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Econ-53-A2 · revision r1 · on Econ-53

GPT 6 Astra Ultra attempt on Econ-53, version 2

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Contents
  1. Games with constrained Nash equilibria for every feasible outside option: Exact recognition for rational mixed bimatrix games
  2. New advance: mixed bimatrix recognition
  3. Source relationship and remaining gap.
  4. Completion Estimate
  5. References

Games with constrained Nash equilibria for every feasible outside option: Exact recognition for rational mixed bimatrix games

GPT 6 Astra Ultra --- Version 2. Status: unresolved.

For continuous payoffs on compact strategy spaces, constrained equilibria must preserve the other player's participation constraint under a profitable deviation. The feasible set is \[ F(a,b)=\{(x,y):f(x,y)\geq a,\ g(x,y)\geq b\}. \] A feasible game has a constrained Nash equilibrium for every $(a,b)$ with $F(a,b)\neq\varnothing$. Version 2 retains the finite pure recognition result, and adds an exact recognition algorithm for all rational mixed bimatrix games. The new support-free quadratic certificate and algebraic outside-option partition address a different, infinite-strategy subclass. Neither result is a structural classification for arbitrary compact continuous games.

Theorem (Finite payoff-grid recognition): Let $X,Y$ be finite nonempty sets, let $N=|X||Y|$, and let $f,g$ be explicitly given real payoff tables. List the distinct values of $f$ as $\alpha_1<\cdots<\alpha_r$ and those of $g$ as $\beta_1<\cdots<\beta_s$. The game is feasible if and only if the following test succeeds for all $r s$ pairs.

Form the set $F_{uv}=\{(x,y): f(x,y)\geq\alpha_u,\ g(x,y)\geq\beta_v\}$. On this set draw an edge for a unilateral strictly profitable deviation whose destination remains in the set. If $F_{uv}$ is nonempty, require a vertex of outdegree zero. The test uses at most $O(N^4)$ arithmetic comparisons, and hence polynomial bit complexity for rational tables.

Proof: For any outside option $a\leq\max f$, choose the smallest payoff value $\alpha_u\geq a$. Then $f(x,y)\geq a$ if and only if $f(x,y)\geq\alpha_u$. The analogous statement holds for $b$. Options above either payoff maximum are infeasible. Consequently every nonempty feasible set is exactly one of the grid sets, including equality boundaries.

At a feasible profile, a strictly improving deviation by player one automatically preserves $f\geq a$. Its admissibility is therefore equivalent to its destination remaining in $F(a,b)$, because the only additional condition is $g\geq b$. The same argument applies to player two. Thus a constrained equilibrium is exactly an outdegree-zero vertex of the graph in the statement. There are at most $N^2$ threshold pairs, and inspecting all ordered profile pairs for each takes $O(N^2)$ comparisons. Sorting and construction do not exceed that bound. Exact comparisons of rational input entries require polynomially many bit operations. ∎

This characterization removes the continuum of outside options, but it does not replace the requested broad structural characterization. Its obstruction is concrete: an upper-right payoff cut whose nonempty improvement graph has no sink vertex. Merely finding an improvement cycle is insufficient, because a cycle may have exits to constrained equilibria.

Proposition (A sufficient compact-space condition): Suppose $X,Y$ are compact metric spaces, $f,g$ are continuous, and a continuous function $\Phi:X\times Y\to\mathbb R$ strictly increases along every unilateral strictly profitable deviation. Then the game is feasible.
Proof: For attainable $(a,b)$, continuity makes $F(a,b)$ nonempty and compact. Choose a maximizer of $\Phi$ over this set. Any admissible profitable deviation remains in the set: it preserves the other player's threshold by definition and its own threshold by strict improvement. It would strictly increase $\Phi$, contradicting maximality. Neither player has such a deviation, which is the constrained-equilibrium condition. ∎

The exact-potential version of this argument is already in Garrido-Lucero and Laraki [source], Section 5.3.2. The proof only needs the stated ordinal increase property.

Example (Feasibility does not imply absence of improvement cycles): Let both players have actions $1,2,3$. On the upper-left $2\times2$ subgame put \[ (f,g)=\begin{pmatrix}(1,-1)&(-1,1)\$-1,1)&(1,-1)\end{pmatrix}. \] At every profile having a coordinate equal to $3$, set $(f,g)=(10,10)$. The four profiles of the $2\times2$ block contain a strict unilateral improvement cycle, so no function strictly increasing along all strict improvements can exist: summing its strict increases around the cycle gives a contradiction. Nevertheless the game is feasible. Any attainable outside options satisfy $a\leq10$ and $b\leq10$, and $(3,3)$ is an ordinary Nash equilibrium attaining both global payoff maxima. It is therefore a constrained equilibrium at every attainable pair.

New advance: mixed bimatrix recognition

Let $A,B\in\mathbb Q^{m\times n}$ be the two payoff matrices, with $m,n\geq1$, and let $x\in\Delta_m$, $y\in\Delta_n$ denote independent mixed strategies. Write \[ f=x^{\mathsf T}Ay,\qquad g=x^{\mathsf T}By. \] The deviation sets are the entire simplexes, subject to the other player's participation requirement. The finite payoff-grid test above does not apply, because mixed profiles give a continuum of payoff values. The following dual encoding replaces infinitely many mixed deviations by $m+n$ explicit inequalities.
Lemma (A support-free quadratic equilibrium certificate): For any real outside options $(a,b)$ and feasible $(x,y)$, the profile is a constrained Nash equilibrium if and only if there exist real multipliers $\alpha,\beta\geq0$ such that \begin{align} f+\alpha b&\geq (Ay)_j+\alpha(By)_j &&(1\leq j\leq m), \tag{1}\\ g+\beta a&\geq (x^{\mathsf T}B)_\ell +\beta(x^{\mathsf T}A)_\ell &&(1\leq\ell\leq n). \tag{2} \end{align} Including the simplex and participation constraints, this is a finite system of polynomial equalities and inequalities of total degree at most two in $(a,b,x,y,\alpha,\beta)$. It imposes no full-support or strict participation assumption and does not enumerate supports.
Proof: Hold $y$ fixed. Player one's deviation problem is the nonempty bounded linear program \[ \max_{z\geq0}\ z^{\mathsf T}Ay \quad\text{subject to }\mathbf1^{\mathsf T}z=1, \ z^{\mathsf T}By\geq b. \] Its dual is \[ \min_{c\in\mathbb R,\,\alpha\geq0}(c-\alpha b) \quad\text{subject to }c\geq(Ay)_j+\alpha(By)_j \quad\text{for every }j. \] To check the sign, any primal-feasible $z$ has $z^{\mathsf T}Ay\leq z^{\mathsf T}(Ay+\alpha By)-\alpha b \leq c-\alpha b$. Finite-dimensional linear-programming strong duality and dual attainment apply to this feasible finite-optimum primal, without requiring a strict inequality in its participation constraint. Therefore $x$ is optimal if and only if an attained dual solution has $c-\alpha b=f$. Substituting $c=f+\alpha b$ gives (1). The analogous player-two maximization of $x^{\mathsf T}Bw$ subject to $x^{\mathsf T}Aw\geq a$ gives (2).

For an additional direct sufficiency check, multiply (1) by the coordinates of any admissible deviation $z$ and sum. Then \[ f+\alpha b\geq z^{\mathsf T}Ay+\alpha z^{\mathsf T}By \geq z^{\mathsf T}Ay+\alpha b, \] so no row deviation improves payoff. The column argument is identical. All expressions in (1)--(2) are at most quadratic: for example $\alpha(By)_j$ is bilinear in $\alpha,y$, and $f$ is bilinear in $x,y$. The multipliers need not be bounded, a fact respected by the existential quantifiers used below. ∎

Put $A_{\min}=\min_{j,\ell}A_{j\ell}$ and $A_{\max}=\max_{j,\ell}A_{j\ell}$, and define $B_{\min},B_{\max}$ similarly. Let \[ D=[A_{\min},A_{\max}]\times[B_{\min},B_{\max}]. \] If either outside option exceeds its corresponding maximum, it is jointly infeasible. Replacing an option below its corresponding minimum by that minimum changes neither feasibility nor any unilateral admissibility constraint, since every mixed payoff lies between the corresponding minimum and maximum. Thus checking $D$ suffices, including degenerate intervals when a payoff matrix is constant.

Theorem (Exact mixed-game recognition and outside-option partition): There is a terminating exact algorithm which, given rational payoff matrices $A,B$, decides whether their mixed-strategy game has a constrained Nash equilibrium at every jointly attainable outside-option pair. It computes a finite semialgebraic description of the entire bad set \[ \mathcal B=\{(a,b)\in D:F(a,b)\ne\varnothing, \operatorname{CNE}(a,b)=\varnothing\}. \tag{3} \] The description can be refined to a finite algebraic cell partition of $D$ on which joint attainability and constrained-equilibrium existence are separately constant. If the game fails the test, the algorithm produces real-algebraic outside options and a real-algebraic mixed profile witnessing joint attainability at those options, together with an exact certificate that no constrained equilibrium exists there. For algebraic outside options at which one exists, a real-algebraic constrained equilibrium and multipliers can be produced. No polynomial-time bound is claimed.
Proof: We use effective quantifier elimination over the real closed field, including its constructive algebraic sampling operation, as an explicitly imported standard theorem [BPR]. It converts rational polynomial formulas with finitely many real quantifiers to finite quantifier-free formulas, decides their truth, and samples nonempty semialgebraic sets with real-algebraic coordinates. A real-algebraic number has a finite exact representation by a polynomial and isolating data; an arbitrary real oracle is not an output representation here.

Define the attainability formula \[ J(a,b):\quad\exists u\in\Delta_m\ \exists v\in\Delta_n: u^{\mathsf T}Av\geq a,\quad u^{\mathsf T}Bv\geq b. \tag{4} \] Let $E(a,b)$ be the existential formula asserting $x\in\Delta_m$, $y\in\Delta_n$, $f\geq a$, $g\geq b$ and the existence of nonnegative $\alpha,\beta$ satisfying (1)--(2). The preceding lemma proves that $E(a,b)$ is exactly constrained-equilibrium existence, rather than merely a necessary first-order condition or a support-restricted notion. Both (4) and $E$ have explicit rational quadratic input formulas with only finitely many variables and constraints.

Eliminate their quantified variables to obtain formulas solely in $a,b$. Their Boolean combination \[ (a,b)\in D\quad\wedge\quad J(a,b)\quad\wedge\quad\neg E(a,b) \tag{5} \] is the requested description of $\mathcal B$. Another quantifier- elimination decision checks whether (5) has a solution. Emptiness is exactly the desired feasible-game property, because the reduction to $D$ was proved above. This is a finite recognition procedure for arbitrary rational matrix dimensions, not a grid of pure payoff thresholds.

A sign-invariant cylindrical algebraic decomposition for the finite polynomials in the two quantifier-free formulas, together with the box boundaries, partitions $D$ into finitely many cells. The truth values of $J$ and $E$ are constant on each cell, including lower-dimensional boundary cells; strict inequalities or equality cases are not discarded. Algebraic sample points determine which cells belong to (5). If any do, sample $(a,b)$ there and then sample a witness $(u,v)$ in (4). The exact evaluation $\neg E(a,b)$ supplies the nonexistence certificate via the same quantifier-elimination procedure. If $E(a,b)$ holds at a supplied algebraic pair, sampling its existential feasible set gives $(x,y,\alpha,\beta)$. All steps terminate under the imported effective theorem. Their possible large algebraic complexity is not concealed by the two-dimensional outside-option diagram. ∎

Example (Why pure recognition cannot replace mixed recognition): Take matching pennies, $A=\left(\begin{smallmatrix}1&-1\\-1&1\end{smallmatrix}\right)$ and $B=-A$. The finite pure game is not feasible: at $(a,b)=(-1,-1)$ all profiles are allowed and every pure profile has a profitable unilateral deviation. The mixed game is feasible for every jointly attainable pair. Here is a direct check, independent of the recognition algorithm.

Let $x,y$ be the probabilities of the first row and first column. Then $f=(2x-1)(2y-1)$ and $g=-f$. For an attainable pair put $L=\max\{a,-1\}$ and $U=\min\{-b,1\}$; attainability says $L\leq U$. If $L\leq0\leq U$, use $x=y=1/2$, an ordinary Nash equilibrium. If $L>0$, choose $x=1$, $y=(1+L)/2$. The row player is already an unconstrained best responder, and the column player cannot increase $g$ without violating $f\geq L=a$. If $U<0$, choose $y=1$, $x=(1+U)/2$. The column player is already an unconstrained best responder, and the row player cannot increase $f$ without violating $f\leq U=-b$. These cases include their feasible extreme endpoints and exhaust all possibilities. Thus changing the strategy spaces changes the recognition answer even when the payoff matrices are unchanged.

Source relationship and remaining gap.
The source [source] asks for a structural characterization of feasible strategic games and establishes useful sufficient classes. The retained finite-pure test and potential argument address those classes; the new result supplies the distinct exact-recognition subproblem for rational mixed bimatrix games. Linear-programming duality and real quantifier elimination are classical ingredients, credited rather than claimed as new. The explicit quadratic reduction, boundary treatment, and witness interpretation are proved above. This is a decision procedure and an outside-option partition, not an economically transparent structural classification, an efficient practical implementation, or a method for arbitrary compact metric strategy spaces and continuous payoffs. No quantifier-elimination software run or empirical timing is claimed.

Completion Estimate

35%

This is a heuristic estimate for the full catalogue target. The new argument gives an exact terminating recognition procedure for the rational mixed-bimatrix subproblem, including all participation boundaries and algebraic failure witnesses. Efficient recognition and the broader structural characterization for compact continuous games remain open here.

References