The competitive equilibrium, and the ladder that reaches it.

1The contest, and what the last round is worth

Two contestants meet over several rounds. In each round they simultaneously add effort, having seen everything spent so far; at the end the larger accumulated total takes the prize. Contestant i values it at , each unit of effort costs , and the cost is strictly convex with . Convexity is the engine: it makes spreading effort across rounds better than dumping it into one, which is what makes the round before last worth studying at all.

Contestant i's reach is the effort whose cost exactly exhausts the prize — a ceiling nobody ever passes. Write for contestant 1's lead going into a round. Everything here sits in the competitive range, where neither contestant holds an overwhelming lead.

The final round is a static contest with head starts, and it is solved. What it hands back to the round before is a continuation value: enter the last round with contestant 1 holding lead , and she is worth

Zero while she is the marginal player, the prize net of the cost of her margin across the competitive middle, and the whole prize once she is past her rival's reach. The three branches agree at both junctions, so is continuous. At the upper junction it is even smooth, because . At the lower one, , where favourite and marginal player exchange roles, it is kinked — and that kink is the subject of section 2.

The penultimate round is now a static game: contestant 1 choosing against earns

Note what this is not. The rival's effort does not reach you only through a probability of winning; it reaches you directly, through the lead you carry forward. Crossing a threshold here moves your payoff continuously — it is in the final round, not this one, that overtaking turns a loss into a win and makes the payoff jump. So this is not an all-pay auction, and no argument in this note turns on shading a rival by a hair or on a tie-breaking rule.

2The kink, and why one game has many equilibria

Fix the rival's effort and watch your own payoff as your effort rises. It has two regimes, divided by the junction where you stop being the marginal player and become the favourite.

• Behind. While the lead you would carry forward stays below the junction, the continuation value is zero, so your payoff is just . Strictly concave, falling, best at zero: effort is pure waste.
• Leading. Past the junction your payoff is . Also strictly concave — the first cost term has an argument that falls linearly in your effort, and a convex function of a linear argument, negated, is concave; the second is strictly concave outright.

Two strictly concave pieces meeting continuously. But the left piece is falling as it arrives and the right piece climbs away, so the join is an upward kink and the payoff is not concave overall. A function like that has two local maxima: zero effort, and an interior peak out on the leading branch.

rival plays 0.15 0.30 √2−1 0.55 0.75

Two polar equilibria, and a ladder between them

What the kink produces is an equilibrium set with two poles and a ladder joining them.

• At one end, collusive equilibria: one contestant inactive, the other taking the best payoff she can credibly hold. These are in pure strategies, and rent dissipation is minimal.
• At the other, the competitive equilibrium: both contestants draw effort from a continuous distribution and both earn exactly zero. As in Bertrand competition, rent dissipation is total.
• In between, the ranked equilibria: mixed, with finite support, the two supports never overlapping and always interleaving. The more effort levels in play, the lower the payoffs.

Write for a ranked equilibrium in which the contestants use and effort levels and the second earns zero. The paper tabulates these for quadratic costs, and the ladder descends steadily toward the competitive value of zero.

3What the competitive equilibrium is

Take the symmetric case: equal values and no lead, , so both contestants share one reach . This is where the competitive equilibrium lives, and not merely for convenience: at general convex cost an equilibrium in which both contestants randomize continuously requires the reaches to be aligned, and off the symmetric point they are not.

We are after a pair of atomless distributions — no atom anywhere, so every single effort level has probability zero — under which both contestants earn nothing. The paper gives it in closed form for a power cost with an integer: the distribution function is a sum of exponentials,

and the two smallest cases are worth carrying in your head, because everything later gets checked against them:

Both have the same shape — a curve rising from zero, cut off flat where it reaches probability one. Below the cut-off the contestants randomize; above it they never go.

4The ranked equilibria: combs, and the march that builds them

Before the limit, the things approaching it. Write contestant 1's effort levels as and contestant 2's as . In any finite-support equilibrium they strictly alternate, starting from contestant 2:

Hence the name: two combs pushed together, teeth meshing.

Why they interleave — the kink again

On any interval carrying none of her rival's atoms, a contestant's payoff is strictly concave, for exactly the reason section 2 gave. A strictly concave function has at most one maximizer; every effort she actually plays must be a maximizer; so at most one of her atoms fits in any such gap. What breaks the concavity, and so creates the gaps, is a rival atom. The value of overtaking one sitting at is

negative below it, exactly zero at it, positive and rising above it — overtaking by nothing is worth nothing. That is the kink of section 2 again, now sitting at each rival atom instead of at the junction.

Interleaving follows in a line. Two atoms of the same contestant adjacent in the merged order would both sit inside one gap of the other's support, and a gap admits at most one. So the merged support alternates. It starts with contestant 2, whose bottom atom overtakes nothing and so is worth to her — which can only be an equilibrium payoff if it is zero. Her earning zero is a conclusion, not a labelling convention.

The march: one number fixes the whole comb

Now the constructive part. Fix the primitives and fix one number — the mass that contestant 2 puts on zero effort, called the dial. Every remaining atom and mass is then forced, one at a time, by indifference and the first-order conditions: a step for contestant 2, a step for contestant 1, alternating. That recursion is the march, and it is well defined at every depth for every dial in .

The march knows nothing about equilibrium — it runs forever, and the masses it generates keep accumulating past one. An equilibrium is a place to stop it, and there are two ways to stop, which alternate:

Each is one scalar equation in the dial. Its root is that rung's equilibrium, and the rungs line up in the ladder of figure 2,

starting at with the dial pinned at — which is the collusive equilibrium, contestant 2 asleep at zero — and running down through finer and finer combs.

The same march, written against the rival's distribution

Stated tooth by tooth, the march carries two conditions of different kinds at every atom: a first-order condition in , and an indifference condition in . Write both against the rival's distribution function rather than against her list of atoms, and they turn out to be one object. Nothing in what follows is special to any cost: it holds for every strictly convex cost in the class, at any values and any lead.

Let be the rival's distribution function. Then contestant 1's payoff at effort is exactly

No approximation and no limit taken. is a staircase, and integrating it tooth by tooth returns the finite sum over rival atoms that the indifference condition is built from. Differentiating (P) gives the marginal return

and now both equilibrium conditions are statements about that one function:

The first is the first-order condition. The second is indifference — and it is nothing but the fundamental theorem of calculus applied to the first. Indifference is not a second kind of condition; it is the integral of the first one. Three pieces of bookkeeping disappear with it.

• The truncation asymmetry goes. Contestant 1 counts rival atoms at or below her effort, contestant 2 strictly below — an easy sign to get wrong. Against a distribution function it is only a matter of which one-sided limit of the staircase you use, right-continuous for one contestant and left-continuous for the other. Nothing to carry in an index.
• The value needs no identity of its own. Contestant 1's payoff is read straight off (P) at her lowest atom, rather than being derived separately and then fed forward.
• The top tooth stops being a special case. The last own atom simply has no successor to balance against. There is no half-step to run and then discard.

And the picture is one you can hold. falls wherever the payoff is strictly concave, and jumps up by exactly at each rival atom — that is the kink of section 2 in derivative form. So the marginal return is a sawtooth: your atoms are its zeros, and the next one sits where the areas cancel.

cost exponent 2 2.5 3 4

Now read the two rules once more with the far pole in mind. A comb sets to zero at its own atoms and balances the areas in between. The competitive equilibrium, built in section 6, sets it to zero everywhere — and then the areas balance for free. Same function; the only difference is where you insist it vanish. Section 8 turns that sentence into the whole story of the limit.

5The first-order conditions are an integral equation

This is the pivot, and it is one line of calculus.

Contestant 1 is indifferent across her support, so her payoff is flat there and its derivative vanishes at each of her atoms. That first-order condition is a finite sum over her rival's atoms below the point:

Apply the fundamental theorem of calculus to each term, integrating from the rival's atom up to the evaluation point:

Multiply by , sum over , and exchange the sum with the integral. Inside, the sum that appears is

which is precisely contestant 2's distribution function, and the collected constant is that same function evaluated at . So the first-order condition reads

with standing for her distribution function and the equation holding at . The same three lines applied to contestant 2 give the identical equation for contestant 1's distribution, holding at each of contestant 2's atoms.

Equation (V) is a Volterra equation of the second kind: the unknown appears once alone and once inside an integral whose upper limit is the point of evaluation. That last feature is what makes it friendly. Such an equation is really a forward recursion — the value at a point depends only on values strictly below it — so it can be built up from the left, the way one solves a difference equation forward in time. It has exactly one continuous solution on , with no side condition and no smallness assumption on the kernel; call it . The machinery is the Neumann series, which for a Volterra kernel converges always.

Differentiating (V) gives the equivalent equation for the density alone, which is the form most of the work is done in:

6Building it, at any convex cost

We now have one canonical function attached to any cost in the class, with no power, polynomial or analyticity assumption anywhere. The claim is that the competitive equilibrium is that function, cut off at height one. Three short steps.

One: the solution overshoots

Integrating (D) once gives, for every ,

Put . The shift inside the integrand collapses, and (★) becomes

Read that as an average. The density integrates against , which is strictly below everywhere on , and yet the result is exactly . A weight strictly smaller than can only average out to if the total weight exceeds one:

So climbs past one strictly before the reach.

Two: the cut-off

It is continuous, starts at zero, is strictly increasing, and ends above one — so it crosses one exactly once. Call the crossing and define

an atomless distribution function supported on . That is the shape of the paper's two examples: a rising curve, flat after a cut-off.

Three: it is an equilibrium, and the only continuous one

Against the truncation, a contestant's payoff is identically zero on the support and strictly negative above it:

The first half is (★) again — differentiate the payoff, watch the moving-boundary term vanish because (overtaking by nothing is worth nothing, once more), and what is left is (★), which is zero. The second half is convexity: above the cut-off the marginal return is bounded by its value at the cut-off while marginal cost keeps rising.

So the truncation is a Nash equilibrium — atomless, both payoffs zero — and it is the only equilibrium in which both contestants' strategies are atomless. That answers the first of section 3's questions: the competitive equilibrium exists and is unique for every cost in the class, power or not. What is special about powers is not the equilibrium. It is the closed form.

7Why the closed form stops at integer powers

An integral equation is harder to work with than a differential one: no phase plane, no comparison theorems, rarely a closed form. So it matters where (D) collapses into an ordinary differential equation — and the answer is sharp.

Take and normalize , so the reach is . Write . Equation (D) becomes

Name the integrals that appear as you differentiate,

Because the integrand equals at the upper limit, differentiating gives

Each differentiation lowers the kernel exponent by exactly one. So the chain walks down and terminates — closing into a finite-order linear ODE — if and only if it can reach in finitely many steps, which happens exactly when

For non-integer there is no finite-order ODE at all. That is the paper's integer restriction, and it is not a convenience: it is the exact boundary of the method.

The two you can write down

At the chain terminates at once — with — so

At one more differentiation is needed, giving with . Its characteristic roots are , a conjugate pair, hence a trigonometric solution:

and the cut-off condition reduces to , giving . These are the paper's two examples, now with a reason attached.

8The ladder converges, and what the finite march really is

Section 3's third question. The ranked equilibria of figure 2 have rents falling to zero; do their distributions converge to the competitive one?

Fine masses force a fine mesh

Subtracting the first-order condition at consecutive atoms, and using that marginal cost is increasing, gives in any closed rung

In words: the comb's teeth cannot be far apart unless some tooth is fat. So if the largest mass in a rung tends to zero, the largest gap does too — the mesh is controlled by the masses.

Subtract the two equations

The comb's distribution function satisfies (V) at the rival's atoms. The competitive distribution satisfies (V) everywhere, so in particular at those same atoms. Subtract them. The error at a node becomes an integral of the errors below it, which is a discrete Grönwall inequality, and Grönwall turns it into an explicit bound:

uniformly in effort, for both contestants — a bound in Kolmogorov distance, which also controls convergence in distribution and Wasserstein-1. At every constant is computable: the bound is , the dial satisfies , and so the rung with teeth sits within order of the competitive equilibrium.

The relation between the two objects

Now put the two facts from section 5 side by side.

• The competitive equilibrium satisfies the Volterra identity (V) at every point.
• A ranked equilibrium satisfies the same identity (V) at its own atoms only, with the leftover closure mass parked on the top atom, above every point at which the identity is checked.

Those two lines are about one function, and section 5 already named it. Write the amount by which (V) fails at a point — its left side minus its right side — as the residual:

That last equality is not an approximation. The residual is the marginal return of section 5, for a staircase and for a density alike: the sawtooth of figure 3 and the Volterra equation of section 6 are one object drawn twice. So the competitive equilibrium is the statement that the residual vanishes identically, and the two rules that generate a rung are two statements about that same residual:

Two equations at each tooth, for the two unknowns that tooth introduces: where the atom sits, and how much mass it carries. The competitive equilibrium satisfies both for free, because a function that vanishes everywhere vanishes at any node and has zero mean on any cell.

And this is where the word Volterra earns its keep. The kernel integrates over , not over the whole interval, so the residual at depends on the distribution only below . The two conditions at the k-th tooth therefore involve only atoms and masses that are already fixed. The system is triangular; the march is forward substitution on it, one atom-and-mass pair at a time, with the dial as the single constant a causal recursion needs in order to start. Replace the kernel by a Fredholm one, integrating over the whole interval, and every tooth couples to every other: there is no march at all, only a global solve. So the claim is stronger than the march can be written in terms of (V). The march exists because (V) is Volterra.

It is worth being exact about which numerical scheme this is, because the obvious name is only half of it. Collocation imposes an equation at chosen nodes. The march imposes it at the nodes and imposes the cell means, and it needs that second condition precisely because the nodes are not chosen: they are unknowns, solved for jointly with the masses, so the mesh is an output of the scheme rather than an input to it. That is the one structural difference from every textbook version, and it is not a technicality. Every convergence theorem in that literature assumes a mesh you control, which is exactly what is unavailable here.

9What is settled

Short version: the competitive equilibrium is characterized at full generality of the cost class, with no auxiliary hypotheses — the closed form is special to integer powers, the object is not. The convergence machinery is built and is sharp at the quadratic. Missing is the one input that would let it run at general cost, and the sign of the collocation residual.

AAppendix

Each entry stands alone. Every one carries a button back to wherever you jumped from, and a list of the sections that cite it.