). ◻

Proof of Proposition 11. Since \(\boldsymbol{\sigma}^{*}\) is a pure strategy equilibrium and \(\pi(s_{h}|H)\geq \pi(s_{h}|L)\), for every agent \(i\) and every history of observed actions \(\boldsymbol{a}\in A^{i-1}\), the equilibrium action rule of agent \(i\) must be one of the following: (i) choose \(a_1\) after observing \(s_{h}\) and \(a_0\) after observing \(s_{l}\), (ii) choose \(a_0\) regardless of the realized signal, or (iii) choose \(a_1\) regardless of the realized signal.

In cases (ii) and (iii), agent \(i\)’s action is independent of her private signal and thus provides no information to subsequent agents. In case (i), agent \(i\)’s action perfectly reveals her private signal, so every subsequent agent can infer that signal from the observed action. Hence, for every agent \(i\), her posterior belief coincides with the posterior that would arise from observing \(k\) private signals for some \(k\leq i\). Consequently, the posterior must be an element of the set above. ◻

Proof of Lemma 9. Since the equilibrium is essentially unique under both \(\pi\) and \(\pi'\) in \(\mathcal{D}\), and since each agent’s optimal action as a function of her posterior belief is identical in \(\mathcal{D}\) and \(\mathcal{D}_r\), the equilibrium strategy profile under \(\pi\) in \(\mathcal{D}\) must coincide with the equilibrium strategy profile under \(\pi\) in \(\mathcal{D}_r\). Likewise, the equilibrium strategy profile under \(\pi'\) in \(\mathcal{D}\) must coincide with the equilibrium strategy profile under \(\pi'\) in \(\mathcal{D}_r\). Therefore, for every agent \(i\), her equilibrium strategy is the same in \(\mathcal{D}\) and \(\mathcal{D}_r\) under both information structures.

Fix an arbitrary agent \(i\). Let \(p\) and \(q\) be the equilibrium probabilities that agent \(i\) chooses action \(a_1\) under \(\pi\) when the true state is \(H\) and \(L\), respectively. Likewise, let \(p'\) and \(q'\) be the corresponding equilibrium probabilities under \(\pi'\). Then, \(V_i^{\mathcal{D}_{r}}(\pi)=\mu_0p(1-r)-(1-\mu_0)qr\), \(V_i^{\mathcal{D}_{r}}(\pi')=\mu_0p'(1-r)-(1-\mu_0)q'r\), \(V_{i}^{\mathcal{D}}(\pi)=\mu_0[pu(a_1,H)+(1-p)u(a_0,H)]+(1-\mu_0)[qu(a_1,L)+(1-q)u(a_0,L)]\), and \(V_{i}^{\mathcal{D}}(\pi')=\mu_0[p'u(a_1,H)+(1-p')u(a_0,H)]+(1-\mu_0)[q'u(a_1,L)+(1-q')u(a_0,L)]\). Since \(V_{i}^{\mathcal{D}_r}(\pi)\geq V_{i}^{\mathcal{D}_r}(\pi')\), we have \[\begin{align} V_i^{\mathcal{D}_{r}}(\pi)-V_i^{\mathcal{D}_{r}}(\pi') = \mu_0(1-r)(p-p') - (1-\mu_0)r(q-q')\geq 0. \end{align}\] By the definition of \(r\), we have \[\begin{align} (1-r)\bigl[u(a_0,L)-u(a_1,L)\bigr]=r\bigl[u(a_1,H)-u(a_0,H)\bigr]. \end{align}\] Hence, \[\begin{align} V_i^{\mathcal{D}}(\pi)-V_i^{\mathcal{D}}(\pi')&=\mu_0(p-p')\bigl[u(a_1,H)-u(a_0,H)\bigr] \\ &\quad\quad+(1-\mu_0)(q-q')\bigl[u(a_1,L)-u(a_0,L)\bigr]\\ &= \bigl[u(a_0,L)-u(a_1,L)\bigr] \left( \mu_0\frac{1-r}{r}(p-p') -(1-\mu_0)(q-q') \right) \\ &\geq 0 \end{align}\] ◻

Proof of Lemma 10. Without loss of generality, we assume that each agent takes \(a_0\) whenever a tie occurs. We prove this claim by mathematical induction. First, consider \(i=2\). Note that agent 2 can perfectly infer agent 1’s private signal from agent 1’s action. Indeed, since \(y<K_r<x\), agent 1 strictly prefers to follow her private signal: she chooses \(a_1\) upon receiving \(s_h\) and \(a_0\) upon receiving \(s_l\). Then, if \(xy\leq K_r\), agent 2 always chooses \(a_0\) if \(s_1=s_l\), and agent 2 follows her private signal if \(s_1=s_h\) because \(x^2>x>K_r\). If \(xy>K_r\), agent 2 always chooses \(a_1\) if \(s_1=s_h\), and agent 2 follows her private signal if \(s_1=s_l\) because \(y^2<y<K_r\).

Next, suppose that the statement holds at \(i=j\). Consider the case of \(i=j+1\). If agent \(j\) chooses an action independently of her private signal, then her action is uninformative. As a result, agent \(j+1\) faces exactly the same situation as \(j\), and therefore chooses the same action as \(j\) regardless of her own private signal. Hence, whenever agent \(j+1\) follows her private signal, agent \(j\) must also follow hers. In this case, by the induction hypothesis, the signal profile \((s_1,\ldots,s_{j-1})\) is uniquely determined. Thus, when this signal profile is realized, agent \(j\) and all subsequent agents can infer that the signals received by agents \(1,\ldots,j-1\) were \((s_1,\ldots,s_{j-1})\).

Take this signal profile and let \[n_j=\left|\{k\in \{1,\dots,j-1\}\mid s_k=s_h\}\right| \quad \text{ and } \quad m_j=\left|\{k\in \{1,\dots,j-1\}\mid s_k=s_l\}\right|.\] Note that \(n_j+m_j=j-1\). If \(x^{n_j+1}y^{m_j+1}\leq K_r\), agent \(j+1\) always chooses \(a_0\) if \(s_j=s_l\) because \(x^{n_j}y^{m_j+2}<x^{n_j+1}y^{m_j+1}\leq K_r\), and agent \(j+1\) follows his private signal if \(s_j=s_h\) because \(x^{n_j+1}y^{m_j+1}\leq K_r\) and \(x^{n_j+2}y^{m_j}>x^{n_j+1}y^{m_j}> K_r\), where the last inequality holds by the fact that agent \(j\) chooses \(a_1\) if she receives \(s_h\) after \((s_1,\dots,s_{j-1})\). Similarly, if \(x^{n_j+1}y^{m_j+1}> K_r\), agent \(j+1\) always chooses \(a_1\) if \(s_j=s_h\) because \(x^{n_j+2}y^{m_j}>x^{n_j+1}y^{m_j+1}> K_r\), and agent \(j+1\) follows her private signal if \(s_j=s_l\) because \(x^{n_j+1}y^{m_j+1}> K_r\) and \(x^{n_j}y^{m_j+2}<x^{n_j}y^{m_j+1}\leq K_r\), where the last inequality holds by the fact that agent \(j\) chooses \(a_0\) if she receives \(s_l\) after \((s_1,\dots,s_{j-1})\). In either case, there exists a unique signal profile under which agent \(j\) follows her private signal. This completes the induction. ◻

Proof of Lemma 11. Prove is by induction. First, consider the case of \(i=1\). Then, \(b_1=1\), \(c_1=0\), and \(f(p)=p\). Thus, \((1-p)f'(p)=1-p<K_r(1-q)=K_r(1-q)f'(q)\).

Next, fix \(n\in\mathbb{N}\) with \(n\geq 2\) and suppose that the statement holds for all \(i=1,\dots,n-1\). Consider the case \(i=n\). Take an arbitrary sequence \(\boldsymbol{b}\) of length \(n\) such that \(b_j\in\{0,1\}\) for all \(j=1,\ldots,n-1\) and \(b_n=1\). Let \(a\) be the largest integer satisfying \(b_n=b_{n-1}=\cdots=b_{n-a}=1.\) By definition, we have \(0\leq a\leq n-1\). Note that, by the definition of \(c_{j}\), we have \(j-c_j=n-c_n\) for all \(j=n,n-1,\dots,n-a\). Then, it follows that \[\begin{align} f(p)&=\sum_{j=1}^{n-1-a}b_jp^{j-c_j}(1-p)^{c_j}+\sum_{j=n-a}^n b_jp^{j-c_j}(1-p)^{c_j}\\ &=\sum_{j=1}^{n-1-a}b_jp^{j-c_j}(1-p)^{c_j}+\sum_{j=n-a}^n p^{n-c_n}(1-p)^{c_n-n+j}\\ &=\sum_{j=1}^{n-1-a}b_jp^{j-c_j}(1-p)^{c_j}+ \left(p^{n-c_n-1}(1-p)^{c_n-a}[1-(1-p)^{a+1}]\right). \end{align}\] Let \(h(p)=\sum_{j=1}^{n-1-a}b_jp^{j-c_j}(1-p)^{c_j}\).³⁶ Taking the derivative with respect to \(p\) and multiplying by \(1-p\) imply that \((1-p)f'(p)\) can be written as \[\begin{align} & (1-p)h'(p) +p^{n-c_n-2}(1-p)^{c_n-a} \Bigl( \bigl[(n-c_n-1)(1-p)-(c_n-a)p\bigr] \bigl[1-(1-p)^{a+1}\bigr] \\ & +(a+1)p(1-p)^{a+1} \Bigr)\\ &=(1-p)h'(p) \\ &\quad +p^{n-c_n-1}(1-p)^{c_n-a} \Bigl(-(c_n-a)+(n-c_n-1)\sum_{j=1}^{a}(1-p)^j+n(1-p)^{a+1}\Bigr). \end{align}\] Since \(n-1-a\leq n-1\), the induction hypothesis implies that \(h(p)=0\) or \((1-p)h'(p)<K_r(1-q)h'(q)\). We will later show that the following inequalities hold. (i): \(p^{n-c_n-1}(1-p)^{c_n-a}>K_rq^{n-c_n-1}(1-q)^{c_n-a}\) if \(c_n\neq a\) and (ii): \(p^{n-c_n-1}(1-p)^{c_n-a+j}<K_rq^{n-c_n-1}(1-q)^{c_n-a+j}\) for all \(j\geq 1\). Assuming for the moment that these inequalities hold, and noting that \(-(c_n-a)=0\) if \(c_n=a\) and \(-(c_n-a)<0\) if \(c_n\neq a\), \(n-c_n-1\geq 0\), and \(n>0\), we obtain \[(1-p)f'(p)<K_r(1-q)f'(q).\]

Therefore, it remains to verify inequalities (i) and (ii). We begin with (i). Since \(c_n\neq a\), we have \(a<n-1\) and \(c_{n-a-1}=c_{n-a}=c_n-a\neq 0\).³⁷ Hence, there exists at least one entry equal to \(1\) among \(b_{n-a-2}, b_{n-a-3},\dots.\) Let \(b_m\) be the first such entry when moving backward from \(n-a-2\). By assumption, \[p^{m-c_m}(1-p)^{c_m+1} > K_rq^{m-c_m}(1-q)^{c_m+1}.\] Since \(m\) is the last index prior to \(n-a-1\) at which \(b_m=1\), we have \(c_m=c_{n-a-1}-1=c_n-a-1\). Hence, \[p^{m-c_n+a+1}(1-p)^{c_n-a} > K_rq^{m-c_n+a+1}(1-q)^{c_n-a}.\] \(1-p<K_r(1-q)\) and \(p>K_r q\) implies \(p>q\). Since \(n-m-a-2\geq n-(n-a-2)-a-2=0\) and \(p>q\), it follows that \[\begin{align} p^{n-c_n-1}(1-p)^{c_n-a} & = p^{n-m-a-2}p^{m-c_n+a+1}(1-p)^{c_n-a}\\ & \geq q^{n-m-a-2}p^{m-c_n+a+1}(1-p)^{c_n-a}\\ & > K_r q^{n-m-a-2}q^{m-c_n+a+1}(1-q)^{c_n-a} = K_r q^{n-c_n-1}(1-q)^{c_n-a}. \end{align}\] This establishes (i).

Next, show (ii). Take \(j\geq 1\). By assumption, \(b_{n-a-1}=0\) implies³⁸ \[p^{(n-a-1)-c_{n-a-1}}(1-p)^{c_{n-a-1}+1}<K_rq^{(n-a-1)-c_{n-a-1}}(1-q)^{c_{n-a-1}+1}.\] Since \(c_{n-a-1}=c_n-a\), we have \[p^{n-c_n-1}(1-p)^{c_{n}-a+1}<K_rq^{n-c_n-1}(1-q)^{c_{n}-a+1}.\] Note that \(1-p<1-q\), then, it follows that \[\begin{align} p^{n-c_n-1}(1-p)^{c_n-a+j} &= p^{n-c_n-1}(1-p)^{c_n-a+1}(1-p)^{j-1}\\ &\leq p^{n-c_n-1}(1-p)^{c_n-a+1}(1-q)^{j-1}\\ &<K_rq^{n-c_n-1}(1-q)^{c_n-a+1}(1-q)^{j-1} = K_r q^{n-c_n-1}(1-q)^{c_n-a+j}. \end{align}\] Therefore, (ii) holds.

Hence, \((1-p)f'(p)<K_r(1-q)f'(q),\) as desired. ◻

Proof of Lemma 12. Prove is by induction. First, consider the case of \(i=1\). Then, \(b_1=1\), \(c_1=0\), and \(g(p)=1-p\). Thus, \(pg'(p)=-p<-K_rq=K_rqg'(q)\).

Next, fix \(n\in\mathbb{N},n\geq 2\) and suppose that the statement holds for all \(i=1,\dots,n\). Consider the case \(i=n+1\). Take an arbitrary sequence \(\boldsymbol{b}\) of length \(n\) such that \(b_j\in\{0,1\}\) for all \(j=1,\ldots,n-1\) and \(b_n=1\). Let \(a\) be defined as in Lemma 11. Then, it follows that \[\begin{align} g(p)&=\sum_{j=1}^{n-1-a}b_jp^{c_j}(1-p)^{j-c_j}+ p^{c_n-a}(1-p)^{n-c_n-1}[1-p^{a+1}]. \end{align}\] Let \(\overline{h}(p)=\sum_{j=1}^{n-1-a}b_jp^{c_j}(1-p)^{j-c_j}\).³⁹ Taking the derivative with respect to \(p\) and multiplying by \(p\) yields \[pg'(p) = p\overline{h}'(p) + p^{c_n-a}(1-p)^{n-c_n-1} \Bigl( (c_n-a)-(n-c_n-1)\sum_{j=1}^{a}p^j-np^{a+1} \Bigr).\] Since \(n-1-a\leq n-1\), the induction hypothesis implies that \(\overline{h}(p)=0\) or \(p\overline{h}'(p)<K_rq\overline{h}'(q)\). We will later show that the following inequalities hold. (i): \(p^{c_n-a}(1-p)^{n-c_n-1}<K_rq^{c_n-a}(1-q)^{n-c_n-1}\) if \(c_n\neq a\) and (ii): \(p^{c_n-a+j}(1-p)^{n-c_n-1}>K_rq^{c_n-a+j}(1-q)^{n-c_n-1}\) for all \(j\geq 1\). Assuming for the moment that these inequalities hold, and noting that \(c_n-a>0\) if \(c_n\neq a\), \(-(n-c_n-1)\leq 0\), and \(-n<0,\) we obtain \[\begin{align} pg'(p)<K_rqg'(q). \end{align}\]

Therefore, it remains to verify inequalities (i) and (ii). We begin with (i). Since \(c_n\neq a\), we can take \(b_m\) in the same way as Lemma 11. By assumption, \(b_m=1\) implies \[\begin{align} p^{c_m+1}(1-p)^{m-c_m} < K_rq^{c_m+1}(1-q)^{m-c_m}. \end{align}\] Since \(c_m=c_{n-a-1}-1=c_n-a-1\), \[\begin{align} p^{c_n-a}(1-p)^{m-c_n+a+1} < K_rq^{c_n-a}(1-q)^{m-c_n+a+1}. \end{align}\] Since \(n-m-a-2\geq n-(n-a-2)-a-2=0\) and \(0<1-p<1-q\), it follows that \[\begin{align} p^{c_n-a}(1-p)^{n-c_n-1}&=p^{c_n-a}(1-p)^{m-c_n+a+1}(1-p)^{n-m-a-2}\\ &\leq p^{c_n-a}(1-p)^{m-c_n+a+1}(1-q)^{n-m-a-2}\\ &<K_r q^{c_n-a}(1-q)^{m-c_n+a+1}(1-q)^{n-m-a-2} = K_r q^{c_n-a}(1-q)^{n-c_n-1}. \end{align}\] This establishes (i).

Next, show (ii). Take \(j\geq 1\). By assumption, \(b_{n-a-1}=0\) implies⁴⁰ \[\begin{align} p^{c_{n-a-1}+1}(1-p)^{(n-a-1)-c_{n-a-1}}>K_rq^{c_{n-a-1}+1}(1-q)^{(n-a-1)-c_{n-a-1}}. \end{align}\] Since \(c_{n-a-1}=c_n-a\), we have \[\begin{align} p^{c_{n}-a+1}(1-p)^{n-c_n-1}>K_rq^{c_{n}-a+1}(1-q)^{n-c_n-1}. \end{align}\] Note that \(p>q\), then, it follows that \[\begin{align} p^{c_n-a+j}(1-p)^{n-c_n-1}&=p^{j-1}p^{c_n-a+1} (1-p)^{n-c_n-1}\\ &\geq q^{j-1}p^{c_n-a+1} (1-p)^{n-c_n-1}\\ &>K_rq^{j-1}q^{c_n-a+1} (1-q)^{n-c_n-1} = K_r q^{c_n-a+j}(1-q)^{n-c_n-1}. \end{align}\] Therefore, (ii) holds.

Hence, \(pg'(p)<K_rqg'(q),\) as desired. ◻

Proof of Proposition 12. First, show the continuity of \(V_i^{\mathcal{D}_{r}}(\pi)\). Suppose that \(\pi \in \overline{\Pi}^{B}\) and \(\pi' \in \overline{\Pi}^{B}\) belong to the same \(r\)-region. Let \(\boldsymbol{\sigma}^{*}\) be an equilibrium under \((\mathcal{D}_{r},\pi)\). Since equilibrium is essentially unique, every signal profile induces a unique action profile. Let \(\tau^{*}=(\tau_1^{*},\ldots,\tau_N^{*})\) denote the induced mapping from signal histories to actions. In particular, \(\tau_i^{*}:S^i\to A\) gives the action chosen by agent \(i\) as a function of the realized signal history \((s_1,\ldots,s_i)\). Then, we have \[\begin{align} V_i^{\mathcal{D}_{r}}(\pi)=\mu_0(1-r)\sum_{(s_1,\dots,s_i)\in S^i}\prod_{j=1}^i \pi(s_j|H)\chi_{T_i^{*}}-(1-\mu_0)r\sum_{(s_1,\dots,s_i)\in S^i}\prod_{j=1}^i \pi(s_j|L)\chi_{T_i^{*}}, \end{align}\] where \(T_i^{*}=\{(s_1,\dots,s_i)\in S^i\mid \tau^{*}_i(s_1,\dots,s_i)=a_1\}\), and \(\chi\) is an indicator function. As noted above, the equilibrium strategy profile is unchanged on the equilibrium path within a given \(r\)-region. Hence, the induced mapping \(\tau^{*}\) is constant throughout the region. Since \(V_i^{\mathcal{D}_{r}}(\pi)\) is determined by a fixed finite sum whose coefficients depend continuously on \(\pi\), it follows that \(V_i^{\mathcal{D}_{r}}(\pi)\) is continuous on each \(r\)-region. In addition, since the mapping \(\varphi:(\pi(s_h|H),\pi(s_h|L))\mapsto\left(\frac{\pi(s_h|H)}{\pi(s_h|L)}, \frac{\pi(s_l|H)}{\pi(s_l|L)}\right)\) is a diffeomorphism, \(V_i^{\mathcal{D}_r}(\pi)\) can be viewed as a differentiable function of \((x,y)\).

Next, show that \(V_i^{\mathcal{D}_r}(\pi)\geq V_i^{\mathcal{D}_r}(\pi')\) if \(\pi \in \overline{\Pi}^{B}\) and \(\pi' \in \overline{\Pi}^{B}\) belong to the same \(r\)-region, and \(\pi \succsim_B \pi'\). Suppose that \(x<K_r\). In this case, under both \(\pi\) and \(\pi'\), every agent always chooses action \(a_0\). Therefore, the expected payoff is zero under the two information structures, and the desired inequality holds trivially. Similarly, when \(y>K_r\), every agent always chooses action \(a_1\), and hence the expected payoff is the same under both \(\pi\) and \(\pi'\). Therefore, it suffices to consider the case \(y<K_r<x\).

Since \(\pi \succsim_B\pi'\) is equivalent to \(x\geq x'\) and \(y\leq y'\), it is enough to show that \(V_i^{\mathcal{D}_r}(\pi)\) is non-decreasing with respect to \(x\) and non-increasing with respect to \(y\). First, show that \(V_i^{\mathcal{D}_r}(\pi)\) is non-decreasing with respect to \(x\). Note that \(\mathcal{D}_r\) induces no tie-break under \(\pi\). Hence, by Lemma 10, there exists a unique signal profile \((s_1,\ldots,s_{i-1})\) under which agent \(i\) follows her private signal. Fix this signal profile. Define a binary sequence \((b_1,\ldots,b_i)\) by \[\begin{align} b_j= \begin{cases} 0 & \text{ if } s_j=s_h,\\ 1 & \text{ if } s_j=s_l, \end{cases} \qquad j=1,\ldots,i-1, \end{align}\] and let \(b_i=1\). For each \(j\), define \(c_j=\left|\{l\in \mathbb{N}\mid l<j,b_l=1\,\}\right|,\) \(p=\pi(s_h|H)\), and \(q=\pi(s_h|L)\). Note that \(1>p>q>0\). Agent \(i\) chooses \(a_1\) in either of the following cases. (I): the realized signal profile \(\boldsymbol{s}^*\) satisfies \(s_j^*=s_j\) for all \(j=1,\ldots,i-1\), and \(s_i^*=s_h\). The probability of this event is \((1-p)^{c_i}p^{i-c_i}\) at the state \(H\) and \((1-q)^{c_i}q^{i-c_i}\) at \(L\). (II): there exists a smallest index \(j\in\{1,\ldots,i-1\}\) such that \(s_j^*\neq s_j\), and this first deviation is favorable, namely \(s_j^*=s_h\). Equivalently, \(b_j=1\). The probability of this event is \(b_j(1-p)^{c_j}p^{j-c_j}\) at \(H\) and \(b_j(1-q)^{c_j}q^{j-c_j}\) at \(L\). Therefore, \[\mathbb{P}(a_i=a_1\mid H)=\sum_{j=1}^{i} b_j(1-p)^{c_j}p^{j-c_j} \,\text{ and }\, \mathbb{P}(a_i=a_1\mid L)=\sum_{j=1}^{i} b_j(1-q)^{c_j}q^{j-c_j}.\] Let \(f(p)=\sum_{j=1}^{i} b_jp^{j-c_j}(1-p)^{c_j}\). Then, it follows that \[\begin{align} V_i^{\mathcal{D}_r}(\pi)=\mu_0(1-r)f(p)-(1-\mu_0)rf(q). \end{align}\]

Next, we verify that the conditions of Lemma 11 are satisfied. If \(b_1=0\), agent 2 always chooses \(a_0\) if agent 1 receives \(s_l\). Hence, \(xy<K_r\), or \(p(1-p)<K_r q(1-q)\). Since \(c_1=0\), this is equivalent to \(p^{1-c_1}(1-p)^{c_1+1}<K_rq^{1-c_1}(1-q)^{c_1+1}\). If \(b_1=1\), agent 2 always chooses \(a_1\) if agent 1 receives \(s_h\). Hence, \(xy>K_r\), or \(p(1-p)>K_r q(1-q)\). Since \(c_1=0\), this is equivalent to \(p^{1-c_1}(1-p)^{c_1+1}>K_rq^{1-c_1}(1-q)^{c_1+1}\). Take \(j\in\{2,\dots,i-1\}\). Suppose that \(s_n^*=s_n\) for all \(n=1,\dots,j-1\) and \(s_j^*\neq s_j\). In this case, agent \(j\) is the first agent at whom a cascade occurs. If \(s_j^*=s_h\), then agent \(j+1\) chooses \(a_1\) even when her private signal is \(s_l\). Since among agents \(1,\ldots,j-1\), exactly \(c_j\) have received \(s_l\) and the remaining \(j-1-c_j\) have received \(s_h\), the posterior likelihood ratio of agent \(j+1\) after observing \(s_l\) is \(x^{\,j-1-c_j}y^{\,c_j}xy,\) which must exceed \(K_r\). Hence, \(x^{\,j-1-c_j}y^{\,c_j}xy>K_r.\) Equivalently, we have \[\begin{align} p^{j-c_j}(1-p)^{c_j+1}>K_rq^{j-c_j}(1-q)^{c_j+1}. \end{align}\] If \(s_j^*=s_l\), the analogous argument yields \[\begin{align} p^{j-c_j}(1-p)^{c_j+1}<K_rq^{j-c_j}(1-q)^{c_j+1}. \end{align}\] Since \(y<K_r<x\), we also have \(1-p<K_r(1-q)\) and \(p>K_rq\). Hence, all the assumptions of Lemma 11 are satisfied.

Since \(p=\frac{x(1-y)}{x-y}\) and \(q=\frac{1-y}{1-x}\), we have \(\frac{\partial p}{\partial x}=-\frac{y(1-y)}{(x-y)^2}\) and \(\frac{\partial q}{\partial x}=-\frac{1-y}{(x-y)^2}\). Hence, \[\begin{align} \frac{\partial }{\partial x}V_i^{\mathcal{D}_r}&=\mu_0(1-r)f'(p)\frac{\partial p}{\partial x}-(1-\mu_0)rf'(q)\frac{\partial q}{\partial x}\\ &=\frac{1-y}{(x-y)^2}\frac{1}{1-q}\mu_0(1-r)[K_r(1-q)f'(q)-(1-p)f'(p)]\geq 0. \end{align}\] The last inequality comes from Lemma 11.

Next, show that \(V_i^{\mathcal{D}_r}(\pi)\) is non-increasing with respect to \(y\). Take the same signal profile \((s_1,\dots,s_{i-1})\). Define a binary sequence \((b_1,\ldots,b_i)\) by \[\begin{align} b_j= \begin{cases} 0 & \text{ if } s_j=s_l,\\ 1 & \text{ if } s_j=s_h, \end{cases} \qquad j=1,\ldots,i-1, \end{align}\] and let \(b_i=1\). For each \(j\), define \(c_j=\left|\{l\in \mathbb{N}\mid l<j,b_l=1\,\}\right|.\) Agent \(i\) chooses \(a_0\) in either of the following cases. (I): the realized signal profile \(\boldsymbol{s}^*\) satisfies \(s_j^*=s_j\) for all \(j=1,\ldots,i-1\), and \(s_i^*=s_l\). The probability of this event is \((1-p)^{i-c_i}p^{c_i}\) at the state \(H\) and \((1-q)^{i-c_i}q^{c_i}\) at \(L\). (II): there exists a smallest index \(j\in\{1,\ldots,i-1\}\) such that \(s_j^*\neq s_j\), and \(s_j^*=s_l\). Equivalently, \(b_j=1\). The probability of this event is \(b_j(1-p)^{j-c_j}p^{c_j}\) at \(H\) and \(b_j(1-q)^{j-c_j}q^{c_j}\) at \(L\). Therefore, \[\mathbb{P}(a_i=a_0\mid H)=\sum_{j=1}^{i} b_j(1-p)^{j-c_j}p^{c_j} \;\text{ and }\; \mathbb{P}(a_i=a_0\mid L)=\sum_{j=1}^{i} b_j(1-q)^{j-c_j}q^{c_j}.\] Let \(g(p)=\sum_{j=1}^{i} b_jp^{c_j}(1-p)^{j-c_j}\). Then, it follows that \[\begin{align} V_i^{\mathcal{D}_r}(\pi)=\mu_0(1-r)[1-g(p)]-(1-\mu_0)r[1-g(q)]. \end{align}\]

A symmetric argument to that in Lemma 11 shows that all the assumptions of Lemma 12 are satisfied. Since \(\frac{\partial p}{\partial y}=-\frac{x(x-1)}{(x-y)^2}\) and \(\frac{\partial q}{\partial y}=-\frac{x-1}{(x-y)^2}\), we have \[\begin{align} \frac{\partial }{\partial y}V_i^{\mathcal{D}_r}&=-\mu_0(1-r)g'(p)\frac{\partial p}{\partial y}+(1-\mu_0)rg'(q)\frac{\partial q}{\partial y}\\ &=\frac{x-1}{(x-y)^2}\frac{1}{q}\mu_0(1-r)[pg'(p)-K_rq g'(q)]\leq 0. \end{align}\] The last inequality comes from Lemma 12.

Therefore, the payoff is weakly increasing in \(x\) and weakly decreasing in \(y\). Hence, the Blackwell order is respected within each \(r\)-region. ◻

Proof of Proposition 13. Without loss of generality, we assume that all agents break ties in favor of \(a_0\) under \(\boldsymbol{\sigma}^0\) and in favor of \(a_1\) under \(\boldsymbol{\sigma}^1\). If \(\mathcal{D}_r\) induces no tie-break, the statement holds since there is an essentially unique equilibrium. Therefore, we consider the case \(\{(n,m)\in \mathbb{Z}_+^2 \mid x^n y^m=K_r, 1\leq n+m\leq i\}\neq \emptyset\). Among the elements of this set, let \((n_1,m_1)\) be one that minimizes \(n+m\). Then, the strategies of agents \(1,\ldots,n+m-1\) coincide on the equilibrium path under both \(\boldsymbol{\sigma}^0\) and \(\boldsymbol{\sigma}^1\). If \(n_1+m_1=i\), then agent \(i\) is indifferent between \(a_0\) and \(a_1\) at the tie, so her expected payoff is independent of the tie-breaking rule. Therefore, \(V_i^{\mathcal{D}_r}(\pi,\boldsymbol{\sigma}^0)=V_i^{\mathcal{D}_r}(\pi,\boldsymbol{\sigma}^{1})\). Thus, in what follows, we restrict attention to the case \(n_1+m_1\leq i-1\). Note that \(n_1+m_1\geq 2\) since \(y<K_r<x\).

By Lemma 10, there exists a unique signal profile \((s_1,\ldots,s_{n_1+m_1-2})\) under which agent \(n_1+m_1-1\) follows her private signal. We first show that among these \(n_1+m_1-2\) signals, exactly \(n_1-1\) are equal to \(s_h\) and exactly \(m_1-1\) are equal to \(s_l\). That is, \(\left|\{\,j<n_1+m_1-1 \mid s_j=s_h\,\}\right|=n_1-1\) and \(\left|\{\,j<n_1+m_1-1 \mid s_j=s_l\,\}\right|=m_1-1.\) Suppose that \(s_h\) appears at least \(n_1\) times. Then, when agent \(n_1+m_1-1\) receives \(s_l\), her likelihood ratio is at least \(x^{n_1}y^{m_1-1}\). Since \(x^{n_1}y^{m_1}=K_r\) and \(y<1\), we have \(x^{n_1}y^{m_1-1}=\frac{K_r}{y}>K_r.\) Hence, agent \(n_1+m_1-1\) would choose \(a_1\) regardless of her private signal, contradicting the fact that she follows her private signal. Therefore, \(s_h\) appears fewer than \(n_1\) times. By a symmetric argument, it is also impossible for \(s_l\) to appear at least \(m_1\) times.

Under \(\boldsymbol{\sigma}^0\), if the signal profile \((s_1,\ldots,s_{n_1+m_1-2})\) is realized and agent \(n_1+m_1-1\) receives \(s_l\), then even if agent \(n_1+m_1\) receives \(s_h\), the likelihood ratio is \(x^{n_1-1}y^{m_1-1}yx=x^{n_1}y^{m_1}=K_r.\) Since ties are broken in favor of \(a_0\) under \(\boldsymbol{\sigma}^0\), all agents from agent \(n_1+m_1\) onward choose \(a_0\). On the other hand, if the signal profile \((s_1,\ldots,s_{n_1+m_1-2})\) is realized and both agents \(n_1+m_1-1\) and \(n_1+m_1\) receive \(s_h\), then even if agent \(n_1+m_1+1\) receives \(s_l\), the likelihood ratio is \(x^{n_1+1}y^{m_1}>K_r.\) Hence, all agents from agent \(n_1+m_1+1\) onward choose \(a_1\). Therefore, under \(\boldsymbol{\sigma}^0\), agent \(i\) chooses \(a_1\) if and only if one of the following events occurs.

1. A cascade toward \(a_1\) occurs before agent \(n_1+m_1-1\).

2. The signal profile \((s_1,\ldots,s_{n_1+m_1-2})\) is realized, and both agents \(n_1+m_1-1\) and \(n_1+m_1\) receive \(s_h\).

3. The signal profile \((s_1,\ldots,s_{n_1+m_1-2})\) is realized, agent \(n_1+m_1-1\) receives \(s_h\), agent \(n_1+m_1\) receives \(s_l\), and thereafter a sequence of signals is realized under which agent \(i\) chooses \(a_1\).

Under \(\boldsymbol{\sigma}^1\), if the signal profile \((s_1,\ldots,s_{n_1+m_1-2})\) is realized and agent \(n_1+m_1-1\) receives \(s_h\), then even if agent \(n_1+m_1\) receives \(s_l\), the likelihood ratio is \(x^{n_1-1}y^{m_1-1}xy=x^{n_1}y^{m_1}=K_r.\) Since ties are broken in favor of \(a_1\) under \(\boldsymbol{\sigma}^1\), all agents from agent \(n_1+m_1\) onward choose \(a_1\). On the other hand, if the signal profile \((s_1,\ldots,s_{n_1+m_1-2})\) is realized and both agents \(n_1+m_1-1\) and \(n_1+m_1\) receive \(s_l\), then even if agent \(n_1+m_1+1\) receives \(s_h\), the likelihood ratio is \(x^{n_1}y^{m_1+1}<K_r.\) Hence, all agents from agent \(n_1+m_1+1\) onward choose \(a_0\). Therefore, under \(\boldsymbol{\sigma}^1\), agent \(i\) chooses \(a_1\) if and only if one of the following events occurs.

1. A cascade toward \(a_1\) occurs before agent \(n_1+m_1-1\).

2. The signal profile \((s_1,\ldots,s_{n_1+m_1-2})\) is realized, and agent \(n_1+m_1-1\) receives \(s_h\).

3. The signal profile \((s_1,\ldots,s_{n_1+m_1-2})\) is realized, agent \(n_1+m_1-1\) receives \(s_l\), agent \(n_1+m_1\) receives \(s_h\), and thereafter a sequence of signals is realized under which agent \(i\) chooses \(a_1\).

Since no tie occurs before agent \(n_1+m_1\), the probabilities of event 0-(i) and 1-(i) are the same.

Moreover, note that the probability that \((s_1,\ldots,s_{n_1+m_1-2})\) is realized is \(p^{n_1-1}(1-p)^{m_1-1}\) under state \(H\), and \(q^{n_1-1}(1-q)^{m_1-1}\) under state \(L\), where \(p=\pi(s_H|H)\) and \(q=\pi(s_l|L)\). Therefore, under \(\boldsymbol{\sigma}^0\), the probability of event 0-(ii) is \(p^{n_1+1}(1-p)^{m_1-1}\) conditional on state \(H\), and \(q^{n_1+1}(1-q)^{m_1-1}\) conditional on state \(L\). Similarly, under \(\boldsymbol{\sigma}^1\), the probability of event 1-(ii) is \(p^{n_1}(1-p)^{m_1-1}\) conditional on state \(H\), and \(q^{n_1}(1-q)^{m_1-1}\) conditional on state \(L\).

Next, consider the probabilities of events 0-(iii) and 1-(iii). In either case, after the realization of \((s_1,\ldots,s_{n_1+m_1-2})\), the next two agents receive \(s_h\) once and \(s_l\) once. Consequently, by agent \(n_1+m_1\), the signal \(s_h\) has occurred exactly \(n_1\) times, and the signal \(s_l\) has occurred exactly \(m_1\) times. Furthermore, all agents from agent \(n_1+m_1+1\) onward can infer this signal history from the observed action history. Indeed, under both 0-(iii) and 1-(iii), no cascade occurs before agent \(n_1+m_1\). Since \(x^{n_1}y^{m_1}=K_r,\) the posterior likelihood ratio after these observations is exactly equal to the cutoff \(K_r\). Therefore, after conditioning on the history up to agent \(n_1+m_1\) described above, the continuation game is equivalent to a social learning process with cutoff \(K_r=1\). Hence, the conditional probability that agent \(i\) chooses \(a_1\) is equal to the probability that agent \(i-(n_1+m_1)\) chooses \(a_1\) in that process. Let \(\alpha^0\) and \(\beta^0\) denote the probabilities that agent \(i-(n_1+m_1)\) chooses action \(a_1\) under the equilibrium that is consistent with \(\boldsymbol{\sigma}^0\), that is, breaking ties in favor of \(a_0\), conditional on the state being \(H\) and \(L\), respectively, in the social learning process with cutoff \(K_r=1\). Similarly, let \(\alpha^1\) and \(\beta^1\) denote the corresponding probabilities under the equilibrium that is consistent with \(\boldsymbol{\sigma}^1\).

Then, it follows that, conditional on state \(H\), the probability that agent \(i\) chooses \(a_1\) under \(\boldsymbol{\sigma}^1\) exceeds the corresponding probability under \(\boldsymbol{\sigma}^0\) by \[\begin{align} & p^{n_1}(1-p)^{m_1-1}-p^{n_1+1}(1-p)^{m_1-1}+p^{n_1}(1-p)^{m_1}\alpha^1-p^{n_1}(1-p)^{m_1}\alpha^0\\ & = p^{n_1}(1-p)^{m_1}(1+\alpha^1-\alpha^0). \end{align}\] Similarly, conditional on state \(L\), the probability that agent \(i\) chooses \(a_1\) under \(\boldsymbol{\sigma}^1\) exceeds the corresponding probability under \(\boldsymbol{\sigma}^0\) by \[\begin{align} q^{n_1}(1-q)^{m_1}(1+\beta^1-\beta^0). \end{align}\] Hence, we have \[\begin{align} &V_i^{\mathcal{D}_r}(\pi,\boldsymbol{\sigma}^1)-V_i^{\mathcal{D}_r}(\pi,\boldsymbol{\sigma}^0)\\ &=\mu_0(1-r)p^{n_1}(1-p)^{m_1}[1+\alpha^1-\alpha^0]-(1-\mu_0)rq^{n_1}(1-q)^{m_1}[1+\beta^1-\beta^0]\\ &=\mu_0(1-r)q^{n_1}(1-q)^{m_1}[x^{n_1}y^{m_1}(1+\alpha^1-\alpha^0)-K_r(1+\beta^1-\beta^0)]\\ &=\mu_0(1-r)q^{n_1}(1-q)^{m_1}K_r(\alpha^1-\alpha^0-\beta^1+\beta^0). \end{align}\] The last equality holds by \(x^{n_1}y^{m_1}=K_r\). Hence, what we need to show is \(\alpha^1-\alpha^0=\beta^1-\beta^0\).

Now suppose that \(\{(n,m)\in \mathbb{Z}_+^2 : x^n y^m=K_r,\; n_1+m_1<n+m\le i\}=\emptyset.\) Then no tie occurs in the new social learning process starting from agent \(n_1+m_1+1\). Hence, \(\alpha^0=\alpha^1\) and \(\beta^0=\beta^1\), and the desired claim follows immediately. We therefore consider the case in which \(\{(n,m)\in \mathbb{Z}_+^2 : x^n y^m=K_r,\; n_1+m_1<n+m\leq i\}\neq\emptyset.\) Let \((n_2,m_2)\) be an element of this set that minimizes \(n+m\). If \(n_2+m_2=i\), then only agent \(i-(n_1+m_1)\) in the new social learning problem may face a tie. Since the tie-breaking rule does not affect the payoff of an agent who is indifferent, we have \[\begin{align} \mu_0(1-r')\alpha^1-(1-\mu_0)r'\beta^1=\mu_0(1-r')\alpha^0-(1-\mu_0)r'\beta^0, \end{align}\] where \(r'\in \mathbb{R}\) is chosen so that \(\frac{(1-\mu_0)r'}{\mu_0(1-r')}=1.\) Then, it follows that \(\alpha^1-\alpha^0=\beta^1-\beta^0.\)

If \(n_2+m_2<i\), applying the same argument as above to the new social learning process, we obtain \[\begin{align} & \alpha^1-\alpha^0=p^{\,n_2-n_1}(1-p)^{m_2-m_1} \bigl(1+\alpha^{\prime 1}-\alpha^{\prime 0}\bigr)\\ &\beta^1-\beta^0=q^{\,n_2-n_1}(1-q)^{m_2-m_1} \bigl(1+\beta^{\prime 1}-\beta^{\prime 0}\bigr), \end{align}\] where \(\alpha^{\prime 0}\) and \(\beta^{\prime 0}\) denote the probabilities that agent \(i-(n_2+m_2)\) chooses action \(a_1\) under the equilibrium that is consistent with \(\boldsymbol{\sigma}^0\) conditional on the state is \(H\) and \(L\), respectively, in the social learning process with cutoff \(K_r=1\), and \(\alpha^{\prime 1}\) and \(\beta^{\prime 1}\) are the corresponding probabilities under the equilibrium that is consistent with \(\boldsymbol{\sigma}^1\). Since \(x^{n_2-n_1}y^{m_2-m_1}=x^{n_2}y^{m_2}(x^{n_1}y^{m_1})^{-1}=1\), we have \[\begin{align} & (\alpha^1-\alpha^0)-(\beta^1-\beta^0)\\ &=q^{n_2-n_1}(1-q)^{m_2-m_1}[x^{n_2-n_1}y^{m_2-m_1}\bigl(1+\alpha^{\prime 1}-\alpha^{\prime 0}\bigr)-\bigl(1+\beta^{\prime 1}-\beta^{\prime 0}\bigr)]\\ &= q^{n_2-n_1}(1-q)^{m_2-m_1}(\alpha^{\prime 1}-\alpha^{\prime 0}-\beta^{\prime 1}+\beta^{\prime 0}). \end{align}\] Hence, what we need to show is \(\alpha^{\prime 1}-\alpha^{\prime 0}=\beta^{\prime 1}-\beta^{\prime 0}\).

By repeatedly applying the same argument, we obtain analogous expressions at each subsequent tie. Since only finitely many agents can face a tie, this procedure must terminate after finitely many steps. Consequently, we have \[\begin{align} V_i^{\mathcal{D}_r}(\pi,\boldsymbol{\sigma}^1)-V_i^{\mathcal{D}_r}(\pi,\boldsymbol{\sigma}^0)=0. \end{align}\] ◻

Proof of Theorem 4. Take arbitrary \(\pi,\pi' \in \Pi^B\) and \(\mathcal{D} \in \mathscr D^B\) satisfying the conditions of Theorem 4. Let \(r=\frac{u(a_0,L)-u(a_1,L)}{u(a_0,L)-u(a_1,L)+u(a_1,H)-u(a_0,H)}\). Then, by Lemma 9, it suffices to show that \(V_i^{\mathcal{D}_r}(\pi)\geq V_i^{\mathcal{D}_r}(\pi')\).

We first consider the case in which \(\pi\) contains a conclusive signal. If \(\pi\) is fully informative, that is, \(\pi(s_h|H)=1\) and \(\pi(s_h|L)=0\), then the claim is immediate. Next, suppose that \(\pi(s_h|H)=1\) and \(\pi(s_h|L)\in (0,1).\) In this case, at the equilibrium under \(\pi\), each agent’s behavior must be one of the following: (i) The agent always chooses \(a_0\), regardless of both the history and her private signal. (ii) The agent chooses \(a_1\) if and only if all preceding agents and the agent herself have received \(s_h\). If case (i) occurs, then every agent also always chooses \(a_0\) under \(\pi'\) by \(\pi\succsim_B \pi'\). Hence, the expected payoff is the same under \(\pi\) and \(\pi'\). If case (ii) occurs, then agent \(i\) chooses \(a_1\) if and only if the first \(i\) agents all receive \(s_h\). Therefore, agent \(i\)’s expected payoff under \(\pi\) is equal to \(\overline{V}_i^{D_r}(\pi).\) Thus, \(V_i^{\mathcal{D}_r}(\pi)=\overline{V}_i^{D_r}(\pi)\geq \overline{V}_i^{D_r}(\pi')\geq V_i^{\mathcal{D}_r}(\pi').\) A symmetric argument establishes the case in which \(\pi(s_h|H) \in (0,1)\) and \(\pi(s_h|L)=0\).

Moreover, if agent \(1\) chooses an action independently of her private signal under \(\pi'\), then every subsequent agent, including agent \(i\), also chooses an action independently of her private signal and histories. Hence, \(V_i^{\mathcal{D}_r}(\pi')\) is equal to the payoff obtained under no information. It therefore follows immediately that \(V_i^{\mathcal{D}_r}(\pi)\geq V_i^{\mathcal{D}_r}(\pi')\).

Therefore, it remains to consider the case in which \(\pi,\pi' \in \overline{\Pi}^{B}\) and \((x,y)=\varphi(\pi)\) and \((x',y')=\varphi(\pi')\) satisfy \(y\leq y' < K_r < x'\leq x\). We henceforth focus on this case.

Define a continuous function \(c:[0,2]\to \{(a,b)\in \mathbb{R}^2 \mid 1<a,0<b<1\}\) by \(c(t)=(x'+t(x-x'),y')\) and \(c(t+1)=(x,y'+t(y-y'))\) for \(t\in [0,1]\). Let \(\pi^t=\varphi^{-1}(c(t))\) for \(t\in [0,2]\). By construction, we have \(\pi^s\succsim_B \pi^u\) if \(s\geq u\). For each \((n,m)\in\mathbb{N}^2\), the curve \(a^n b^m = K_r\) intersects the path \(c(t)\) at most once on \([0,1]\) and at most once on \([1,2]\), since the slope of the curve is negative. Hence, it has at most two intersections with \(c(t)\). Therefore, \[\begin{align} \left\{(a,b)\in\mathbb{R}^2 \mid a>1,\,0<b<1,\,a^n b^m = K_r\text{ for some }n,m\in\mathbb{N}\text{ with }n+m\leq N\right\} \end{align}\] and the path \((a,b)=c(t)\) have only finitely many intersection points. Let \(0<\underline{t}_1<\cdots<\underline{t}_n<1\) be all values of \(t\) corresponding to such intersection points with \(0<t<1\), and let \(1<\overline{t}_1<\cdots<\overline{t}_m<2\) be all values of \(t\) corresponding to such intersection points with \(1<t<2\). Finally, define \(\underline{t}_0:=0\), \(\underline{t}_{n+1}:=1\), \(\overline{t}_{0}=1\), and \(\overline{t}_{m+1}:=2\).

Note that if \(t^1\geq t^2\) satisfies \((t^1,t^2)\in (\underline{t}_j,\underline{t}_{j+1})^2\cup(\overline{t}_j,\overline{t}_{j+1})^2\cup[0,\underline{t}_{1})^2\cup(\overline{t}_m,2]^2\) for some \(j\), \(\pi^{t^1}\) and \(\pi^{t^2}\) are in the same \(r\)-region. Then, by Proposition 12, \(V_i^{\mathcal{D}_r}(\pi^{t^1})\geq V_i^{\mathcal{D}_r}(\pi^{t^2})\) holds.

Since each \(r\)-region is an open set, there exists \(\underline{t}^*\in (\underline{t}_n,1)\) and \(\overline{t}^*\in (1,\overline{t}_1)\) such that \(\underline{t}^*\) and \(\overline{t}^*\) is in the same \(r\)-region. Then, it follows that \(V_i^{\mathcal{D}_r}(\pi^{\overline{t}^*})\geq V_i^{\mathcal{D}_r}(\pi^{\underline{t}^*})\) by Proposition 12.

By construction, for all \(t<\underline{t}_j\) sufficiently close to \(\underline{t}_j\), the equilibrium action profile under \(\pi^t\) coincides with the equilibrium under \(\pi^{\underline{t}_j}\) in which ties are broken in favor of \(a_0\). Therefore, \[\begin{align} \lim_{t\nearrow \underline{t}_j}V_i^{\mathcal{D}_r}(\pi^{t})=V_i^{\mathcal{D}_r}(\pi^{\underline{t}_j},\boldsymbol{\sigma}^0), \end{align}\] where \(\boldsymbol{\sigma}^0\) is an equilibrium under \(\pi^{\underline{t}_j}\) that breaks ties in favor of \(a_0\). Similarly, we have \[\begin{align} \lim_{t\searrow \underline{t}_j}V_i^{\mathcal{D}_r}(\pi^{t})=V_i^{\mathcal{D}_r}(\pi^{\underline{t}_j},\boldsymbol{\sigma}^1), \end{align}\] where \(\boldsymbol{\sigma}^1\) is an equilibrium under \(\pi^{\underline{t}_j}\) that breaks tie in favor of \(a_1\). Thus, Proposition 13 implies \[\begin{align} \lim_{t\nearrow \underline{t}_j}V_i^{\mathcal{D}_r}(\pi^{t})= \lim_{t\searrow \underline{t}_j}V_i^{\mathcal{D}_r}(\pi^{t}) \end{align}\] for all \(j\). Analogously, for all \(j\), \[\begin{align} \lim_{t\nearrow \overline{t}_j}V_i^{\mathcal{D}_r}(\pi^{t})= \lim_{t\searrow \overline{t}_j}V_i^{\mathcal{D}_r}(\pi^{t}) \end{align}\]

Repeatedly applying the monotonicity within each \(r\)-region and the continuity at each boundary point, we obtain \[V_i^{\mathcal{D}_r}(\pi^{0}) \leq V_i^{\mathcal{D}_r}(\pi^{\underline{t}^*}) \leq V_i^{\mathcal{D}_r}(\pi^{\overline{t}^*}) \leq V_i^{\mathcal{D}_r}(\pi^{2}).\] Therefore, we obtain \(V_i^{\mathcal{D}_r}(\pi)\geq V_i^{\mathcal{D}_r}(\pi')\), which completes the proof. ◻

Proof of Proposition 8. It suffices to show that \(\overline{V}_{i}^{\mathcal{D}}(\pi)\geq V_{i}^{\mathcal{D}}(\pi,\boldsymbol{\sigma})\) holds for all \(i\), \(\mathcal{D}\), \(\pi\), and \(\boldsymbol{\sigma}\) because the remaining part of the proof is identical to that of Theorem 1. Take any \(\mathcal{D},\pi\), and \(\boldsymbol{\sigma}\). Note that \(\overline{V}_{1}^{\mathcal{D}}(\pi)= V_{1}^{\mathcal{D}}(\pi,\boldsymbol{\sigma})\). Fix \(i\geq 2\). For each \(\boldsymbol{s}\in S_{1}\times \dots \times S_{i-1}\), define \(f_{i-1}(\boldsymbol{s})\in \Delta (A^{i-1})\) as \[\begin{align} f_{i-1}(\boldsymbol{a}|\boldsymbol{s})=\prod_{k=1}^{i-1}\sigma_{k}(a_{k}|a_{1},\dots,a_{k-1},s_{k}). \end{align}\] Thus, \(f_{i-1}(\boldsymbol{a}|\boldsymbol{s})\) is the probability that agents \(1,\dots,i-1\) take action \(\boldsymbol{a}=(a_{1},\dots,a_{i-1})\) when they receive private signals \(\boldsymbol{s}=(s_{1},\dots,s_{i-1})\). Let \(g^{\omega}_{i}(\boldsymbol{a},s_{i}'|\pi,\boldsymbol{\sigma})\) denote the probability that agent \(i\) observes \((\boldsymbol{a},s_{i}')\) when the state is \(\omega\) and agents follow strategy profile \(\boldsymbol{\sigma}\). Then, \(g_{i}(\cdot|\pi,\boldsymbol{\sigma})\) is a garbling of \(\pi_{\leq i}\), where \(g_{i}(\cdot|\pi,\boldsymbol{\sigma})\) is the ex-ante distribution induced by \(g_{i}^{\omega}(\cdot|\pi,\boldsymbol{\sigma})\).⁴¹ ◻