Cheap Talk Under Partial Conflicts: A Dynamic Analysis of Pragmatic Meaning

Abstract

In natural language, meanings of words often deviate from their literal meanings under pragmatic reasoning. As is shown in game-theoretical pragmatics, when players do not have aligned benefits, communication with non-literal meaning is even more frequent. In these situations, the pragmatic inference under iterated best response plays the essential role for building the pragmatic meaning. The paper provides a systematic analysis of the deviation from literal meaning to pragmatic meaning when the interlocutors have partial conflicts. We apply the classical Cheap Talk game and Iterated Best Response reasoning to demonstrate the result.

Appendix

Theorem 2 Given a game $G_I$ and the IBR model $I^k$ , there exists a natural number n such that for any natural number $m \geqslant n$ , $I^m_S=I^m_R=[0,1]$ , and $(s_m, r_m)$ yields a pooling equilibrium.

Proof

Case 1: Suppose that $b>= 1/4$, $N(b)=1$, the required result holds trivially.

Case 2: Suppose that $0<b<1/4$.

The idea is to find the connection of the sender’s inutility functions and the changes of the meanings between different levels. Then construct the convergence result of the dynamic.

First, given any $b\in (0, \frac{1}{4})$, we can calculate $N(b)=N$ which is the largest number of the partitions for the partition equilibrium, then the total number of the signals are $M=\frac{(N+1)N}{2}$.

From the constructions of the IBR model, for any signal $I_{ij}^k$, the sender’s utility function has the form $U_{S}=-(r^k-t)^2$, where $r^k=r-b$. List all these functions as r increases as $U_{S}=-(r_1^k-t)^2, U_{S}=-(r_2^k-t)^2, \dots , U_{S}=-(r_M^k-t)^2$ where $r_1^k<r_2^k<\dots <r_M^k$.

All the interval meanings of the signals at each level are derived from those utility functions, as we illustrated in Fig. 2.

Fig. 2.

Expanded signal structure for b = 1/20

$p_j^k$ is used to indicate those interval points. And for any level k, assume that $p_{0}^k=0$ and $p_{M}^k=1$. And for any $0<i<j<M$, $p_i<p_j$. Then by the structure of the game, r and p has the following relationships.

$$\begin{aligned} \begin{array}{ c } p_{1}^0 =\frac{1}{2}(r_{1}^0+r_{2}^0) \\[2mm] p_{2}^0 =\frac{1}{2}(r_{2}^0+r_{3}^0) \\[2mm] \cdots \\[2mm] p_{M-1}^0=\frac{1}{2}(r_{M-1}^0+r_{M}^0) \end{array} \end{aligned}$$

Moreover, we have that

$$\begin{aligned} \begin{array}{ c r } r_{1}^{k+1}=\frac{1}{2}(0+p_{1}^k)-b &{} \\[2mm] r_{2}^{k+1}=\frac{1}{2}(p_{1}^k+p_{2}^k)-b &{} \qquad [1] \\[2mm] \cdots &{} \\[2mm] r_{M}^{k+1}=\frac{1}{2}(p_{M-1}^k+1)-b &{} \\ \end{array} \end{aligned}$$

In addition, we have the following equations:

$$\begin{aligned} \begin{array}{ c r } p_{1}^{k+1}=\frac{1}{2}(r_{1}^{k+1}+r_{2}^{k+1}) &{} \\[2mm] p_{2}^{k+1}=\frac{1}{2}(r_{2}^{k+1}+r_{3}^{k+1}) &{} \qquad [2] \\[2mm] \cdots &{} \\[2mm] p_{M-1}^{k+1}=\frac{1}{2}(r_{M-1}^{k+1}+r_{M}^{k+1}) &{} \\ \end{array} \end{aligned}$$

By substituting all the $r_{i}^{k+1}$s in formula series [2] with all the formulas in [1], we can obtain the following equations:

$$\begin{aligned} \begin{array}{ c r } p_{1}^{k+1}=\frac{1}{4}p_{0}^k+\frac{1}{2}p_{1(n)}+\frac{1}{4}p_{2^k}-b &{} \\[2mm] p_{2}^{k+1}=\frac{1}{4}p_{1}^k+\frac{1}{2}p_{2(n)}+\frac{1}{4}p_{3^k}-b &{} \\[2mm] p_{3}^{k+1}=\frac{1}{4}p_{2}^k+\frac{1}{2}p_{3(n)}+\frac{1}{4}p_{4}^k-b &{} \qquad [3] \\[2mm] \cdots &{} \\[2mm] p_{M-1}^{k+1}=\frac{1}{4}p_{M-2}^k+\frac{1}{2}p_{M-1}^k+\frac{1}{4}p_{M}^k-b &{} \\ \end{array} \end{aligned}$$

Therefore, the formulas in [3] can be rewritten in the metric form as follows:

$$p(n+1)=Ap(n)-B$$

where

$$ A = \left( \begin{array}{cccccc} \frac{1}{2} &{}\frac{1}{4} &{} 0 &{} \cdots &{} 0 &{} 0 \\ \frac{1}{4} &{} \frac{1}{2} &{}\frac{1}{4} &{} 0 &{} \cdots &{} 0 \\ 0 &{} \frac{1}{4} &{} \frac{1}{2} &{}\frac{1}{4} &{} \cdots &{} 0 \\ \vdots &{}&{}\ddots &{}\ddots &{}\ddots &{}\vdots \\ 0 &{} 0&{} \cdots &{} \frac{1}{4} &{} \frac{1}{2}&{} \frac{1}{4}\\ 0 &{} 0&{} \cdots &{} 0&{} \frac{1}{4} &{} \frac{1}{2}\\ \end{array} \right) $$

$$B=\left( \begin{array}{c} b\\ \cdots \\ b\\ \cdots \\ \frac{1}{4}-b\\ \end{array} \right) $$

Observe that A has the exact form as the matrix known as Toeplitz Matrix. Thus its eigenvalues are given by

$$\lambda _{n}=\frac{1}{2}+\frac{1}{4}\cos (\frac{n\pi }{k+1}), n=1, 2, \cdots , M-1$$

Therefore, $|\lambda |<1$ for all n. According to the following mathematical result: Assuming that A is any $k \times k$ matrix, then $\lim \limits _{n \rightarrow \infty }A^n=0$ iff $|\lambda |<1$ for all eigenvalues $\lambda$ of A,⁶ we thus have that $\lim \limits _{n \rightarrow \infty }A^n=0$. It follows that $p_{n+1}<0$ as $n \rightarrow \infty$. As all the $p_i$s are becoming negative, there will be no divided point in the interval [0, 1]. That is to say, there is only one signal that is considered from the sender’s point of view for $t \in [0, 1]$. Thus, the babbling equilibrium eventually occurs.