Patent Licensing and Technological Catch-Up in a Heterogeneous Duopoly

Abstract

A cost-reducing innovation is available to Cournot duopolists through licensing. The firms are ex ante heterogeneous and adoption of the innovation impacts them differently. Is it possible for the inefficient duopolist to catch up with its efficient rival? Under certain conditions, yes. The conditions, however, are stringent: It is not sufficient, for instance, that the innovation promotes a change in the efficiency rank of the firms.

Appendix: Omitted Proof

Before we prove Proposition 1, let us remark that the second and third assertions of the Proposition were proved by Zhao (2001). The following proof is included here because that paper uses slightly different assumptions (instead of log-concavity, it is required that $P' + q_i P'' < 0$) and its equivalent to our third statement is formulated in terms of market shares—following Kimmel (1992)—as opposed to costs.

Proof of Proposition 1

Let $q_i^{\circ } = q_i(\kappa _1, \kappa _2)$ denote the equilibrium output of firm i corresponding to cost profile $(\kappa _1, \kappa _2)$ and $Q^{\circ }$ the associated industry output. Because $q_i^{\circ } > 0$—the technology is non-drastic—, it satisfies

$$\begin{aligned} P(Q^{\circ }) + q_i^{\circ } P'(Q^{\circ }) = \kappa _i. \end{aligned}$$

Adding these equations yields (1). Since $P(\cdot )$ is log-concave, it follows that $\eta (\cdot )$ is non-decreasing. The LHS of (1) is therefore increasing, and we conclude that a unique p satisfies the equation. This establishes (i).

Let us next prove the second and third parts of the Proposition. Denote by $\pi _i(q_1, q_2)$ firm i’s Cournot profit. Then differentiating the system of FOCs yields

$$\begin{aligned} {{\mathrm{d}}} q^{\circ } = \left( \frac{\partial ^2 \pi (q_1^{\circ }, q_2^{\circ })}{\partial q_j \partial q_i}\right) ^{-1} {{\mathrm{d}}} c \end{aligned}$$

where

$$\begin{aligned} \frac{\partial ^2 \pi (q_1^{\circ }, q_2^{\circ })}{\partial q_j \partial q_i} = \begin{pmatrix} \dfrac{\partial ^2 \pi _1(q_1^{\circ }, q_2^{\circ })}{\partial q_1^2} &{} \dfrac{\partial ^2 \pi _1(q_1^{\circ }, q_2^{\circ })}{\partial q_2 \partial q_1} \\ \dfrac{\partial ^2 \pi _2(q_1^{\circ }, q_2^{\circ })}{\partial q_1 \partial q_2} &{} \dfrac{\partial ^2 \pi _2(q_1^{\circ }, q_2^{\circ })}{\partial q_2^2} \end{pmatrix} \end{aligned}$$

is the matrix of derivatives of $\partial \pi _i/\partial q_i$, $i = 1,2$, evaluated at the equilibrium outputs. Its determinant $\Delta = P'[(P' + q_i^{\circ } P'') + (2P' + q_j^{\circ } P'')]$ is positive because (i) $P' < 0$, and (ii) $P' + q_i^{\circ } P'' < 0$, which is implied by log-concavity of P. Hence,

$$\begin{aligned} \frac{\partial q_i^{\circ }}{\partial c_i} = \frac{1}{\Delta }\left( 2 P' + q_j^{\circ } P'' \right) < 0 \quad {{\text{and}}}\quad \frac{\partial q_i^{\circ }}{\partial c_j} = - \frac{1}{\Delta }\left( P' + q_i^{\circ } P''\right) > 0. \end{aligned}$$

The industry reduced-form total profit is given by

$$\begin{aligned} \Pi (c_1, c_2) = (P(Q^{\circ }) - c_1) q_1^{\circ } + (P(Q^{\circ }) - c_2) q_2^{\circ }. \end{aligned}$$

Using the firms’ FOCs, we then get

$$\begin{aligned} \frac{\partial \Pi (c_1, c_2)}{\partial c_1} = \left( P' \frac{\partial q_2^{\circ }}{\partial c_1} - 1\right) q_1^{\circ } + P'\frac{\partial q_1^{\circ }}{\partial c_1} q_2^{\circ }. \end{aligned}$$

Substituting for the partial derivatives in the above expression and rearranging, we arrive at

$$\begin{aligned} \frac{\partial \Pi (c_1, c_2)}{\partial c_1} = - P' \cdot \left( \frac{P' + q_2^{\circ } P''}{\Delta } + \frac{P' + q_1^{\circ } P''}{\Delta }\right) q_1^{\circ } - P' \cdot \left( \frac{2P' + q_2^{\circ } P''}{\Delta }\right) (q_1^{\circ } - q_2^{\circ }) < 0 \end{aligned}$$

since $q_1^{\circ } - q_2^{\circ } = - (c_2 - c_1)/P' > 0$. This proves the second statement of the Proposition.

To prove the third statement, we first compute

$$\begin{aligned} \frac{\partial \Pi (c_1, c_2)}{\partial c_2} = \left( P' \frac{\partial q_1^{\circ }}{\partial c_2} - 1\right) q_2^{\circ } + P'\frac{\partial q_2^{\circ }}{\partial c_2} q_1^{\circ }, \end{aligned}$$

again making use of firms’ FOCs. Substituting for the partial derivatives and rearranging we arrive at

$$\begin{aligned} \frac{\partial \Pi (c_1, c_2)}{\partial c_2} = - P' \cdot \left( \frac{P' + q_1^{\circ } P''}{\Delta } + \frac{P' + q_2^{\circ } P''}{\Delta }\right) q_2^{\circ } + P' \cdot \left( \frac{2P' + q_1^{\circ } P''}{\Delta }\right) (q_1^{\circ } - q_2^{\circ }) \end{aligned}$$

or, substituting for $q_1^{\circ } - q_2^{\circ }$,

$$\begin{aligned} \frac{\partial \Pi (c_1, c_2)}{\partial c_2} = - P' \cdot \left( \frac{P' + q_1^{\circ } P''}{\Delta } + \frac{P' + q_2^{\circ } P''}{\Delta }\right) q_2^{\circ } - \left( \frac{2P' + q_1^{\circ } P''}{\Delta }\right) (c_2 - c_1). \end{aligned}$$

Now at $c_2 = c_1$ the derivative is negative whereas for sufficiently large $c_2$ its is positive (since $q_2^{\circ }$ is eventually zero). Thus, there must be $c_{*}$, with $\Pi (c_1, c_{*}) > 0$, such that $\partial \Pi (c_1, c_2)/ \partial c_2 > 0$ whenever $c_2 \ge c_{*}$. This concludes the proof of the Proposition. $\square$