The role of domestic savings in outward-oriented growth strategies

Abstract

This paper presents a theory of technological catching-up in which local savings plays a key complementary role to international finance and foreign technology. Until now, the literature has primarily emphasized “outward orientation” as the key ingredient of catching-up success. It has indeed been argued that countries which have relied intensively on foreign technologies, either through capital goods imports or foreign direct investment inflows, have been successful while countries which have opted for inward-oriented growth strategy relying on domestic investment and import-substitution strategies have been unsuccessful. In this paper, we develop a sequential model of industrialization in which domestic savings is key to the success of outward-oriented growth strategies. Indeed, internal finance helps to overcome time-to-adjustment constraints which occur in the early phases of the catching-up process when both advanced foreign technologies and backward domestic ones co-exist. In this model, external finance, though international borrowing, and domestic savings are complementary, not substitutable, in the course of technological catching-up.

Appendix

In order to compare the number of starts of new processes on the traverse path with those on the reference path, a variable that will represent the resulting gap is needed. Taking Eq. 8 and replacing $C*\left( t \right)$ with its value $\int\limits_{t - n}^t {q*\left( {t - u} \right)x*e^{gu} du}$, we obtain: $\int\limits_{t - \inf \left( {t,n - k_1 } \right)}^t {q*\left( {t - u} \right)x*e^{gu} } du = \int\limits_0^t {\left[ {q\left( {t + k_1 - u} \right) - v} \right]x\left( u \right)du}$ which is equivalent to: $\begin{array}{*{20}l} {\int\limits_{t - \inf \left( {t,n - k_1 } \right)}^t {\left[ {q\left( {t + k_1 - u} \right) - v} \right]x\left( u \right)du - \quad \int\limits_{t - \inf \left( {t,n - k_1 } \right)}^t {\left[ {q\left( {t + k_1 - u} \right) - v} \right]x^* e^{gu} du} } } \hfill \\ { = \int\limits_0^t {q^* \left( {t - u} \right)x^* e^{gu} du - \quad \int\limits_{t - \inf \left( {t,n - k_1 } \right)}^t {\left[ {q\left( {t + k_1 - u} \right) - v} \right]x^* e^{gu} du} \quad } } \hfill \\ \end{array}$

Integrating by parts the left-hand side of the latter equation and transforming the right-hand side by a change in variable (u becomes t − u), we obtain a formulation that is akin to Eq. 9:

For t < n − k ₁

$$Y\left( t \right) + \int\limits_0^t {\frac{{q\prime \left( {t + k_1 - u} \right)}}{{q\left( {k_1 } \right) - v}}Y\left( u \right)du = \frac{1}{{q\left( {k_1 } \right) - v}}\left( {\int\limits_0^t {\left[ {q*\left( u \right) - \left( {q\left( {u + k_1 } \right) - v} \right)} \right]x*e^{g\;\left( {t - u} \right)} du} } \right)} $$

(10)

For $n - k_1 < t < n*$

$$Y\left( t \right) + \int\limits_0^t {\frac{{q\prime \left( {t + k_1 - u} \right)}}{{q\left( {k_1 } \right) - v}}Y\left( u \right)du = } \frac{1}{{q\left( {k_1 } \right) - v}}\left( {\begin{array}{*{20}c} {\int\limits_0^{n - k_1 } {\left( {q*\left( u \right) - \left[ {q\left( {u + k_1 } \right) - v} \right]} \right)x*e^{g\left( {t - u} \right)} du + } } \\ {\int\limits_{n - k_1 }^t {q*\left( u \right)x*e^{g\left( {t - u} \right)} du + q\left( {n - k_1 } \right)Y\left( {t - n + k_1 } \right)} } \\ \end{array} } \right)$$

This Volterra integral equation of the second kind has the same convolution kernel as Eq. 9. The unknown variable is Y(t) and we are, in fact, interested in is studying its sign. As the resolved kernel is positive, this study boils down to finding the sign of the right-hand side of Eq. 8. Indeed, by expressing the latter by h(t), we have:

$$Y\left( t \right) = h\left( t \right) + \int\limits_0^t {S\left( {t + k_1 - u} \right)h\left( u \right)du} $$

The expression of h(t) in Eq. 10 is:

$$h\left( t \right) = \left( {\int\limits_0^t {\frac{{\left[ {q*\left( u \right) - \left( {q\left( {u + k_1 } \right) - v} \right)} \right]}}{{q\left( {k_1 } \right) - v}}x*e^{g\left( {t - u} \right)} du} } \right)$$

It is positive whatever t < n − k ₁ and, thus, Y(t) > 0 whatever t < n − k ₁.

Moreover, we have $h\prime \left( \tau \right) = gh\left( t \right) + \frac{{q*\left( t \right) - q\left( {k_{1 + t} } \right)}}{{q\left( {k_1 } \right) - v}}x* > 0\quad \forall t < n - k_1$ and, then, $Y\prime \left( t \right) = x\left( t \right) - x*e^{gt} > 0\,\forall t < n - k_1$.

For t ∈ [n − k ₁, n*], h(t) takes another form:

$$\begin{array}{*{20}c} {h\left( t \right) = \frac{1}{{q\left( {k_1 } \right) - v}}\int\limits_0^{n - k_1 } {\frac{{\left( {q*\left( u \right) - \left[ {q\left( {u + k_1 } \right) - v} \right]} \right)x*e^{g\left( {t - u} \right)} du}}{{q\left( {k_1 } \right) - v}}} } \\ { + \int\limits_{n - k_1 }^t {\frac{{q*\left( u \right)x*e^{g\left( {t - u} \right)} du}}{{q\left( {k_1 } \right) - v}} + \frac{{q\left( {n - k_1 } \right)}}{{q\left( {k_1 } \right) - v}}Y\left( {t - n + k_0 } \right)} } \\ \end{array} $$

The two last terms of the right-hand side are negative. It is then possible that h(t), and consequently Y(t), becomes negative between t = n − k ₁ and t = n*.