Endogenous structural change, aggregate balanced growth, and optimality

Abstract

Developed countries have experienced labor employment shifts from the manufacturing sector to the services sector, which is a process termed structural change. We construct a two-sector endogenous growth model of structural change in which knowledge spillover sustains growth. Our model is consistent with the empirically observed pattern of structural change as well as aggregate balanced growth. In both centralized and decentralized economies, labor employment shifts from a high productivity growth sector to a low productivity growth sector. Because of knowledge spillover, excessive labor is allocated to productivity improvements in the high productivity growth sector. Hence, the decentralized economy is characterized by rapid structural change compared to the optimal allocation.

Appendices

Appendices

1.1 Appendix 1: Optimization problem of the representative household

We solve the optimization problem of the representative household in three steps. We first maximize (2) subject to (3), which yields the following demand function for brand j of good i:

$$\begin{aligned} x_{i,j}(t) = \frac{E_i(t) {p_{i,j}(t)}^{-\frac{1}{1-\alpha }} }{ \int ^{n_i(t)}_0 { p_{i,j^\prime }(t) }^{-\frac{\alpha }{1-\alpha }} \mathrm{d}j^\prime }. \end{aligned}$$

(24)

Substituting (24) into (2) yields $D_i(t) = E_i(t)/P_i(t)$, where $P_i(t) \equiv {n_i(t)}^{\frac{1-\alpha }{\alpha }-\sigma } \left[ \int ^{n_i(t)}_0 { p_{i,j}(t) }^{-\frac{\alpha }{1-\alpha } } \mathrm{d}j \right] ^{-\frac{1-\alpha }{\alpha } }$ is the price index of the composite good i. The next step maximizes (1) subject to $E(t) = E_1(t) + E_2(t)$ and $D_i(t) = E_i(t)/P_i(t)$ to obtain

$$\begin{aligned} \frac{D_2(t)}{D_1(t)} = \left( \frac{1-\gamma }{\gamma } \right) ^\varepsilon \left( \frac{P_1(t)}{P_2(t)} \right) ^\varepsilon , \quad \text{ or } \quad \frac{E_2(t)}{E_1(t)} = \left( \frac{1-\gamma }{\gamma } \right) ^\varepsilon \left( \frac{P_2(t)}{P_1(t)} \right) ^{1-\varepsilon }. \end{aligned}$$

(25)

Using $E(t)=P_1(t)D_1(t)+P_2(t)D_2(t)$ and the first equation of (25), we can rewrite (1) as $D(t) = E(t)/P(t)$, where $P(t) \equiv [ \gamma ^\varepsilon {P_1(t)}^{1-\varepsilon } + (1-\gamma )^\varepsilon {P_2(t)}^{1-\varepsilon } ]^\frac{1}{1-\varepsilon }$ represents the price index for all consumption goods. We normalize $P_t=1$. We finally maximize U(t) subject to (4) and $D(t) = E(t)/P(t)$, which yields

$$\begin{aligned} \frac{D(t+1)}{D(t)}=\frac{E(t+1)}{E(t)} = \beta ( 1+ r(t) ). \end{aligned}$$

(26)

The transversality condition is $\lim _{t \rightarrow +\infty } \beta ^t W(t)/E(t)=0$.

1.2 Appendix 2: Solution for decentralized economy

We begin with the derivation of (12). Using (7a) and the definition of $P_i(t)$ [see the sentence just below (24)], we obtain

$$\begin{aligned} \frac{P_2(t)}{P_1(t)} = \frac{ b_1(t) }{ b_2(t) } N(t)^{-\sigma }. \end{aligned}$$

(27)

We substitute the first equation of (10) and (27) into the second equation of (25) to obtain

$$\begin{aligned} \frac{a_2(g_2)}{a_1(g_1) }N(t) = \left( \frac{1-\gamma }{\gamma } \right) \left[ \frac{b_1(t)}{b_2(t)}N(t)^{-\sigma } \right] ^{1-\varepsilon } = \left( \frac{1-\gamma }{\gamma } \right) \left[ \frac{b_{1,0}}{b_{2,0}} \left( \frac{g_{1}}{g_{2}} \right) ^t N(t)^{-\sigma } \right] ^{1-\varepsilon }. \end{aligned}$$

In the second equality, we use $b_i(t)=b_{i,0}{g_i}^t$. Solving this equation for N(t) yields (12).

The remainder of this appendix provides the solutions of variables other than N(t) and checks the transversality condition. We first derive $L^I(t)$ and $L^P(t)$. We use (26) and (8) to obtain

$$\begin{aligned} L^I(t)=a_1(g_1)n_1(t+1) + a_2(g_2)n_2(t+1) = \frac{(1-\alpha ) E(t+1)}{ (1+r(t))w(t) }=\frac{(1-\alpha ) \beta E(t)}{ w(t) }. \end{aligned}$$

(28)

Using (9) and $E(t) = E_1(t) + E_2(t)$, we obtain

$$\begin{aligned} L^P(t) = \alpha \frac{E_1(t)+ E_2(t)}{w(t)} =\alpha \frac{E(t)}{w(t)}. \end{aligned}$$

(29)

From (28) and (29), we have $L^I(t)/L^P(t)= (1-\alpha )\beta /\alpha$. Using this equation and $L=L^I(t)+L^P(t)$, we can derive

$$\begin{aligned} L^I(t)=\frac{(1-\alpha )\beta }{\alpha + (1-\alpha )\beta }L \equiv L^I \quad \text{ and } \quad L^P(t)=\frac{\alpha }{\alpha + (1-\alpha )\beta }L \equiv L^P. \end{aligned}$$

(30)

We next derive E(t) , $E_1(t)$, and $E_2(t)$. (29) and (30) imply

$$\begin{aligned} E(t)=\frac{Lw(t)}{\alpha + (1-\alpha )\beta }. \end{aligned}$$

(31)

From the first equation of (10), (12), and $E(t) = E_1(t) + E_2(t)$, we have

$$\begin{aligned} E_1(t) = \frac{ 1 }{ 1+ \frac{a_2(g_2) }{a_1(g_1) } \Phi G^{t} } E(t) \quad \hbox { and }\quad E_2(t) = \frac{ \frac{a_2(g_2) }{ a_1(g_1) } \Phi {G}^{t} }{1+ \frac{a_2(g_2) }{a_1(g_1) } \Phi {G}^t } E(t). \end{aligned}$$

(32)

Because $L^I(t)$ is constant, W(t) is also constant.

$$\begin{aligned} W(t) = a_(g_1) n_1(t+1) w(t) + a(g_2) n_2(t+1) w(t) = L^I w(t). \end{aligned}$$

(33)

Here, we use the first equality of (28). From (31) and (33), W(t) / E(t) becomes constant. Then, the transversality condition, $\lim _{t \rightarrow +\infty } \beta ^t W(t)/E(t)=0$, is satisfied.

From (12) and the first equality of (28), we derive

$$\begin{aligned} n_1(t) = \frac{L^I}{a_1(g_1)+a_2(g_2) \Phi {G}^t} \quad \text{ and } \quad n_2(t) = \frac{\Phi {G}^tL^I}{a_1(g_1)+a_2(g_2) \Phi {G}^t}. \end{aligned}$$

(34)

These two equations give the sequences of $n_1(t)$ and $n_2(t)$. If $G=1$, $n_1(t)$ and $n_2(t)$ are constant over time. If $G>1$, $n_1(t)$ decreases over time and satisfies $\lim _{t \rightarrow \infty } n_1(t)=0$. In contrast, $n_2(t)$ increases over time and satisfies $\lim _{t \rightarrow \infty } n_2(t)=L^I/a_2(g_2)$. Because of $L^I_i(t)=a_i(g_i)n_i(t+1)$ [see (8)], (34) drives the dynamics of $L^I_i(t)$. Substituting (34) and $b_{i,t}= b_{i,0} {g_i}^t$ into $A_i(t) = b_i(t) n_i(t)^\sigma$, we obtain the dynamics of $A_i(t)$.

The sequence of $p_i(t)$ is given by $p_i(t) =w(t) [\alpha b_i(t)]^{-1} = w(t) [\alpha b_{i,0} {g_i}^t]^{-1}$ [see (7a)]. The sequence of $P_i(t)$ is given by $P_i(t) =w(t) [\alpha b_i(t) {n_i(t)}^\sigma ]^{-1}= w(t)[\alpha b_{i,0} {g_i}^t {n_i(t)}^\sigma ]^{-1}$. Substituting $P_i(t)$ into $P(t) = [ \gamma ^\varepsilon {P_1(t)}^{1-\varepsilon } + (1-\gamma )^\varepsilon {P_2(t)}^{1-\varepsilon } ]^\frac{1}{1-\varepsilon } =1$ yields

$$\begin{aligned} w(t) = [ \gamma ^\varepsilon \{\alpha b_1(t) n_1(t)^\sigma \}^{\varepsilon -1} + (1-\gamma )^\varepsilon \{\alpha b_2(t) n_2(t)^\sigma \}^{\varepsilon -1} ]^\frac{1}{\varepsilon -1}. \end{aligned}$$

(35)

Given the sequences of $b_i(t)$ and $n_i(t)$, the above equation gives the sequence of w(t) . (31) and (33) give the sequences of E(t) and W(t) , respectively. The sequences of $E_1(t)$ and $E_2(t)$ are given by (32). The sequences of $p_i(t)$ and $P_i(t)$ are given by $p_i(t) = w(t) [\alpha b_{i,0} {g_i}^t]^{-1}$ and $P_i(t) = w(t)[\alpha b_{i,0} {g_i}^t {n_i(t)}^\sigma ]^{-1}$, respectively. Now, we have the sequences of $n_i(t)$, $E_i(t)$, and $p_i(t)$. Then, the sequences of $x_i(t)$ and $\pi _i(t)$ are given by (7b) and (7c), respectively.

From (26) and (31), we have $D(t+1)/D(t)=E(t+1)/E(t) = w(t+1)/w(t)$. Using $P(t) = [ \gamma ^\varepsilon {P_1(t)}^{1-\varepsilon } + (1-\gamma )^\varepsilon {P_2(t)}^{1-\varepsilon } ]^\frac{1}{1-\varepsilon } =1$, we obtain the following relation:

$$\begin{aligned} 1=\left( \frac{P(t+1)}{P(t)} \right) ^{1-\varepsilon }&=\frac{ 1+\left( \frac{1-\gamma }{\gamma } \right) ^\varepsilon \left( \frac{P_2(t+1)}{P_1(t+1)} \right) ^{1-\varepsilon } }{ 1+\left( \frac{1-\gamma }{\gamma } \right) ^\varepsilon \left( \frac{P_2(t)}{P_1(t)} \right) ^{1-\varepsilon } } \times \left( \frac{P_1(t+1)}{P_1(t)} \right) ^{1-\varepsilon } \nonumber \\&=\frac{ 1+ \frac{E_2(t+1)}{E_1(t+1)} }{ 1+\frac{E_2(t)}{E_1(t)} } \times \left( \frac{P_1(t+1)}{P_1(t)} \right) ^{1-\varepsilon } \nonumber \\&=\frac{ 1+ \frac{a_2(g_2)}{a_1(g_1)} \Phi {G}^{t+1} }{ 1+\frac{a_2(g_2)}{a_1(g_1)} \Phi {G}^{t} } \times \left[ \frac{1}{g_1} \left( \frac{n_1(t)}{n_1(t+1)} \right) ^\sigma \frac{w(t+1)}{w(t)} \right] ^{1-\varepsilon } \nonumber \\&= \frac{1}{g_1^{1-\varepsilon }} \left( \frac{ 1+ \frac{a_2(g_2)}{a_1(g_1)} \Phi {G}^{t+1} }{ 1+\frac{a_2(g_2)}{a_1(g_1)} \Phi {G}^{t} } \right) ^{1+\sigma (1-\varepsilon )} \left[ \frac{w(t+1)}{w(t)} \right] ^{1-\varepsilon } . \end{aligned}$$

(36)

The second equality uses the second equation of (25). The third equality uses the first equation of (10), (12), and $P_1(t) = w(t) [\alpha b_{1,0} {g_1}^t {n_1(t)}^\sigma ]^{-1}$. The last equality uses the first equation of (34). From (36) and $D(t+1)/D(t)=w(t+1)/w(t)$, we obtain

$$\begin{aligned} \frac{D(t+1)}{D(t)} =\frac{w(t+1)}{w(t)} = g_1 \times \left( \frac{ 1+ \frac{a_2(g_2)}{a_1(g_1)}\Phi {G}^{t} }{ 1+\frac{a_2(g_2)}{a_1(g_1)}\Phi {G}^{t+1} }\right) ^{1/\phi }, \end{aligned}$$

(37)

where $\phi = \frac{ 1-\varepsilon }{1+\sigma (1-\varepsilon ) }$ is defined just below (12). If $G=1$, $D(t+1)/D(t)$ is constant. In contrast, if ${G}>1$, $D(t+1)/D(t)$ decreases with t and satisfies $\lim _{t \rightarrow +\infty } D(t+1)/D(t) = g_2$. As in Acemoglu and Guerrieri (2008), the asymptotic constant growth path exists.

1.3 Appendix 3: Centralized economy

The first-order conditions for the central planner’s problems are

$$\begin{aligned}&\lambda _P(t) = \lambda _I(t), \end{aligned}$$

(38)

$$\begin{aligned}&\lambda _P(t) \frac{n_i^*(t)x_i^*(t)}{b_i^*(t)}=v(t)^\frac{1-\varepsilon }{\varepsilon } \gamma _i \left( n_i^*(t)^{\sigma +1} x_i^*(t) \right) ^\frac{\varepsilon -1}{\varepsilon }, \end{aligned}$$

(39)

$$\begin{aligned}&v(t)^\frac{1-\varepsilon }{\varepsilon } (\sigma +1) \gamma _i \left( n_i^*(t)^{\sigma +1} x_i^*(t) \right) ^\frac{\varepsilon -1}{\varepsilon }\nonumber \\&\quad = n_i^*(t) \left[ \lambda _P(t) \frac{x_i^*(t)}{b_i^*(t)} + \frac{\lambda _I(t-1)}{\beta } a_i\left( \frac{b_i^*(t)}{b_i^*(t-1)} \right) \right] , \end{aligned}$$

(40)

$$\begin{aligned}&\frac{\lambda _I(t-1)}{\beta } \frac{b_i^*(t)}{b_i^*(t-1)} a_i^\prime \left( \frac{b_i^*(t)}{b_i^*(t-1)} \right) n_i^*(t)\nonumber \\&\quad =\lambda _P(t) \frac{n_i^*(t)x_i^*(t)}{b_i^*(t)} + \lambda _I(t) \frac{b_i^*(t+1)}{b_i^*(t)} a_i^\prime \left( \frac{b_i^*(t+1)}{b_i^*(t)} \right) n_i^*(t+1), \end{aligned}$$

(41)

where $i=$1 or 2, $\gamma _1=\gamma$, $\gamma _2=1-\gamma$, and $\lambda _P(t)$ and $\lambda _I(t)$ are the Lagrangian multipliers associated with (15) and (16), respectively. We must have $L^{P*}(t) \in (0, L)$ because if $L^{P*}(t) = L$, no intermediate firms produce in period t and if $L^{P*}(t) = 0$, there are no intermediate goods firms in period $t+1$. Then, (38) holds. We denote $\lambda (t) \equiv \lambda _P(t) = \lambda _I(t)$.

Using (15), (39), and the definition of v(t) , we obtain ${L^P}^*(t) = \frac{v(t)^\frac{1-\varepsilon }{\varepsilon }}{\lambda (t)} \sum _{i} \gamma _i \left( n_i^*(t)^{\sigma +1} x_i^*(t) \right) ^\frac{\varepsilon -1}{\varepsilon } = 1/\lambda (t)$.

From (15), (16), (40), the definition of v(t) , and ${L^P}^*(t)= 1/\lambda (t)$, we can derive $\sigma \beta = \lambda (t) {L^I}^*(t)$.

We substitute ${L^P}^*(t)= 1/\lambda (t)$ and $\sigma \beta = \lambda (t) {L^I}^*(t)$ into $L={L^P}^*(t) + {L^I}^*(t)$ to obtain

$$\begin{aligned} \lambda (t) = \frac{1+\sigma \beta }{L} \equiv \lambda , \quad {L^P}^*(t) = \frac{L}{1+\sigma \beta } \equiv {L^P}^*, \quad \text{ and } \quad {L^I(t)}^* = \frac{\sigma \beta L}{1+\sigma \beta } \equiv {L^I}^*. \end{aligned}$$

(42)

We denote $g_{i}^*(t) \equiv b_i^*(t)/b_i^*(t-1)$ and $G^*(t) =[g_{1}^*(t)/g_{2}^*(t)]^\phi$. To simplify our expressions, we denote $a_i \left( g_{i}^*(t) \right) =a_{i}^*(t)$. From (39) and (40), the following two relations are obtained:

$$\begin{aligned} \beta \sigma \frac{n_i^*(t)x_i^*(t)}{b_i^*(t)}&= a_{i}^*(t)n_i^*(t), \end{aligned}$$

(43)

$$\begin{aligned} \frac{x_2^*(t)}{x_1^*(t)}&= \frac{a_{2}^*(t)}{a_{1}^*(t)}\frac{b_2^*(t)}{b_1^*(t)}. \end{aligned}$$

(44)

In addition, from (39), (40), and (43), we obtain

$$\begin{aligned} \frac{1-\gamma }{\gamma } \left[ \frac{ n_2^*(t)^{\sigma +1} x_2^*(t) }{ n_1^*(t)^{\sigma +1} x_1^*(t) } \right] ^\frac{\varepsilon -1}{\varepsilon } = \frac{a_{2}^*(t)}{a_{1}^*(t)}\frac{n_2^*(t)}{n_1^*(t)}. \end{aligned}$$

(45)

Using (44) and (45), we obtain

$$\begin{aligned} \frac{n_2^*(t)}{n_1^*(t)} = \left[ \left( \frac{1-\gamma }{\gamma }\right) ^\varepsilon \frac{a_{1}^*(t)}{a_{2}^*(t)}\right] ^\frac{1}{1+\sigma (1-\varepsilon )} \left( \frac{b_1^*(t)}{b_2^*(t)}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )} . \end{aligned}$$

(46)

From (16) and (46), we have

$$\begin{aligned} n_1^*(t)&=\frac{ {L^I}^* }{ a_{1}^*(t) + a_{2}^*(t)\left[ \left( \frac{1-\gamma }{\gamma }\right) ^\varepsilon \frac{a_{1}^*(t)}{a_{2}^*(t)}\right] ^\frac{1}{1+\sigma (1-\varepsilon )} \left( \frac{b_1^*(t)}{b_2^*(t)}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )} }, \end{aligned}$$

(47)

$$\begin{aligned} n_2^*(t)&=\frac{\left[ \left( \frac{1-\gamma }{\gamma }\right) ^\varepsilon \frac{a_{1}^*(t)}{a_{2}^*(t)}\right] ^\frac{1}{1+\sigma (1-\varepsilon )} \left( \frac{b_1^*(t)}{b_2^*(t)}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )} {L^I}^* }{ a_{1}^*(t) + a_{2}^*(t)\left[ \left( \frac{1-\gamma }{\gamma }\right) ^\varepsilon \frac{a_{1}^*(t)}{a_{2}^*(t)}\right] ^\frac{1}{1+\sigma (1-\varepsilon )} \left( \frac{b_1^*(t)}{b_2^*(t)}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )} }. \end{aligned}$$

(48)

Using (41), (43), and $a_i(g)= f_i e^{\xi _i g}$, we derive

$$\begin{aligned} g_{i}^*(t) = \frac{1}{\sigma \xi _i} + \beta \frac{a_{i}^*(t+1) n_i^*(t+1) }{a_{i}^*(t) n_i^*(t)}g_{i}^*(t+1). \end{aligned}$$

(49)

This equation shows $g_{i}^*(t) \ge (\sigma \xi _i)^{-1} >0$. (47) and (48) show $n_i^*(t)>0$. If $g_{i}^*(t)=+\infty$, the resource constraint implies that $n_i^*(t)$ must be strictly equal to zero. To ensure $n_i^*(t)>0$, $g_{i}^*(t)<+\infty$ must hold.

We now prove a series of lemmas to prove Proposition 2 in Sect. 4.

Lemma 1

Suppose that Assumption (i) is satisfied ($0<\varepsilon <1$) and that $\xi _1=\xi _2$. There exists a solution for the central planner’s problem such that $g^*_1(t)=g^*_2(t)= [ (1-\beta ) \sigma \xi ]^{-1}$ and $G^*(t) =1$ for all $t=0,1,\ldots$.

Proof

We guess that $g_{1}^*(t) =g_{2}^*(t) =g^*(t)$ holds for all $t=0,1,\ldots$. Then, we verify this guess and derive the expression for $g^*(t)$.

Because $g_{1}^*(t) =g_{2}^*(t) =g^*(t)$ holds for all $t=0,1,\ldots$, we have $a_{1}^*(t)/a_{2}^*(t) = f_1/f_2$ and $b_{1}^*(t)/b_{2}^*(t) = b_{1,0}/b_{2,0}$. Consequently, (47) and (48) show that $a_{i}^*(t)n_i^*(t)$ becomes constant. We can rewrite (49) as $g_{i}^*(t) = 1/(\sigma \xi ) + \beta g_{i}^*(t+1)$. This difference equation has a unique and unstable steady state, where $g_{i}^*(t)= [ (1-\beta ) \sigma \xi ]^{-1}$. Because $0<g_{i}^*(t) < +\infty$, $g_{i}^*(t)$ always stays at the steady state, which implies $g_{1}^*(t) =g_{2}^*(t)= [ (1-\beta ) \sigma \xi ]^{-1}$ and $G^*(t) =1$ for all $t=0,1,\ldots$. $\square$

Lemma 2

Suppose that Assumption (i) is satisfied ($0<\varepsilon <1$) and $\xi _1=\xi _2$. There exists a solution for the central planner’s problem such that $G_{n}^*(t)=G_{L^I}^*(t)=G_{L^P}^*(t) = G^*(t) =1$ for all $t=0,1,\ldots$.

Proof

We have ${L^I}^*_i(t) = a_i^*(t+1)n_i^*(t+1)$ and ${L^p}^*_i(t) = n_i^*(t)x_i^*(t)/b_i^*(t)$. From (43), we have ${L^I_2}^*(t-1)/{L^I_1}^*(t-1)={L^P_2}^*(t)/{L^P_1}^*(t) = a_2^*(t)n_2^*(t) / [ a_1^*(t)n_1^*(t) ]$. If $\xi _1=\xi _2$, we have $g_{1}^*(t) =g_{2}^*(t)= [ (1-\beta ) \sigma \xi ]^{-1}$ and $G^*(t)=1$ from Lemma 1. Then, $a_i^*(t)$ and $n_2^*(t)/n_1^*(t)$ become constant. Consequently, ${L^I_2}^*(t-1)/{L^I_1}^*(t-1)$ and ${L^P_2}^*(t)/{L^P_1}^*(t)$ are constant. $\square$

Lemma 3

Suppose that Assumption (i) is satisfied ($0<\varepsilon <1$) and that $\xi _1<\xi _2$. There exists a solution for the central planner’s problem such that $\lim _{t\rightarrow \infty } g^*_1(t)=g^*_1$ and $\lim _{t\rightarrow \infty } g^*_2(t)= [ (1-\beta ) \sigma \xi _2 ]^{-1} \equiv g^*_2$, where $g^*_1$ is defined by (17) and satisfies $g^*_2< g_{1}^* < [ (1-\beta ) \sigma \xi _1 ]^{-1}$.

Proof

We guess $\lim _{t\rightarrow \infty } g^*_1(t)=g^*_1> g^*_2=\lim _{t\rightarrow \infty } g^*_2(t)$. Then, we verify this guess and derive the expressions for $g^*_1$ and $g^*_2$.

Because of $\lim _{t\rightarrow \infty } g^*_1(t)=g^*_1> g^*_2=\lim _{t\rightarrow \infty } g^*_2(t)$, we have $\lim _{t\rightarrow \infty } a_{1}^*(t)/a_{2}^*(t)$ is constant and $\lim _{t\rightarrow \infty } b_{1}^*(t)/b_{2}^*(t) = +\infty$. Consequently, (48) shows $\lim _{t\rightarrow \infty } a_{2}^*(t)n_2^*(t)= {L^I}^*$. Then, for $i=2$ and a sufficiently large t, (49) can be approximated as $g_{2}^*(t) = 1/(\sigma \xi _2) + \beta g_{2}^*(t+1)$. This difference equation has a unique and unstable steady state, where $g_2^*(t)= [ (1-\beta ) \sigma \xi _2 ]^{-1}$. Because $0<g_2^*(t) < +\infty$, $g_2^*(t)$ must stay at the steady state for large t, which implies $\lim _{t\rightarrow \infty } g^*_2(t)= [ (1-\beta ) \sigma \xi _2 ]^{-1} \equiv g^*_2$.

We rewrite (46) as

$$\begin{aligned} \frac{a_{2}^*(t) n_2^*(t)}{a_{1}^*(t) n_1^*(t)} = \left( \frac{1-\gamma }{\gamma }\right) ^\frac{\varepsilon }{1+\sigma (1-\varepsilon )} \left[ \frac{a_{2}^*(t)}{a_{1}^*(t)}\right] ^\frac{\sigma (1-\varepsilon )}{1+\sigma (1-\varepsilon )} \left( \frac{b_1^*(t)}{b_2^*(t)}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )}. \end{aligned}$$

(50)

For sufficiently large t, we have $a_{2}^*(t) n_2^*(t)=a_{2}^*(t+1) n_2^*(t+1)= {L^I}^*$ and $g_{2}^*(t) = [ (1-\beta ) \sigma \xi _2 ]^{-1}$. Furthermore, we have $b_i^*(t+1)=g_{i}^*(t+1)b_i^*(t)$. Using these equations and (50), we obtain

$$\begin{aligned} \frac{a_{1}^*(t+1) n_1^*(t+1)}{a_{1}^*(t) n_1^*(t)} = \left[ \frac{a_{1}^*(t+1)}{a_{1}^*(t)}\right] ^\frac{\sigma (1-\varepsilon )}{1+\sigma (1-\varepsilon )} \left( \frac{[ (1-\beta ) \sigma \xi _2 ]^{-1}}{g_{1}^*(t+1)}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )}. \end{aligned}$$

for a sufficiently large t. Using this equation and (49), we obtain

$$\begin{aligned}&a_{1}^*(t)^\frac{\sigma (1-\varepsilon )}{1+\sigma (1-\varepsilon )} \left( g_{1}^*(t)-\frac{1}{\sigma \xi _1} \right) \\&\quad = \beta a_{1}^*(t+1)^\frac{\sigma (1-\varepsilon )}{1+\sigma (1-\varepsilon )} \left( [ (1-\beta ) \sigma \xi _2 ]^{-1}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )} {g_{1}^*(t+1)}^\frac{\varepsilon +\sigma (1-\varepsilon )}{1+\sigma (1-\varepsilon )}, \end{aligned}$$

for sufficiently large t. Because $a_{1}^*(t)=a_1 \left( g_{1}^*(t) \right) = f_1 e^{\xi _1 g_{1}^*(t) }$, the above equation is the first-order difference equation for $g_{1}^*(t)$ and it has a unique and unstable steady state. Because $0<g_{1}^*(t) < +\infty$, $g_{1}^*(t)$ must stay at the steady state for large t. Because the above equation reduces to (17) in the steady state, $g^*_1$ is defined by (17).

We evaluate the RHS of (17) at $g_{1}^* = [ (1-\beta ) \sigma \xi _2 ]^{-1}$ and $g_{1}^* = [ (1-\beta ) \sigma \xi _1 ]^{-1}$, where $[ (1-\beta ) \sigma \xi _2 ]^{-1} < [ (1-\beta ) \sigma \xi _1 ]^{-1}$.

$$\begin{aligned} \left. RHS \right| _{g_{1}^* = [ (1-\beta ) \sigma \xi _2 ]^{-1}}&=\frac{1}{\sigma \xi _1} + \beta \left( [ (1-\beta ) \sigma \xi _1 ]^{-1}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )} \left( [ (1-\beta ) \sigma \xi _2 ]^{-1} \right) ^\frac{\varepsilon +\sigma (1-\varepsilon )}{1+\sigma (1-\varepsilon )} \\&>\frac{1}{\sigma \xi _2} + \beta \left( [ (1-\beta ) \sigma \xi _2 ]^{-1}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )} \left( [ (1-\beta ) \sigma \xi _2 ]^{-1} \right) ^\frac{\varepsilon +\sigma (1-\varepsilon )}{1+\sigma (1-\varepsilon )}\\&= [ (1-\beta ) \sigma \xi _2 ]^{-1} = g^*_2, \\ \left. RHS \right| _{g_{1}^* = [ (1-\beta ) \sigma \xi _1 ]^{-1}}&=\frac{1}{\sigma \xi _1} + \beta \left( [ (1-\beta ) \sigma \xi _1 ]^{-1}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )} \left( [ (1-\beta ) \sigma \xi _2 ]^{-1} \right) ^\frac{\varepsilon +\sigma (1-\varepsilon )}{1+\sigma (1-\varepsilon )}\\&<\frac{1}{\sigma \xi _1} + \beta \left( [ (1-\beta ) \sigma \xi _1 ]^{-1}\right) ^\frac{1-\varepsilon }{1+\sigma (1-\varepsilon )} \left( [ (1-\beta ) \sigma \xi _1 ]^{-1} \right) ^\frac{\varepsilon +\sigma (1-\varepsilon )}{1+\sigma (1-\varepsilon )}\\&= [ (1-\beta ) \sigma \xi _1 ]^{-1}, \end{aligned}$$

Then, $g_{1}^*$ is between $[ (1-\beta ) \sigma \xi _2 ]^{-1}$ and $[ (1-\beta ) \sigma \xi _1 ]^{-1}$. $\square$

Lemma 4

Suppose that Assumption (i) is satisfied ($0<\varepsilon <1$) and that $\xi _1<\xi _2$. There exists a solution for the central planner’s problem such that $\lim _{t\rightarrow \infty } G^*(t)=\lim _{t\rightarrow \infty } G_{n}^*(t)=\lim _{t\rightarrow \infty } G_{L^I}^*(t)=\lim _{t\rightarrow \infty } G_{L^P}^*(t) =\left( g^*_1/g^*_2 \right) ^\phi \equiv G^* >1$.

Proof

The proof is similar to that of Lemma 2. Lemma 3 shows that if $\xi _1<\xi _2$, we have $\lim _{t\rightarrow \infty } G^*(t)=\left( g^*_1/g^*_2 \right) ^\phi \equiv G^* >1$. We know from (46) that $\lim _{t\rightarrow \infty } G_{n}^*(t)= \lim _{t\rightarrow \infty } G^*(t)$. In addition, (43) implies that $\lim _{t\rightarrow \infty } G_{n}^*(t)=\lim _{t\rightarrow \infty } G_{L^I}^*(t)=\lim _{t\rightarrow \infty } G_{L^P}^*(t)$.