2
$\begingroup$

Consider a game with $N$ players, each indexed by $i=1,...,N$. Every player $i$ has to choose a $J\times 1$ vector of actions $a_i\equiv (a_{i,1},...,a_{i,J})$ where each $a_{i,j}$ can be zero or one. The payoff of each player $i$ is $u_i(a_i, a_{-i})$, where $a_{-i}$ denotes the actions of the other players.

A pure strategy Nash equilibrium (PSNE) of the game is $a^*\equiv (a_1^*,...,a_N^*)$ solving $$ (1) \quad a_i^*\in argmax_{a_i\in \{0,1\}^J} u_i(a_i, a^*_{-i}) \quad \forall i=1,...,N $$

Note that if $a^*$ is a PSNE, then $$ (2) \quad \text{If $a^*_{i,j}$=0, then } u_i(a^*_i, a^*_{-i})\geq u_i(a^*_i+\{a_{i,j}=1\}, a^*_{-i})\\ \quad \quad \text{If $a^*_{i,j}$=1, then } u_i(a^*_i, a^*_{-i})\geq u_i(a^*_i+\{a_{i,j}=0\}, a^*_{-i})\\ \forall i=1,...,N \quad \forall j=1,...,J $$ where $a^*_i+\{a_{i,j}=1\}$ denotes $a^*$ where $a^*_{i,j}=0$ is replaced by one; $a^*_i+\{a_{i,j}=0\}$ denotes $a^*$ where $a^*_{i,j}=1$ is replaced by zero.

Claim: Assume that ties are zero probability events. Then, an action profile $a$ satisfies (2) if and only if $$ (3) \quad a_{i,j}=1\{u_i(a_i+\{a_{i,j}=1\}, a_{-i})- u_i(a_i+\{a_{i,j}=0\}, a_{-i})\geq 0\}\\ \forall i=1,...,N \quad \forall j=1,...,J $$

Question: Let $A$ be the set of PSNE. Let $B$ be the set of $a$ satisfying (3). Is $A\subseteq B$? If $B$ is empty, then $A$ is empty?

$\endgroup$
4
  • 1
    $\begingroup$ The last $N$ should be a $J$, right? $\endgroup$ – VARulle Apr 30 at 22:30
  • 1
    $\begingroup$ I edited this (and a missing bracket). $\endgroup$ – VARulle Apr 30 at 22:34
  • 1
    $\begingroup$ @user3285148 For you question, are you also assuming that ties never occur? $\endgroup$ – tdm May 1 at 11:01
  • $\begingroup$ Yes, ties never occur. $\endgroup$ – user3285148 May 1 at 11:32
2
$\begingroup$

Condition 3 can be written in the following way: $a^\ast \in B$ iff for all $i$ and all $j$ $$ a^\ast_{i,j} = 1 \iff u_i(a_i^\ast + \{a_{i,j} = 1\}, a^\ast_{-i}) \ge u_i(a_i^\ast + \{a_{i,j} = 0\}, a^\ast_{-i}) $$ As there are no ties, this can be written as: $$ a^\ast_{i,j} = 1 \to u_i(a_i^\ast + \{a_{i,j} = 1\}, a^\ast_{-i}) \ge u_i(a_i^\ast + \{a_{i,j} = 0\}, a^\ast_{-i}),\\ a^\ast_{i,j} = 0 \to u_i(a_i^\ast + \{a_{i,j} = 0\}, a^\ast_{-i}) \ge u_i(a_i^\ast + \{a_{i,j} = 1\}, a^\ast_{-i}) $$

If $a^\ast$ is a Nash equilibrium (i.e. $\in A$), then for all $i$ and all $a_i$: $$ u_i(a_i^\ast, a_{-i}^\ast) \ge u_i(a_i, a_{-i}^\ast) $$

Is $A \subseteq B$?

Yes. Let $a^\ast \in A$. We want to show that $a^\ast \in B$. To show that $a^\ast \in B$, we only need to look at deviations where $a_{i,j}^\ast$ changes for a particular $j$. Towards this end, define $a_i = (a_{i,\ell}|\ell \le J)$ such that: $$ a_{i,\ell} = \left\{\begin{array}{ll} a^\ast_{i,l} &\text{ if } \ell \ne j\\ 1- a_{i,j}^\ast &\text{ if } \ell = j \end{array}\right. $$ So $a_i$ differs from $a_i^\ast$ only because $a_{i,j}$ switched value (from 1 to 0 or vice versa). To show that $a^\ast \in B$, we show the two conditions.

  1. If $a_{i,j}^\ast = 1$ , we have:

$$ u_i(a_i^\ast, a_{-i}^\ast) = u_i(a_i^\ast + \{a_{i,j} = 1\}, a_{-i}^\ast),\\ \ge u_i(a_i, a_{-i}^\ast) = u_i(a_i^\ast + \{a_{i,j} = 0\}, a_{-i}^\ast). $$ The inequality follows from the definition of a Nash equilibrium. The equalities follow from how we defined $a_i$.

  1. If $a^\ast_{i,j} = 0$, we have:

$$ u_i(a_i^\ast, a_{-i}^\ast) = u_i(a_i^\ast + \{a_{i,j} = 0\}, a_{-i}^\ast),\\ \ge u_i(a_i, a_{-i}^\ast) = u_i(a_i^\ast + \{a_{i,j} = 1\}, a_{-i}^\ast). $$ This shows that $a^\ast \in B$.

If $B$ is empty, then $A$ is empty?

This follows from $A \subseteq B$. If $B = \emptyset$ and $A$ is a subset of $B$ then immediately $A = \emptyset$.

Notice that the inclusion can be is strict, i.e. there are cases where $A \ne B$. To see this, consider the case where $J = 2$ and the payoff table is the following.

(00) (01) (10) (11)
(00) 4,4 1,1 2,2 4,0
(01) 2,2 2,3 1,4 1,0
(10) 1,4 3,2 4,3 2,0
(11) 0,4 4,0 0,2 3,3

We have that $(11,11)$ is in $B$ but not in $A$ as it is not a Nash equilibrium. The only Nash equilibrium is $(00,00)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.