“How powerful is demography? The serendipity theorem revisited” comment on De la Croix et al. (2012)

Abstract

Samuelson’s (Int Econ Rev 16(3):531-538, 1975) serendipity theorem states that the “goldenest golden rule” steady-state equilibrium can be obtained by a competitive two-period overlapping generation economy with capital accumulation, provided that the optimal growth rate prevails. De la Croix et al. (J Popul Econ 25:899-922, 2012) extended the scope of the theorem by showing that it also holds for risky lifetime. With this note, we introduce medical expenditure as a determinant of the probability of surviving to old age to prove the theorem. The original as well as all extended versions of the serendipity theorem, however, fail to prove that second-order conditions are satisfied in general. Still, unlike De la Croix et al. (J Popul Econ 25:899-922, 2012), we can exclude the existence of corner solutions where the probability of reaching old age is zero or one. The zero survival probability case becomes irrelevant if the option to randomize between death and life utility is taken into account. Survival with certainty is ruled out if the marginal cost of survival is increasing. Hence, the optimal survival probability represents an interior solution. Furthermore, we show for the optimal survival probability that the value of a statistical life is positive and equal to its marginal cost.

Appendix

If we assume that the population growth rate is exogenous, three first-order conditions remain for the planner’s problem, Eqs. (3), (4), and (12) in the model with costly survival. Substituting the resource constraint (11) for c, the Hessian matrix of the second-order derivatives associated with this problem is

$$ H=\left[\begin{array}{ccc}\hfill \frac{\partial^2\mathrm{E}\mathrm{U}}{\partial {d}^2}\hfill & \hfill \frac{\partial^2\mathrm{E}\mathrm{U}}{\partial d\partial k}\hfill & \hfill \frac{\partial^2\mathrm{E}\mathrm{U}}{\partial d\partial m}\hfill \\ {}\hfill \frac{\partial^2\mathrm{E}\mathrm{U}}{\partial k\partial d}\hfill & \hfill \frac{\partial^2\mathrm{E}\mathrm{U}}{\partial {k}^2}\hfill & \hfill \frac{\partial^2\mathrm{E}\mathrm{U}}{\partial k\partial m}\hfill \\ {}\hfill \frac{\partial^2\mathrm{E}\mathrm{U}}{\partial m\partial d}\hfill & \hfill \frac{\partial^2\mathrm{E}\mathrm{U}}{\partial m\partial k}\hfill & \hfill \frac{\partial^2\mathrm{E}\mathrm{U}}{\partial {m}^2}\hfill \end{array}\right] $$

Hence, $\left[\begin{array}{ccc}\hfill {\left[\frac{\pi (m)}{1+n}\right]}^2u^{\prime\prime }(c)+\pi (m)u^{\prime\prime }(d)\hfill & \hfill -u^{\prime\prime }(c)\left(f^{\prime }(k)-n\right)\frac{\pi (m)}{1+n}\hfill & \hfill \begin{array}{l}u^{\prime\prime }(c)\frac{\pi (m)}{{\left(1+n\right)}^2}\left[1+n+\pi^{\prime }(m)d\right]\\ {}-u^{\prime }(c)\frac{\pi^{\prime }(m)}{1+n}+u^{\prime }(d)\pi^{\prime }(m)\end{array}\hfill \\ {}\hfill -u^{\prime\prime }(c)\left(f^{\prime }(k)-n\right)\frac{\pi (m)}{1+n}\hfill & \hfill \begin{array}{l}u^{\prime\prime }(c){\left[\left(f\prime (k)-n\right)\right]}^2\\ {}+u^{\prime }(c)f^{\prime\prime }(k)\end{array}\hfill & \hfill -\frac{u^{\prime\prime }(c)}{1+n}\left[1+n+\pi^{\prime }(m)d\right]\left({f}^{\prime }(k)-n\right)\hfill \\ {}\hfill \begin{array}{l}u^{\prime\prime }(c)\frac{\pi (m)}{{\left(1+n\right)}^2}\left[1+n+\pi^{\prime }(m)d\right]\\ {}-u^{\prime }(c)\frac{\pi^{\prime }(m)}{1+n}+{u}^{\prime }(d){\pi}^{\prime }(m)\end{array}\hfill & \hfill -\frac{u^{\prime\prime }(c)}{1+n}\left(f^{\prime }(k)-n\right)\left[1+n+\pi^{\prime }(m)d\right]\hfill & \hfill \begin{array}{l}\frac{u^{\prime\prime }(c)}{{\left(1+n\right)}^2}{\left[1+n+\pi \prime (m)d\right]}^2\\ {}-u^{\prime }(c)\frac{\pi^{\prime\prime }(m)d}{1+n}+\pi^{\prime\prime }(m)u(d)\end{array}\hfill \end{array}\right]$ Substituting the first-order conditions yields

$$ \left[\begin{array}{ccc}\hfill {\left[\frac{\pi (m)}{1+n}\right]}^2u^{\prime\prime }(c)+\pi (m)u^{\prime\prime }(d)\hfill & \hfill 0\hfill & \hfill u^{\prime\prime }(c)\frac{\pi (m)}{{\left(1+n\right)}^2}\frac{\pi^{\prime }(m)u(d)}{u^{\prime }(d)}\hfill \\ {}\hfill 0\hfill & \hfill u^{\prime }(c)f^{\prime \prime }(k)\hfill & \hfill 0\hfill \\ {}\hfill u^{\prime\prime }(c)\frac{\pi (m)}{{\left(1+n\right)}^2}\frac{\pi^{\prime }(m)u(d)}{u^{\prime }(d)}\hfill & \hfill 0\hfill & \hfill \frac{u^{\prime\prime }(c)}{{\left(1+n\right)}^2}{\left[\frac{\pi \prime (m)u(d)}{u\prime (d)}\right]}^2+u^{\prime }(c)\frac{\pi^{\prime\prime }(m)}{\pi^{\prime }(m)}\hfill \end{array}\right] $$

A sufficient condition for (c, d, k, m) to be a maximum is that the Hessian matrix is negative definite. The second-order conditions (19)–(21) in the text are the three conditions concerning the determinants of the three submatrices of size 1 × 1, 2 × 2, and 3 × 3 that guarantee the negative definiteness of H.