Macroeconomic Effects of Changes in the Eligibility Age for Payments of Social Security Benefits Under Cash-in-Advance Constraints

Abstract

This chapter examines the effects of changes in the eligibility age on balanced growth and inflation rates through the adjustments of portfolio choices of individuals, assuming that individuals have to hold money for consumption during the earlier years of retirement before the eligibility age for public pensions (i.e., the cash-in-advance (CIA) constraint). There are still countries which have institutional or compulsory retirement age, although some Anglo-Saxon countries have recently abandoned it. In such countries there can be an estrangement between these ages.

This chapter is based on the manuscript “Macroeconomic effects of changes in the eligibility age for payments of social security benefits under cash-in-advance constraints,” co-authored with Makoto Hirazawa.

Appendix 1

1.1 1.1 Case of Labor Supply by Retirees

Making use of savings plans of individuals (5.33), the labor productivity (5.8), and ${w}_t={A}_t\omega$, we obtain

$$ \begin{array}{ll}n\left(\frac{a}{\omega}\right)\frac{A_{t+1}}{A_t}=& \left(1-B\right)\left(1-\tau \right)\left[1+b\overline{e}\left(1+{\pi}_{t+1}\right)\right]\hfill \\ {}& -\left(1+{\pi}_{t+1}\right)\left\{B\left[\frac{1-\theta }{\left(1+r\right)\left(1+{\pi}_{t+1}\right)}+\theta \right]-\theta \right\}\frac{T_{t+1}}{w_t}\hfill \end{array} $$

and from (5.38)

$$ \begin{array}{ll}\frac{T_{t+1}}{w_t}=& \frac{\frac{1+\beta }{\beta}\left(1-\tau \right)n}{G_t}\frac{A_{t+1}}{A_t}\; \hfill \\ {}& +\frac{\frac{1-\beta }{\beta}\frac{1-\tau }{\tau}\left\{\mu B\frac{1-\tau }{1+{\pi}_{t+1}}+\left[\tau -\mu \left(1-B\right)\left(1-\tau \right)\right]b\overline{e}\right\}-\frac{1-\tau }{1+{\pi}_{t+1}}-\left(1-\tau \right)b\overline{e}}{G_t}\hfill \end{array} $$

where ${G}_t=\left[\frac{1-\theta }{\left(1+r\right)\left(1+{\pi}_{t+1}\right)}+\theta \right]+\frac{1+\beta }{\beta}\frac{1-\tau }{\tau}\left\{1-\mu \left[\frac{1-\theta }{\left(1+r\right)\left(1+{\pi}_{t+1}\right)}B-\left(1-B\right)\theta \right]\right\}$.

Therefore, together with (5.20), we obtain the balanced growth rate of inflation which satisfies the following condition:

$$ \begin{array}{l}\frac{1+\mu }{1+\pi}\left(\frac{a}{\omega }+\frac{\left(1+\pi \right)\left\{B\left[\frac{1-\theta }{\left(1+r\right)\left(1+\pi \right)}+\theta \right]-\theta \right\}\frac{1+\beta }{\beta}\frac{1-\tau }{\tau }}{G_t}\right)\\ {}\; =\left(1-B\right)\left(1-\tau \right)\left[1+b\overline{l}\left(1+\pi \right)\right]\\ {}\; -\frac{\left(1+\pi \right)\left\{B\left[\frac{1-\theta }{\left(1+r\right)\left(1+\pi \right)}+\theta \right]-\theta \right\}}{G_t}\\ {}\; \cdot \left[\frac{1+\beta }{\beta}\frac{1-\tau }{\tau}\left\{\mu B\frac{1-\tau }{1+\pi }+\left[\tau -\mu \left(1-B\right)\left(1-\tau \right)\right]b\overline{e}\right\}-\frac{1-\tau }{1+\pi }-\left(1-\tau \right)b\overline{e}\right].\end{array} $$

1.2 1.2 The Discriminant of Eq. (5.22)

Denoting $\varPi =1+\pi$, (5.22) can be rewritten as

$$ \begin{array}{ll}\hfill & \left(1+\mu \right)\left[1+\left(1-\alpha \right)\theta \mu +\frac{\tau \omega }{a}\frac{\alpha \left(1-\theta \right)}{1+r}\right]\varPi -\left(1+\mu \right)\mu \frac{\alpha \left(1-\theta \right)}{1+r}\\ {}& -\frac{\tau \omega }{\alpha}\left(1+\mu \right)\left(1-\alpha \right)\theta {\varPi}^2\; =\frac{\left(1-\alpha \right)\left(1-\tau \right)\omega }{a}\left(1+\theta \mu \right){\varPi}^2\hfill \\ {}& -\frac{\left(1-\alpha \right)\left(1-\tau \right)\omega }{a}\frac{1-\theta }{1+r}\frac{\alpha }{1-\alpha}\mu \varPi .\hfill \end{array} $$

Rearranging it, we obtain

$$ \begin{array}{ll}\hfill & \left[\frac{\left(1-\alpha \right)\left(1-\tau \right)\omega }{a}\left(1+\theta \mu \right)+\frac{\tau \omega }{\alpha}\left(1+\mu \right)\left(1-\alpha \right)\theta \right]{\varPi}^2\\ {}& \; -\left\{\frac{\left(1-\alpha \right)\left(1-\tau \right)\omega }{a}\frac{1-\theta }{1+r}\frac{\alpha }{1-\alpha}\mu +\left(1+\mu \right)\left[1+\left(1-\alpha \right)\theta \mu +\frac{\tau \omega }{a}\frac{\alpha \left(1-\theta \right)}{1+r}\right]\right\}\varPi \; \hfill \\ {}& +\left(1+\mu \right)\mu \frac{\alpha \left(1-\theta \right)}{1+r}=0.\hfill \end{array} $$

The discriminant of the equation is

$$ \begin{array}{l}D\equiv {\left\{\frac{\left(1-\alpha \right)\left(1-\tau \right)\omega }{a}\frac{1-\theta }{1+r}\frac{\alpha }{1-\alpha}\mu +\left(1+\mu \right)\left[1+\left(1-\alpha \right)\theta \mu +\frac{\tau \omega }{a}\frac{\alpha \left(1-\theta \right)}{1+r}\right]\right\}}^2\\ {}\; -4\left[\frac{\left(1-\alpha \right)\left(1-\tau \right)\omega }{a}\left(1+\theta \mu \right)+\frac{\tau \omega }{\alpha}\left(1+\mu \right)\left(1-\alpha \right)\theta \right]\left(1+\mu \right)\mu \frac{\alpha \left(1-\theta \right)}{1+r}.\end{array} $$

We cannot determine the sign of the discriminant a priori. However, when $\mu >0$ is sufficiently small, e.g., when μ is close to but >0, we have $D>0$ and, therefore, two real solutions, i.e., $\varPi =0$ and $\varPi =\frac{1+\left(1-\alpha \right)\theta \mu +\frac{\tau \omega }{a}\frac{\alpha \left(1-\theta \right)}{1+r}}{\frac{\left(1-\alpha \right)\left(1-\tau \right)\omega }{a}+\frac{\tau \omega }{a}\left(1-\alpha \right)\theta }>0$ or $\pi =-1$ and $\pi =\frac{1+\left(1-\alpha \right)\theta \mu +\frac{\tau \omega }{a}\frac{\alpha \left(1-\theta \right)}{1+r}}{\frac{\left(1-\alpha \right)\left(1-\tau \right)\omega }{a}+\frac{\tau \omega }{a}\left(1-\alpha \right)\theta }-1>-1$. Thus, at least, for sufficiently small $\mu >0$, we will have two real solutions.