Expected effort in a rank-order contest ($\mathbb{E}{\sum }_{i}b(V_i)={\sum }_{k=1}^{n-1}k(w_k-w_{k+1})\mathbb{E}{V}_{[k+1]}$)

Let $V_{[1]}\ge \cdots \ge V_{[n]}$ be descending order statistics, with $w_n=0$ and nonnegative decreasing prizes. Decompose the prize vector into awards of $w_k-w_{k+1}$ to each of the best $k$ players. The corresponding truthful multi-unit auction charges each winner the next value $V_{[k+1]}$. Its total payment is $k(w_k-w_{k+1})V_{[k+1]}$. Revenue equivalence transfers the expected payment to the all-pay contest, because the interim allocations and lowest-type utilities coincide. Summing the layers proves the formula.

With $n$ independent uniform types, $m<n$ equal prizes of scale $w(m/n)$ and unit effort costs, a symmetric Bayesian Nash equilibrium has total expected effort $R_m=m(n-m)w(m/n)/(n+1)$. The all-pay effort identity yields $b(v)=w(m/n){\int }_0^{v}t{q}_m^{\prime }(t)dt$. The allocation derivative is a Beta distribution density with parameters $n-m,m$, so integration gives the formula.

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.