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My question: is the set of potential games closed under convex combinations?

An n player game with action set $A = A_1 \times \ldots \times A_n$ and payoff functions $u_i$ is called an exact potential game if there exists a potential function $\Phi$ such that: $$\forall_{a\in A} \forall_{a_{i},b_{i}\in A_{i}} \Phi(b_{i},a_{-i})-\Phi(a_{i},a_{-i}) = u_{i}(b_{i},a_{-i})-u_{i}(a_{i},a_{-i})$$

A game is a general (ordinal) potential game if there exists a potential function $\Phi$ such that: $$\forall_{a\in A} \forall_{a_{i},b_{i}\in A_{i}} sgn(\Phi(b_{i},a_{-i})-\Phi(a_{i},a_{-i})) = sgn(u_{i}(b_{i},a_{-i})-u_{i}(a_{i},a_{-i}))$$

Say that we have two games on the same action set, with utility functions $u_i$ and $u'_i$ respectively, for each player $i$. For any $0 \leq p \leq 1$, there is a convex combination of these two games, again on the same action set, where the utility function for each player $i$ is now $u^p_i(\cdot) = (1-p)u_i(\cdot) + pu'_i(\cdot)$.

Clearly, the convex combination of two exact potential games is also an exact potential game: just take the same convex combination of the two potential functions.

But is it possible to have two (general) potential games such that their convex combination is not a potential game?

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NO.

The following two-player game, with e very close to 0, is a potential game:

(e,0) (e,1)

(0,0) (0,e)

Taking the 8 equivalent games (switching the roles of the players / rows / columns), then a convex combination of them gets you as close as you want to any game with positive utilities, like to matching pennies with a positive constant added to each utility. But that is not a potential game.

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