Automation, growth, and factor shares in the era of population aging

Abstract

How does population aging affect economic growth and factor shares in times of increasingly automatable production processes? The present paper addresses this question in a new macroeconomic model of automation where competitive firms perform tasks to produce output. Tasks require labor and machines as inputs. New machines embody superior technological knowledge and substitute for labor in the performance of tasks. Automation is labor-augmenting in the reduced-form aggregate production function. If wages increase then the incentive to automate becomes stronger. Moreover, the labor share declines even though the aggregate production function is Cobb–Douglas. Population aging due to a higher longevity reduces automation in the short and promotes it in the long run. It boosts the growth rate of absolute and per-capita GDP in the short and the long run, lifts the labor share in the short and reduces it in the long run. Population aging due to a decline in fertility increases automation, reduces the growth rate of GDP, and lowers the labor share in the short and the long run. In the short run, it may or may not increase the growth rate of per-capita GDP, in the long run it unequivocally accelerates per-capita GDP growth.

Appendix: Proofs

The proofs of Proposition 2.2, 2.3, as well as of Corollary 2.2, 2.3, 4.6, and 5.3 are given in the main text.

1.1 Proof of Proposition 2.1

Given $q\left( \omega _{t}\right)$ of (2.12), Eq. (2.8) delivers $h_t=1/\left( A_{t-1}\left( 1+ q\left( \omega _{t}\right) \right) \right) \equiv h\left( \omega _{t}\right) /A_{t-1}$. From (2.4) $i_t= i\left( \omega _{t}\right)$. Since the wage cost per task is $w_t h_t=\omega _t h\left( \omega _{t}\right)$, we have $c_t=\omega _t h\left( \omega _{t}\right) +i\left( \omega _{t}\right) \equiv c\left( \omega _{t}\right)$. Continuity of these functions follows since $\lim _{\omega _t\downarrow \alpha } q\left( \omega _{t}\right) =0$. The remaining arguments that complete the proof are straightforward or given in the main text. $\square$

1.2 Proof of Corollary 2.1

If $\omega _t>\alpha$ then $q_t>0$ and the rationalization effect follows since $\left( A_{t-1}\left( 1+q_t\right) \right) ^{-1}<A_{t-1}^{-1}$. The productivity effect follows since $c_t$ is the solution to (2.10) and $c\left( \omega _{t}\right) |_{\omega _t=\alpha }=\omega _t$. $\square$

1.3 Proof of Proposition 2.4

For ease of notation I shall most often suppress the time argument. Consider problem (2.21). Since preferences are increasing in $c^o$ both periodic budget constraints will hold as equalities and can be merged. Accordingly, the Lagrangian of this problem is

$$\begin{aligned} \mathcal {L}=\ln c^y+\ln \left( 1-\phi \left( 1- l\right) \left( c^y\right) ^{\frac{\nu }{1-\nu }}\right) +\mu \beta \ln c^o+\lambda \left[ w\left( 1-l\right) -c^y-\frac{\mu c^o}{R}\right] . \end{aligned}$$

(7.1)

Corner solutions involving $c^y=c^o=0$ and $l=1$ can be excluded since U satisfies the Inada conditions and $l=1$ implies no income. Hence, with $x\equiv \left( 1-l\right) \left( c^y\right) ^\frac{\nu }{1-\nu }$ the respective first-order Kuhn-Tucker conditions read as follows:

$$\begin{aligned} \frac{\partial \mathcal {L}}{\partial c^y}= & {} \frac{1-\nu -\phi x}{c^y(1-\nu )(1-\phi x)}-\lambda =0, \end{aligned}$$

(7.2)

$$\begin{aligned} \frac{\partial \mathcal {L}}{\partial l}= & {} \frac{\phi \left( c^y\right) ^{\frac{\nu }{1-\nu }}}{1-\phi x}-\lambda w\le 0,\quad \text{ with } \text{ strict } \text{ inequality } \text{ if } l_t=0, \end{aligned}$$

(7.3)

$$\begin{aligned} \frac{\partial \mathcal {L}}{\partial c^o}= & {} \frac{\beta }{c^o}-\frac{\lambda }{R}=0, \end{aligned}$$

(7.4)

$$\begin{aligned} \frac{\partial \mathcal {L}}{\partial \lambda }= & {} w\left( 1-l\right) -c^y-\frac{\mu c^o}{R}=0. \end{aligned}$$

(7.5)

Suppose $l>0$. Then, upon multiplication by $(1-l)$, condition (7.3) may be written as

$$\begin{aligned} \frac{\phi x}{\left( 1-\phi x\right) w \left( 1-l\right) }=\lambda . \end{aligned}$$

(7.6)

Using the latter to replace $\lambda$ in (7.2) and (7.4) delivers, respectively,

$$\begin{aligned} c^y=\left( \frac{1}{\phi x}-\frac{1}{1-\nu }\right) w(1-l) \end{aligned}$$

(7.7)

and

$$\begin{aligned} c^o= \beta R\left( \frac{1}{\phi x}-1\right) w (1-l). \end{aligned}$$

(7.8)

With (7.7) and (7.8) in the budget constraint (7.5) I obtain

$$\begin{aligned} \phi x_t=\phi x=\frac{(1+\mu \beta )(1-\nu )}{1+\left( 1+\mu \beta \right) \left( 1-\nu \right) }\in (0,1). \end{aligned}$$

(7.9)

Using (7.9) in (7.7), (7.8), and (2.20) delivers (2.22). Since the optimal plan satisfies Assumption 1 I have $1>\nu (1+\mu \beta )$, hence, $c^y>0$.

From the definition of x with $h^s=1-l$ it holds that $c^y= \left( x \left( h^s\right) ^{-1}\right) ^\frac{1-\nu }{\nu }$. Replacing $c^y$ with this expression in (2.22) and solving for $h^s$ delivers $h^s_t$. Using the latter in (2.22) delivers $c_t$ and $s_t$. Then, $c^o_{t+1}$ is obtained from the budget when old. Clearly, $h^s_t\le 1$ as long as $w_t\ge w_c$. In accordance with this, $w< w_c$ implies a strict inequality in (7.3).

To see that the solution identified by the Lagrangian (7.1) is indeed a global maximum if $\nu <\bar{\nu }\left( \mu \beta \right)$ consider first the leading principal minors of the Hessian matrix of $U\left( c^y,l,c^o\right)$, i.e.,

$$\begin{aligned} D_1\left( c^y,l,c^o\right)= & {} -\frac{\left( 1-\nu -\phi x\right) ^2+\nu \phi x\left( 1-\phi x\right) }{\left( c^y(1-\nu )\left( 1-\phi x\right) \right) ^2},\\ D_2\left( c^y,l,c^o\right)= & {} \frac{\phi ^2 \left( 1-2 \nu -(1-\nu )\phi x\right) }{\left( c^y\right) ^{\frac{2 (1-2\nu )}{1-\nu }}\left( 1-\nu \right) ^2\left( 1-\phi x\right) ^3},\\ D_3\left( c^y_t,l_t,c^o_{t+1}\right)= & {} -\frac{\mu \beta }{\left( c^o\right) ^2}D_2\left( c^y_t,l_t,c^o_{t+1}\right) . \end{aligned}$$

First, we have $-D_1\left( c^y,l,c^o\right) >0$. Second, observe that $D_2\left( c^y,l,c^o\right) >0$ and $-D_3\left( c^y,l,c^o\right) >0$ hold if and only if condition (2.19) holds. Hence, U is strictly concave for all $\left( c^y,l,c^o\right) \in \mathcal {P}$.

What remains to be shown is that the solution identified by the Lagrangian satisfies condition (2.19). With $\phi x$ of (7.9) this is the case if and only if

$$\begin{aligned} \frac{1-2\nu }{1-\nu }>\frac{(1+\mu \beta )(1-\nu )}{1+(1+\mu \beta )(1-\nu )} \end{aligned}$$

or

$$\begin{aligned} \nu ^2(1+\mu \beta )-\nu (3+\mu \beta )+1>0. \end{aligned}$$

It is not difficult to show that the latter condition is satisfied if and only if $\nu <\bar{\nu }\left( \mu \beta \right)$ as stated in Assumption 1.

Finally, observe that surviving members of cohort 0 satisfy their budget constraint when old as equality, i.e., we have $c^o_1=R_1 s_0/\mu >0$. $\square$

1.4 Proof of Corollary 2.4

Some algebra reveals that

$$\begin{aligned} \frac{\partial w_c}{\partial \mu }=\frac{\beta w_c}{\nu (1+\mu \beta ) (1+(1-\nu )(1+\mu \beta )) (1-\nu (1+\mu \beta )}>0. \end{aligned}$$

It follows that $\partial h^s_t/\partial \mu >0$. From the definition of $w_c$ and Proposition 2.4, $c_t^y$ may be written as

$$\begin{aligned} c_t^y=\left( \frac{1-\nu \left( 1+\mu \beta \right) }{\phi \left( 1+\left( 1+\mu \beta \right) (1-\nu )\right) }\right) ^{1-\nu }w_t^{1-\nu }. \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\partial c^y_t}{\partial \mu }=\frac{-\left( 1-\nu \right) \beta w_t^{1-\nu }}{\phi ^{1-\nu } (1+(1-\nu )(1+\mu \beta ))^{2-\nu }(1-\nu (1+\mu \beta ))^{\nu }}<0. \end{aligned}$$

The sign of $\partial s_t/\partial \mu >0$ follows since the marginal propensity to save in (2.22) increases in $\mu$ and $\partial h^s_t/\partial \mu >0$. Finally, using $s_t$ in the budget constraint of a surviving old delivers

$$\begin{aligned} c^o_{t+1}=\frac{\beta R_{t+1} w_t^{1-\nu } }{\phi ^{1-\nu }(1+(1-\nu )(1+\mu ) )^{1-\nu }(1-\nu (1+\mu ) )^{\nu }}. \end{aligned}$$

Hence, by Assumption 1

$$\begin{aligned} \frac{\partial c^o_{t+1}}{\partial \mu }=-\frac{ \left( \nu ^2(1+\mu \beta ) -\nu (3+\mu \beta ) +1\right) \beta ^2 R_{t+1}w_t^{1-\nu } }{\phi ^{1-\nu } (1 +(1+\nu )(1+\mu \beta ) )^{2-\nu } (1-\nu (1+\mu \beta ) )^{1+\nu }}<0. \end{aligned}$$

$\square$

1.5 Proof of Proposition 3.1

Under Assumption 2, $H^{d}_t$ is given by (3.2) whereas $H^s_t$ is given by (3.3). Hence, (3.4) delivers (3.5). Denote the right-hand side of (3.5) by $RHS(\omega _t)$ where $RHS:[\alpha ,\infty )\rightarrow [\underline{k}_c,\infty )$. Then, $RHS(\alpha )=\underline{k}_c>0$. Moreover, since $\nu <1/2$ we have $RHS'(\omega _t)>0$ for $\omega _t>\alpha$ and $\lim _{\omega _t\rightarrow \infty }RHS(\omega _t)=\infty$. Hence, for Eq. (3.5) to be satisfied for any value $\omega _t>\alpha$ it is necessary and sufficient to have $RHS(\alpha )<k_t$ or $k_t>\underline{k}_c$. Then, the above-mentioned properties of $RHS(\omega _t)$ assure that there is indeed a unique $\hat{\omega }_t>\alpha$ that satisfies (3.5). By the implicit function theorem, the function $\omega \left( k_t\right)$ has the indicated properties. $\square$

1.6 Proof of Proposition 3.2

First, observe that $k_t$ is a state variable of the inter-temporal general equilibrium. Indeed, given $k_t>\underline{k}_c$, the labor market determines $\hat{\omega }_t=A_{t-1}\hat{w}_t>\alpha$. Hence, Proposition 2.1 delivers $q_t$, $a_t$, $i_t,$ and $c_t$. Proposition 2.2 and (3.2) determine $N_t$, $Y_t$, $I_t$, and $H^d_t$. Hence, (2.13) delivers $R_t$. On the household side, Proposition 2.4 gives $h_t^s$, $l_t$, $c_t^y$, $c_{t}^o$, $s_t$. Finally, $K_{t+1}$ follows from (3.6).

Second, consider (3.5) and replace $\hat{\omega }_t$ by $\omega _t=\left( k_{t+1}/\Omega \right) ^{1/(1-\nu )}$ from (3.7). Then, the equilibrium difference equation (3.8) is given by

$$\begin{aligned} k_t=\frac{\underline{k}_c}{\alpha ^{\frac{1}{2}-\nu }}\left( \frac{k_{t+1}}{\Omega }\right) ^\frac{1-2\nu }{2(1-\nu )}\left( \frac{2}{\sqrt{\alpha }}\left( \frac{k_{t+1}}{\Omega }\right) ^\frac{1}{2(1-\nu )}-1\right) ^\frac{1}{\gamma }. \end{aligned}$$

(7.10)

Denote the right-hand side of (7.10) by RHS(k). The latter satisfies $\lim _{k\downarrow \overline{k}_c}RHS(k)=\underline{k}_c$, is continuous, and, since $\nu <1/2$, increasing with $\lim _{k\rightarrow \infty }RHS(k)=\infty$. Hence, (7.10) assigns to each $k_t>\underline{k}_c$ a unique $k_{t+1}>\overline{k}_c$. To see that there is a unique fixed point $k=RHS(k)$ write (7.10) for $k_t=k_{t+1}=k$ as

$$\begin{aligned} k=Z_1 k^\frac{1-2\nu }{2(1-\nu )}\left( Z_2 k^\frac{1}{2(1-\nu )}-1\right) ^\frac{1}{\gamma } \end{aligned}$$

or

$$\begin{aligned} k^\frac{\gamma }{2(1-\nu )}=Z^\gamma _1 Z_2 k^\frac{1}{2(1-\nu )}-Z^\gamma _1, \end{aligned}$$

(7.11)

where

$$\begin{aligned} Z_1\equiv \frac{\underline{k}_c}{\alpha ^{\frac{1}{2}-\nu }}\left( \frac{1}{\Omega }\right) ^\frac{1-2\nu }{2(1-\nu )}>0\quad \text{ and }\quad Z_2\equiv \frac{2}{\sqrt{\alpha }}\left( \frac{1}{\Omega }\right) ^\frac{1}{2(1-\nu )}>0. \end{aligned}$$

Since $0<\gamma <1$ and $0<\nu <1/2$ it holds that $0<\gamma /(2(1-\nu ))<1/(2(1-\nu ))<1$. Hence, the left and the right-hand side of (7.11) are concave with a unique intersection at some $k>0$. Moreover, a straightforward graphical argument in $(k_{t+1},k_t)$ - space reveals that $k>\overline{k}_c$ and that k is stable for all $k_1>\underline{k}_c$ (see, e.g., Galor 2007). $A_{0}>w_c/\alpha$ ensures Assumption 2. $\square$

1.7 Proof of Corollary 4.1

Since the right-hand side of (3.5) is increasing in $\hat{\omega }_t$ and $\partial w_c/\partial \mu >0$ there is by the implicit function theorem a function $\hat{\omega }_t=\hat{\omega }(\mu )$ with $d \hat{\omega }_t/d \mu <0$. Hence, $d\hat{w}_t/d\mu =\left( \partial \hat{w}_t/\partial \omega _t\right) \left( d \hat{\omega }_t/d \mu \right) <0$. The sign of $d\hat{q}_t/d\mu$ follows with Proposition 2.1. $\square$

1.8 Proof of Corollary 4.2

Let $\hat{H}_t=H^d(\hat{\omega }_t)=H^d(\hat{\omega }(\mu ))$ where $\hat{\omega }_t=\hat{\omega }(\mu )$ is defined in the Proof of Corollary 4.1. With $\hat{N}_t=A_{t-1}\left( 1+q\left( \hat{\omega }(\mu )\right) \right) H^d(\hat{\omega }(\mu ))$ define $\hat{GDP}_t=F\left( K_t,\hat{N}_t\right) -\hat{N}_t i\left( \hat{\omega }(\mu )\right) \equiv GDP\left( \hat{\omega }(\mu )\right)$. Then, it holds that $d\hat{GDP}_t/d\mu =\left( \partial GDP\left( \hat{\omega }_t\right) /\partial \omega _t \right) \left( d\hat{\omega }_t/d\mu \right)$ where Corollary 4.1 delivers $d\hat{\omega }_t/d\mu <0$. Some manipulations reveal that

$$\begin{aligned} \frac{\partial GDP\left( \hat{\omega }_t\right) }{\partial \omega _t}=\, & {} A_{t-1}H^d\left( \hat{\omega }_t\right) \frac{\partial q\left( \hat{\omega }_t\right) }{\partial \omega _t}\left[ F_2-\left( 1+q\left( \hat{\omega }_t\right) \right) \frac{\partial i\left( \hat{\omega }_t\right) }{\partial q_t}-i\left( \hat{\omega }_t\right) \right] \nonumber \\+\, & {} A_{t-1}\left( 1+ q\left( \hat{\omega }_t\right) \right) \frac{\partial H^d\left( \hat{\omega }_t\right) }{\partial \omega _t}\left[ F_2-i\left( \hat{\omega }_t\right) \right] , \end{aligned}$$

(7.12)

where F is evaluated at $\left( K_t,\hat{N}_t\right)$. In equilibrium, the bracketed expression in the first line vanishes. To see this, observe that the cost per task in a symmetric configuration is $w_t h_t +i(q_t)=\omega _t/(1+q_t) +i(q_t)$. Minimizing this expression with respect to $q_t$ gives the first-order condition $\omega _t/(1+q_t)=(1+q_t)\partial i(q_t)/\partial q_t$. Hence, the minimized cost per task can be written as $c_t=(1+q_t)\partial i(q_t)/\partial q_t+i(q_t)$. Profit maximization requires conditions (2.13) to hold. Hence, $F_2=c_t=(1+q_t)\partial i(q_t)/\partial q_t+i(q_t)$ holds in equilibrium. As $\partial H^d\left( \hat{\omega }_t\right) /\partial \omega _t<0$ and $F_2-i\left( \hat{\omega }_t\right) >0$ it follows that $\partial GDP\left( \hat{\omega }_t\right) /\partial \omega _t<0$, hence, $d\hat{GDP}_t/d\mu >0$. $\square$

1.9 Proof of Corollary 4.3

Corollary 2.3 proves $\partial LS_t/\partial \omega _t<0$. Hence, $d\hat{LS}_t/d\mu =\left( \partial LS_t/\partial \omega _t\right) \left( d \hat{\omega }_t/d\mu \right) >0$. $\square$

1.10 Proof of Corollary 4.4

Consider (3.5) with $k_{t+1}=K_{t+1}/\left( A_{t}^{1-\nu }L_t(1+g_L)\right)$. Since the right-hand side of (3.5) is increasing in $\hat{\omega }_{t+1}$ there is by the implicit function theorem a function $\hat{\omega }_{t+1}=\hat{\omega }(g_L)$ with $d \hat{\omega }_{t+1}/d g_L<0$. Hence, $d\hat{w}_{t+1}/dg_L=\left( \partial \hat{w}_{t+1}/\partial \omega _{t+1}\right) \left( d \hat{\omega }_{t+1}/d g_L\right) <0$. The sign of $d\hat{q}_{t+1}/dg_L$ follows with Proposition 2.1. $\square$

1.11 Proof of Corollary 4.5

Let $\hat{H}_{t+1}=H^d(\hat{\omega }_{t+1})=H^d(\hat{\omega }(g_L))$ where $\hat{\omega }_{t+1}=\hat{\omega }(g_L)$ is defined in the proof of Corollary 4.4. With $\hat{N}_{t+1}=A_{t}\left( 1+q\left( \hat{\omega }(g_L)\right) \right) H^d(\hat{\omega }(g_L))$ let $\hat{GDP}_{t+1}=F\left( K_{t+1},\hat{N}_{t+1}\right) -\hat{N}_{t+1} i\left( \hat{\omega }(g_L)\right) \equiv GDP\left( \hat{\omega }(g_L)\right)$. Then, it holds that $d\hat{GDP}_{t+1}/dg_L=\left( \partial GDP\left( \hat{\omega }_{t+1}\right) /\partial \omega _{t+1} \right) \left( d\hat{\omega }_{t+1}/d g_L\right)$. Some manipulations reveal that

$$\begin{aligned} \frac{\partial GDP\left( \hat{\omega }_{t+1}\right) }{\partial \omega _{t+1}}=\, & {} A_{t}H^d\left( \hat{\omega }_{t+1}\right) \frac{\partial q\left( \hat{\omega }_{t+1}\right) }{\partial \omega _{t+1}}\left[ F_2-\left( 1+q\left( \hat{\omega }_{t+1}\right) \right) \frac{\partial i\left( \hat{\omega }_{t+1}\right) }{\partial q_{t+1}}-i\left( \hat{\omega }_{t+1}\right) \right] \nonumber \\&+A_{t}\left( 1+ q\left( \hat{\omega }_{t+1}\right) \right) \frac{\partial H^d\left( \hat{\omega }_{t+1}\right) }{\partial \omega _{t+1}}\left[ F_2-i\left( \hat{\omega }_{t+1}\right) \right] , \end{aligned}$$

(7.13)

where F is evaluated at $\left( K_{t+1},\hat{N}_{t+1}\right)$. For the same reason as in the proof of Corollary 4.2, the bracketed expression in the first line vanishes. Hence, as $\partial H^d\left( \hat{\omega }_{t+1}\right) /\partial \omega _{t+1}<0$ and $F_2-i\left( \hat{\omega }_{t+1}\right) >0$ it follows that $\partial GDP\left( \hat{\omega }_{t+1}\right) /\partial \omega _{t+1}<0$, hence, $d\hat{GDP}_t/d g_L>0$.

Turning to the effect of $g_L$ on $\hat{gdp}_{t+1}$, observe that $\hat{gdp}_{t+1}\equiv \hat{GDP}_{t+1}/\left( L_t\left( \mu +1+g_L\right) \right)$. Hence,

$$\begin{aligned} \frac{d \hat{gdp}_{t+1}}{d g_L}\gtreqless 0\quad \Leftrightarrow \quad \frac{d\hat{GDP}_{t+1}}{d g_L}\gtreqless \frac{\hat{GDP}_{t+1}}{\mu +1+g_L}. \end{aligned}$$

With (7.13) we have

$$\begin{aligned} \frac{d \hat{GDP}_{t+1}}{d g_L}=\, & {} A_{t}\left( 1+ q\left( \hat{\omega }_{t+1}\right) \right) \left[ F_2-i\left( \hat{\omega }_{t+1}\right) \right] \frac{\partial H^d\left( \hat{\omega }_{t+1}\right) }{\partial \omega _{t+1}}\frac{d\hat{\omega }_{t+1}}{d g_L},\nonumber \\=\, & {} \hat{w}_{t+1} \left( \frac{\partial H^d\left( \hat{\omega }_{t+1}\right) }{\partial \omega _{t+1}}\frac{d\hat{\omega }_{t+1}}{d g_L}\right) , \end{aligned}$$

(7.14)

where the second line uses the first-order condition (2.13) for $N_{t+1}$, i.e., $F_2(K_{t+1},N_{t+1})=c_{t+1}=w_{t+1}h_{t+1}+i_{t+1}$. The latter implies $w_{t+1}=\left[ F_2(K_{t+1},N_{t+1})-i_{t+1}\right] /h_{t+1}=A_t\left( 1+q_{t+1}\right) \left[ F_2(K_{t+1},N_{t+1})-i_{t+1}\right]$.

With (3.5) and (4.2) one finds

$$\begin{aligned} \frac{\partial H^d\left( \hat{\omega }_{t+1}\right) }{\partial \omega _{t+1}}\frac{d\hat{\omega }_{t+1}}{d g_L}=\, & {} \frac{H^d\left( \hat{\omega }_{t+1}\right) }{1+g_L} \left[ \frac{\frac{\hat{\omega }_{t+1}}{h\left( \hat{\omega }_{t+1}\right) }\frac{\partial h\left( \hat{\omega }_{t+1}\right) }{\partial \omega _{t+1}}+\frac{\hat{\omega }_{t+1}}{N(\hat{c}_{t+1})}\frac{\partial N(\hat{c}_{t+1})}{\partial c_{t+1}}\frac{\partial c\left( \hat{\omega }_{t+1}\right) }{\partial \omega _{t+1}}}{\frac{\hat{\omega }_{t+1}}{h\left( \hat{\omega }_{t+1}\right) }\frac{\partial h\left( \hat{\omega }_{t+1}\right) }{\partial \omega _{t+1}}+\frac{\hat{\omega }_{t+1}}{N(\hat{c}_{t+1})}\frac{\partial N(\hat{c}_{t+1})}{\partial c_{t+1}}\frac{\partial c\left( \hat{\omega }_{t+1}\right) }{\partial \omega _{t+1}}+\nu }\right] \nonumber \\=\, & {} \frac{H^d\left( \hat{\omega }_{t+1}\right) }{1+g_L}\times \Psi \left( \hat{\omega }_{t+1}\right) , \end{aligned}$$

(7.15)

where

$$\begin{aligned} \Psi \left( \hat{\omega }_{t+1}\right) \equiv \left( \frac{\partial H^d\left( \hat{\omega }_{t+1}\right) }{\partial \hat{\omega }_{t+1}}\frac{ \omega _{t+1}}{H^d\left( \hat{\omega }_{t+1}\right) }\right) \left( \frac{d\hat{\omega }_{t+1}}{d g_L}\frac{1+g_L}{\hat{\omega }_{t+1}}\right) >1. \end{aligned}$$

With (7.15) and $L_{t+1}=(1+g_L)L_t$ one obtains (4.4). $\square$

1.12 Proof of Proposition 5.1

From Proposition 3.2 the steady state has $k_t=k^*>\overline{k}_c>\underline{k}_c$ so that Proposition 3.1 implies $\omega _t=\hat{\omega }^*=\omega \left( k^*\right) >\alpha$. Then, from (2.12) I have $q_t=q^*=q(\hat{\omega }^*)>0$, and the results listed under a) - d) follow from Proposition 2.1, Proposition 2.2, Proposition 2.4, Proposition 3.1, and Eqs. (2.13), (3.2) and (3.6). $\square$

1.13 Proof of Corollary 5.1

I show how a change in $\mu$ and $g_L$ affects $\hat{\omega }^*$. Then, the corollary follows since $\partial q(\hat{\omega }^*)/\partial \omega _t>0$ (see Proposition 2.1).

Consider the labor market equilibrium condition (3.5) and the capital market condition (3.7) in steady state. Solving both equations for $k^*$ and substitution delivers

$$\begin{aligned} \frac{\mu \beta \left( \Gamma (1-\gamma )\right) ^\frac{1}{\gamma } }{(1+\mu \beta )(1-\nu )(1+g_L)}=\left( \alpha \hat{ \omega }^*\right) ^\frac{-1}{2}\left( 2\sqrt{\alpha \hat{ \omega }^*}-\alpha \right) ^\frac{1}{\gamma }. \end{aligned}$$

(7.16)

The right-hand side of Eq. (7.16) defines a continuous function $RHS(\omega )$ with $RHS'(\omega )>0$ for all $\hat{\omega }^*>\alpha$. Then, total differentiation of (7.16) delivers $d \hat{\omega }^*/d \mu >0$ and $d \hat{\omega }^*/d g_L<0$. $\square$

1.14 Proof of Corollary 5.2

The sign of $dg_{GDP}^*/d \mu$, $d g_{gdp}^*/d \mu$, and $d g_{gdp}^*/d g_L$ follow immediately from Corollary 5.1. To show that $d g_{GDP}^*/d g_L>0$ observe that

$$\begin{aligned} \frac{d g_{GDP}^*}{d g_L}=(1-\nu )\left( 1+q^*\right) ^{-\nu }\frac{d q\left( \hat{\omega }^*\right) }{d g_L}(1+g_L)+\left( 1+q^*\right) ^{1-\nu }, \end{aligned}$$

where $d q\left( \hat{\omega }^*\right) /d g_L=\left( \partial q\left( \hat{\omega }^*\right) /\partial \omega \right) \cdot \left( d \hat{\omega }^*/d g_L\right) <0$. Some algebraic manipulations using

$$\begin{aligned} \frac{d \hat{\omega }^*}{d g_L}=-\frac{\hat{\omega }^*}{1+ g_L}\left[ \frac{1}{2}+\frac{1}{\gamma }\left( \frac{\sqrt{\hat{\omega }^*}}{2\sqrt{\hat{\omega }^*}-\sqrt{\alpha }}\right) -\nu \right] ^{-1}<0 \end{aligned}$$

obtained from (3.5) deliver the desired result. $\square$