<p>NO.</p>

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

<p>(e,0)   (e,1)</p>

<p>(0,0)   (0,e)</p>

<p>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.</p>

CC BY-SA 2.5
2010-08-23T19:14:22.650Z
2010-08-23T19:14:22.650Z
Noam
<p>My question: is the set of potential games closed under convex combinations?</p>

<p>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:
$$\forall_{a\in A} \forall_{a_{i},b_{i}\in A_{i}} \Phi(b_{i},a_{-i})-\Phi(a_{i},a_{-i}) = u_{i}(b_{i},a_{-i})-u_{i}(a_{i},a_{-i})$$</p>

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

<p>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 $u^p_i(\cdot) = (1-p)u_i(\cdot) + pu'_i(\cdot)$. </p>

<p>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. </p>

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

CC BY-SA 4.0
2018-12-06T14:43:22.257Z
Here is the same question on MathOverflow: [Is the convex combination of two potential games a potential game?](https://mathoverflow.net/q/36457) (Which is probably what the previous comment is about.)
Martin
CC BY-SA 2.5
2010-08-24T06:06:02.087Z
Aaron, please see http://meta.cstheory.stackexchange.com/questions/25/material-to-supplement-the-faq/36#36 for guidelines about crossposting. Also, it might help to share the answers on either site.
Suresh Venkat
<gt.game-theory>
Is the convex combination of two potential games a potential game?
Aaron Roth
# Is the convex combination of two potential games a potential game?

(Asked by: username_0 on 2010-08-23)

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?

**Tags:** gt.game-theory
> **Comments:**
> **username_2**: Aaron, please see http://meta.cstheory.stackexchange.com/questions/25/material-to-supplement-the-faq/36#36 for guidelines about crossposting. Also, it might help to share the answers on either site.  
> **username_1**: Here is the same question on MathOverflow: [Is the convex combination of two potential games a potential game?](https://mathoverflow.net/q/36457) (Which is probably what the previous comment is about.)  

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## 1 answers

### Answer by username_3 on 2010-08-23. Score: 12 | Accepted answer

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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