= Expected effort in a rank-order contest
{title2=$\mathbb E\sum_i b(V_i)=\sum_{k=1}^{n-1} k(w_k-w_{k+1})\mathbb E V_{[k+1]}$}
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> $V_{[1]}\ge\cdots\ge V_{[n]}$,
$$
\mathbb E\sum_i b(V_i)=\sum_{k=1}^{n-1}k(w_k-w_{k+1})\mathbb EV_{[k+1]}.
$$
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.
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