Path dependence in evolving R&D networks

Abstract

This paper models the formation of R&D networks in an oligopolistic industry. In particular, it focuses on the coevolutionary process involving firms’ technological capabilities, market structure and the network of interfirm technological agreements.

The main result of the paper is that the R&D network can work as a strong selection mechanism in the industry, creating ex post asymmetries among ex ante similar firms. This is due to a self-reinforcing, path-dependent process, in which events in the early stages of the industry affect firms’ survival in the long run. In this framework, both market and technological externalities created by the formation of cooperative agreements play a role. Although the R&D network creates profound differences at the beginning, which are reflected by an unequal distribution of links, it tends to eliminate them as it becomes denser and denser. The nature of the technological environment affects the speed of the transition and some of the characteristics of the industry in the long run.

Appendix

Proof of Proposition 1

Simple derivations show that:

$$\frac{\partial F}{\partial d(i,j)}=\lambda ({a}_{2}-{a}_{3}d(i,j))c[\frac{n{\gamma }_{j}(1-{\gamma }_{i})({e}^{-\lambda ({\gamma }_{j}f(d(i,j))}-1)-{\gamma }_{i}(1-{\gamma }_{j})(1-{e}^{-\lambda f(d(i,j){\gamma }_{i}})}{n+1}]$$

As long as firms are close enough, the second factor is positive.¹⁴ The sign of the derivative is then determined by ${a}_{2}-{a}_{3}d\left(i,j\right)$, which is positive if firms’ distance is lower than the optimal one, and negative otherwise.

Proof of Proposition 2

Deriving one obtains:

$$\begin{array}{c}\frac{\partial F}{\partial {\gamma }_{j}}=c[\frac{n\lambda f(d(i,j))(1-{\gamma }_{i}){e}^{-\lambda {\gamma }_{j}f(d(i,j))}+1-{e}^{-\lambda {\gamma }_{i}f(d(i,j))}}{n+1}][{q}_{i}(+ij)+{q}_{i}(-ij)]\\ +c[\frac{(1-{\gamma }_{j})({e}^{-\lambda f(d(i,j){\gamma }_{i}}-1)-n(1-{\gamma }_{i})({e}^{-\lambda {\gamma }_{j}f(d(i,j))}-1)}{n+1}][\frac{d{q}_{i}(+ij)}{d{\gamma }_{j}}+\frac{d{q}_{i}(-ij)}{d{\gamma }_{j}}]\end{array}$$

The quantities in the first two square brackets are positive, so it is the first addend. The sign of the second addend depends on

$$\frac{d{q}_{i}(+ij)}{d{\gamma }_{j}}+\frac{d{q}_{i}(-ij)}{d{\gamma }_{j}}=\frac{c[\lambda f(d(i,j))n(1-{\gamma }_{i}){e}^{-\lambda {\gamma }_{j}f\;(d(i,j))}-{e}^{-\lambda {\gamma }_{i}f\;({d}(i,j))}-1]}{n+1}$$

which is negative for $\lambda$ sufficiently small.

From the study of the second derivative, it can be shown that it is negative for $\lambda$ sufficiently small. Then the point (if any) where the derivative becomes 0 must be a maximum point. If gains from the collaboration are positive, there are consequently three possible cases: the increase in ${\gamma }_{j}$ 1) has always a positive effect; 2) has always a negative effect; 3) has a positive effect initially, and then has a negative effect.

Proof of Proposition 3

Deriving one obtains:

$$\begin{array}{c}\frac{\partial F}{\partial {\gamma }_{i}}=c[\frac{-\lambda f(d(i,j))(1-{\gamma }_{j}){e}^{-\lambda {\gamma }_{i}f\;(d(i,j)}+n({e}^{-{\lambda }^{\mathrm{^{\prime}}}{\gamma }_{j}f\;(d(i,j)}-1)}{n+1}][{q}_{i}(+ij)+{q}_{i}(-ij)]\\ +c[\frac{(1-{\gamma }_{j})({e}^{-\lambda f\;(d(i,j){\gamma }_{i}}-1)-n(1-{\gamma }_{i})({e}^{-\lambda {\gamma }_{j}f\;(d(i,j))}-1)}{n+1}][\frac{d{q}_{i}(+ij)}{d{\gamma }_{i}}+\frac{d{q}_{i}(-ij)}{d{\gamma }_{i}}]\end{array}$$

The first addend is negative, while, if the necessary condition for positive gain holds, the sign of the second addend depends on $[\frac{d{q}_{i}(+ij)}{d{\gamma }_{i}}+\frac{d{q}_{i}(-ij)}{d{\gamma }_{i}}]$.

It can be shown that:

$$\left[\frac{{dq}_{i}\left(+ij\right)}{{d\gamma }_{i}}+\frac{{dq}_{i}\left(-ij\right)}{{d\gamma }_{i}}\right]=c\left[\frac{n\left(1+{e}^{-\lambda \gamma {j}^{f}\left({d}_{ij}\right)}\right)-\left(1-\gamma j\right)\lambda f\left(d\left(i,j\right)\right){e}^{-\lambda {\gamma }_{i}f\left(d\left(i,j\right)\right)}}{n+1}\right]$$

The first quantity in square brackets is larger than 1, while the second is smaller than 1 for $\lambda f(d(i,j))$ small. Their difference is then positive.

The overall effect is ambiguous. Studying the second derivative, one gets $\frac{{\partial }^{2}F}{\partial {\gamma }_{i}{}^{2}}<0$ for $\lambda$ sufficiently small. Then the point (if any) where the derivative becomes 0 must be a maximum point. There are consequently three possible cases: the increase in ${\gamma }_{j}$ 1) has always a positive effect; 2) has always a negative effect; 3) has a positive effect initially, and then a negative effect.

Proof of Proposition 4

The proposition comes directly from:

$$\frac{\partial F}{\partial \sum\limits_{k\ne i,j}{c}_{k}}=\frac{2}{n+1}[\frac{(1-{\gamma }_{j})({e}^{-\lambda f(d(i,j){\gamma }_{i}}-1)-n(1-{\gamma }_{i})({e}^{-\lambda {\gamma }_{j}f(d(i,j))}-1)}{n+1}]$$

Proof of Proposition 5

I consider the situation where a stable oligopolistic structure has emerged, in the sense that the number of firms will remain constant in the future (the market structure at time t will be maintained in all the periods if $\frac{A-{n}_{t}{c}_{it}}{{n}_{t}+1}>0\forall i\in {N}_{t}$). I have to prove that $\underset{t\to \infty }{\mathrm{lim}}\mathrm{Pr}({g}_{ijt}=1)=0\forall ij\in {N}_{t}^{2}$. If $\underset{t\to \infty }{\mathrm{lim}}\mathrm{Pr}({g}_{ijt}=1)\ne 0$ I would have $\underset{t\to \infty }{\mathrm{lim}}{\gamma }_{i}=\underset{t\to \infty }{\mathrm{lim}}{\gamma }_{j}=1$. By continuity of $F(\cdot )$ (which is the gain function defined in Section 4.1), this implies $\underset{t\to \infty }{\mathrm{lim}}{F}_{i}({\alpha }_{j},{\gamma }_{j})=\underset{t\to \infty }{\mathrm{lim}}{F}_{j}({\alpha }_{i},{\gamma }_{i})=0$. But then, since E > 0, the link will asymptotically become unprofitable. Given that each link is updated with a positive probability, it will be severed with probability 1 as $t\to \infty$, and then I have the initial claim.