Endogenous heterogeneity in duopoly with deterministic one-way spillovers

Abstract

This paper examines the standard symmetric two-period R&D duopoly model, but with a deterministic one-way spillover structure. Though the two firms are ex-ante identical, one obtains a unique pair of asymmetric equilibria of R&D investments, leading to inter-firm heterogeneity in the industry, in R&D roles as well as in unit costs. We analyze the impact of a change in the spillover parameter and R&D costs on firms’ levels of R&D and profits. We find that higher spillovers need not lead to lower R&D investments for both firms. In addition, equilibrium profits may improve due to the presence of spillovers, and it may be advantageous to be the R&D imitator rather than the R&D innovator.

Appendix (Proofs)

1.1 Proof of Lemma 1

The reaction function r as given by Eq. (2) is not continuous since, letting $x^{S1}=r_{1}(x^{S1})$ for $x_{1}\ge x_{2}$ and $x^{S2}=r_{1}(x^{S2})$ for $x_{1}\le x_{2}$, one obtains

$$\begin{aligned} x^{S1}= & {} \frac{2\left( a-c\right) \left( 2-\beta \right) }{9\gamma -2\left( 2-\beta \right) }, \\ x^{S2}= & {} \frac{4\left( a-c\right) \left( 1-\beta \right) }{9\gamma -4\left( 1-\beta \right) } \end{aligned}$$

with $x^{S1}>x^{S2}$. Hence, the reaction function has a downward jump, and letting $\widehat{x}$ be the solution to $U(r_{1}(\widehat{x}), \widehat{x})=L(r_{1}(\widehat{x}),\widehat{x})$, we have that

$$\begin{aligned} \widehat{x}=\frac{\left( a-c\right) \left( \sqrt{1+\frac{2\beta \left( 4-3\beta \right) }{9\gamma -2\left( \beta -2\right) ^{2}}}-1\right) }{2\beta -1+(1-\beta )\sqrt{1+\frac{2\beta \left( 4-3\beta \right) }{9\gamma -2\left( \beta -2\right) ^{2}}}} \end{aligned}$$

(5)

It is easy to check that $r_{1}(\widehat{x})\ne \widehat{x}$. Hence, there cannot exist any symmetric Nash equilibria in pure strategies.

Furthermore, $\hat{x}$ is unique since both U and L are monotonic in $x_{2}$. U is decreasing in $x_{2}$ for all $\beta \in [0,1]$, while L either increases with $x_{2}$ for $\beta >1/2$ or decreases with $x_{2}$ slower than U.

1.2 Proof of Proposition 1

A lengthy but simple computation establishes that $\bar{x},\underline{x}$ as given by Eqs. (3) and (4) satisfy $\overline{x}>\widehat{x}$ if $9\gamma >I_{1}$ and $\underline{x}<\widehat{x}$ if $9\gamma >I_{2}$, with $\hat{x}$ as defined by Eq. (5) and

$$\begin{aligned} I_{1}= & {} \left( 5\beta ^{2}-12\beta +8\right) +\sqrt{13\beta ^{4}-48\beta ^{3}+68\beta ^{2}-48\beta +16}, \\ I_{2}= & {} \left( 5\beta ^{2}-12\beta +8\right) +\sqrt{73\beta ^{4}+224\beta ^{2}-216\beta ^{3}-96\beta +16}, \end{aligned}$$

Straightforward computations then establish that $I(\beta )>I_{1}$ and $I(\beta )>I_{2}$. Hence, if assumptions (A1) through (A4) hold, the pair of PSNE $(\overline{x},\underline{x})$ and $\left( \underline{x},\overline{x} \right)$, with $\bar{x},\underline{x}$ as given by Eqs. (3) and (4), is unique.

1.3 Proof of Proposition 2

• (i) Differentiating $\bar{x}$ with respect to $\beta$ and collecting terms yields

  $$\begin{aligned} \frac{d\bar{x}}{d\beta }=-\frac{6\left( a-c\right) \gamma L_{1}}{\left( 3\gamma \left[ 9\gamma -2\left( 5\beta ^{2}-12\beta +8\right) \right] +8\left( 2-\beta \right) \left( \beta -1\right) ^{2}\right) ^{2}} \end{aligned}$$

  where $L_{1}=27\gamma ^{2}-6\gamma \left( 13\beta ^{2}-28\beta +14\right) +8\left( 5\beta ^{2}-12\beta +8\right) \left( \beta -1\right) ^{2}$. Thus, $\frac{d\bar{x}}{d\beta }>0$ if and only if $L_{1}<0$, which holds for $(\beta ,9\gamma )$ such that $\beta <0.04$ and $9\gamma \in \bar{\Gamma } =\left( \gamma _{1}(\beta ),\gamma _{2}(\beta )\right)$, where

  $$\begin{aligned} \gamma _{1}(\beta )=&13\beta ^{2}-28\beta -\sqrt{260\beta ^{2}-112\beta -200\beta ^{3}+49\beta ^{4}+4}+14 \\ \gamma _{2}(\beta )=&13\beta ^{2}-28\beta +\sqrt{260\beta ^{2}-112\beta -200\beta ^{3}+49\beta ^{4}+4}+14 \end{aligned}$$

• (ii) In a similar fashion, differentiating $\underline{x}$ with respect to $\beta$ and collecting terms yields

  $$\begin{aligned} \frac{d\underline{x}}{d\beta }=-\frac{12\left( a-c\right) \gamma L_{2}}{ \left( 3\gamma \left[ 9\gamma -2\left( 5\beta ^{2}-12\beta +8\right) \right] +8\left( 2-\beta \right) \left( \beta -1\right) ^{2}\right) ^{2}}, \end{aligned}$$

  where $L_{2}=27\gamma ^{2}-6\gamma \left( 4\beta ^{2}-14\beta +11\right) +4\left( \beta -1\right) ( 18\beta -15\beta ^{2}+5\beta ^{3}-10)$. Thus, $\frac{d\underline{x}}{d\beta }>0$ if and only if $L_{2}<0$, which holds if $(\beta ,9\gamma )$ is such that $9\gamma \in \underline{\Gamma } =\left( \gamma _{3}(\beta ),\gamma _{4}(\beta )\right)$, where

  $$\begin{aligned} \gamma _{3}(\beta )= & {} \left( 4\beta ^{2}-14\beta +11\right) -\sqrt{28\beta -112\beta ^{2}+128\beta ^{3}-44\beta ^{4}+1} \\ \gamma _{4}(\beta )= & {} \left( 4\beta ^{2}-14\beta +11\right) +\sqrt{28\beta -112\beta ^{2}+128\beta ^{3}-44\beta ^{4}+1} \end{aligned}$$

• (iii) The sum of autonomous cost reductions is given by

  $$\begin{aligned} \frac{d}{d\beta }(\overline{x}+\underline{x})=\frac{6\gamma \left( a-c\right) L_{3}}{\left( 3\gamma \left[ 9\gamma -2\left( 5\beta ^{2}-12\beta +8\right) \right] +8\left( 2-\beta \right) \left( \beta -1\right) ^{2}\right) ^{2}} \end{aligned}$$

  where

  $$\begin{aligned} L_{3}= & {} -81\gamma ^{2}+6\gamma \left( 21\beta ^{2}-56\beta +36\right) +16\left( 1-\beta \right) \left( 19\beta -16\beta ^{2}+5\beta ^{3}-9\right) \\&<0\text { for all }\beta \in \left[ 0,1\right] \end{aligned}$$

  It is easy to verify that the sum $(\overline{x}+\underline{x})$ is decreasing in $\beta$.

• (iv) As to the fact that $(1+\beta )\overline{x}+(1-\beta )\underline{ x}$ is also decreasing in $\beta$, a direct computation shows that$\frac{d}{d\beta }((1+\beta )\overline{x}+(1-\beta )\underline{x})<0$ (the details are left out).

1.4 Proof of Proposition 3

• (i) From Eq. (1), the innovator’s equilibrium profit is given by

  $$\begin{aligned} F(\overline{x},\underline{x})=\frac{\left( a-c+\overline{x}\left( 2-\beta \right) -\underline{x}\left( 1-\beta \right) \right) ^{2}}{9}-\frac{\gamma }{ 2}\overline{x}^{2} \end{aligned}$$

  Plugging $\bar{x}$ and $\underline{x}$ as given by Eqs. (3) and (4) and differentiating totally $F(\overline{x},\underline{x})$ with respect to $\beta$ yields

  $$\begin{aligned} \frac{d}{d\beta }F(\overline{x},\underline{x})=-\frac{12\left( 3\gamma -4\left( 1-\beta \right) ^{2}\right) \gamma ^{2}\left( a-c\right) ^{2}L_{4}}{ \left( 3\gamma \left[ 9\gamma -2\left( 5\beta ^{2}-12\beta +8\right) \right] +8\left( 2-\beta \right) \left( \beta -1\right) ^{2}\right) ^{3}} \end{aligned}$$

  where $L_{4}=27\gamma ^{2}\left( 3\beta -2\right) -6\gamma \left( 13\beta ^{3}-48\beta ^{2}+58\beta -22\right) +8( 5\beta ^{3}-16\beta ^{2}+20\beta -10) \left( 1-\beta \right) ^{2}.$ Under assumption (A4), the denominator is strictly positive and $3\gamma >4\left( 1-\beta \right) ^{2}$ for any $\beta \in [0,1]$. Therefore, we have that $\frac{\text { d}}{\text {d}\beta }F(\overline{x},\underline{x})>0$ if and only if $L_{4}<0$ , which holds for $\beta <2/3$ and $9\gamma \in \bar{\Gamma }^{\prime }=(\gamma _{5}(\beta ),\infty )$, where

  $$\begin{aligned} \gamma _{5}(\beta )=\frac{22-58\beta +48\beta ^{2}-13\beta ^{3}+\sqrt{ 4+88\beta -572\beta ^{2}+1348\beta ^{3}-1540\beta ^{4}+864\beta ^{5}-191\beta ^{6}}}{\left( 2-3\beta \right) }. \end{aligned}$$

• (ii) Likewise, the imitator’s equilibrium profit is given by

  $$\begin{aligned} F(\underline{x},\bar{x})=\frac{\left( a-c+2\underline{x}\left( 1-\beta \right) +\bar{x}\left( 2\beta -1\right) \right) ^{2}}{9}-\frac{\gamma }{2} \underline{x}^{2} \end{aligned}$$

  Differentiating totally $F(\underline{x},\bar{x})$ with respect to $\beta$ yields

  $$\begin{aligned} \frac{d}{d\beta }F(\underline{x},\overline{x})=\frac{12\left( a-c\right) ^{2}\left( 3\gamma -2\left( 1-\beta \right) \left( 2-\beta \right) \right) \gamma ^{2}L_{5}}{\left( 3\gamma \left[ 9\gamma -2\left( 5\beta ^{2}-12\beta +8\right) \right] +8\left( 2-\beta \right) \left( \beta -1\right) ^{2}\right) ^{3}}, \end{aligned}$$

  where $L_{5}=$$27\gamma ^{2}+6\gamma \left( -14+40\beta -45\beta ^{2}+16\beta ^{3}\right) -8( -8+24\beta -27\beta ^{2}+10\beta ^{3}) \left( 1-\beta \right) ^{2}$. Again, if (A4) holds, the denominator is strictly positive and $3\gamma >2\left( 1-\beta \right) \left( 2-\beta \right)$ for any $\beta \in [0,1]$. Hence, the imitator’s profit is strictly decreasing in $\beta$ if and only if $L_{5}<0$ , which holds for $9\gamma \in \underline{\Gamma }^{\prime }=\left( \gamma _{6}(\beta ),\gamma _{7}(\beta )\right)$, where

  $$\begin{aligned}&\gamma _{6}(\beta ) =45\beta ^{2}-40\beta -16\beta ^{3}\\&\quad +14-\sqrt{868\beta ^{2}-160\beta -1936\beta ^{3}+2177\beta ^{4}-1200\beta ^{5}+256\beta ^{6}+4}\\&\gamma _{7}(\beta ) =45\beta ^{2}-40\beta -16\beta ^{3}\\&\quad +14+\sqrt{868\beta ^{2}-160\beta -1936\beta ^{3}+2177\beta ^{4}-1200\beta ^{5}+256\beta ^{6}+4} \end{aligned}$$

1.5 Proof of Proposition 4

Under (A1)–(A4), computations show that the inequality $F(\overline{x}, \underline{x})>F(\underline{x},\overline{x})$ in the light red area is defined by the inequalities $9\gamma >I(\beta )$, $0<\beta <2/3$, and $9\gamma > \displaystyle \frac{6(1-x)^2(4-5x)}{2-3x}.$ Likewise, $F(\overline{x}, \underline{x})<F(\underline{x},\overline{x})$ in the light blue area is defined by the inequalities $9\gamma >I(\beta )$ and either (1) $0<\beta <2/3$, and $9\gamma <\displaystyle \frac{6(1-x)^2(4-5x)}{2-3x}$ or (2) $\beta \ge 2/3$.

1.6 Proof of Proposition 5

Consumer surplus at the noncooperative interior equilibrium reduces to

$$\begin{aligned} CS(\bar{x},\underline{x})=\frac{18\left( 3\gamma +\left( 1-\beta \right) \left( 3\beta -4\right) \right) ^{2}\left( a-c\right) ^{2}\gamma ^{2}}{ \left( 27\gamma ^{2}-6\gamma \left( 5\beta ^{2}-12\beta +8\right) -8\left( \beta -2\right) \left( \beta -1\right) ^{2}\right) ^{2}} \end{aligned}$$

Differentiating $CS(\bar{x},\underline{x})$ with respect to $\beta$ yields

$$\begin{aligned} \frac{d}{d\beta }CS(\bar{x},\underline{x})=\frac{36\gamma ^{2}\left( a-c\right) ^{2}K_{1}K_{2}}{\left( 27\gamma ^{2}-6\gamma \left( 5\beta ^{2}-12\beta +8\right) -8\left( \beta -2\right) \left( \beta -1\right) ^{2}\right) ^{3}} \end{aligned}$$

where

$$\begin{aligned} K_{1}=9\gamma ^{2}\left( 2\beta -3\right) +6\gamma \left( 11\beta ^{2}-24\beta +12\right) -8\left( 3\beta ^{2}-8\beta +6\right) \left( \beta -1\right) ^{2} \end{aligned}$$

and

$$\begin{aligned} K_{2}=3\gamma -\left( 1-\beta \right) \left( 4-3\beta \right) . \end{aligned}$$

We have that $K_{1}.K_{2}<0$ since $K_{1}<0$ and $K_{2}>0$ for all $9\gamma >I(\beta )$. Since the denominator is strictly positive, it follows that $\frac{d}{d\beta }CS(\bar{x},\underline{x})<0$.

Likewise, we have that

$$\begin{aligned} \frac{d}{d\gamma }CS(\bar{x},\underline{x})=-\frac{36\left( a-c\right) ^{2}\gamma K_{2}K_{3}}{\left( 27\gamma ^{2}-6\gamma \left( 5\beta ^{2}-12\beta +8\right) -8\left( \beta -2\right) \left( \beta -1\right) ^{2}\right) ^{3}}, \end{aligned}$$

where

$$\begin{aligned} K_{3}=9\gamma ^{2}\left( \beta ^{2}-3\beta +4\right) -48\gamma \left( 2-\beta \right) \left( 1-\beta \right) ^{2}-8\left( 3\beta -4\right) \left( 2-\beta \right) \left( 1-\beta \right) ^{3}. \end{aligned}$$

Observe that $K_{2}.K_{3}>0$ since $K_{2}>0$ and $K_{3}>0$ for all $9\gamma >I(\beta ).$ Hence, $\frac{d}{d\gamma }CS(\bar{x},\underline{x})<0$.