R&D Subsidies and Foreign Direct Investment

Abstract

We devise a model in which domestic firms do applied R&D, which can be subsidized by the government, and foreign firms with superior technology can enter in the domestic market. Foreign Direct Investment can act as a substitute of subsidies to improve domestic R&D, the share of domestic leading firms and consumption. Relatively closed economies may benefit from R&D subsidization while relatively open economies may not. For relatively low growth of the technological frontier, it is optimal to subsidize R&D and close the economy to foreign investment but the opposite happens for relatively high growth. Numerical simulations show the economy dynamics after policy experiments.

Appendix A: Proofs

Proof of Proposition 1

Equation 20 is obtained from Eq. 19, using that $s_{1,t}=s_{1,t-1}=\bar s_{1}$. Equation 21 and (22) result from Eqs. 15 and (16). Substituting (20) into (18), and rearranging terms, we get Eq. 23, which is a generalized polynomial whose coefficients are ordered in decreasing order.¹⁴ If the coefficient of $L_{A_{1}}$ is nonnegative, there is only one sign change in the sequence of coefficients and, therefore, there is exactly one root in (0, ∞ ) (see Jameson2006). If the coefficient of $L_{A_{1}}$ is negative, there are two changes of sign and, therefore, there are zero or two roots. Given that p(0) < 0 and

$$ p((1/\theta)^{1/\lambda})=\eta^{\frac{1}{1-\lambda }} \theta^{\frac{\lambda -1}{\lambda }} \left[(1-\alpha ) \lambda (\gamma-1 +\sigma) (1-\theta^{1/\lambda }\bar L)+\alpha \gamma (1-s_{r})\right]>0, $$

(31)

there is exactly one feasible solution $\bar L_{A_{1}}\in (0,(1/\theta )^{1/\lambda })$ to Eq. 23—and, as $\bar L_{A_{2}}=\eta ^{1/(1-\lambda )} \bar L_{A_{1}}$, we also have that $\bar L_{A_{2}}\in (0,(1/(\eta \theta ))^{1/\lambda })$. Now, Eq. 20 entails that $\bar s_{1}\in (0,1)$, which proves that there is a unique feasible steady state.

To prove stability, let us differentiate s _1,t in Eq. 19 with respect to s _1,t−1 to obtain that

$$\begin{array}{@{}rcl@{}} \frac{d s_{1,t}}{d s_{1,t-1}}=\theta(1-\eta^{\frac{1}{1-\lambda}})L_{A_{1}}(s_{1,t-1})^{\lambda}&+& \lambda\theta\left[\eta^{\frac{1}{1-\lambda}}+(1-\eta^{\frac{1}{1-\lambda}}) s_{1,t-1} \right]\\&\times&{L_{A_{1}}(s_{1,t-1})^{\lambda-1}}\frac{d L_{A_{1}}}{d s_{1,t-1}}. \end{array} $$

Implicit differentiation of Eq. 18 entails that

$$ \frac{d L_{A_{1}}}{d s_{1,t-1}}=-\frac{\frac{\partial h}{\partial s_{1,t-1}}}{\frac{\partial h}{\partial L_{A_{1}}}}<0, $$

where

$$\frac{\partial h}{\partial s_{1,t-1}}= \left( 1-\eta^{\frac{1}{1-\lambda }}\right) \left[1+\frac{ (1-s_{r})\alpha (\gamma -1) (1-\sigma )}{(1-\alpha ) \lambda (\gamma -1+\sigma)}\right]L_{A_{1}}>0, $$

and, using Eq. 18,

$$ \frac{\partial h}{\partial L_{A_{1}}}\,=\, \frac{(1-\lambda )\bar L}{L_{A_{1}}}\,+\,\left[\eta^{\frac{1}{1-\lambda }}\,+\, \left( 1\,-\,\eta^{\frac{1}{1-\lambda }}\right)s_{1,t-1}\right] \left[\lambda \,+\,\frac{ (1-s_{r})\alpha (\gamma -1) (1-\sigma )}{(1-\alpha) (\gamma -1+\sigma)}\right] >0. $$

(32)

Evaluating d s _1,t /d s _1,t−1 at the steady state, after simplification, we get that

$$\begin{array}{@{}rcl@{}} \frac{d s_{1,t}}{d s_{1,t-1}}&=& \left( 1-\eta^{\frac{1}{1-\lambda }}\right) \theta \bar L_{A_{1}}^{\lambda }\\ \\&&\!\times\!\left[ \!1\!-\frac{\eta^{\frac{1}{1-\lambda}}+(1-\eta^{\frac{1}{1-\lambda}}) \bar s_{1}}{\eta^{\frac{1}{1-\lambda}}+(1-\eta^{\frac{1}{1-\lambda}}) \bar s_{1} +\frac{(1-\lambda) (1-\alpha) (\gamma-1 +\sigma) \bar L}{\ \left[\lambda(1-\alpha ) (\gamma-1 +\sigma) +\alpha (\gamma -1) (1-\sigma) (1-s_{r})\right]\bar L_{A_{1}}}}\right]. \end{array} $$

Hence, it must be that d s _1,t /d s _{1,t − 1}∈(0, 1) at the steady state and, therefore, the steady state is stable. □

Proof Proof of Proposition 2

Differentiating Eq. 24 and evaluating at the steady state we have that

$$\begin{array}{@{}rcl@{}} \frac{\partial h}{\partial s_{1}}&=& (1-\eta^{\frac{1}{1-\lambda}}) \left[ 1+\frac{ (1-s_{r})\alpha (\gamma -1) (1-\sigma )}{(1-\alpha )\lambda (\gamma -1+\sigma)} \right] \bar L_{A_{1}}>0, \end{array} $$

(33)

$$\begin{array}{@{}rcl@{}} \frac{\partial h}{\partial s_{r}}&=&- \frac{\bar L-\left[\eta^{\frac{1}{1-\lambda }}+ \left( 1-\eta^{\frac{1}{1-\lambda }}\right)\bar s_{1}\right]L_{A_{1}}}{1-s_{r}}<0, \end{array} $$

(34)

$$\begin{array}{@{}rcl@{}} \frac{\partial h}{\partial\sigma}&=&-\frac{\alpha \gamma (1-s_{r})[1-(\gamma-1) (1-\bar s_{1})] \bar L_{A_{1}}^{1-\lambda }}{(1-\alpha ) \theta \lambda (\gamma+\sigma -1)^{2}}, \end{array} $$

(35)

and $\partial h/\partial L_{A_{1}}>0$ is given by Eq. 32 after replacing s _1,t−1 with $\bar s_{1}$. Differentiating (25) we get

$$\begin{array}{@{}rcl@{}} \frac{\partial f}{\partial L_{A_{1}}}&=& \frac{\lambda \bar s_{1}}{\bar L_{A_{1}}}>0,\\ \frac{\partial f}{\partial s_{1}}&=&-\left[ 1-(1-\eta^{\frac{1}{1-\lambda}})\theta \bar L_{A_{1}}^{\lambda} \right]<0,\\ \frac{\partial f}{\partial s_{r}}&=& \frac{\partial f}{\partial \sigma}=0. \end{array} $$

Let A denote the matrix

$$A= \left( \begin{array}{cc} \partial h/\partial L_{A_{1}} & \partial h/\partial s_{1}\\ \partial f/\partial L_{A_{1}} & \partial f/\partial s_{1} \end{array} \right). $$

Using the former results on the sign of the derivatives, we have that detA < 0.

Let us first study the effect on the skilled labor devoted to research. Using the implicit function theorem, we get that

$$\begin{array}{@{}rcl@{}} \frac{d\bar L_{A_{1}}}{ds_{r}}&=& -\frac{\left| \begin{array}{lllll} \partial h/\partial s_{r} & \partial h/\partial s_{1}\\ 0 & \partial f/\partial s_{1} \end{array} \right|}{\det A}=- \frac{(\partial h/\partial s_{r})(\partial f/\partial s_{1})}{\det A}>0, \end{array} $$

(36)

$$\begin{array}{@{}rcl@{}} \frac{d\bar{L}_{A_{1}}}{d\sigma}&=& -\frac{\left| \begin{array}{lllll} \partial h/\partial \sigma & \partial h/\partial s_{1}\\ 0 & \partial f/\partial s_{1} \end{array} \right|}{\det A}=- \frac{(\partial h/\partial \sigma)(\partial f/\partial s_{1})}{\det A}. \end{array} $$

(37)

Hence, an increase in the subsidy to R&D increases the skilled labor devoted to applied research. From Eq. 37 we have that sign dL̄ _{A 1}/dσ) = −sign(∂h/∂σ) = sign [1 − (γ − 1) (1 − s̄₁)]. If γ ≤ 2, we immediately have that $d\bar L_{A_{1}}/d\sigma >0$. If γ > 2, using Eq. 20, we have that $d\bar L_{A_{1}}/d\sigma =0$ —i.e., $(\gamma -1)(1-\bar s_{1})=1$— if

$$\bar L_{A_{1}}=(\gamma-2)^{\frac{1}{\lambda} } \eta^{-\frac{1}{(1-\lambda) \lambda }} \theta^{-\frac{1}{\lambda} } \left[1+(\gamma-2) \eta^{\frac{1}{\lambda -1}} \right]^{-1/\lambda}=\tilde L_{A_{1}}. $$

Substituting $\tilde L_{A_{1}}$ into Eq. 23, and denoting ${\Theta }=p(\tilde L_{A_{1}})$, we get Eq. 26. Hence, we immediately have that $d\bar L_{A_{1}}/d\sigma >0$ if γ ≤ 2 or if γ > 2 and Θ < 0; $d\bar L_{A_{1}}/d\sigma =0$ if γ > 2 and Θ=0, and $d\bar L_{A_{1}}/d\sigma <0$ if γ > 2 and Θ > 0.

The effects of the subsidy to R&D and openness on the steady-state fraction of state 1 can be easily obtained by differentiation of Eq. 20:

$$\begin{array}{@{}rcl@{}} \frac{d\bar s_{1}}{ds_{r}}&=&\frac{d\bar s_{1}}{dL_{A_{1}}}\frac{d\bar L_{A_{1}}}{ds_{r}} =\frac{\lambda \eta^{\frac{1}{\lambda -1}} \bar L_{A_{1}}^{-\lambda -1}\bar {s_{1}^{2}}}{\theta} \frac{d\bar L_{A_{1}}}{ds_{r}}>0, \end{array} $$

(38)

$$\begin{array}{@{}rcl@{}} \frac{d\bar s_{1}}{d\sigma}&=&\frac{d\bar s_{1}}{dL_{A_{1}}}\frac{d\bar L_{A_{1}}}{d\sigma} =\frac{\lambda \eta^{\frac{1}{\lambda -1}} \bar L_{A_{1}}^{-\lambda -1}\bar {s_{1}^{2}}}{\theta} \frac{d\bar L_{A_{1}}}{d\sigma}, \end{array} $$

(39)

which entails that sign(ds̄ ₁/d) = sign (dL̄ _A1 dσ). In any case, we have that $\bar s_{1}$ and $\bar L_{A_{1}}$ evolve in the same direction. Skilled labor devoted to R&D is given by

$$ \bar L_{R}=\bar s_{1} \bar L_{A_{1}}+ (1-\bar s_{1}) \bar L_{A_{2}}= \left[ \bar s_{1}+ \eta^{1/(1-\lambda)}(1-\bar s_{1})\right]\bar L_{A_{1}}, $$

(40)

where we have used Eq. 21 and that $L_{A_{2}}=\eta ^{1/(1-\lambda )}L_{A_{1}}$. Differentiating the former expression with respect to x, where x can stand for s _r or σ, we get that

$$ \frac{d\bar L_{R}}{dx}= \left[ \bar s_{1}+ \eta^{1/(1-\lambda)}(1-\bar s_{1})\right] \frac{d\bar L_{A_{1}}}{dx}+ \left( 1- \eta^{1/(1-\lambda)}\right)\bar L_{A_{1}}\frac{d \bar s_{1}}{dx}. $$

(41)

Given that $\bar s_{1}$ and $\bar L_{A_{1}}$ evolve in the same direction, skilled labor devoted to research $\bar L_{R}$ does so. This completes the proof. □

Proof Proof of Lemma 1

Noting that Eq. 36 entails that

$$\frac{d\bar L_{A_{1}}}{ds_{r}}= \frac{(\partial h/\partial s_{r})(\partial f/\partial s_{1})}{-(\partial h/\partial L_{A_{1}})(\partial f/\partial s_{1})+(\partial h/\partial s_{1}) (\partial f/\partial L_{A_{1}})}\\<- \frac{\partial h/\partial s_{r}}{\partial h/\partial L_{A_{1}}}, $$

after simplification, we have that

$$\begin{array}{@{}rcl@{}} \frac{\partial w}{\partial L_{A_{1}}} \frac{d\bar L_{A_{1}}}{d s_{r}} +\frac{\partial w}{\partial s_{r}}=\frac{(1-\alpha)w}{(1-s_{r})} \left[1-\frac{(1-s_{r})(1-\lambda)}{\bar L_{A_{1}}}\frac{d\bar L_{A_{1}}}{ds_{r}} \right]\\ >\frac{(1-\alpha)w}{(1-s_{r})} \left[1+\frac{(1-s_{r})(1-\lambda)}{\bar L_{A_{1}}} \frac{\partial h/\partial s_{r}}{\partial h/\partial L_{A_{1}}} \right]=\frac{(1-\alpha)w}{(1-s_{r})}\\ \times \left\{1-\frac{-(1-\lambda)\frac{\partial h}{\partial s_{r}}} {-(1-\lambda)\frac{\partial h}{\partial s_{r}}+\left[\eta^{\frac{1}{1-\lambda}}+(1-\eta^{\frac{1}{1-\lambda}}) \bar s_{1}\right]\bar L_{A_{1}}\left[ \frac{1}{1-s_{r}}+\frac{\alpha(\gamma-1)(1-\sigma)}{(1-\alpha)(\gamma-1+\sigma)} \right]} \right\}>0. \end{array} $$

Furthermore, we have that ∂Υ/∂ s ₁ = (1 − τ _f )(1 − α)α σ+(γ−1)(1 − σ)/γ > 0, and thus, using Eq. 38, we get that

$$\frac{\bar c}{\Upsilon}\frac{\partial {\Upsilon}}{\partial s_{1}} \frac{d\bar s_{1}}{d s_{r}}>0. $$

□