Codes of conduct, private information and repeated games

Abstract

We examine self-referential games in which there is a chance of understanding an opponent’s intentions. Even when this source of information is weak, we are able to prove a folk-like theorem for repeated self-referential games with private monitoring. Our main focus is on the interaction of two sources of information about opponents’ play: direct observation of the opponent’s intentions, and indirect observation of the opponent’s play in a repeated setting.

Appendix

We assume that in the base game all players have access to N individual randomizing devices ${\varTheta }=\{\theta ^{1},\ldots ,\theta ^{N}\}$ each of which has an independent probability $\varepsilon _{R}$ of an outcome we call punishment, $\theta _{p}$. Suppose that $s^{0}$ is an $\varepsilon _{0}$-Nash equilibrium giving utility $u_{i}(s^{0})$ for each i. For any $s\in S$ and strategies $s_{j}^{i}$ for any pair of players $i\ne j$ suppose that $s^{i}=(s_{j}^{i},s_{-j})$ are $\varepsilon _{1}$-Nash equilibria. Define $P_{i}=u_{i}(s^{0})-u_{i}(s^{i})$. We assume that $P_{i}\ge \underline{P}\ge 0$ and for some $\varepsilon _{p}\ge 0$ that $|u_{j}(s^{i})-u_{j}(s^{0})|\le \varepsilon _{p}$. Let $\underline{u}=\min _{i,s}u_{i}(s)$ and $\bar{u}=\max _{i,s}u_{i}(s)$ be the lowest and highest payoffs, respectively. Define $\varepsilon =\varepsilon _{0}+(N+\bar{u}-\underline{u})(\varepsilon _{1}+\varepsilon _{p})E$, and $K=\max \{(N+\bar{u}-\underline{u})3N^{4}(1+\bar{u}-\underline{u}),(N^{4}(\bar{u}-\underline{u})+1)(\bar{u}-\underline{u})\}$.

Theorem 3

If $(D(\underline{P}-\varepsilon _{1}))^{2}>4K\varepsilon$, then there is an $\varepsilon _{R}$ and strict Nash equilibrium codes of conduct $\hat{r}\in R$ such that for all players i, $\left| u_{i}(s^{0})-U_{i}(\hat{r})\right| \le \varepsilon +D(\underline{P}-\varepsilon _{1})-\sqrt{(D(\underline{P}-\varepsilon _{1}))^{2}-4K\varepsilon }$.

Proof

There are $\left| {\varTheta }\right| ^{\left| I\right| }=N^{2}$ independent randomization devices in operations. From $P_{\theta }(\theta _{p}=0)=(1-\varepsilon _{R})^{N^{2}}$ and $P_{\theta }(\theta _{p}=1)=N^{2}\varepsilon _{R}(1-\varepsilon _{R})^{N^{2}-1}$ we find $P_{\theta }(\theta _{p}\ge 2)\le N^{4}\varepsilon _{R}^{2}$.

Take $\hat{r}^{i}$ such that: for all players i, if $\bar{y}_{i}\in \bar{Y}_{i}$ and $\theta _{p}\ge 1$ play $\hat{r}_{i}^{i}(y_{i})=s_{i}^{j}$ and $\hat{r}_{j}^{i}(y_{j})=s_{j}^{0}$ for any $y_{j}\in Y_{j}$ and all players $j\ne i$, otherwise play $\hat{r}_{i}^{i}(y_{i})=s_{i}^{0}$ for all $y_{i}\notin \bar{Y}_{i}$ and for all $j\ne i$ choose $\hat{r}_{j}^{i}(y_{j})=s_{j}^{0}$ for any $y_{j}\notin \bar{Y}_{j}$. The following mutually exclusive events can occur to player i when all players $j\ne i$ choose $\hat{r}^{j}$, but he chooses $\tilde{r}^{i}$ defined below and $\tilde{r}_{j}^{i}=\hat{r}_{j}^{i}$ for all $j\ne i$: (1) Nobody is punished: if player i chooses $\hat{r}^{i}$ he gets $u_{i}(s^{0})$, while if i chooses $\tilde{r}^{i}_i\ne s_{i}^{0}$ he gets at most $u_{i}(s^{0})+\varepsilon _{0}$, (2) Player j is the only player punished: by following $\hat{r}^{i}$ i gets $u_{i}(s^{j})$, if i chooses $\tilde{r}_{i}^{i}(\bar{y}_{i})\ne s_{i}^{j}$, he gets at most $u_{i}(s^{j})+\varepsilon _{1}$, and (3) Two or more players are punished: if $\hat{r}^{i}$ is followed player i he gets at worst $\underline{u}$, if i deviates while choosing $\tilde{r}^{i}(y_{i})=\tilde{s}_{i}$ with $\tilde{s}_{i}\in S_{i}$ and $\tilde{s}_{i}\ne s_{i}^{j}\ne s_{j}^{0}$ he gets at most $\bar{u}$.

We can bound expected payoffs $U_i(\hat{r})\le \overline{U}_i(\hat{r})=u_{i}(s^{0})+(1-(1-E)^{N})\left[ \varepsilon _{p}+N^{4}\varepsilon _{R}^{2}(\bar{u}-\underline{u})\right]$ and $U_i(\hat{r})\ge \underline{U}_i(\hat{r})=u_{i}(s^{0})-(1-(1-E)^{N})\big [\varepsilon _{p}+N^{4}\varepsilon _{R}^{2}(\bar{u}-\underline{u})\big ]-\pi _{j}(\bar{y}_{j}|\hat{r})\varepsilon _{R}P_{i}$. Suppose player i chooses the alternative code of conduct $\tilde{r}^{i}$, he gets $U_i(\tilde{r}^i,\hat{r}^{-i})\le \overline{U}_i(\tilde{r}^i,\hat{r}^{-i})$ where

$$\begin{aligned} \overline{U}_i(\tilde{r}^i,\hat{r}^{-i})= & {} u_{i}(s^{0})+\varepsilon _{0}+(1-(1-E)^{N})\left[ \varepsilon _{1}+N^{4}\varepsilon _{R}^{2}(\bar{u}-\underline{u})\right] \\&-\left[ (\pi _{j} (\bar{y}_{j}|\hat{r})+D)\varepsilon _{R}-N^{4}\varepsilon _{R}^{2}\right] (P_{i}+\varepsilon _{1}). \end{aligned}$$

Consequently, the gain to choosing $\tilde{r}^i$ is $U_i(\tilde{r}^i,\hat{r}^{-i})-U_i(\hat{r})$ and bounded by

$$\begin{aligned}&\varepsilon _{0} +(1-(1-E)^{N})\left[ \varepsilon _{1}+\varepsilon _{p}+2N^{4}\varepsilon _{R}^{2}(\bar{u}-\underline{u})\right] +\pi _{j}(\bar{y}_{j}|\hat{r})\varepsilon _{R}P_{i}\\&\qquad -\left[ (\pi _{j}(\bar{y}_{j}|\hat{r})+D)\varepsilon _{R}-N^{4}\varepsilon _{R}^{2}\right] (P_{i}-\varepsilon _{1})\\&\quad \le \varepsilon _{0}+(N+\bar{u}-\underline{u})E\left[ \varepsilon _{1}+\varepsilon _{p}+3N^{4}\varepsilon _{R}^{2}(1+\bar{u}-\underline{u})\right] \\&\qquad -D\varepsilon _{R}(\underline{P}-\varepsilon _{1}) \le \varepsilon +K\varepsilon _{R}^{2}-D\varepsilon _{R}(\underline{P}-\varepsilon _{1}). \end{aligned}$$

Hence if $D\varepsilon _{R}(\underline{P}-\varepsilon _{1})>\varepsilon +K\varepsilon _{R}^{2}$ then there is a strict Nash equilibrium with

$$\begin{aligned} \left| u_{i}(s^{0})-U_{i}(\hat{r})\right| \le NE\varepsilon _{p}+(N^{4}(\bar{u}-\underline{u})+1)(\bar{u}-\underline{u})\varepsilon _{R}\le \varepsilon +2K\varepsilon _{R}. \end{aligned}$$

We conclude by solving the inequality for $\varepsilon _{R}=[D(\underline{P}-\varepsilon _{1})\pm \sqrt{(D(\underline{P}-\varepsilon _{1}))^{2}-4K\varepsilon }]/2K$, which gives two real roots since $(D(\underline{P}-\varepsilon _{1}))^{2}>4K\varepsilon$, implying the existence of an $\varepsilon _{R}$ for which $\hat{r}$ is a strict Nash equilibrium of the self-referential game. Plugging the lower root into the inequality for the utility difference $\left| u_{i}(s^{0})-U_{i}(\hat{r})\right|$ gives the remainder of the result. $\square$