# Why all games are not Potential?

A definition given in wikipedia of an exact potential game as follow:

A game $$G=(N,A=A_{1}\times\ldots\times A_{N}, u: A \rightarrow \mathbb{R}^N)$$ is: an exact potential game if there is a function $$\Phi: A \rightarrow \mathbb{R}$$ such that $$\forall {a_{-i}\in A_{-i}},\ \forall {a'_{i},\ a''_{i}\in A_{i}}$$,

$$\Phi(a'_{i},a_{-i})-\Phi(a''_{i},a_{-i}) = u_{i}(a'_{i},a_{-i})-u_{i}(a''_{i},a_{-i})$$

That is: when player $$i$$ switches from action $$a'$$ to action $$a''$$, the change in the potential equals the change in the utility of that player.

A well know result related to potential games is that:

• Potential games have at least one Nash equilibrium.
• All Nash equilibrium are the maximizers of the potential function, either locally or globally.
• There are several algorithms available which guaranteed to converge to a Nash equilibrium.

My question is simple: It could be wrong. So please correct me.

Why not all games are potential games? If one chooses $$\Phi=u_i$$, then we have the previous condition verified. What is worng in this?

• The potential $\Phi$ must have the defining property for all players simultaneously. This puts conditions on how the players' utilities must relate to one another. Apr 16 '14 at 19:27
mjqxxxx already answered in the comments: $i$ is indexing the players, so different players may have different payoffs, $u_i$. The potential condition requires that the gain (or loss) from changing the action of one player is the same for the potential function. Maybe the differential case is easy to illustrate: we require that $$\dfrac{\partial}{a_i}\Phi(a)=\dfrac{\partial}{a_i}u_i(a).$$
• That means if a game which has a payoff function that does not depend on the player, i.e., $u_i=u\;,\forall\, i$, then it is a potential game? Apr 16 '14 at 19:44
• another example: if payoffs are separable $u_i(a)=f^i_1(a_1)+...+f^i_I(a_I)$ then $\Phi(a)=\sum_{i=1}^I f^i_i(a_i)$ is a potential. Apr 16 '14 at 20:12