All-pay effort identity ($b(v)={\int }_{\underline{v}}^{v}t{a}^{\prime }(t)dt$)

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.