Secular satiation

Abstract

Satiation of need is generally ignored by growth theory. I study a model where consumers may be satiated in any given good but new goods may be introduced. A social planner will never elect a trajectory with long-run satiation. Instead, he will introduce enough new goods to avoid such a situation. In contrast, the decentralized equilibrium may involve long run satiation. This, despite that the social costs of innovation are second order compared to their social benefits. Multiple equilibria may arise: depending on expectations, the economy may then converge to a satiated steady state or a non satiated one. In the latter equilibrium, capital and the number of varieties are larger than in the former, while consumption of each good is lower. This multiplicity comes from the following strategic complementarity: when people expect more varieties to be introduced in the future, this raises their marginal utility of future consumption, inducing them to save more. In turn, higher savings reduces interest rates, which boosts the rate of innovation. When TFP grows exogenously and labor supply is endogenized, the satiated equilibrium generically survives. For some parametrer values, its growth rate is positive while labor supply declines over time to zero. Its growth rate is then lower than that of the non satiated equilibrium. Hence, the economy may either coordinate on a high leisure, low growth, satiated "leisure society" or a low leisure, high growth, non satiated "consumption society".

Appendix

1.1 Proof of Lemma 1

Since utility does not satisfy the Inada conditions at $C=0,$ the nonnegativity constraint $C\ge 0$ has to be imposed. Consequently, the Hamiltonian is (see for example Büskens and Maurer (2000))

$$\begin{aligned} H(C_{t},K_{t},\lambda _{t},\mu _{t},t)=u(C_{t})e^{-\rho t}+\lambda _{t}e^{-\rho t}(AK_{t}^{\alpha }-C_{t}-\delta K_{t})+\mu _{t}e^{-\rho t}C_{t}. \end{aligned}$$

The FOCs are

$$\begin{aligned} u^{\prime }(C_{t})= & {} \lambda _{t}-\mu _{t} \nonumber \\ -{\dot{\lambda }}_{t}+\rho \lambda _{t}= & {} \lambda _{t}r(K_{t}) \nonumber \\ \mu _{t}C_{t}= & {} 0 \end{aligned}$$

(20)

Therefore, either $\mu _{t}=0$ and $C_{t}\ge 0,$ implying that $u^{\prime }(C_{t})=\lambda _{t},$ i.e. $C_{t}=\frac{1-\lambda _{t}}{\beta },$ or $\mu _{t}\ge 0$ and $C_{t}=0,$ implying that $1=u^{\prime }(0)\le \lambda _{t}.$ This proves that $C_{t}=\max (\frac{1-\lambda _{t}}{\beta },0),$ while (20) clearly coincides with (5). $\square$

1.2 Proof of Proposition 1

A. Remember that $C^{*}=\max _{K}AK^{\alpha }-\delta K.$ Therefore if $1/\beta >C^{*}$ the equation $AK^{\alpha }-\delta K=1/\beta$ has no solution and the only steady states are non satiated. Consequently, $K=K_{MGR \text { }}$ in steady state.

B. (i). Assume $C_{t}=1/\beta$ throughout. Then ${\dot{K}}_{t}=AK^{\alpha }-\delta K-1/\beta .$ Over $(K_{S1},K_{S2}],$ ${\dot{K}}>0,$ so that K clearly converges to $K_{S2}.$ Furthermore, this trajectory is optimal as it delivers the maximum possible utility level.

(ii) By (5), the consumer is either always satiated or never satiated. Since K would become equal to zero in finite time if he were always satiated, the consumer is never satiated. Assume the economy converges to a satiated steady state, implying $\lim _{t\rightharpoonup \infty }K=K_{S,i}.$ Since $K_{S2}>K_{S1}>K_{MGR},$ for some T and $t>T,$ $K>K_{MGR}.$ Then ${\dot{C}}<0$ for $t>T,$ which contradicts the fact that $\lim _{t\rightarrow \infty }C=1/\beta .$ Therefore the economy converges to a non satiated steady state, implying $\lim K=K_{MGR}.$

C. (i) Same proof as B (i)

(ii) Again the consumer is never satiated. Assume the economy converges to a non satiated steady state. Then $\lim _{t\rightharpoonup \infty }K=K_{MGR}.$ Since $K_{MGR}>K_{S1},$ for some T and $t>T,$ $K_{S2}>K>K_{S1}.$ Let $t_{0}>T.$ Consider the following deviation ${\tilde{C}}_{t}$ from the actual trajectory: ${\tilde{C}}_{t}=C_{t},$ $t<t_{0},$ ${\tilde{C}}_{t}=1/\beta ,$ $t\ge t_{0}.$ Then the corresponding capital stock ${\tilde{K}}_{t}$ is such that ${\tilde{K}}_{t}=K_{t},$ $t<t_{0}.$ Therefore $K_{S2}>K_{t_{0}}>K_{S1}.$ Clearly, then $\lim {\tilde{K}}=K_{S2}.$ Hence the deviation is feasible. Furthermore, as $C_{t}<1/\beta$ throughout, ${\tilde{C}}_{t}>C_{t}$ for $t\ge t_{0}.$ It follows that the deviation delivers a strictly higher utility level than the initial trajectory, which therefore cannot be optimal. Consequently, the economy necessarily converges to a satiated steady state. $\square$

[Observe that, as in the standard case with a single steady state, one can show that the economy necessarily converges to a steady state. To see this, note that $\lambda$ is monotonic throughout. If not, there exists t such that ${\dot{\lambda }}_{t}=0,$ implying $K_{t}=K_{MGR},$ unless $\lambda =0,$ in which case $C=1/\beta$ throughout and convergence is straightforward. If $C_{t}=C_{MGR},$ this is clearly a steady state. If $C_{t}>C_{MGR},$ ${\dot{K}} <0$ locally, implying that for $s>t,$ ${\dot{\lambda }}_{s}<0,$ ${\dot{C}}_{s}>0$ and therefore ${\dot{K}}_{s}<0.$ Since furthermore ${\dot{K}}_{s}<AK_{s}^{\alpha }-\delta K_{s}-C_{t}<AK_{MGR}^{\alpha }-\delta K_{MGR}-C_{t}=C_{MGR}-C_{t}<0,$ K becomes equal to zero in finite time, which is impossible. If $C_{t}<C_{MGR},$ ${\dot{K}}>0$ locally, so that for $s>t,$ ${\dot{\lambda }}_{s}>0,$ ${\dot{K}}_{s}>0,$ ${\dot{C}}_{s}\le 0.$ This cannot be optimal since one could instead choose $C=C_{MGR}$ over $[t,\infty )$ and get a higher utility.

Since $\lambda$ is monotonic, C converges monotonically to a limit $C_{\infty }.$ Finally, one can show that K is monotonic too at least for t greater than some T. Assume to fix ideas that C is nondecreasing. Assume there exists t such that ${\dot{K}}_{t}=0.$ (Otherwise K is clearly monotonic). Then by continuity ${\dot{K}}_{s}\le 0$ for $s\in [t,t+\varepsilon ]$ and $\varepsilon >0$ not too large. Since ${\dot{K}}$ is continuous, one can never have ${\dot{K}}_{s}>0$ for $s>t.$ Indeed, let $s_{\max }=\max \{u<s,{\dot{K}}_{u}\le 0\}$. One must have $s_{\max }<s.$ On the other hand, for some $\varepsilon >0$ small enough ${\dot{K}}_{s_{\max }+\varepsilon }\le 0,$ a contradiction. Therefore ${\dot{K}}\le 0,$ implying K is nonincreasing throughout. Hence K is eventually monotonic in all cases. Since K is bounded, K has a limit.]

1.3 Proof of Proposition 2

The Hamiltonian is given by

$$H = \left( {C_{t} - \beta \frac{{C_{t}^{2} }}{{2N_{t} }}} \right)e^{{ - \rho t}} + \lambda _{t} e^{{ - \rho t}} (AK_{t}^{\alpha } - C_{t} - \delta K_{t} - R_{t} ) + \varepsilon _{t} e^{{ - \rho t}} \gamma R_{t} + \mu _{t} e^{{ - \rho t}} C_{t}$$

The representative consumer maximizes H subject to $R_{t}\ge 0$ and $C_{t}\ge 0.$ Let $c_{t}=C_{t}/N_{t},$ $r(K)=\alpha AK^{\alpha -1}-\delta .$ Using the same steps as in the proof of Lemma 1 to eliminate $\mu _{t},$ the first order conditions are

$$\begin{aligned} c_{t}= & {} \max \left( 0,\frac{1-\lambda _{t}}{\beta }\right) \\ -{\dot{\lambda }}_{t}+\rho \lambda _{t}= & {} \lambda _{t}r(K_{t}); \\ \lambda _{t}\ge & {} \gamma \varepsilon _{t}\text { and }\left( \lambda _{t}-\gamma \varepsilon _{t}\right) R_{t}=0; \\ -{\dot{\varepsilon }}_{t}+\rho \varepsilon _{t}= & {} \frac{\beta c_{t}^{2}}{2}. \end{aligned}$$

Consider any steady state. Then either $\lambda >\gamma \varepsilon$ or $\lambda =\gamma \varepsilon .$

Assume $\lambda >\gamma \varepsilon .$ Since $\varepsilon \ge 0$, $\lambda >0.$ Therefore, the steady state is non satiated. Then clearly $\rho =r(K),$ implying $K=K_{MGR}.$ Since ${\dot{K}}={\dot{N}}=0,$ it must be that $C=C_{MGR}.$ In particular, $c=C/N>0,$ so that $\lambda =1-\beta c.$ Since $\rho \varepsilon =\frac{\beta c^{2}}{2},$ $\lambda =1-\beta c$ and $\lambda >\gamma \varepsilon ,$ it must be that $\rho (1-\beta c)>\frac{\gamma \beta c^{2}}{2},$ or equivalently $c<c^{*}.$ Therefore $N=C/c=C_{MGR}/c>C_{MGR}/c^{*}=N^{*}.$

Assume $\lambda =\gamma \varepsilon .$ Since $\rho \varepsilon =\beta c^{2}/2,$ one cannot have $c=0.$ Otherwise one would have $\lambda =\varepsilon =0,$ implying $c=1/\beta ,$ a contradiction. Therefore $c>0,$ implying $\varepsilon >0$ and therefore $\lambda >0$ and satiation cannot hold. Again $\rho =r(K),$ $K=K_{MGR}$ and $C=C_{MGR}.$ Since $\lambda =1-\beta c=\gamma \varepsilon =\gamma \beta c^{2}/(2\rho ),$ $c=c^{*}$ and $N=C_{MGR}/c^{*}=N^{*}.$ $\square$

1.4 The problem with no price cap

Assume output is the numéraire. Let $p_{it}$ be the price of good i at date t. At date t the demand for good i must be solution to the following problem

$$\begin{aligned}&\max \int _{0}^{N_{t}}u(c_{it})di, \\&s.t.\int _{0}^{N_{t}}p_{it}c_{it} \le S_{t}, \end{aligned}$$

where $S_{t}$ is the consumer’s total spending at date t. While $S_{t}$ is endogenous, its optimal level does not change when an infinitesimal producer changes its price, since such a change only has negligible effects on the consumer’s budget constraint. Therefore, $S_{t}$ is treated as given by the producer of good i when setting $p_{i}.$

The solution to this problem is

$$\begin{aligned} c_{it}=\frac{1-\lambda _{t}p_{it}}{\beta }, \end{aligned}$$

where the Lagrange multiplier is

$$\begin{aligned} \lambda _{t}=\max \left( \frac{\int _{0}^{1}p_{it}di-\beta S_{t}}{ \int _{0}^{1}p_{it}^{2}di},0\right) . \end{aligned}$$

Each individual monopoly sets its price $p_{it}$ so as to solve

$$\begin{aligned} \max _{p_{it}}\pi (p_{it},\lambda _{t})=\frac{1-\lambda _{t}p_{it}}{\beta } (p_{it}-1). \end{aligned}$$

Again, $\lambda$ is treated as fixed by the firm, since the firm is infinitesimal. Clearly, if $\lambda =0,$ the optimal price is infinite, for all i. But this contradicts the fact that if all prices are above $\beta S_{t},\lambda >0.$ Therefore, one must have $\lambda >0$ in equilibrium, implying satiation does not arise. The optimal price is the same across all firms and equal to

$$\begin{aligned} p_{it}=p_{t}=\frac{1+\lambda }{2\lambda } \end{aligned}$$

In such a setting, one can compute steady states and compare it with the first best characterized in Proposition 2.

In steady state, p is constant, and the consumer’s problem is the same as in Sect. 3.3.Therefore the same Euler equation holds, and since there is no satiation, $r=\rho ,$ $K=K_{MGR}$ and $C=C_{MGR}.$

In the innovation regime, the steady state is characterized by the following three conditions

$$\begin{aligned} \lambda p= & {} 1-\beta c \\ p= & {} \frac{1+\lambda }{2\lambda } \\ V= & {} \frac{\pi }{\rho }=\frac{(p-1)c}{\rho }=\frac{1}{\gamma }. \end{aligned}$$

The last expression is the zero profit condition for innovation.

Solving for c,  we find that it is the unique solution to

$$\begin{aligned} \beta c^{2}=\frac{\rho (1-2\beta c)}{\gamma }. \end{aligned}$$

(21)

Comparing with (10), we find that this steady state has a lower c than in the first best. Since in both cases C is the same and equal to $C_{MGR}$, it follows that N is larger in the decentralized equilibrium.

While even at the monopoly price innovators only appropriate a fraction of the social surplus of their invention, at the same time monopoly pricing distorts the price of consumption goods upwards. Markets convey the wrong signal that the opportunity cost of one unit of output in terms of final consumption goods is $1/p<1,$ whereas it should be equal to 1. This tends to generate too much innovation. While under Dixit and Stiglitz preferences the two effects exactly cancel each other (Dixit and Stiglitz 1977), here the latter effect dominates.

1.5 Proof of Proposition 3

A. We first construct a trajectory such that $R=0,$ $N=N_{0},$ and $c=1/\beta .$ Along this trajectory, the law of motion for K is

$$\begin{aligned} {\dot{K}}_{t}=AK_{t}^{\alpha }-\frac{N_{0}}{\beta }-\delta K_{t}. \end{aligned}$$

Since $K_{0}>{\hat{K}},$ one has ${\dot{K}}>0$ throughout and the trajectory converges to ${\tilde{K}},$ which is the larger solution to $N_{0}=\beta (AK^{\alpha }-\delta K).$ In particular, $r({\tilde{K}})<0,$ implying $\tilde{K }>K_{S}.$ Let T be the date at which, along this trajectory, $K_{t}=K_{S}.$

Our equilibrium trajectory coincides with that trajectory for $t\le T.$ Thereafter, we assume that $c_{t}=1/\beta ,$ $K_{t}=K_{S},$ and therefore

$$\begin{aligned} {\dot{N}}_{t}=\gamma \left( AK_{S}^{\alpha }-\frac{N_{t}}{\beta }-\delta K_{S}\right) . \end{aligned}$$

Clearly, N converges monotonically to $N_{S}>N_{0}.$

We now show that this trajectory satisfies all the equilibrium conditions.

First, the Euler condition (14) is always satisfied since ${\dot{c}}=0$ and $c=1/\beta .$ The law of motion (6) is also satisfied. For $t\ge T,$ we have that $r_{t}=r(K_{S})=\frac{\gamma (\mu -1)}{\beta }$ and $\pi _{t}=(\mu -1)/\beta .$ Therefore, from (11), $V_{t}=1/\gamma .$ The economy is in the innovation regime, consistent with our assumption that ${\dot{N}}>0.$ For $t<T,$ integrating (11) yields

$$\begin{aligned} V_{t}=\int _{t}^{T}\frac{\mu -1}{\beta }e^{-\int _{t}^{s}r_{u}du}ds+\frac{1}{ \gamma }e^{-\int _{t}^{T}r_{u}du}. \end{aligned}$$

Since $K_{t}<K_{S},$ $r_{s}>\frac{\gamma (\mu -1)}{\beta }.$ Straightforward algebra then implies that $V_{t}<1/\gamma .$ Consequently, the economy is in the no innovation regime for $t<T,$ consistent with ${\dot{N}}=0.$

Thus, the constructed trajectory satisfies all the equilibrium conditions.

B. Assume now that $K_{0}<{\hat{K}}$ and consider the following system

$$\begin{aligned} {\dot{K}}_{t}= & {} AK_{t}^{\alpha }-N_{0}c_{t}-\delta K_{t}. \end{aligned}$$

(22)

$$\begin{aligned} {\dot{c}}_{t}= & {} \left( \frac{1}{\beta }-c_{t}\right) \left( r(K_{t})-\rho \right) . \end{aligned}$$

(23)

This is a standard system and there exists a unique saddle-path converging to its steady state such that $c=1/\beta$ and $K={\hat{K}}.$ To see this, note that since $N_{0}<\beta C_{MGR},$ ${\hat{K}}<K_{MGR}.$ Check that the eigenvalues of the (triangular) linearized system around the steady state are $r({\hat{K}})>0$ and $-(r({\hat{K}})-\rho )<0.$ Then apply Theorem 7.15 in Acemoglu (2009).

Consider this trajectory for $K<{\hat{K}}.$ In this portion, $c<1/\beta .$ Otherwise, ${\dot{K}}<A{\hat{K}}^{\alpha }-N_{0}/\beta -\delta {\hat{K}}=0$ and the path cannot converge to ${\hat{K}},$ a contradiction. Consequently, since $r(K_{t})>r({\hat{K}})>r(K_{MGR})=\rho ,$ ${\dot{c}}>0.$ Also, ${\ddot{K}}=r(K)\dot{ K}-N_{0}{\dot{c}}.$ If ${\dot{K}}\le 0$ for some $K<{\hat{K}},$ then ${\ddot{K}}<0,$ implying, by induction, that K is decreasing after that date along the trajectory, which again would preclude convergence. Therefore, ${\dot{K}}>0.$ Finally, by using the Cauchy-Lipschitz theorem this saddle path can always be prolonged backwards to the minimum feasible value of K,  $K=0.$ Therefore we can express this trajectory as $c=c_{SP0}(K),$ where $c_{SP0}^{\prime }>0,$ for all $K\in (0,{\hat{K}}]$

Let $c_{SP}(K)=\max (c_{SP0}(K),0).$ Clearly, $c_{SP}(K)=0$ for K lower than some critical $K_{c}$ (If $c_{SP0}$ is always positive, take $K_{c}=0$ ). Since $c>0$ locally around the steady state, clearly $K_{c}<{\hat{K}}.$ It is then straightforward to check that the trajectory defined by $c_{t}=c_{SP}(K)$ and ${\dot{K}}_{t}=AK_{t}^{\alpha }-N_{0}c_{SP}(K)-\delta K_{t}$ satisfies the consumer’s optimality conditions throughout and converges to the steady state such that $c=1/\beta$ and $K={\hat{K}}.$ This new trajectory coincides with the preceding one for $K\ge K_{c},$ and is clearly such that ${\dot{K}}>0$ for $0<K<K_{c}$.

Assume that the trajectory followed by the economy is such that (c, K) are along this saddle-path and $N=N_{0}.$ To prove that this is an equilibrium trajectory, we only need to show that $V\le 1/\gamma$ throughout. From ( 11) we have that

$$\begin{aligned} V_{t}=\int _{t}^{+\infty }(\mu -1)c_{t}e^{-\int _{t}^{s}r_{u}du}ds. \end{aligned}$$

Since $N_{0}<N_{S},$ ${\hat{K}}<K_{S}.$ Therefore $r_{t}=r(K_{t})>r({\hat{K}})> \frac{\gamma (\mu -1)}{\beta }$ and $c_{t}<1/\beta .$ Clearly, then, $V_{t}<1/\gamma .$ $\square$

1.6 Proof of Proposition 4

First, we can construct a unique saddle path $(K_{t},c_{t})$ such that (i) $(K_{t},c_{t})$ satisfies the system (22-23), (ii) $K=K_{0}$ initially and (iii) $c=\frac{\rho }{\gamma (\mu -1)}$ at $K=K_{MGR}.$²⁵ The same steps as in the proof of Proposition 3, B,. allow us to construct a trajectory which satisfies the consumer’s optimality conditions and coincides with that saddle path for $c>0$, as a continuous function $c_{SP}(K)$, defined over $(0,K_{MGR})$ such that $\ c_{SP}\ge 0,$ $c_{SP}^{\prime }>0$ if $c_{SP}>0,$ $c_{SP}(K_{MGR})=\frac{\rho }{\gamma (\mu -1)}$, as well as to prove that ${\dot{K}}>0$ for all $K\in (0,K_{MGR}).$ Furthemore, over the compact set $[K_{0},K_{MGR}],$ the continuous function $AK^{\alpha }-N_{0}c_{SP}(K)-\delta K$ reaches its minimum, denoted by $\varepsilon$. Since this expression defines ${\dot{K}},$ $\varepsilon$ is strictly positive except perhaps if it is reached at $K=K_{MGR}$. But at $K=K_{MGR}$, $c=\frac{\rho }{\gamma (\mu -1)}$ and therefore $\frac{dK}{dt} =(N_{N}-N_{0})\frac{\rho }{\gamma (\mu -1)}>0.$ Therefore $\varepsilon >0,$ implying that the point $c=\frac{\rho }{\gamma (\mu -1)}$, $K=K_{MGR}$ is reached in finite time. Denote by T the date when it is reached.

For $t\le T$ we assume the economy follows this trajectory.

For $t\ge T$ we assume that c and K remain constant at $c=\frac{\rho }{ \gamma (\mu -1)},$ $K=K_{MGR},$ while

$$\begin{aligned} {\dot{N}}=C_{MGR}-N\frac{\rho }{\gamma (\mu -1)}. \end{aligned}$$

Since $r=\rho$ and c is constant, the consumer’s local optimality conditions are clearly satisfied for $t\ge T.$ Since c is C1 as a function t,  they are also satisfied locally around T.

Since the point $c=\frac{\rho }{\gamma (\mu -1)}=c_{N},$ $K=K_{MGR}$ lies on II,  i.e. $r(K_{MGR})=\rho =\gamma (\mu -1)c_{N},$ we clearly have, for $t\ge T,$ $V=\frac{(\mu -1)c_{N}}{r}=\frac{1}{\gamma },$ consistent with the free entry condition for innovation.

Next, we prove that $V<1/\gamma$ for $t<T.$ As in the proof of Proposition 3, A, we have that for $t<T$

$$\begin{aligned} V_{t}=\int _{t}^{T}(\mu -1)c_{t}e^{-\int _{t}^{s}r_{u}du}ds+\frac{1}{\gamma } e^{-\int _{t}^{T}r_{u}du}. \end{aligned}$$

Since $K_{t}<K_{T}=K_{MGR},$ $r_{t}>\rho .$ Furthermore, $c_{t}<\frac{\rho }{ \gamma (\mu -1)}.$ Therefore $V_{t}<\int _{t}^{T}\frac{\rho }{\gamma } e^{-\rho (s-t)}ds+\frac{1}{\gamma }e^{-\rho (T-t)}=1/\gamma .$

To complete the proof, we have to show that in partial equilibrium the consumer will prefer to pick the constructed equilibrium trajectory for $c_{t}$ as opposed to $c_{t}=1/\beta$ throughout. This latter choice needs to be ruled out, since it satisfies the local necessary conditions for optimality and, if feasible, would clearly deliver a higher utility level. Consider a consumer who would pick $c_{t}=1/\beta .$ His wealth $W_{t},$ which includes productive capital as well as patents, i.e. claims on existing producers of differentiated goods, evolves as

$$\begin{aligned} {\dot{W}}_{t}=r_{t}W_{t}+w_{t}-N_{t}/\beta . \end{aligned}$$

A sufficient condition for this strategy to violate the consumer’s No Ponzi Game condition is that in steady state

$$\begin{aligned} w<N/\beta . \end{aligned}$$

(24)

To check this, note that $N/\beta =N_{N}/\beta =C_{MGR}\frac{\gamma (\mu -1) }{\rho \beta }.$ By assumption, $\frac{\gamma (\mu -1)}{\rho \beta }>1.$ Since the modified golden rule has less capital than the golden rule, profits are larger than investment (see Abel et al. (1989)), hence consumption is greater than the wage bill. Therefore $C_{MGR}>w,$ so that ( 24) holds.

Finally, the property that $N_{N}>N_{S}$ is straightforward. By assumption, $K_{S}=r^{-1}(\frac{\gamma (\mu -1)}{\beta })<K_{MGR}=r^{-1}(\rho )<K^{*},$ implying $C_{S}<C_{MGR}.$ Therefore $N_{N}=C_{MGR}\frac{\gamma (\mu -1) }{\rho }>\beta C_{MGR}>\beta C_{S}=N_{S}.$ $\square$

1.7 Proof of Proposition 7

Consider a trajectory such that (i), (ii) and (v) hold. It follows from (i) that the net marginal product of capital satisfies (iii). It is also true from (ii) that the consumption Euler equation (14) holds. Property (iv) holds by construction and from the production function $Y_{t}=A_{t}K_{t}^{\alpha }.$ In particular Y is proportional to K,  $Y= \frac{r+\delta }{\alpha }K=yA.$ From (iii) and (ii) we get that the equilibrium condition for innovation $V=\pi /r=1/\gamma$ holds. Given the equilibrium values of r and g,  the property $n>0$ follows from the assumption that $g_{A}<\frac{1-\alpha }{\alpha }\left( \frac{\gamma (\mu -1) }{\beta }+\delta (1-\alpha )\right) .$ To complete the proof, just substitute the values of $N=nA,$ $K=kA,Y=yA$ into the capital accumulation equation (6), which can be rewritten as

$$\begin{aligned} {\dot{K}}_{t}=A_{t}K_{t}^{\alpha }-C_{t}-\delta K_{t}-{\dot{N}}_{t}/\gamma , \end{aligned}$$

(25)

and check that it is satisfied. $\square$

1.8 Proof of Proposition 8

The proof is similar to that of Proposition 7, with the following differences. The Euler equation now holds because $r=\rho .$ The equilibrium consumption level is the one which guarantees equilibrium innovation at $r=\rho .$ The assumption $\rho <\frac{\gamma (\mu -1)}{\beta }$ guarantees that this level is below satiation. In addition, the consumer’s transversality condition has to hold (it always holds under satiation). For this to be the case we need that $r>g,$ which is implied by the assumption that $g_{A}<(1-\alpha )\rho .$ This in turn guarantees that $(\rho +\delta )/\alpha -g-\delta >0,$ i.e. that N is positive along the trajectory. $\square$

1.9 Proof of Proposition 9

We construct a trajectory based on a path $\{c_{t}\}.$ Clearly, the only degree of freedom is in picking the initial value of c,  $c_{0}$, since the subsequent trajectory for c is pinned down by the Euler equation. Let us impose that the economy is in the innovation regime throughout. Then $r_{t}=\gamma (\mu -1)c_{t},$ implying that

$$\begin{aligned} K_{t}=\left( \frac{\alpha A_{t}}{\gamma (\mu -1)c_{t}+\delta }\right) ^{ \frac{1}{1-\alpha }}. \end{aligned}$$

(26)

Since K is a state variable, this pins down the initial c,  $c_{0}.$ From the Euler equation (14), provided $c_{0}>\frac{\rho }{\gamma (\mu -1)} ,$ $r>\rho$ throughout and $c_{t}\rightarrow 1/\beta .$ Furthermore, from the expression for k in (15), for $k_{0}=K_{0}/A_{0}^{\frac{1}{ 1-\alpha }}$ above and close enough to k,  $c_{0}$ will be below and close to $1/\beta ,$ and therefore above $\rho /(\gamma (\mu -1)),$ implying indeed that $c_{t}\longrightarrow 1/\beta .$

To have $L_{t}=1,$ throughout, we need that $w_{t}(1-\beta c_{t})>\mu \tau .$ Note that

$$\begin{aligned} w_{t}= & {} (1-\alpha )A_{t}K_{t}^{\alpha } \\= & {} \left( \frac{\alpha }{\gamma (\mu -1)c_{t}+\delta }\right) ^{\frac{\alpha }{1-\alpha }}(1-\alpha )A_{t}^{\frac{1}{1-\alpha }}. \end{aligned}$$

I now show that $w_{t}(1-\beta c_{t})$ grows over time. Let $\ Q_{t}=A_{t}^{ \frac{1}{1-\alpha }}\left( \gamma (\mu -1)c_{t}+\delta \right) ^{-\frac{ \alpha }{1-\alpha }}(1-\beta c_{t})$. From the fact that $r_{t}=\gamma (\mu -1)c_{t}$ and the Euler condition (14) we get that

$$\begin{aligned} \frac{{\dot{Q}}}{Q}= & {} \frac{1}{1-\alpha }g_{A}-\frac{\alpha }{1-\alpha }\left( \frac{1}{\beta }-c_{t}\right) \frac{\gamma (\mu -1)c_{t}}{\gamma (\mu -1)c_{t}+\delta }\left( \gamma (\mu -1)-\rho /c_{t}\right) \\&-\frac{\beta }{ 1-\beta c_{t}}\left( \frac{1}{\beta }-c_{t}\right) (\gamma (\mu -1)c_{t}-\rho ) \end{aligned}$$

A sufficient condition for Q to grow over time is therefore

$$\begin{aligned} \frac{1}{1-\alpha }g_{A}-\frac{\alpha }{1-\alpha }\left( \frac{1}{\beta } -c_{t}\right) \left( \gamma (\mu -1)-\rho /c_{t}\right) -(\gamma (\mu -1)c_{t}-\rho )>0. \end{aligned}$$

Clearly, since we have assumed that $g_{A}>(1-\alpha )\left[ \frac{\gamma (\mu -1)}{\beta }-\rho \right] ,$ this holds if c close enough to $1/\beta ,$ i.e. again $K/A^{\frac{1}{1-\alpha }}$ close enough to k. Along our trajectory c goes up and $K/A^{\frac{1}{1-\alpha }}$ falls. Therefore, we can always pick initial conditions so that these two quantities are arbitrarily close to $1/\beta$ and k,  respectively. In such a case, Q and therefore $w(1-\beta c)$ grow over time, so that the inequality $w_{t}(1-\beta c_{t})>\mu \tau$ will hold for all t,  provided it holds initially, that is

$$\begin{aligned} \left( \frac{\alpha }{\gamma (\mu -1)c_{0}+\delta }\right) ^{\frac{\alpha }{ 1-\alpha }}(1-\alpha )A_{0}^{\frac{1}{1-\alpha }}(1-\beta c_{0})>0. \end{aligned}$$

Clearly, given the initial value of $k_{0}$ and hence of $c_{0},$ the preceding inequality holds for $A_{0}$ large enough.

To summarize, there exists $\eta$ such that we can choose $k_{0}\in (k,k+\eta )$ and $A_{0}>A(k_{0})$ in such a way that for $K_{0}=k_{0}A_{0}^{ \frac{1}{1-\alpha }},$ the trajectory for $K_{t}$ defined by (26), with $\{c_{t}\}$ the unique solution to (14) such that (26) holds at $t=0,$ is such that the representative consumer would pick $L=1$ throughout. Furthermore, by construction the economy is in the innovation regime throughout, and since $c_{t}\longrightarrow 1/\beta ,$ $k_{t}=K_{t}/A_{t}^{\frac{1}{1-\alpha }}\longrightarrow k.$

To complete our construction, we have to check the feasibility of the implied trajectory for $N_{t}.$ Let $g=g_{A}/(1-\alpha )$ and $n_{t}=N_{t}/A_{t}^{\frac{1}{1-\alpha }}.$ Using (25) we get that

$$\begin{aligned} {\dot{n}}_{t}/\gamma =k_{t}^{\alpha }-\delta k_{t}-{\dot{k}} _{t}-gk_{t}-c_{t}n_{t}-gn_{t}/\gamma . \end{aligned}$$

Note that $n=\frac{k^{\alpha }-(g+\delta )k}{1/\beta +g/\gamma }.$ Clearly, then, as $c_{t}\rightarrow 1/\beta$ and $k_{t}\rightarrow k,n_{t}\rightarrow n.$ Since $\frac{1}{\alpha }\left( \frac{\gamma (\mu -1)}{ \beta }+\delta (1-\alpha )\right) >\frac{g_{A}}{1-\alpha },$ by (16) $n>0.$ For the trajectory for N to be feasible, we need that ${\dot{N}} _{t}\ge 0$ throughout. Note that we can always choose $\eta$ small enough such that the quantity $k_{t}^{\alpha }-(g+\delta )k_{t}$ is arbitrarily close to $(1/\beta +g/\gamma )n$ and $c_{t}$ is arbitrarily close to $1/\beta .$ Then

$$\begin{aligned} \frac{{\dot{n}}_{t}}{\gamma n_{t}}=-\frac{{\dot{k}}_{t}}{n_{t}}+\left( \frac{ k_{t}^{\alpha }-(g+\delta )k_{t}}{n_{t}}-\frac{(1/\beta +g/\gamma )n}{n_{t}} \right) +\left( \frac{(1/\beta +g/\gamma )n}{n_{t}}-(c_{t}+\frac{g}{\gamma } )\right) . \end{aligned}$$

If $n_{t}$ is sufficiently close to n,  the two terms in brackets are arbitrarily small. Since the first term is $>0,$ this can be made larger than any negative number. In particular, larger than $-g/\gamma .$ Thus we can pick $k_{0}$ and $n_{0}$ simultaneously close enough to k and n to make sure that ${\dot{N}}/N={\dot{n}}/n+g>0.$ $\square$

1.10 Proof of Proposition 10

We construct our trajectory in a similar fashion as for Proposition  9. Given a path for $c_{t},$ we make sure that we are in the innovation regime and that there is an interior solution for labor supply. That is

$$\begin{aligned} r_{t}= & {} \alpha A_{t}K_{t}^{\alpha -1}L_{t}^{1-\alpha }-\delta \\= & {} \gamma (\mu -1)c_{t} \end{aligned}$$

and

$$\begin{aligned} L_{t}= & {} \frac{w_{t}}{\tau \mu }(1-\beta c_{t}) \\= & {} \left[ \frac{(1-\alpha )A_{t}}{\tau \mu }\right] ^{\frac{1}{1+\alpha } }K_{t}^{\frac{\alpha }{1+\alpha }}(1-\beta c_{t})^{\frac{1}{1+\alpha }}, \end{aligned}$$

where the marginal productivity condition $w_{t}=(1-\alpha )A_{t}K_{t}^{\alpha }L_{t}^{-\alpha }$ has been used to derive this last expression. In turn, we can solve for K and L as a function of c : 

$$\begin{aligned} K_{t}=\varphi A_{t}^{\frac{2}{1-\alpha }}(1-\beta c_{t})(\gamma (\mu -1)c_{t}+\delta )^{-\frac{1+\alpha }{1-\alpha }}, \end{aligned}$$

(27)

$$\begin{aligned} L_{t}=\psi A_{t}^{\frac{1}{1-\alpha }}(1-\beta c_{t})(\gamma (\mu -1)c_{t}+\delta )^{-\frac{\alpha }{1-\alpha }}. \end{aligned}$$

(28)

The composite parameters $\varphi$ and $\psi$ are defined as follows:

$$\begin{aligned} \varphi= & {} \alpha ^{\frac{1+\alpha }{1-\alpha }}\left( \frac{1-\alpha }{\tau \mu }\right) , \\ \psi= & {} \left( \frac{1-\alpha }{\tau \mu }\right) ^{\frac{1}{1+\alpha } }\varphi ^{\frac{\alpha }{1+\alpha }}. \end{aligned}$$

We note that (27) defines K as a decreasing function of c which maps $(0,1/\beta )$ onto $(0,K_{\max }),$ where $K_{\max }=\varphi A_{t}^{ \frac{2}{1-\alpha }}\delta ^{-\frac{1+\alpha }{1-\alpha }}.$ Therefore, for $K_{0}<K_{\max }$ there exists a unique $c_{0}$ associated with the initial capital stock $K_{0}.$

Since (14) here is equivalent to

$$\begin{aligned} {\dot{c}}_{t}=\left( \frac{1}{\beta }-c_{t}\right) \left( \gamma (\mu -1)c_{t}-\rho \right) , \end{aligned}$$

(29)

if $\kappa _{0}=K_{0}/A_{0}^{\frac{2}{1-\alpha }}$ is not too large,  i.e. $\kappa _{0}<\varphi (1-\beta \frac{\rho }{\gamma (\mu -1)})(\rho +\delta )^{- \frac{1+\alpha }{1-\alpha }},$ then $c_{0}>\rho /(\gamma (\mu -1))$ and we clearly have that ${\dot{c}}>0$ throughout and $c_{t}\rightarrow 1/\beta .$ Then, using (29) and (27), we have that

$$\begin{aligned} \frac{{\dot{K}}_{t}}{K_{t}}=\frac{2}{1-\alpha }g_{A}-(\gamma (\mu -1)c_{t}-\rho )-\frac{1+\alpha }{1-\alpha }\frac{\left( \gamma (\mu -1)c_{t}-\rho \right) (1/\beta -c_{t})}{\gamma (\mu -1)c_{t}+\delta }. \end{aligned}$$

(30)

Therefore

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{{\dot{K}}_{t}}{K_{t}}=\frac{2}{1-\alpha } g_{A}-\left( \frac{\gamma (\mu -1)}{\beta }-\rho \right) =g>0, \end{aligned}$$

(31)

where the inequality comes from the assumption that $g_{A}>\frac{1-\alpha }{2 }(\gamma (\mu -1)/\beta -\rho ).$ A corollary is that for $0<g^{\prime }<g,$ if $c_{t}$ is close enough to $1/\beta ,$ the RHS of (30) is $>g^{\prime }.$ Since, from (27), for $\kappa _{0}$ small enough $c_{0}$ is arbitrarily close to $1/\beta$, we can always pick $\kappa _{0}$ small enough so that $\frac{{\dot{K}}_{t}}{K_{t}}>g^{\prime }>0$ throughout our entire constructed trajectory.

It is also possible to compute $Y_{t}=A_{t}K_{t}^{\alpha }L_{t}^{1-\alpha },$ and we get

$$\begin{aligned} Y_{t}=\chi A_{t}^{\frac{2}{1-\alpha }}(1-\beta c_{t})(\gamma (\mu -1)c_{t}+\delta )^{-\frac{2\alpha }{1-\alpha }}, \end{aligned}$$

where

$$\begin{aligned} \chi =\varphi ^{\alpha }\psi ^{1-\alpha }. \end{aligned}$$

Again $\lim _{t\rightarrow \infty }\frac{{\dot{Y}}_{t}}{Y_{t}}=g.$

Turning now to labor supply, we have to make sure that $L_{t}$ as defined by (28) remains $<1$ throughout. We have that

$$\begin{aligned} \frac{{\dot{L}}_{t}}{L_{t}}=\frac{1}{1-\alpha }g_{A}-(\gamma (\mu -1)c_{t}-\rho )-\frac{\alpha }{1-\alpha }\frac{\left( \gamma (\mu -1)c_{t}-\rho \right) (1/\beta -c_{t})}{\gamma (\mu -1)c_{t}+\delta }. \end{aligned}$$

(32)

Consequently,

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{{\dot{L}}_{t}}{L_{t}}=\frac{1}{1-\alpha } g_{A}-\left( \frac{\gamma (\mu -1)}{\beta }-\rho \right) =g_{L}<0, \end{aligned}$$

where the inequality comes from the assumption that $g_{A}<(1-\alpha )(\gamma (\mu -1)/\beta -\rho ).$

Clearly, then, if $c_{t}$ is close enough to $1/\beta ,$ the RHS of (32 ) is negative. Since $c_{t}$ grows over time, all we need for this to be the case for all t is that $c_{0}$ is close enough to $1/\beta ,$ or equivalently, again, that $\kappa _{0}$ is not too large. Then, for $L_{t}$ to be lower than 1, we just need it to be lower than 1 initially. Substituting (27) into (28), we get that

$$\begin{aligned} L_{t}=\frac{\psi }{\varphi }A_{t}^{-\frac{1}{1-\alpha }}K_{t}(\gamma (\mu -1)c_{t}+\delta )^{\frac{1}{1-\alpha }}. \end{aligned}$$

Thus we have $L_{0}<1$ provided $\lambda _{0}=K_{0}A_{0}^{-\frac{1}{1-\alpha }\text { }}$is small enough. Denoting by $\eta$ and $\varsigma$ the upper bounds for $\kappa _{0}$ and $\lambda _{0}$ respectively, we can summarize our sufficient conditions on initial capital as $K_{0}\le \min (\eta A_{0}^{ \frac{2}{1-\alpha }},\varsigma A_{0}^{\frac{1}{1-\alpha }}).$

To conclude the proof, we need to check that the implied trajectory for $N_{t}$ is feasible. Let $z_{t}=N_{t}/K_{t}.$ The law of motion (25) is now replaced with

$$\begin{aligned} {\dot{K}}_{t}=A_{t}K_{t}^{\alpha }L_{t}^{1-\alpha }-C_{t}-\delta K_{t}-{\dot{N}} _{t}/\gamma . \end{aligned}$$

(33)

Using (27) and (28), it can be rewritten as

$$\begin{aligned} \frac{{\dot{z}}_{t}}{\gamma }=\frac{1}{\alpha }(\gamma (\mu -1)c_{t}+\delta )-z_{t}\left( c_{t}+\frac{1}{\gamma }\frac{{\dot{K}}_{t}}{K_{t}}\right) -\delta -\frac{\dot{ K}_{t}}{K_{t}}. \end{aligned}$$

Clearly, then $z_{t}\rightarrow \frac{\frac{1}{\alpha }(\frac{\gamma (\mu -1) }{\beta }+\delta )-g-\delta }{1/\beta +g/\gamma }=z.$ From the definition of g in (31), this expression is positive iff

$$\begin{aligned} g_{A}<\frac{1-\alpha }{2}\left[ \frac{1+\alpha }{\alpha }\frac{\gamma (\mu -1)}{\beta }+\frac{1-\alpha }{\alpha }\delta -\rho \right] . \end{aligned}$$

The expression on the RHS is always larger than $(1-\alpha )\left( \frac{ \gamma (\mu -1)}{\beta }-\rho \right) ,$ which is $>g_{A}$ by (19).

Recall that by picking $\kappa _{0}$ small enough, we can have $c_{t}$ arbitrarily close to $1/\beta$ and $\frac{{\dot{K}}_{t}}{K_{t}}$ arbitrarily close to g throughout the entire trajectory. If in addition to that, we pick $z_{0}$ close enough to z,  i.e. $N_{0}\in (K_{0}z(1-\varepsilon ),K_{0}z(1+\varepsilon ))$ for some $\varepsilon >0,$ then ${\dot{z}} _{t}/z_{t}$ will be arbitrarily close to zero. Consequently, $N_{t}=z_{t}K_{t}$ will grow at a strictly positive rate throughout. This proves that the trajectory for N is feasible. $\square$