Let $p:A \times B \to \mathbb{R}$ be a nonnegative real-valued function on $A \times B$, where $A$ and $B$ are arbitrary set. Assume $f:A \to B$ and $g:B \to A$ are such that \begin{align*} f(a) &= \operatorname*{arg\,min}_b~p(a,b) \\ g(b) &= \operatorname*{arg\,max}_a~p(a,b) \end{align*}

Does the map $f \circ g:B \to B$ have a fixed point?
Which conditions are necessary for this fixed point to exists?

This question arised while searching an equilibrium point for a game. For my game, I empirically noticed that repeated iteration of the map $f \circ g$ eventually gives me a fixed point. So I start looking for some theoretical justification of this observation. I dig hard, but I couldn't really find one.

My game is the fair pricing of an insurance product.
The set $A$ is a convex subset of $\mathbb{R}^n$, and is the set of action of the policyholder. While the set $B=[0,1]$ is the set of premium of the insurer.
The function $f$ represents the objective function of a fair insurer who wants the price to be zero. While the function $g$ is the objective function of a rational policyholder who wants to arbitrage the insurer.
The function $p$ is such that $p(a,b)=\operatorname{E}(X \mid \text{Action=a, Premium=b})$ where $X=|\text{claim}-\text{premium}|$ is the random variable of the product cash flow.

I looked into Banach fixed point theorem, Nash equilibrium existence theorem and Kakutani's fixed point theorem.

Any help or pointer appreciated.


I'm not sure if this works or not, but I think your general problem can basically just be turned into a normal form game.

You have two players with strategy sets $A$ and $B$ where both players have the same utility function $p:A\times B\rightarrow\mathbb{R}$. The players best response functions are $f$ and $g$ respectively as you defined them above.

Right off the bat one could point out that these aren't in general functions but a correspondence (eg. what if it's a constant function?), and take for example if you defined $p:\mathbb{R}-\{0\}\times\mathbb{R}-\{0\}\rightarrow\mathbb{R}$ by $p(a,b) = \frac{1}{a} + \frac{1}{b}$, what's $f(2)$ for example? It may not even be a correspondence.

Now, if $g \circ g$ has a fixed point iff $f \circ f$ has a fixed point then I think $g \circ g$ will have a fixed point iff the game given above has a Nash equilibrium (the game's best response correspondence has a fixed point) using the strategies considered above (ie. if A and B are pure strategy sets then it would require the game has a pure strategy Nash equilibrium which is not in general true, whereas if A and B were probability distributions over some other arbitrary sets (mixed strategies) then by Nash's theorem there would exist at least one Nash equilibria).

So basically you need to consider what game is formed more specifically with your example, at least I think you could argue it this way.

  • $\begingroup$ I'm not clear on how to interpret mixed strategies in my case. The strategy sets $A$ and $B$ are both convex subset of $\mathbb{R}^n$. Let $\Pi(A)$ be the set of possible probability distribution on $A$. My understanding is that the mixed strategies are $\Pi(A) \times \Pi(B)$. Aren't these different? $\endgroup$ – user14108 Jun 14 '12 at 12:22

I think the conditions necessary for this fixed point to exists are: A and B both are Banach space and p is a continuous function.

But these may not be enough.

  • $\begingroup$ which theorem do you have in mind? $\endgroup$ – user14108 Jun 14 '12 at 12:09

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