Rank-order contest

A rank-order contest awards a sequence of prizes according to the ranks of efforts, while every player incurs its own effort cost. With an independent private values model, an increasing symmetric bidding function orders efforts in the same way as valuations.

Suppose values are nonnegative, the rank-order expected prize allocation $a$ is increasing and differentiable, and $a(\underline{v})=0$. With unit effort cost and zero effort at the lowest type, the symmetric equilibrium effort is $b(v)=va(v)-{\int }_{\underline{v}}^{v}a(t)dt$. A true type $v$ imitating type $z$ receives utility $va(z)-b(z)$, whose derivative in $z$ is $(v-z)a^{\prime }(z)$. It increases up to $v$ and decreases afterwards, proving the best response property. This is the interim payment identity specialized to an all-pay contest.

Assume $n$ risk-neutral players have independent identically distributed atomless nonnegative values and use an increasing symmetric equilibrium in a rank-order contest with unit effort costs. Let prizes satisfy $w_1\ge \cdots \ge w_n=0$, and normalize the lowest type's equilibrium utility to zero. For descending order statistics $V_{[1]}\ge \cdots \ge V_{[n]}$,To derive it, decompose the prize vector into layers: layer $k$ gives $w_k-w_{k+1}$ to each of the best $k$ players. A truthful multi-unit auction implementing that allocation charges each winner $(w_k-w_{k+1})V_{[k+1]}$, hence total $k(w_k-w_{k+1})V_{[k+1]}$. Revenue equivalence transfers its expected payment to the all-pay contest, since the interim allocation and lowest-type utility agree. Summing the layers proves the formula. The symmetry and common-value-distribution assumptions are needed to identify rank allocation from the bid ranking.

Let $n\ge 2$ players have independent values uniform on $[0,1]$, unit effort costs, and $1\le m<n$ equal prizes of size $w(m/n)>0$. In the symmetric Bayesian Nash equilibrium, total expected effort isThe support $[0,1]$ fixes the scale; a different uniform interval requires adjusting the order-statistic calculation.

One proof uses expected effort in a rank-order contest: only prize layer $m$ is nonzero, so total effort is $mw(m/n)\mathbb{E}{V}_{[m+1]}$. The corresponding descending order statistic of $n$ uniform values has mean $(n-m)/(n+1)$. Equivalently, the all-pay effort identity gives $b(v)=w(m/n){\int }_0^{v}t{q}_m^{\prime }(t)dt$, where the allocation derivative is a Beta distribution density with parameters $n-m,m$.

For prize scale $w(x)=x^{-\alpha }$, total effort is proportional to $h(x)=x^{1-\alpha }(1-x)$ on the feasible grid $x=m/n$. If $\alpha \ge 1$, one prize is optimal. For $0<\alpha <1$, $h$ increases up to $x_{\ast }=(1-\alpha )/(2-\alpha )$ and then decreases, so the optimal feasible integer is among $\lfloor n{x}_{\ast }\rfloor$ and $\lfloor n{x}_{\ast }\rfloor +1$. Remove infeasible candidates and compare their objective values. No definition at $x=0$ is needed.

For $n$ players with independent valuations having a continuous distribution function $F$, a type $v$ occupies rank $k$ when exactly $k-1$ rivals have higher values. Multiplying this binomial distribution by the rank prize $w_k$ and summing gives $a(v)$. Ties occur only on zero-probability events when the valuation law is atomless.