New technology adoption in a Cournot oligopoly with spillovers

Abstract

In a two-stage Cournot oligopoly where a subset of firms first make a choice between two alternative production technologies independently and then all firms compete in quantity, the effect of information spillovers is analyzed when the outcome of R&D is uncertain. It is shown that the range of parameter values that support heterogeneous firms in equilibrium will diminish as information spillovers become larger. Particularly, when the spillover effect is so strong that the investment by one firm is beneficial to its R&D active rivals, all active firms will choose the same technology. A similar result can be derived from a socially desirable point of view except that the cut-off magnitude of spillovers is different. By introducing a positive success probability to characterize the uncertainty of the R&D outcome, it is found that when information spillovers are not too small, there will be underinvestment in equilibrium relative to the social optimum.

Appendices

Appendix A: The proof of Proposition 1

(1) Consider $\beta <{\hat{\beta }}$, which implies that $H(m)$ is strictly decreasing with respect to $m$. Hence, the equilibrium number of investing firms if exists, is unique. By virtue of the assumption (1), some simple derivations can yield $f_{n-k}>0$. Thus, we have

$$\begin{aligned} 0<f_{n-k} <f_{n-k-1}<\cdots <f_2 <f_1. \end{aligned}$$

It only suffices to show part (ii), since part (i) and part (iii) are obvious. For later convenience, define $g(x)$ as the largest integer equal or smaller than $x$. In the case of $f_{n-k} <f\le f_1$, i.e. $H(n-k)<0\le H(1),{\tilde{m}}\in [1,n-k)$ is the unique solution of $H(m)=0$. It follows immediately from the monotonicity property of $H$ that $H( {g({\tilde{m}})+1})<0=H({\tilde{m}})\le H({g({\tilde{m}})})$, which means that $g({\tilde{m}})$ is the equilibrium number of investing firms.

(2) Noting that $f_1=\cdots =f_{n-k} >0$ when $\beta ={\hat{\beta }}$, one can easily finish this proof.

(3) Consider $\beta >{\hat{\beta }}$, which implies that $H(m)$ is strictly increasing with respect to $m$. Accordingly, there is no number of investing firms which satisfies the equilibrium condition $H({\hat{m}}+1)<0\le H({\hat{m}})$. That is, ${\hat{m}}$ can only take either $0$ or $n-k$. Besides, it is obvious that

$$\begin{aligned} 0<f_1 <f_2 <\cdots <f_{n-k-1} <f_{n-k}. \end{aligned}$$

If $f\le f_1$, i.e. $H(1)\ge 0$, it follows from the monotonicity property of $H$ that $H(n-k)>H(n-k-1)>\cdots >H(2)>H(1)\ge 0$. As a result, ${\hat{m}}=n-k$ is the unique equilibrium number of investing firms. Likewise, it can be concluded that ${\hat{m}}=0$ is the unique equilibrium number of investing firms if $f>f_{n-k}$. Finally, in the case of $f_1 <f\le f_{n-k}$, i.e. $H(1)<0\le H(n-k)$, by virtue of the definition, both $0$ and $n-k$ can constitute Nash equilibria.

Appendix B: The proof of Proposition 2

(1) Consider $\beta <{\hat{\beta }}^*$, which implies that $G(m)$ is strictly decreasing with respect to $m$. By the definition of ${\hat{m}}^*,{\hat{m}}^*=0$ is the socially optimal size of innovating firms if and only if $G(1)<0;{\hat{m}}^*\in \{1,\ldots ,n-k-1\}$ is the socially optimal size of innovating firms if and only if $G({\hat{m}}^*+1)<0\le G({\hat{m}}^*)$; and ${\hat{m}}^*=n-k$ is the socially optimal size of innovating firms if and only if $G(n-k)\ge 0$. By virtue of the assumption (1), straightforward derivations can lead to $f_{n-k}^*>0$. Thus, one can see that

$$\begin{aligned} 0<f_{n-k}^*<f_{n-k-1}^*<\cdots <f_2^*<f_1^*. \end{aligned}$$

Since claim (i) and claim (iii) are immediate, it only needs to validate (ii). In the case of $f_{n-k}^*<f\le f_1^*$, i.e. $G(n-k)<0\le G(1),{\tilde{m}}^*\in [1,n-k)$ is the unique solution of $G(m)=0$. Given that $G(m)$ is strictly decreasing with respect to $m$, we have $G({g({\tilde{m}}^*)+1})<0=G({\tilde{m}}^*)\le G( {g({\tilde{m}}^*)})$, which means that $g({\tilde{m}}^*)$ is the socially optimal size of innovating firms.

(2) Consider $\beta \ge {\hat{\beta }}^*$, which implies $G(m)$ is weakly increasing with respect to $m$. Since $\phi (\beta )>0$, it is clear that $f^*>0$. It can be concluded that ${\hat{m}}^*$ can only take $0$ or $n-k$. Otherwise, $r\in \{1,\ldots ,n-k-1\}$ is the socially optimal size of innovating firms. Then $G(r)\ge 0>G(r+1)$, which is in contradiction to the monotonicity of $G$. Noting that

$$\begin{aligned} SW(n-k)-SW(0)=(n-k)(f^*-f), \end{aligned}$$

one can easily finish the proof.

Appendix C: Some lemmas

In this part, some necessary preliminaries are provided in order to prove Proposition 3 and Proposition 4. Lemma 1 (Lemma 2) is a direct applications of Proposition 1 (Proposition 2) in the simple stochastic version. It should be noted that the results are presented in a more explicit way.

Lemma 1

Let

$$\begin{aligned} f_i =\frac{[n-(n-1)\beta ]\theta \Delta }{(n+1)^2}\left\{ {2(a-c)+[(n+1)(1-\beta )-(2i-1)(1-2\beta )]\Delta } \right\} . \end{aligned}$$

• (1) If $\beta <1/2$,

  $$\begin{aligned} {\hat{m}}= {\left\{ \begin{array}{ll} n, &{} \mathrm{{if}}\ f\le f_n\\ j,&{}\mathrm{{if}}\ f_{j+1} <f\le f_j,\,j\in \{1,\ldots ,n-1\}\\ 0,&{}\mathrm{{if}}\ f>f_1 \end{array}\right. }. \end{aligned}$$

• (2) If $\beta \ge 1/2$,

  $$\begin{aligned} {\hat{m}}= {\left\{ \begin{array}{ll} n,&{}\mathrm{{if}}\ f\le f_{n}\\ 0,&{}\mathrm{{if}}\ f>f_{n} \end{array}\right. }. \end{aligned}$$

Lemma 2

Let $\phi (\beta )=2n+3-2(5+3n)\beta -(n^2-2n-7)\beta ^2$.

• (1) If $\beta <{\hat{\beta }}^*$,

  $$\begin{aligned} {\hat{m}}^*={\left\{ \begin{array}{ll} n, &{} \mathrm{{if}}\ f\le f_n^*\\ j, &{} \mathrm{{if}}\ f_{j+1}^*<f\le f_j^*,\,j\in \{1,\ldots ,n-1\}\\ 0, &{} \mathrm{{if}}\ f>f_{1}^*\end{array}\right. }, \end{aligned}$$

  where $f_i^*\!=\!\frac{\theta \Delta }{2(n+1)^2}\big \{ 2(n+2)[1\!+\!(n-1)\beta ](a-c)\!+\![2(n+1)^2(1\!-\!\beta )^2 -(2i-1) \phi (\beta )]\Delta \big \}$.

• (2) If $\beta \ge {\hat{\beta }}^*$,

  $$\begin{aligned} {\hat{m}}^*= {\left\{ \begin{array}{ll} 0, &{} \mathrm{{if}}\ f>f^*\\ n, &{} \mathrm{{if}}\ f\le f^*\end{array}\right. }, \end{aligned}$$

  where

  $$\begin{aligned} f^*=\theta \Delta \left\{ {(1-\beta )^2\Delta +\frac{2(n+2)[1+(n-1)\beta ](a-c)-n\phi (\beta )\Delta }{2(n+1)^2}} \right\} . \end{aligned}$$

Lemma 3

Consider $\beta <{\hat{\beta }}^*$.

• (1) If $\delta _1 (\beta )>0$ and $\delta _i (\beta )\ge 0$  for all $i\in \{2,\ldots ,n\}$, then ${\hat{m}}\ge {\hat{m}}^*$. Furthermore, when $f_1^*<f\le f_1,{\hat{m}}>{\hat{m}}^*$.

• (2) If $\delta _1 (\beta )<0$ and $\delta _i (\beta )\le 0$  for all $i\in \{2,\ldots ,n\}$, then ${\hat{m}}\le {\hat{m}}^*$. Furthermore, when $f_1 <f\le f_1^*,{\hat{m}}<{\hat{m}}^*$.

• (3) If there exists $r\in \{1,\ldots ,n-1\}$ such that

  $$\begin{aligned} \delta _1 (\beta )<\cdots <\delta _r (\beta )\le 0<\delta _{r+1} (\beta )<\cdots <\delta _n (\beta ), \end{aligned}$$

  then ${\hat{m}}\ge {\hat{m}}^*$ when $f\le f_r$, and ${\hat{m}}\le {\hat{m}}^*$ when $f>f_r$. Furthermore, there always exist two ranges of $f$ to respectively support ${\hat{m}}>{\hat{m}}^*$ and ${\hat{m}}<{\hat{m}}^*$.

Proof

We only prove part (1) and part (3), since part (2) can be obtained following the same line as the proof of part (1). It follows from Lemma 1 that

$$\begin{aligned} {\hat{m}}\ge i\quad \mathrm{{for~any}}\ f\le f_i\ \mathrm{{and}}\ {\hat{m}}\le i-1\quad \mathrm{{for~any}}\ f>f_i \end{aligned}$$

(4)

Likewise,

$$\begin{aligned} {\hat{m}}^*\ge i\quad \mathrm{{for~any}}\ f\le f_i^*\ \mathrm{{and}}\ {\hat{m}}^*\le i-1\quad \mathrm{{for~any}}\ f>f_i^*\end{aligned}$$

(5)

• (1) Noting

  $$\begin{aligned} 0<f_n^*<f_{n-1}^*<\cdots <f_1^*<f_1, \end{aligned}$$

  we can complete the proof by considering the following several cases.

• (1a) If $0\le f\le f_n^*$, noting that $f_n^*\le f_n$, we have ${\hat{m}}={\hat{m}}^*=n$.

• (1b) If $f_{j+1}^*<f\le f_j^*$, where $j\in \{1,\ldots ,n-1\}$, it is clear that ${\hat{m}}^*=j$. Given that $f\le f_j^*\le f_j,{\hat{m}}\ge j$ according to (4).

• (1c) If $f_1^*<f\le f_1$, by means of Lemma 2 and (4), we can easily obtain that ${\hat{m}}^*=0$ and ${\hat{m}}\ge 1$.

• (1d) If $f>f_1$, it is evident that ${\hat{m}}={\hat{m}}^*=0$.

• (3) Since

  $$\begin{aligned} f_n^*<f_{n-1}^*<\cdots <f_{r+2}^*<f_{r+1}^*<f_{r+1} <f_r <f_{r-1} <\cdots <f_1 <f_1^*, \end{aligned}$$

  by virtue of Lemmas 1, 2, (4) and (5), we can finish this proof by Table 1, where $j\in \{r+1,\ldots ,n-1\}$ and $t\in \{1,\ldots ,r-1\}$.

Table 1 The comparison between ${\hat{m}}$ and ${\hat{m}}^*$ for case (3)

Appendix D: The proof of Proposition 3

Setting $n=2$, we have ${\hat{\beta }}^*=\frac{11-6\sqrt{2}}{7}\approx 0.359$ and

$$\begin{aligned} \left\{ \begin{array}{l} \delta _1 (\beta )=-3[(1+3\beta ^2)\Delta +2(2a-2c-\Delta )\beta ]<0\\ \delta _2 (\beta )=-3[2(2a-2c+3\Delta )\beta -(1-\beta ^2)\Delta \end{array} \right. . \end{aligned}$$

It follows from our basic assumption that

$$\begin{aligned} \delta _2 ({\hat{\beta }}^*)<-3\Delta [({\hat{\beta }}^*)^2+10{\hat{\beta }}^*-1]<0, \end{aligned}$$

which means that ${\bar{\beta }}<{\hat{\beta }}^*$.

• (1) If $\beta <{\bar{\beta }},\delta _1 (\beta )<0<\delta _2 (\beta )$. By virtue of Lemma 3(3), this result is clear.

• (2) Noticing ${\bar{\beta }}<{\hat{\beta }}^*<1/2$, this proof can be done by considering three cases.

• (2a) If $\bar{\beta }\le \beta <{\hat{\beta }}^*$, $\delta _1 (\beta )<0$ and $\delta _2 (\beta )\le 0$. This result is obvious following Lemma 3 (1).

• (2b) If ${\hat{\beta }}^*\le \beta <1/2$, noting $\phi (\beta )\le 0$ and

  $$\begin{aligned} f^*-f_1 =\frac{\theta \Delta \left\{ {6\beta (a-c)+[(1-2\beta )(5-4\beta )-\phi (\beta )]\Delta } \right\} }{9}, \end{aligned}$$

  we can obtain that $f^*>f_1 >f_2$. By virtue of Lemma 1 and Lemma 2, one can easily derive Table 2, which implies our result.

• (2c) If $\beta \ge 1/2,f^*>f_2$ since

  $$\begin{aligned} f^*-f_2 =\frac{\theta \Delta \left\{ {6\beta (a-c)+[5\beta ^2+2(1-\beta )]\Delta } \right\} }{9}>0. \end{aligned}$$

  The result follows immediately from Table 3.

Table 2 The comparison between ${\hat{m}}$ and ${\hat{m}}^*$ for case (2b)

Table 3 The comparison between ${\hat{m}}$ and ${\hat{m}}^*$ for case (2c)

Appendix E: The proof of Proposition 4

Obviously, for any $i\in \{1,\ldots ,n-1\}$,

$$\begin{aligned} \delta _{i+1} (\beta )-\delta _i (\beta )=2[3-12\beta -(n^2+2n-11)\beta ^2]\Delta :=2d(\beta )\Delta . \end{aligned}$$

It is easy to see that

$$\begin{aligned} d\left( {\frac{n-2}{(n-1)(n+4)}}\right) = \frac{2[n^4+4n^3+n(n-2)+2(n^2-1)]}{(n-1)^2(n+4)^2}>0. \end{aligned}$$

Besides, since

$$\begin{aligned} \phi (\beta )-d(\beta )=2(1-2\beta )[\beta +(1-\beta )n], \end{aligned}$$

we have $d({\hat{\beta }}^*)=d({\hat{\beta }}^*)-\phi ({\hat{\beta }}^*)=-2(1-2{\hat{\beta }}^*)[{\hat{\beta }}^*+(1-{\hat{\beta }}^*)n]<0$. Noting that $d(\beta )$ is strictly decreasing, there exists a unique solution for $d(\beta )$ with respect to $\beta \in ( {\frac{n-2}{(n-1)(n+4)},{\hat{\beta }}^*})$, denoted by $\beta ^d$. Hence, if $\beta <\beta ^d,d(\beta )>0$, which results in

$$\begin{aligned} \delta _1 (\beta )<\delta _2 (\beta )<\cdots <\delta _{n-1} (\beta )<\delta _n (\beta ). \end{aligned}$$

(6)

Noting that $\frac{n-2}{(n-1)(n+4)}<\beta ^d<{\hat{\beta }}^*<\frac{1}{2}$ and $d(\beta ^d)=0$, it is not difficult to verify that

$$\begin{aligned} \delta _1 (\beta ^d)=\delta _n (\beta ^d)&= -2\Bigg \{ (n-1)(n+4)\left[ {\beta ^d-\frac{n-2}{(n-1)(n+4)}} \right] \\&\quad \times (a-c)+(n+1)(1-\beta ^d)(1-2\beta ^d)\Delta \Bigg \}<0. \end{aligned}$$

Consequently, $\beta ^d>{\underline{\beta }}$ and $\beta ^d>{\bar{\beta }}$. Since ${\underline{\beta }}<\beta ^d$, it follows directly from (6) that $\delta _n ({\underline{\beta }})>\delta _1 ({\underline{\beta }})\!=\!0$, leading to the fact that ${\underline{\beta }}<\bar{\beta }$. In sum,

$$\begin{aligned} {\underline{\beta }}<{\bar{\beta }}<\beta ^d<{\hat{\beta }}^*<1/2. \end{aligned}$$

• (1) If $\beta <{\underline{\beta }},\delta _1 (\beta )>0$. Since ${\underline{\beta }}<\beta ^d$, it follows from (6) that

  $$\begin{aligned} 0<\delta _1 (\beta )<\delta _2 (\beta )<\ldots <\delta _{n-1} (\beta )<\delta _n (\beta ). \end{aligned}$$

  By virtue of Lemma 3(1), we have ${\hat{m}}\ge {\hat{m}}^*$.

• (2) If ${\underline{\beta }}\le \beta <{\bar{\beta }},\delta _n (\beta )>0$ and $\delta _1 (\beta )\le 0$. Hence, there exists $r\in \{1,\ldots ,n-1\}$ such that

  $$\begin{aligned} \delta _1 (\beta )<\cdots <\delta _r (\beta )\le 0<\delta _{r+1} (\beta )<\ldots <\delta _n (\beta ). \end{aligned}$$

  The proof can be completed by Lemma 3(3).

• (3) We can finish this proof by the following three cases.

• (3a) If ${\bar{\beta }}\le \beta <{\hat{\beta }}^*, \delta _1 (\beta )<0$ and $\delta _n (\beta )\le 0$, which implies that $\delta _i (\beta )<0$ for all $i\in \{2,\ldots ,n-1\}$. By virtue of Lemma 3(2), we have ${\hat{m}}\le {\hat{m}}^*$.

• (3b) Consider ${\hat{\beta }}^*\le \beta <1/2$. One can easily obtain that the sigh of $f^*-f_1$ is determined by

  $$\begin{aligned} 2[(n-1)(n+4)\beta -(n-2)](a-c)+[2(1-2\beta )(2n-2n\beta +1)-n\phi (\beta )]\Delta . \end{aligned}$$

  Given $\beta \ge {\hat{\beta }}^*>\frac{n-2}{(n-1)(n+4)}$, the coefficient of $a-c$ is positive. Since $\phi (\beta )\le 0$ and $\beta <1 / 2$, the coefficient of $\Delta$ is also positive. Consequently, $f^*>f_1$. In sum,

  $$\begin{aligned} f_n <f_{n-1} <\cdots <f_2 <f_1 <f^*. \end{aligned}$$

  By virtue of Lemmas 1 and 2, Table 4 is evident, which yields our result.

• (3c) Consider $\beta \ge 1/2$. Some simple calculations give

  $$\begin{aligned} \frac{2(n+1)^2(f^*-f_n )}{\theta \Delta }=2[(n-1)(n+4)\beta -(n-2)](a-c)+\psi (\beta )\Delta , \end{aligned}$$

  where

  $$\begin{aligned} \psi (\beta )=(n^3+6n^2-15n+8)\beta ^2-2(3n^2-7n+4)\beta +2n^2-3n+2. \end{aligned}$$

  Since

  $$\begin{aligned} \left\{ \begin{array}{l} {{\psi }^{\prime }(\beta )=2(n^3+6n^2-15n+8)\beta -2(3n^2-7n+4)}\\ {{\psi }^{\prime \prime }(\beta )=2(n^3+6n^2-15n+8)>0} \end{array}\right. , \end{aligned}$$

  ${\psi }^{\prime }(\beta )\ge {\psi }^{\prime }(1 / 2)=n(n^2-1)>0,$ which implies $\psi (\beta )\ge \psi (1 / 2)={n(n+1)^2} / 4>0$. Obviously, the coefficient of $a-c$ is positive. Marshaling these facts lead to $f^*>f_n$. The result can be thus indicated by Table 5.

Table 4 The comparison between ${\hat{m}}$ and ${\hat{m}}^*$ for case (3b)

Table 5 The comparison between ${\hat{m}}$ and ${\hat{m}}^*$ for case (3c)