Incentive contracts under product market competition and R&D spillovers

Abstract

This paper studies cost-reducing R&D incentives in a principal-agent model with product market competition. It argues that moral hazard does not necessarily decrease firms’ profits in this setting. In highly competitive industries, firms are driven by business-stealing incentives and exert such high levels of R&D that they burn up their profits. In the presence of moral hazard, underprovision of R&D incentives due to risk sharing can generate considerable cost savings, implying higher profits for both rivals. This result indicates firms’ incentives to adopt collusive-like behavior in the R&D market. We also examine the agents’ contracts and the profits-risk relationship when cross-firm R&D spillovers occur.

Appendix

1.1 Proof of Lemma 1

By Eqs. (1) and (2), agent $i$’s expected utility is²⁴

$$\begin{aligned} E\left\{ U_{i}\left( x_{i}\right) \mid \varepsilon _{i}\in \Theta \right\} =-e^{-r\left[ \alpha _{i}+\beta _{i}x_{i}-\psi (x_{i})\right] }E\left\{ e^{-r\beta _{i}\varepsilon _{i}}\mid \varepsilon _{i}\in \Theta \right\} \text {.} \end{aligned}$$

(13)

The conditional density of $\varepsilon _{i}$ is

$$\begin{aligned} f\left( \varepsilon _{i}\mid \Theta \right) =\frac{\frac{1}{\sigma }\phi \left( \frac{\varepsilon _{i}}{\sigma }\right) }{\varPhi \left( \frac{\theta }{ \sigma }\right) -\varPhi \left( \frac{-\theta }{\sigma }\right) }\text {, } -\theta \le \varepsilon _{i}\le \theta \text { where }\phi \left( \frac{ \varepsilon _{i}}{\sigma }\right) =\frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2} \left( \frac{\varepsilon _{i}}{\sigma }\right) ^{2}}\text {.} \end{aligned}$$

(14)

By Eqs. (13) and (14), and letting $\widehat{r}\equiv -r$, we have

$$\begin{aligned} \int \limits _{-\theta }^{\theta }e^{\widehat{r}\beta _{i}\varepsilon _{i}}f\left( \varepsilon _{i}\right) \mathrm{d}\varepsilon _{i}&= \frac{1}{\sigma \sqrt{2\pi }} \int \limits _{-\theta }^{\theta }e^{\widehat{r}\beta _{i}\varepsilon _{i}}e^{-\frac{1 }{2}\left( \frac{\varepsilon _{i}}{\sigma }\right) ^{2}} \mathrm{d} \varepsilon _{i}\\&= \frac{1}{\sigma \sqrt{2\pi }}\int \limits _{-\theta }^{\theta }e^{-\frac{\left( \varepsilon _{i}^{2}-2\sigma ^{2}\widehat{r}\beta _{i}\varepsilon _{i}\right) }{2\sigma ^{2}}} \mathrm{d} \varepsilon _{i}\\&= \frac{1}{\sigma \sqrt{2\pi }}\int \limits _{-\theta }^{\theta }e^{-\frac{\left[ \varepsilon _{i}^{2}-2\varepsilon _{i}\left( \sigma ^{2}\widehat{r}\beta _{i}\right) +\left( \sigma ^{2}\widehat{r}\beta _{i}\right) ^{2}-\left( \sigma ^{2}\widehat{r}\beta _{i}\right) ^{2}\right] }{2\sigma ^{2}} } \mathrm{d}\varepsilon _{i}\\&= \frac{1}{\sigma \sqrt{2\pi }}\int \limits _{-\theta }^{\theta }e^{- \frac{\left( \varepsilon _{i}-\sigma ^{2}\widehat{r}\beta _{i}\right) ^{2}}{ 2\sigma ^{2}}}e^{\frac{\left( \sigma ^{2}\widehat{r}\beta _{i}\right) ^{2}}{ 2\sigma ^{2}}} \mathrm{d} \varepsilon _{i}\\&= e^{\frac{\widehat{r}^{2}\beta _{i}^{2}\sigma ^{2}}{2}}\frac{1}{\sigma \sqrt{2\pi }}\int \limits _{-\theta }^{\theta }e^{-\frac{1}{2}\left( \frac{ \varepsilon _{i}-\sigma ^{2}\widehat{r}\beta _{i}}{\sigma }\right) ^{2}} \mathrm{d} \varepsilon _{i}\\&= e^{\frac{\widehat{r}^{2}\beta _{i}^{2}\sigma ^{2}}{2} }\int \limits _{-\theta }^{\theta }\frac{1}{\sigma }\phi \left( \frac{\varepsilon _{i}-\sigma ^{2}\widehat{r}\beta _{i}}{\sigma }\right) \mathrm{d} \varepsilon _{i}\\&= e^{\frac{\widehat{r}^{2}\beta _{i}^{2}\sigma ^{2}}{2}}\left[ \varPhi \left( \frac{\theta -\sigma ^{2}\widehat{r}\beta _{i}}{\sigma }\right) -\varPhi \left( \frac{-\theta -\sigma ^{2}\widehat{r}\beta _{i}}{\sigma }\right) \right] . \end{aligned}$$

$$\begin{aligned} \hbox {Thus, }E\left\{ e^{\widehat{r}\beta _{i}\varepsilon _{i}}\mid \varepsilon _{i}\in \Theta \right\}&= \frac{\int _{-\theta }^{\theta }e^{ \widehat{r}\beta _{i}\varepsilon _{i}}f\left( \varepsilon _{i}\right) \mathrm{d}\varepsilon _{i}}{\varPhi \left( \frac{\theta }{\sigma }\right) -\varPhi \left( \frac{-\theta }{\sigma }\right) }\\&= e^{\frac{\widehat{r}^{2}\beta _{i}^{2}\sigma ^{2}}{2}}\frac{\varPhi \left( \frac{\theta -\sigma ^{2}\widehat{r }\beta _{i}}{\sigma }\right) -\varPhi \left( \frac{-\theta -\sigma ^{2}\widehat{ r}\beta _{i}}{\sigma }\right) }{\varPhi \left( \frac{\theta }{\sigma }\right) -\varPhi \left( \frac{-\theta }{\sigma }\right) }=e^{\frac{\widehat{r}^{2}\beta _{i}^{2}\sigma ^{2}}{2}}\Omega _{i}\text {,} \end{aligned}$$

where $\Omega _{i}\equiv \frac{\varPhi \left( \frac{\theta -\sigma ^{2}\widehat{r} \beta _{i}}{\sigma }\right) -\varPhi \left( \frac{-\theta -\sigma ^{2}\widehat{r }\beta _{i}}{\sigma }\right) }{\varPhi \left( \frac{\theta }{\sigma }\right) -\varPhi \left( \frac{-\theta }{\sigma }\right) }$. By Eq. (13), we get

$$\begin{aligned} E\left\{ U_{i}\left( w{i},x_{i}\right) \mid \varepsilon _{i}\in \Theta \right\}&= -\Omega _{i}e^{-r\left[ \alpha _{i}+\beta _{i}x_{i}-\psi (x_{i}) \right] }e^{\frac{r^{2}\beta _{i}^{2}\sigma ^{2}}{2}}\\&= -\Omega _{i}e^{-r\left[ \alpha _{i}+\beta _{i}x_{i}-\frac{r}{2}\beta _{i}^{2}\sigma ^{2}-\psi (x_{i}) \right] }\text {.} \end{aligned}$$

Given that $\Omega _{i}$ is positive and independent of $x_{i}$, agent $i$’s optimization problem is equivalent to choosing

$$\begin{aligned} x_{i}\in \arg \max \widetilde{U}_{i}\left( x_{i}\right) \equiv \alpha _{i}+\beta _{i}x_{i}-\frac{r}{2}\beta _{i}^{2}\sigma ^{2}-\psi (x_{i})\text {, } \end{aligned}$$

where $\widetilde{U}_{i}\left( x_{i}\right)$ is the certainty equivalent of agent $i$’s utility. Thus, given CARA preferences and linear contracts, agent $i$’s problem has a closed form solution even if $\varepsilon _{i}$ follows a truncated normal distribution which is symmetric around the mean.

1.2 Proof of Lemma 2

The existence of an equilibrium in R&D requires showing that each firm’s R&D reaction function—denoted by $r_{i}\left( x_{j}\right)$ for any $i$ and $j$—is a (monotone) contraction, and then applying the Contraction Mapping Theorem. Assumption $\left( A.2\right)$ suffices to guarantee that $\pi _{i}$ is strictly concave in $x_{i}$, and thus $r_{i}\left( x_{j}\right)$ is single-valued and continuous. Given that the action set is compact, an equilibrium exists. The interiority of this equilibrium requires $\pi _{i}$ to be strictly increasing when $x_{i}=0$ for all $x_{_{j}}\in X$; i.e., $\frac{\partial \pi _{i}\left( 0,x_{j}\right) }{\partial x_{i}}=\frac{4}{\left( 4-b^{2}\right) \left( 2+b\right) }\left( A- \overline{c}-\frac{bx_{j}}{2-b}\right) -\left[ 1+r\sigma ^{2}\psi ^{\prime \prime }\left( 0\right) \right] \psi ^{\prime }\left( 0\right) >0$. Given that $\psi ^{\prime }(0)=0$ and $\frac{\partial \pi _{i}\left( 0,x_{j}\right) }{\partial x_{i}}$ decreases with $x_{j}$, if the latter inequality holds for $x_{j}=\overline{c}+\theta$, it will also hold for all $x_{j}\in X$, which is guaranteed by assumption $\left( A.1\right)$. Firm $i$’s reaction function, $r_{i}\left( x_{j}\right)$, must also be strictly decreasing. Let

$$\begin{aligned} H\equiv \frac{\partial ^{2}\pi _{i}\left( r_{i}\left( x_{j}\right) ,x_{j}\right) }{\partial x_{i}^{2}}=\frac{8}{\left( 4-b^{2}\right) ^{2}}- \left[ 1+r\sigma ^{2}\psi ^{\prime \prime }\left( x_{i}\right) \right] \psi ^{\prime \prime }\left( x_{i}\right) \text {,} \end{aligned}$$

(15)

which is negative by assumption $\left( A.2\right)$. Then, taking $\frac{d\left( \partial \pi _{i}/\partial x_{i}\right) }{dx_{j}}=0,$ the equation $\frac{ \partial ^{2}\pi _{i}\left( r_{i}\left( x_{j}\right) ,x_{j}\right) }{ \partial x_{i}\partial x_{j}}+\frac{\partial ^{2}\pi _{i}\left( r_{i}\left( x_{j}\right) ,x_{j}\right) }{\partial x_{i}^{2}}r_{i}^{\prime }\left( x_{j}\right) =0$ implies

$$\begin{aligned} r_{i}^{\prime }\left( x_{j}\right) =-\frac{\frac{\partial ^{2}\pi _{i}\left( r_{i}\left( x_{j}\right) ,x_{j}\right) }{\partial x_{i}\partial x_{j}}}{ \frac{\partial ^{2}\pi _{i}\left( r_{i}\left( x_{j}\right) ,x_{j}\right) }{ \partial x_{i}^{2}}}=\frac{4b}{\left( 4-b^{2}\right) ^{2}H}\text {,} \end{aligned}$$

(16)

which is also negative for any $b>0$ since $H<0$.

The uniqueness of this equilibrium requires $-1<r_{i}^{\prime } \left( x_{j}\right) <1$ for all $x_{j}\in X$. Given that $r_{i}^{\prime }\left( x_{j}\right) <0$, it suffices to show that $-1<r_{i}^{\prime }\left( x_{j}\right)$ which is also implied by assumption $\left( A.2\right)$. Therefore, given that the game in the product market has a unique equilibrium, the subgame perfect equilibrium of the overall game is also unique.

1.3 Proof of Proposition 1

We first need to examine how the optimal effort $x^{*}$ changes with $r\sigma ^{2}$. We have

$$\begin{aligned} \frac{d\left( \partial \pi _{i}/\partial x_{i}\right) }{d\left( r\sigma ^{2}\right) }=0\Leftrightarrow \frac{\partial \left( \partial \pi _{i}/\partial x_{i}\right) }{\partial \left( r\sigma ^{2}\right) }+\frac{\partial ^{2}\pi _{i}}{\partial x_{i}^{2}}\frac{\mathrm{d} x^{*}}{\mathrm{d} \left( r\sigma ^{2}\right) }=0\text {.} \end{aligned}$$

(17)

To find the effect of $r\sigma ^{2}$ on the marginal profitability of R&D, we take the derivative $\frac{\partial \pi _{i}}{\partial x_{i}}$ and substitute $x_{j}$ with the optimal value $x^{*}$. We differentiate with respect to $r\sigma ^{2}$, taking $x_{i}$ as constant, and obtain

$$\begin{aligned} \frac{\partial \left( \partial \pi _{i}/\partial x_{i}\right) }{\partial \left( r\sigma ^{2}\right) }=-\frac{4b}{\left( 4-b^{2}\right) ^{2}}\frac{ d x^{*}}{d \left( r\sigma ^{2}\right) }-\psi ^{\prime \prime }\left( x_{i}\right) \psi ^{\prime }\left( x_{i}\right) . \end{aligned}$$

(18)

Then, we substitute $x_{i}$ with $x^{*}$. By (15) and (18), Eq. (17) gives

$$\begin{aligned} \frac{\mathrm{d} x^{*}}{\mathrm{d} \left( r\sigma ^{2}\right) }=\frac{\psi ^{\prime \prime }\left( x^{*}\right) \psi ^{\prime }\left( x^{*}\right) }{H-\frac{4b}{\left( 4-b^{2}\right) ^{2}}}\text {,} \end{aligned}$$

(19)

which is negative since $H<0$ and $\psi \left( .\right)$ is convex; i.e., $\frac{\mathrm{d} x^{*}}{\mathrm{d} \left( r\sigma ^{2}\right) }<0$ for all $b,r$ and $\sigma ^{2}$. Having $\frac{\mathrm{d}\psi (x^{*})}{\mathrm{d} (r \sigma ^2)}=\psi ^{\prime }(x^{*})\frac{\mathrm{d} x^{*}}{\mathrm{d} (r\sigma ^{2})}$, Eq. (7) implies that $\pi ^{*}$ increases with $r\sigma ^{2}$, if and only if,

$$\begin{aligned} \frac{2}{\left( 2+b\right) ^{2}}\left( A-\overline{c}+x^{*}\right) \frac{ \mathrm{d} x^{*}}{\mathrm{d} \left( r\sigma ^{2}\right) }>\left[ \frac{1}{2} \psi ^{\prime }\left( x^{*}\right) +r\sigma ^{2}\frac{\mathrm{d} \psi ^{\prime }\left( x^{*}\right) }{\mathrm{d} \left( r\sigma ^{2}\right) }+ \frac{\mathrm{d} x^{*}}{\mathrm{d} \left( r\sigma ^{2}\right) }\right] \psi ^{\prime }\left( x^{*}\right) \text {.} \end{aligned}$$

(20)

Given also that $\frac{2}{\left( 2+b\right) ^{2}}\left( A-\overline{c}+x^{*}\right) =\frac{2-b}{2}\left[ 1+r\sigma ^{2}\psi ^{\prime \prime }\left( x^{*}\right) \right] \psi ^{\prime }\left( x^{*}\right)$ by Eq. (6) and $\frac{d \psi ^{\prime }\left( x^{*}\right) }{d \left( r\sigma ^{2}\right) }=\psi ^{\prime \prime }\left( x^{*}\right) \frac{d x^{*}}{d \left( r\sigma ^{2}\right) }$, inequality (20) becomes

$$\begin{aligned} -b\left[ 1+r\sigma ^{2}\psi ^{\prime \prime }\left( x^{*}\right) \right] \frac{\mathrm{d} x^{*}}{\mathrm{d} \left( r\sigma ^{2}\right) }>\psi ^{\prime }\left( x^{*}\right) \text {.} \end{aligned}$$

(21)

In (21), we substitute $\frac{\mathrm{d} x^{*}}{\mathrm{d} \left( r\sigma ^{2}\right) }$ with Eq. (19). Using also Eq. (15), we obtain the condition in Proposition 1.

1.4 Proof of Proposition 2

      The interiority of the equilibrium requires, if efforts are strategic substitutes, $\pi _{i}$ to be strictly increasing when $x_{i}=0$; $\frac{\partial \pi _{i}\left( 0,x_{j}\right) }{\partial x_{i}}=\frac{2\left( 2-bh\right) }{\left( 4-b^{2}\right) \left( 2+b\right) } \left[ A-\overline{c}+\frac{\left( 2h-b\right) x_{j}}{2-b}\right] >0$. If this inequality holds for $x_{j}=\overline{c}+\theta$, it will also hold for all $x_{_{j}}$ since $h<\frac{b}{2}$. This is guaranteed by assumption $\left( A.3\right)$. If $h> \frac{b}{2}$, $\frac{\partial \pi _{i}\left( 0,x_{j}\right) }{\partial x_{i}}$ is always positive. The uniqueness of the equilibrium requires $-1<r_{i}^{\prime }\left( x_{j}\right) <1$. Let

$$\begin{aligned} M\equiv \frac{\partial ^{2}\pi _{i}\left( r_{i}\left( x_{j}\right) ,x_{j}\right) }{\partial x_{i}^{2}}=\frac{2\left( 2-bh\right) ^{2}}{\left( 4-b^{2}\right) ^{2}}-\left[ 1+r\sigma ^{2}\psi ^{\prime \prime }\left( x_{i}\right) \right] \psi ^{\prime \prime }\left( x_{i}\right) \text {,} \end{aligned}$$

(22)

which is negative by assumption $\left( A.4\right)$. The slope of firm $i$’s reaction function is $r_{i}^{\prime }\left( x_{j}\right) =\frac{2\left( 2-bh\right) \left( b-2h\right) }{\left( 4-b^{2}\right) ^{2}M}$. If $h<\frac{b}{2}$, given that $r_{i}^{\prime }\left( x_{j}\right) <0$, it suffices to show that $r_{i}^{\prime }\left( x_{j}\right) >-1$. This requires $\frac{2\left( 2-bh\right) \left( 2-bh+b-2h\right) }{\left( 4-b^{2}\right) ^{2}}<\left[ 1+r\sigma ^{2}\psi ^{\prime \prime }\left( x_{i}\right) \right] \psi ^{\prime \prime }\left( x_{i}\right)$ which is implied by assumption $\left( A.4\right)$. If $h>\frac{b}{2}$, given that $r_{i}^{\prime }\left( x_{j}\right) >0$, it suffices to show that $r_{i}^{\prime }\left( x_{j}\right) <1$. This requires $\frac{2\left( 2-bh\right) \left( 2-bh+2h-b \right) }{\left( 4-b^{2}\right) ^{2}}<\left[ 1+r\sigma ^{2}\psi ^{\prime \prime }\left( x_{i}\right) \right] \psi ^{\prime \prime }\left( x_{i}\right)$ which is also implied by assumption $\left( A.4\right)$. Thus, a unique interior equilibrium exists.

Given Eqs. (10) and (11), and that the mean of the random shocks is zero, we find the optimal R&D incentives and effort by considering the Lagrange function of principal $i$’s problem in expected terms:

$$\begin{aligned} L_{i}&= \left( q_{i}^{*}\right) ^{2}-\alpha _{i}-(\beta _{i}+h\gamma _{i})x_{i}-(h\beta _{i}+\gamma _{i})x_{j}+\lambda _{i}\left[ \beta _{i}+h\gamma _{i}-\psi ^{\prime }\left( x_{i}\right) \right] \nonumber \\&+\mu _{i}\left[ \alpha _{i}+(\beta _{i}+h\gamma _{i})x_{i}+(h\beta _{i}+\gamma _{i})x_{j}-\psi (x_{i})\right. \nonumber \\&-\left. \frac{r}{2}\left[ \left( \beta _{i}+h\gamma _{i}\right) ^{2}+\left( h\beta _{i}+\gamma _{i}\right) ^{2}\right] \sigma ^{2}\right] . \end{aligned}$$

(23)

The expected output is given by Eq. (12). Omitting details, the Kuhn–Tucker condition with respect to $\alpha _{i}$ gives $-1+\mu _{i}=0\Leftrightarrow \mu _{i}=1$, implying that the $IR_{i}$ constraint binds at the optimum. Given also that the equilibrium is interior, the profit-maximizing conditions become

$$\begin{aligned} \frac{\partial L_{i}}{\partial \lambda _{i}}&= \beta _{i}+h\gamma _{i}-\psi ^{\prime }\left( x_{i}\right) =0\text {, }\forall i \end{aligned}$$

(24)

$$\begin{aligned} \frac{\partial L_{i}}{\partial \beta _{i}}&= \lambda _{i}-r\left[ \beta _{i}+h\gamma _{i}+h\left( h\beta _{i}+\gamma _{i}\right) \right] \sigma ^{2}=0, \forall i \end{aligned}$$

(25)

$$\begin{aligned} \frac{\partial L_{i}}{\partial \gamma _{i}}&= h\lambda _{i}-r\left[ h\left( \beta _{i}+h\gamma _{i}\right) +h\beta _{i}+\gamma _{i}\right] \sigma ^{2}=0, \forall i \end{aligned}$$

(26)

$$\begin{aligned} \frac{\partial L_{i}}{\partial x_{i}}&= \frac{2\left( 2-bh\right) }{\left( 4-b^{2}\right) \left( 2+b\right) }\left[ A-\overline{c} +\frac{ \left( 2-bh\right) x_{i}-\left( b-2h\right) x_{j}}{2-b}\right] \nonumber \\&-\lambda _{i}\psi ^{\prime \prime }\left( x_{i}\right) -\psi ^{\prime }\left( x_{i}\right) =0\text {, }\forall i,j \end{aligned}$$

(27)

By Eqs. (25) and (26), we have

$$\begin{aligned} \frac{\beta _{i}+h\gamma _{i}+h\left( h\beta _{i}+\gamma _{i}\right) }{ h\left( \beta _{i}+h\gamma _{i}\right) +h\beta _{i}+\gamma _{i}}=\frac{1}{h} . \end{aligned}$$

(28)

Solving with respect to $\gamma _{i}$, we obtain $\gamma _{i}^{*}=-h\beta _{i}^{*}$. By Eq. (24), we also have $\left( 1-h^{2}\right) \beta _{i}^{*}=\psi ^{\prime }\left( x_{i}^{*}\right)$. Since $\lambda _{i}=r\sigma ^{2}\psi ^{\prime }\left( x_{i}\right)$ by Eq. (25), the optimal effort level $x^{*}$ solves Eq. (27) which implies the form in Proposition 2.

1.5 Effect of competition on optimal effort

The effect of $b$ on the marginal profitability of R&D is given by

$$\begin{aligned} \frac{\partial \left( \partial \pi _{i}/\partial x_{i}\right) }{\partial b}&= \frac{2\left[ 4b-h(4+b^{2})\right] }{\left( 4-b^{2}\right) ^2\left( 2+b\right) }[A\!-\!\bar{c}\!+\!(1+h)x^{*}]\\&\quad +\frac{2(2-bh)}{4-b^2}\left[ \frac{2h-b}{4-b^2}\left( \frac{{d}x^*}{{d}b}\right) -\frac{A-\bar{c}}{(2+b)^2}-\frac{1+h}{(2+b)^2}x^*\right] . \end{aligned}$$

Given also Eq. (22), the decomposition $\frac{\partial \left( \partial \pi _{i}/\partial x_{i}\right) }{\partial b}+\frac{\partial ^{2}\pi _{i}}{\partial x_{i}^{2}}\frac{\mathrm{d} x^{*}}{\mathrm{d} b}=0$ implies

$$\begin{aligned} \frac{d x^{*}}{db}\!=\!\left( \frac{4}{4-b^{2}}\right) \frac{ \left[ 3b-2-\left( 2-b+b^{2}\right) h\right] \left[ A-\overline{c}+\left( 1+h\right) x^{*}\right] }{\left( 4-b^{2}\right) \left( 2+b\right) \left[ 1+r\sigma ^{2}\psi ^{\prime \prime }\left( x^{*}\right) \right] \psi ^{\prime \prime }\left( x^{*}\right) -2\left( 2-bh\right) \left( 1+h\right) }\text {.} \end{aligned}$$

The denominator is positive by assumption $\left( A.4\right)$. Thus, the derivative $\frac{d x^{*}}{db}$ is positive, if and only if, $3b-2-\left( 2-b+b^{2}\right) h>0\Leftrightarrow b>\frac{3+h-\left( 9-2h-7h^{2}\right) ^{1/2}}{2h}$. By L’Hôpital’s rule, it is $\lim _{h\rightarrow 0^{+}}\left\{ \frac{3+h-\left( 9-2h-7h^{2}\right) ^{1/2} }{2h}\right\} =\lim _{h\rightarrow 0^{+}}\left\{ \frac{1}{2}\left( 1+\frac{ 1+7h}{\left[ \left( 1-h\right) \left( 9+7h\right) \right] ^{1/2}}\right) \right\} =\frac{2}{3}$.

1.6 Effect of spillovers on optimal effort

The effect of spillovers on the marginal probability of R&D is

$$\begin{aligned} \frac{\partial \left( \partial \pi _{i}/\partial x_{i}\right) }{\partial h}&= - \frac{2b}{\left( 4-b^{2}\right) \left( 2+b\right) }\left[ A-\overline{c} +\left( 1+h\right) x^{*}\right] \\&+\frac{2\left( 2-bh\right) }{4-b^{2}} \left[ \frac{2h-b}{4-b^2} \left( \frac{{d}x^*}{{d}b}\right) +\frac{1}{2+b}x^*\right] \text {.} \end{aligned}$$

Thus, given Eq. (22), the decomposition of the effect of spillovers on effort implies

$$\begin{aligned} \frac{\mathrm{d} x^{*}}{\mathrm{d} h}=2\frac{\left[ 2-b\left( 1+2h\right) \right] x^{*}-b\left( A-\overline{c}\right) }{\left( 4-b^{2}\right) \left( 2+b\right) \left[ 1+r\sigma ^{2}\psi ^{\prime \prime }\left( x^{*}\right) \right] \psi ^{\prime \prime }\left( x^{*}\right) -2\left( 2-bh\right) \left( 1+h\right) } \end{aligned}$$

(29)

where $x^{*}$ satisfies Proposition 2. The denominator in Eq. (29) is positive by assumption $\left( A.4\right)$. Thus, the derivative $\frac{\mathrm{d} x^{*}}{\mathrm{d} h}$ is positive only if $h< \frac{2-b}{2b}$ and $\frac{b\left( A-\overline{c}\right) }{2-b\left( 1+2h\right) }<x^{*}$, which becomes $\frac{2\left( 2-bh\right) ^{2}\left( A-\overline{c}\right) }{\left[ 2-b\left( 1+2h\right) \right] \left( 4-b^{2}\right) \left( 2+b\right) }< \left[ 1+r\sigma ^{2}\psi ^{\prime \prime }\left( x^{*}\right) \right] \psi ^{\prime }\left( x^{*}\right)$ by Proposition 2.