Robust mechanisms: the curvature case

Abstract

This paper considers the problem of a Principal who faces a privately informed Agent and only knows one moment of the type’s distribution. Preferences are nonlinear in the allocation, and the Principal maximizes her worst-case expected profits. The robustness property of the optimal mechanism imposes restrictions on the Principal’s ex-post payoff function: subject to the allocation being nonzero, ex-post payoffs are linear in the Agent’s type. The robust mechanism entails exclusion of low types, distortions at the intensive margin and efficiency at the top. We show that, under some conditions, distortions in the optimal mechanism are decreasing in types. This monotonicity has relevant consequences for several applications discussed. Our characterization uses an auxiliary zero-sum game played by the Principal and an adversarial Nature who seeks to minimize her expected payoffs which also gives us a characterization of the worst-case distribution from the Principal’s perspective. Applications of our framework to insurance provision, optimal taxation, nonlinear pricing and regulation are discussed.

Appendix

1.1 Proof of Lemma 2

Proof

(i) and (ii) The differential system (5) is equivalent to $\theta \left( \cdot \right)$ solving

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d}\theta }{\mathrm{d}q}(q)=\xi ^{-1}s_{q}(q,\theta (q)),\text { for all }q\in [0,q_{1}]\\&\theta (q_{1})=1. \end{aligned} \end{aligned}$$

(10)

The function $\xi ^{-1}s_{q}(q,\theta )$ is continuously differentiable and defined on a compact set in $\mathbb {R}^{2}$ and, hence, is Lipschitz in the second argument. By assumption it is defined on a neighborhood of its domain. In particular, it has a unique solution of the ODE on a neighborhood of $\left( q,\theta \right) =\left( q_{1},1\right)$$\textendash$ see for instance Theorem 3.1 (p. 18) of Hale (1969). Let $(q^{*},q_{1}]$ be the left maximal interval of definition of the ODE. We claim that $\theta \left( \cdot \right)$ is strictly increasing in this interval. First notice that $\theta ^{\prime }\left( q\right) =0$ implies that

$$\begin{aligned} \theta ^{\prime \prime }\left( q\right) =\xi ^{-1}\left( s_{qq}\left( q,\theta \left( q\right) \right) +s_{q\theta }\left( q,\theta \left( q\right) \right) \theta ^{\prime }\left( q\right) \right) =\xi ^{-1}s_{qq}\left( q,\theta \left( q\right) \right) <0. \end{aligned}$$

In particular, since $\theta '(q_{1})=0$, the function $\theta (\cdot )$ is strictly concave at $q_{1}$, which implies that it must be increasing slightly below of $q_{1}$. Now if $x\in (q^{*},q_{1})$ and y is the minimum of $\theta (\cdot )$ in $\left[ x,q_{1}\right]$, then $y\notin (x,q_{1})$. Indeed, $\theta ^{\prime }\left( y\right) =0$ and $\theta ^{\prime \prime }\left( y\right) <0$ imply that $\theta (\cdot )$ decreases on the right of y. The minimum cannot be $q_{1}$ since $\theta (\cdot )$ is increasing on the left of $q_{1}$. Thus, the minimum must be $y=x$. The same reasoning implies that $\theta \left( \cdot \right)$ is strictly increasing. We now define $\theta \left( q^{*}\right) =\inf \,\theta \left( (q^{*},q_{1}]\right)$. Thus, $\theta ^{\prime }\left( q^{*}\right) =\lim \limits _{q\downarrow q^{*}}\xi ^{-1}s_{q}\left( q,\theta \left( q\right) \right) =\xi ^{-1}s_{q}\left( q^{*},\theta \left( q^{*}\right) \right)$. By the first part of A2, the maximality of the interval $(q^{*},q_{1}]$ now ensures that $q^{*}=0$.

(iii) For any $\xi >0$, define $\theta ^{\xi }\left( q\right)$ as the solution of the ODE (10). Let $\underline{\theta }^{\xi }=\theta ^{\xi }\left( 0\right)$ and $q^{\xi }:\left[ \underline{\theta }^{\xi },1\right] \rightarrow \mathbb {R}_{+}$ denote the inverse of $\theta ^{\xi }\left( \cdot \right)$. The continuity will be also obvious by the argument that follows from Theorem 3.2 (p. 20) of Hale (1969). We now prove monotonicity in $\xi$. For any $\xi >0$ and $\left( q,\theta \right) \in \mathbb {R}_{+}\times \left[ 0,1\right]$, the mapping $\xi \mapsto \xi ^{-1}s_{q}(q,\theta )$ is differentiable $\textendash$ see Theorem 3.3 (p. 21) of Hale (1969). Hence, the solution $\theta ^{\xi }\left( q\right)$ is differentiable in $\xi$. Consider a fixed $\xi >0$ and if $k\left( q\right) \equiv \frac{\partial \theta ^{\xi }\left( q\right) }{\partial \xi }$, then

$$\begin{aligned} k^{\prime }\left( q\right)&=-\xi ^{-2}s_{q}\left( q,\theta ^{\xi }\left( q\right) \right) +\xi ^{-1}s_{q\theta }\left( q,\theta ^{\xi }\left( q\right) \right) k\left( q\right) ,\\ k\left( q_{1}\right)&=0. \end{aligned}$$

Define $a\left( q\right) =s_{q}\left( q,\theta ^{\xi }\left( q\right) \right)$ and $b\left( q\right) =s_{q\theta }\left( q,\theta ^{\xi }\left( q\right) \right)$. Thus,

$$\begin{aligned} k^{\prime }\left( q\right) =-\xi ^{-2}a\left( q\right) +\xi ^{-1}b\left( q\right) k\left( q\right) . \end{aligned}$$

We prove that $k\left( q\right) \ge 0$. Let $P\left( q\right) \equiv e^{\int _{q}^{q_{1}}\xi ^{-1}b\left( x\right) \mathrm{d}x}>0$. Then,

$$\begin{aligned} \left( Pk\right) ^{\prime }&=P^{\prime }k+k^{\prime }P=-\xi ^{-1}b\left( q\right) kP+k^{\prime }P=P(-\xi ^{-1}b\left( q\right) k+k^{\prime })\\&=-\xi ^{-2}a\left( q\right) P\le 0. \end{aligned}$$

Thus, $P\left( q\right) k\left( q\right) \ge P\left( q_{1}\right) k\left( q_{1}\right) =0$.

(iv) Let $\theta ^{0}\left( q\right) =\lim _{\xi \downarrow 0}\theta ^{\xi }\left( q\right) \ge 0$. Integrating the differential equation (10) we get

$$\begin{aligned} 1\ge 1-\theta ^{\xi }\left( 0\right) =\theta ^{\xi }\left( q_{1}\right) -\theta ^{\xi }\left( 0\right) =\int _{0}^{q_{1}}\xi ^{-1}s_{q}\left( q,\theta ^{\xi }\left( q\right) \right) \mathrm{d}q. \end{aligned}$$

(11)

Since $s_{q}\left( q,\theta ^{\xi }\left( q\right) \right) >0$, we have that as $\xi \rightarrow 0$, $s_{q}\left( q,\theta ^{\xi }\left( q\right) \right) \rightarrow s_{q}\left( q,\theta ^{0}\left( q\right) \right) =0$, for almost all $q\in \left[ 0,1\right]$. The condition $s_{q\theta }>0$ implies that $q^{FB}\left( \theta ^{0}\left( q\right) \right) =q$, for almost all $q\in \left[ 0,1\right]$. Since $\theta ^{0}\left( q\right)$ is strictly increasing and $\theta ^{\xi }\left( \cdot \right)$ is strictly increasing for any $\xi >0$, the convergence occurs for all $q\in \left[ 0,1\right]$. Finally, notice that, for all $q\in \left[ 0,1\right]$ and $\xi >0$,

$$\begin{aligned} s_{q}\left( 0,1\right) \ge s_{q}\left( q,1\right)>s_{q}\left( q,\theta ^{\xi }\left( q\right) \right) >0. \end{aligned}$$

Hence, (11) implies that $\frac{\mathrm{d}\theta ^{\xi }\left( q\right) }{\mathrm{d}q}=\xi ^{-1}s_{q}\left( q,\theta ^{\xi }\left( q\right) \right)$ converges uniformly to zero as $\xi \rightarrow \infty$, and hence $\theta ^{\xi }\left( \cdot \right)$ converges uniformly to 1.

1.2 Proof of Proposition 2

Proof

Consider an arbitrary $q\in \mathscr {Q}$. From Tonelli’s theorem,

$$\begin{aligned} \begin{aligned}&\int _{0}^{1}\left[ \int _{0}^{\theta }s_{\theta }\left( q\left( \tau \right) ,\tau \right) \mathrm{d}\tau \right] \mathrm{d}F^{\xi }\left( \theta \right) = \int 1_{\left[ 0,\theta \right] }\left( \tau \right) s_{\theta }\left( q\left( \tau \right) ,\tau \right) \mathrm{d}\tau \mathrm{d}F^{\xi }\left( \theta \right) \\&\quad =\int \left( 1-F^{\xi }\left( \tau \right) \right) s_{\theta }\left( q\left( \tau \right) ,\tau \right) \mathrm{d}\tau = \int _{0}^{\underline{\theta }^{\xi }}s_{\theta }\left( q\left( \tau \right) ,\tau \right) \mathrm{d}\tau \\&\qquad +\int _{\underline{\theta }^{\xi }}^{1}\frac{1-F^{\xi }\left( \tau \right) }{f^{\xi }\left( \tau \right) }s_{\theta }\left( q\left( \tau \right) ,\tau \right) f^{\xi }\left( \tau \right) \mathrm{d}\tau . \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned}&\int _{0}^{1}\left[ s\left( q\left( \theta \right) ,\theta \right) -\int _{0}^{\theta }s_{\theta }\left( q\left( \tau \right) ,\tau \right) \mathrm{d}\tau \right] \mathrm{d}F^{\xi }\left( \theta \right) \\&\quad =s\left( q\left( 1\right) ,1\right) \left( 1-F^{\xi }\left( 1-\right) \right) +\int _{\underline{\theta }^{\xi }}^{1}s\left( q\left( \theta \right) ,\theta \right) f^{\xi }\left( \theta \right) \mathrm{d}\theta \\&\qquad -\int _{0}^{\underline{\theta }^{\xi }}s_{\theta }\left( q\left( \tau \right) ,\tau \right) \mathrm{d}\tau -\int _{\underline{\theta }^{\xi }}^{1}\frac{1-F^{\xi }\left( \tau \right) }{f^{\xi }\left( \tau \right) }s_{\theta }\left( q\left( \tau \right) ,\tau \right) f^{\xi }\left( \tau \right) \mathrm{d}\tau \\&\quad =s\left( q\left( 1\right) ,1\right) \left( 1-F^{\xi }\left( 1-\right) \right) -\int _{0}^{\underline{\theta }^{\xi }}s_{\theta }\left( q\left( \tau \right) ,\tau \right) \mathrm{d}\tau \\&\qquad +\int _{\underline{\theta }^{\xi }}^{1}\left[ s\left( q\left( \theta \right) ,\theta \right) -\frac{1-F^{\xi }\left( \theta \right) }{f^{\xi }\left( \theta \right) }s_{\theta }\left( q\left( \theta \right) ,\theta \right) \right] f^{\xi }\left( \theta \right) \mathrm{d}\theta . \end{aligned} \end{aligned}$$

Note that $s\left( q\left( 1\right) ,1\right) \le s\left( q_{1},1\right) =s\left( q^{\xi }\left( 1\right) ,1\right)$. Since $s_{q\theta }>0,$$q(\theta )=0$ pointwise maximizes the expression above for each $\theta \in [0,\underline{\theta ^{\xi }]}$. Define, for each $\theta \in \left( \underline{\theta }^{\xi },1\right)$, the function $h\left( q,\theta \right) \equiv s(q,\theta )-\frac{1-F^{\xi }(\theta )}{f^{\xi }(\theta )}s_{\theta }(q,\theta )$. Then,

$$\begin{aligned} h_{q}\left( q,\theta \right) =s_{q}\left( q,\theta \right) -\frac{1-F^{\xi }(\theta )}{f^{\xi }(\theta )}s_{q\theta }(q,\theta ) =s_{q}\left( q,\theta \right) -\frac{s_{q}(q^{\xi }(\theta ),\theta )}{s_{q\theta }(q^{\xi }(\theta ),\theta )}s_{q\theta }(q,\theta ). \end{aligned}$$

If $q<q^{\xi }\left( \theta \right)$ then $s_{q\theta }(q,\theta )>0$ and $\frac{s_{q}(q,\theta )}{s_{q\theta }(q,\theta )}\ge \frac{s_{q}(q^{\xi }(\theta ),\theta )}{s_{q\theta }(q^{\xi }(\theta ),\theta )}$, which implies $h_{q}\left( q,\theta \right) \ge 0$. Analogously, if $q\in \left( q^{\xi }\left( \theta \right) ,q^{FB}\left( \theta \right) \right]$, then $h_{q}\left( q,\theta \right) \le 0$. For any $q>q^{FB}\left( \theta \right)$, it is easy to see that $h_{q}\left( q,\theta \right) <0$. Hence, $q^{\xi }\left( \theta \right)$ maximizes $h\left( \cdot ;\theta \right)$ pointwise. Since $q^{\xi }\in \mathscr {Q}$ and $q^{\xi }\left( \theta \right) \le q^{FB}\left( \theta \right) \le q_{1}$, for all $\theta \in \left[ 0,1\right]$, then $q^{\xi }$ is an optimal allocation.

1.3 Proof of Lemma 3

Proof

(i) From Lemma 2 (iii), this function is continuous and strictly increasing. Using (9), we have that

$$\begin{aligned} \int _{0}^{1}\theta \mathrm{d}F^{\xi }(\theta )=\int _{0}^{1}\left( 1-F^{\xi }\left( \theta \right) \right) \mathrm{d}\theta =\underline{\theta }^{\xi } +\int _{\underline{\theta }^{\xi }}^{1}e^{-\int _{\underline{\theta }^{\xi }}^{\theta }\gamma ^{\xi }(\tau )\mathrm{d}\tau }\mathrm{d}\theta . \end{aligned}$$

(12)

Lemma 2 implies that $\underline{\theta }^{\xi }$ converges to 1 as $\xi \rightarrow \infty$. This implies $\int _{0}^{1}\theta \mathrm{d}F^{\xi }(\theta )\rightarrow 1$, as $\xi \rightarrow \infty$. On the other hand, by Lemma 2, if $\xi \rightarrow 0$ then $q^{\xi }\left( \theta \right) \rightarrow q^{FB}\left( \theta \right)$ and hence $\underline{\theta }^{\xi }\rightarrow \underline{\theta }$. Convergence to the first-best allocation also implies that $\gamma ^{\xi }(\cdot )\rightarrow \infty$ as $\xi \rightarrow 0$. Therefore, $\int _{0}^{1}\theta \mathrm{d}F^{\xi }(\theta )\rightarrow \underline{\theta }$.

(ii) Since $\theta ^{\xi }\left( q\right)$ is continuous in $\xi$, for any q and $k>\underline{\theta }$, by the intermediate value theorem, there must exist $\xi ^{*}>0$ such that $\int _{0}^{1}\theta \mathrm{d}F^{\xi ^{*}}(\theta )=k$ .

1.3.1 Proof of Proposition 3

Proof

First notice that $F^{*}$ has support $\left[ \underline{\theta }^{*},1\right]$. Proposition 1 implies that $F^{*}$ minimizes expected profits, given ex-post payoff function $\varPi ^{*}$. Proposition 2 implies that $q^{*}$ maximizes the Principal’s payoff, given distribution $F^{*}$.

1.4 Proof of Proposition 4

Proof

The wedge $w\left( \theta \right) =s_{q}\left( q\left( \theta \right) ,\theta \right)$ is decreasing if and only if

$$\begin{aligned} \left( s_{q}\left( q,\theta \left( q\right) \right) \right) ^{\prime }=s_{qq}\left( q,\theta \right) +s_{q\theta }\theta ^{\prime }\left( q\right) <0. \end{aligned}$$

Using (10) this condition is equivalent to

$$\begin{aligned} s_{qq}\left( q,\theta ^{\xi }(q)\right) +\xi ^{-1}s_{q\theta }\left( q,\theta ^{\xi }\left( q\right) \right) s_{q}\left( q,\theta ^{\xi }\left( q\right) \right) <0, \end{aligned}$$

where we are making the dependence on $\xi$ explicit. Since $s_{q}\left( q_{1},\theta ^{\xi }\left( q_{1}\right) \right) =s_{q}\left( q_{1},1\right) =0$ and $s_{qq}\left( q_{1},1\right) <0$ (by A1), the inequality is true for $\theta \ge \underline{\theta }^{\xi }$ when $\xi \rightarrow \infty$. By Lemma 3 (i), we have that the function $\xi \rightarrow \int _{0}^{1}\theta \mathrm{d}F^{\xi }(\theta )$ is continuous and strictly increasing from $(0,\infty )$ into $(\underline{\theta },1)$. Therefore, the result immediately follows.