Ex-ante Agreements and FRAND Commitments in a Repeated Game of Standard-Setting Organizations

Abstract

I study licensing and technology choice in standard setting. I find that there may be inefficient adoption of technologies, even when firms commit to a maximum royalty or price cap for the use of their patents. When firms interact repeatedly to develop standards, a commitment to set fair, reasonable and non-discriminatory (FRAND) royalty fees may lead to more efficient technologies and higher surplus for all parties. This result can explain why standard-setting organizations favor FRAND commitments over more structured licensing commitments—such as price caps—and why there are been relatively few cases of hold-up in practice, even though such opportunistic behavior has been a primary cause of concern for innovation economists.

Appendix: Proofs

Proof of Proposition 1

Suppose technology 0 is used: Then, consumer surplus is $\underline{v}$, firm profit is 0, and SSO welfare is $W = {\underline{v}}$.

Suppose technology 1 is used: In the third stage, given realization v, firm chooses price $p = v$. In the second stage, the firm chooses x to maximize $x \, {\overline{v}} + (1-x) \, {\underline{v}} - \frac{1}{2} \, x^2$, which yields $x^* = ({\overline{v}} - \underline{v})$. Expected quality is ${\underline{v}} + ({\overline{v}} - \underline{v})^2$, firm profits are ${\underline{v}} + \frac{1}{2} \, ({\overline{v}} - {\underline{v}})^2$, consumer surplus is 0, and SSO welfare is $U = \alpha \, \left( {\underline{v}} + \frac{1}{2} \, ({\overline{v}} - {\underline{v}})^2 \right)$.

The SSO chooses technology 1 if and only if

$$\begin{aligned} \alpha \, \left( {\underline{v}} + \frac{1}{2} \, ({\overline{v}} - {\underline{v}})^2 \right) \ge {\underline{v}} \quad \Leftrightarrow \quad \frac{({\overline{v}} - {\underline{v}})^2}{{\underline{v}}} \ge 2 \, \frac{1-\alpha }{\alpha }. \end{aligned}$$

Proof of Proposition 2

Suppose $\varDelta < 2 \, \frac{1-\alpha }{\alpha }$ so technology 0 is chosen in the absence of price caps. We want to find an equilibrium with ${\underline{v}} \le {\overline{p}} \le {\overline{v}}$.

Suppose the SSO chooses technology 1: In the fourth stage, if the realization is ${\underline{v}}$, the firm charges $p = {\underline{v}}$; and if the realization is ${\overline{v}}$, the firm charges $p = {\overline{p}}$. In the third stage, the firm chooses x to maximize $x \, {\overline{p}} + (1-x) \, {\underline{v}} - \frac{1}{2} \, x^2$, which yields $x^* = {\overline{p}} - {\underline{v}}$. Expected quality is ${\underline{v}} + ({\overline{p}} - {\underline{v}}) \, ({\overline{v}} - {\underline{v}})$, firm profits are ${\underline{v}} + \frac{1}{2} \, ({\overline{p}} - {\underline{v}})^2$, consumer surplus is $({\overline{p}} - {\underline{v}}) \, ({\overline{v}} - {\overline{p}})$, and SSO welfare is $U = ({\overline{p}} - {\underline{v}}) \, ({\overline{v}} - {\overline{p}}) + \alpha \, \left({\underline{v}} + \frac{1}{2} \, ({\overline{p}} - \underline{v})^2 \right)$.

In the second stage, the SSO will choose technology 1 if

$$\begin{aligned} ({\overline{p}} - {\underline{v}}) \, ({\overline{v}} - {\overline{p}}) + \alpha \, \left( {\underline{v}} + \frac{1}{2} \, ({\overline{p}} - \underline{v})^2 \right) \ge {\underline{v}}. \end{aligned}$$

In the first stage, the firm will choose the largest price cap possible so that the SSO chooses technology 1, which is

$$\begin{aligned} {{\overline{p}}}^* = \frac{{\overline{v}} + (1-\alpha ) \, {\underline{v}} + \sqrt{({\overline{v}} - {\underline{v}})^2 - 2 {\underline{v}} \left( 2 - \alpha (3-\alpha ) \right) }}{2-\alpha }. \end{aligned}$$

If $({\overline{v}} - {\underline{v}})^2 - 2 \, {\underline{v}} \, \left( 2 - \alpha \, (3-\alpha ) \right) < 0$, then there is no price cap such that the SSO chooses technology 1. Therefore, for technology 1 to be adopted, we need

$$\begin{aligned} ({\overline{v}} - {\underline{v}})^2 - 2 \, {\underline{v}} \, \left( 2 - \alpha \, (3-\alpha ) \right) \ge 0 \quad \Leftrightarrow \quad \frac{({\overline{v}} - {\underline{v}})^2}{{\underline{v}}} \ge 2 \left( 2 - \alpha (3-\alpha ) \right) . \end{aligned}$$

Proof of Proposition 3

The firm can deviate in the second or third stages. In the third stage, if the realization is ${\overline{v}}$, the firm does not deviate as long as

$$\begin{aligned} {\overline{r}} + \frac{\delta }{1-\delta } \, \left( {\hat{x}} \, \overline{r} + (1-{\hat{x}}) \, {\underline{r}} - \frac{1}{2} \, {\hat{x}}^2 \right) \ge {\overline{v}}. \end{aligned}$$

(1)

If the realization is ${\underline{v}}$, the firm does not deviate as long as

$$\begin{aligned} {\underline{r}} + \frac{\delta }{1-\delta } \, \left( {\hat{x}} \, \overline{r} + (1-{\hat{x}}) \, {\underline{r}} - \frac{1}{2} \, {\hat{x}}^2 \right) \ge {\underline{v}}. \end{aligned}$$

(2)

Multiplying the left- and right-hand sides of (1) by $\widehat{x}$ and the left and right hand sides of (2) by $1-{\widehat{x}}$, and adding the resulting inequalities, I obtain

$$\begin{aligned}&{\widehat{x}} \, {\overline{r}} + \left(1-{\widehat{x}}\right) \, {\underline{r}} + \frac{\delta }{1-\delta } \, \left( {\widehat{x}} \, {\overline{r}} + \left(1-{\widehat{x}}\right) \, {\underline{r}} - \frac{1}{2} \, {\widehat{x}}^2 \right) \ge {\widehat{x}} \, {\overline{v}} + \left(1-{\widehat{x}}\right) \, {\underline{v}}, \nonumber \\&{\widehat{r}} \ge (1-\delta ) \, \left( {\widehat{x}} \, {\overline{v}} + \left(1-{\widehat{x}}\right) \, {\underline{v}} \right) + \delta \, \frac{1}{2} \, {\widehat{x}}^2, \nonumber \\&{\widehat{r}} \ge (1-\delta ) \, {\widehat{v}} + \delta \, \frac{1}{2} \, \left({\overline{v}} - {\underline{v}} \right)^2,\end{aligned}$$

(3)

$$\begin{aligned}&{\widehat{r}} \ge (1-\delta ) \, {\underline{v}} + \left( 1- \frac{\delta }{2} \right) \, \left({\overline{v}} - {\underline{v}} \right)^2. \end{aligned}$$

(4)

In the second stage, the firm does not deviate as long as

$$\begin{aligned}&\frac{1}{1-\delta } \, \left( {\widehat{x}} \, {\overline{r}} + \left(1-{\widehat{x}} \right) \, {\underline{r}} - \frac{1}{2} \, {\widehat{x}}^2 \right) \ge \max _x \, \left( x \, {\overline{v}} + (1-x) \, {\underline{v}} - \frac{1}{2} \, x^2 \right) \\&\frac{1}{1-\delta } \, \left( {\widehat{x}} \, {\overline{r}} + \left(1-{\widehat{x}} \right) \, {\underline{r}} - \frac{1}{2} \, {\widehat{x}}^2 \right) \ge {\widehat{x}} \, {\overline{v}} + \left(1-{\widehat{x}} \right) \, {\underline{v}} - \frac{1}{2} \, {\widehat{x}}^2, \end{aligned}$$

which is equivalent to (3).

The SSO does not deviate as long as

$$\begin{aligned}&\frac{1}{1 - \delta } \left( {\widehat{x}} \, ({\overline{v}} - {\overline{r}}) + \left(1-{\widehat{x}} \right) \, ({\underline{v}} - {\underline{r}}) + \alpha \, \left( \widehat{x} \, {\overline{r}} + \left(1-{\widehat{x}} \right) \, {\underline{r}} - \frac{1}{2} \, \widehat{x}^2 \right) \right) \ge \frac{1}{1 - \delta } \, {\underline{v}} \nonumber \\&{\widehat{r}} \le ({\overline{v}} - {\underline{v}})^2 \frac{1-\frac{\alpha }{2}}{1-\alpha } \end{aligned}$$

(5)

There exists ${\widehat{r}}$ that is consistent with (4) and (5) as long as

$$\begin{aligned}&(1-\delta ) \, {\underline{v}} + \left( 1- \frac{\delta }{2} \right) \, ({\overline{v}} - {\underline{v}})^2 \le ({\overline{v}} - {\underline{v}})^2 \frac{1-\frac{\alpha }{2}}{1-\alpha }, \\&\frac{({\overline{v}} - {\underline{v}})^2}{{\underline{v}}} \ge \frac{2 \, (1-\alpha ) \, (1-\delta )}{\alpha + \delta \, (1-\alpha )}. \end{aligned}$$