A politico-economic model of aging, technology adoption and growth

Abstract

This paper provides a politico-economic theory that explains how an economy evolves when the longevity of its citizens is jointly determined with the process of economic development. We propose a three-period overlapping generation model where agents’ decisions embrace two dimensions: a private choice about education and a public one on innovation policy. We find that (a) poverty traps can emerge in human capital accumulation, (b) higher life expectancy increases the incentive to innovate for both young and adults, (c) different political configurations can arise depending on endogenous demographic structures and (d) the steady state can entertain both innovation and its absence.

Appendix

Proof of Proposition 1

For simplicity, we drop the time index and substitute H with h. Let $p\left( h\right) =\frac{p^{L}+p^{H}\left( \frac{h}{h^{F}}\right) ^{\sigma }\left(\frac{\sigma -1}{\sigma +1}\right)}{1+\left( \frac{h}{h^{F}}\right) ^{\sigma }\left( \frac{\sigma -1}{\sigma +1}\right)}$, with σ > 1 and 0 < h ^F < ∞. Straightforward algebra leads to $p^{\prime }\left( h\right) \geq 0$, $p^{\prime \prime }\left(h\right) \gtreqless 0$ for $h\lesseqgtr h^{F}$, and $p^{\prime }\left( h^{F}\right) =\frac{\left( p^{H}-p^{L}\right)}{4h^{F}}\left(\frac{\sigma ^{2}-1}{\sigma }\right)$. h ^F is therefore the value of h such that $p\left( h\right)$ shows an inflection point. Note that $p^{\prime }\left( h\right) |_{h=h^{F}}>0$ and $\frac{\partial p^{\prime }\left( h\right) |_{h=h^{F}}}{\partial \sigma }>0$. From Eq. 10, we build the function $\tilde{\Gamma} \left( p\left( h\right) ;h\right) =\Gamma \left( h_{t};i_{t}\right) h_{t}^{\epsilon }$, where we (a) separate the dependency from human capital and life expectancy and (b) drop the innovation variable i _t . Varying λ we obtain that for h = h ^F, the limiting functions $\tilde{\Gamma}\left(p^{L};h\right)$ and $\tilde{\Gamma}\left(p^{H};h\right)$ take values below and above h ^F, respectively, i.e. $\tilde{\Gamma}\left( p^{L};h^{F}\right) <h^{F}$ and $\tilde{\Gamma}\left( p^{H};h^{F}\right) >h^{F}$. Since $\lim_{\sigma \rightarrow +\infty }p^{\prime}\left( h^{F}\right) \rightarrow +\infty$, the function $\tilde{\Gamma}\left( p\left(h\right);h\right)$ takes values $\tilde{\Gamma}\left( p\left( h^{F}-\bigtriangleup h\right) ;\left( h^{F}-\bigtriangleup h\right) \right) <h^{F}$ and $\tilde{\Gamma}( p( h^{F}+\bigtriangleup h) ;( h^{F}+\bigtriangleup h) ) >h^{F}$ for any ∆ h = o(h) > 0 and $1<\sigma ^{M}\left( \bigtriangleup h\right) <\sigma <+\infty$, where $\sigma ^{M}\left( \bigtriangleup h\right)$ is a threshold related to the $\left(\text{arbitrary small}\right)$ magnitude of ∆ h. For continuity of $\tilde{\Gamma}\left( p\left( h\right) ;h\right)$, there is a steady state at h ^F where function $\tilde{\Gamma}\left( p\left( h\right);h\right)$ crosses the 45 degrees line from below. This steady state is therefore unstable. Calculus inspection shows that $\frac{\partial \left( \tilde{\Gamma}\left( p\left( h\right) ;h\right) \right) }{\partial h} >0$ in [0; ∞ ), $\lim_{h\rightarrow 0^{+}}\frac{\partial \left( \tilde{ \Gamma}\left( p\left( h\right) ;h\right) \right) }{\partial h}\rightarrow +\infty$ and $\lim_{h\rightarrow +\infty }\frac{\partial \left( \tilde{ \Gamma}\left( p\left( h\right) ;h\right) \right) }{\partial h}\rightarrow 0^{+}$. With $\tilde{\Gamma}\left( p\left( 0\right) ;0\right) =0$, by adopting the same logic, it is straightforward to prove that $\tilde{\Gamma} \left( p\left( h\right) ;h\right)$ shows four steady states, alternatively unstable and stable: h ^U0 = 0, 0 < h ^S1 < h ^F, h ^U1 = h ^F and h ^F < h ^S2 < + ∞.□

Proof of Lemma 1

Adults get the absolute majority if and only if their share is bigger than $\frac{1}{2}:$ imposing $\frac{1}{\eta +1+p_{t}}>\frac{1}{2}$ we obtain p _t  < 1 − η. For similar considerations, it is easy to show that both $\frac{\eta }{\eta +1+p_{t}}$ and $\frac{p_{t}}{\eta +1+p_{t}}$ can not exceed $\frac{1}{2}$.□

Proof of Lemma 2

The expression of p ^A is obtained from Eq. 16 solving $\Delta u_{t}^{A}\left( p_{t+1}\right) =0$ for p _t + 1. Given Eq. 17 and i > 0, the graph of $\Delta u_{t}^{A}\left( p_{t+1}\right)$ has a negative intercept and crosses the $\Delta u_{t}^{A}=0$ axis from below.□

Proof of Lemma 3

The expression of p ^Y is obtained from Eq. 21 solving $\Delta u_{t}^{Y}\left( p_{t+2}\right) =0$ for p _t + 2.□

Proof of Proposition 2

The strategy we follow to prove the first part of the Proposition is to break the two inequalities and to show that both cannot simultaneously hold for any parametrization of the model. Let us define

$$ \Psi \left(\alpha \right) _{1}=\alpha ^{2}\left( 1-p^{Y}\right) =\alpha ^{2}+\alpha \frac{\log \left( \left( 1+\theta \right) \left(1-\delta \right) ^{\gamma }\right) }{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\epsilon \gamma }\right) }+\frac{\log \left( 1-i\right) }{ \log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma \epsilon }\right)} $$

and

$$ \begin{aligned} \Psi \left( \alpha \right) _{2} &=\alpha ^{2}\left( p^{A}-p^{Y}\right) \\ &=\alpha \left( \frac{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) }{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\epsilon \gamma }\right) }+\frac{\log \left( \frac{1-s}{ 1-s-i}\right) }{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) }\right) \cr &+ \frac{\log \left( 1-i\right) }{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\epsilon \gamma }\right) } \end{aligned} $$

using Eqs. 19 and 23 and substituting β = α ². We write the dependency of both Ψ₁ and Ψ₂ on α for brevity. It turns out that the two inequalities p ^A < p ^Y and p ^Y < 1 are both satisfied if both Ψ₂ < 0 and Ψ₁ > 0 hold, respectively. In Fig. 6, we show the shapes of these two functions in terms of α.□

Fig. 6

Shapes of Ψ₁ and Ψ₂, their intersections with the axis and their crossing points

Proof

The first order derivatives of Ψ₁ and Ψ₂ with respect to α are $\Psi _{1}^{\prime }=\frac{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) }{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\epsilon \gamma }\right) }+2\alpha$ and $\Psi _{2}^{\prime }=\frac{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) }{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\epsilon \gamma }\right) }+\frac{\log \left( \frac{ 1-s}{1-s-i}\right) }{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) }$. The second derivatives are $\Psi _{1}^{\prime \prime }=2$ and $\Psi _{2}^{\prime \prime }=0$, implying that Ψ₂ is a quadratic function of α and for α > 0 it is increasing at an increasing rate. Ψ₁ is a positively sloped straight line. For α = 0, both Ψ₁ and Ψ₂ take the same value $\Psi \left( 0\right) _{1}=\Psi \left( 0\right) _{2}=\Psi \left( 0\right) _{n}=- \frac{\log \left( \frac{1}{1-i}\right) }{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\epsilon \gamma }\right) }<0$, as shown in the graph. Moreover, Ψ₂ is steeper than Ψ₁, for any values of the parameters and for some small values of α $\left( \text{from an inspection of }\Psi _{1}^{\prime }\text{ and }\Psi _{2}^{\prime }, \text{until }\frac{\log \left( \frac{1-s}{1-s-i}\right) }{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) }>2\alpha \right)$. Therefore, for α:0 ≤ α < α ₂, Ψ₂ < 0 holds, while it is not the case for Ψ₁ > 0. Because of their shapes and their crossing in α = 0, Ψ₁ and Ψ₂ cross again for some positive value of α. If the crossing point $\left( \Psi \left( \alpha _{c}\right) _{n}\right)$ of Ψ₁ and Ψ₂ lies below the α-axis and α _c  < 1, then there are values of α ∈ (α ₁;α ₂) such that both Ψ₂ < 0 and Ψ₁ > 0 hold at the same time. This cannot be the case because equating Ψ₁ to Ψ₂ gives $\alpha _{c}=\frac{\log \left( \frac{1-s}{1-s-i}\right) }{\log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) }$, that plugged into Ψ₁ or Ψ₂ gives $\Psi \left( \alpha _{c}\right) _{1}=\Psi \left( \alpha _{c}\right) _{2}=\Psi \left( \alpha _{c}\right) _{n}=\frac{\log \left( \frac{1-s}{1-s-i}\right) ^{2}}{ \left( \log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) \right) ^{2}}+\frac{\log \left( \frac{1-s-i+is}{1-s-i}\right) }{ \log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\epsilon \gamma }\right) }>0$, for any values of θ,s,i,δ,γ and ε in their supports. The first part of the Proposition is therefore proved. The second part relies on splitting the inequality p ^A < p ^Y < 1 in two. We rewrite p ^A < p ^Y as $\frac{m^{Y}}{m^{A}}<-\frac{\log \left( 1-i\right) +\alpha \log \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\epsilon \gamma }\right) }{\alpha \log \left( \frac{1-s}{1-s-i} \right) }$ that is satisfied for α→0 since the left-hand side is constant and independent from α while the right-hand side goes to + ∞. This result does not depend on β while it depends on i: Although for α→0 the inequality holds, for a range of α such that $i<1-\left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) ^{-\alpha }$, the right-hand side numerator is negative. For some positive values of α, the investment cost cannot be too small. On the other hand, we rewrite p ^Y < 1 as $\left( 1-i\right) ^{-1}<\left( 1+\theta \right) ^{\alpha +\beta }\left( 1-\delta \right) ^{\gamma \left( \alpha +\epsilon \beta \right) }$. With α and β approaching their lower and upper bounds $\left( 0 \text{ and }1\text{, respectively}\right)$, the inequality holds for small i and it also holds for some right interval of α that is compatible with the inequality p ^A < p ^Y. This proves the second part of the Proposition.□

Proof Corollary 1

We need that p ^Y < 0 for some small values of i. Under assumption 17, it is enough to show that q ^Y > 0, by graphical considerations based on Fig. 6. This is true if and only if $\left( 1-i\right) \left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) ^{\alpha }>1.$ By simple algebra, the Corollary is proved.□

Proof of Proposition 3

β = α ² ensures that p ^Y < p ^A. For $i<1-( \left( 1+\theta \right)$ $\left( 1-\delta \right) ^{\gamma }) ^{-\alpha }$, young are in favour of innovation ∀ p ≥ 0. This leads to both a share of voters $\frac{\eta }{\eta +1+p_{t}}$ in favour of innovation for $p_{t+1}<p^{A}$ and a share of voters $\frac{\eta +1}{\eta +1+p_{t+1}}$ in favour of innovation for $p_{t+1}>p^{A}$. Since p ^O = 1 − η, it is straightforward to show that $\frac{\eta }{\eta +1+p_{t}}<\frac{1}{2} ,\forall \eta \in (0;1]$ and $\frac{\eta +1}{\eta +1+p_{t}}>\frac{1}{2} ,\forall \eta \in (0;1]$. In the case of $i>1-\left( \left( 1+\theta \right) \left( 1-\delta \right) ^{\gamma }\right) ^{-\alpha }$ for the positive values of life expectancy $0<p_{t+2}<p^{Y}$, nobody is in favour of innovation, while for $p^{Y}<p_{t+2}$ and $p_{t+1}<p^{A}$, only young are in favour. Being them, a minority of the political outcome is unchanged with respect to the case of nobody backing innovation. Once $p_{t+1}>p^{A}$ is achieved, the previous analysis applies.□

Proof of Lemma 4

When $p_{t+1}<p^{A}$ < p ^Y < 1, there is no way for the economy to support innovation, while with $p_{t+1}>p^{A}$ at least adults vote for innovation. If η < 1 − p ^A, using the definition p ^O ≡ 1 − η, then p ^O is larger than p ^A. In a right interval of p ^A, innovation takes place because adults alone want it: Their voting share is $\frac{1}{ \eta +1+p_{t+1}}$ > 1/2 until $p_{t+1}<p^{O}$. In case p ^A < p ^Y < p ^O, innovation takes again place with $p_{t+1}>p^{A}$, but in the interval {p ^Y;p ^O}, it would arise even if adults had not an absolute majority. Above p ^Y, also young contribute in backing innovation.□

Proof of Proposition 4

The Proof is divided in four points. The assumption of small δ ensures that whenever the economy switches between no innovation to innovation (or vice versa), the ordering of p ^O;p ^A;p ^Y and p ^S does not change.

• 1. $\left( 1.a\right)$ β = α ² ensures, from Proposition (2), that p ^Y < p ^A. Sufficiently large s make p ^A to be larger than p ^S. In turns, small γ lower the steady state level of human capital and therefore p ^S. Independently from η, the orderings p ^Y < p ^S < p ^A and p ^S < p ^Y < p ^A are consistent with a conservative policy outcome for any values of p < p ^S. $\left( 1.b\right)$ β/α ² substantially larger then 1 leads to p ^A < p ^Y: In this case, the inequality p ^S < p ^A, ensured by small γ,ε and/or λ, is a sufficient condition for the case of constant no innovation to be in place. $\left( 1.c\right)$ The strong political power of young impedes adults to decide for innovation alone, since p ^O < p ^A < p ^Y. The limited effectiveness of human capital production ensures p ^S < p ^Y.

• 2. As $\left( 1.a\right)$, in $\left( 2.a\right)$ β = α ² ensures that p ^Y < p ^A. At the same time, large λ ensures that p ^A < p ^S holds. For values of life expectancy above p ^A, output is increasing at rate θ because both adults and young vote for innovation. As in case $\left( 1.b\right)$, β/α ² substantially larger then 1 leads to p ^A < p ^Y also in $\left( 2.b\right)$. Moreover, small α implies a small p ^A, so that for value of life expectancy larger than p ^A innovation is always chosen.

• 3. This case takes place if and only if p ^A < p ^O < p ^S < p ^Y holds. This configuration requires β/α ² larger then 1 as a necessary condition so to have p ^A < p ^Y. Moreover, young’s weight η must be small in order to have p ^O < p ^S. Small α makes the inequality p ^A < p ^O to be satisfied.

• 4. This case takes place if and only if p ^A < p ^O < p ^Y < p ^S holds. Conditions on the parameters are the same as in point $\left( 3\right),$ but α must be slightly larger in order to have young in favour of innovation for values of life expectancy in the interval (p ^Y,p ^S).□