Language and coordination games

Abstract

I formalize the role of the self-signaling condition in guaranteeing coordination for pre-play cheap talk games with one-sided communication. I model a preexisting common language by assuming that the Receiver either ignores or follows cheap talk recommendations, but never inverts them. This assumption creates asymmetry between messages, which captures the essence of a common language. It does not rule out any outcome at hand in that every equilibrium outcome of the original game remains an equilibrium outcome in this transformed game. However, applying iterative admissibility to the transformed game yields sharp predictions. If the stage game satisfies a certain self-signaling condition, then the Sender gets her Stackelberg payoff in every iteratively admissible outcome. On the other hand, if the stage game violates a weaker self-signaling condition, miscoordination can happen in an iteratively admissible outcome.

Appendices

Appendix 1: Preliminaries

Define $Literal:M\rightarrow A^{R}$ to respond to every message $m=$ “$A_{1}\ldots \left\{ a^{R}\right\}$” with its final level of recommendation $a^{R}$. Then Literal follows every conjugate pair and thus is a language-based response.

Lemma 4

Let E be the union of conjugate pair $M\left( A_{1}\ldots A_{k}A_{k+1}\right)$ and $M\left( A_{1}\ldots A_{k}\left( A_{k}\backslash A_{k+1}\right) \right)$ and F be the union of conjugate pair $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }A_{l+1}^{\prime }\right)$ and $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\left( A_{l}^{\prime }\backslash A_{l+1}^{\prime }\right) \right)$. Suppose $k\ge l$. Then, either $E\cap F=\emptyset$, or $E=F$, or E is contained in either $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }A_{l+1}^{\prime }\right)$ or $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\left( A_{l}^{\prime }\backslash A_{l+1}^{\prime }\right) \right)$.

Proof

Suppose $A_{1}\ldots A_{l}\ne A_{1}^{\prime }\ldots A_{l}^{\prime }$ or $A_{1}\ldots A_{l}=A_{1}^{\prime }\ldots A_{l}^{\prime }$, but $A_{l+1}\notin \left\{ A_{l+1}^{\prime },A_{l}^{\prime }\backslash A_{l+1}^{\prime }\right\}$. Because $k\ge l$, $A_{1}\ldots A_{l+1}$ is a subsequence of $A_{1}\ldots A_{k+1}$. So $F\cap M\left( A_{1}\ldots A_{k+1}\right) =\emptyset$ because every message in $M\left( A_{1}\ldots A_{k+1}\right)$ begins with $A_{1}\ldots A_{l+1}$ but no message in F does. Similarly $F\cap M\left( A_{1}\ldots \left( A_{k}\backslash A_{k+1}\right) \right) =\emptyset$. So $E\cap F=\emptyset$.

Otherwise, $A_{1}\ldots A_{l}=A_{1}^{\prime }\ldots A_{l}^{\prime }$, and $A_{l+1}\in \left\{ A_{l+1}^{\prime },A_{l}^{\prime }\backslash A_{l+1}^{\prime }\right\}$. If $k=l$, then $E=F$. Otherwise, $k\ge l+1$, then

$$\begin{aligned} E\subset & {} M\left( A_{1}\ldots A_{k}\right) \subset M\left( A_{1}\ldots A_{l+1}\right) \\& = {} \left\{ \begin{array}{ll} M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }A_{l+1}^{\prime }\right) &{}\quad {\text {if }}A_{l+1}=A_{l+1}^{\prime } \\ M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\left( A_{l}^{\prime }\backslash A_{l+1}^{\prime }\right) \right) &{}\quad {\text {if }}A_{l+1}=A_{l}^{\prime }\backslash A_{l+1}^{\prime } \end{array} \right. . \end{aligned}$$

$\hfill\square$

For message subset $E\subset M$, language-based responses $s_{E}^{R}$ and $s^{R}$, define $T\left( s_{E}^{R},s^{R},E\right) :M\rightarrow A^{R}$ to be equal to $s_{E}^{R}$ on E and equal to $s^{R}$ elsewhere.

For example, consider $s_{Z}^{R}$ in Fig. 1, which ignores the conjugate pair of “$\left\{ Y\right\}$” and $M\left( \left\{ X,Z\right\} \right)$, but follows the conjugate pair of “$\left\{ X\right\}$” and $M\left( \left\{ Y,Z\right\} \right)$.

$T\left( Literal,s_{Z}^{R},M\left( \left\{ X,Z\right\} \right) \right)$ is not a language-based response: It does not follow the conjugate pair “$\left\{ Y\right\}$” and $M\left( \left\{ X,Z\right\} \right)$ because it maps “$\left\{ Y\right\}$” into $\left\{ Z\right\}$, but does not ignore the aforementioned conjugate pair because it is equal to Literal on $M\left( \left\{ X,Z\right\} \right)$ and thus not constant on $M\left( \left\{ X,Z\right\} \right)$.

On the other hand, $T\left( Literal,s_{Z}^{R},M\left( \left\{ Y,Z\right\} \right) \right)$ follows the conjugate pair of “$\left\{ X\right\}$” and $M\left( \left\{ Y,Z\right\} \right)$ because (1) $s_{Z}^{R}$ follows the same conjugate pair and thus maps “$\left\{ X\right\}$” into X, and (2) Literal maps $M\left( \left\{ Y,Z\right\} \right)$ into $\left\{ Y,Z\right\}$. For other conjugate pairs, $T\left( Literal,s_{Z}^{R},M\left( \left\{ Y,Z\right\} \right) \right)$ can be shown to either follow or ignore them, using the same argument as in the proof for Claim 42. So $T\left( Literal,s_{Z}^{R},M\left( \left\{ Y,Z\right\} \right) \right)$ is a language-based response.

Definition 11

Let E be the union of one or more conjugate pair(s) succeeding $M\left( A_{1}\ldots A_{k}\right)$. A language-based response $s^{R}$ is language-based on E if it maps E into $A_{k}$.⁹

Lemma 5 follows immediately from the definition.

Lemma 5

$s^{R}$ is language-based on the union of a conjugate pair succeeding $M\left( A_{1}\ldots A_{k}\right)$ if it follows the conjugate pair $M\left( A_{1}\ldots A_{k}\right)$ and $M\left( A_{1}\ldots \left( A_{k}\backslash A_{k-1}\right) \right)$.

Lemma 6 gives a sufficient condition for $T\left( s_{E}^{R},s^{R},E\right)$ to be a language-based response.

Lemma 6

Let E be the union of one or more conjugate pair(s) succeeding $M\left( A_{1}\ldots A_{k}\right)$. Let $s_{E}^{R},s^{R}$ be language-based responses. If $s_{E}^{R}$ is language-based on E, and $s^{R}$ follows the conjugate pair $M\left( A_{1}\ldots A_{k}\right)$ and $M\left( A_{1}\ldots \left( A_{k-1}\backslash A_{k}\right) \right)$, then $T\left( s_{E}^{R},s^{R},E\right)$ is a language-based response.

Proof

Consider any conjugate pair $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\right)$ and $M\left( A_{1}^{\prime }\ldots \left( A_{l-1}^{\prime }\backslash A_{l}^{\prime }\right) \right)$, with its union denoted by F.

If $F\subset E$ or $F\subset E^{c}$, then $T\left( \tilde{s} ^{R},s^{R},E\right)$ either ignores or follows the conjugate pair because it responds in the same way as $\tilde{s}^{R}$ in the former case and $s^{R}$ in the latter case and because $\tilde{s}^{R}$ and $s^{R}$ are both language-based responses.

Otherwise, by Lemma 4, F strictly contains E and either $E\subset M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\right)$ or $E\subset M\left( A_{1}^{\prime }\ldots \left( A_{l-1}^{\prime }\backslash A_{l}^{\prime }\right) \right)$. W.l.o.g. assume the former. Then

$$\begin{aligned} M\left( A_{1}\ldots A_{k}\right) \subset M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\right) . \end{aligned}$$

(9)

I show that $T\left( s_{E}^{R},s^{R},E\right)$ follows conjugate pair $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\right)$ and $M\left( A_{1}^{\prime }\ldots \left( A_{l-1}^{\prime }\backslash A_{l}^{\prime }\right) \right)$. Because $s^{R}$ follows conjugate pair $M\left( A_{1}\ldots A_{k}\right)$ and $M\left( A_{1}\ldots \left( A_{k-1}\backslash A_{k}\right) \right)$, it is non-constant on its union by Definition 3, and thus non-constant on $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\right) \cup M\left( A_{1}^{\prime }\ldots \left( A_{l-1}^{\prime }\backslash A_{l}^{\prime }\right) \right)$ by (9). Then, by Definitions 3 and 4, $s^{R}$ maps $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\right)$ into $A_{l}^{\prime }$ and $M\left( A_{1}^{\prime }\ldots \left( A_{l-1}^{\prime }\backslash A_{l}^{\prime }\right) \right)$ into $A_{l-1}^{\prime }\backslash A_{l}^{\prime }$. Then $T\left( s_{E}^{R},s^{R},E\right)$ maps $M\left( A_{1}^{\prime }\ldots \left( A_{l-1}^{\prime }\backslash A_{l}^{\prime }\right) \right)$ into $A_{l-1}^{\prime }\backslash A_{l}^{\prime }$ because it responds the same way as $s^{R}$ on $M\left( A_{1}^{\prime }\ldots \left( A_{l-1}^{\prime }\backslash A_{l}^{\prime }\right) \right)$. $T\left( s_{E}^{R},s^{R},E\right)$ maps E into $A_{l}^{\prime }$ because it responds the same way as $s_{E}^{R}$ on E and $s_{E}^{R}$ is language-based on E. It maps $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\right) \backslash E$ into $A_{l}^{\prime }$ because it responds the same way as $s^{R}$ on $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\right) \backslash E$. So $T\left( s_{E}^{R},s^{R},E\right)$ maps $M\left( A_{1}^{\prime }\ldots A_{l}^{\prime }\right)$ into $A_{l}^{\prime }$. $\hfill\square$

Next we give sufficient conditions for the combination of two strategies in $S^{R}\left( n\right)$ to give rise to another strategy in $S^{R}\left( n\right)$.

Lemma 7

Let E be the union of one or more conjugate pair(s) succeeding $M\left( A_{1}\ldots A_{k}\right)$. Suppose $s^{R}\in S^{R}\left( n+1\right)$ follows the conjugate pair $M\left( A_{1}\ldots A_{k}\right)$ and $M\left( A_{1}\ldots \left( A_{k-1}\backslash A_{k}\right) \right)$, and $s_{E}^{R}\in S^{R}\left( n+1\right)$ is optimal against $\sigma _{E}^{S}\in \varDelta ^{+}S^{S}\left( n\right) \cap \left( E\times A^{S}\right)$ among strategies language-based on E. Then $T\left( s_{E}^{R},s^{R},E\right) \in S^{R}\left( n+1\right)$ if either

• (A) every best response to $\sigma _{E}^{S}$ follows $M\left( A_{1}\ldots A_{k}\right)$ and $M\left( A_{1}\ldots \left( A_{k-1}\backslash A_{k}\right) \right)$, or

• (B) $s^{R}$ is a best response to $\sigma ^{S}\in \varDelta S^{S}\left( n\right)$ with support in $E^{c}\times A^{S}$ to which every best response follows $M\left( A_{1}\ldots A_{k}\right)$ and $M\left( A_{1}\ldots \left( A_{k-1}\backslash A_{k}\right) \right)$.

Proof

By Lemma 6, $T\left( s_{E}^{R},s^{R},E\right)$ is a language-based response. Suppose to the contrary that $T\left( s_{E}^{R},s^{R},E\right) \notin S^{R}\left( n+1\right)$. Then, for some $s_{*}^{R}\in S_{L}^{R}$, some $k\le n$ and all $s^{S}\in S^{S}\left( k\right)$, $u^{R}\left( s^{S},s_{*}^{R}\right) \ge u^{R}\left( s^{S},T\left( s_{E}^{R},s^{R},E\right) \right)$. In addition, strict inequality holds for some $\left( \hat{m},\hat{a}^{S}\right) \in S^{S}\left( k\right)$. So $T\left( s_{E}^{R},s^{R},E\right)$ is weakly dominated by $s_{*}^{R}$ w.r.t. either (case 1) $S^{S}\left( k\right) \cap E\times A^{S}$, or (case 2) $S^{S}\left( k\right) \cap \left( M\backslash E\right) \times A^{S}$, where case 1 holds if $\hat{m}\in E$ and case 2 holds if $\hat{m}\notin E$.

Claim 71

$s_{*}^{R}$ is follows $M\left( A_{1}\ldots A_{k}\right)$ and $M\left( A_{1}\ldots \left( A_{k-1}\backslash A_{k}\right) \right)$.

Suppose condition (A) holds. Because $T\left( s_{E}^{R},s^{R},E\right)$ responds the same way as $s_{E}^{R}$ on E and the support of $\sigma _{E}^{S}$ is contained in $E\times A^{S}$, $T\left( s_{E}^{R},s^{R},E\right)$ is a best response to $\sigma _{E}^{S}$. So $s_{*}^{R}$ is a best response to $\sigma _{E}^{S}$ and the claim follows. Otherwise, condition (B) holds. Because $T\left( s_{E}^{R},s^{R},E\right)$ responds the same way as $s^{R}$ on $M\backslash E$ and the support of $\sigma ^{S}$ is contained in $\left( M\backslash E\right) \times A^{S}$, $T\left( s_{E}^{R},s^{R},E\right)$ is a best response to $\sigma ^{S}$. So $s_{*}^{R}$ is a best response to $\sigma ^{S}$ and the claim follows.

Suppose case 1 holds. Then $u^{R}\left( \sigma _{E}^{S},T\left( s_{E}^{R},s^{R},E\right) \right) <u^{R}\left( \sigma _{E}^{S},s_{*}^{R}\right)$. By Lemma 5 and the claim, $s_{*}^{R}$ is language-based on E. Then $u^{R}\left( \sigma _{E}^{S},T\left( s_{E}^{R},s^{R},E\right) \right) =u^{R}\left( \sigma _{E}^{S},s_{E}^{R}\right) \ge u^{R}\left( \sigma _{E}^{S},s_{*}^{R}\right)$ where the equality holds because $T\left( s_{E}^{R},s^{R},E\right)$ responds in the same way as $s_{E}^{R}$ on the support of $\sigma _{E}^{S}$ and the inequality holds because $s_{E}^{R}$ is optimal against $\sigma _{E}^{S}$ among strategies language-based on E. Contradiction.

So case 2 must hold. Because $T\left( s_{E}^{R},s^{R},E\right)$ responds the same way as $s^{R}$ on $M\backslash E$, $s_{*}^{R}$ weakly dominates $s^{R}$ w.r.t. $S^{S}\left( k\right) \cap \left( M\backslash E\right) \times A^{S}$. Since $T\left( s^{R},s_{*}^{R},E\right)$ responds the same way as $s^{R}$ on E and as $s_{*}^{R}$ on $\left( M\backslash E\right)$, $T\left( s^{R},s_{*}^{R},E\right)$ weakly dominates $s^{R}$ w.r.t. $S^{S}\left( k\right)$. By Lemma 5 and because $s^{R}$ follows the conjugate pair $M\left( A_{1}\ldots A_{k}\right)$ and $M\left( A_{1}\ldots \left( A_{k-1}\backslash A_{k}\right) \right)$, $s^{R}$ is language-based on E. So $T\left( s^{R},s_{*}^{R},E\right)$ is a language-based response. Thus $s^{R}\notin S^{R}\left( k+1\right)$. Contradiction because $s^{R}\in S^{R}\left( n+1\right)$ and $n\ge k$. $\hfill\square$

Given any $\tilde{m}={\text{``}}A_{1}\ldots A_{n}$”, define $X\left( \tilde{m},\tilde{a}^{S}\right)$ to be the union of the conjugate pair $M\left( A_{1}\ldots .A_{k}A_{k+1}\right)$ and $M\left( A_{1}\ldots .A_{k}\left( A_{k}\backslash A_{k+1}\right) \right)$ where $A_{k}$ is the smallest set in $\left\{ A_{0},A_{1},\ldots ,A_{n}\right\}$ that contains $b^{R}\left( \tilde{a}^{S}\right)$. That is, $b^{R}\left( \tilde{a} ^{S}\right) \in A_{k}\backslash A_{k+1}$ and $X\left( \tilde{m},\tilde{a} ^{S}\right)$ is the union of a conjugate pair succeeding $M\left( A_{1}\ldots A_{k}\right)$. So any $s^{R}\in S_{L}^{R}$ with $s^{R}\left( \tilde{ m}\right) =b^{R}\left( \tilde{a}^{S}\right)$ must be constant on $X\left( \tilde{m},\tilde{a}^{S}\right)$ at $b^{R}\left( \tilde{a}^{S}\right)$ and $Constant-b^{R}\left( \tilde{a}^{S}\right)$ is language-based on $X\left( \tilde{m},\tilde{a}^{S}\right)$. For any $s^{R}\in S_{L}^{R}$,

$$\begin{aligned} u^{R}\left( \left( \tilde{m},\tilde{a}^{S}\right) ,s^{R}\right)& = {} g^{R}\left( \tilde{a}^{S},s^{R}\left( \tilde{m}\right) \right) \\\le & {} g^{R}\left( \tilde{a}^{S},b^{R}\left( \tilde{a}^{S}\right) \right) =u^{R}\left( \left( \tilde{m},\tilde{a}^{S}\right) ,Constant-b^{R}\left( \tilde{a}^{S}\right) \right) . \end{aligned}$$

By BM self-signaling, the inequality holds and it holds with equality if and only if $s^{R}\left( \tilde{m}\right) =b^{R}\left( \tilde{a}^{S}\right)$. We thus obtain Lemma 8.

Lemma 8

A language-based response $s^{R}$ is a best response to $\left( m,a^{S}\right)$ if and only if $s^{R}$ is constant on $X\left( m,a^{S}\right)$ at $b^{R}\left( a^{S}\right)$. In addition, $Constant-b^{R}\left( a^{S}\right)$ is language-based on $X\left( m,a^{S}\right)$.

Say that $\left( m_{1},a_{1}^{S}\right)$ conflicts with $\left( m_{2},a_{2}^{S}\right)$ if $X\left( m_{1},a_{1}^{S}\right) \cap X\left( m_{2},a_{2}^{S}\right) \ne \emptyset$ and $b^{R}\left( a_{1}^{S}\right) \ne b^{R}\left( a_{2}^{S}\right)$. If $\left( m_{1},a_{1}^{S}\right)$ conflicts with $\left( m_{2},a_{2}^{S}\right)$, then no language-based response is a best response to both $\left( m_{1},a_{1}^{S}\right)$ and $\left( m_{2},a_{2}^{S}\right)$ by Lemma 8. Lemma 9 shows that $\left( m,a^{S}\right)$ guarantees its commitment payoff if no Sender strategy conflicts with it.

Lemma 9

Let F be the union of one or more conjugate pairs succeeding $M\left( A_{1}\ldots A_{k}\right)$ and $X^{S}$ be a subset of Sender strategies using messages in F. Suppose, for any $\left( m_{1},a_{1}^{S}\right) ,\left( m_{2},a_{2}^{S}\right)$ in $X^{S}$, either $X\left( m_{1},a_{1}^{S}\right) \cap X\left( m_{2},a_{2}^{S}\right) =\emptyset$ or $b^{R}\left( a_{1}^{S}\right) =b^{R}\left( a_{2}^{S}\right)$ . Then a language-based response exists that is optimal against every $\left( m,a^{S}\right) \in X^{S}$.

Proof

We can write $X^{S}=\left\{ \left( m_{1},a_{1}^{S}\right) ,\ldots ,\left( m_{L},a_{L}^{S}\right) \right\}$ since M and $A^{S}$ are both finite. Define $s_{0}^{R}$ to be the Literal strategy. Suppose $s_{l-1}^{R}$ is a language-based response and is a best response to $\left( m_{j},a_{j}^{S}\right)$ for $1\le j\le l-1$. Define

$$\begin{aligned} I_{l}=\left\{ j=1,\ldots ,l-1:X\left( m_{j},a_{j}^{S}\right) \cap X\left( m_{l},a_{l}^{S}\right) \ne \emptyset \right\} . \end{aligned}$$

Then, for every $j\in I_{l}$, $b^{R}\left( a_{j}^{S}\right) =b^{R}\left( a_{l}^{S}\right)$.

If $X\left( m_{l},a_{l}^{S}\right) \subset X\left( m_{\hat{\jmath }},a_{\hat{ \jmath }}^{S}\right)$ for some $\hat{\jmath }\in I_{l}$, define $s_{l}^{R}=s_{l-1}^{R}$. By the induction hypothesis, $s_{l-1}^{R}$ responds to $m_{\hat{\jmath }}$ with $b^{R}\left( a_{\hat{\jmath }}^{S}\right)$. Thus, by the Lemma 8, $s_{l-1}^{R}$ is constant on $X\left( m_{\hat{\jmath } },a_{\hat{\jmath }}^{S}\right)$ at $b^{R}\left( a_{\hat{\jmath }}^{S}\right)$ . Because $b^{R}\left( a_{\hat{\jmath }}^{S}\right) =b^{R}\left( a_{l}^{S}\right)$ and $X\left( m_{l},a_{l}^{S}\right) \subset X\left( m_{ \hat{\jmath }},a_{\hat{\jmath }}^{S}\right)$, $s_{l-1}^{R}$ is a best response to $\left( m_{l},a_{l}^{S}\right)$. Then $s_{l}^{R}=s_{l-1}^{R}$ is a best response to $\left( m_{j},a_{j}^{S}\right)$ for $j=1,\ldots ,l$.

Otherwise, by Lemma 4, $X\left( m_{j},a_{j}^{S}\right)$ $\subsetneq X\left( m_{l},a_{l}^{S}\right)$ for all $j\in I_{l}$. Then $s_{l-1}^{R}$ responds the same way as Literal on $X\left( m_{l},a_{l}^{S}\right) \backslash \cup _{j\in I_{l}}X\left( m_{j},a_{j}^{S}\right)$, which is a non-empty union of conjugate pairs. So $s_{l-1}^{R}$ is non-constant on $X\left( m_{l},a_{l}^{S}\right)$.

Define $s_{l}^{R} = T\left( Constant-b^{R}\left( a_{l}^{S}\right) ,s_{l-1}^{R},X\left( m_{l},a_{l}^{S}\right) \right)$. By Lemma 8, $Constant-b^{R}\left( a_{l}^{S}\right)$ is language-based on $X\left( m_{l},a_{l}^{S}\right)$. By Lemma 6, $s_{l}^{R}$ is a language-based response. For all $j\in I_{l}$, $s_{l}^{R}\left( m_{j}\right) =b^{R}\left( a_{l}^{S}\right) =b^{R}\left( a_{j}^{S}\right)$ where the first equality holds because $X\left( m_{j},a_{j}^{S}\right) \subset X\left( m_{l},a_{l}^{S}\right)$ and the second equality holds by construction of $I_{l}$. For $j\in \left\{ 1,\ldots ,l-1\right\} \backslash I_{l}$, $s_{l}^{R}\left( m_{j}\right) =s_{l-1}^{R}\left( m_{j}\right) =b^{R}\left( a_{j-1}^{S}\right)$ where the first equality holds because $m_{j}\notin X\left( m_{l},a_{l}^{S}\right)$ by construction of $I_{l}$ and the second equality holds by the induction hypothesis. So $s_{l}^{R}$ is a best response to $\left( m_{j},a_{j}^{S}\right)$ for $j=1,\ldots ,l$. By induction, there exists $s_{*}^{R}\in S_{L}^{R}$ that is a best response to every $\left( m,a^{S}\right) \in X^{S}$. $\hfill\square$

Appendix 2: Order of deletion does not matter for generic stage games

Consider a finite game G with players in I, strategy set $S=\times _{i\in I}S^{i}$, and payoff function $u=\left( u^{i}\right) _{i\in I}$.

Definition 12

$W\subset S$ is a full reduction of S by weak dominance if (1) there exists $S=S_{0}\supset S_{1}\supset \ldots \supset S_{K}=W$ for some integer $K\ge 0$ such that, for each player $i\in I$, $S_{k}^{i}$ is obtained from $S_{k-1}^{i}$ by deleting strategies weakly dominated for i w.r.t. $S_{k-1}$, and (2) no player i has any strategy in $W^{i}$ weakly dominated w.r.t. W.

Marx and Swinkels (1997) identifies a condition under which order of deletion does not matter.

Definition 13

(TDI by Marx and Swinkels (1997)) A game satisfies transference of decision-maker indifference (TDI) if $u^{i}\left( s_{1}^{i},s^{-i}\right) =u^{-i}\left( s_{2}^{i},s^{-i}\right)$ implies that $u^{j}\left( s_{1}^{i},s^{-i}\right) =u^{j}\left( s_{2}^{i},s^{-i}\right)$, for every player $i,j\in I$, every $s_{1}^{i},s_{2}^{i}\in S^{i}$ and $s_{-i}\in S_{-i}$.

Definition 14

A strategy $s_{i}\in S_{i}$ is redundant on $W\subset S$ if there is $r_{i}\in W_{i}\backslash s_{i}$ such that $u\left( r_{i},s_{-i}\right) =u\left( s_{i},s_{-i}\right)$ for all $s_{-i}\in W_{-i}$.

Theorem 1

(Marx and Swinkels 1997) Suppose $G=\left( I,S,u\right)$ satisfies TDI. Let X and Y be full reductions of S by weak dominance. Then, X and Y are the same up to the addition or removal of redundant strategies and a renaming of strategies.

Lemma 10

If g is generic, then the language talk game $G_{L}$ satisfies TDI.

Proof

If g is generic, then a unilateral change leaves a player indifferent only if it does not change the resulting action profile. Since messages are costless, this change leaves every player indifferent. $\hfill\square$

Appendix 3: Omitted Proof for Proposition 2

1.1 Iteration 1 and the preliminary step

For any bijective function $\phi :\left\{ 1,2,\ldots ,N\right\} \rightarrow A^{R}$, define $A_{\phi ,k}=:\left\{ \phi \left( k+1\right) ,\ldots ,\phi \left( N\right) \right\}$ for $k=0,\ldots ,N-1$. Define

$$\begin{aligned} M_{\phi }\left( k\right)\equiv & {} M\left( A_{\phi ,1}\ldots A_{\phi ,k}\right) \text {;} \\ m_{\phi ,k}\equiv & {} {\text{``}}A_{\phi ,1}\ldots A_{\phi ,k-1}\left\{ \phi \left( k\right) \right\} {\text{''}}. \end{aligned}$$

In words, $M_{\phi }\left( k\right)$ is the set of messages that say “Do not take action $\phi \left( 1\right)$. Among the rest, do not take action $\phi \left( 2\right)$. ...Among $\left\{ \phi \left( k\right) ,\ldots ,\phi \left( N\right) \right\}$, do not take action $\phi \left( k\right)$.” That is, these messages recommend, one by one, against $\phi \left( 1\right)$, $\phi \left( 2\right)$, ..., up to $\phi \left( k\right)$. Compared to $M_{\phi }\left( k\right)$, $m_{\phi ,k}$ recommends $\phi \left( k\right)$ among $\left\{ \phi \left( k\right) ,\ldots ,\phi \left( N\right) \right\}$ though it shares the first $k-1$ levels of recommendations with all messages in $M_{\phi }\left( k\right)$. Their union, $m_{\phi ,k}\cup M_{\phi }\left( k\right)$, is decreasing in k in terms of set inclusion. We can see the set of all hierarchical recommendations as $M_{\phi }\left( 0\right)$ or as the conjugate pair $m_{\phi ,0}\cup M_{\phi }\left( 0\right)$ where $m_{\phi ,0}=\emptyset$.

Lemma 11

If $m\in M_{\phi }\left( k\right)$ and $\left( m,a^{S}\right) \in S^{S}\left( 1\right)$, then $b^{R}\left( a^{S}\right) \in A_{\phi ,k}$.

Proof

Suppose $m\in M_{\phi }\left( k\right)$ and $b^{R}\left( a^{S}\right) =\phi \left( q\right)$ where $q\le k$. Because $m\in M_{\phi }\left( k\right) \subset M_{\phi }\left( q\right)$ given that $q\le k$, by the language assumption (Definitions 3 and 4), for every $s^{R}=S_{L}^{R}$, either (case 1) $s^{R}\left( m_{\phi ,q}\right) =s^{R}\left( m\right)$, or (case 2) $s^{R}\left( m_{\phi ,q}\right) =\phi \left( q\right) \ne s^{R}\left( m\right)$ because $s^{R}\left( m\right) \in A_{\phi ,q}$. Hence, for all $s^{R}\in S_{L}^{R}$,

$$\begin{aligned} u^{S}\left( \left( m_{\phi ,q},a^{S}\right) ,s^{R}\right) =g^{S}\left( a^{S},s^{R}\left( m_{\phi ,q}\right) \right) \ge g^{S}\left( a^{S},s^{R}\left( m\right) \right) =u^{S}\left( \left( m,a^{S}\right) ,s^{R}\right) , \end{aligned}$$

where the inequality holds with equality in case (1) and with strict inequality in case (2) by BM self-signaling. In addition, $Literal\in S_{L}^{R}$ and case (2) holds for Literal because $Literal\left( m\right) \in A_{\phi }\left( k\right)$ and $Literal\left( m_{\phi ,q}\right) =\phi \left( q\right) \notin A_{\phi ,k}$. So $\left( m_{\phi ,q},a^{S}\right)$ weakly dominates $\left( m,a^{S}\right)$ w.r.t. $S_{L}^{R}$. $\hfill\square$

Corollary 2

If $\left( m,a^{S}\right) \in S^{S}\left( 1\right)$ where $m\in m_{\phi ,N-1}\cup M_{\phi ,N-1}$ for some permutation $\phi$, then $X\left( m,a^{S}\right) =\left\{ m\right\}$.

Proof

If $m\in M_{\phi ,N-1}=\left\{ m_{\phi ,N}\right\}$, then it follows by Lemma 11. If $m=m_{\phi ,N-1}$, then $m=m_{\phi ^{\prime },N}$ where $\phi ^{\prime }\left( N-1\right) =\phi \left( N\right)$, $\phi ^{\prime }\left( N\right) =\phi \left( N-1\right)$ and $\phi ^{\prime }\left( q\right) =\phi \left( q\right)$ for $q\le N-2$. The conclusion then follows. $\hfill\square$

Given any $\sigma ^{R}\in \varDelta S^{R}\left( 0\right)$, any $a^{S}$ where $b^{R}\left( a^{S}\right) \in A_{\phi ,j}$, define

$$\begin{aligned}\chi \left( a^{S};\sigma ^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) &\equiv \sum _{\begin{array}{c} s^{R}\text { constant } \\ \text {on }m_{\phi ,j}\cup M_{\phi }\left( j\right) \end{array}}\sigma ^{R}\left( s^{R}\right) g^{S}\left( a^{S},s^{R}\left( m_{\phi ,j}\right) \right)\\& +\sum _{\begin{array}{c} s^{R}\text { non-constant } \\ \text {on }m_{\phi ,j}\cup M_{\phi }\left( j\right) \end{array}}\sigma ^{R}\left( s^{R}\right) g^{S}\left( a^{S},b^{R}\left( a^{S}\right) \right) \end{aligned}$$

where $\sigma ^{R}\left( s^{R}\right)$ denotes the probability $\sigma ^{R}$ places on $s^{R}$.

Lemma 12

For any $m\in M_{\phi }\left( j\right)$ and $b^{R}\left( a^{S}\right) \in A_{\phi ,j}$,

$$\begin{aligned} u^{S}\left( \left( m,a^{S}\right) ,\sigma ^{R}\right) \le \chi \left( a^{S};\sigma ^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) . \end{aligned}$$

The inequality holds strictly if and only if there exists $s^{R}$ in the support of $\sigma ^{R}$ which is non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ with $s^{R}\left( m\right) \ne b^{R}\left( a^{S}\right)$.

Proof

Note that

$$\begin{aligned}u^{S}\left( \left( m,a^{S}\right) ,\sigma ^{R}\right) & =\sum _{\begin{array}{c} s^{R}\text { constant } \\ \text {on }m_{\phi ,j}\cup M_{\phi }\left( j\right) \end{array}}\sigma ^{R}\left( s^{R}\right) g^{S}\left( a^{S},s^{R}\left( m\right) \right)\\&\quad +\sum _{\begin{array}{c} s^{R}\text { non-constant } \\ \text {on }m_{\phi ,j}\cup M_{\phi }\left( j\right) \end{array}} \sigma ^{R}\left( s^{R}\right) g^{S}\left( a^{S},s^{R}\left( m\right) \right) \\& =\sum _{\begin{array}{c} s^{R}\text { constant } \\ \text {on }m_{\phi ,j}\cup M_{\phi }\left( j\right) \end{array}}\sigma ^{R}\left( s^{R}\right) g^{S}\left( a^{S},s^{R}\left( m_{\phi ,j}\right) \right)\\&\quad +\sum _{\begin{array}{c} s^{R}\text { non-constant } \\ \text {on }m_{\phi ,j}\cup M_{\phi }\left( j\right) \end{array}} \sigma ^{R}\left( s^{R}\right) g^{S}\left( a^{S},s^{R}\left( m\right) \right) \end{aligned}$$

where the equality holds because $s^{R}\left( m_{\phi ,j}\right) =s^{R}\left( m\right)$ for $s^{R}$ constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ given that $m\in M_{\phi }\left( j\right)$. The Lemma follows by BM self-signaling. $\hfill\square$

Define

$$\begin{aligned} B_{\phi ,j}^{S}\left( \sigma ^{R}\right) :=\arg \max _{a^{S}:b^{R}\left( a^{S}\right) \in A_{\phi ,j}}\chi \left( a^{S};\sigma ^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) . \end{aligned}$$

Suppose $\left( \hat{m},\hat{a}^{S}\right)$ is a best response to $\hat{ \sigma }^{R}$ and $b^{R}\left( \hat{a}^{S}\right) \in A_{\phi ,j}$. Then Sender strategies in $M_{\phi }\left( j\right) \times B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right)$ can potentially weakly dominate $\left( \hat{m}, \hat{a}^{S}\right)$.

Lemma 13

Consider $\hat{m}\in m_{\phi ,j}\cup M_{\phi }\left( j\right)$. Suppose $\left( \hat{m},\hat{a} ^{S}\right) \in S^{S}\left( n+1\right)$, and $s_{d}^{R}\left( \hat{m} \right) \ne b^{R}\left( \hat{a}^{S}\right)$ where $s_{d}^{R}\in S^{R}\left( n\right)$ is non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$. Then, for the union F of any conjugate pair succeeding $M_{\phi }\left( j\right)$, and any $\hat{\sigma }^{R}\in \varDelta ^{+}S^{R}\left( n\right)$ to which $\left( \hat{m},\hat{a}^{S}\right)$ is a best response,

• 1. there exists no $\left( m_{F},a_{F}^{S}\right)$ where $m_{F}\in F$ and $a_{F}^{S}\in B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right) \cup \left\{ \hat{a}^{S}\right\}$ such that $s^{R}\left( m_{F}\right) =b^{R}\left( a_{F}^{S}\right)$ for all $s^{R}\in S^{R}\left( n\right)$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$, and

• 2. for any $\left( m_{F},a_{F}^{S}\right) \in F\times B_{\phi ,j}^{S}\left( \hat{\sigma } ^{R}\right) \cup \left\{ \hat{a}^{S}\right\}$, there exists $\sigma _{\left( m_{F},a_{F}^{S}\right) }^{S}\in \varDelta S^{S}\left( n-1\right)$ with support in $F\times A^{S}$ against which Receiver’s optimal strategy among those language-based on F does NOT respond to $m_{F}$ with $b^{R}\left( a_{F}^{S}\right)$.

Proof

Suppose to the contrary of part 1 that there exists $m_{F}\in F$ and $a_{F}^{S}\in B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right) \cup \left\{ a_{1}^{S}\right\}$ such that $s^{R}\left( m_{F}\right) =b^{R}\left( a_{F}^{S}\right)$ for all $s^{R}\in S^{R}\left( n\right)$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$. Then

$$\begin{aligned} u^{S}\left( \left( m_{F},a_{F}^{S}\right) ,\hat{\sigma }^{R}\right)& = {} \chi \left( a_{F}^{S};\hat{\sigma }^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) \\& \ge {} \chi \left( \hat{a}^{S};\hat{\sigma }^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) >u^{S}\left( \left( \hat{m},\hat{a}^{S}\right) , \hat{\sigma }^{R}\right) , \end{aligned}$$

where the equality holds by Lemma 12 and because $s^{R}\left( m_{F}\right) =b^{R}\left( a_{F}^{S}\right)$ for all $s^{R}\in S^{R}\left( n\right)$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$, the first inequality holds because $a_{F}^{S}$ $\in B_{\phi ,j}^{S}\left( \sigma _{1}^{R}\right) \cup \left\{ \hat{a}^{S}\right\}$, and the last inequality holds by Lemma 12 and because $s_{d}^{R}\left( \hat{m}\right) \ne b^{R}\left( \hat{a}^{S}\right)$ for some $s_{d}^{R}$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ in the support of $\hat{\sigma } ^{R}$. Contradiction because $\left( \hat{m},\hat{a}^{S}\right)$ is a best response to $\hat{\sigma }^{R}$. So part 1 holds.

Suppose to the contrary of part 2 that there exists $m_{F}\in F$ and $a_{F}^{S}\in B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right) \cup \left\{ a_{1}^{S}\right\}$ such that, for any $\sigma ^{S}\in \varDelta S^{S}\left( n-1\right)$ with support in $F\times A^{S}$, some $s_{\sigma ^{S}}^{R}$ language-based on F is optimal against both $\left( m_{F},a_{F}^{S}\right)$ and $\sigma ^{S}$. By part 1, there exists $\tilde{s}^{R}\in S^{R}\left( n\right)$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ where $\tilde{s}^{R}\left( m_{F}\right) \ne b^{R}\left( a_{F}^{S}\right)$. For any $\sigma ^{S}\in \varDelta ^{+}S^{S}\left( n-1\right)$, write

$$\begin{aligned} \sigma ^{S}=p_{\left( m_{F},a_{F}^{S}\right) }\cdot \left( m_{F},a_{F}^{S}\right) +p_{F}\cdot \sigma _{F}^{S}+p_{F^{c}}\cdot \sigma _{F^{c}}^{S} \end{aligned}$$

where $\sigma _{F}^{S}$ is $\sigma ^{S}$ conditional on $F\times A^{S}\left( m,a^{S}\right) \backslash \left\{ \left( \hat{m},\hat{a}^{S}\right) \right\}$, $\sigma _{F^{c}}^{S}$ is $\sigma ^{S}$ conditional on $F^{c}\times A^{S}$, and $p_{\left( m_{F},a_{F}^{S}\right) }$, $p_{F}$ and $p_{F^{c}}$ are the probability $\sigma ^{S}$ puts on $\left( m_{F},a_{F}^{S}\right)$, $\sigma _{F}^{S}$, and $\sigma _{F^{c}}^{S}$, respectively.

By construction, $s_{\sigma ^{S}}^{R}$ is language-based on F and $\tilde{s }^{R}$ is non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$. Then, $T\left( s_{\sigma ^{S}}^{R},\tilde{s}^{R},F\right)$ is a language-based response by Lemma 6 and $\tilde{s}^{R}$ is language based on F by Lemma 5. So

$$\begin{aligned}&u^{R}\left( \sigma ^{S},T\left( s_{\sigma ^{S}}^{R},\tilde{s}^{R},F\right) \right) -u^{R}\left( \sigma ^{S},\tilde{s}^{R}\right) \\&\quad =p_{\left( m_{F},a_{F}^{S}\right) }\times \left[ g^{S}\left( a_{F}^{S},b^{R}\left( a_{F}^{S}\right) \right) -g^{S}\left( a_{F}^{S},\tilde{ s}^{R}\left( m_{F}\right) \right) \right] \\&\qquad +p_{F}\times \left[ u^{R}\left( \sigma _{F}^{S},s_{\sigma ^{S}}^{R}\right) -u^{R}\left( \sigma _{F}^{S},\tilde{s}^{R}\right) \right] +p_{F^{c}}\times 0 \\&\quad \ge p_{\left( m_{F},a_{F}^{S}\right) }*\left[ g^{S}\left( a_{F}^{S},b^{R}\left( a_{F}^{S}\right) \right) -g^{S}\left( a_{F}^{S},\tilde{ s}^{R}\left( m_{F}\right) \right) \right] \\&\quad > 0, \end{aligned}$$

where the equality holds because $T\left( s_{\sigma ^{S}}^{R},\tilde{s} ^{R},F\right)$ responds the same way as $\tilde{s}^{R}$ on $F^{c}$, the first inequality holds because $s_{\sigma ^{S}}^{R}$ is optimal against $\sigma _{F}^{S}$ among strategies language-based on F and $\tilde{s}^{R}$ is language-based on F, and the last inequality holds because $p_{\left( m_{F},a_{F}^{S}\right) }>0$ and by BM self-signaling because $\tilde{s} ^{R}\left( m_{F}\right) \ne b^{R}\left( a_{F}^{S}\right)$. So $\tilde{s} ^{R}\notin S^{R}\left( n\right)$ because it is not a best response to any $\sigma ^{S}\in \varDelta ^{+}S^{S}\left( n-1\right)$. Contradiction to the construction of $\tilde{s}^{R}$. Part 2 is thus established. $\hfill\square$

1.2 The key step analogous to Lemma 46 in Section 4.1.1

We prove the equivalent of Lemma 46 for the general case.

1.2.1 The case of $j=N-2$

The case of $j=N-2$ parallels Claim 46.

Lemma 14

Suppose there exists $\left( m_{1},a_{1}^{S}\right) ,\left( m_{2},a_{2}^{S}\right) \in S^{S}\left( 3\right)$ with $m_{1},m_{2}\in m_{\phi ,N-2}\cup M_{\phi }\left( N-2\right)$, $X\left( m_{2},a_{2}^{S}\right) \subset X\left( m_{1},a_{1}^{S}\right)$ and $b^{R}\left( a_{1}^{S}\right) \ne b^{R}\left( a_{2}^{S}\right)$. Then

• 1. there exists $\left( \tilde{m}_{1},a_{1}^{S}\right) \in S^{S}\left( 1\right)$ where $\tilde{m}_{1}\in M_{\phi }\left( N-2\right)$, and

• 2. there exists $s_{d}^{R}\in S^{R}\left( 2\right)$ that is non-constant on $m_{\phi ,N-2}\cup M_{\phi }\left( N-2\right) .$

Proof

By Lemma 11,

$$\begin{aligned} \text {for }l=1,2\text {, }b^{R}\left( a_{l}^{S}\right) \in \left\{ \phi \left( N-2\right) ,\phi \left( N-1\right) ,\phi \left( N\right) \right\} . \end{aligned}$$

(10)

To highlight the parallel with Claim 46, label $\phi \left( N-2\right) =X$, $\phi \left( N-1\right) =Y$ and $\phi \left( N\right) =Z$.

We first see that $m_{1}=m_{\phi ,N-2}=$“...$\left\{ X,Y,Z\right\} \left\{ X\right\}$” and $b^{R}\left( a_{1}\right) \in \left\{ \phi \left( N-1\right) ,\phi \left( N\right) \right\} =\left\{ Y,Z\right\}$. Otherwise, either (1) $m_{1}\in M_{\phi }\left( N-2\right)$ and thus $X\left( m_{1},a_{1}^{S}\right) =\left\{ m_{1}\right\}$ by Corollary 2, or (2) $b^{R}\left( a_{1}\right) =\phi \left( N-2\right)$, and thus $m_{1}=m_{\phi ,N-2}$ by Lemma 11. In both cases, $X\left( m_{1},a_{1}^{S}\right) =\left\{ m_{1}\right\}$. Then $X\left( m_{2},b^{R}\left( a_{2}^{S}\right) \right) =\left\{ m_{2}\right\} =\left\{ m_{1}\right\}$ since $X\left( m_{2},a_{2}^{S}\right) \subset X\left( m_{1},a_{1}^{S}\right)$. Then $b^{R}\left( a_{2}^{S}\right) =b^{R}\left( a_{1}^{S}\right)$, contradiction to the assumption.

W.l.o.g. assume that $b^{R}\left( a_{1}^{S}\right) =\phi \left( N-1\right)$ . Then, by (10), $b^{R}\left( a_{2}^{S}\right) \in \left\{ \phi \left( N-2\right) ,\phi \left( N\right) \right\}$.

Suppose $b^{R}\left( a_{2}^{S}\right) =\phi \left( N-2\right)$. Then $m_{2}=m_{\phi ,N-2}$ by Lemma 11.Both parts follow from the same arguments as in Claim 46, with “$\left\{ X\right\}$” replaced with $m_{\phi ,N-2}={\text{``}}\ldots \left\{ X,Y,Z\right\} \left\{ X\right\}$” and “$\left\{ Y,Z\right\} \left\{ Y\right\}$” replaced with $m_{\phi ,N-1}={\text{``}}\ldots \left\{ X,Y,Z\right\} \left\{ Y,Z\right\} \left\{ Y\right\}$” .

Otherwise, $b^{R}\left( a_{2}^{S}\right) =\phi \left( N\right)$. We can show that $\left( m_{\phi ,N-1},a_{1}^{S}\right) \in S^{S}\left( 1\right)$, using the same arguments as in Claim 46, with “$\left\{ Y,Z\right\} \left\{ Y\right\}$” replaced by $m_{\phi ,N-1}$. Because $b^{R}\left( a_{2}^{S}\right) =\phi \left( N\right)$, by Corollary 2, $m_{2}\ne m_{\phi ,N-1}$. If $m_{2}=m_{\phi ,N-2}$, then $X\left( m_{2},a_{2}^{S}\right) =m_{\phi ,N-2}\cup M_{\phi }\left( N-2\right) =X\left( m_{1},a_{1}^{S}\right)$. We can show that $\left( m_{\phi ,N},a_{2}^{S}\right) \in S^{S}\left( 1\right)$, using the same arguments for showing that $\left( m_{\phi ,N-1},a_{1}^{S}\right) \in S^{S}\left( 1\right)$. Otherwise, $m_{2}=m_{\phi ,N}$ and $\left( m_{\phi ,N},a_{2}^{S}\right) =\left( m_{2},a_{2}^{S}\right) \in S^{S}\left( 1\right)$. Then, $\frac{1}{2}\left( m_{\phi ,N-1},a_{1}^{S}\right) +\frac{1}{2} \left( m_{\phi ,N},a_{2}^{S}\right) \in \varDelta S^{S}\left( 1\right)$ and thus at least one best response to it belongs to $S^{R}\left( 2\right)$. Moreover, it must respond to $m_{\phi ,N-1}$ with $\phi \left( N-1\right)$ and $m_{\phi ,N}$ with $\phi \left( N\right)$ because $X\left( m_{\phi ,N-1},a_{1}^{S}\right) =\left\{ m_{\phi ,N-1}\right\}$ and $X\left( m_{\phi ,N},a_{2}^{S}\right) =\left\{ m_{\phi ,N}\right\}$ and by Lemma 9. Part 2 then follows. $\hfill\square$

1.2.2 Existence of Key Sender Strategies for the case of $j \le N-3$

This is first the equivalence of part 1 in Lemma 46.

Lemma 15

Consider $j\le N-3$ and $n\in \mathbb {N}$. Suppose there exists $\left( \hat{m},\hat{a}^{S}\right) \in S^{S}\left( n+1\right)$ where $\hat{m}\in m_{\phi ,j}\cup M_{\phi }\left( j\right)$ and $b^{R}\left( \hat{a}^{S}\right) \in A_{\phi ,j}$, and $s_{d}^{R}\in S^{R}\left( n-1\right)$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ where $s_{d}^{R}\left( \hat{m}\right) \ne b^{R}\left( \hat{a} ^{S}\right)$. Then,

• 1. there exists a conjugate pair succeeding $M_{\phi }\left( j\right)$ whose union E does not contain $\hat{m}$;

• 2. for any such set E, and any $\hat{\sigma }^{R}\in \varDelta ^{+}S^{R}\left( n-1\right)$ to which $\left( \hat{m},\hat{a}^{S}\right)$ is a best response, there exists $\left( m_{E},a_{E}^{S}\right) \in S^{S}\left( n\right)$ where $m_{E}\in E$ and $a_{E}^{S}\in B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right)$;

• 3. if $s_{d}^{R}\in S^{R}\left( n\right)$, then for the union F of any conjugate pair succeeding $M_{\phi }\left( j\right)$, and any $\hat{\sigma } _{F}^{S},\tilde{\sigma }_{F}^{S}\in \varDelta S^{S}\left( n-1\right)$ with support in $F\times A^{S}$, there exists $\hat{s}^{R},\tilde{s}^{R}\in S^{R}\left( n\right)$ where $\hat{s}^{R}$ responds in the same way as $\tilde{s}^{R}$ on $F^{c}$, and, among strategies language-based on F, $\hat{s}^{R}$ is optimal against $\hat{\sigma }_{F}^{S}$ whereas $\tilde{s} ^{R}$ is optimal against $\tilde{\sigma }_{F}^{S}$.

I now prove Lemma 15.

Since $j\le N-3$, there are at least three actions in $A_{\phi }\left( j\right)$ and thus at least three different ways to partition $A_{\phi ,j}$ into two. Claim 72 follows immediately.

Claim 72

There are at least three different conjugate pairs succeeding $M_{\phi }\left( j\right)$.

Part 1 follows from Claim 72 and Lemma 4.

Part 2 holds for $n=0$ because $S^{S}\left( 0\right) =M\times A^{S}$. We now show that it holds for $n=k\ge 1$.

Pick any $\hat{m}_{E}\in E$ and $\hat{a}_{E}^{S}\in B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right)$. I will construct $\varphi _{\left( \hat{m}_{E}, \hat{a}_{E}^{S}\right) }\left( \hat{\sigma }^{R}\right) \in \varDelta S^{R}\left( k-1\right)$ to which every best response belongs to $E\times B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right)$. Part 2 then follows for $n=k$ because one such best response must belong to $S^{S}\left( k\right)$.

For the union G of any conjugate pair succeeding $M_{\phi }\left( j\right)$, define

$$\begin{aligned} D_{G}\left( k-1\right) :=\left\{ \begin{array}{ll} S^{S}\left( k-1\right) \cap G\times B_{\phi ,j}^{S}\left( \hat{\sigma } ^{R}\right) &{} \text {if }\hat{m}\notin G \\ S^{S}\left( k-1\right) \cap G\times B_{\phi ,j}^{S}\left( \hat{\sigma } ^{R}\right) \cup \left\{ \hat{a}^{S}\right\} &{} \text {if }\hat{m}\in G \end{array} \right. . \end{aligned}$$

By the induction hypothesis,

$$\begin{aligned} D_{G}\left( k-1\right) \ne \emptyset . \end{aligned}$$

(11)

Claim 73

For $\left( m_{G},a_{G}^{S}\right) \in D_{G}\left( k'\right)$, there exists $\sigma _{\left( m_{G},a_{G}^{S}\right) }^{S}\in \varDelta S^{S}\left( k^{\prime }\right)$ with support in $G\times A^{S}$ against which Receiver’s optimal strategy among those language-based on G must respond to $m_{G}$ NOT with $b^{R}\left( a_{G}^{S}\right)$ for $k^{\prime }=k-1$, and for $k^{\prime }=k$ if $s_{d}^{R}\in S^{R}\left( k\right)$.

Proof

It follows by Lemma 13 because $s_{d}^{R}\in S^{R}\left( k-1\right)$ and by (11). $\hfill\square$

Claim 74

Part 3 holds for $n=k-1$. It holds for $n=k$ if $D_{G}\left( k\right) \ne \emptyset$ for the union G of any conjugate pair succeeding $M_{\phi }\left( j\right)$. Moreover, for any $\left( m,a^{S}\right) \in S^{S}\left( k-1\right)$ where $m\in M\backslash E$ and $a^{S}\in B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right)$, there exists $s_{\left( m,a^{S}\right) }^{R}\in S^{R}\left( k-1\right)$ such that (1) it responds to $\hat{m}_{E}$ with $b^{R}\left( \hat{a}_{E}^{S}\right)$ and to m NOT with $b^{R}\left( a^{S}\right)$, and (2) for any $s^{R}\in S^{R}\left( k-1\right)$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$, we have $T\left( s_{\left( m,a^{S}\right) }^{R},s^{R},M_{\phi }\left( j\right) \right) \in S^{R}\left( k-1\right)$.

Proof

Consider $k^{\prime }=k-1,k$ where $s_{d}^{R}\in S^{R}\left( k^{\prime }\right)$ and $D_{G}\left( k^{\prime }\right) \ne \emptyset$ for the union G of any conjugate pair succeeding $M_{\phi }\left( j\right)$.

By Claim 72, there exists two conjugate pairs succeeding $M_{\phi }\left( j\right)$ whose unions are $F^{\prime },F^{\prime \prime }\ne F$. For $G=F,F^{\prime },F^{\prime \prime }$, there exists $\left( m_{G},a_{G}^{S}\right) \in D_{G}\left( k^{\prime }\right)$ since $D_{G}\left( k^{\prime }\right) \ne \emptyset$.

Define

$$\begin{aligned} \sigma _{F^{\prime \prime }}^{S}:=\left\{ \begin{array}{ll} \left( m_{F^{\prime \prime }},a_{F^{\prime \prime }}^{S}\right) &{} \text {if } b^{R}\left( a_{F^{\prime \prime }}^{S}\right) \ne b^{R}\left( a_{F^{\prime }}^{S}\right) \text { } \\ \sigma _{\left( m_{F^{\prime \prime }},a_{F^{\prime \prime }}^{S}\right) }^{S} &{} \text {otherwise} \end{array} \right. \end{aligned}$$

where $\sigma _{\left( m_{F^{\prime \prime }},a_{F^{\prime \prime }}^{S}\right) }^{S}$ is given by Claim 73. Define $\sigma _{D}^{S}=\left( 1-\varepsilon \right) \left( m_{F^{\prime }},a_{F^{\prime }}^{S}\right) +\varepsilon \sigma _{F^{\prime \prime }}^{S}$.

Because of finite strategy space, for $\varepsilon >0$ sufficiently small, every best response to $\sigma _{D}^{S}$ must be a best response to $\left( m_{F^{\prime }},a_{F^{\prime }}^{S}\right)$, and optimal against $\sigma _{F^{\prime \prime }}^{S}$ among best responses to $\left( m_{F^{\prime }},a_{F^{\prime }}^{S}\right)$. Construct a strategy to be equal to $s_{d}^{R}$ except it responds to all messages in $F^{\prime }$ with $b^{R}\left( a_{F^{\prime }}^{S}\right)$ and responds the same way on $F^{\prime \prime }$ as a strategy optimal against $\sigma _{F^{\prime \prime }}^{S}$ among strategies language-based on $F^{\prime \prime }$. This strategy can be shown to be a language-based response by applying 7 twice because $s_{d}^{R}$ is non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ and $b^{R}\left( a_{F^{\prime }}^{S}\right) \in A_{\phi ,j}$. So every best response to $\sigma _{D}^{S}$ must respond to $m_{F^{\prime }}$ with $b^{R}\left( a_{F^{\prime }}^{S}\right)$ and, by construction of $\sigma _{F^{\prime \prime }}^{S}$, responds to $m_{F^{\prime \prime }}$ NOT with $b^{R}\left( a_{F^{\prime }}^{S}\right)$. So

$$\begin{aligned} \text {every best response to }\sigma _{D}^{S}\text { follows the conjugate pair }m_{\phi ,j}\cup M_{\phi }\left( j\right) \text { } \end{aligned}$$

(12)

Because $\sigma _{D}^{S}\in \varDelta S^{S}\left( k^{\prime }-1\right)$, some $s_{D}^{R}\in S^{R}\left( k^{\prime }\right)$ must be a best response to $\sigma _{D}^{S}$.

Define $\hat{\sigma }_{D}^{S}:=\left( 1-\varepsilon \right) \sigma _{d}^{S}+\varepsilon \hat{\sigma }_{F}^{S}$ and $\tilde{\sigma } _{D}^{S}:=\left( 1-\varepsilon \right) \sigma _{d}^{S}+\varepsilon \tilde{ \sigma }_{F}^{S}$. Then $\hat{\sigma }^{S}_D,\tilde{\sigma }^{S}_D\in \varDelta S^{S}\left( k^{\prime }-1\right)$. Thus some $\hat{s}_{F}^{R},\tilde{s} _{F}^{R}\in S^{R}\left( k^{\prime }\right)$ must be best responses to $\hat{ \sigma }_{D}^{S}$ and $\tilde{\sigma }_{D}^{S}$, respectively. For $\varepsilon >0$ sufficiently small, $\hat{s}^{R}$ and $\tilde{s}^{R}$ must both be best responses to $\sigma _{D}^{S}$. Define $\hat{s}_{1}^{R}$ ($\tilde{s}_{1}^{R}$ ) to be equal to $\hat{s}_{F}^{R}$ ($\tilde{s}_{F}^{R}$) except it responds the same way on F as a strategy optimal against $\hat{\sigma }_{F}^{S}$ ($\tilde{\sigma }_{F}^{S}$) among strategies language-based on F. By Lemma 6, $\hat{s}_{1}^{R}$ and $\tilde{s}_{1}^{R}$ are language-based responses. So $\hat{s}_{D}^{R}$ and $\tilde{s}_{D}^{R}$ must also be optimal against $\hat{\sigma }_{F}^{S}$ and $\tilde{\sigma }_{F}^{S}$, respectively, among strategies language-based on F. Define $\hat{s} ^{R}:=T\left( \hat{s}_{F}^{R},s_{D}^{R},F\right)$ and $\tilde{s} _{F}^{R}:=T\left( \tilde{s}_{F}^{R},s_{D}^{R},F\right)$. By (12) and Lemma 7, $\hat{s}^{R},\tilde{s}^{R}$ both belong to $S^{R}\left( k^{\prime }\right)$. They have the desired properties.

By construction, $\left( m,a^{S}\right) \in D_{F}\left( k-1\right)$ for the union F of some conjugate pair succeeding $M_{\phi }\left( j\right)$ with $m_{E}\notin F$. So $F\ne E$ by Lemma 4. Let $F^{\prime }=E$, $\left( m_{F^{\prime }},a_{F^{\prime }}^{S}\right) =\left( \hat{m}_{E},\hat{a}_{E}^{S}\right)$ and $\hat{\sigma } ^{S}=\sigma _{\left( m,a^{S}\right) }^{S}$ where $\sigma _{\left( m,a^{S}\right) }^{S}\in \varDelta S^{S}\left( k-1\right)$ is given by Claim 73. Define $s_{\left( m,a^{S}\right) }^{R}=\hat{s}^{R}$. It is immediate that $s_{\left( m,a^{S}\right) }^{R}$ satisfies property (1). It satisfies property (2) by Lemma 7. $\hfill\square$

I now construct $\varphi _{\left( \hat{m}_{E},\hat{a}_{E}^{S}\right) }\left( \hat{\sigma }^{R}\right)$. Consider $s^{R}\in S^{R}\left( k-1\right)$. If $s^{R}$ is non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$, define $\varphi _{\left( \hat{m}_{E},\hat{a}_{E}^{S}\right) }\left( s^{R}\right)$ to be totally mixed on

$$\begin{aligned} \left\{ T\left( s_{\left( m,a^{S}\right) }^{R},s^{R},M_{\phi }\left( j\right) \right) :\left( m,a^{S}\right) \in S^{S}\left( k-1\right) \cap \left( M_{\phi }\left( j\right) \backslash E\right) \times B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right) \right\} , \end{aligned}$$

(13)

which is a subset of $S^{R}\left( k-1\right)$ by Claim 74. By Claim 72 and Lemma 4, there exists a conjugate pair succeeding $M_{\phi }\left( j\right)$ whose union F does not contain either $\hat{m}_{E}$ or $\hat{m}$. By (11), (13) is not empty. If $s^{R}$ is constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$, define $\varphi _{\left( \hat{m}_{E},\hat{a}_{E}^{S}\right) }\left( s^{R}\right) =s^{R}$. Extend the definition of $\varphi _{\left( \hat{m}_{E},\hat{a} _{E}^{S}\right) }$ to mixed strategies by defining $\varphi _{\left( \hat{m} _{E},\hat{a}_{E}^{S}\right) }\left( \hat{\sigma }^{R}\right)$ to place probability $\hat{\sigma }^{R}\left( s^{R}\right)$ on $\varphi _{\left( \hat{ m}_{E},\hat{a}_{E}^{S}\right) }\left( s^{R}\right)$. Then $\varphi _{\left( \hat{m}_{E},\hat{a}_{E}^{S}\right) }\left( \hat{\sigma }^{R}\right) \in \varDelta S^{R}\left( k-1\right)$.

Claim 75

A best response to $\varphi _{\left( \hat{m}_{E}, \hat{a}_{E}^{S}\right) }\left( \hat{\sigma }^{R}\right)$ must belong to $E\times B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right)$

Proof

For $m\notin M_{\phi }\left( j\right)$ and any $a^{S}\in A^{S}$,

$$\begin{aligned} u^{S}\left( \left( m,a^{S}\right) ,\varphi _{\left( \hat{m}_{E},\hat{a} _{E}^{S}\right) }\left( \hat{\sigma }^{R}\right) \right)& = {} u^{S}\left( \left( m,a^{S}\right) ,\hat{\sigma }^{R}\right) \\\le & {} u^{S}\left( \left( \hat{m},\hat{a}^{S}\right) ,\hat{\sigma }^{R}\right) \\< & {} \chi \left( \hat{a}^{S};\hat{\sigma }^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) \\\le & {} \chi \left( \hat{a}_{E}^{S};\hat{\sigma }^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) . \end{aligned}$$

where the first equality holds because $\varphi _{\left( \hat{m}_{E},\hat{a} _{E}^{S}\right) }\left( s^{R}\right) \left( m\right) = s^{R}\left( m\right)$ for all $s^{R}$ if $m\notin M_{\phi }\left( j\right)$, the first inequality holds because $\left( \hat{m},\hat{a}^{S}\right)$ is a best response to $\hat{\sigma }^{R}$, the second inequality holds because $\hat{ \sigma }^{R}$ puts positive probability on $s_{d}^{R}$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ with $s_{d}^{R}\left( \hat{m} \right) \ne b^{R}\left( \hat{a}^{S}\right)$, and the last inequality holds because $\hat{a}_{E}^{S}\in B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right)$.

Consider $\left( m,a^{S}\right) \in S^{S}\left( k\right)$ where $m\in M_{\phi }\left( j\right)$. Since $S^{S}\left( k\right) \subset S^{S}\left( 1\right)$, by lemma 11, $b^{R}\left( a^{S}\right) \subset A_{\phi ,j}$ and thus $\chi \left( a^{S};\hat{\sigma } ^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right)$ is well-defined. Then

$$\begin{aligned} u^{S}\left( \left( m,a^{S}\right) ,\varphi _{\left( \hat{m}_{E},\hat{a} _{E}^{S}\right) }\left( \hat{\sigma }^{R}\right) \right)\le & {} \chi \left( a^{S};\hat{\sigma }^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) \nonumber \\\le & {} \chi \left( \hat{a}_{E}^{S};\hat{\sigma }^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) \end{aligned}$$

(14)

where the first inequality holds by Lemma 12, and the second inequality holds because $\hat{a}_{E}^{S}\in B_{\phi ,j}^{S}\left( \hat{ \sigma }^{R}\right)$ and strictly so if $a^{S}\notin B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right)$ by definition of $B_{\phi ,j}^{S}\left( \hat{\sigma }^{R}\right)$. Moreover, if $a^{S}\in B_{\phi ,j}^{S}\left( \hat{\sigma } ^{R}\right)$ but $m\in M_{\phi }\left( j\right) \backslash E$, then the first inequality holds strictly because $\varphi _{\left( \hat{m}_{E},\hat{a} _{E}^{S}\right) }\left( \hat{\sigma }^{R}\right)$ puts probability $\hat{ \sigma }^{R}\left( s_{d}^{R}\right) >0$ on $T\left( s_{\left( m,a^{S}\right) }^{R},s_{d}^{R},M_{\phi }\left( j\right) \right)$, which is non-constant on $M_{\phi }\left( j\right)$ and responds to m NOT with $b^{R}\left( a^{S}\right)$. Lastly,

$$\begin{aligned} \chi \left( \hat{a}_{E}^{S};\hat{\sigma }^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) =u^{S}\left( \left( \hat{m}_{E},\hat{a}_{E}^{S}\right) ,\varphi _{\left( \hat{m}_{E},\hat{a}_{E}^{S}\right) }\left( \hat{\sigma } ^{R}\right) \right) \end{aligned}$$

because $\varphi _{\left( \hat{m}_{E},\hat{a}_{E}^{S}\right) }\left( s^{R}\right) \left( \hat{m}_{E}\right) =b^{R}\left( \hat{a}_{E}^{S}\right)$ for all $s^{R}$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$. $\hfill\square$

Since $\varphi _{\left( \hat{m}_{E},\hat{a}_{E}^{S}\right) }\left( \hat{ \sigma }^{R}\right) \in \varDelta S^{R}\left( k-1\right)$, at least one best response to $\varphi _{\left( \hat{m}_{E},\hat{a}_{E}^{S}\right) }\left( \hat{\sigma }^{R}\right)$ must belong to $S^{S}\left( k\right)$. Part 2 then follows by Claim 75. Let G be the union of a conjugate pair succeeding $M_{\phi }\left( j\right)$. Then $D_{G}\left( k\right) \ne \emptyset$, by part 2 if $\hat{m}\notin G$ and by definition of $D_{G}\left( k\right)$ if $\hat{m}\in G$ because $\left( \hat{ m},\hat{a}^{S}\right) \in S^{S}\left( k+1\right)$. Then Part 3 holds by Claim 74.

1.2.3 Existence of non-Constant Receiver Strategies for the case of $j \leq N-3$

Lemma 16

Suppose there exists $\left( \hat{m},\hat{a}^{S}\right) ,\left( \tilde{m},\tilde{a}^{S}\right) \in S^{S}\left( k+1\right)$ where $\hat{m},\tilde{m}\in M_{\phi }\left( j\right)$ for some $j\le N-3$, $b^{R}\left( \hat{a}^{S}\right) \ne b^{R}\left( \tilde{a} ^{S}\right)$ and $X\left( \hat{m},\hat{a}^{S}\right) \cap X\left( \tilde{m}, \tilde{a}^{S}\right) \ne \emptyset$, and there exists $s_{d}^{R}\in S^{R}\left( k-1\right)$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ that responds to $\hat{m}$ NOT with $b^{R}\left( \hat{a} ^{S}\right)$. Then there exists $\hat{s}_{d}^{R}\in S^{R}\left( k\right)$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ that responds to $\hat{m}$ NOT with $b^{R}\left( \hat{a}^{S}\right)$.

Proof

Since $\hat{m},\tilde{m}\in M_{\phi }\left( j\right)$, and $\left( \hat{m}, \hat{a}^{S}\right) ,\left( \tilde{m},\tilde{a}^{S}\right) \in S^{S}\left( k+1\right) \subset S^{S}\left( 1\right)$, by Lemma 11, $b^{R}\left( \hat{a}^{S}\right) ,b^{R}\left( \tilde{a} ^{S}\right) \in A_{\phi ,j}$. Then $X\left( \hat{m},\hat{a}^{S}\right)$ and $X\left( \tilde{m},\tilde{a}^{S}\right)$ are both contained in the union of the same conjugate pair succeeding $M_{\phi }\left( j\right)$, denoted by $F_{1}$, because the union of one conjugate pair succeeding $M_{\phi }\left( j\right)$ is disjoint from that of another.

Suppose there exists $\left( m_{2},a_{2}^{S}\right) \in S^{S}\left( k-1\right)$ where $m_{2}\in M_{\phi }\left( j\right) \backslash F_{1}$ and $b^{R}\left( a_{2}^{S}\right) \in A_{\phi }\left( j\right) \backslash \left\{ b^{R}\left( \tilde{a}^{S}\right) \right\}$. Then, $X\left( m_{2},a_{2}^{S}\right) \subset M_{\phi }\left( j\right) \backslash F_{1}$ while $X\left( \tilde{m},\tilde{a}^{S}\right) \subset F_{1}$. Define $\left( m_{1},a_{1}^{S}\right) =\left( \tilde{m},\tilde{a}^{S}\right)$ and $\sigma _{12}^{S}=\frac{1}{2}\left( m_{1},a_{1}^{S}\right) +\frac{1}{2}\left( m_{2},a_{2}^{S}\right)$. Then $X\left( m_{1},a_{1}^{S}\right) \cap X\left( m_{2},a_{2}^{S}\right) =\emptyset$. By Lemma 9, every best response to $\sigma _{12}^{S}$ must respond to $m_{1}$ with $b^{R}\left( a_{1}^{S}\right)$ and $m_{2}$ with $b^{R}\left( a_{2}^{S}\right)$, and thus is non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ since $b^{R}\left( a_{1}^{S}\right) \ne b^{R}\left( a_{2}^{S}\right)$. By construction, $\sigma _{12}^{S}\in \varDelta S^{S}\left( k-1\right)$. Thus some strategy $s_{12}^{R}\in S^{R}\left( k\right)$ must be a best response to $\sigma _{12}^{S}$. Because $X\left( \hat{m},\hat{a}^{S}\right) \cap X\left( \tilde{m},\tilde{a}^{S}\right) \ne \emptyset$ and because $s_{12}^{R}$ responds to $m_{1}=\tilde{m}$ with $b^{R}\left( a_{1}^{S}\right) =b^{R}\left( \tilde{a}^{S}\right) \ne b^{R}\left( \hat{a}^{S}\right)$, $s_{12}^{R}$ responds to $\hat{m}$ NOT with $b^{R}\left( \hat{a}^{S}\right)$. We are done.

Otherwise,

$$\begin{aligned} b^{R}\left( a^{S}\right) =b^{R}\left( \tilde{a}^{S}\right) \text { for every } \left( m,a^{S}\right) \in S^{S}\left( k-1\right) \cap \left( M_{\phi }\left( j\right) \backslash F_{1}\right) \times A_{\phi }\left( j\right) . \end{aligned}$$

(15)

I will show that this is not possible. By applying Lemma 15 to $\left( \hat{m},\hat{a}^{S}\right)$, there exists $\left( m_{2},a_{2}^{S}\right) \in S^{S}\left( k-1\right)$ where $m_{2}\in M_{\phi }\left( j\right) \backslash F_{1}$ and $b^{R}\left( a_{2}^{S}\right) \in A_{\phi ,j}$. By (15), $b^{R}\left( a_{2}^{S}\right) =b^{R}\left( \tilde{a}^{S}\right)$. Define $\left( m_{1},a_{1}^{S}\right) =\left( \hat{m},\hat{a}_{1}^{S}\right)$. Using the same arguments, we can show that some $s_{12}^{R}\in S^{R}\left( k\right)$ must respond to $m_{1}$ with $b^{R}\left( a_{1}^{S}\right)$ and to $m_{2}$ with $b^{R}\left( a_{2}^{S}\right)$. Since $b^{R}\left( a_{1}^{S}\right) \ne b^{R}\left( a_{2}^{S}\right)$ by construction, $s_{12}^{R}$ must be non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$. Since $X\left( \hat{m},\hat{a} ^{S}\right) \cap X\left( \tilde{m},\tilde{a}^{S}\right) \ne \emptyset$ and since $s_{12}^{R}$ is a best response to $\left( m_{1},a_{1}^{S}\right) =\left( \hat{m},\hat{a}_{1}^{S}\right)$, $s_{12}^{R}\left( \tilde{m}\right) \ne b^{R}\left( \tilde{a}^{S}\right)$. Because $\left( \tilde{m},\tilde{a} ^{S}\right) \in S^{S}\left( k+1\right)$ and $s_{12}^{R}\in S^{R}\left( k\right)$, by applying Lemma 15 to $\left( \tilde{m}, \tilde{a}^{S}\right)$, we can show that there exists $\left( \tilde{m}_{1}, \tilde{a}_{1}^{S}\right) \in S^{S}\left( k-1\right)$ where $\tilde{m} _{1}\in M_{\phi }\left( j\right) \backslash F_{1}$ and $\tilde{a}_{1}^{S}\in B_{\phi ,j}^{S}\left( \tilde{\sigma }^{R}\right)$ where $\left( \tilde{m}, \tilde{a}^{S}\right)$ is a best response to $\tilde{\sigma }^{R}\in \varDelta ^{+}S^{R}\left( k-1\right)$. Since $\left( \tilde{m},\tilde{a}^{S}\right) \in S^{S}\left( k+1\right)$, $s_{12}^{R}\in S^{R}\left( k\right)$ is non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ and responds to $\tilde{m}$ NOT with $b^{R}\left( \tilde{a}^{S}\right)$, and $\left( \tilde{m }_{1},\tilde{a}_{1}^{S}\right) \in S^{S}\left( k-1\right) \cap \left( M_{\phi }\left( j\right) \backslash F_{1}\right) \times B_{\phi ,j}^{S}\left( \tilde{\sigma }^{R}\right)$, (15) contradicts with Lemma 13. $\hfill\square$

Lemma 17

Consider $j\le N-3$. Suppose there exists $\left( m_{1},a_{1}^{S}\right) ,\left( m_{2},a_{2}^{S}\right) \in S^{S}\left( n+1\right)$ with $m_{1},m_{2}\in m_{\phi ,j}\cup M_{\phi }\left( j\right)$, $b^{R}\left( a_{1}^{S}\right) \ne b^{R}\left( a_{2}^{S}\right)$, and $X\left( m_{2},a_{2}^{S}\right) \subset X\left( m_{1},a_{1}^{S}\right)$. Then there exists $s_{d}^{R}\in S^{R}\left( n\right)$ which is non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ with $s_{d}^{R}\left( m_{1}\right) \ne b^{R}\left( a_{1}^{S}\right)$.

Proof

We first show that

$$\begin{aligned} X\left( m_{1},a_{1}^{S}\right) \ne \left\{ m_{1}\right\} . \end{aligned}$$

(16)

Otherwise, because $X\left( m_{2},a_{2}^{S}\right) \subset X\left( m_{1},a_{1}^{S}\right)$, we have $m_{2}=m_{1}$ and $b^{R}\left( a_{2}^{S}\right) =b^{R}\left( a_{1}^{S}\right)$. Contradiction.

Therefore, the lemma holds for $n=0$ because Literal follows every conjugate pair and thus $Literal\left( m_{1}\right) \ne b^{R}\left( a_{1}^{S}\right)$.

Suppose the lemma holds for $n=0,1,\ldots ,k-1$ for all $j\le N-3$. By the induction hypothesis,

$$\begin{aligned} \exists s_{d}^{R}\in S^{R}\left( k-1\right) \text { non-constant on }m_{\phi ,j}\cup M_{\phi }\left( j\right) \text { with }s_{d}^{R}\left( m_{1}\right) \ne b^{R}\left( a_{1}^{S}\right) . \end{aligned}$$

(17)

Because $\left( m_{1},a_{1}^{S}\right) \in S^{S}\left( k+1\right)$, it is a best response to some $\sigma _{1}^{R}\in \varDelta ^{+}S^{R}\left( k-1\right)$ . Then, by (17) and Lemma 15,

$$\begin{aligned} \exists \left( \tilde{m}_{1},\tilde{a}_{1}^{S}\right) \in S^{S}\left( k\right) \text { where }\tilde{m}_{1}\in M_{\phi }\left( j\right) \text { and } \tilde{a}_{1}^{S}\in B_{\phi ,j}\left( \sigma _{1}^{R}\right) \subset A_{\phi }\left( j\right) . \end{aligned}$$

(18)

Suppose $m_{1}\in M_{\phi }\left( j\right)$. Since $\left( m_{1},a_{1}^{S}\right) \subset S^{S}\left( k+1\right) \subset S^{S}\left( 1\right)$, $b^{R}\left( a_{1}^{S}\right) \in A_{\phi ,j}$. Thus, $X\left( m_{2},a_{1}^{S}\right) \subset X\left( m_{1},a_{1}^{S}\right) \subset M_{\phi }\left( j\right)$. This lemma then follows from Lemma 16 by (17).

Suppose $m_{1}=m_{\phi ,j}$. By (16)

$$\begin{aligned} b^{R}\left( a_{1}^{S}\right) \ne \phi \left( j\right) . \end{aligned}$$

(19)

It suffices to show existence of $\tilde{s}_{d}^{R}\in S^{R}\left( k\right)$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$, because in that case, $\tilde{s}_{d}^{R}\left( m_{\phi ,j}\right) =\phi \left( j\right)$ by Definitions 3 and 4 and thus $\tilde{s}_{d}^{R}\left( m_{\phi ,j}\right) \ne b^{R}\left( a_{1}^{S}\right)$ by (19).

There are three cases.

Case 1

Suppose there exists $\left( \hat{m},\hat{a}^{S}\right) ,\left( \tilde{m},\tilde{a}^{S}\right) \in S^{S}\left( k+1\right)$ where $\hat{m},\tilde{m}\in M_{\phi }\left( j\right)$ and $b^{R}\left( \hat{a} ^{S}\right) \ne b^{R}\left( \tilde{a}^{S}\right)$. If $X\left( \hat{m}, \hat{a}^{S}\right) \cap X\left( \tilde{m},\tilde{a}^{S}\right) =\emptyset$, then by Lemma 9, a best response to $\sigma _{d}^{S}=\frac{1}{2}\left( \hat{m},\hat{a}^{S}\right) + \frac{1}{2}\left( \tilde{m},\tilde{a}^{S}\right)$ must respond to $\hat{m}$ with $b^{R}\left( \hat{a}^{S}\right)$ and to $\tilde{m}$ with $b^{R}\left( \tilde{a}^{S}\right)$, and thus is non-constant on $M_{\phi }\left( j\right)$. Because $\sigma _{d}^{S}\in \varDelta S^{S}\left( k-1\right)$, at least one such best response belongs to $S^{R}\left( k\right)$. Otherwise, by Lemma 16, there exists $\tilde{s} _{d}^{R}\in S^{R}\left( k\right)$ which is non-constant on $M_{\phi }\left( j\right)$.

Case 2

Suppose there exists $\left( m_{\phi ,j}a_{\phi ,j}^{S}\right) \in S^{S}\left( k-1\right)$ where $b^{R}\left( a_{\phi ,j}^{S}\right) =\phi \left( j\right)$. Define $\sigma _{d}^{S}=\frac{1}{2} \left( m_{\phi ,j},a_{\phi ,j}^{S}\right) +\frac{1}{2}\left( \tilde{m}_{1}, \tilde{a}_{1}^{S}\right) \in \varDelta S^{S}\left( k-1\right)$. Then some $\tilde{s}_{d}^{R}\in S^{R}\left( k\right)$ must be a best response to $\sigma _{d}^{S}$. Because $b^{R}\left( \tilde{a}_{1}^{S}\right) \in A_{\phi }\left( j\right)$ by (18), $X\left( \tilde{m}_{1},\tilde{a} _{1}^{S}\right) \subset M_{\phi }\left( j\right)$. Then $X\left( m_{\phi ,j},a_{\phi ,j}^{S}\right) =\left\{ m_{\phi ,j}\right\}$ is disjoint from $X\left( \tilde{m}_{1},\tilde{a}_{1}^{S}\right)$. By Lemma 9, $\tilde{s}_{d}^{R}$ must respond to $m_{\phi ,j}$ with $\phi \left( j\right)$ and to $\tilde{m}_{1}$ with $b^{R}\left( \tilde{a}_{1}^{S}\right) \in A_{\phi ,j}$, and thus is non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$.

Case 3

Suppose Cases 1 and 2 do not hold. Assume to the contrary that every $s^{R}\in S^{R}\left( k\right)$ is constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$.

We first show that we can w.l.o.g. assume $b^{R}\left( a_{1}^{S}\right) \ne b^{R}\left( \tilde{a}_{1}^{S}\right)$. If $m_{2}\in M_{\phi }\left( j\right)$, then $b^{R}\left( \tilde{a}_{1}^{S}\right) =b^{R}\left( a_{2}^{S}\right) \ne b^{R}\left( a_{1}^{S}\right)$. Otherwise, $m_{2}=m_{\phi ,j}$. Since $\left( m_{1},a_{1}^{S}\right) ,\left( m_{2},a_{2}^{S}\right) \in S^{S}\left( k+1\right) \subset S^{S}\left( 1\right)$, by Lemma 11, $b^{R}\left( a_{1}^{S}\right) ,b^{R}\left( a_{2}^{S}\right) \in A_{\phi ,j-1}$. Because Case 2 does not hold and by (19), $b^{R}\left( a_{1}^{S}\right) ,b^{R}\left( a_{2}^{S}\right) \in A_{\phi }\left( j\right)$. So $X\left( m_{1},a_{1}^{S}\right) =m_{\phi ,j}\cup M_{\phi }\left( j\right) =X\left( m_{2},a_{2}^{S}\right)$. Because $b^{R}\left( a_{1}^{S}\right) \ne b^{R}\left( a_{2}^{S}\right)$, either $b^{R}\left( a_{1}^{S}\right) \ne b^{R}\left( \tilde{a}_{1}^{S}\right)$ or $b^{R}\left( a_{2}^{S}\right) \ne b^{R}\left( \tilde{a}_{1}^{S}\right)$. By symmetry between $\left( m_{1},a_{1}^{S}\right)$ and $\left( m_{2},a_{2}^{S}\right)$, assume w.l.o.g. $b^{R}\left( a_{1}^{S}\right) \ne b^{R}\left( \tilde{a}_{1}^{S}\right)$.

Because $m_{\phi ,j}$ and $M_{\phi }\left( j\right)$ forms a conjugate pair succeeding $M_{\phi }\left( j-1\right)$, the condition for Case 1 holds if j is replaced with $j-1$. Thus there exists $\tilde{s} _{d,j-1}^{R}\in S^{R}\left( k\right)$ non-constant on $m_{\phi ,j-1}\cup M_{\phi }\left( j-1\right)$ with $\tilde{s}_{d,j-1}^{R}\left( m_{1}\right) \ne b^{R}\left( a_{1}^{S}\right)$. By Lemma 15 and because $\left( m_{1},a_{1}^{S}\right) ,\left( \tilde{m}_{1},\tilde{a} _{1}^{S}\right) \in S^{S}\left( k\right) \subset S^{S}\left( k-1\right)$, there exists $s_{1}^{R},\tilde{s}_{1}^{R}\in S^{R}\left( k\right)$ which respond in the same way outside of $m_{\phi ,j}\cup M_{\phi }\left( j\right)$, but $s_{1}^{R}\left( m_{1}\right) =b^{R}\left( a_{1}^{S}\right)$, while $\tilde{s}_{1}^{R}\left( \tilde{m}_{1}\right) =b^{R}\left( \tilde{a} _{1}^{S}\right)$.

Define $\tilde{\sigma }^{R}_1$ to place the same probability as $\sigma _{1}^{R}\in \varDelta ^{+}S^{R}\left( k\right)$ on all $s^{R}\ne s_{1}^{R}, \tilde{s}_{1}^{R}$, to place probability $\sigma _{1}^{R}\left( s_{1}^{R}\right) +\varepsilon$ on $s_{1}^{R}$, and probability $\sigma _{1}^{R}\left( \tilde{s}_{1}^{R}\right) -\varepsilon$ on $\tilde{s}_{1}^{R}$ . For $\varepsilon >0$ sufficiently small, $\tilde{\sigma }_{1}^{R}\in \varDelta ^{+}S^{R}\left( k\right)$. Then,

$$\begin{aligned} \text {for }m\notin m_{\phi ,j}\cup M_{\phi }\left( j\right) \text {, } u^{S}\left( \left( m,a^{S}\right) ,\tilde{\sigma }_{1}^{R}\right) =u^{R}\left( \left( m,a^{S}\right) ,\sigma _{1}^{R}\right) , \end{aligned}$$

(20)

while for $m\in M_{\phi }\left( j\right)$,

$$\begin{aligned}&u^{S}\left( \left( m,a^{S}\right) ,\tilde{\sigma }_{1}^{R}\right) \end{aligned}$$

(21a)

$$\begin{aligned}&\quad =u^{S}\left( \left( m,a^{S}\right) ,\sigma _{1}^{R}\right) +\varepsilon \left( g^{S}\left( a^{S},b^{R}\left( a_{1}^{S}\right) \right) -g^{S}\left( a^{S},b^{R}\left( \tilde{a}_{1}^{S}\right) \right) \right) \end{aligned}$$

(21b)

$$\begin{aligned}&\quad < u^{S}\left( \left( m,a^{S}\right) ,\sigma _{1}^{R}\right) \end{aligned}$$

(21c)

where the equality holds by the hypothesis that every $s^{R}\in S^{R}\left( k\right)$ is constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ and the inequality holds because $b^{R}\left( a^{S}\right) =b^{R}\left( \tilde{a} ^{S}_1\right)$ for all $\left( m,a^{S}\right) \in S^{S}\left( k-1\right)$ where $m\in M_{\phi }\left( j\right)$ and by BM self-signaling. And

$$\begin{aligned}&u^{S}\left( \left( m_{1},a_{1}^{S}\right) ,\tilde{\sigma }_{1}^{R}\right) \end{aligned}$$

(22a)

$$\begin{aligned}&\quad =u^{S}\left( \left( m_{1},a_{1}^{S}\right) ,\sigma _{1}^{R}\right) +\varepsilon \left( g^{S}\left( a_{1}^{S},b^{R}\left( a_{1}^{S}\right) \right) -g^{S}\left( a_{1}^{S},b^{R}\left( \tilde{a}_{1}^{S}\right) \right) \right) \end{aligned}$$

(22b)

$$\begin{aligned}&\quad > u^{S}\left( \left( m_{1},a_{1}^{S}\right) ,\sigma _{1}^{R}\right) \ge u^{S}\left( \left( m,a^{S}\right) ,\sigma _{1}^{R}\right) \end{aligned}$$

(22c)

for all $\left( m,a^{S}\right) \in S^{S}$ where the equality holds by the hypothesis that every $s^{R}\in S^{R}\left( k\right)$ is constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$, the first inequality holds by BM self-signaling, and the last inequality holds because $\left( m_{1},a_{1}^{S}\right)$ is a best response to $\sigma _{1}^{R}$.

Let $\left( m_{*},a_{*}^{S}\right)$ be a best response $\tilde{ \sigma }_{1}^{R}$. Then $\left( m_{*},a_{*}^{S}\right) \in S^{S}\left( k+1\right)$ because $\tilde{\sigma }_{1}^{R}\in \varDelta ^{+}S^{R}\left( k\right)$. By (20), (21) and ( 22), $m_{*}=m_{\phi ,j}$. Because case 2 does not hold, $b^{R}\left( a_{*}^{S}\right) \ne \phi \left( j\right)$. Since $s_{d}^{R}$ is non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$, $s_{d}^{R}\left( m_{*}\right) =s_{d}^{R}\left( m_{\phi ,j}\right) =\phi \left( j\right) \ne b^{R}\left( a_{*}^{S}\right)$. By Lemma 15, there exists $\left( \hat{m}_{1},\hat{a} _{1}^{S}\right) \in S^{S}\left( k\right)$ where $\hat{m}_{1}\in M_{\phi }\left( j\right)$ and $\hat{a}_{1}^{S}\in B_{\phi ,j}^{S}\left( \tilde{ \sigma }_{1}^{R}\right)$. Then

$$\begin{aligned} u^{S}\left( \left( \hat{m}_{1},\hat{a}_{1}^{S}\right) ,\tilde{\sigma } ^{R}\right)& = {} \chi \left( \hat{a}_{1}^{S};\tilde{\sigma }_{1}^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) \\& \ge {} \chi \left( a_{*}^{S}; \tilde{\sigma }_{1}^{R},m_{\phi ,j}\cup M_{\phi }\left( j\right) \right) = u^{S}\left( \left( m_{*},a_{*}^{S}\right) ,\tilde{\sigma } _{1}^{R}\right) \end{aligned}$$

where the two equalities holds by Lemma 12 and the hypothesis that every $s^{R}\in S^{R}\left( k\right)$ is constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$ and the inequality holds because $\hat{a} _{1}^{S}\in B_{\phi ,j}^{S}\left( \tilde{\sigma }_{1}^{R}\right)$. Since $\left( m_{*},a_{*}^{S}\right)$ is a best response to $\tilde{\sigma } _{1}^{R}$, $\left( \hat{m}_{1},\hat{a}_{1}^{S}\right)$ is a best response to $\tilde{\sigma }_{1}^{R}$, contradiction to (21) and (22) because $\hat{m}_{1}\in M_{\phi }\left( j\right)$. So there exists $\tilde{s}_{d}^{R}\in S^{R}\left( k\right)$ non-constant on $m_{\phi ,j}\cup M_{\phi }\left( j\right)$. $\hfill\square$

1.2.4 When commitment payoff is guaranteed

I first show when Sender gets her commitment payoff from her action.

Claim 76

Suppose, for any $\left( \hat{m},\hat{a} ^{S}\right) ,\left( \tilde{m},\tilde{a}^{S}\right) \in S^{S}\left( 2n-1\right)$ where $\hat{m},\tilde{m}\in m_{\phi ,N-n}\cup M_{\phi }\left( N-n\right)$, either $X\left( \hat{m},\hat{a}^{S}\right) \cap X\left( \tilde{ m},\tilde{a}^{S}\right) =\emptyset$ or $b^{R}\left( \hat{a}^{S}\right) =b^{R}\left( \tilde{a}^{S}\right)$. Then $s^{R}\left( m\right) =b^{R}\left( a^{S}\right)$ for any $\left( \left( m,a^{S}\right) ,s^{R}\right) \in S\left( 2n\right)$ where $m\in m_{\phi ,N-n}\cup M_{\phi }\left( N-n\right)$ and $s^{R}$ non-constant on $m_{\phi ,N-n-1}\cup M_{\phi }\left( N-n-1\right)$

Proof

Let $X^{S}=S^{S}\left( 2n-1\right) \cap \left( m_{\phi ,N-n}\cup M_{\phi }\left( N-n\right) \right) \times A^{S}$. By Lemma 9, there exists $s_{*}^{R}$ that is a best response to every $\left( m,a^{S}\right) \in X^{S}$. Because $S^{S}\left( 2n-1\right) \subset S^{S}\left( 1\right)$, by Lemma 11, $b^{R}\left( a^{S}\right) \in A_{\phi ,N-n-1}$ for all $\left( m,a^{S}\right) \in X^{S}$. So $s_{*}^{R}$ is language-based on $m_{\phi ,N-n}\cup M_{\phi }\left( N-n\right)$. Consider any $s^{R}\in S^{R}\left( 2n\right)$ non-constant on $m_{\phi ,N-n-1}\cup M_{\phi }\left( N-n-1\right)$. By Lemma 6, $T\left( s_{*}^{R},s^{R},m_{\phi ,N-n}\cup M_{\phi }\left( N-n\right) \right)$ is a language-based response. Consider any $\left( m,a^{S}\right) \in S^{S}\left( 2n-1\right)$. For $m\notin m_{\phi ,N-n}\cup M_{\phi }\left( N-n\right)$,

$$\begin{aligned} u^{R}\left( \left( m,a^{S}\right) ,T\left( s_{*}^{R},s^{R},m_{\phi ,N-n}\cup M_{\phi }\left( N-n\right) \right) \right)& =g^{R}\left( a^{S},s^{R}\left( m\right) \right)\\& =u^{R}\left( \left( m,a^{S}\right) ,s^{R}\right) . \end{aligned}$$

For $m\in m_{\phi ,N-n}\cup M_{\phi }\left( N-n\right)$,

$$\begin{aligned}&u^{R}\left( \left( m,a^{S}\right) ,T\left( s_{*}^{R},s^{R},m_{\phi ,N-n}\cup M_{\phi }\left( N-n\right) \right) \right) \\&\quad =g^{R}\left( a^{S},s_{*}^{R}\left( m\right) \right) =g^{R}\left( a^{S},b^{R}\left( a^{S}\right) \right) \ge g^{R}\left( a^{S},s^{R}\left( m\right) \right) =u^{R}\left( \left( m,a^{S}\right) ,s^{R}\right) . \end{aligned}$$

By BM self-signaling, the inequality holds and strictly so if $s^{R}\left( m\right) \ne b^{R}\left( a^{S}\right)$. If $s^{R}$ is non-constant on $m_{\phi ,N-n-1}\cup M_{\phi }\left( N-n-1\right)$ and $s^{R}\left( m\right) \ne b^{R}\left( a^{S}\right)$ for some $\left( m,a^{S}\right) \in X^{S}$, then $s^{R}\notin \left( 2n\right)$. $\hfill\square$

By seeing the entire message space as the union of the conjugate pair $m_{\phi ,0}\cup M_{\phi ,0}$ where $m_{\phi ,0}=\emptyset$ and $M_{\phi ,0}=M$, Lemma 18 establishes that every strategy profile in $S\left( 2n\right)$ attains coordination.

Lemma 18

$\forall n=1,\ldots ,N$, for every permutation $\phi$, for every $\left( \left( m,a^{S}\right) ,s^{R}\right) \in S\left( 2n\right)$ where $m\in m_{\phi ,N-n}\cup M_{\phi }\left( N-n\right)$ and $s^{R}$ is non-constant on $m_{\phi ,N-n-1}\cup M_{\phi }\left( N-n-1\right)$, Sender gets her commitment payoff in that $s^{R}\left( m\right) =b^{R}\left( a^{S}\right)$.

Proof

For $n=1$, it follows by Corollary 2 and Claim 76.

Suppose the lemma holds for $n=1,\ldots ,k-1$, but not for $n=k$. By Claim 76, there exists $\left( m_{1},a_{1}^{S}\right) ,\left( m_{2},a_{2}^{S}\right) \in S^{S}\left( 2k-1\right)$ with $m_{1},m_{2}\in m_{\phi ,N-k}\cup M_{\phi }\left( N-k\right)$ where $X\left( m_{2},a_{2}^{S}\right) \subset X\left( m_{1},a_{1}^{S}\right)$ and $b^{R}\left( a_{1}^{S}\right) \ne b^{R}\left( a_{2}^{S}\right)$. Then there exists $\hat{\sigma }^{R}\in \varDelta ^{+}S^{R}\left( 2k-2\right)$ to which $\left( m_{1},a_{1}^{S}\right)$ is a best response. By lemma 17, there exists $s_{1}^{R}\in S^{R}\left( 2k-2\right)$ non-constant on $m_{\phi ,N-k-1}\cup M_{\phi }\left( N-k-1\right)$ where $s_{1}^{R}\left( m_{1}\right) \ne b^{R}\left( a_{1}^{S}\right)$. Define $E=m_{\phi ,N-k+1}\cup M_{\phi }\left( N-k+1\right)$. By the induction hypothesis and Lemma 17, $m_{1},m_{2}\notin E$. By Lemma 15, there exists $\left( m_{E},a_{E}^{S}\right) \in S^{S}\left( 2k-2\right)$ where $m_{E}\in E$ and $b^{R}\left( a_{E}^{S}\right) \in B_{\phi ,N-k}^{S}\left( \hat{\sigma }^{R}\right)$. By the induction hypothesis, for every $s^{R}$ non-constant on $m_{\phi ,N-k}\cup M_{\phi }\left( N-k\right)$, $s^{R}\left( m_{E}\right) =b^{R}\left( a_{E}^{S}\right)$. Contradiction to Lemma 13. So the lemma here must hold for $n=k$. $\hfill\square$

The following lemma says Sender strategy using a Stackelberg action can only be weakly dominated by another one using a Stackelberg action.

Lemma 19

Suppose $a_{1}^{S}$ is a Stackelberg action for Sender. If $\left( m_{1},a_{1}^{S}\right) \notin S^{S}\left( n\right)$ for some $n\in \mathbb {N}$, then it must be weakly dominated by some $\left( m_{*},a_{*}^{S}\right) \in S^{S}\left( k+1\right)$ w.r.t. $S^{R}\left( k\right)$ for some $k\le n-1$ where $a_{*}^{S}$ is a Stackelberg action for Sender.

Proof

Since $\left( m_{1},a_{1}^{S}\right) \in S^{S}\left( 0\right)$, there exists $k\in \left\{ 0,\ldots ,n-1\right\}$ such that $\left( m_{1},a_{1}^{S}\right) \in S^{S}\left( k\right)$ is weakly dominated by some $\left( m_{*},a_{*}^{S}\right)$ w.r.t. $S^{R}\left( k\right)$ . Then, at least one Receiver best response to $\left( m_{1},a_{1}^{S}\right)$ belongs to $S^{R}\left( k+1\right)$, say $s_{B1}^{R}$, and it must respond to $m_{1}$ with action $b^{R}\left( a_{1}^{S}\right)$. Then

$$\begin{aligned} g^{S}\left( a_{*}^{S},s_{B1}^{R}\left( m_{*}\right) \right) =u^{S}\left( \left( m_{*},a_{*}^{S}\right) ,s_{B1}^{R}\right) \ge u^{S}\left( \left( m_{1},a_{1}^{S}\right) ,s_{B1}^{R}\right) =g^{S}\left( a_{1}^{S},b^{R}\left( a_{1}^{S}\right) \right) \end{aligned}$$

where the inequality holds because $\left( m_{*},a_{*}^{S}\right)$ weakly dominates $\left( m_{1},a_{1}^{S}\right)$ w.r.t. $S^{R}\left( k\right)$ and $s_{B1}^{R}\in S^{R}\left( k+1\right) \subset S^{R}\left( k\right)$. By Lemma 1, Sender’s Stackelberg payoff is her highest stage game payoff. Thus, every Sender best response to $s_{B1}^{R}$ must use a Stackelberg action. Since $s_{B1}^{R}\in S^{R}\left( k\right)$, at least one Sender strategy taking a Stackelberg action belongs to $S^{R}\left( k+1\right)$. $\hfill\square$

Now I can finish the proof for Proposition 2.

Proof

(of Proposition 2) By Lemma 19, there exists $\left( m_{*},a_{*}^{S}\right) \in S^{S}\left( 2N\right)$ where $a_{*}^{S}$ is a Stackelberg action for Sender. By Lemma 18, $s^{R}\left( m_{*}\right) =b^{R}\left( a_{*}^{S}\right)$ for all $s^{R}\in S^{R}\left( 2N\right)$. It follows that Sender must obtain her Stackelberg payoff in every strategy profile in $S\left( 2N+1\right)$. $\hfill\square$

Proof

(of Claim 47) Because x is the unique Stackelberg action for Sender, by Lemma 19, there exists $k\in \left\{ 0,\ldots ,n-1\right\}$ such that $\left( {\text{``}} \left\{ X\right\} {\text{''}},x\right) \in S^{S}\left( k\right)$ is weakly dominated by some $\left( m_{*},x\right)$ w.r.t. $S^{R}\left( k\right)$. Since Claim 41 holds for non-generic stage games as well, $m^{*}\notin M\left( \left\{ Y,Z\right\} \right)$. Then $m^{*}\notin E_{X}$ because $\left( m^{*},a_{*}^{S}\right) \ne \left( {\text{``}}\left\{ X\right\} {\text{''}},x\right)$.

Consider any $s^{R}\in S^{R}\left( k\right)$. By Lemma 7, $T\left( s_{X}^{R},s^{R},E_{X}\right) \in S^{R}\left( k\right)$. Then,

$$\begin{aligned} g^{S}\left( a_{*}^{S},s^{R}\left( m_{*}\right) \right)& = {} u^{S}\left( \left( m^{*},a_{*}^{S}\right) ,T\left( s_{X}^{R},s^{R},E_{X}\right) \right) \\& \ge {} u^{S}\left( \left( {\text{``}}\left\{ X\right\} {{\text{''}} },x\right) ,T\left( s_{X}^{R},s^{R},E_{X}\right) \right) \\& = {} g^{S}\left( x,s_{X}^{R}\left( {\text{``}}\left\{ X\right\} {\text{''}}\right) \right) =g^{S}\left( x,X\right) . \end{aligned}$$

Thus, $s^{R}\left( m_{*}\right) =X$ for all $s^{R}\in S^{R}\left( k\right)$. $\hfill\square$

Appendix 4: Omitted proofs in Sect. 4.2

1.1 Proof of Proposition 3

Proof

Say that $a_{1}^{R}\ge a_{2}^{R}$ if $g^{S}\left( a^{S},a_{1}^{R}\right) \ge g^{S}\left( a^{S},a_{2}^{R}\right)$. The definition does not depend on $a^{S}$ by the assumption on g. Define $M^{*}$ to be the set of messages $m={\text{``}}A_{1}\ldots A_{n}$” (for any integer $n\in \left[ 1,N\right]$) such that, for every $j=1,\ldots ,n$, there exists $\bar{a}_{j}\left( m\right) \in A_{j}$ and $\underline{a}_{j}\left( m\right) \in A_{j-1}\backslash A_{j}$ where $\bar{a}_{j}\left( m\right) > \underline{a}_{j}\left( m\right)$. I will show that $S^{S}\left( 1\right) =M^{*}\times A^{S}$.

I first show that $M^{*}\times A^{S}\subset S^{S}\left( 1\right)$. Fix $\hat{m}={\text{``}}\hat{A}_{1}\ldots \hat{A}_{n}{{\text{''}} }\in M^{*}$. For $k=1,\ldots ,n$, iteratively define $s_{ \hat{m},k}^{R}$ to respond with action $\bar{a}_{k}\left( \hat{m}\right)$ to messages in $M\left( \hat{A}_{1}\ldots \hat{A}_{k-1}\hat{A}_{k}\right)$, to respond with action $\underline{a}_{k}\left( \hat{m}\right)$ to the remaining messages in $M\left( \hat{A}_{1}\ldots \hat{A}_{k-1}\right)$, and to respond the same way as $s_{\hat{m},k-1}^{R}$ on all other messages. Given the conjecture $s_{\hat{m},1}^{R}$, the Sender prefers messages in $M\left( \hat{A}_{1}\right)$ to those outside. For $k=1,\ldots ,n-1$, within $M\left( \hat{A}_{1}\ldots \hat{A}_{k}\right)$, the conjecture $s_{\hat{m},k+1}^{R}$ renders messages in $M\left( \hat{A}_{1}\ldots \hat{A}_{k}\hat{A}_{k+1}\right)$ preferable to messages outside of $M\left( \hat{A}_{1}\ldots \hat{A} _{k+1}\right)$. Define

$$\begin{aligned} \hat{\sigma }_{\hat{m},\varepsilon }^{R}=\sum _{j=1}^{n-1}\varepsilon ^{j-1}\left( 1-\varepsilon \right) s_{\hat{m},j}^{R}+\varepsilon ^{n}s_{\hat{ m},n}^{R}. \end{aligned}$$

Then for $\varepsilon >0$ sufficiently small, $\left( \hat{m},a^{S}\right)$ is the unique best response to the belief $\left( 1-\varepsilon \right) *Constant-b^{R}\left( a^{S}\right) +\varepsilon *\hat{\sigma }_{\hat{m},\varepsilon }^{R}$, for any $a^{S}$.

Next I show that $S^{S}\left( 1\right) \subset M^{*}\times A^{S}$. If $\hat{m}={\text{``}}\hat{A}_{1}\ldots \hat{A}_{n}{{\text{''}} }\notin M^{*}$, then $\exists j\in \left\{ 1,\ldots ,n\right\}$ such that $\max \hat{A}_{j}<\min \left( \hat{A} _{j-1}\backslash \hat{A}_{j}\right)$. Consider any message $m^{\prime }$ in $M\left( \hat{A}_{1}\ldots \hat{A}_{j-1}\left( \hat{A}_{j-1}\backslash \hat{A} _{j}\right) \right)$. If a language-based response $s^{R}$ responds to $\hat{m}$ and $m^{\prime }$ with different actions, then $s^{R}\left( \hat{m} \right) \le \max \hat{A}_{j}<\min \left( \hat{A}_{j-1}\backslash \hat{A} _{j}\right) \le s^{R}\left( m^{\prime }\right)$. Thus, for any $a^{S}$, $\left( \hat{m},a^{S}\right)$ is weakly dominated by $\left( m^{\prime },a^{S}\right)$.

I now show that $S^{R}\left( 2\right) =S^{R}\left( 0\right)$. Pick any $\hat{s}^{R}\in S^{R}\left( 0\right)$. Every best response to the conjecture that is uniform on

$$\begin{aligned} \left\{ \left( m,b^{S}\left( \hat{s}^{R}\left( m\right) \right) \right) :m\in M^{*}\right\} \subset S^{R}\left( 1\right) \end{aligned}$$

must equal $\hat{s}^{R}$ on $M^{*}$ because $b^{R}=\left( b^{S}\right) ^{-1}$. Because no message outside of $M^{*}$ is used in $S^{S}\left( 1\right)$, all such best responses give the same payoff to the Receiver against any $\left( m,a^{S}\right) \in S^{S}\left( 1\right)$. Thus, $\hat{s} ^{R}\in S^{R}\left( 2\right)$. So $S^{R}\left( 2\right) =S^{R}\left( 0\right)$ and thus $S^{R}\left( 1\right) =S^{R}\left( 0\right)$. Then $S^{S}\left( 2\right) =S^{S}\left( 1\right)$ and thus $S\left( 2\right) =S\left( 1\right)$. By induction, $S\left( \infty \right) =\left( M^{*}\times A^{S},S^{R}\left( 0\right) \right)$. $\hfill\square$

1.2 Proof of Proposition 4

Order Sender actions so that $\left( a_{i}^{S},b^{R}\left( a_{i}^{S}\right) \right)$ gives Sender the $i$th highest payoff among all stage game Nash equilibria. Define $M_{i}^{n}=:\left\{ m\in M:\left( m,a_{i}^{S}\right) \in S^{S}\left( n\right) \right\}$ for $i=1,2,\ldots N$ and $n=1,2,\ldots$.

Because g is not weak self-signaling, there exists $j\in \left\{ 2,\ldots ,N\right\}$ such that $g^{S}\left( a_{j}^{S},b^{R}\left( a_{j}^{S}\right) \right) <g^{S}\left( a_{j}^{S},b^{R}\left( a_{i}^{S}\right) \right)$ for every $i=1,\ldots ,j-1$.

Lemma 20

For every $n=0,1,2,\ldots$,

• 1. $M_{j}^{n}\ne \emptyset$

• 2. $\exists s_{n}^{R}\in S^{R}\left( n\right)$ where $s_{n}^{R}\left( m\right) =b^{R}\left( a_{j}^{S}\right)$ for all $m\in M_{j}^{n-1}$ and $s_{n}^{R}\left( m\right) \ne b^{R}\left( a_{i}^{S}\right)$ for any $i=1,\ldots ,j-1$ and any m,

• 3. for any $i=1,\ldots ,j-1$, any $m\in M_{i}^{n}$ , $\exists m^{\prime }\in M_{j}^{n}$ such that there exists no language-based best response to both $\left( m,a_{i}^{S}\right)$ and $\left( m^{\prime },a_{j}^{S}\right)$.

Proof

For $n=0,1,\ldots$, at least one best response to $\frac{1}{\#M_{j}^{n}} \sum _{m\in M_{j}^{n}}\left( m,a_{j}^{S}\right) \in \varDelta S^{S}\left( n\right)$ belongs to $S^{R}\left( n+1\right)$. Denote it by $s_{n+1}^{R}$. Then $s_{n+1}^{R}\left( m\right) =b^{R}\left( a_{j}^{S}\right)$ for all $m\in M_{j}^{n}$.

The statements are true for $n=0$ because $M_{i}^{0}=M$ for all $i=1,\ldots ,n$ and because $Constant-b^{R}\left( a_{j}^{S}\right) \in S^{R}\left( 0\right)$ .

Suppose the statements are true for $n=k$. Suppose

$$\begin{aligned} M_{i}^{\prime }=:\left\{ m:s_{k+1}^{R}\left( m\right) =b^{R}\left( a_{i}^{S}\right) \right\} \ne \emptyset \end{aligned}$$

for some $i=1,\ldots j-1$. Let q be the smallest such i. A Sender best response to $s_{k+1}^{R}\in S^{R}\left( k+1\right)$ exists in $S^{S}\left( k\right)$, which must be $\left( m^{\prime },a_{q}^{S}\right)$ for some $m^{\prime }\in M_{q}^{\prime }$. Then the language-based response $s_{k+1}^{R}$ is a best response to $\left( m^{\prime },a_{q}^{S}\right)$ and to every strategy in $M_{j}^{k}\times \left\{ a_{j}^{S}\right\}$, contradiction to the assumption that statement 3 holds for $n=k$. Thus $M_{i}^{\prime }=\emptyset$ for all $i=1,\ldots ,j-1$. This established statement 2. Statement 1 follows because every best response to $s_{k+1}^{R}$ must take action $a_{j}^{S}$.

Suppose to the contrary of statement 3 that for some $i=1,\ldots ,j-1$, a language-based response exists that is a best response to some $\left( \hat{m},a_{i}^{S}\right) \in S^{S}\left( k+1\right)$ and to every strategy in $M_{j}^{k+1}\times \left\{ a_{j}^{S}\right\}$. Then a best response $s_{d,k+1}^{R}\in S^{R}\left( k+1\right)$ to

$$\begin{aligned} \frac{1}{\#M_{j}^{k+1}+1}\left( \left( \hat{m},a_{i}^{S}\right) +\sum _{m\in M_{j}^{k+1}}\left( m,a_{j}^{S}\right) \in \varDelta S^{S}\left( k\right) \right) \end{aligned}$$

(23)

exists, which responds to $\hat{m}$ with $b^{R}\left( a_{i}^{S}\right)$ and to every $m\in M_{j}^{k+1}$ with $b^{R}\left( a_{j}^{S}\right)$. For $\varepsilon >0$ sufficiently small, a best response to $\left( 1-\varepsilon \right) s_{k}^{R}+\varepsilon s_{d,k+1}^{R}\in \varDelta S^{S}\left( k\right)$ must be optimal against $s_{d,k+1}^{R}$ among all best responses to $s_{k}^{R}$, and thus must use action $a_{j}^{S}$ and a message in $\left( s_{k}^{R}\right) ^{-1}\left( b^{R}\left( a_{j}^{S}\right) \right)$. At least one best response belongs to $S^{S}\left( k+1\right)$. Denote it by $\left( m^{*},a_{j}^{S}\right)$. Because statement 3 holds for $n=k$ by hypothesis, we have $s_{k}^{R}\left( \hat{m} \right) =b^{R}\left( a_{j}^{S}\right)$. By the construction of $s_{d,k+1}^{R}$, $u^{S}\left( \left( \hat{m},a_{j}^{S}\right) ,s_{d,k+1}^{R}\right) =g^{S}\left( a_{j}^{S},b^{R}\left( a_{i}^{S}\right) \right) >g^{S}\left( a_{j}^{S},b^{R}\left( a_{j}^{S}\right) \right) =u^{S}\left( \left( m,a_{j}^{S}\right) ,s_{d,k+1}^{R}\right)$ for any $m\in M_{j}^{k+1}$. Thus $m^{*}\notin M_{j}^{k+1}$, contradiction to the definition of $M_{j}^{k+1}$. $\hfill\square$

Now we can prove Proposition 4. By finiteness of the strategy space, $S\left( \infty \right) =S\left( K\right)$ for some K. Because a strategy using the Stackelberg action $a_{1}^{S}$ cannot be weakly dominated by one that uses any $a^{S}\ne a_{1}^{S}$, $\exists m^{*}\in M_{1}^{K}$. The proposition follows immediately from statement 2 of Lemma 20 because miscoordination happens at $\left( \left( m^{*},a_{1}^{S}\right) ,s_{K}^{R}\right) \in S^{R}\left( \infty \right)$.

Appendix 5: Omitted proofs in Sect. 5

1.1 Crude language

Proof

(of Proposition 5 For $a_{S}\ne x$, $\left( {\text{``}}\left\{ X\right\} {{\text{''}} },a^{S}\right)$ is weakly dominated by $\left( { {\text{``}}}A^{R}\backslash \left\{ X\right\} {{\text{''}} },a^{S}\right)$ by the strong self-signaling condition and because, if Receiver responds to the two messages with different actions, then he responds to “$\left\{ X\right\}$” with X and to “$A^{R}\backslash \left\{ X\right\}$” with some $a^{R}\ne X$. Moreover, $\left( {\text{``}}\left\{ X\right\} {\text{''}},x\right) \in S^{S}\left( 1\right)$ because it is the unique best response to $s_{X}^{R}$ that responds to “$\left\{ X\right\}$” with X and to all other messages with the same action in $A^{R}\backslash \left\{ X\right\}$.

Consider $\hat{s}^{R}\left( {\text{``}}\left\{ X\right\} {{\text{''}} }\right) \ne X$. Then $s^{R}\left( { {\text{``}}}A^{R}\backslash \left\{ X\right\} {{\text{''}} }\right) =s^{R}\left( {\text{``}}\left\{ X\right\} {\text{''}}\right)$. Define $s_{D}^{R}$ by replacing the response of $s^{R}$ on “$\left\{ X\right\}$” to X. Then $s_{D}^{R}$ follows the conjugate pair of “$\left\{ X\right\}$” and “$A^{R}\backslash \left\{ X\right\}$”, and does what $\hat{s}^{R}$ does for other conjugate pairs. So $s_{D}^{R}$ is a language-based response. And $s_{D}^{R}$ weakly dominates $\hat{s}^{R}$ w.r.t. $S^{S}\left( 1\right)$. Thus $s^{R}\left( {\text{``}} \left\{ X\right\} {\text{''}}\right) =X$ for every $s^{R}\in S^{R}\left( 2\right)$. So every $s^{S}\in S^{S}\left( 3\right)$ must give Sender her Stackelberg payoff too. $\square$