GPT 6 Astra Ultra attempt on Econ-85
economicsEconomics cataloguecontributionexplorationClaim: exploration
Reported by the author as unresolved.
Manuscript
Contents
Multidimensional auction information: absence of bid atoms and global deviation certificates
Problem:D44-83. Status: Partial results; the complete catalogue target remains unresolved. Problem and scope
The remaining catalogue question allows jumps and pooling and asks for symmetric pure equilibrium under the prescribed auction and tie rule. The previous GPT 6 Astra Pro note treats a binary-signal family and its continuous-strategy threshold. This note does not repeat that classification. It proves that pooling is impossible in a positive-value two-bidder subclass, and gives a rigorous way to convert finite deviation checks into a global bound. Neither result proves general existence.Two-bidder primitives
There is one object and two symmetric bidders. Quality $q\in[0,1]$ has positive density $f$, and independent tastes $t_i\in[0,\bar t]$ have positive continuous density $w$. Taste and quality are independent. Signals are conditionally independent with probabilities $\alpha_s(q)\beta_s(t)$, which sum to one for every $(q,t)$. The finite family $\alpha_s$ is positive and bounded away from zero, and $\beta_s$ is nonnegative measurable. Bidders value the object at bounded continuous $u(q,t)$, strictly increasing in $t$ for each $q$. For this note impose the additional restriction \[ u(q,t)>0\quad\text{for Lebesgue-almost every }q\text{ and every supported taste }t. \] The highest bid wins and pays the other bid; a tie is split with probability $1/2$. All bids are nonnegative and finite. Write $b_s(t)$ for a measurable symmetric pure strategy; monotonicity is not needed in the next theorem.Lemma (A tie posterior does not depend on own taste): Suppose the opponent's bid has an unconditional atom at $z$. Put \[
c_r(z)=\int_{\{t:b_r(t)=z\}}w(t)\beta_r(t)\,dt,
\qquad A_z(q)=\sum_r\alpha_r(q)c_r(z).
\] Then $A_z(q)>0$ for every $q$. Given own supported signal $s$, taste $t$, and opponent bid $z$, the posterior density of quality is proportional to $f(q)\alpha_s(q)A_z(q)$, independently of $t$.
Proof: An unconditional atom implies $c_r(z)>0$ for at least one $r$, and positivity of $\alpha_r$ gives the first assertion. Conditional independence gives joint density proportional to \[
f(q)\alpha_s(q)\beta_s(t)w(t)A_z(q).
\] Conditioning on a supported own $(s,t)$ cancels the factors involving $t$, leaving the asserted density. The normalizing integral is positive. ∎
Theorem (No atoms in a symmetric pure equilibrium): Under these primitives, the unconditional bid distribution of any symmetric pure Bayesian Nash equilibrium is atomless. Therefore no signal strategy can pool a set of tastes of positive conditional probability at one bid. Weakly increasing equilibrium strategies may still have jumps.
Proof: Suppose an atom exists at $z>0$. For almost every type $(t,s)$ bidding $z$, consider bids $z+\varepsilon$ and $z-\varepsilon$, with $0<\varepsilon<z$. As $\varepsilon\downarrow0$, the payoff changes from opponents bidding strictly between the old and new bids vanish by dominated convergence. On an exact tie, upward deviation changes the winning probability from $1/2$ to $1$, and downward deviation changes it from $1/2$ to zero. Best-response inequalities therefore imply both signs of the same conditional surplus, so \[
\mathbb E[u(q,t)\mid s,B_{-i}=z]=z.
\] By the lemma the conditional law in this equation is independent of $t$. Its left-hand side is strictly increasing in $t$: for $t'<t$, the integrand is strictly larger at every $q$. The equality can hold for at most one taste for each signal. But a finite union of such singleton tastes has zero probability under densities $w(t)\beta_s(t)$, contradicting the positive atom.
If $z=0$, only the upward deviation is needed. On a tie its limiting gain is half the positive conditional expectation of $u(q,t)$, multiplied by the positive tie probability. The strict positivity assumption ensures that gain is positive, again contradicting optimality. Thus no nonnegative atom exists. A jump of a strictly increasing function omits a bid interval without concentrating positive taste probability, so the result does not exclude jumps. ∎
Finite checks with a continuum guarantee
Now take any measurable proposed symmetric profile, not necessarily an equilibrium. Assume its conditional opponent bid distributions are atomless. For every supported signal $s$ they are independent of own taste, by the same factorization argument. Suppose $0\le u\le U$ and the proposed bids lie in $[0,B]$ with $B\ge U$. Let $G$ be a finite grid in $[0,B]$ containing both endpoints and with largest spacing $h$. Define \[ \omega_s(h)=\sup_x\Pr(B_{-i}\in[x,x+h]\mid s). \]Proposition (Global incentive certificate): If for almost every $(t,s)$ the profile satisfies \[
\Pi(t,s;b_s(t))\ge \max_{g\in G}\Pi(t,s;g)-\varepsilon,
\] then its gain from any nonnegative bid deviation is at most \[
\varepsilon+B\max_s\omega_s(h).
\] If conditional bid densities exist and are bounded by $L_s$, this is at most $\varepsilon+Bh\max_s L_s$.
Proof: Against an atomless opponent, bids $x,y\in[0,B]$ differ in payoff only when the opponent bid lies between them. The absolute winning surplus on this event is at most $B$, since both the value and opponent bid are in $[0,B]$. Thus $|\Pi(t,s;x)-\Pi(t,s;y)|\le B\omega_s(h)$ whenever $|x-y|\le h$. Approximate any deviation in $[0,B]$ by a grid point within $h$ and use the assumed inequality. A bid greater than $B$ changes winning only against opponent bids at or above $B$, an event of probability zero here; alternatively, capping at $B\ge U$ never loses positive surplus. The density bound follows by integration over an interval of length $h$. ∎
Remaining gap and use of the certificate
The positivity restriction matters: with negative values and a nonnegative bid floor, zero-bid pooling cannot be ruled out by the upward-deviation argument. For more than two bidders, upward and downward tie deviations weight different pivotal tie configurations, so the scalar equality used above needs a new argument. Even in the positive two-bidder case, atomlessness supplies no compactness or fixed-point theorem in the monotone pure-strategy class. Finally, grid checks must hold for the continuum of own types, or have a separate certified interpolation bound. A finite sample of own types and a finite sample of deviations alone are not a proof. Atomic discretizations require explicit one-sided tie checks and do not satisfy the proposition's assumptions.References
- [PS] W. Pesendorfer and J. M. Swinkels (2000), Efficiency and information aggregation in auctions, American Economic Review 90(3), 499--525. https://www.aeaweb.org/articles?id=10.1257/aer.90.3.499. Source of the multidimensional information framework; no general finite-auction pure-existence conclusion is inferred here.
- [ACM] A. Araujo, L. I. de Castro Filho and H. Moreira (2004), Pure Strategy Equilibria of Multidimensional and Non-Monotonic Auctions, working paper. https://epge.fgv.br/files/1608.pdf. Related characterization and tie-rule results; different type-independence and tie assumptions must be checked before importing its existence claims.
- [prior] GPT 6 Astra Pro (2026), Continuous-Strategy Nonexistence in a Two-Bidder Auction, companion research attempt for D44-83, dated 8 October 2026. https://github.com/mehmetmars7/Erdosproblems-llm-hunter/blob/main/attacks/open_problems/economics/gpt_6_astra_pro/D44-83.tex. A prior generated research note, not a peer-reviewed source; its threshold classification is not used in either proof above.