Separated TZ Induced Optimal Growth Through Virtual Trade

Abstract

The purpose of this chapter is to propose a model where trade has a direct and positive impact on growth rate of two trading nations beyond the level effect. We use the idea of virtual trade in intermediates induced by non-overlapping time zones and show how trade can increase the equilibrium optimal rate of growth. In this structure the trade impact goes beyond the level effect and directly causes growth. Typically standard models of trade cannot generate an automatic growth impact. Virtual trade may allow production to continue for 24 × 7 in separated time zones such as between US and India and that can lead to higher growth for both countries. Later we extend the model to incorporate accumulation of skill which becomes necessary for sustaining steady state growth.

This chapter draws on Marjit and Mandal [23].

Appendices

Appendix 1

1.1 Optimum Demand for $\varvec{m}_{1}$ and $\varvec{m}_{2}$

Total input cost to produce one unit of Y is $\left\{ {m_{1} \left( {1 + \mu } \right) + m_{2} } \right\}$ as $m_{1}$ needs to be carried over for final use which has cost of $\left( {1 + \mu } \right)$. Note that $P_{y}$ is normalized to unity. Therefore, the profit equation becomes:

$$ \pi = P_{y} Y - \left\{ {m_{1} \left( {1 + \mu } \right) + m_{2} } \right\} = A m_{1}^{{\frac{1 - \alpha }{2}}} m_{2}^{{\frac{1 - \alpha }{2}}} K^{\alpha } - \left\{ {m_{1} \left( {1 + \mu } \right) + m_{2} } \right\} $$

Setting $\pi_{{m_{1} }}$ and $\pi_{{m_{2} }} = 0$ (First Order Condition for profit maximization):

$\frac{1 - \alpha }{2}Ym_{1}^{ - 1} = \left( {1 + \mu } \right)$ or $m_{1} = \frac{1 - \alpha }{{2\left( {1 + \mu } \right)}}Y$ (we have denoted it by $m_{10}$ in the main text). Similarly, $\frac{1 - \alpha }{2}Ym_{2}^{ - 1} = 1$ or $m_{2} = \frac{1 - \alpha }{2}Y$ (we have denoted it by $m_{20}$ in the main text).

Appendix 2

The Lagrangian for Utility maximization:

$$L\left( {c_{t} , k_{t + 1}, \lambda_t } \right) = u\left( {c_{t} } \right) + \beta V \left( {k_{t + 1} } \right) + \lambda_{t} \left[ {y_{t} - c_{t} - \left( {k_{t + 1} - k_{t} } \right)} \right]$$

First order conditions for optimization, i.e. $\frac{\delta L}{{\delta c_{t} }} = \frac{\delta L}{{\delta k_{t + 1} }} = \frac{\delta L}{{\delta \lambda_{t} }} = 0\,{\text{yield}}$

$$ u^{{\prime }} \left( {c_{t} } \right) = \lambda_{t} $$

(8.21)

$$ \beta V^{{\prime }} \left( {k_{t + 1} } \right) = \lambda_{t} $$

(8.22)

$$ y_{t} = c_{t} + \left( {k_{t + 1} - k_{t} } \right) $$

(8.23)

Also by definition (from the budget constraint)

$$ V^{{\prime }} \left( {k_{t} } \right) = \lambda_{t} \left( {\tilde{A} + 1} \right) $$

(8.24)

Updating Eq. (8.24), Eq. (8.22) is modified as

$$ \beta \lambda_{t + 1} \left( {\tilde{A} + 1} \right) = \lambda_{t} $$

(8.25)

Substituting the value of $\beta = \frac{1}{1 + \rho }$, and comparing Eqs. (8.21) and (8.25)

$$ \frac{{\left( {\tilde{A} + 1} \right)}}{1 + \rho } = \frac{{u^{{\prime }} \left( {c_{t} } \right)}}{{u^{{\prime }} \left( {c_{t + 1} } \right)}} $$

(8.26)

Given $u^{\prime \prime } < 0,$ it is obvious that $c_{t + 1} > c_{t}$ if $\tilde{A} > \rho$.

We shall work with a log-linear utility function. Therefore, $\frac{{u^{{\prime }} \left( {c_{t} } \right)}}{{u^{{\prime }} \left( {c_{t + 1} } \right)}} = \frac{{c_{t + 1} }}{{c_{t} }}$.

From Eq. (8.26)

$$ \frac{{c_{t + 1} - c_{t} }}{{c_{t} }} = g = \frac{{\left( {\tilde{A} - \rho } \right)}}{1 + \rho } \simeq \left( {\tilde{A} - \rho } \right) $$

(8.27)

Therefore, $\left( {\tilde{A} - \rho } \right)$ means the growth rate, g. We worked with this formulation for the rest of the chapter. It is the distance between $\tilde{A}$ and $\rho$ that determines the magnitude of the growth rate.

Appendix 3

From Eq. (8.12), we have $B = A^{1/\alpha } \left( {\frac{1 - \alpha }{2}} \right)^{{\frac{1}{\alpha }}} \cdot \left( {\frac{1}{1 + \mu }} \right)^{{\frac{1 - \alpha }{2\alpha }}} \cdot \left( {1 + \frac{1}{1 + \mu }} \right).$ Simple algebraic manipulation provides

$$ B^{2\alpha } = A^{2} \left( {\frac{1 - \alpha }{2}} \right)^{2} \cdot \left( {\frac{1}{1 + \mu }} \right)^{1 - \alpha } \cdot \left( {1 + \frac{1}{1 + \mu }} \right)^{2\alpha } $$

Assuming $A^{2} \left( {\frac{1 - \alpha }{2}} \right)^{2} = X$ and $\left( {1 + \mu } \right) = \varphi,$

$$ \begin{aligned} B^{2\alpha } \left( {1 + \mu } \right) & = X\left( {\frac{1}{\varphi }} \right)^{1 - \alpha } \cdot \left( {1 + \frac{1}{\varphi }} \right)^{2\alpha } \cdot \varphi \\ & = X \cdot \varphi^{ - 1 + \alpha } \cdot \left( {1 + 1/\varphi } \right)^{2\alpha } \cdot \varphi = X \cdot \varphi^{\alpha - 1} \left( {1 + 1/\varphi } \right)^{2\alpha } \cdot \varphi = X\varphi^{\alpha } \cdot \left( {\frac{\varphi + 1}{\varphi }} \right)^{2\alpha } \\ & = X\varphi^{\alpha } \cdot \left( {\varphi + 1} \right)^{2\alpha } \cdot \varphi^{ - 2\alpha } = X\varphi^{ - \alpha } \left( {\varphi + 1} \right)^{2\alpha } \\ \end{aligned} $$

Differentiating the above result, we get

$$ \frac{{d\left[ {B^{2\alpha } \left( {1 + \mu } \right)} \right]}}{{d\left( {1 + \mu } \right)}} = \frac{{d\left( {B^{2\alpha } \cdot \varphi } \right)}}{d\varphi } > 0 $$

• iff $\left[ {2\alpha \left( {\varphi + 1} \right)^{2\alpha - 1} \cdot X\varphi^{ - \alpha } - \alpha X\varphi^{ - \alpha - 1} \cdot \left( {\varphi + 1} \right)^{2\alpha } } \right] > 0$

• or $X\alpha \left( {\varphi + 1} \right)^{2\alpha - 1} \cdot \varphi^{ - \alpha } \left[ {2 - \varphi^{ - 1} \cdot \left( {\varphi + 1} \right)} \right] > 0$

• or $2 - \left( {1 + \frac{1}{\varphi }} \right) > 0$

Substituting $\left( {1 + \mu } \right) = \varphi,$

$$2 - \left( {1 + \frac{1}{\varphi }} \right) = 2 - \left( {\frac{1 + \varphi + 1}{1 + \varphi }} \right) = \frac{\varphi }{1 + \varphi } = \frac{1}{1 + 1/\varphi }.$$

Since $0 < \varphi < 1$, we obtain $\frac{1}{1 + 1/\varphi } > 0$. This implies $\frac{{d\left[ {B^{2\alpha } \left( {1 + \mu } \right)} \right]}}{{d\left( {1 + \mu } \right)}} > 0$.