Bilateral trading with contingent contracts

Abstract

We study the bilateral trading problem under private information. We characterize the range of possible mechanisms which satisfy ex-post efficiency, incentive compatibility, individual rationality, and budget balance. In particular, we show that the famous Myerson–Satterthwaite impossibility result no longer holds when contingent contracts are allowed.

Appendix

In this appendix, we consider contingent payments that depend on the value of the asset to the seller as well as to the buyer. Thus, the seller’s payoff is $p(t_1, t_2) \bigl ( \alpha (t_1,t_2)t_2-(1-\beta (t_1,t_2))t_1 \bigr ) + x(t_1, t_2)$ and the buyer’s payoff is $p(t_1, t_2) \bigl ( (1-\alpha (t_1, t_2))t_2 - \beta (t_1, t_2) t_1 \bigr ) - x(t_1, t_2)$, where $\alpha (t_1, t_2)$ is the royalty rate and $\beta (t_1, t_2)$ is the cost-sharing rate.

We defined $y_1(s_1)$, $y_2(s_2)$, $q_1(s_1)$ and $q_2(s_2)$ in the text. In addition, define

$$\begin{aligned} \begin{aligned} r_1^{\alpha }(s_1)&{\mathop {=}\limits ^{\mathrm{def}}}\int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} p(s_1, t_2) \alpha (s_1, t_2) t_2 f_2(t_2) dt_2, \\ r_2^{\alpha }(s_2, t_2)&{\mathop {=}\limits ^{\mathrm{def}}}t_2 {\hat{r}}_2^{\alpha }(s_2) {\mathop {=}\limits ^{\mathrm{def}}}\int _{{\underline{t}}_{1}}^{{\overline{t}}_{1}} p(t_1, s_2) \alpha (t_1, s_2) t_2 f_1(t_1) dt_1, \\ r_1^{\beta }(s_1, t_1)&{\mathop {=}\limits ^{\mathrm{def}}}t_1 {\hat{r}}_1^{\beta }(s_1) {\mathop {=}\limits ^{\mathrm{def}}}\int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} p(s_1, t_2) \beta (s_1, t_2) t_1 f_2(t_2) dt_2, \\ r_2^{\beta }(s_2)&{\mathop {=}\limits ^{\mathrm{def}}}\int _{{\underline{t}}_{1}}^{{\overline{t}}_{1}} p(t_1, s_2) \beta (t_1, s_2) t_1 f_1(t_1) dt_1, \\ {\hat{q}}_1(s_1)&{\mathop {=}\limits ^{\mathrm{def}}}q_1(s_1)- {\hat{r}}_1^{\beta }(s_1)=\int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} p(s_1, t_2) \bigl (1-\beta (s_1, t_2)\bigr ) f_2(t_2) dt_2, \\ {\hat{q}}_2(s_2)&{\mathop {=}\limits ^{\mathrm{def}}}q_2(s_2) - {\hat{r}}_2^{\alpha }(s_2) = \int _{{\underline{t}}_{1}}^{{\overline{t}}_{1}} p(t_1, s_2) \bigl (1-\alpha (t_1, s_2)\bigr ) f_1(t_1) dt_1. \end{aligned} \end{aligned}$$

If the seller believes that the buyer will report truthfully, and he reports $s_1$ when his true type is $t_1$, then his expected payoff is

$$\begin{aligned} U_1(s_1, t_1) {\mathop {=}\limits ^{\mathrm{def}}}r_1^{\alpha }(s_1)-{\hat{q}}_1(s_1)t_1+y_1(s_1). \end{aligned}$$

Likewise, if the buyer believes that the seller will report truthfully, and she reports $s_2$ when her true type is $t_2$, then her expected payoff is

$$\begin{aligned} U_2(s_2, t_2) {\mathop {=}\limits ^{\mathrm{def}}}{\hat{q}}_2(s_2) t_2 - r_2^{\beta }(s_2) - y_2(s_2). \end{aligned}$$

Define, as in the text, $U_i(t_i) {\mathop {=}\limits ^{\mathrm{def}}}U_i(t_i,t_i)$ for $i=1,2$ as well as (IC) and (IR) conditions. We can establish the following propositions.

Proposition A1

The mechanism $(p,\alpha , \beta , x)$ is incentive compatible if and only if

• (i) ${\hat{q}}_1(t_1)$ is decreasing,

• (ii) ${\hat{q}}_2(t_2)$ is increasing,

• (iii) $U_1(t_1)=U_1({\overline{t}}_1)+\int _{t_{1}}^{{\overline{t}}_{1}} {\hat{q}}_1(\tau ) d\tau$,

• (iv) $U_2(t_2)=U_2({\underline{t}}_2)+ \int _{{\underline{t}}_{2}}^{t_{2}} {\hat{q}}_2(\tau ) d\tau$.

Proposition A2

Given any probability of trade $p: T \rightarrow [0,1]$, royalty rate $\alpha : T \rightarrow [0,1]$ and cost-sharing rate $\beta : T \rightarrow [0,1]$, we can find a cash payment $x: T \rightarrow {\mathbb {R}}$ such that $(p,\alpha ,\beta ,x)$ is incentive compatible as long as ${\hat{q}}_1(t_1)$ is decreasing and ${\hat{q}}_2(t_2)$ is increasing.

Proposition A3

An incentive compatible mechanism $(p,\alpha ,\beta ,x)$ is individually rational if and only if

$$\begin{aligned} \begin{aligned} 0&\le \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{{\overline{t}}_{1}} \left[ \left\{ t_2- \left( 1-\alpha (t_1,t_2)\right) \frac{1-F_2(t_2)}{f_2(t_2)}\right\} - \left\{ t_1 + \left( 1-\beta (t_1,t_2)\right) \frac{F_1(t_1)}{f_1(t_1)} \right\} \right] \\&\quad \times p(t_1,t_2) f_1(t_1) f_2(t_2) dt_1 dt_2. \end{aligned} \end{aligned}$$

With the probability of trade $p^0(t_1,t_2)$ defined in the text, we have

$$\begin{aligned} \begin{aligned} {\hat{q}}^0_1(t_1)&= \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} p^0(t_1, t_2) \bigl (1-\beta (t_1, t_2)\bigr ) f_2(t_2) dt_2 = \int _{t_{1}}^{{\overline{t}}_{2}} \bigl (1-\beta (t_1, t_2)\bigr ) f_2(t_2) dt_2; \\ {\hat{q}}^0_2(t_2)&= \int _{{\underline{t}}_{1}}^{{\overline{t}}_{1}} p^0(t_1, t_2) \bigl (1-\alpha (t_1, t_2)\bigr ) f_1(t_1) dt_1 = \int _{{\underline{t}}_{1}}^{t_{2}} \bigl (1-\alpha (t_1, t_2)\bigr ) f_1(t_1) dt_1. \end{aligned} \end{aligned}$$

A sufficient condition for ${\hat{q}}^0_1(t_1)$ to be decreasing and ${\hat{q}}^0_2(t_2)$ to be increasing is:

$$\begin{aligned} \alpha (t_1, t'_2) \le \alpha (t_1, t_2) \quad \hbox {for} \quad t_2< t'_2 \quad \hbox {and}\quad \beta (t'_1, t_2) \ge \beta (t_1, t_2) \quad \hbox {for} \, t_1 < t'_1. \end{aligned}$$

That is, the royalty rate is decreasing in player 2’s type and the cost-sharing rate is increasing in player 1’s type. In particular, this condition is satisfied when both $\alpha (t_1, t_2)$ and $\beta (t_1, t_2)$ are constant functions, i.e., $\alpha (t_1, t_2) = \alpha$ and $\beta (t_1, t_2)=\beta$ for all $(t_1, t_2) \in T$. Let us choose any royalty rate $\alpha : T \rightarrow [0,1]$ and cost-sharing rate $\beta : T \rightarrow [0,1]$ that make ${\hat{q}}^0_1(t_1)$ decreasing in $t_1$ and ${\hat{q}}^0_2(t_2)$ increasing in $t_2$. Then, there exists an incentive compatible mechanism $(p^0, \alpha , \beta , x)$ by Proposition A2.

Define

$$\begin{aligned} \begin{aligned} \Delta&= \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{{\overline{t}}_{1}} \left[ \left\{ t_2- \left( 1-\alpha (t_1,t_2)\right) \frac{1-F_2(t_2)}{f_2(t_2)}\right\} - \left\{ t_1 + \left( 1-\beta (t_1,t_2)\right) \frac{F_1(t_1)}{f_1(t_1)} \right\} \right] \\&\quad \times p^0(t_1,t_2) f_1(t_1) f_2(t_2) dt_1 dt_2. \end{aligned} \end{aligned}$$

Then,

$$\begin{aligned} \Delta&= \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{{\overline{t}}_{1}} \left[ \left\{ t_2- \frac{1-F_2(t_2)}{f_2(t_2)}\right\} - \left\{ t_1 + \frac{F_1(t_1)}{f_1(t_1)} \right\} + \alpha (t_1,t_2) \frac{1-F_2(t_2)}{f_2(t_2)} \right. \\&\quad \left. +\beta (t_1,t_2) \frac{F_1(t_1)}{f_1(t_1)} \right] p^0(t_1,t_2) f_1(t_1) f_2(t_2) dt_1 dt_2 \\&=\int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{\min \{t_{2}, {\overline{t}}_{1}\}} \bigl ( t_2 f_2(t_2) + F_2(t_2) -1 \bigr ) f_1(t_1) dt_1 dt_2 \\&\quad - \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{\min \{t_{2}, {\overline{t}}_{1}\}} \bigl ( t_1 f_1(t_1) + F_1(t_1) \bigr ) f_2(t_2) dt_1 dt_2 \\&\quad + \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{\min \{t_{2}, {\overline{t}}_{1}\}} \alpha (t_1, t_2)\bigl (1-F_2(t_2)\bigr ) f_1(t_1) dt_1 dt_2 \\&\quad + \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{\min \{t_{2}, {\overline{t}}_{1}\}} \beta (t_1, t_2)F_1(t_1) f_2(t_2) dt_1 dt_2 \\&=\int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \bigl ( t_2 f_2(t_2) + F_2(t_2) -1 \bigr ) F_1(t_2) dt_2 - \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \min \{t_2 F_1(t_2), {\overline{t}}_1 \} f_2(t_2) dt_2 \\&\quad + \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{\min \{t_{2}, {\overline{t}}_{1}\}} \alpha (t_1, t_2)\bigl (1-F_2(t_2)\bigr ) f_1(t_1) dt_1 dt_2 \\&\quad + \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{\min \{t_{2}, {\overline{t}}_{1}\}} \beta (t_1, t_2)F_1(t_1) f_2(t_2) dt_1 dt_2 \\&= - \int _{{\underline{t}}_{2}}^{{\overline{t}}_{1}} \bigl (1-F_2(t_2)\bigr ) F_1(t_2) dt_2 \\&\quad + \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{\min \{t_{2}, {\overline{t}}_{1}\}} \alpha (t_1, t_2)\bigl (1-F_2(t_2)\bigr ) f_1(t_1) dt_1 dt_2 \\&\quad + \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \int _{{\underline{t}}_{1}}^{\min \{t_{2}, {\overline{t}}_{1}\}} \beta (t_1, t_2)F_1(t_1) f_2(t_2) dt_1 dt_2. \end{aligned}$$

Assume that $\alpha (t_1, t_2)=\alpha (t_2)$ and $\beta (t_1, t_2)=\beta (t_1)$. Then, $\Delta$ becomes

$$\begin{aligned}&\qquad - \int _{{\underline{t}}_{2}}^{{\overline{t}}_{1}} \bigl (1-F_2(t_2)\bigr ) F_1(t_2) dt_2 + \int _{{\underline{t}}_{2}}^{{\overline{t}}_{2}} \alpha (t_2)\bigl (1-F_2(t_2)\bigr ) F_1(t_2) dt_2 \\&\qquad + \int _{{\underline{t}}_{1}}^{{\overline{t}}_{1}} \beta (t_1)F_1(t_1) \bigl (1-F_2(t_1)\bigr ) dt_1 \\&= - \int _{{\underline{t}}_{2}}^{{\overline{t}}_{1}} \bigl (1-F_2(x)\bigr ) F_1(x) dx + \int _{{\underline{t}}_{2}}^{{\overline{t}}_{1}}\bigl (\alpha (x)+\beta (x)\bigr ) \bigl (1-F_2(x)\bigr ) F_1(x) dx \\&\qquad + \int _{{\overline{t}}_{1}}^{{\overline{t}}_{2}} \alpha (x)\bigl (1-F_2(x)\bigr ) dx + \int _{{\underline{t}}_{1}}^{{\underline{t}}_{2}} \beta (x) F_1(x) dx \\&= \int _{{\underline{t}}_{2}}^{{\overline{t}}_{1}} \bigl (\alpha (x)+\beta (x)-1\bigr ) \bigl (1-F_2(x)\bigr ) F_1(x) dx \\&\qquad + \int _{{\overline{t}}_{1}}^{{\overline{t}}_{2}} \alpha (x)\bigl (1-F_2(x)\bigr ) dx + \int _{{\underline{t}}_{1}}^{{\underline{t}}_{2}} \beta (x) F_1(x) dx. \end{aligned}$$

Since the last two terms are non-negative, $\Delta$ is positive as long as $\alpha (x)+\beta (x)\ge 1$.¹⁹