Technology choice, externalities in production, and a chaotic middle-income trap

Abstract

We incorporate the external effects of capital in production and endogenous technology choice into the standard overlapping generations model. We demonstrate that our model can exhibit a poverty trap, a middle-income trap, and perpetual growth paths. We also show that, under some economic conditions, an economy exhibits all three of these phenomena, depending on its initial capital level, and that the economy caught in the middle-income trap can exhibit chaotic fluctuations in the long run. This is a stark contrast to the result obtained by Umezuki and Yokoo (J Econ Dyn Control 100:164-175, 2019a) that we cannot observe any chaotic fluctuations. Because the model of Umezuki and Yokoo (2019) is a special case of our model in the sense that there is no externality in production, our result means that we need the combination of technology choice and externalities in production to obtain chaotic fluctuations in the standard overlapping generations model with Cobb-Douglas technologies.

Appendix

Proof of Claim 2

Let $\alpha _1A_1/\alpha _2A_2=1$ and $1>\alpha _2>\alpha _1$. Then, all we need to show is that the following inequalities are possible:

$$\begin{aligned} s(1-\alpha _2)A_2<1<s(1-\alpha _1)\alpha _2 A_2/\alpha _1. \end{aligned}$$

Rewriting the above expression as:

$$\begin{aligned} \frac{\alpha _1}{\alpha _2(1-\alpha _1)}<sA_2<\frac{1}{1-\alpha _2}, \end{aligned}$$

we notice that $\alpha _2/\alpha _1(1-\alpha _1)<1/(1-\alpha _2)$ always holds because $\alpha _2>\alpha _1$. As $sA_2$ can take any positive value, the claim is proven. $\square$

Proof of Lemma 2

Let $\alpha _1A_1/\alpha _2A_2=1$ and $1>\alpha _2>\alpha _1$. Then, the first inequality in condition (9) can be rewritten as:

$$\begin{aligned} \left( \frac{1}{1-\alpha _2}\right) ^{\frac{\beta _1-1}{\beta _1}}\left( \frac{\alpha _1}{\alpha _2(1-\alpha _1)}\right) ^{\frac{1}{\beta _1}}<sA_2. \end{aligned}$$

Similarly, the second inequality in condition (9) is expressed as

$$\begin{aligned} sA_2<\left( \frac{1}{1-\alpha _2}\right) ^{\frac{1}{\beta _2}}\left( \frac{\alpha _1}{\alpha _2(1-\alpha _1)}\right) ^\frac{\beta _2-1}{\beta _2}. \end{aligned}$$

Letting $V=1/(1-\alpha _2)$ and $W=\alpha _1/\alpha _2(1-\alpha _1)$, we observe that $V>W$ because $\alpha _2>\alpha _1$. Because $sA_2$ can take any positive value, it suffices to show that $V^{1-1/\beta _1}W^{1/\beta _1}<V^{\beta _2}W^{1-1/\beta _2}$ or $V^\gamma <W^\gamma$, where $\gamma =1-1/\beta _1-1/\beta _2$. Thus, the last inequality holds if $\gamma <0$ or $1/\beta _1+1/\beta _2>1$. $\square$

Proof of Lemma 3

Using the same notations as in the proof of Claim 2, it suffices to show that $V^{1-1/\beta _1}W^(1/\beta _1)>V^{\beta _2}W^{1-1/\beta _2}$ or $(V/W)^\gamma >1$, where $\gamma =1-1/\beta _1-1/\beta _2$. Because $V/W>1$, the last inequality holds if we take $\beta _1$ and $\beta _2$ ($\beta _2>\beta _1>1$) such that $\gamma >0$ or $1/\beta _1+1/\beta _2<1$. $\square$

Proof of Lemma 4

Let $s\in (0,1)$ and $\alpha _2\in (0,1)$ (hence, $\eta _2=\beta _2-\alpha _2$) be fixed. Let $a_i$ ($i=1, 2$) be any numbers such that $1<a_1<a_2$. Let $\alpha _1A_1/\alpha _2A_2=1$. Then, inequalities (15) and (16) can be reduced to:

$$\begin{aligned} 1<s(1-\alpha _1)\alpha _2A_2/\alpha _1<\left( s(1-\alpha _2)A_2\right) ^{\frac{1}{1-\beta _2}}. \end{aligned}$$

Solving the following simultaneous equations for $A_2$ and $\alpha _1$,

$$\begin{aligned} a_1= \,& {} s(1-\alpha _1)\alpha _2A_2/\alpha _1,\\ a_2= \,& {} \left( s(1-\alpha _2)A_2\right) ^{\frac{1}{1-\beta _2}}, \end{aligned}$$

we obtain:

$$\begin{aligned} A_2=\frac{1}{\left( s(1-\alpha _2)a_2^{\beta _2-1}\right) }>0 \quad \text {and} \quad \alpha _1=\frac{1}{1+\left( \frac{1-\alpha _2}{\alpha _2}\right) a_1a_2^{\beta _2-1}}\in (0,1), \end{aligned}$$

which verifies the assertion. $\square$

Proof of Lemma 8

Let us begin with $\tau _1(0)=1-c\beta _1>c$, which implies from Lemma 7 that any trajectory visits $I_R$ at least once immediately after visiting $I_L$. To ensure that the trajectory stays successively twice in $I_R$, we require that:

$$\begin{aligned} \tau _2(\tau _1(0))=\beta _2(1-c\beta _1-c)=\beta _2-c\beta _1\beta _2-c\beta _2>c. \end{aligned}$$

To ensure that the trajectory stays in $I_R$ successively at least three times, we have:

$$\begin{aligned} \tau ^2_2(\tau _1(0))=\beta _2(\beta _2-c\beta _1\beta _2-c\beta _2-c) =\beta ^2_2-c\beta _1\beta ^2_2-c\beta ^2_2-c\beta _2>c. \end{aligned}$$

Repeating this up to n times, we obtain:

$$\begin{aligned} \tau ^{n-1}_2(\tau _1(0))= \,& {} \beta ^{n-1}_2-c\beta _1\beta ^{n-1}_2-c\beta ^{n-1}_2-c\beta ^{n-2}_2 -\cdots -c\beta ^2_2-c\beta _2\\=\, & {} \beta ^{n-1}_2-c\beta _1\beta ^{n-1}_2-c\beta _2 \left( \sum _{j=0}^{n-2}\beta _2^{j}\right) >c. \end{aligned}$$

Solving the last inequality for c yields the result. $\square$

Proof of Proposition 8

Note that for the value of the threshold c in the assumption of the proposition to be taken, it must hold that:

$$\begin{aligned} \frac{\beta _2-1}{\beta _2}<\frac{1}{1+\beta _1}. \end{aligned}$$

This inequality is equivalent to $\beta _1\beta _2<1+\beta _1$, which is assured by assumption. Because Lemma 7 indicates that any trajectory (i.e., irrelevant to the initial conditions) of $\tau$ visits $I_L$ successively at most once, it follows for any initial condition $x_0\in (0,1)$ that:

$$\begin{aligned} (\tau ^2)'(x_0)\ge \beta _1 \beta _2>1, \end{aligned}$$

where the last inequality follows by assumption. Thus, $\tau$ is eventually expanding and hence chaotic in the sense of Lasota and Yorke (1973). By conjugacy h, T is also chaotic. $\square$