Symmetric equilibrium existence with multidimensional auction information

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.

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 \[ \omega_s(h)=\sup_x\Pr(B_{-i}\in[x,x+h]\mid s). \]

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.

Completion Estimate

15%

Tools. Web browsing: YES. Computation: YES (Python, exact integer/rational checks, and explicitly identified numerical diagnostics). Status convention. A full proof of a restricted theorem does not resolve the full input. Literature results are marked as imported; no novelty or exhaustive current-literature certification is claimed. Computational searches certify only their stated finite domains.

FORMAL RESTATEMENT

An admissible primitive tuple $\mathcal A$ consists of integers $n\ge2$, $1\le k<n$, $S\ge1$, a taste bound $\bar t>0$, strictly positive continuous probability densities $f$ on $[0,1]$ and $w$ on $[0,\bar t]$, and a $C^1$ function $u(q,t)$ strictly increasing in each variable. The common quality $q$ has density $f$. Independently conditional on $q$, bidders draw tastes with density $w$ and signals satisfying \[ \Pr(s_i=s\mid q,t_i)=\alpha_s(q)\beta_s(t_i),\qquad \sum_{s=1}^S\alpha_s(q)\beta_s(t)=1. \] Here $\alpha_s$ is $C^1$ and bounded away from zero, $\beta_s\ge0$ is measurable, and $\alpha_r/\alpha_s$ is strictly increasing in $q$ when $r>s$. The conditional laws are proper, not unnormalized likelihoods. Bidders are unit-demand and risk-neutral: a winner's utility is $u(q,t)$ minus its payment; a loser receives zero. The $k$ highest nonnegative bids win and each pays the $(k+1)$st highest bid; a tie for the remaining units is resolved uniformly among the tied bidders.

For a symmetric profile $b=(b_s)$ define its conditional payoff $\Pi_s(t,x;b)$ at each positive-probability conditional type. A pure Bayesian equilibrium requires, for almost every $(t,s)$ and every $x\ge0$, \[ \Pi_s(t,b_s(t);b)\ge\Pi_s(t,x;b). \] Write $\mathcal E^{\mathrm c}$ for continuous strictly taste-increasing profiles, $\mathcal E^{\mathrm{A2}}\subseteq\mathcal E^{\mathrm c}$ when each function is differentiable except at finitely many points, and $\mathcal E^{\mathrm{mon}}$ for measurable weakly taste-increasing profiles, allowing jumps and pooling. Null supported types impose no equilibrium restriction. An $\varepsilon$ equilibrium replaces the right side by $\Pi_s(t,x;b)-\varepsilon$.

The four questions are: (a) whether $\forall\mathcal A$, $\mathcal E_{\mathcal A}^{\mathrm{A2}}\ne\varnothing$; (b) classify both continuous and monotone existence for every $T>0$ in the binary family below; (c) whether $\forall\mathcal A$, $\mathcal E_{\mathcal A}^{\mathrm{mon}}\ne\varnothing$; and (d) give primitive restrictions and fixed-$\mathcal A$, every-$\varepsilon>0$ approximation results. A disproof of (a) is not a disproof of (c).

In (b), $n=2$, $k=1$, $q\sim U[0,1]$, $t_i\sim U[0,T]$, $u(q,t)=q+t$, $\beta_1=\beta_2=1$, and \[ \alpha_1(q)=\frac34-\frac q2,\qquad \alpha_2(q)=\frac14+\frac q2. \] All logarithms are natural. The degenerate $T=0$ is excluded because the stated positive-density taste model has $T>0$. Stress points are global rather than local optimality, possible singular continuous bid functions, common empty bid intervals, and keeping the auction rule unchanged.

QUICK LITERATURE/CONTEXT CHECK

The input names an 8 October 2026 research note, Continuous-Strategy Nonexistence in a Two-Bidder Auction. A separate copy of that note was not available in the eleven attachments, and the search did not independently retrieve it. Its reported threshold is therefore checked below directly from the supplied primitives; no unseen proof is imported.

Jackson's accessible author manuscript [jackson] exhibits nonexistence with discrete tastes, binary quality, and signals that can reveal quality completely. Those features do not satisfy this input's positive continuous densities and everywhere-positive signal likelihoods. It is relevant context, not a counterexample to the present primitive class. No general finite-auction monotone existence theorem is assumed here. The binary-family conclusions below follow from the displayed calculations and global incentive verification.

ATTACK PLAN

This is an equilibrium-existence and endogenous-information problem. The proof track derives posterior moments, obtains an inverse-bid system without assuming differentiability, and verifies candidate bids by the sign of every payoff increment. The construction track first separates the two bid supports and then removes a middle inverse-curve segment to create a common bid gap. The counterexample track tests the interval between the two reported thresholds and checks whether an argument rules out continuous strategies only, or all monotone strategies. The selected path completely treats the binary family, disproves universal A2 existence, and gives two restricted answers to (d). General question (c) is kept separate.

WORK

Posterior moments and a payoff formula

Set \[ a=\frac9{26},\qquad d=\frac{17}{26},\qquad \delta=\frac2{13}, \qquad a=\frac12-\delta,\quad d=\frac12+\delta. \] Direct polynomial integration gives \[ \left(\int_0^1\alpha_s(q)\alpha_r(q)\,dq\right)_{s,r} =\frac1{48}\begin{pmatrix}13&11\\11&13\end{pmatrix},\qquad \left(\int_0^1q\alpha_s(q)\alpha_r(q)\,dq\right)_{s,r} =\frac1{96}\begin{pmatrix}9&11\\11&17\end{pmatrix}. \] Each own signal has probability $1/2$. Consequently, with \[ (c_{sr})=\begin{pmatrix}13&11\\11&13\end{pmatrix},\qquad (\mu_{sr})=\begin{pmatrix}a&1/2\\1/2&d\end{pmatrix}, \] we have $\Pr(s_{-i}=r\mid s_i=s)=c_{sr}/24$ and ${\mathbb{E}}[q\mid s_i=s,s_{-i}=r]=\mu_{sr}$. The opponent's taste remains uniform and independent of these signals.

Against strictly increasing strategies, including strategies with jumps but no flat taste intervals, the opponent's bid distribution has no atoms. For every $t$ and $x\ge0$, \begin{equation}\label{wsref-0-65-71-3a-61-75-63-74-69-6f-6e-70-61-79} \Pi_s(t,x;b)=\frac1{24T}\sum_{r=1}^2 c_{sr} \int_{\{z:b_r(z)<x\}}\left(t+\mu_{sr}-b_r(z)\right)\,dz. \end{equation} This follows by conditioning on the opponent's signal and taste; the second price is exactly its bid. Ties have probability zero. The formula is continuous in $(t,x)$ when $b_r$ is strictly increasing, since the induced bid laws have no atoms and all their bids are bounded on the compact taste interval. Therefore, for continuous candidate strategies, an almost-everywhere best-response inequality extends to every type: first use a countable dense set of deviations $x$, extend in $t$ by continuity, and then extend in $x$.

A derivative-free necessary condition for continuous equilibrium

Construction and global verification for every taste range

Define, for $r\in[0,\infty)$, \[ B(r)=\frac12+H(r),\qquad X(r)=H(r)+\frac2{13+11r},\qquad Y(r)=H(r)-\frac{2r}{11+13r}. \] Differentiating and putting terms over a common denominator gives \begin{equation}\label{wsref-0-65-71-3a-58-59-64-65-72-69-76-61-74-69-76-65-73} X'(r)=\frac{1056(r+1)}{(13+11r)^2(11+13r)^2}>0, \qquad Y'(r)=rX'(r). \end{equation} Thus $X$ and $Y$ are strictly increasing on their ranges (the single point $Y'(0)=0$ does not create a constant interval). Their endpoint values are \[ (X(0),Y(0))=(\delta,0),\qquad (X(\infty),Y(\infty))=(T_c,T_c-\delta). \] The parameterized curve $(b,t_1,t_2)=(B(r),X(r),Y(r))$ satisfies both inverse-bid identities.

Two fully proved, but restricted, answers to variant (d)

VERIFICATION

Exact algebra. The accompanying Python program checks the pair-signal integrals, the identities $H'=h$, (eq:XYderivatives), and the reciprocal taste-gap relation using symbolic arithmetic. They can also be verified directly by clearing the displayed positive denominators. The signs $X'>0$, $Y'>0$ for $r>0$, and $h>0$ control every inverse and uniqueness claim.

Numerical diagnostic, not a certificate of exact equilibrium. The program obtains \[ T_c\approx0.1794277425784363,\qquad \int_0^\infty h(r)\,dr\approx0.17942774257843636. \] For $T=1/6$, solving $T(z)=T$ gives approximately \[ z=0.3328174256,\quad D=0.0127610759,\quad G=0.0766109090. \] The cutoff tastes are approximately $0.1650687527$ for signal 1 and $0.0015979140$ for signal 2. A discretized diagnostic using 10,000 opponent tastes per signal and 61 grid types plus two types immediately adjacent to each signal's cutoff found a maximum positive computed deviation residual of about $4.09\cdot10^{-6}$. This nonzero finite-grid residual is not rounded into an exact-zero claim; the exact certificate is Proposition Proposition (Global, not merely local, incentives). Its bound does not rely on quadrature or on an inability to find a profitable grid deviation.

\begin{lstlisting} compute posterior pair-signal probabilities and means exactly verify H' = h, X' > 0, and Y' = r*X' if T <= 2/13:     use the two separated affine strategies elif T >= Tc:     use the inverse curve, inserting length T-Tc at r=1 else:     solve Tc - (X(1/z)-X(z)) = T for z in (0,1)     delete the middle branch; shift the upper branch by -D for each signal and tested taste:     compare the prescribed payoff to all grid bid thresholds # Exact global incentives are proved by signed payoff increments. \end{lstlisting}

Adversarial checks. The continuous nonexistence proof does not divide by a potentially zero ordinary bid derivative: its inverse equations are Stieltjes identities, and the $r=1$ exceptional set is retained as a possible interval rather than discarded. The possibilities $r=0,\infty$ in the overlap interior are separately excluded. Endpoint best responses follow from continuity and full taste support. The discontinuous profile has no pooling and no bid atoms; assigning a cutoff type either gap endpoint is valid because the entire common gap gives the same payoff. The weak and strict inequalities at $T=\delta$ and $T=T_c$ are checked by the explicit constructions. The approximation proof varies the strategy only, not the primitives, the number of bidders, or the tie rule. The one-signal sufficient condition explicitly adds the nonnegative expected-value hypothesis missing from the general primitive list.

FINAL

Overall LABEL: UNRESOLVED.

Question (a): LABEL: FULL SOLUTION. SUBLABEL: COUNTEREXAMPLE/DISPROOF. The exact primitives in Corollary Corollary (Explicit disproof of question (a)), with $T=1/6$, disprove universal existence in A2, and even in the larger continuous strictly increasing class.

Question (b): LABEL: FULL SOLUTION. SUBLABEL: FULL PROOF. Theorem Theorem (Complete classification of the binary family) proves the reported continuous threshold classification without assuming differentiable strategies and proves monotone pure existence for every $T>0$. Both the continuous and discontinuous constructions satisfy global Bayesian incentive inequalities.

Question (d), restricted subclaims: FULL PROOF. Every fixed member of the binary family has continuous A2 $\varepsilon$ equilibria for every $\varepsilon>0$. Uninformative one-signal primitives with nonnegative expected values have an exact A2 equilibrium for every allowed $n,k$. These results do not characterize all primitives.

Exact first gap. For an arbitrary admissible informative primitive tuple, no argument here proves or disproves the existence of a symmetric measurable weakly taste-increasing pure equilibrium under the unchanged uniform-price and symmetric tie rules. The general fixed-auction approximation and primitive-characterization variants are also not completed.

Three next moves. (1) Extend the sign-of-increments construction to more than two signal classes, allowing endogenous common gaps and possible atoms. (2) Prove a monotone-strategy compactness result that controls tie payoffs at atoms instead of changing the tie rule. (3) Search three-signal or multi-unit primitives for an incompatibility among inverse-bid branches and check all nonlocal deviations, not only first-order conditions.

What a counterexample to (c) would need. It cannot be any member of the fully solved binary family, regardless of its positive taste bound. It must use genuinely different informative primitives, a different number of signals, or a different $n,k$, and must rule out jump and pooling equilibria rather than just continuous profiles. No claim that three signals or more bidders are necessary is made: another two-signal family might behave differently.

Completion Estimate

60%

The continuous-existence subquestion is disproved and the entire specified binary family, including monotone existence and fixed-auction continuous approximation, is treated. The general monotone pure-existence question and a general primitive characterization remain unresolved.

This is a subjective estimate of the fraction of the whole requested mathematical scope addressed by this report, including the named variants. It is not a probability of correctness, an objectively measured fraction of a proof, or an estimate that the remaining work takes the complementary fraction of the time. Only the eleven supplied files are assessed; no missing rank-1--200 statements are inferred.

References

• [jackson] M. O. Jackson. Non-Existence of Equilibrium in Vickrey, Second-Price, and English Auctions. Review of Economic Design 13 (2009), 137--145. Author manuscript dated 21 September 2005, especially Section 2. https://web.stanford.edu/~jacksonm/nonexist.pdf.
• [inputnote] Continuous-Strategy Nonexistence in a Two-Bidder Auction, 8 October 2026. Source note named in the supplied problem statement; a separate copy was not independently retrieved. The proof in this report uses the supplied explicit primitives, not an unseen note.

Problem ID: D44-83 Model: GPT 6 Astra Pro Version: 1

Status: Unresolved for the complete catalogue problem; this manuscript records partial results and research attempts.

Continuous-Strategy Nonexistence in a Two-Bidder Auction A sharp threshold classification, removal of the singular-inverse caveat, and an exact discontinuous equilibrium

Abstract

We verify the proposed counterexample to continuous, strictly increasing symmetric equilibrium existence in the auction environment of Pesendorfer and Swinkels (2000). For the specified binary signals, uniform quality, additive valuation, and tastes uniform on $[0,T]$, such an equilibrium exists exactly when $0<T\le 2/13$ or $T\ge{T_\star}$, where \[ {T_\star}=\frac16+\frac{11}{144}\log\!\frac{13}{11} =0.1794277425784363\ldots. \] Thus $T=1/6$ is a valid counterexample, whereas $T=1$ is not. We derive the pivotal equations as identities of Stieltjes measures, rather than presupposing bid densities. This closes the singular-inverse loophole: every continuous, strictly increasing equilibrium in this family necessarily has absolutely continuous inverse bids. A particularly short argument proves nonexistence at $T=1/6$ without evaluating the integral defining ${T_\star}$. We also give an important qualification absent from the earlier draft: for every $2/13<T<{T_\star}$, an exact symmetric, strictly taste-increasing equilibrium exists with a jump in each direct bid function. The result therefore refutes existence under the paper's maintained continuity restriction, not unrestricted symmetric equilibrium existence. All constructions are verified by global best-response inequalities.

Scope and attribution. The model and the historical existence question come from Pesendorfer and Swinkels (2000) [PS2000]. The proofs and additional constructions below are derived in this note; the cited papers are not represented as containing these particular results. No claim of publication priority or external peer review is made.

The question, the answer, and the source

What Pesendorfer and Swinkels (2000) leave unresolved

The article is Efficiency and Information Aggregation in Auctions, by Wolfgang Pesendorfer and Jeroen M. Swinkels, published in the American Economic Review in June 2000, volume 90(3), pages 499--525.

The paper studies $n$ bidders competing for $k$ identical objects, with $1\le k<n$. Each bidder knows a taste $t$ and receives a finite-valued signal about common quality $q$. The highest $k$ bids win and pay the $(k+1)$st-highest bid; ties are resolved by symmetric randomization. Its basic primitives impose positive continuous densities for quality and tastes and a continuously differentiable utility $u(q,t)$ strictly increasing in both arguments. Assumption 1 imposes a factorized signal structure, limited information, and strict monotone likelihood ratios. Assumption 2 is a restriction on the equilibrium strategies, not another restriction on the primitives: for each signal, bids are continuous, strictly increasing in taste, and differentiable except at finitely many points [PS2000, pp. 502--503].

For a self-contained formulation, write a primitive tuple as \[ \mathcal A=(n,k,S,\bar t,f,w,u,\alpha,\beta), \quad n\ge2,\quad 1\le k<n,\quad S\ge1,\quad\bar t>0. \] Here $f\in C([0,1])$ and $w\in C([0,\bar t])$ are strictly positive densities of total integral one; quality and the independent tastes are independent. The utility $u\in C^1([0,1]\times[0,\bar t])$ has $\partial_qu>0$ and $\partial_tu>0$. For $s=1,\dots,S$, $\alpha_s\in C^1([0,1])$, $\beta_s$ is nonnegative and measurable, and \begin{gather*} p(s\mid q,t)=\alpha_s(q)\beta_s(t),\qquad \alpha_s(q)\ge\eta>0,\\ \sum_s\alpha_s(q)=1,\qquad \sum_s\alpha_s(q)\beta_s(t)=1,\qquad \int_0^{\bar t}\beta_s(t)w(t){\,\mathrm d} t>0. \end{gather*} For every $r>s$, $\alpha_r(q)/\alpha_s(q)$ is strictly increasing in $q$. Signals are drawn independently across bidders conditional on quality and their tastes. These are the basic model and Assumption 1, with conditional-probability normalization made explicit.

For a bid profile $\mathbf b$, let $p(\mathbf b)$ be its $(k+1)$st-highest coordinate and let $\chi_i(\mathbf b,\omega)\in\{0,1\}$ indicate whether bidder $i$ receives an object, with $\omega$ implementing the stated symmetric tie-breaking. The interim payoff used below is \[ \Pi_{\mathcal A}(t,s;x,b_{-i}) ={\mathbb{E}}\!\left[\chi_i((x,b_{-i}),\omega) \{u(q,t)-p((x,b_{-i}))\}\mid t_i=t,s_i=s\right]. \]

The authors expressly say that they have not proved symmetric equilibrium existence in this environment. Their asymptotic results are conditional on equilibria satisfying the maintained restrictions, and Section IV constructs $\varepsilon$-equilibria for sufficiently large markets [PS2000, pp. 501, 505, 510--511]. The historical question can be formulated precisely as follows.

This note gives a negative answer to Question Question (The finite-auction continuity question). It does not contradict the conditional asymptotic theorems in the paper.

Main result

Fix the following family of auctions: \begin{equation} \label{wsref-0-65-71-3a-70-72-69-6d-69-74-69-76-65-73} n=2,\quad k=1,\quad q\sim U[0,1],\quad t_i\sim U[0,T],\quad u(q,t)=q+t, \end{equation} with independent tastes, independent of quality, and signals conditionally independent given quality, where \begin{equation} \label{wsref-0-65-71-3a-73-69-67-6e-61-6c-73} \pi_1(q)=\frac34-\frac q2, \qquad \pi_2(q)=\frac14+\frac q2. \end{equation} Let ${\mathcal E_T^{\mathrm{c}}}$ be the set of symmetric pure equilibria whose two direct bid functions $b_1,b_2:[0,T]\to{\mathbb{R}}_+$ are continuous and strictly increasing. No differentiability or absolute continuity is included in the definition of ${\mathcal E_T^{\mathrm{c}}}$.

The discontinuous existence assertion is proved in Section sec:jumps. It is essential to distinguish it from (eq:classification).

Bayesian calculations and the payoff formula

Verification of the assumptions

Here $f(q)=1$ and $w(t)=1/T$ are positive continuous densities on their respective compact intervals. Both partial derivatives of $u(q,t)=q+t$ equal one. In the factorization of Assumption 1, take $p_1(s\mid q)=\pi_s(q)$ and $p_2(s\mid t)=1$. Both signal probabilities are at least $1/4$, and \[ \frac{\pi_2(q)}{\pi_1(q)}=\frac{1+2q}{3-2q},\qquad \frac{\mathrm d}{\mathrm dq}\frac{1+2q}{3-2q} =\frac8{(3-2q)^2}>0. \] Thus limited information and strict MLRP both hold. The informative binary probabilities coincide with those described in Example 1 of Pesendorfer and Swinkels (2000); retaining only the two informative realizations gives exactly (eq:signals) [PS2000, p. 503].

Signal probabilities and conditional quality means

Both signals have unconditional probability $1/2$. Direct integration gives \begin{align} \left(\int_0^1\pi_s(q)\pi_r(q){\,\mathrm d} q\right)_{s,r} &=\frac1{48}\begin{pmatrix}13&11\\11&13\end{pmatrix}, \label{wsref-0-65-71-3a-6a-6f-69-6e-74-73-69-67-6e-61-6c-73}\\ \left(\int_0^1 q\pi_s(q)\pi_r(q){\,\mathrm d} q\right)_{s,r} &=\frac1{96}\begin{pmatrix}9&11\\11&17\end{pmatrix}. \label{wsref-0-65-71-3a-6a-6f-69-6e-74-71-75-61-6c-69-74-79} \end{align} Consequently, define \begin{equation} \label{wsref-0-65-71-3a-61-6d} \begin{aligned} a_{sr}&={\mathbb{P}}(s_j=r\mid s_i=s) =\frac1{24}\begin{pmatrix}13&11\\11&13\end{pmatrix}_{sr},\\[4pt] m_{sr}&={\mathbb{E}}[q\mid s_i=s,s_j=r] =\begin{pmatrix}9/26&1/2\\1/2&17/26\end{pmatrix}_{sr}. \end{aligned} \end{equation} In particular, ${\mathbb{E}}[q\mid s_i=1]=5/12$ and ${\mathbb{E}}[q\mid s_i=2]=7/12$.

Inverse bids as measures

Suppose for now that $b_1,b_2$ are continuous and strictly increasing. Write \[ I_s=[\ell_s,h_s]=[b_s(0),b_s(T)]. \] On $I_s$, let $t_s$ be the inverse of $b_s$; extend it by zero below $\ell_s$ and by $T$ above $h_s$. Then $t_s$ is continuous, nondecreasing, and of bounded variation. Its Stieltjes measure $\mathrm dt_s$ has total mass $T$ and no atoms. It has full support on $I_s$, whether or not it has a Lebesgue density.

For own type $(t,s)$, the expected payoff from a bid $x$ is \begin{equation} \label{wsref-0-65-71-3a-70-61-79-6f-66-66} V_s(t,x)=\frac1T\sum_{r=1}^2 a_{sr} \int_{(-\infty,x)}(t+m_{sr}-b)\,\mathrm dt_r(b). \end{equation} Indeed, conditional on the opponent's signal $r$, the opponent's taste is uniform and independent of quality. The measure $\mathrm dt_r/T$ is the conditional bid distribution, and the mean quality given both signals is $m_{sr}$. Ties have probability zero.

A continuous equilibrium that is optimal for almost every taste is optimal for every taste: $V_s(t,x)$ is continuous in $(t,x)$, and a strict violation at one taste would persist on a positive-measure neighborhood. We may therefore use the equilibrium inequalities for the endpoint tastes as well.

Pivotal equations without assuming bid densities

The pivotal-value equation in Proposition 1 of [PS2000, p. 504] motivates the following calculation. To avoid using the inverse-regularity assertion in its Appendix A, we derive the required equations directly from (eq:payoff).

A compact mixture parameter

Suppose that the interiors of the two supports overlap, and denote their intersection by $J=(L,U)$. Let \[ \mu=\mathrm dt_1+\mathrm dt_2,\qquad \lambda=\frac{\mathrm dt_2}{\mathrm d\mu}\in[0,1] \] initially as a Radon--Nikodym derivative, defined $\mu$-almost everywhere on $J$. Applying Lemma Lemma (Stieltjes pivotal identity) to both signals gives \begin{equation} \label{wsref-0-65-71-3a-63-6f-6d-70-61-63-74-70-69-76-6f-74-61-6c} t_s(b)=b-M_s(\lambda(b)),\qquad s=1,2, \end{equation} $\mu$-almost everywhere, where \begin{equation} \label{wsref-0-65-71-3a-4d-73} M_1(\lambda)=\frac{9+2\lambda}{2(13-2\lambda)}, \qquad M_2(\lambda)=\frac{11+6\lambda}{2(11+2\lambda)}. \end{equation} Both maps are smooth and strictly increasing on $[0,1]$, with \begin{equation} \label{wsref-0-65-71-3a-4d-73-70-72-69-6d-65} M_1'(\lambda)=\frac{22}{(13-2\lambda)^2}, \qquad M_2'(\lambda)=\frac{22}{(11+2\lambda)^2}. \end{equation}

The difference between the two pivotal tastes is particularly informative: \begin{equation} \label{wsref-0-65-71-3a-6d-65-61-6e-67-61-70} t_1(b)-t_2(b) =M_2(\lambda)-M_1(\lambda) =\frac{2[11+4\lambda(1-\lambda)]}{143+4\lambda(1-\lambda)}. \end{equation} It lies between $2/13$ and $1/6$, attaining its maximum $1/6$ exactly at $\lambda=1/2$.

Support geometry and the counterexample at $T=1/6$

Separated supports: incentives, not just geometry

Ordered overlap and boundary values

An elementary proof of nonexistence at the proposed width

This proof already closes the singular-inverse caveat for the specific counterexample. The following section proves the stronger automatic regularity result needed for the full sharp threshold ${T_\star}$.

Automatic regularity and the exact moving--sticking decomposition

Define the positive smooth function \begin{equation} \label{wsref-0-65-71-3a-4b} K(\lambda)= \frac{22(169-4\lambda+4\lambda^2)} {(13-2\lambda)^2(11+2\lambda)^2},\qquad 0\le\lambda\le1, \end{equation} and its strictly increasing primitive \begin{equation} \label{wsref-0-65-71-3a-48} H(\lambda)=\int_0^\lambda K(v){\,\mathrm d} v. \end{equation}

The proof is included in Appendix app:bv; continuity is important in both statements.

Relation to the original slope ratio

The parameter cannot equal zero or one at an interior bid: by monotonicity this would create a nondegenerate constant interval at an endpoint level, contradicting (eq:lambdaODE). Thus $0<\lambda<1$ in the interior. The derivative formulas below show that both inverse slopes are positive almost everywhere there. Writing \[ \rho=\frac{t_2'}{t_1'}=\frac{\lambda}{1-\lambda}, \] the pivotal means become exactly the ones in the draft: \[ Q_1(\rho)=\frac{11\rho+9}{2(11\rho+13)},\qquad Q_2(\rho)=\frac{17\rho+11}{2(13\rho+11)}. \] The compact kernel satisfies \[ K(\lambda)=\frac{G(\lambda/(1-\lambda))}{(1-\lambda)^2}, \quad G(\rho)=\frac{22(169\rho^2+334\rho+169)} {(11\rho+13)^2(13\rho+11)^2}. \] The differential equation in the original notation is \begin{equation} \label{wsref-0-65-71-3a-72-68-6f-4f-44-45} (1-\rho)\,[G(\rho)\rho'-1]=0. \end{equation} The moving branch has $\mathrm db/\mathrm d\rho=G(\rho)$, but $\rho\equiv1$ on a bid interval is a distinct valid branch. Cancelling $1-\rho$ discards that branch. A removable singularity in a quotient is not a license to discard solutions of the original degenerate equation.

On moving portions, with $N(\lambda)=169-4\lambda+4\lambda^2$, \begin{equation} \label{wsref-0-65-71-3a-69-6e-76-65-72-73-65-73-6c-6f-70-65-73} t_1'(b)=\frac{48(1-\lambda)}{N(\lambda)},\qquad t_2'(b)=\frac{48\lambda}{N(\lambda)}. \end{equation} Both equal $1/7$ at $\lambda=1/2$ on a moving branch; on a genuine sticking interval they both equal one. This is compatible with finitely many nondifferentiable junctions in the direct bids.

Construction and global verification of the continuous equilibria

Construction for every $T\ge{T_\star}$

Put $\ell=T-{T_\star}\ge0$ and \[ a=\frac12+\frac{{T_\star}}{2},\qquad c=a+\ell=\frac12+T-\frac{{T_\star}}{2}. \] On the common-support interval $[1/2,T+1/2]$, construct inverse bids as follows: \begin{align} b&=\frac12+H(\lambda),\quad t_s(b)=b-M_s(\lambda), &&0\le\lambda\le\tfrac12,\label{wsref-0-65-71-3a-63-6f-6e-73-74-72-75-63-74-6c-6f-77-65-72}\\ t_1(b)&=b-\frac5{12},\quad t_2(b)=b-\frac7{12}, &&a\le b\le c,\label{wsref-0-65-71-3a-63-6f-6e-73-74-72-75-63-74-73-74-69-63-6b}\\ b&=\frac12+\ell+H(\lambda),\quad t_s(b)=b-M_s(\lambda), &&\tfrac12\le\lambda\le1.\label{wsref-0-65-71-3a-63-6f-6e-73-74-72-75-63-74-75-70-70-65-72} \end{align} For $\ell=0$, the middle interval is degenerate. Complete the direct bids outside the overlap by \begin{align} b_1(t)&=t+\frac9{26},&&0\le t\le\frac2{13},\label{wsref-0-65-71-3a-6c-6f-77-65-72-65-78-74-65-72-69-6f-72}\\ b_2(t)&=t+\frac{17}{26},&&T-\frac2{13}\le t\le T.\label{wsref-0-65-71-3a-75-70-70-65-72-65-78-74-65-72-69-6f-72} \end{align} The values match at all joins by (eq:tasteendpoints) and $M_1(1/2)=5/12$, $M_2(1/2)=7/12$.

The moving inverse slopes in (eq:inverseslopes) are positive in the interior. On the sticking and exterior portions they equal one for the active signal. Thus each inverse is strictly increasing on its own support; a zero one-sided derivative at an endpoint does not create a constant interval. The direct strategies are continuous and strictly increasing, smooth on the finitely many interior pieces, and differentiable except at finitely many points. Their supports are \begin{equation} \label{wsref-0-65-71-3a-6f-76-65-72-6c-61-70-73-75-70-70-6f-72-74-73} I_1=[9/26,T+1/2],\qquad I_2=[1/2,T+17/26]. \end{equation}

A global best-response criterion

For the constructed continuous strategies, let \[ \tau_s(b)=b-{\mathbb{E}}[q\mid B_j=b,s_i=s] \] on the occupied bid set. On the overlap, the construction and its derivative ratios give $\tau_s=t_s$. Below the overlap only signal 1 can generate an opponent bid, and above it only signal 2 can do so. Therefore \begin{equation} \label{wsref-0-65-71-3a-67-6c-6f-62-61-6c-74-61-75} \begin{array}{c|cc} \text{bid range}&\tau_1(b)&\tau_2(b)\\ \hline \bigl[9/26,1/2\bigr)&b-9/26&b-1/2\\ \bigl[1/2,T+1/2\bigr]&t_1(b)&t_2(b)\\ (T+1/2,T+17/26]&b-1/2&b-17/26 \end{array} \end{equation} These functions fit continuously and increase across the entire occupied bid range. Below signal 2's support, $\tau_2\le0$; above signal 1's support, $\tau_1\ge T$. Each prescribed bid crosses its own taste threshold. Lemma Lemma (Monotone surplus cutoffs) proves global optimality, not merely stationarity.

Completion of the continuous classification

Proposition Proposition (Separated-support threshold) gives existence and necessity for separated supports. Propositions Proposition (Forced overlap endpoints) and Proposition (Exact overlap length) show that overlap requires $T\ge{T_\star}$. The construction above provides equilibrium for every such $T$.

There is also uniqueness within the continuous class. In the separated case the pivotal equations force (eq:separatedbids). In the overlapping case the endpoints are fixed, the lower moving branch is $b=1/2+H(\lambda)$, the upper moving branch is fixed by its upper endpoint, and (eq:length) fixes the intervening sticking length. Hence (eq:constructlower)--(eq:upperexterior) are forced.

For $T=1$, the sticking length is \[ 1-{T_\star}=0.8205722574215637\ldots, \] and its endpoints are \[ a=0.5897138712892182\ldots, \qquad c=1.4102861287107818\ldots. \] This is why the original $T=1$ nonexistence claim fails. At $T={T_\star}$ there is no sticking interval; this is not the only taste width admitting an exact equilibrium.

An exact discontinuous equilibrium in the alleged nonexistence region

The previous results concern continuous direct bids. There is another escape from the differential path restriction besides atoms or singular inverses: both direct bids can jump across a common interval that no opponent ever bids in. This section constructs such an equilibrium for every $2/13<T<{T_\star}$.

Selecting the skipped mixture range

The posterior means satisfy \begin{equation} \label{wsref-0-65-71-3a-72-65-66-6c-65-63-74-69-6f-6e} M_1(1-\lambda)=1-M_2(\lambda),\qquad K(1-\lambda)=K(\lambda). \end{equation} For $0\le\alpha\le1/2$, set \begin{align} g(\alpha)&=M_1(1-\alpha)-M_1(\alpha) =M_2(1-\alpha)-M_2(\alpha),\label{wsref-0-65-71-3a-67}\\ \Gamma(\alpha)&=2H(\alpha)+g(\alpha).\label{wsref-0-65-71-3a-47-61-6d-6d-61} \end{align} Here $g(\alpha)>0$ for $\alpha<1/2$, and \begin{equation} \label{wsref-0-65-71-3a-47-61-6d-6d-61-62-6f-75-6e-64-73} \Gamma(0)=\frac2{13},\qquad \Gamma(1/2)={T_\star}, \qquad \Gamma'(\alpha)= \frac{1056}{(13-2\alpha)^2(11+2\alpha)^2}>0. \end{equation} The derivative follows by differentiating (eq:Gamma) and using $M_1'(1-\alpha)=M_2'(\alpha)$. Hence for each $T\in(2/13,{T_\star})$ there is a unique $\alpha\in(0,1/2)$ with $\Gamma(\alpha)=T$.

Generalized inverse cutoffs and direct bid jumps

Fix this $\alpha$ and define \begin{equation} \label{wsref-0-65-71-3a-67-61-70-62-6f-75-6e-64-73} j_- =\frac12+H(\alpha),\qquad j_+=j_-+g(\alpha),\qquad \kappa_s=j_--M_s(\alpha). \end{equation} Equation (eq:g) ensures \begin{equation} \label{wsref-0-65-71-3a-6a-75-6d-70-6d-61-74-63-68} j_+-M_s(1-\alpha)=\kappa_s,\qquad s=1,2. \end{equation} Define two continuous nondecreasing cutoff functions on $[1/2,T+1/2]$ by \begin{align} b&=\frac12+H(\lambda),\quad t_s(b)=b-M_s(\lambda), &&0\le\lambda\le\alpha,\label{wsref-0-65-71-3a-6a-75-6d-70-6c-6f-77-65-72}\\ t_s(b)&=\kappa_s, &&j_-\le b\le j_+,\label{wsref-0-65-71-3a-6a-75-6d-70-67-61-70}\\ b&=\frac12+T-{T_\star}+H(\lambda),\quad t_s(b)=b-M_s(\lambda), &&1-\alpha\le\lambda\le1.\label{wsref-0-65-71-3a-6a-75-6d-70-75-70-70-65-72} \end{align} At the start of the upper portion, \[ \frac12+T-{T_\star}+H(1-\alpha) =\frac12+T-H(\alpha)=j_+, \] using $T=2H(\alpha)+g(\alpha)$. Thus the pieces match exactly. The upper endpoint is $T+1/2$. Complete the exterior portions exactly as in (eq:lowerexterior)--(eq:upperexterior) and extend the cutoffs by zero and $T$ outside their ranges.

The cutoff functions are strictly increasing on the occupied pieces and constant on $[j_-,j_+]$. They are absolutely continuous. They correspond to direct strategies as follows: invert the lower occupied portions for $t<\kappa_s$ and the upper occupied portions for $t>\kappa_s$, and choose any bid in $[j_-,j_+]$ for the single taste $t=\kappa_s$. Each direct strategy is strictly increasing in taste, but has a jump from the left limit $j_-$ to the right limit $j_+$. The cutoff tastes lie strictly between zero and $T$. For example, $\kappa_2>0$ follows from the positive inverse slope on the initial moving portion; by reflection, \begin{equation} \label{wsref-0-65-71-3a-63-75-74-6f-66-66-73-75-6d} \kappa_1+\kappa_2=T, \end{equation} so $\kappa_1<T$ as well.

No positive mass of types bids at the selected isolated cutoff bids. Thus the bid distributions are atomless, and $(j_-,j_+)$ is an empty bid interval, up to irrelevant zero-probability assignments.

The explicit discontinuous equilibrium when $T=1/6$

The scalar equation $\Gamma(\alpha)=1/6$ has solution \[ \alpha=0.2497096895488589\ldots. \] The common skipped bid interval and the two cutoff tastes are \begin{align*} j_-&=0.5450278788374292\ldots,& j_+&=0.6216387878292374\ldots,\\ \kappa_1&=0.1650687526503747\ldots,& \kappa_2&=0.0015979140162920\ldots. \end{align*} These decimals are only illustrations; equations (eq:Gamma)--(eq:jumpupper) define the equilibrium exactly.

Audit of the previous draft and the remaining general question

What survives and what changes

The exact calculations survive.

The posterior means, the derivatives of $Q_s$, the factorization in (eq:rhoODE), the integral ${T_\star}$, its symmetry, the antiderivative, and the moving inverse slopes are correct. The separated-support incentive threshold is $2/13$, not the geometric interval-disjointness threshold $4/13$. All these quantities have been derived explicitly above or in Appendix app:algebra.

The original $T=1$ counterexample fails.

The missing $\rho=1$ sticking branch supplies length $1-{T_\star}$. Both bid functions remain strictly increasing there; it is the slope ratio, not either bid function, that is constant.

The revised $T=1/6$ counterexample is valid for the maintained class.

It refutes Question Question (The finite-auction continuity question), and Corollary Corollary (The counterexample needs no inverse-regularity assumption) does so without an absolute-continuity assumption. The full threshold classification follows from the measure argument and automatic regularity. Thus a singular inverse is not an unresolved loophole for this family.

The measure-theoretic patch can be simplified and strengthened.

Working directly with $\rho\in[0,\infty]$ requires special care at zero, infinity, and bids of zero aggregate density. Proving that exceptional sets have Lebesgue measure zero alone does not automatically exclude every global continuation pathology. The compact parameter $\lambda$ and the signed-measure equation (eq:DAEmeasure) address all these issues at once, including genuinely singular inverse measures.

Unrestricted symmetric nonexistence does not follow.

Proposition Proposition (Discontinuous existence throughout the gap) supplies an exact discontinuous symmetric equilibrium for the same $T=1/6$ primitives. Thus the natural object of the negative result is continuous-strategy existence. It is not necessary to change tie-breaking, introduce bid atoms, or invoke mixed strategies to obtain equilibrium in this particular example. Footnote 6 of [PS2000, p. 503] says that continuity was unnecessary for the authors' earlier binary-signal analysis; this should not be read as a theorem that every binary-signal equilibrium must itself have continuous bids.

The $\varepsilon$-equilibrium statement is an asymptotic one.

Section IV of [PS2000, pp. 510--511] constructs continuous, strictly increasing approximate equilibria for sufficiently large $n$. It is not a theorem asserting such approximate equilibria for each fixed finite auction at every tolerance. The length mismatch here is not itself a proof or a complete explanation of their asymptotic construction.

The abstract inverse-regularity issue

Appendix A of the paper states that Assumption 2 implies the claimed differentiability and absolute continuity of inverse bids [PS2000, p. 513]. Such an implication is not valid for arbitrary strictly increasing smooth functions. For instance, let $C\subset[0,1]$ be a closed nowhere-dense set of positive Lebesgue measure, and set \[ f(x)=\int_0^x\operatorname{dist}(u,C){\,\mathrm d} u. \] Then $f$ is continuously differentiable and strictly increasing, with $f'=0$ on $C$. Since $f$ is Lipschitz and $f'=0$ on $C$, $f(C)$ has measure zero, while $f^{-1}(f(C))=C$ has positive measure. Thus $f^{-1}$ fails the null-set preservation property of absolutely continuous maps. This is an issue with a generic inference about inverses; it does not invalidate Proposition Proposition (Singular inverse bids are impossible in this family), which uses equilibrium optimality to prove regularity for the present family.

Relation to later nonexistence work

Jackson's nonexistence example [Jackson2009] concerns common and private components, but uses finite private types, a binary common state, and signals some of which reveal the state exactly. It therefore is not a direct verification of nonexistence under the continuous-quality, positive-density, limited-information primitives used here. Its conclusion about symmetric equilibrium nonexistence is stronger in a different environment.

Araujo, de Castro, and Moreira [ACM2004] study conditions for regular symmetric equilibria in their own class and, in another part of their analysis, use a modified second-price tie-breaking/payment procedure. Their mechanism-change result should not be read as an existence theorem for the unchanged mechanism in this note. Conversely, Proposition Proposition (Discontinuous existence throughout the gap) does not change the mechanism at all.

A precise broader question left by this analysis

The singular-inverse question for the explicit family is answered here, and so is the existence of possibly discontinuous pure equilibria for every $T$ in that family. The remaining general existence issue is different.

The focus on pure, weakly taste-increasing strategies follows the reduction and best-response ordering discussed in footnotes 5 and 6 of [PS2000, p. 503]; it is not an assertion that such a reduction alone proves existence.

This is a continuation question not settled by the present proofs, not a claim that an exhaustive search has established its current literature status. The paper identifies the historical existence difficulty, and the related sources above explain why standard tie-breaking and multidimensional information matter. The explicit family here is not a counterexample to Question Question (Symmetric existence without the continuity restriction); it has pure symmetric equilibria for every $T$.

A complementary mathematical task is to characterize additional primitive conditions guaranteeing a continuous equilibrium in the general model. For the uniform two-signal family, (eq:classification) is such a necessary-and-sufficient characterization.

What the general differential-algebraic system says

For reference, write the paper's factorization as \[ p(s\mid q,t)=\alpha_s(q)\beta_s(t). \] With clipped inverse tastes $t_s(b)$ and an absolutely continuous profile, define \[ Y(b\mid q)=\sum_r\alpha_r(q)\int_0^{t_r(b)}\beta_r(x)w(x){\,\mathrm d} x. \] The density of the $k$th-highest of the $n-1$ opponents' bids, conditional on $q$, is a positive constant times \[ Y^{n-k-1}(1-Y)^{k-1}\,\partial_bY. \] For a set of active signals $\mathcal R(b)$, at regular bids for which the corresponding own signal weights $\beta_s(t_s(b))$ are positive, pivotal optimality implies \[ A_{\mathcal R}(b,t)\,t'_{\mathcal R}(b)=0, \] where the active rows and columns have entries \begin{align*} A_{sr}(b,t)={}&\beta_r(t_r)w(t_r)\\[-2pt] &\times\int_0^1[u(q,t_s)-b]f(q)\alpha_s(q)\alpha_r(q) Y^{n-k-1}(1-Y)^{k-1}{\,\mathrm d} q. \end{align*} The factors involving the bidder's own taste and signal cancel from the posterior when they are positive. For a square active system with a nonzero derivative vector, singularity of $A_{\mathcal R}$ is necessary. One must use the correct active submatrix at a support face; inactive columns have zero derivative and cannot be treated as interior directions. A nonnegative null vector, consistent boundaries, and global incentive compatibility are additional requirements. None of these local conditions alone establishes a global continuous path. Empty bid intervals, as in Section sec:jumps, require a separate treatment and do not obey density-based pivotal equations in their interiors.

Proof of the bounded-variation facts

We provide the elementary one-dimensional facts used to remove the inverse-regularity assumption.

Chain rule

Let $v$ be continuous and of bounded variation on $[a,b]$, and let $F\in C^1$ near its compact range. The derivative $F'$ is uniformly continuous there. For a partition $a=x_0<\cdots<x_m=b$, Taylor's formula with a uniform modulus of continuity gives \[ F(v(x_i))-F(v(x_{i-1})) =F'(v(x_{i-1}))[v(x_i)-v(x_{i-1})]+R_i, \] with \[ \sum_i|R_i| \le\omega\!\left(\max_i|v(x_i)-v(x_{i-1})|\right) {\operatorname{Var}}(v;[a,b]), \] where $\omega(\varepsilon)\to0$. Uniform continuity of $v$ makes the right side vanish as the partition mesh tends to zero. The main sum converges to the Stieltjes integral of $F'(v)$ against $\mathrm dv$. Applying this on every subinterval gives \[ \mathrm d(F(v))=F'(v)\,\mathrm dv. \]

Derivative measures do not charge a level set

Fix $c\in{\mathbb{R}}$. Choose continuous cutoffs $\chi_\varepsilon$ with $0\le\chi_\varepsilon\le1$, equal to one on $[c-\varepsilon,c+\varepsilon]$ and zero outside $[c-2\varepsilon,c+2\varepsilon]$. Define \[ F_\varepsilon(z)=\int_c^z\chi_\varepsilon(u){\,\mathrm d} u. \] Then $F_\varepsilon\in C^1$ and $\|F_\varepsilon\|_\infty\le2\varepsilon$. The chain rule gives \[ \mathrm d(F_\varepsilon(v))=\chi_\varepsilon(v)\,\mathrm dv. \] The left side tends to zero as a distribution because $F_\varepsilon(v)$ tends uniformly to zero. On the right side, dominated convergence with respect to the finite total variation measure $|\mathrm dv|$ gives convergence to ${\mathbf 1}_{\{v=c\}}\mathrm dv$, even in total variation. Hence that restricted signed measure is zero. Its total variation on the level set is therefore zero as well. This proves Lemma Lemma (Two elementary bounded-variation facts).

Algebra, integration, and endpoint checks

Conversion to the unbounded slope ratio

With $\rho=\lambda/(1-\lambda)$, $M_s(\lambda)=Q_s(\rho)$. Differentiation gives \[ Q_1'(\rho)=\frac{22}{(11\rho+13)^2},\qquad Q_2'(\rho)=\frac{22}{(13\rho+11)^2}. \] The exact polynomial identity \[ (11\rho+13)^2-\rho(13\rho+11)^2 =(1-\rho)(169\rho^2+334\rho+169) \] proves \[ Q_2'(\rho)-\rho Q_1'(\rho)=(1-\rho)G(\rho). \] The original slope-ratio equation follows by differentiating $t_s=b-Q_s(\rho)$ at regular bids and using $t_2'=\rho t_1'$.

Integral and symmetry

An antiderivative of $G$ is \begin{equation} \label{wsref-0-65-71-3a-61-6e-74-69-64-65-72-69-76-61-74-69-76-65} F(\rho)=\frac{11}{288}\log\!\left(\frac{13\rho+11}{11\rho+13}\right) -\frac{11}{12(13\rho+11)}-\frac{13}{12(11\rho+13)}. \end{equation} Differentiating directly verifies $F'=G$. Its endpoint values give \begin{align*} \int_0^\infty G(\rho){\,\mathrm d}\rho &=F(\infty)-F(0)\\ &=\frac16+\frac{11}{144}\log\!\frac{13}{11}={T_\star}. \end{align*} Since $G(1/\rho)/\rho^2=G(\rho)$, \[ \int_0^1G(\rho){\,\mathrm d}\rho =\int_1^\infty G(\rho){\,\mathrm d}\rho =\frac1{12}+\frac{11}{288}\log\!\frac{13}{11}. \] Equivalently, $K(1-\lambda)=K(\lambda)$ and $H(1/2)={T_\star}/2$.

A compact closed form for $H$ that avoids the endpoint $\rho=\infty$ is \begin{align} H(\lambda)={}&\frac{11}{288} \log\!\left(\frac{11+2\lambda}{13-2\lambda}\right) -\frac{11(1-\lambda)}{12(11+2\lambda)} -\frac{13(1-\lambda)}{12(13-2\lambda)}\nonumber\\ &-\frac{11}{288}\log\!\frac{11}{13}+\frac16. \label{wsref-0-65-71-3a-48-63-6c-6f-73-65-64} \end{align} This is valid on all of $[0,1]$.

Inverse slopes and the critical taste gap

On a moving portion, $\lambda'=1/K(\lambda)$, so \[ t_s'=1-\frac{M_s'(\lambda)}{K(\lambda)}. \] Simplification gives (eq:inverseslopes), or in the original parameter, \[ t_1'=\frac{48(\rho+1)}{169\rho^2+334\rho+169},\qquad t_2'=\frac{48\rho(\rho+1)}{169\rho^2+334\rho+169}. \] Also, with $z=4\lambda(1-\lambda)\in[0,1]$, \[ M_2-M_1=\frac{2(11+z)}{143+z}, \] which is strictly increasing in $z$. This establishes the extrema $2/13$ and $1/6$ used in Corollary Corollary (The counterexample needs no inverse-regularity assumption).

Finally, \[ \frac16-\frac2{13}=\frac1{78}>0,\qquad {T_\star}-\frac16=\frac{11}{144}\log\!\frac{13}{11}>0. \] Thus the proposed taste width lies strictly inside the continuous-equilibrium nonexistence interval.

Reproducible checks and limits of the verification

The research audit recorded symbolic checks of the polynomial factorization, inverse slopes, antiderivative, reflection symmetry, kernel symmetry, and derivative of $\Gamma$. The recorded numerical diagnostics also constructed the separated, continuous-overlap, and discontinuous profiles and evaluated interim payoffs by quadrature with respect to their inverse-bid measures. The diagnostic code is not included with this manuscript.

Numerical best-response diagnostics were run at \[ T\in\{0.1,\ 2/13,\ 1/6,\ {T_\star},\ 0.3,\ 1\}. \] For each signal, the program checks a grid of 31 tastes, adds the relevant jump-cutoff taste in the discontinuous case, and compares the prescribed payoff with 601 alternative bids. The moving-portion integrals use 64-point Gauss--Legendre quadrature; the affine portions are integrated analytically. No positive sampled deviation gain above $2\times10^{-10}$ was found.

These computations are diagnostics, not proofs over a continuum of types or bids. The mathematical justification is the signed-measure pivotal argument, the bounded-variation regularity proof, and the global sign-of-surplus verification in the text. The manuscript is not a proof-assistant formalization. The literature references establish the original model and related results, not priority for the calculations here.

References

• [PS2000] Wolfgang Pesendorfer and Jeroen M. Swinkels (2000). Efficiency and Information Aggregation in Auctions. American Economic Review 90(3), 499--525. DOI: 10.1257/aer.90.3.499. Source used: the published article. Relevant locations: existence discussion, pp. 501 and 505; assumptions, pp. 502--503; Proposition 1, p. 504; $\varepsilon$-equilibria, pp. 510--511; Appendix A, p. 513.
• [Jackson2009] Matthew O. Jackson (2009). Non-existence of equilibrium in Vickrey, second-price, and English auctions. Review of Economic Design 13, 137--145. DOI: 10.1007/s10058-008-0059-2. Author-hosted manuscript consulted: https://web.stanford.edu/~jacksonm/nonexist.pdf. The hosted version is dated 21 September 2005 and identifies itself as a revision of Jackson (1999). The journal citation above is to the published version.
• [ACM2004] Aloisio Araujo, Luciano I. de Castro Filho, and Humberto Moreira (2004). Pure Strategy Equilibria of Multidimensional and Non-Monotonic Auctions. Working paper, March 2004, IMPA and EPGE/FGV. https://epge.fgv.br/files/1608.pdf. Relevant locations in the consulted manuscript: regular strategies, Definition 1, p. 11; modified second-price tie-breaking/payment procedure, pp. 22--23.

Completion Estimate

50%