Equilibrium selection through $\mathbf {p}_{u}$-dominance

Abstract

The paper introduces and discusses the concept of $\mathbf {p}_{u}$-dominance in the context of finite games in normal form. It then presents the $\mathbf { p}_{u}$-dominance criterion for equilibrium selection. The $\mathbf {p}_{u}$-dominance criterion is inspired by and closely related to the $\mathbf {p}$-dominance criterion originally proposed by Morris et al. (Econometrica 63:145–157, 1995). However, there are games in which the two criteria lead to different refinements. We provide sufficient conditions under which equilibrium selection through $\mathbf {p}_{u}$-dominance is weakly finer (respectively, coarser) than equilibrium selection through $\mathbf {p}$-dominance.

Appendix

1.1 Proof of Proposition 2

If $a^{*}$ is a Nash equilibrium of $G,$ then there exists at least one vector for which $a^{*}$ is $\mathbf {p}$-dominant. In particular, $a^{*}$ is certainly $\mathbf {p}$-dominant for $\mathbf {p}( a^{*}) =(1,\ldots ,1)$. It follows that the vector $\bar{\mathbf {p}}( a^{*}) =( \bar{p}_{1}(a^{*}),\ldots ,\bar{p}_{n}(a^{*}))$ also exists where, for any $i$ and any $j\ne i$, $\bar{p} _{i}(a^{*})$ is the smallest probability for which action $a_{i}^{*}$ best responds to any conjecture $\lambda (a^{*}) \in \Delta ( A_{-i})$ according to which $\lambda (a_{j}^{*}) \ge \bar{p}_{i}(a^{*})$ while the remaining probability $( 1-\lambda ( a_{j}^{*}) )$ can follow any distribution on actions $a_{j}\ne a_{j}^{*}$. The conjecture thus includes the situation in which $\lambda (a_{j}) =\frac{( 1-\lambda ( a_{j}^{*}) ) }{k_{j}-1}$ for any $j\ne i$, $a_{j}\ne a_{j}^{*}$ and $k_{j}=\left| A_{j}\right|$. Therefore, if $a^{*}$ is $\mathbf {p}$-dominant with $\bar{\mathbf {p}} ( a^{*})$ then $a^{*}$ is also $\mathbf {p}_{u}$-dominant with $\mathbf {p}_{u}( a^{*}) =\bar{\mathbf {p}}( a^{*})$. Notice that if $k_{j}=2$ for any $j\in N$, then $\bar{\mathbf {p}} _{u}(a^{*}) =\bar{\mathbf {p}}( a^{*})$ as both $\mathbf {p}$-dominance and $\mathbf {p}_{u}$-dominance assign probability $( 1-\lambda ( a_{j}^{*}))$ to the event of $j$ playing action $a_{j}\ne a_{j}^{*}$. Now consider the case in which $k_{j}>2$ for some $j$. Assume there exists at least one action $a_{i}\in A_{i}$ that is not (weakly or strictly) dominated by $a_{i}^{*}$ .⁴ Let $\tilde{a}(a_{i})=(a_{i},\tilde{a}_{-i})$ be the action profile that supports the (non-necessarily unique) outcome for which $u_{i}(a_{i},a_{-i})-u_{i}(a_{i}^{*},a_{-i})>0$ is maximal. Given that $u_{i}(a_{i},a_{-i}^{*})-u_{i}(a_{i}^{*},a_{-i}^{*})\le 0$ ($a^{*}$ is a Nash equilibrium), it must be the case that in $\tilde{a} (a_{i})$ there is at least one player $j\ne i$ that plays $\tilde{a} _{j}\ne a_{j}^{*}$. The equilibrium $a^{*}$ is $\mathbf {p}$-dominant with $\bar{\mathbf {p}}\left( a^{*}\right) =\left( \bar{p} _{1}(a^{*}),\ldots ,\bar{p}_{n}(a^{*})\right)$ if $a_{i}^{*}$ dominates any $a_{i}$ even under the specific conjecture that assigns probability $\left( 1-\bar{p}_{i}(a^{*})\right)$ to the event of $j$ playing $\tilde{a}_{j}\ne a_{j}^{*}$. On the other hand, $a^{*}$ is $\mathbf {p}_{u}$-dominant with $\bar{\mathbf {p}}_{u}\left( a^{*}\right) =\left( \bar{p}_{1}^{\prime }(a^{*}),\ldots ,\bar{p}_{n}^{\prime }(a^{*})\right)$ if $a_{i}^{*}$ dominates any $a_{i}$ under the conjecture that assigns probability $\left( \frac{1-\bar{p}_{i}^{\prime }(a^{*})}{ k_{j}-1}\right)$ to the event of $j$ playing $\tilde{a}_{j}\ne a_{j}^{*}$. Equilibrium $a^{*}$ cannot be $\mathbf {p}$-dominant with $\mathbf {p} \left( a^{*}\right) =\bar{\mathbf {p}}_{u}\left( a^{*}\right)$ given that $( 1-\bar{p}_{i}^{\prime }(a^{*})) > \left( \frac{1-\bar{p }_{i}^{\prime }(a^{*})}{k_{j}-1}\right)$ and the conjecture would thus assign too much probability to the occurrence of the profile $\tilde{a} (a_{i})$. It then must be the case that $\bar{p}_{i}^{\prime }(a^{*})< \bar{p}_{i}(a^{*})$. We can thus conclude that $\bar{\mathbf {p}} _{u}\left( a^{*}\right) \le \bar{\mathbf {p}}\left( a^{*}\right)$.

1.2 Proof of Lemma 1

Without loss of generality, assume the $\mathbf {p}_{u}$-dominance criterion selects two equilibria, i.e., let $A^{**}=\{a^{**1},a^{**2}\}$. By expression (2) this means:

$$\begin{aligned} \sum _{i=1}^{n}\bar{p}_{i}(a^{**1})=\sum _{i=1}^{n}\bar{p}_{i}(a^{**2}) \end{aligned}$$

(3)

The fact that $G$ is symmetric implies that (3) reduces to $\bar{\mathbf {p}}_{u}(a^{**1})=\bar{\mathbf {p}}_{u}(a^{**2})$, i.e., $\bar{p}_{i}(a^{**1})=\bar{p}_{i}(a^{**2})$ for any $i\in N$. Now define the conjecture $\bar{\lambda }( a^{**\phi })$ with $\phi \in \left\{ 1,2\right\}$ such that $\bar{\lambda } ( a_{j}^{**\phi }) =\bar{p}_{i}(a^{**\phi })$ and $\bar{\lambda }( a_{j}) =\frac{( 1-\bar{p}_{i}(a^{**\phi }) ) }{k-1}$ for all $a_{j}\ne a_{j}^{**\phi }$ and for $j\in N_{-i}$. Let $\lambda ^{(\phi )}( a_{-i}) =\times _{j\in N_{-1}}\bar{\lambda }( a^{**\phi })$. With the probability distribution $\lambda ^{(\phi )}$, any player is by construction indifferent between playing action $a_{i}^{**\phi }$ and action $a_{i}^{**\xi }$ with $\xi \in \{ 1,2\}$ and $\xi \ne \phi$. This implies that the following two conditions must simultaneously hold:

$$\begin{aligned} \sum \limits _{a_{-i}\in A_{-i}}\lambda ^{(1)}\left( a_{-i}\right) u_{i}\left( a_{i}^{**1},a_{-i}\right) =\sum \limits _{a_{-i}\in A_{-i}}\lambda ^{(1)}\left( a_{-i}\right) u_{i}\left( a_{i}^{**2},a_{-i}\right) \end{aligned}$$

(4)

$$\begin{aligned} \sum \limits _{a_{-i}\in A_{-i}}\lambda ^{(2)}\left( a_{-i}\right) u_{i}\left( a_{i}^{**2},a_{-i}\right) =\sum \limits _{a_{-i}\in A_{-i}}\lambda ^{(2)}\left( a_{-i}\right) u_{i}\left( a_{i}^{**1},a_{-i}\right) \end{aligned}$$

(5)

Therefore, the following condition also holds:

$$\begin{aligned}&\sum \limits _{a_{-i}\in A_{-i}}\lambda ^{(1)}\left( a_{-i}\right) u_{i}\left( a_{i}^{**1},a_{-i}\right) -\sum \limits _{a_{-i}\in A_{-i}}\lambda ^{(1)}\left( a_{-i}\right) u_{i}\left( a_{i}^{**2},a_{-i}\right) \nonumber \\&\quad =\sum \limits _{a_{-i}\in A_{-i}}\lambda ^{(2)}\left( a_{-i}\right) u_{i}\left( a_{i}^{**1},a_{-i}\right) -\sum \limits _{a_{-i}\in A_{-i}}\lambda ^{(2)}\left( a_{-i}\right) u_{i}\left( a_{i}^{**2},a_{-i}\right) \end{aligned}$$

(6)

Now notice that, given that by assumption $A^{**}=\{ a^{**1},a^{**2}\}$ then it must be the case that $\bar{p}_{i}(a^{**1})=\bar{p}_{i}(a^{**2})$ for any $i\in N$. Therefore, $\frac{\left( 1-\bar{p}_{i}\left( a^{**1}\right) \right) }{ k-1}=\frac{\left( 1-\bar{p}_{i}\left( a^{**2}\right) \right) }{k-1}$ for any $i\in N$. Expression (6) thus boils down to:

$$\begin{aligned}&\left( \left( \bar{p}_{i}(a^{**1})\right) ^{n-1}-\left( \frac{ \left( 1-\bar{p}_{i}\left( a^{**1}\right) \right) }{k-1}\right) ^{n-1}\right) \left( u_{i}\left( a_{i}^{**1},a_{-i}^{**1}\right) -u_{i}\left( a_{i}^{**2},a_{-i}^{**1}\right) \right) \nonumber \\&\quad =\left( \!\!\left( \bar{p}_{i}(a^{**1})\right) ^{n-1}-\left( \frac{\left( 1-\bar{p} _{i}\left( a^{**1}\right) \right) }{k-1}\right) ^{n-1}\right) \left( u_{i}\left( a_{i}^{**1},a_{-i}^{**2}\right) -u_{i}\left( a_{i}^{**2},a_{-i}^{**2}\!\right) \!\right) \nonumber \\ \end{aligned}$$

(7)

The fact that $a^{**1}$ and $a^{**2}$ are Nash equilibria implies that $( u_{i}( a_{i}^{**1},a_{-i}^{**1}) -u_{i}( a_{i}^{**2},a_{-i}^{**1}) ) \ge 0$ and $(u_{i}( a_{i}^{**1},a_{-i}^{**2}) -u_{i}( a_{i}^{**2},a_{-i}^{**2}) ) \le 0$ with at least one strict inequality if at least one equilibrium is strict. Then, for expression (7) to always hold [notice that both $\bar{p}_{i}(a^{**1})$ and $\left( \frac{1-\bar{p}_{i}(a^{**1})}{k-1}\right)$ are non-negative and cannot be simultaneously equal to zero], it must be the case that $\bar{p}_{i}(a^{**1})=\left( \frac{1-\bar{p}_{i}(a^{**1})}{k-1}\right)$. But since $\bar{p}_{i}(a^{**1})=\bar{p}_{i}(a^{**2})$ then the following result holds:

$$\begin{aligned} \bar{p}_{i}(a^{**\phi })=\left( \frac{1-\bar{p}_{i}(a^{**\phi })}{k-1}\right) \quad \text { for }\phi \in \left\{ 1,2\right\} \end{aligned}$$

(8)

which implies $\bar{p}_{i}(a^{**\phi })=\frac{1}{k}$ for any $\phi \in \left\{ 1,2\right\}$. Given that the choice of $\left| A^{**}\right| =2$ was made without loss of generality as one can replicate the passages above for any possible couple of equilibria that belong to $A^{**}$, the more general result that $\bar{\mathbf {p}} _{u}(a^{**\phi })=\left( \frac{1}{k},\ldots ,\frac{1}{k}\right)$ for any $a^{**\phi }\in A^{**}=\left\{ a^{**1},\ldots ,a^{**\Phi }\right\}$ and $\Phi \ge 2$ easily follows.

1.3 Proof of Proposition 3

Assume the $\mathbf {p}_{u}$-dominance criterion selects multiple equilibria. Without loss of generality, let $A^{**}=\left\{ a^{**1},a^{**2}\right\}$ be the set of selected equilibria. Then:

$$\begin{aligned} \sum \limits _{a_{-i}\in A_{-i}}\lambda ^{(1)}\left( a_{-i}\right) u_{i}\left( a_{i}^{**1},a_{-i}\right) =\sum \limits _{a_{-i}\in A_{-i}}\lambda ^{(1)}\left( a_{-i}\right) u_{i}\left( a_{i}^{**2},a_{-i}\right) \end{aligned}$$

(9)

where $\lambda ^{(1)}\left( a_{-i}\right) =\times _{j\in N_{-1}}\lambda ^{1}\left( a_{j}\right)$ and $\lambda ^{1}\left( a_{j}\right) =\bar{p} _{i}(a^{**1})$ if $a_{j}=a_{j}^{**1}$ while $\lambda ^{1}\left( a_{j}\right) =\frac{\left( 1-\bar{p}_{i}\left( a^{**1}\right) \right) }{k-1}$ if $a_{j}\ne a_{j}^{**1}$ for any $j\in N_{-i}$. But given that $a^{**1}\in A^{**}$ and $\left| A^{**}\right| >1$, Lemma 1 implies that $\bar{p}_{i}(a^{**1})=\frac{\left( 1-\bar{p}_{i}\left( a^{**1}\right) \right) }{ k-1}=\frac{1}{k}$. Therefore, (9) becomes:

$$\begin{aligned} \left( \frac{1}{k}\right) ^{n-1}\sum _{a_{-i}\in A_{-i}}u_{i}(a_{i}^{**1},a_{-i})=\left( \frac{1}{k}\right) ^{n-1}\sum _{a_{-i}\in A_{-i}}u_{i}(a_{i}^{**2},a_{-i}) \end{aligned}$$

(10)

which necessarily requires:

$$\begin{aligned} \sum _{a_{-i}\in A_{-i}}u_{i}(a_{i}^{**1},a_{-i})=\sum _{a_{-i}\in A_{-i}}u_{i}(a_{i}^{**2},a_{-i}). \end{aligned}$$

(11)

It follows that if the above condition does not hold then $a^{**1}$ and $a^{**1}$ cannot simultaneously belong to $A^{**}$. Therefore, a sufficient condition for the $\mathbf {p}_{u}$-dominance criterion to select a unique equilibrium is that the sum of an agent’s individual payoffs across any equilibrium action differs. More formally:

$$\begin{aligned} \sum \limits _{a_{-i}\in A_{-i}}u_{i}(a_{i}^{*\psi },a_{-i})\ne \sum \limits _{a_{-i}\in A_{-i}}u_{i}(a_{i}^{*\xi },a_{-i})\quad \text { for any } a_{i}^{*\psi },a_{i}^{*\xi }\in A^{*}\text { and }a_{i}^{*\psi }\ne a_{i}^{*\xi } \end{aligned}$$