50 is the new 30—long-run trends of schooling and retirement explained by human aging

Abstract

Workers in the US and other developed countries retire no later than a century ago and spend a significantly longer part of their life in school, implying that they stay less years in the work force. The facts of longer schooling and simultaneously shorter working life are seemingly hard to square with the rationality of the standard economic life cycle model. In this paper we propose a novel theory, based on health and aging, that explains these long-run trends. Workers optimally respond to a longer stay in a healthy state of high productivity by obtaining more education and supplying less labor. Better health increases productivity and amplifies the return on education. The health accelerator allows workers to finance educational efforts with less forgone labor supply than in the previous state of shorter healthy life expectancy. When both life-span and healthy life expectancy increase, the health effect is dominating and the working life gets shorter if the intertemporal elasticity of substitution for leisure is sufficiently small or the return on education is sufficiently large. We calibrate the model and show that it is able to predict the historical trends of schooling and retirement.

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Notes

  1. 1.

      There is no contradiction between Lee’s (2001) observation of a (mild) increase in age of retirement and the observed decline in labor force participation of the elderly (Costa 1998). The reason is that the labor force participation rate captures individuals who manage to survive until age 65; individuals who retired and died before the age of 65 do not count. During the period of observation improvements in health allowed more people to survive until the age of 65. This implies that the expected age of retirement increased while labor force participation declined.

  2. 2.

      There is also a direct link from better health in childhood to performance at school (Bleakley 2007; Bharadwaj et al. 2013) and to parental investment in education (Hazan and Zoabi 2006) and thus to adult health and productivity. Andersen et al. (2015) find a strong impact of eye disease on contemporaneous income across countries and argue that it results partly from the delay of the fertility transition caused by the strong exposure to UV light in some countries and the reduced return on human capital due to the entailed high prevalence of eye disease. The trade-off between fertility and education is not explored in the present paper but it is potentially amplifying the positive association between life-expectancy, healthy life expectancy, and productivity.

  3. 3.

      Recent research by Hamermesh (2013) suggests that this is even true for economists. Hamermesh observes a trend towards higher average age of authors publishing in the top five journals and explains this by a trend that made the profession less like pure mathematics (requiring mostly fluid abilities, which decline early in life) and more like a humanistic field (requiring mostly crystallized abilities, which are more persistent).

  4. 4.

      The assumption that \(T>\lambda \) is plausible but not necessary for our results, which hold for \(\lambda \rightarrow T\) as well. Moreover, health could decrease non-linearly without affecting our results. The linearity assumption is made to obtain a closed-form solution. The crucial assumption is that there exists a point in life after which health and productivity decline, which allows us to disentangle the effects of increasing life expectancy and increasing healthy life expectancy.

  5. 5.

    In Appendix F we compare the treatment of depreciation in the standard human capital model vs. our approach in more detail.

  6. 6.

    Better health also increases the hours worked per time increment because of reduced sickness absence. In our model a reduction of sickness absence would be captured by an increase of supply of human capital per time increment, i.e. an upward shift of the a(t)-curve. In the Appendix we show that such a shift leaves all results unaffected.

  7. 7.

    For the benchmark calibration of the model we assume a mild increase of \(\lambda \) together with T. The implied \(\beta \) is given by \( [{\lambda (1970)-\lambda (1850)]}/[{T(1970)-T(1850)}]\) = 0.119. Thus \(\alpha > 0.119+0.03=0.22\) is sufficient for life time labor supply to decline. For the benchmark calibration we use \(\alpha =0.84\), based upon the estimate provided with Fig. 3.

  8. 8.

    Keeping a zero interest rate does not much affect the predicted trends but it implies a too high level of education for all cohorts. The reason is that the opportunity costs of education would be too low when consumption during the schooling period is financed at a zero interest rate.

  9. 9.

    In the Appendix we extend the model by a constant trend of wage growth (due to technological progress). In this case the best fit of the data is obtained for \(\theta =0.0715\), a value slightly smaller than in the benchmark case.

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Acknowledgments

We would like to thank Lothar Banz, Carl Johan Dalgaard, David de la Croix, Paula Gobbi, Moshe Hazan, Ben Heijdra, Gregory Ponthiere, Alexia Prskwetz, and four anonymous referees for helpful comments.

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Appendices

Appendix A: derivation of the optimal solution

The interior solution of the optimization problem of the household is derived from the first order conditions (5) and (6), i.e.

$$\begin{aligned} 0&\mathop {=}\limits ^{!} \theta + \frac{2(\lambda -\tau )}{R^2-2 \lambda R + 2s(\lambda -\tau )+\tau ^2}, \end{aligned}$$
(13)
$$\begin{aligned} 0&\mathop {=}\limits ^{!} \frac{2 T (R-\lambda )}{R^2-2R\lambda +2s(\lambda -\tau )+\tau ^2} - (\lambda -R)^{-\eta } \omega . \end{aligned}$$
(14)

Rearranging Eq. (13) we obtain

$$\begin{aligned} \frac{2}{R^2-2R\lambda + 2s(\lambda -\tau ) + \tau ^2} = -\frac{\theta }{\lambda -\tau }. \end{aligned}$$

Inserting this into Eq. (14) we arrive at

$$\begin{aligned} \omega (\lambda -R)^{-\eta } = - \frac{T \theta (R-\lambda )}{\lambda -\tau }\quad \Leftrightarrow \quad R = \lambda - \left( \frac{(\lambda -\tau )\omega }{T\theta } \right) ^{\frac{1}{1+\eta }}. \end{aligned}$$
(15)

Optimal schooling follows from inserting the optimal retirement age R from (15) into Eq. (14), i.e.

$$\begin{aligned} s = \frac{\lambda + \tau }{2} - \frac{1}{\theta } - \frac{\left( \frac{(\lambda -\tau )\omega }{T\theta } \right) ^{\frac{2}{1+\eta }}}{2(\lambda -\tau )}. \end{aligned}$$
(16)

Therefore, optimal labor supply is, given by \(L=R-s\), is

$$\begin{aligned} L = \frac{1}{\theta } + \frac{1}{2(\lambda -\tau )} \left( \tau -\lambda +\left( \frac{(\lambda -\tau ) \omega }{T \theta } \right) ^{\frac{1}{1+\eta }}\right) ^2. \end{aligned}$$
(17)

If the Hessian of U is negative definite at the critical point (sR), then this point is a local maximum. The Hessian of U is given by

$$\begin{aligned} H_U(s,R):= \begin{pmatrix} \frac{\partial ^2 U}{\partial s^2} &{} \frac{\partial ^2U}{\partial s \partial R}\\ \frac{\partial ^2 U}{\partial R \partial s} &{} \frac{\partial ^2 U}{\partial R^2} \end{pmatrix} =: \begin{pmatrix} H_{1,1} &{} H_{1,2} \\ H_{2,1} &{} H_{2,2} \\ \end{pmatrix}, \end{aligned}$$
(18)

where

$$\begin{aligned} H_{1,1}&= \frac{2 T (-R^2 + 2 R \lambda + 2(s-\lambda )\lambda -2s\tau +\tau ^2)}{(R^2-2R\lambda + 2s(\lambda -\tau )+\tau ^2)^2} - \eta \omega (\lambda -R)^{-1-\eta },\\ H_{1,2}&= - \frac{4 T (R-\lambda )(\lambda -\tau )}{(R^2-2R \lambda + 2s(\lambda -\tau )+\tau ^2)^2},\\ H_{2,1}&= H_{1,2},\\ H_{2,2}&= -\frac{4 T (\lambda -\tau )^2}{(R^2-2R\lambda +2s(\lambda -\tau )+\tau ^2)^2}. \end{aligned}$$

The two principal minors of H evaluated in the critical point are

$$\begin{aligned} H_{U,1}&= -\eta \frac{T \theta }{(\lambda -\tau )} - \frac{T \theta \left( \lambda -\tau +\theta \left( \frac{(\lambda -\tau )\omega }{T \theta } \right) ^{\frac{2}{1+\eta }} \right) }{(\lambda -\tau )^2} <0, \end{aligned}$$
(19)
$$\begin{aligned} H_{U,2}&= \det H(R,s) = \frac{T^2 \theta ^3}{\lambda -\tau } (1 + \eta ) >0. \end{aligned}$$
(20)

Hence, the Hessian is negative definite and the critical point is a maximum.

Appendix B: proofs of the propositions

Proof of Proposition 1

The partial derivatives of Rs and L, c.f. (7) – (9), with respect to T are

$$\begin{aligned} \frac{\partial s}{\partial T}&= \frac{\omega \left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{1-\eta }{1+\eta }}}{T^2 (1+\eta )\theta } >0,\\ \frac{\partial R}{\partial T}&= \frac{\left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{1}{1+\eta }}}{T(1+\eta )} >0,\\ \frac{\partial L}{\partial T}&=-\frac{\omega \left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{-\eta }{1+\eta }}}{T^2(1+\eta )\theta } \left( -\lambda + \tau + \left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{1}{1+\eta }} \right) . \end{aligned}$$

We have \({\partial L}/{\partial T}>0\) because

$$\begin{aligned} \frac{\omega \left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{-\eta }{1+\eta }}}{T^2(1+\eta )\theta } \ge 0 \ \text {and} \ \left( -\lambda + \tau + \left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{1}{1+\eta }} \right) = \tau - R \le 0. \end{aligned}$$

\(\square \)

Proof of Proposition 2

The partial derivative of s, c.f. (8), with respect to \(\tau \) is

$$\begin{aligned} \frac{\partial s}{\partial \tau } = \frac{1}{2} \left( 1+ \frac{(1-\eta )\left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{2}{1+\eta }}}{(1+\eta )(\lambda -\tau )^2} \right) . \end{aligned}$$

For \(\eta \le 1\) it is easy to see that \(\frac{\partial s}{\partial \tau } >0\). If \(\eta >1\), it holds that

$$\begin{aligned} 1+ \frac{(1-\eta )\left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{2}{1+\eta }}}{(1+\eta )(\lambda -\tau )^2} >0 \ \Leftrightarrow \ (1+\eta ) (\lambda -\tau )^2 > (\eta -1)\left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{2}{1+\eta }} \end{aligned}$$

This inequality holds because

$$\begin{aligned} (\eta -1)\left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{2}{1+\eta }} \le (\eta -1)(\lambda -\tau )^2\mathop {<}\limits ^{!}(1+\eta )(\lambda -\tau )^2 \ \Leftrightarrow \ \eta -1<1+\eta . \end{aligned}$$

The partial derivatives of R and L, c.f. (7) and (9), with respect to \(\tau \) are

$$\begin{aligned} \frac{\partial R}{\partial \tau }&= \frac{\left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{1}{1+\eta }}}{(1+\eta )(\lambda -\tau )}>0,\\ \frac{\partial L}{\partial \tau }&= \frac{\left( -\lambda + \tau + \left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{1}{1+\eta }}\right) \left( (1+\eta )(\lambda -\tau ) + (\eta -1)\left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{1}{1+\eta }} \right) }{2(1+\eta )(\lambda -\tau )^2}. \end{aligned}$$

It holds

$$\begin{aligned} \frac{\partial L}{\partial \tau } <0&\Leftrightarrow \ \left( (1+\eta )(\lambda -\tau ) + (\eta -1)\left( \frac{(\lambda -\tau )\omega }{\theta T} \right) ^{\frac{1}{1+\eta }} \right) >0 \\&\Leftrightarrow (1+\eta )(\lambda -\tau ) + (\eta -1)(\lambda -R) >0\\&\Leftrightarrow \frac{\lambda -\tau }{\lambda -R}>\frac{1-\eta }{1+\eta }, \end{aligned}$$

which holds \(\forall \ \eta \ge 0\). \(\square \)

Proof of Proposition 3

An equal increase in \(\tau \) and T is captured by the directional derivative \({\partial L}/{\partial T} + {\partial L}/{\partial \tau }\), i.e.

$$\begin{aligned} \frac{\partial L}{\partial T} + \frac{\partial L}{\partial \tau } = \frac{\left( -\lambda +\tau + \left( \frac{(\lambda -\tau )\omega }{T\theta }\right) ^{\frac{1}{1+\eta }} \right) }{2 T^2(1+\eta )\theta (\lambda -\tau )^2}\cdot \Lambda \end{aligned}$$
(21)

with

$$\begin{aligned} \Lambda :=- \omega \left( \frac{(\lambda -\tau )\omega }{T\theta }\right) ^{\frac{-\eta }{1+\eta }} 2 (\lambda -\tau )^2 + T^2\theta \left( (1+\eta )(\lambda -\tau )+(\eta -1)\left( \frac{(\lambda -\tau )\omega }{T\theta }\right) ^{\frac{1}{1+\eta }} \right) . \end{aligned}$$

For the numerator of the first term of Eq. (21) it holds that

$$\begin{aligned} \left( -\lambda +\tau + \left( \frac{(\lambda -\tau )\omega }{T\theta }\right) ^{\frac{1}{1+\eta }} \right) = \tau -R <0. \end{aligned}$$

The denominator is greater than 0. Therefore, the problem simplifies to

$$\begin{aligned}&- \omega \left( \frac{(\lambda -\tau )\omega }{T\theta }\right) ^{\frac{-\eta }{1+\eta }} 2 (\lambda -\tau )^2 + T^2\theta \left( (1+\eta )(\lambda -\tau )+(\eta -1)\left( \frac{(\lambda -\tau )\omega }{T\theta }\right) ^{\frac{1}{1+\eta }} \right) \mathop {>}\limits ^{!}0 \\&\quad \Leftrightarrow \ \frac{\omega }{T^2 \theta } \left( \frac{(\lambda -\tau )\omega }{T\theta }\right) ^{\frac{-\eta }{1+\eta }} 2(\lambda -\tau )^2 - (\eta -1)\left( \frac{(\lambda -\tau )\omega }{T\theta }\right) ^{\frac{1}{1+\eta }} < (1+\eta )(\lambda -\tau )\\&\quad \Leftrightarrow \ \left( \frac{(\lambda -\tau )\omega }{T\theta }\right) ^{\frac{1}{1+\eta }} \left[ \frac{2(\lambda -\tau )^2 \omega }{T^2\theta } \frac{T^2 \theta }{(\lambda -\tau )\omega }-(\eta -1) \right] < (1+\eta )(\lambda -\tau )\\&\quad \Leftrightarrow \ \left( \frac{(\lambda -\tau )\omega }{T}\right) ^{\frac{1}{1+\eta }} \left[ \frac{2(\lambda -\tau )+T(1-\eta )}{T(1+\eta )(\lambda -\tau )} \right] < \theta ^{\frac{1}{1+\eta }}, \end{aligned}$$

which is always fulfilled for \(\eta \ge \frac{2(\lambda -\tau )}{T}+1\). If \(\eta < \frac{2(\lambda -\tau )}{T}+1\), the last inequality becomes

$$\begin{aligned} \theta > \frac{(\lambda -\tau )\omega }{T} \left[ \frac{2(\lambda -\tau )+T(1-\eta )}{T(1+\eta )(\lambda -\tau )}\right] ^{1+\eta }. \end{aligned}$$

\(\square \)

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Strulik, H., Werner, K. 50 is the new 30—long-run trends of schooling and retirement explained by human aging. J Econ Growth 21, 165–187 (2016). https://doi.org/10.1007/s10887-015-9124-1

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Keywords

  • Healthy life expectancy
  • Longevity
  • Education
  • Retirement
  • Labor supply
  • Compression of morbidity

JEL Classification

  • E20
  • I25
  • J22
  • O10
  • O40