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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 the usual continuous nonnegative valuation distribution, so ties occur only on zero-probability events. Put $t=F(v)$. In a monotone symmetric Bayesian Nash equilibrium, a type $v$ is first with probability $t^{n-1}$ and second with probability $(n-1)(1-t)t^{n-2}$. Its rank-order expected prize allocation in $G_1$ is thereforeThe all-pay effort identity gives $b_1(v)={\int }_{\underline{v}}^{v}s{a}_1^{\prime }(s)ds$. It also verifies equilibrium directly: a type $v$ imitating type $z$ has utility $v{a}_1(z)-b_1(z)$, whose derivative is $(v-z)a_1^{\prime }(z)$, so the true type is a best response.

In the first version of $G_2$, the two contests have expected allocationsThere is no common effort budget, and quasilinear utility makes the two effort choices separable. Since $a_{21}+a_{22}=a_1$, adding their all-pay effort identities yieldsThe equality holds type by type for aggregate effort, rather than only after taking expectations. The within-player correlation of the two efforts does not enter these additive expected payoffs.

For the second version of $G_2$, let $V_{[1]}\ge \cdots \ge V_{[n]}$ denote descending order statistics. The expected effort in a rank-order contest with prize vector $(2,1,0,\dots )$ isEquivalently, decompose the allocation into a unit award to the best player and a unit award to each of the best two players, then use revenue equivalence: the corresponding total auction payments are $V_{[2]}$ and $2V_{[3]}$. Two separate first-place contests with prize values one and two instead generateConsequentlyFor a nondegenerate continuous distribution, the inequality is strict. No regularity of virtual valuations is needed for this comparison. With a uniform distribution on $[0,1]$, the two totals are $(3n-5)/(n+1)$ and $3(n-1)/(n+1)$, giving a difference of $4/(n+1)$.