\(\bar b^{FPAP}\) 严格递增 / \(\bar b^{FPAP}\) is strictly increasing 由 §26.1.2 可知 \(\bar b^{FPAP}(\theta_i)=G(\theta_i)\bar b^{FPA}(\theta_i)\)，其中 \(\bar b^{FPA}(\theta_i)\) 是标准一价拍卖的竞价函数。§24.3 已证 \(\bar b^{FPA}(\theta_i)\) 严格递增，又 \(G(\theta_i)\) 与 \(\bar b^{FPA}(\theta_i)\) 在 \((\underline\theta,\overline\theta)\) 上都严格为正且递增，故二者之积也严格递增。From §26.1.2 we see \(\bar b^{FPAP}(\theta_i)=G(\theta_i)\bar b^{FPA}(\theta_i)\), where \(\bar b^{FPA}(\theta_i)\) is the bidding function of a standard first-price auction. §24.3 showed \(\bar b^{FPA}(\theta_i)\) is strictly increasing, and both \(G(\theta_i)\) and \(\bar b^{FPA}(\theta_i)\) are strictly positive on \((\underline\theta,\overline\theta)\) and increasing, so their product is also strictly increasing.

全支付下出价更低 / Bid less under all-pay 最后一步因 \(G(\theta_i)=F(\theta_i)^{n-1}\le1\)。事实上 \(F(\theta_i)=1\) 概率为 0（因 \(\mathbb{P}(\theta_i=\overline\theta)=0\)），故以概率 1 有 \(\bar b^{FPAP}(\theta_i)<\bar b^{FPA}(\theta_i)\)，即买家在一价全支付拍卖中比在标准一价拍卖中出价更低——直觉上是因为在全支付拍卖中无论输赢都要付出价。The last line holds because \(G(\theta_i)=F(\theta_i)^{n-1}\le1\). In fact \(F(\theta_i)=1\) has probability 0 (since \(\mathbb{P}(\theta_i=\overline\theta)=0\)), so with probability 1 \(\bar b^{FPAP}(\theta_i)<\bar b^{FPA}(\theta_i)\), i.e. bidders bid less in a first-price all-pay auction than in a standard first-price auction — intuitively because they have to pay the bid in an all-pay auction regardless of win or loss.

与标准一价相同 + RET / Same as standard first-price + RET §24.3.3 中 \(R_i(\theta_i)=G(\theta_i)\times\bar b^{FPA}(\theta_i)\)，标准一价的期望收入 \(ER^{FPA}=\mathbb{E}[\sum_i R_i(\theta_i)]=\mathbb{E}[\sum_i G(\theta_i)\bar b^{FPA}(\theta_i)]\) (26.5)，与 (26.4) 完全相同，故二者期望收入相同。也可诉诸 RET 25.1：对称分布；标准一价与一价全支付都把物品分给最高类型（\(\bar b^{FPAP}\) 与 \(\bar b^{FPA}\) 严格递增）；最低类型 \(\underline\theta\) 在两者中支付都为零（标准一价中中标概率为零、付零；一价全支付中不参与、付零）。由 RET 25.1，期望收入相同。In §24.3.3, \(R_i(\theta_i)=G(\theta_i)\times\bar b^{FPA}(\theta_i)\), and the standard first-price expected revenue \(ER^{FPA}=\mathbb{E}[\sum_i R_i(\theta_i)]=\mathbb{E}[\sum_i G(\theta_i)\bar b^{FPA}(\theta_i)]\) (26.5), exactly the same as (26.4), so the two have the same expected revenue. We can also invoke RET 25.1: symmetric distribution; both standard first-price and first-price all-pay award the good to the highest type (\(\bar b^{FPAP}\) and \(\bar b^{FPA}\) strictly increasing); the lowest type \(\underline\theta\) has zero payoff in both (zero winning probability and pays zero in standard first-price; does not participate and pays zero in first-price all-pay). By RET 25.1, the expected revenue is the same.

剩余 \(U_i(\hat\theta_i\mid\theta_i)\) 的化简 / Simplification of \(U_i(\hat\theta_i\mid\theta_i)\) 在两买家下，第二高出价即两出价之最小值 \(\min\{\bar b^{SPAP}(\hat\theta_i),\bar b^{SPAP}(\theta_j)\}\)。逐步化简（用 \(\bar b^{SPAP}\) 严格递增、\(F\) 为 \(\theta_j\) 分布）：With two bidders, the second-highest bid is the minimum of the two bids \(\min\{\bar b^{SPAP}(\hat\theta_i),\bar b^{SPAP}(\theta_j)\}\). Simplify step by step (using \(\bar b^{SPAP}\) strictly increasing and \(F\) the distribution of \(\theta_j\)):

$$ > \begin{aligned} > U_i(\hat\theta_i\mid\theta_i)&=\underbrace{F(\hat\theta_i)}_{\equiv\phi_i}\theta_i-\underbrace{\mathbb{E}_{\theta_j}[\min\{\bar b^{SPAP}(\hat\theta_i),\bar b^{SPAP}(\theta_j)\}\mid\hat\theta_i]}_{\equiv t_i}\\ > &=F(\hat\theta_i)\theta_i-\bar b^{SPAP}(\hat\theta_i)(1-F(\hat\theta_i))-\int_{\underline\theta}^{\hat\theta_i}\bar b^{SPAP}(\theta_j)f(\theta_j)\,d\theta_j > \end{aligned} > $$

\(\bar b^{SPAP}\) 严格递增 / \(\bar b^{SPAP}\) is strictly increasing 在 \((\underline\theta,\overline\theta)\) 上 \(\frac{d\bar b^{SPAP}(\theta_i)}{d\theta_i}=\frac{f(\theta_i)\theta_i}{1-F(\theta_i)}>0\)，在正则条件 \(\underline\theta\ge0\)（故 \(\theta_i>0\)）与 \(f(\cdot)\) 在 \((\underline\theta,\overline\theta)\) 上严格为正时成立。On \((\underline\theta,\overline\theta)\), \(\frac{d\bar b^{SPAP}(\theta_i)}{d\theta_i}=\frac{f(\theta_i)\theta_i}{1-F(\theta_i)}>0\), which holds under the regularity condition \(\underline\theta\ge0\) (so \(\theta_i>0\)) and \(f(\cdot)\) strictly positive on \((\underline\theta,\overline\theta)\).

\(ER^{SPAP}=\mathbb{E}[\theta\mid\theta=\) 2nd highest$]$ 的推导 / Derivation §24.3.3 证 \(n(n-1)f(\theta)F(\theta)^{n-2}(1-F(\theta))\) 是第二高类型的密度；因 \(\bar b^{SPAP}(\theta)\) 严格递增，\(\min\{\bar b^{SPAP}(\theta_i),\bar b^{SPAP}(\theta_j)\}\) 的密度也是它（\(n=2\) 时为 \(2f(\theta)(1-F(\theta))\)）。代入并分部积分：§24.3.3 showed \(n(n-1)f(\theta)F(\theta)^{n-2}(1-F(\theta))\) is the density of the second-highest type; since \(\bar b^{SPAP}(\theta)\) is strictly increasing, the density of \(\min\{\bar b^{SPAP}(\theta_i),\bar b^{SPAP}(\theta_j)\}\) is also it (for \(n=2\), \(2f(\theta)(1-F(\theta))\)). Substitute and integrate by parts:

$$ > \begin{aligned} > ER^{SPAP}&=2\times\int_{\underline\theta}^{\overline\theta}\bar b^{SPAP}(\theta)\cdot2f(\theta)(1-F(\theta))\,d\theta=2\times\int_{\underline\theta}^{\overline\theta}\left(\int_{\underline\theta}^\theta\frac{f(x)x}{1-F(x)}\,dx\right)2f(\theta)(1-F(\theta))\,d\theta\\ > &=2\times\int_{\underline\theta}^{\overline\theta}\theta f(\theta)(1-F(\theta))\,d\theta=\int_{\underline\theta}^{\overline\theta}\theta\cdot n(n-1)f(\theta)F(\theta)^{n-2}(1-F(\theta))\,d\theta=\mathbb{E}[\theta\mid\theta=\text{2nd highest type}] > \end{aligned} > $$

与标准一价、标准二价、一价全支付相同。这在两买家情形下不诉诸 RET 25.1 即证得收入等价。The same as standard first-price, standard second-price, and first-price all-pay. This proves revenue equivalence in the two-bidder case without appealing to RET 25.1.

一般情形 + 注 26.1 / General case + Remark 26.1 一般可诉诸 RET 25.1：对称分布；SPAP 把物品分给最高类型（\(\bar b^{SPAP}\) 严格递增）；最低类型 \(\underline\theta\) 在 SPAP 中支付为零。由 RET 25.1，所有拍卖（标准一价、标准二价、一价全支付、二价全支付）期望收入相同。注 26.1：一般 \(n\) 买家下，设 \(\bar b^{SPAP}(\cdot)\) 严格递增，由 RET 25.1 可用隐式函数刻画 \(\bar b^{SPAP}(\cdot)\)：\(\int_{\underline\theta}^{\overline\theta}(n\bar b^{SPAP}(\theta)-\theta)n(n-1)f(\theta)F(\theta)^{n-2}(1-F(\theta))\,d\theta=0\)。看似 \(\bar b^{SPAP}(\theta)=\frac{\theta}{n}\) 是解，但其实不是——因 \(\bar b^{SPAP}(\theta)=\frac{\theta}{n}\) 可能不激励相容（给定他人都用此函数，当 \(\theta_i>\frac{\max_{j\ne i}\theta_j}{n}\) 时某些 \(\theta_i<\max_{j\ne i}\theta_j\) 会偏离改出 \(\theta_i\)），故不能成为均衡竞价函数。In general we appeal to RET 25.1: symmetric distribution; SPAP awards the good to the highest type (\(\bar b^{SPAP}\) strictly increasing); the lowest type \(\underline\theta\) has zero payoff in SPAP. By RET 25.1, all auctions (standard first-price, standard second-price, first-price all-pay, second-price all-pay) have the same expected revenue. Remark 26.1: for general \(n\) bidders, assuming \(\bar b^{SPAP}(\cdot)\) strictly increasing, by RET 25.1 we characterize \(\bar b^{SPAP}(\cdot)\) by the implicit function \(\int_{\underline\theta}^{\overline\theta}(n\bar b^{SPAP}(\theta)-\theta)n(n-1)f(\theta)F(\theta)^{n-2}(1-F(\theta))\,d\theta=0\). It might look obvious that \(\bar b^{SPAP}(\theta)=\frac{\theta}{n}\) is a solution, but actually it is not — because \(\bar b^{SPAP}(\theta)=\frac{\theta}{n}\) might not be incentive compatible (given others all use this function, when \(\theta_i>\frac{\max_{j\ne i}\theta_j}{n}\) some \(\theta_i<\max_{j\ne i}\theta_j\) will deviate by bidding \(\theta_i\)), so it cannot be an equilibrium bidding function.