Population aging and legal retirement age

Abstract

This paper analyzes the effects of population aging on the preferred legal retirement age. What is revealed is the crucial role that the indirect ‘macro’ effects resulting from a change in the legal retirement age play in the optimal decision. Two social security systems are studied. Under a defined contribution scheme, aging lowers the preferred legal retirement age. However, under a defined pension scheme, the retirement age is delayed. This result shows the relevance of correctly choosing the parameter affected by the dependency ratio in the design of the social security programme.

Appendix

Proof proposition 1

1. We have to prove that ∂R _p */∂n is negative. It can easily be checked that ∂² U/∂R ² is negative-evaluated at R=R _p *. Therefore, the sign of ∂R _p */∂n coincides with the sign of ∂ (∂U/∂R)/∂n,

$$\frac{{\partial {\left( {\frac{{\partial U}}{{\partial R}}} \right)}}}{{\partial n}} = {\left( {1 - e^{{ - rT}} } \right)}{\left( {u^{{\prime \prime }} {\left( {c_{i} } \right)}\frac{{\partial c_{i} }}{{\partial R}}\frac{{\partial c_{i} }}{{\partial n}} + u^{\prime } {\left( {c_{i} } \right)}\frac{{\partial {\left( {{\partial c_{i} } \mathord{\left/ {\vphantom {{\partial c_{i} } {\partial R}}} \right. \kern-\nulldelimiterspace} {\partial R}} \right)}}}{{\partial n}}} \right)}$$

(10)

The constant consumption c _p is as follows:

$$c_{p} = {\left( {1 - \frac{{{\left( {e^{{ - nR}} - e^{{ - nT}} } \right)}}}{{{\left( {1 - e^{{ - nR}} } \right)}}}p} \right)}\frac{{{\left( {1 - e^{{ - rR}} } \right)}}}{{{\left( {1 - e^{{ - rT}} } \right)}}} + p\frac{{{\left( {e^{{ - rR}} - e^{{ - rT}} } \right)}}}{{{\left( {1 - e^{{ - rT}} } \right)}}}.$$

(11)

First, ∂c _p /∂R and ∂c _p /∂n are strictly positive: an increase in R augments the working period and reduces the contribution rate, both effects implying higher consumption; an increase in n reduces the dependency ratio and the contribution rate, and this effect also implies higher consumption.

Secondly, ∂ (∂c _p /∂R)/∂n can be reduced to

$$\frac{{\partial {\left( {{\partial c_{p} } \mathord{\left/ {\vphantom {{\partial c_{p} } {\partial R}}} \right. \kern-\nulldelimiterspace} {\partial R}} \right)}}}{{\partial n}} = \frac{1}{{{\left( {1 - e^{{ - rT}} } \right)}}}{\left( {\frac{{e^{{ - nR}} {\left( {1 - e^{{ - nT}} } \right)}}}{{{\left( {1 - e^{{ - nR}} } \right)}^{2} }}{\left( {1 - e^{{ - nR}} - nR} \right)}} \right)}.$$

(12)

It is easy to check that

$$1 - e^{{ - nR}} - nR < 0\forall nR \in {\left( {0,\infty } \right)}$$

(13)

and therefore, ∂ (∂c _p /∂R)/∂n<0.

2. We have to prove that ${\partial R^{*}_{\tau } } \mathord{\left/ {\vphantom {{\partial R^{*}_{\tau } } {\partial n}}} \right. \kern-\nulldelimiterspace} {\partial n} > 0$. It is easy to check that Eq. 2 is strictly concave; therefore, the sign of ${\partial R^{*}_{\tau } } \mathord{\left/ {\vphantom {{\partial R^{*}_{\tau } } {\partial n}}} \right. \kern-\nulldelimiterspace} {\partial n}$ also coincides with the sign of Eq. 10.

The constant consumption $c_{\tau }$ is

$$c_{\tau } = \frac{{{\left( {1 - e^{{ - rR}} } \right)}}} {{{\left( {1 - e^{{ - rT}} } \right)}}}{\left( {1 - \tau } \right)} + \tau \frac{{{\left( {1 - e^{{ - nR}} } \right)}}} {{{\left( {e^{{ - nR}} - e^{{ - nT}} } \right)}}}\frac{{{\left( {e^{{ - rR}} - e^{{ - rT}} } \right)}}} {{{\left( {1 - e^{{ - rT}} } \right)}}}.$$

(14)

Again, ${\partial c_{\tau } } \mathord{\left/ {\vphantom {{\partial c_{\tau } } {\partial R}}} \right. \kern-\nulldelimiterspace} {\partial R}$ and ${\partial c_{\tau } } \mathord{\left/ {\vphantom {{\partial c_{\tau } } {\partial n}}} \right. \kern-\nulldelimiterspace} {\partial n}$ are strictly positive: an increase in R augments the working period and raises pension benefits, both effects implying higher consumption; an increase in n reduces the dependency ratio and thus augments pension benefits, this effect also implying higher consumption.

${\partial {\left( {{\partial c_{\tau } } \mathord{\left/ {\vphantom {{\partial c_{\tau } } {\partial R}}} \right. \kern-\nulldelimiterspace} {\partial R}} \right)}} \mathord{\left/ {\vphantom {{\partial {\left( {{\partial c_{\tau } } \mathord{\left/ {\vphantom {{\partial c_{\tau } } {\partial R}}} \right. \kern-\nulldelimiterspace} {\partial R}} \right)}} {\partial n}}} \right. \kern-\nulldelimiterspace} {\partial n}$ can be reduced to

$$\frac{{\partial {\left( {{\partial c_{\tau } } \mathord{\left/ {\vphantom {{\partial c_{\tau } } {\partial R}}} \right. \kern-\nulldelimiterspace} {\partial R}} \right)}}}{{\partial n}} = \frac{1}{{{\left( {1 - e^{{ - rT}} } \right)}}}\frac{{e^{{ - nR}} {\left( {1 - e^{{ - nT}} } \right)}e^{{ - nT}} }}{{{\left( {e^{{ - nR}} - e^{{ - nT}} } \right)}^{2} }}{\left( {e^{{n{\left( {T - R} \right)}}} - {\left( {n{\left( {T - R} \right)} + 1} \right)}} \right)}.$$

(15)

It is easy to check that

$$e^{{n{\left( {T - R} \right)}}} - {\left( {n{\left( {T - R} \right)} + 1} \right)} > 0\forall n{\left( {T - R} \right)} \in {\left( {0,\infty } \right)}.$$

(16)

Therefore, since ∂ (∂c/∂R)/∂n>0, the next step is to prove that the positive part of Eq. 10 is larger than the negative part.

Taking common factor and rearranging terms, Eq. 10 will be

$$\frac{{\partial {\left( {\frac{{\partial U}}{{\partial R}}} \right)}}}{{\partial n}} = {\left( {1 - e^{{ - rT}} } \right)}\frac{{u^{\prime } {\left( c \right)}}}{c}{\left( {c\frac{{\partial {\left( {{\partial c} \mathord{\left/ {\vphantom {{\partial c} {\partial R}}} \right. \kern-\nulldelimiterspace} {\partial R}} \right)}}}{{\partial n}} - \rho _{r} {\left( c \right)}\frac{{\partial c}}{{\partial R}}\frac{{\partial c}}{{\partial n}}} \right)}.$$

(17)

Since ρ _r (c)<1, Eq. 10 is positive if

$$\frac{{c\frac{{\partial {\left( {{\partial c} \mathord{\left/ {\vphantom {{\partial c} {\partial R}}} \right. \kern-\nulldelimiterspace} {\partial R}} \right)}}}{{\partial n}}}}{{\frac{{\partial c}}{{\partial R}}\frac{{\partial c}}{{\partial n}}}} > 1.$$

(18)

Since we are evaluating at n=r, after some simplifications, the left-hand side (LHS) of Eq. 18 can be reduced to the following expression:

$$LHS = \frac{{{\left( {e^{{nR}} - 1} \right)}{\left( {e^{{nT}} - 1} \right)}{\left( {e^{{nT}} - e^{{nR}} \left( {1 + n} \right.{\left( {T - R} \right)}} \right)}}} {{{\left( {e^{{nT}} - e^{{nR}} } \right)}n{\left( {R{\left( {e^{{nT}} - 1} \right)} - T{\left( {e^{{nR}} - 1} \right)}} \right)}}}.$$

(19)

It can be shown numerically by means of a mathematical programme that LHS>1 for any n∈(0, 1) and for any R∈(0, T). Indeed, the mathematical programme runs results lower than 1 for values of R very close to 0. Therefore, we should impose a lower bound on R, although almost no restrictive, for instance R∈(T/100, T).

Proof proposition 2

We have to prove that with strictly concave indirect utility functions, if τ (n, p, R _p *)=τ, then $R^{*}_{p} > R^{*}_{\tau }$.

Disregarding the scheme, the first-order condition of the maximization problem would be

$$\frac{{\partial U_{i} }} {{\partial R}} = \frac{{{\left( {1 - e^{{ - rT}} } \right)}}} {r}u^{\prime } {\left( {c_{i} } \right)}{\left( {\frac{{\partial c_{i} }} {{\partial R}}} \right)} - e^{{ - rR}} v = 0.$$

(20)

Given that for R=R _p *, c _p will be equal to $c_{\tau }$, we have to prove that at that point,

$$\frac{{\partial c_{p} }} {{\partial R}} >\frac{{\partial c_{\tau } }} {{\partial R}}$$

(21)

If this inequality holds, the single-peakness property of the utility function implies that $R^{*}_{p} > R^{*}_{\tau }$. We calculate ∂c _p /∂R and ${\partial c_{\tau } } \mathord{\left/ {\vphantom {{\partial c_{\tau } } {\partial R}}} \right. \kern-\nulldelimiterspace} {\partial R}$ to compare them:

$$\frac{{\partial c_{p} }}{{\partial R}} = \frac{{re^{{ - rR}} }}{{{\left( {1 - e^{{ - rT}} } \right)}}}{\left( {1 - \tau {\left( {n,p,R^{*}_{p} } \right)}} \right)} + \frac{{pne^{{ - nR}} {\left( {1 - e^{{ - rR}} } \right)}}}{{{\left( {1 - e^{{ - rT}} } \right)}{\left( {1 - e^{{ - nR}} } \right)}}}$$

(22)

$$+ \frac{{pne^{{ - nR}} {\left( {1 - e^{{ - rR}} } \right)}{\left( {e^{{ - nR}} - e^{{ - nT}} } \right)}}} {{{\left( {1 - e^{{ - rT}} } \right)}{\left( {1 - e^{{ - nR}} } \right)}^{2} }} - \frac{{pre^{{ - rR}} }} {{{\left( {1 - e^{{ - rT}} } \right)}}}$$

(23)

$$\frac{{\partial c_{\tau } }} {{\partial R}} = \frac{{re^{{ - rR}} }} {{{\left( {1 - e^{{ - rT}} } \right)}}}{\left( {1 - \tau } \right)} + \frac{{\tau ne^{{ - nR}} {\left( {e^{{ - rR}} - e^{{ - rT}} } \right)}}} {{{\left( {1 - e^{{ - rT}} } \right)}{\left( {e^{{ - nR}} - e^{{ - nT}} } \right)}}}$$

(24)

and

$$ + \frac{{\tau ne^{{ - nR}} {\left( {1 - e^{{ - nR}} } \right)}{\left( {e^{{ - rR}} - e^{{ - rT}} } \right)}}}{{{\left( {1 - e^{{ - rT}} } \right)}{\left( {e^{{ - nR}} - e^{{ - nT}} } \right)}^{2} }} - \frac{{p{\left( {n,\tau ,R^{*}_{\tau } } \right)}re^{{ - rR}} }}{{{\left( {1 - e^{{ - rT}} } \right)}}}$$

(25)

Given that for R=R _p *, τ (n, p, R _p *)=τ and p(n, τ, R _p *)=p, we only have to compare the second and the third addends. In the defined pension scheme

$$\tau {\left( {R,n,p} \right)} = \frac{{{\left( {e^{{ - nR}} - e^{{ - nT}} } \right)}}} {{{\left( {1 - e^{{ - nR}} } \right)}}}p$$

(26)

we know that τ=τ (n, p, R _p *). Therefore, substituting Eq. 26 in Eqs. 24 and 25,that is, the constant contribution τ for its value, and comparing the two derivatives, we obtain that Eq. 21 holds if:

$$\frac{{{\left( {e^{{ - nR}} - e^{{ - nT}} } \right)}}} {{{\left( {1 - e^{{ - nR}} } \right)}}} >\frac{{{\left( {e^{{ - rR}} - e^{{ - rT}} } \right)}}} {{{\left( {1 - e^{{ - rR}} } \right)}}}.$$

(27)

The inequality (Eq. 27) will be true for r>n since the dependency ratio is decreasing with respect to the population growth rate.