Leapfrogging cycles in international competition

Abstract

Technological leadership has shifted at various times from one country to another. We propose a mechanism that explains this perpetual cycle of technological leapfrogging in a two-country model including the dynamic optimization of an infinitely lived consumer. In the model, each country accumulates knowledge stock over time because of domestic innovation and spillovers from foreign innovation. We show that if the international knowledge spillovers are reasonably efficient, technological leadership may shift first from one country to another, and then alternate between countries along an equilibrium path.

Appendix

Proof of Lemma 1 The necessity has been shown in the text. Namely, $\varTheta <1$ must hold if both countries produce, recalling that only the leading country innovates in equilibrium if both countries manufacture. Here, we prove sufficiency, i.e., both countries produce in equilibrium if $\varTheta <1$. For this purpose, let us prove the contrapositive: we show $\varTheta >1$ must hold if only one country manufactures. There are two cases. (a) If only leading country $A$ manufactures in equilibrium, $w^{A} (t)<w^{B}(t)$ must hold. With $K^{A}(t)>K^{B}(t)$, this implies $w^{A} (t)/K^{A}(t)<w^{B}(t)/K^{B}(t)$, under which only leading country $A$ innovates owing to its lower R&D cost. There is no labor demand in country $B$, so (a) cannot be an equilibrium. (b) If only lagging country $B$ manufactures in equilibrium, $w^{A}(t)>w^{B}(t)$ must hold. Case (b) has three sub-cases. (b-i) Only lagging country $B$ innovates, which is inconsistent for the same reason as in (a). (b-ii) If only leading country $A$ innovates in equilibrium, $w^{A}(t)/K^{A}(t)<w^{B}(t)/K^{B}(t)$ must hold. The labor market condition (12) implies $N^{A}(t)=LK^{A}(t)$ and $w^{B}(t)=\left( 1-\theta \right) E(t)/L$.⁴⁰ The free-entry condition requires $V^{A}(t)=0$ because innovation takes place in equilibrium for country $A$, which implies $w^{A}(t)=\beta \theta K^{A}(t)E(t)/N(t+1)$.⁴¹ Noting $N(t+1)=N^{A}(t)$ in this case, $w^{A}(t)>w^{B} (t)\$implies $\varTheta >1$, not $\varTheta <1$. (b-iii) If both countries innovate in equilibrium, it must be by the free-entry condition that $V^{A} (t)=V^{B}(t)=0$ in equilibrium, which implies $w^{A}(t)/K^{A}(t)=w^{B} (t)/K^{B}(t)=\frac{\beta \theta E(t)}{N(t+1)}$.⁴² Using this and $x^{B} (t)=\frac{(1-\theta )E(t)}{N(t)w^{B}(t)}$ from (4), $N^{B}(t)=( \varTheta K^{B}(t)-K^{A}(t) ) L/(1+\varTheta )$ is derived from (12). Since $N^{B}(t)>0$, then $\varTheta >1$ must hold, noting $K^{A}(t)>K^{B}(t)$. We have proven that if $\varTheta <1$, both countries must manufacture in equilibrium; thus, only the leader innovates.

We can easily prove Lemmata 2 and 3 in an analogous way (see Furukawa (2014) for a formal proof).$\square$

Derivations for (37)

In the North–North regime: there are two cases (A) and (B). (A) Assume $\frac{\theta }{1-\theta }>\frac{K^{A}(t)}{K^{B}(t)}>1$ (i.e., $\psi (t)\in \left( 0.5,\theta \right)$), where country $A$ is a leading country. The free-entry condition (27) becomes

$$\begin{aligned} \frac{\theta E(t)}{N^{A}(t)+N^{B}(t)}=\frac{w^{i}(t)}{K^{i}(t)}. \end{aligned}$$

(38)

Given the labor condition for leading country $A$, we have

$$\begin{aligned} N^{A}(t)=LK^{A}(t). \end{aligned}$$

(39)

By the labor condition for the lagging country, $L=N^{B}(t)/K^{B} (t)+(N^{A}(t)+N^{B}(t))x^{B}(t)$, with $x^{B}(t)=\frac{(1-\theta )E(t)}{\left( N^{A}(t)+N^{B}(t)\right) w^{B}(t)}$, we thus have

$$\begin{aligned} N^{B}(t)=\theta LK^{B}(t)-\left( 1-\theta \right) LK^{A}(t), \end{aligned}$$

(40)

in which $w^{B}(t)$ is eliminated using (38). Noting $M^{B}(t)=N^{A}(t)$ here, by (9), we have the dynamic system as follows:

$$\begin{aligned} \psi (t+1)=\frac{\left( L+1\right) \psi (t)}{\mu L\psi (t)+\theta L+1}\quad \text {for }\,\psi (t)\in \left( 0.5,\theta \right) . \end{aligned}$$

(41)

Note that $\theta >0.5$ holds in the North–North regime. (B) Assume $\frac{1-\theta }{\theta }<\frac{K^{A}(t)}{K^{B}(t)}<1$ (i.e., $\psi (t)\in \left( 1-\theta ,0.5\right)$), where country $B$ is a leading country. Due to the symmetry,

$$\begin{aligned} N^{A}(t)=\theta LK^{A}(t)-\left( 1-\theta \right) LK^{B}(t)\ \text {and } N^{B}(t)=LK^{B}(t) \end{aligned}$$

(42)

hold. We can also derive

$$\begin{aligned} \psi (t+1)=\frac{(\theta +\left( \mu -1\right) )L+\left( 1+\left( 1-\mu \right) L\right) \psi (t)}{\theta L+1+\mu L\left( 1-\psi (t)\right) }\quad \text {for }\,\psi (t)\in \left( 1-\theta ,0.5\right) . \end{aligned}$$

(43)

In the full North–South regime, there are two cases (C) and (D). (C) Assume $\frac{K^{A}(t)}{K^{B}(t)}>\frac{\theta }{1-\theta }>1$ (i.e., $\psi (t)\in \left( \theta ,1\right)$). The leading country innovates following

$$\begin{aligned} N^{A}(t)=LK^{A}(t)\ \text {and }N^{B}(t)=0. \end{aligned}$$

(44)

The lagging country receives spillovers $\mu M^{B}(t)=\mu N^{A}(t),\$with $\mu \le 1$.⁴³ Then, the knowledge dynamics are as follows:

$$\begin{aligned} \psi (t+1)=\frac{\left( L+1\right) \psi (t)}{1+\left( 1+\mu \right) L\psi (t)}\quad \text {for }\,\psi (t)\in \left( \theta ,1\right) . \end{aligned}$$

(45)

(D) Assume $\frac{K^{A}(t)}{K^{B}(t)}<\frac{1-\theta }{\theta }<1$ (i.e., $\psi (t)\in \left( 0,1-\theta \right)$). Due to the symmetry, $N^{A}(t)=0$ and $N^{B}(t)=LK^{B}(t)$, with $\mu M^{A}(t)=\mu N^{B}(t)$. We can easily have

$$\begin{aligned} \psi (t+1)=\frac{\mu L+\left( 1-\mu L\right) \psi (t)}{1+\left( 1+\mu \right) L\left( 1-\psi (t)\right) }\quad \text {for }\psi (t)\in \left( 0,1-\theta \right) . \end{aligned}$$

(46)