定理 25.1（RET 特例）/ Theorem 25.1 (RET special case) 设 \(F_i=F_j\) 对 \(\forall i,j\)（对称分布）。若 (1) 两种拍卖都把物品分给最高类型买家、(2) 两种拍卖都给最低类型买家零支付，则拍卖人的期望收入相同。Assume \(F_i=F_j\) for \(\forall i,j\) (symmetric distribution). (1) If two auctions both award the good to the highest-type bidder and (2) both yield zero payoff to the lowest-type bidder, then the expected revenue of the auctioneer is the same.

证明 / Proof 记主体 \(i\) 中标概率为 \(\phi_i(b_i,\bar b_{-i}(\theta_{-i}))\)（\(b_i\) 为 \(i\) 选的出价、\(\bar b_{-i}(\cdot)\) 为其余主体的竞价函数），参与博弈的支付为 \(t_i(b_i,\bar b_{-i}(\theta_{-i}))\)。类型 \(\theta_i\) 的主体 \(i\) 的期望效用（剩余）为 (25.1)：Denote agent \(i\)'s probability of winning by \(\phi_i(b_i,\bar b_{-i}(\theta_{-i}))\) (\(b_i\) the bid chosen by \(i\), \(\bar b_{-i}(\cdot)\) the bidding functions of all other agents), and the payment for participating by \(t_i(b_i,\bar b_{-i}(\theta_{-i}))\). The expected utility (surplus) of agent \(i\) with type \(\theta_i\) is (25.1):

$$U_i(\theta_i)=\max_{b_i}\Big\{\mathbb{E}_{\theta_{-i}}\big[\phi_i(b_i,\bar b_{-i}(\theta_{-i}))\big]\theta_i-\mathbb{E}_{\theta_{-i}}\big[t_i(b_i,\bar b_{-i}(\theta_{-i}))\big]\Big\} \tag{25.1}$$

由包络定理（对两端关于 \(\theta\) 求导），\(U_i'(\theta_i)=\mathbb{E}_{\theta_{-i}}[\phi_i(b_i,\bar b_{-i}(\theta_{-i}))]=G(\theta_i)\)（首式因 \(U_i\) 中涉及 \(\theta_i\) 的只有 \(\theta_i\) 与竞价函数 \(b_i(\theta)\)，\(\partial U_i/\partial b_i(\theta_i)=0\) 由包络成立；次式由显示原理，每个拍卖对应一个讲真话的直接机制 \(\{\phi_i(\cdot,\cdot),t_i(\cdot,\cdot)\}_{i=1}^n\)，故 \(\mathbb{E}_{\theta_{-i}}[\phi_i]=\mathbb{P}(\theta_i\ge\max_{j\ne i}\theta_j\mid\theta_i)=G(\theta_i)\)，由 \(\theta_j\) 的 i.i.d. 与 \(G\) 的定义）。积分并用零最低类型支付定常数 0，得 \(U_i(\theta_i)=\int_0^{\theta_i}G(x)\,dx\)。因双方风险中性，期望收入 \(ER=\) 得物品者（最高类型）的期望估值减全体期望剩余：By the envelope theorem (differentiating both sides w.r.t. \(\theta\)), \(U_i'(\theta_i)=\mathbb{E}_{\theta_{-i}}[\phi_i(b_i,\bar b_{-i}(\theta_{-i}))]=G(\theta_i)\) (the first line because the arguments involving \(\theta_i\) in \(U_i\) are only \(\theta_i\) and the bidding function \(b_i(\theta)\), with \(\partial U_i/\partial b_i(\theta_i)=0\) by the envelope; the second by the revelation principle, every auction corresponds to a truth-telling direct mechanism \(\{\phi_i(\cdot,\cdot),t_i(\cdot,\cdot)\}_{i=1}^n\), so \(\mathbb{E}_{\theta_{-i}}[\phi_i]=\mathbb{P}(\theta_i\ge\max_{j\ne i}\theta_j\mid\theta_i)=G(\theta_i)\), by the i.i.d. of \(\theta_j\) and the definition of \(G\)). Integrating and pinning the constant to 0 via zero lowest-type payoff gives \(U_i(\theta_i)=\int_0^{\theta_i}G(x)\,dx\). Since both sides are risk-neutral, the expected revenue \(ER\) = the expected valuation of the person getting the good (the highest type) minus the total expected surplus:

$$ER=\mathbb{E}\!\left[\max\{\theta_1,\dots,\theta_n\}-\sum_{i=1}^n U_i(\theta_i)\right]=\mathbb{E}\!\left[\max\{\theta_1,\dots,\theta_n\}-\sum_{i=1}^n\left(\int_0^{\theta_i}F(x)^{n-1}\,dx\right)\right]$$

该式只关于主体的特征，不含任何拍卖特异项 \(\phi\) 或 \(t\)。故只要物品归最高类型、最低类型支付为零，\(ER\) 在所有拍卖机制下相同。\(\blacksquare\)This expression is only about the characteristics of the agents and does not involve any auction-specific terms \(\phi\) or \(t\). So as long as the good goes to the highest type and the lowest type's payoff is zero, \(ER\) is the same across all auction mechanisms. \(\blacksquare\)

定理 25.2（显示原理）/ Theorem 25.2 (The revelation principle) 对拍卖博弈 \(\Gamma\) 的任何贝叶斯纳什均衡 \(\mathbf{s}^\star=(s_1^\star,\dots,s_n^\star)\)，都存在一个直接机制（其中 \(\hat{\mathcal{S}}_i=\{\hat s_i:\Theta_i\to\Theta_i\}\) 对 \(\forall i\)）\(\hat\Gamma=\{\hat\phi_1(\theta_1),\dots,\hat\phi_n(\theta_n);\hat t_1(\theta_1),\dots,\hat t_n(\theta_n);\hat{\mathcal{S}}_1,\dots,\hat{\mathcal{S}}_n\}\)，使得 (1) \(\hat s_i^\star(\theta_i)=\theta_i\)（讲真话是均衡）、(2) \(\mathbb{E}_{\theta_{-i}}[\phi_i(\mathbf{s}^\star,\theta_i,\theta_{-i})]=\hat\phi_i(\theta_i)\)、\(\mathbb{E}_{\theta_{-i}}[t_i(\mathbf{s}^\star,\theta_i,\theta_{-i})]=\hat t_i(\theta_i)\) 对 \(\forall\theta_i,i\)（直接机制中对双方与拍卖人的期望结果等价）。For any Bayesian Nash equilibrium \(\mathbf{s}^\star=(s_1^\star,\dots,s_n^\star)\) of the auction game \(\Gamma\), there exists a direct mechanism (with \(\hat{\mathcal{S}}_i=\{\hat s_i:\Theta_i\to\Theta_i\}\) for \(\forall i\)) \(\hat\Gamma=\{\hat\phi_1(\theta_1),\dots,\hat\phi_n(\theta_n);\hat t_1(\theta_1),\dots,\hat t_n(\theta_n);\hat{\mathcal{S}}_1,\dots,\hat{\mathcal{S}}_n\}\) such that (1) \(\hat s_i^\star(\theta_i)=\theta_i\) (truth-telling is the equilibrium); (2) \(\mathbb{E}_{\theta_{-i}}[\phi_i(\mathbf{s}^\star,\theta_i,\theta_{-i})]=\hat\phi_i(\theta_i)\), \(\mathbb{E}_{\theta_{-i}}[t_i(\mathbf{s}^\star,\theta_i,\theta_{-i})]=\hat t_i(\theta_i)\) for \(\forall\theta_i,i\) (the expected outcomes for both agents and the auctioneer are equivalent in the direct mechanism).

证明 / Proof 构造直接机制 \(\{\hat\phi_i(\theta_i),\hat t_i(\theta_i)\}_{i=1}^n\)：令 \(\hat\phi_i(\theta_i)=\mathbb{E}_{\theta_{-i}}[\phi_i(s_i^\star(\theta_i),s_{-i}^\star(\theta_{-i}))]\)，\(\hat t_i\) 同理。条件 (2) 由构造自动成立。只需证条件 (1) 讲真话为均衡，即 \(\{\hat\phi_i,\hat t_i\}_{i=1}^n\) 激励相容。因 \(\mathbf{s}^\star\) 是 BNE，有无偏离条件 (25.2)：Construct the direct mechanism \(\{\hat\phi_i(\theta_i),\hat t_i(\theta_i)\}_{i=1}^n\): let \(\hat\phi_i(\theta_i)=\mathbb{E}_{\theta_{-i}}[\phi_i(s_i^\star(\theta_i),s_{-i}^\star(\theta_{-i}))]\), similarly \(\hat t_i\). Condition (2) holds automatically by construction. We need only condition (1), that truth-telling is the equilibrium, i.e. \(\{\hat\phi_i,\hat t_i\}_{i=1}^n\) is incentive compatible. Since \(\mathbf{s}^\star\) is a BNE, the no-deviation condition (25.2) holds:

$$\mathbb{E}_{\theta_{-i}}[\phi_i(s_i^\star(\theta_i),s_{-i}^\star(\theta_{-i}))]\theta_i-\mathbb{E}_{\theta_{-i}}[t_i(s_i^\star(\theta_i),s_{-i}^\star(\theta_{-i}))]\ge\mathbb{E}_{\theta_{-i}}[\phi_i(\hat s_i(\theta_i),s_{-i}^\star(\theta_{-i}))]\theta_i-\mathbb{E}_{\theta_{-i}}[t_i(\hat s_i(\theta_i),s_{-i}^\star(\theta_{-i}))] \tag{25.2}$$

对 \(\forall i,\theta_i,\hat s_i\in\mathcal{S}_i\)。欲证 \(\hat\phi_i(\theta_i)\theta_i-\hat t_i(\theta_i)\ge\hat\phi_i(\hat\theta_i)\theta_i-\hat t_i(\hat\theta_i)\)（即 (25.3)）。记 \(\mathcal{B}_i^\star=\{s_i^\star(\theta):\theta\in\Theta_i\}\)、\(\tilde{\mathcal{B}}_i=\{\hat s_i(\theta):\hat s_i\in\mathcal{S}_i\}\)。(25.2) 中未对 \(\hat s_i\) 与 \(\mathcal{S}_i\) 的值域施加限制，故 \(\tilde{\mathcal{B}}_i\) 可为 \(\mathbb{R}\)；而 \(\mathcal{B}_i^\star\) 只含某均衡下的均衡出价，显然 \(\mathcal{B}_i^\star\subseteq\tilde{\mathcal{B}}_i\)。既然在 (25.2) 中 \(i\) 可从更大的 \(\tilde{\mathcal{B}}_i\) 选出价都不偏离，那他在只能从更小的 \(\mathcal{B}_i^\star\) 选时更不会偏离。故 (25.3) 成立。\(\blacksquare\)for \(\forall i,\theta_i,\hat s_i\in\mathcal{S}_i\). We want \(\hat\phi_i(\theta_i)\theta_i-\hat t_i(\theta_i)\ge\hat\phi_i(\hat\theta_i)\theta_i-\hat t_i(\hat\theta_i)\) (i.e. (25.3)). Denote \(\mathcal{B}_i^\star=\{s_i^\star(\theta):\theta\in\Theta_i\}\) and \(\tilde{\mathcal{B}}_i=\{\hat s_i(\theta):\hat s_i\in\mathcal{S}_i\}\). In (25.2) we imposed no restriction on the range of \(\hat s_i\) and \(\mathcal{S}_i\), so \(\tilde{\mathcal{B}}_i\) could be \(\mathbb{R}\); but \(\mathcal{B}_i^\star\) only contains equilibrium bids for some equilibrium, so clearly \(\mathcal{B}_i^\star\subseteq\tilde{\mathcal{B}}_i\). Since individual \(i\) didn't deviate in (25.2) when allowed to pick a bid from the larger \(\tilde{\mathcal{B}}_i\), he definitely won't deviate when only allowed to pick from the smaller \(\mathcal{B}_i^\star\). So (25.3) holds. \(\blacksquare\)

定理 25.3（IC 的充要条件）/ Theorem 25.3 (Necessary and sufficient conditions for IC) \(\{\hat\phi_i(\theta_i),\hat t_i(\theta_i)\}_{i=1}^n\) 激励相容当且仅当：(1) \(\hat\phi_i(\theta_i)\) 关于 \(\theta_i\) 非降，对 \(\forall i\)；(2) \(U_i(\theta_i)=U(\underline\theta_i)+\int_{\underline\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\)。\(\{\hat\phi_i(\theta_i),\hat t_i(\theta_i)\}_{i=1}^n\) is incentive compatible iff: (1) \(\hat\phi_i(\theta_i)\) is non-decreasing in \(\theta_i\) for \(\forall i\); (2) \(U_i(\theta_i)=U(\underline\theta_i)+\int_{\underline\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\).

证明 / Proof 必要性：IC 蕴含 \(U_i(\theta_i)\ge U_i(\hat\theta_i\mid\theta_i)=U_i(\hat\theta_i)+\hat\phi_i(\hat\theta_i)(\theta_i-\hat\theta_i)\)，得 \(U_i(\theta_i)-U_i(\hat\theta_i)\ge\hat\phi_i(\hat\theta_i)(\theta_i-\hat\theta_i)\) (25.4)；对称得 (25.5) \(U_i(\theta_i)-U_i(\hat\theta_i)\le\hat\phi_i(\theta_i)(\theta_i-\hat\theta_i)\)。合并 (25.6) 给出 (25.7) \((\hat\phi_i(\theta_i)-\hat\phi_i(\hat\theta_i))(\theta_i-\hat\theta_i)\ge0\)，故 \(\theta_i>\hat\theta_i\Rightarrow\hat\phi_i(\theta_i)\ge\hat\phi_i(\hat\theta_i)\)（非降）。令 \(\hat\theta_i\uparrow\theta_i\)，(25.6) 除以差取极限得 \(U_i'(\theta_i)=\hat\phi_i(\theta_i)\)，从而 (25.8) \(U_i(\theta_i)=U(\underline\theta_i)+\int_{\underline\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\)。充分性：由 (2)，\(U_i(\theta_i)-U_i(\hat\theta_i)=\int_{\hat\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\)，只需 (25.9) \(\int_{\hat\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\ge\int_{\hat\theta_i}^{\theta_i}\hat\phi_i(\hat\theta_i)\,ds\)。不失一般性设 \(\theta_i\ge\hat\theta_i\)，由 (1) 对 \(\forall s\in[\hat\theta_i,\theta_i]\) 有 \(\hat\phi_i(s)\ge\hat\phi_i(\hat\theta_i)\)，故左端被积函数恒 \(\ge\) 右端，证毕。\(\blacksquare\)Necessity: IC implies \(U_i(\theta_i)\ge U_i(\hat\theta_i\mid\theta_i)=U_i(\hat\theta_i)+\hat\phi_i(\hat\theta_i)(\theta_i-\hat\theta_i)\), giving \(U_i(\theta_i)-U_i(\hat\theta_i)\ge\hat\phi_i(\hat\theta_i)(\theta_i-\hat\theta_i)\) (25.4); by symmetry (25.5) \(U_i(\theta_i)-U_i(\hat\theta_i)\le\hat\phi_i(\theta_i)(\theta_i-\hat\theta_i)\). Combining (25.6) gives (25.7) \((\hat\phi_i(\theta_i)-\hat\phi_i(\hat\theta_i))(\theta_i-\hat\theta_i)\ge0\), so \(\theta_i>\hat\theta_i\Rightarrow\hat\phi_i(\theta_i)\ge\hat\phi_i(\hat\theta_i)\) (non-decreasing). Letting \(\hat\theta_i\uparrow\theta_i\), dividing (25.6) and taking the limit gives \(U_i'(\theta_i)=\hat\phi_i(\theta_i)\), hence (25.8) \(U_i(\theta_i)=U(\underline\theta_i)+\int_{\underline\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\). Sufficiency: by (2), \(U_i(\theta_i)-U_i(\hat\theta_i)=\int_{\hat\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\), so we only need (25.9) \(\int_{\hat\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\ge\int_{\hat\theta_i}^{\theta_i}\hat\phi_i(\hat\theta_i)\,ds\). WLOG \(\theta_i\ge\hat\theta_i\); by (1), for \(\forall s\in[\hat\theta_i,\theta_i]\), \(\hat\phi_i(s)\ge\hat\phi_i(\hat\theta_i)\), so the LHS integrand is always \(\ge\) the RHS, which proves it. \(\blacksquare\)

定理 25.4（RET 一般情形）/ Theorem 25.4 (RET general case) 若拍卖 I 与拍卖 II 有相同的中标概率函数 \(\{\hat\phi_i(\cdot)\}_{i=1}^n\)（作为 \(\theta_i\) 的函数），即 \(\hat\phi_i^I(\cdot)=\hat\phi_i^{II}(\cdot)\) 对 \(\forall i,\theta_i\)，且每个个体的剩余下界相同，即 \(U_i^I(\underline\theta_i)=U_i^{II}(\underline\theta_i)\) 对 \(\forall i,\theta_i\)，则两拍卖给拍卖人的期望收入相同，\(ER^I=ER^{II}\)。If two auctions, auction I and auction II, have the same \(\{\hat\phi_i(\cdot)\}_{i=1}^n\), the individual probability function of winning the good as a function of \(\theta_i\), i.e. \(\hat\phi_i^I(\cdot)=\hat\phi_i^{II}(\cdot)\) for \(\forall i,\theta_i\), and have the same lower bound of utility surplus for each individual, i.e. \(U_i^I(\underline\theta_i)=U_i^{II}(\underline\theta_i)\) for \(\forall i,\theta_i\), then the expected revenues of the auctioneer from the two auctions are the same, \(ER^I=ER^{II}\).

证明 / Proof 由显示原理，任何拍卖机制都可由对应直接机制 \(\{\hat\phi_i(\theta_i),\hat t_i(\theta_i)\}_{i=1}^n\) 等价表示。期望收入 \(ER=\mathbb{E}_{\{\theta_i\}}[\sum_{i=1}^n\hat t_i(\theta_i)]=\mathbb{E}[\sum_{i=1}^n(\hat\phi_i(\theta_i)\theta_i-U_i(\theta_i))]\)。由 IC 充要条件 \(U_i(\theta_i)=U_i(\underline\theta_i)+\int_{\underline\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\)，得 (25.10)：By the revelation principle, any auction mechanism can be equivalently represented by the corresponding direct mechanism \(\{\hat\phi_i(\theta_i),\hat t_i(\theta_i)\}_{i=1}^n\). The expected revenue \(ER=\mathbb{E}_{\{\theta_i\}}[\sum_{i=1}^n\hat t_i(\theta_i)]=\mathbb{E}[\sum_{i=1}^n(\hat\phi_i(\theta_i)\theta_i-U_i(\theta_i))]\). By the necessary-sufficient conditions for IC \(U_i(\theta_i)=U_i(\underline\theta_i)+\int_{\underline\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\), we get (25.10):

$$ER=\mathbb{E}_{\{\theta_i\}}\!\left[\sum_{i=1}^n\left(\hat\phi_i(\theta_i)\theta_i-U_i(\underline\theta_i)-\int_{\underline\theta_i}^{\theta_i}\hat\phi_i(s)\,ds\right)\right] \tag{25.10}$$

(25.10) 表明期望收入仅通过 \(\{\hat\phi_i(\theta_i)\}_{i=1}^n\) 与 \(\{U_i(\underline\theta_i)\}_{i=1}^n\) 与拍卖特异性相关，其余只关于 \(\theta_i\) 的分布。故只要两拍卖有相同的 \(\{\hat\phi_i(\theta_i)\}\) 与 \(\{U_i(\underline\theta_i)\}\)，期望收入相同。\(\blacksquare\)(25.10) shows the expected revenue is only related to auction-specific characteristics through \(\{\hat\phi_i(\theta_i)\}_{i=1}^n\) and \(\{U_i(\underline\theta_i)\}_{i=1}^n\), and otherwise is about the distribution of \(\theta_i\)'s. So as long as the two auctions have the same \(\{\hat\phi_i(\theta_i)\}\) and \(\{U_i(\underline\theta_i)\}\), the expected revenue is the same. \(\blacksquare\)

注 25.1 / Remark 25.1 相同的 \(\{\phi_i(\theta_i,\theta_{-i})\}_{i=1}^n\) 蕴含相同的 \(\{\hat\phi_i(\theta_i)\}_{i=1}^n\)：因对同一组主体 \(\{F_i(\cdot)\}\) 相同，故若 \(\{\phi_i(\theta_i,\theta_{-i})\}\) 跨主体相同，则 \(\hat\phi_i(\theta_i)=\mathbb{E}_{\theta_{-i}}[\phi_i(\theta_i,\theta_{-i})]\) 也跨主体相同。综上，相同的 \(\{\phi_i(\theta_i,\theta_{-i})\}\) 与 \(\{U_i(\underline\theta_i)\}\) 也带来相同期望收入，但相同的 \(\{\phi_i(\theta_i,\theta_{-i})\}\) 是比相同 \(\{\hat\phi_i(\theta_i)\}\) 更强（即充分非必要）的条件。The same \(\{\phi_i(\theta_i,\theta_{-i})\}_{i=1}^n\) implies the same \(\{\hat\phi_i(\theta_i)\}_{i=1}^n\), because for the same group of agents \(\{F_i(\cdot)\}\) is the same, so if \(\{\phi_i(\theta_i,\theta_{-i})\}\) is the same across agents, \(\hat\phi_i(\theta_i)=\mathbb{E}_{\theta_{-i}}[\phi_i(\theta_i,\theta_{-i})]\) is also the same across agents. In conclusion, the same \(\{\phi_i(\theta_i,\theta_{-i})\}\) and \(\{U_i(\underline\theta_i)\}\) also lead to the same expected revenue, but the same \(\{\phi_i(\theta_i,\theta_{-i})\}\) is a stronger (i.e. sufficient but not necessary) condition than the same \(\{\hat\phi_i(\theta_i)\}\).