22. Random Mechanism
22. Random Mechanism
本章导读 本章把 Ch 20 的确定性五步法推广到随机直接机制 \(\{\phi^\star(q\mid\theta),t^\star(\theta)\}\)。假设买家对货币 \(t\) 风险中性、但对数量 \(q\) 不必中性，故随机机制是基于报告类型的关于 \(q\) 的彩票。§22.1 定义报告 \(\hat\theta\)、真实 \(\theta\) 的支付 \(U(\hat\theta\mid\theta)\equiv\int_{\underline q}^{\overline q}\phi(q\mid\hat\theta)u(q,\theta)\,dq-t(\hat\theta)\)。§22.2 卖家问题及 IC 充要条件（引理 22.1 必要、22.2 充分）。§22.3 五步法：第三步将目标化为对 \(\Lambda(q,\theta)=u(q,\theta)-c(q)-u_\theta(q,\theta)\frac{1-F(\theta)}{f(\theta)}\) 的期望最大化——由于 \(\Lambda\) 与 Ch 20 (20.16) 完全相同，点态最优的 \(\phi(q\mid\theta)\) 把全部概率质量 1 放在确定性机制的解 \(q^d(\theta)\) 上、其余为 0。结论：随机机制退化为确定性机制，步骤 4、5 与 §20.4.5 相同。无图。
22. Random Mechanism
Overview This chapter extends the deterministic 5-step procedure of Ch 20 to the random direct mechanism \(\{\phi^\star(q\mid\theta),t^\star(\theta)\}\). Assuming the buyer is risk-neutral in money \(t\) but not necessarily in quantity \(q\), the random mechanism is a lottery over \(q\) based on the reported type. §22.1 define the payoff of reporting \(\hat\theta\) at true type \(\theta\) by \(U(\hat\theta\mid\theta)\equiv\int_{\underline q}^{\overline q}\phi(q\mid\hat\theta)u(q,\theta)\,dq-t(\hat\theta)\). §22.2 the seller's problem and the necessary-sufficient conditions for IC (Lemma 22.1 necessity, 22.2 sufficiency). §22.3 the 5 steps: in Step 3 the objective becomes maximizing the expectation of \(\Lambda(q,\theta)=u(q,\theta)-c(q)-u_\theta(q,\theta)\frac{1-F(\theta)}{f(\theta)}\) — since \(\Lambda\) is exactly the same as in Ch 20 (20.16), the point-wise optimal \(\phi(q\mid\theta)\) puts probability mass 1 on the deterministic-mechanism solution \(q^d(\theta)\) and 0 elsewhere. Conclusion: the random mechanism degenerates to the deterministic one, so Steps 4 and 5 are the same as §20.4.5. No figures.
22.1 关于 \(q\) 的彩票 / Lottery over \(q\)
在 §20.4 我们建立了为垄断筛选博弈设计确定性直接机制 \(\{q^\star(\theta),t^\star(\theta)\}_{\theta\in\Theta}\) 的五步法。现把它推广到随机直接机制 \(\{\phi^\star(q,t\mid\theta)\}_{\theta\in\Theta}\)。假设买家对货币 \(t\) 风险中性、但对消费量 \(q\) 不必风险中性，故可把随机直接机制改写为 \(\{\phi^\star(q\mid\theta),t^\star(\theta)\}_{\theta\in\Theta}\)，即报价本身是基于报告类型的、关于 \(q\) 的彩票。其余记号与 §20 一致。在五步法前，定义真实类型 \(\theta\)、报告 \(\hat\theta\) 时买家的支付 \(U(\hat\theta\mid\theta)\)，并记讲真话的支付 \(U(\theta)\equiv U(\hat\theta\mid\theta)|_{\hat\theta=\theta}\)：
22.1 Lottery over \(q\)
In §20.4 we established the 5-step procedure for designing the deterministic direct mechanism \(\{q^\star(\theta),t^\star(\theta)\}_{\theta\in\Theta}\) in the monopolistic screening game. Now we extend that procedure to the random direct mechanism \(\{\phi^\star(q,t\mid\theta)\}_{\theta\in\Theta}\). Assume the buyer is risk-neutral in money \(t\) but not necessarily in quantity \(q\), so we can rewrite the random direct mechanism as \(\{\phi^\star(q\mid\theta),t^\star(\theta)\}_{\theta\in\Theta}\), meaning the offer itself is a lottery over \(q\) based on the reported type. All other notations are consistent with §20. Before the five steps, define the buyer's payoff of reporting \(\hat\theta\) at true type \(\theta\) by \(U(\hat\theta\mid\theta)\), and denote the truthful payoff \(U(\theta)\equiv U(\hat\theta\mid\theta)|_{\hat\theta=\theta}\):
$$ U(\hat\theta\mid\theta)\equiv\int_{\underline q}^{\overline q}\phi(q\mid\hat\theta)u(q,\theta)\,dq-t(\hat\theta) $$
22.2 卖家的问题 / The seller's problem
22.2.1 问题 / The problem
卖家求解下式以设计最优、完全揭示的随机直接机制，受 IC（讲真话）与 IR 约束（\(\underline U\) 为外部机会，设为 0）：
22.2 The seller's problem
22.2.1 The problem
The seller solves the following to design the optimal random direct mechanism featuring full revelation, subject to IC (truth-telling) and IR (\(\underline U\) the outside opportunity, set to 0):
$$ \max_{\{\phi(\cdot\mid\theta),t(\cdot)\}}\ \mathbb{E}_\theta\!\left[t(\theta)-\int_{\underline q}^{\overline q}\phi(q\mid\theta)c(q)\,dq\right] $$
$$ \int_{\underline q}^{\overline q}\phi(q\mid\theta)u(q,\theta)\,dq-t(\theta)\ge\int_{\underline q}^{\overline q}\phi(q\mid\hat\theta)u(q,\theta)\,dq-t(\hat\theta)\quad\forall\theta,\hat\theta \tag{22.1} $$
$$ \int_{\underline q}^{\overline q}\phi(q\mid\theta)u(q,\theta)\,dq-t(\theta)\ge\underline U\quad\forall\theta \tag{IR} $$
22.2.2 激励相容的充要条件 / Necessary and sufficient conditions for IC
22.2.2 Necessary and sufficient conditions for incentive compatibility
引理 22.1（必要）、引理 22.2（充分）/ Lemma 22.1 (Necessity), Lemma 22.2 (Sufficiency) 直接显示机制 \(\{\phi(q\mid\theta),t(\theta)\}_{\theta\in\Theta}\) 激励相容当且仅当：(1) \(\left.\int_{\underline q}^{\overline q}\phi(q\mid\hat\theta)u_\theta(q,\theta)\,dq\right|_{\hat\theta=\theta}\) 关于 \(\theta\) 递增；(2) \(U(\theta)=U(\theta')+\int_{\theta'}^\theta\left(\int_{\underline q}^{\overline q}\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\)。A direct revelation mechanism \(\{\phi(q\mid\theta),t(\theta)\}_{\theta\in\Theta}\) is incentive compatible iff: (1) \(\left.\int_{\underline q}^{\overline q}\phi(q\mid\hat\theta)u_\theta(q,\theta)\,dq\right|_{\hat\theta=\theta}\) is increasing in \(\theta\); (2) \(U(\theta)=U(\theta')+\int_{\theta'}^\theta\left(\int_{\underline q}^{\overline q}\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\).
证明（必要性，引理 22.1）/ Proof (necessity, Lemma 22.1) IC 蕴含 \(U(\theta)\ge\int\phi(q\mid\hat\theta)u(q,\theta)\,dq-t(\hat\theta)\)。定义 \(U(\hat\theta)=\int\phi(q\mid\hat\theta)u(q,\hat\theta)\,dq-t(\hat\theta)\)，得 \(U(\theta)-U(\hat\theta)\ge\int\phi(q\mid\hat\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\) (22.2)。由对称性 \(U(\theta)-U(\hat\theta)\le\int\phi(q\mid\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\) (22.3)。合并 (22.4)：IC implies \(U(\theta)\ge\int\phi(q\mid\hat\theta)u(q,\theta)\,dq-t(\hat\theta)\). Define \(U(\hat\theta)=\int\phi(q\mid\hat\theta)u(q,\hat\theta)\,dq-t(\hat\theta)\), giving \(U(\theta)-U(\hat\theta)\ge\int\phi(q\mid\hat\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\) (22.2). By symmetry \(U(\theta)-U(\hat\theta)\le\int\phi(q\mid\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\) (22.3). Combining (22.4):
$$\int_{\underline q}^{\overline q}\phi(q\mid\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\ge U(\theta)-U(\hat\theta)\ge\int_{\underline q}^{\overline q}\phi(q\mid\hat\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq \tag{22.4}$$
对 \(\hat\theta>\theta'>\theta\) 令 \(\hat\theta\downarrow\theta\)，(22.4) 两端除以差并取极限，得 \(\int\phi(q\mid\theta)u_\theta(q,\theta)\,dq\ge U'(\theta)\ge\left.\int\phi(q\mid\hat\theta)u_\theta(q,\theta)\,dq\right|_{\hat\theta=\theta}\)，故 \(U'(\theta)=\int\phi(q\mid\theta)u_\theta(q,\theta)\,dq\) (22.6)，从而 (22.7) \(U(\theta)=U(\theta')+\int_{\theta'}^\theta\left(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\)；条件 (1) 的单调性由 (22.5) 得出。\(\blacksquare\)For \(\hat\theta>\theta'>\theta\) letting \(\hat\theta\downarrow\theta\), dividing (22.4) and taking the limit gives \(\int\phi(q\mid\theta)u_\theta(q,\theta)\,dq\ge U'(\theta)\ge\left.\int\phi(q\mid\hat\theta)u_\theta(q,\theta)\,dq\right|_{\hat\theta=\theta}\), so \(U'(\theta)=\int\phi(q\mid\theta)u_\theta(q,\theta)\,dq\) (22.6), hence (22.7) \(U(\theta)=U(\theta')+\int_{\theta'}^\theta\left(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\); the monotonicity of condition (1) follows from (22.5). \(\blacksquare\)
证明（充分性，引理 22.2）/ Proof (sufficiency, Lemma 22.2) 欲证 \(U(\theta)-U(\hat\theta)\ge\int\phi(q\mid\hat\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\)。由条件 (2)，\(U(\theta)-U(\hat\theta)=\int_{\hat\theta}^\theta\left(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\)，故只需 (22.8)：\(\int_{\hat\theta}^\theta\left(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\ge\int\phi(q\mid\hat\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\)。因 \(\left.\int\phi(q\mid\hat\theta)u_\theta(q,\theta)\,dq\right|_{\hat\theta=\theta}\) 关于 \(\hat\theta\) 递增，对 \(\theta>\tilde\theta\)（\(\forall\tilde\theta\in[\hat\theta,\theta]\)）有 \(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\ge\int\phi(q\mid\hat\theta)u_\theta(q,\tilde\theta)\,dq\)，积分后右端 \(=\int\phi(q\mid\hat\theta)\left(\int_{\hat\theta}^\theta u_\theta(q,\tilde\theta)\,d\tilde\theta\right)dq=\int\phi(q\mid\hat\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\)，证毕 (22.8)。\(\blacksquare\)We want \(U(\theta)-U(\hat\theta)\ge\int\phi(q\mid\hat\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\). By condition (2), \(U(\theta)-U(\hat\theta)=\int_{\hat\theta}^\theta\left(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\), so we only need (22.8): \(\int_{\hat\theta}^\theta\left(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\ge\int\phi(q\mid\hat\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\). Since \(\left.\int\phi(q\mid\hat\theta)u_\theta(q,\theta)\,dq\right|_{\hat\theta=\theta}\) is increasing in \(\hat\theta\), for \(\theta>\tilde\theta\) (\(\forall\tilde\theta\in[\hat\theta,\theta]\)) we have \(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\ge\int\phi(q\mid\hat\theta)u_\theta(q,\tilde\theta)\,dq\), and after integrating the RHS \(=\int\phi(q\mid\hat\theta)\left(\int_{\hat\theta}^\theta u_\theta(q,\tilde\theta)\,d\tilde\theta\right)dq=\int\phi(q\mid\hat\theta)[u(q,\theta)-u(q,\hat\theta)]\,dq\), which proves (22.8). \(\blacksquare\)
22.3 五步法 / The 5 steps
步骤 1：用充要条件刻画 IC/可实施机制：(1) \(\left.\int_{\underline q}^{\overline q}\phi(q\mid\hat\theta)u_\theta(q,\theta)\,dq\right|_{\hat\theta=\theta}\) 关于 \(\theta\) 递增；(2) \(U(\theta)=U(\theta')+\int_{\theta'}^\theta\left(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\)。
步骤 2：用 \(U(\theta)=U(\underline\theta)+\int_{\underline\theta}^\theta\left(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\) 化简 \(\mathbb{E}_\theta[U(\theta)]\)（分部积分）：
22.3 The 5 steps
Step 1: characterize the IC/implementable mechanism with the necessary-sufficient conditions: (1) \(\left.\int_{\underline q}^{\overline q}\phi(q\mid\hat\theta)u_\theta(q,\theta)\,dq\right|_{\hat\theta=\theta}\) increasing in \(\theta\); (2) \(U(\theta)=U(\theta')+\int_{\theta'}^\theta\left(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\).
Step 2: simplify \(\mathbb{E}_\theta[U(\theta)]\) using \(U(\theta)=U(\underline\theta)+\int_{\underline\theta}^\theta\left(\int\phi(q\mid\tilde\theta)u_\theta(q,\tilde\theta)\,dq\right)d\tilde\theta\) (integration by parts):
$$ \mathbb{E}_\theta[U(\theta)]=U(\underline\theta)+\mathbb{E}_\theta\!\left[\left(\int_{\underline q}^{\overline q}\phi(q\mid\theta)u_\theta(q,\theta)\,dq\right)\frac{1-F(\theta)}{f(\theta)}\right] $$
步骤 3：把化简后的 \(\mathbb{E}_\theta[U(\theta)]\) 代入卖家目标，得：
Step 3: incorporate the simplified \(\mathbb{E}_\theta[U(\theta)]\) into the seller's objective:
$$ \max_{\{\phi(\cdot)\}}\ \mathbb{E}_\theta\!\left[\int_{\underline q}^{\overline q}\phi(q\mid\theta)\underbrace{\left[u(q,\theta)-c(q)-u_\theta(q,\theta)\frac{1-F(\theta)}{f(\theta)}\right]}_{\equiv\Lambda(q,\theta)}dq\right]-U(\underline\theta) $$
退化为确定性机制 / Degeneration to the deterministic mechanism 对 \(\phi(\cdot)\) 逐 \(\theta\) 点态取 f.o.c.（正则条件）。注意 \(\Lambda(q,\theta)\) 与确定性机制目标 (20.16) 完全相同——因目标中的各项与确定性机制相同——故点态最优的 \(\phi(q\mid\theta)\) 应把全部概率质量 1 放在求解确定性机制的那个 \(q\)（记为 \(q^d(\theta)\)）上、其余点放 0：\(\phi(q\mid\theta)=\{1\text{ if }q=q^d(\theta);\ 0\text{ otherwise}\}\)。这意味着问题退化为确定性机制设计问题，故步骤 4、5 与确定性情形（§20.4.5）相同。Take the f.o.c. for \(\phi(\cdot)\) point-wise at each \(\theta\) (under regularity). Note \(\Lambda(q,\theta)\) is exactly the same as the objective (20.16) of the deterministic mechanism — since the terms in the objective are the same as in the deterministic mechanism — so the point-wise maximized \(\phi(q\mid\theta)\) should put probability mass 1 on the \(q\) that solves the deterministic mechanism (denoted \(q^d(\theta)\)) and 0 on all other points: \(\phi(q\mid\theta)=\{1\text{ if }q=q^d(\theta);\ 0\text{ otherwise}\}\). This means the problem degenerates to the deterministic mechanism design problem, so Steps 4 and 5 are the same as in the deterministic case (§20.4.5).
参考文献 / References
- 本章是 [[monopolistic-screening]]（Ch 20）确定性五步法在随机机制上的推广；结论是随机机制并不带来额外收益，退化为确定性机制。
References
- This chapter extends the deterministic 5-step method of [[monopolistic-screening]] (Ch 20) to random mechanisms; the conclusion is that random mechanisms bring no extra benefit and degenerate to the deterministic one.