Elderly empowerment, fertility, and public pensions

Abstract

The economic role of elderly people is underestimated in the context of the debate on pension reforms. This is because, as healthy life expectancy increases, the elderly become more active not only in the labor market but also in household production. Using a three-period overlapping-generations model in which grandparents allocate time between labor and informal childcare (“grandparenting”), we analyze the interaction among grandparenting, fertility, elderly labor, and public pensions. We obtain two analytical results. First, increasing the pension contribution rate increases grandparenting. Second, elderly labor force participation rates are negatively (positively) related to the fertility rates among countries with small (large) public pensions. The nonlinear relationship between elderly labor and fertility is empirically supported.

Appendices

Appendix 1

Proof of Lemma 1

Substituting $w_{t+1}/w_{t}=\gamma _{t}$, and $R_{t+1}=\alpha A$ into (7), we obtain (14).

Combining the market-clearing conditions (12) and (13), the capital–labor ratio in period $t+1$ is given by

$$\begin{aligned} k_{t+1}=\frac{K_{t+1}}{L_{t+1}}=\frac{N_{t}s_{t}}{N_{t+1}+N_{t}\theta l_{t+1} }=\frac{s_{t}}{n_{t}+\theta l_{t+1}}. \end{aligned}$$

Substituting the saving function (8) into this, and using $w_{t}=(1-\alpha )Ak_{t}$, we obtain (15).

Let us denote the relative wage rate of old adults to young adults in period $t+1$ by $p_{t+1}$,

$$\begin{aligned} p_{t+1}=\frac{(1-\tau )w_{t+1}\theta -P_{t+1}}{(1-\tau )w_{t+1}}. \end{aligned}$$

Using this, (9) yields (17).

Meanwhile, Eqs. (11) and (12) in period $t+1$ give the pension benefit in period $t+1$,

$$\begin{aligned} P_{t+1}=\frac{\tau w_{t+1}(n_{t}+\theta l_{t+1})}{{\bar{l}}-l_{t+1}}. \end{aligned}$$

Using this, we obtain

$$\begin{aligned} (1-\tau )w_{t+1}\theta -P_{t+1}=w_{t+1}\left[ \theta -\frac{\tau (\theta {\bar{l}}+n_{t})}{{\bar{l}}-l_{t+1}}\right] \end{aligned}$$

which yields (18). Solving (17) and (18), we obtain (16). $\square$

Appendix 2

Proof of Lemma 3

Without loss of generality, we omit the time script. Let us denote the left-hand side by f(n),

$$\begin{aligned} f(n)=n\left[ 1-\frac{\theta {\bar{l}}(\eta +\beta (1-\alpha ))}{\alpha n+\theta {\bar{l}}}\right] ,\quad n>0. \end{aligned}$$

(31)

Obviously, $f(n)<n$. Furthermore, we know that

$$\begin{aligned} \lim _{n\rightarrow \infty }[f(n)-n]=-\frac{\theta {\bar{l}}}{\alpha }(\eta +\beta (1-\alpha )) \end{aligned}$$

which implies that the curve $z=f(n)$ has an asymptotic line,

$$\begin{aligned} z=n-\frac{\theta {\bar{l}}}{\alpha }(\eta +\beta (1-\alpha )). \end{aligned}$$

(32)

Therefore, the range of f(n) is given by

$$\begin{aligned} n-\frac{\theta {\bar{l}}}{\alpha }(\eta +\beta (1-\alpha ))<f(n)<n. \end{aligned}$$

Next, differentiating (31) with respect to n, we obtain

$$\begin{aligned} f^{\prime }(n)=1-\frac{(\theta {\bar{l}})^{2}(\eta +\beta (1-\alpha ))}{ (\alpha n+\theta {\bar{l}})^{2}}. \end{aligned}$$

Obviously, $f^{\prime \prime }>0$. Furthermore, we know that

$$\begin{aligned} f^{\prime }(0)&=1-\eta -\beta (1-\alpha )>0 \\ \lim _{n\rightarrow \infty }f^{\prime }(n)&=1 \end{aligned}$$

which implies

$$\begin{aligned} 1-\eta -\beta (1-\alpha )<f^{\prime }(n)<1. \end{aligned}$$

(33)

Fig. 4

Fertility function

In Fig. 4, the solid curve is $z=f(n)$ and the solid horizontal line is $z=(1-\tau )\rho \phi (g)$. The dashed lines represent a 45-degree line and the asymptotic line (32). At the intersection of the solid curve and the solid line, the unique solution, $n^{*}=n(g)$, is obtained. When $g$ increases, the horizontal line shifts upward, which increases $n^{*}$.

In Fig. 4, the range of $n^{*}$ is given by

$$\begin{aligned} (1-\tau )\rho \phi (g)<n^{*}<(1-\tau )\rho \phi (g)+\frac{\theta {\bar{l}} }{\alpha }(\eta +\beta (1-\alpha )) \end{aligned}$$

(34)

which is (21) in Lemma 3.

Next, we derive the range of $\varepsilon _{n}=gn^{\prime }(g)/n(g)$. By total-differentiating both sides of (19), we obtain

$$\begin{aligned} f^{\prime }(n)\mathrm{d}n=(1-\tau )\rho \phi ^{\prime }(g)\mathrm{d}g \end{aligned}$$

which yields

$$\begin{aligned} n^{\prime }(g)=\frac{(1-\tau )\rho \phi ^{\prime }(g)}{f^{\prime }(n)}. \end{aligned}$$

Using (33), the range of $n^{\prime }(g)$ is given by

$$\begin{aligned} (1-\tau )\rho \phi ^{\prime }(g)<n^{\prime }(g)<\frac{(1-\tau )\rho \phi ^{\prime }(g)}{1-\eta -\beta (1-\alpha )}. \end{aligned}$$

(35)

Using the elasticity of grandparenting $\varepsilon =g\phi ^{\prime }(g)/\phi (g)$, and the elasticity of fertility $\varepsilon _{n}=gn^{\prime }(g)/n(g)$, (35) can be rewritten as

$$\begin{aligned} \frac{(1-\tau )\rho \phi (g)\varepsilon }{n(g)}<\varepsilon _{n}<\frac{ (1-\tau )\rho \phi (g)\varepsilon }{[1-\eta -\beta (1-\alpha )]n(g)}. \end{aligned}$$

Finally, using (34), we obtain

$$\begin{aligned} \frac{(1-\tau )\rho \phi (g)\varepsilon }{(1-\tau )\rho \phi (g)+\frac{ \theta {\bar{l}}}{\alpha }(\eta +\beta (1-\alpha ))}<\varepsilon _{n}<\frac{ \varepsilon }{1-\eta -\beta (1-\alpha )} \end{aligned}$$

which is (22) in Lemma 3. $\square$

Appendix 3

Proof of Proposition 5

Differentiating (25) with respect to $g_{t}$, we obtain

$$\begin{aligned} \frac{{\mathrm{d}}g_{t+1}}{{\mathrm{d}}g_{t}}=-\frac{[\eta +\beta (1-\alpha )\tau ]{\bar{l}}}{\eta +\beta (1-\alpha )}\cdot \frac{n^{\prime }(g_{t})}{[n(g_{t})]^{2}}. \end{aligned}$$

(36)

The steady state is locally stable if

$$\begin{aligned} -1<\left( \frac{{\mathrm{d}}g_{t+1}}{{\mathrm{d}}g_{t}}\right) ^{*}<0. \end{aligned}$$

(37)

Using the elasticity of fertility in the steady state, $\varepsilon _{n}^{*}=g^{*}n^{\prime }(g^{*})/n(g^{*})$, (37) can be rewritten as

$$\begin{aligned} \varepsilon _{n}^{*}<\frac{\eta +\beta (1-\alpha )}{[\eta +\beta (1-\alpha )\tau ]{\bar{l}}}n^{*}g^{*}. \end{aligned}$$

(38)

Here, the range of $\varepsilon _{n}^{*}$ is given by (22) in Lemma 3. For any pair (n, g) that satisfies (22), the condition (38) is satisfied if

$$\begin{aligned} \frac{\varepsilon ^{*}}{1-\eta -\beta (1-\alpha )}\le \frac{\eta +\beta (1-\alpha )}{[\eta +\beta (1-\alpha )\tau ]{\bar{l}}}n^{*}g^{*}, \end{aligned}$$

(39)

where $\varepsilon ^{*}=g^{*}\phi ^{\prime }(g^{*})/\phi (g^{*})$ is the elasticity of grandparenting in the steady state.

Note that (25) yields

$$\begin{aligned} n^{*}g^{*}=\frac{\eta \alpha +(\beta +\eta )(1-\alpha )\tau }{\theta [\eta +\beta (1-\alpha )]}n^{*}+\frac{[\eta +\beta (1-\alpha )\tau ]{\bar{l}}}{\eta +\beta (1-\alpha )}. \end{aligned}$$

(40)

Substituting (40) into (39), we obtain

$$\begin{aligned} n^{*}\ge \left[ \frac{\varepsilon ^{*}}{1-\eta -\beta (1-\alpha )}-1 \right] \frac{\theta {\bar{l}}[\eta +\beta (1-\alpha )\tau ]}{\eta \alpha +(\beta +\eta )(1-\alpha )\tau }. \end{aligned}$$

One of the sufficient conditions to make this inequality hold is

$$\begin{aligned} \varepsilon ^{*}\le 1-\eta -\beta (1-\alpha ) \end{aligned}$$

which is (26) in Proposition 5. $\square$

Appendix 4

Proof of Proposition 7

By total-differentiating (27) and (28), we obtain

$$\begin{aligned} \left( \begin{array}{cc} \phi ^{\prime } &{}\quad -\Gamma _{n}^{1} \\ 1 &{}\quad -\Gamma _{n}^{2} \end{array} \right) \left( \begin{array}{c} \mathrm{d}g^{*} \\ \mathrm{d}n^{*} \end{array} \right) =\left( \begin{array}{c} \Gamma _{\tau }^{1} \\ \Gamma _{\tau }^{2} \end{array} \right) \mathrm{d}\tau . \end{aligned}$$

The determinant of the $2\times 2$ matrix is $\Delta =-\phi ^{\prime }\Gamma _{n}^{2}+\Gamma _{n}^{1}>0$. Using the inverse matrix, we obtain

$$\begin{aligned} \left( \begin{array}{c} \mathrm{d}g^{*} \\ \mathrm{d}n^{*} \end{array} \right) =\frac{1}{\Delta }\left( \begin{array}{cc} -\Gamma _{n}^{2} &{}\quad \Gamma _{n}^{1} \\ -1 &{}\quad \phi ^{\prime } \end{array} \right) \left( \begin{array}{c} \Gamma _{\tau }^{1} \\ \Gamma _{\tau }^{2} \end{array} \right) \mathrm{d}\tau . \end{aligned}$$

That is,

$$\begin{aligned} \frac{\mathrm{d}g^{*}}{\mathrm{d}\tau }&=\frac{1}{\Delta }(-\Gamma _{n}^{2}\Gamma _{\tau }^{1}+\Gamma _{n}^{1}\Gamma _{\tau }^{2})>0 \end{aligned}$$

(41)

$$\begin{aligned} \frac{\mathrm{d}n^{*}}{\mathrm{d}\tau }&=\frac{1}{\Delta }(-\Gamma _{\tau }^{1}+\phi ^{\prime }\Gamma _{\tau }^{2}). \end{aligned}$$

(42)

In the following, we examine the sign of $\mathrm{d}n^{*}/\mathrm{d}\tau$.

First, differentiating (27) with respect to $\tau$ yields

$$\begin{aligned} \Gamma _{\tau }^{1}=\frac{n^{*}}{\rho (1-\tau )^{2}}\left[ 1-\frac{ \theta {\bar{l}}(\eta +\beta (1-\alpha ))}{\alpha n^{*}+\theta {\bar{l}}} \right] =\frac{\phi (g^{*})}{1-\tau }. \end{aligned}$$

Next, differentiating (28) with respect to $\tau$ yields

$$\begin{aligned} \Gamma _{\tau }^{2}=\frac{1-\alpha }{\eta +\beta (1-\alpha )}\left( \frac{ \beta +\eta }{\theta }+\frac{\beta {\bar{l}}}{n^{*}}\right) . \end{aligned}$$

Meanwhile, (28) gives

$$\begin{aligned} \frac{1}{n^{*}}=\frac{\eta +\beta (1-\alpha )}{[\eta +\beta (1-\alpha )\tau ]{\bar{l}}}g^{*}-\frac{\eta \alpha +(\beta +\eta )(1-\alpha )\tau }{ \theta {\bar{l}}[\eta +\beta (1-\alpha )\tau ]}. \end{aligned}$$

Using this, $\Gamma _{\tau }^{2}$ is rewritten as

$$\begin{aligned} \Gamma _{\tau }^{2}=\frac{\beta (1-\alpha )}{\eta +\beta (1-\alpha )\tau } \left( \frac{\eta }{\beta \theta }+g^{*}\right) . \end{aligned}$$

Substituting $\Gamma _{\tau }^{1}$ and $\Gamma _{\tau }^{2}$ into (42), we obtain

$$\begin{aligned} \frac{\mathrm{d}n^{*}}{\mathrm{d}\tau }=\frac{\phi (g^{*})}{\Delta g^{*}}\left[ - \frac{g^{*}}{1-\tau }+\frac{\beta (1-\alpha )\varepsilon ^{*}}{\eta +\beta (1-\alpha )\tau }\left( \frac{\eta }{\beta \theta }+g^{*}\right) \right], \end{aligned}$$

where $\varepsilon ^{*}=g^{*}\phi ^{\prime }(g^{*})/\phi (g^{*})\ge 0$ represents the elasticity of grandparenting.

Evaluating this at $\tau =0$ yields

$$\begin{aligned} \frac{\mathrm{d}n^{*}}{\mathrm{d}\tau }|_{\tau =0}=\frac{\phi (g^{*})}{\Delta g^{*} }\left[ \frac{(1-\alpha )\varepsilon ^{*}}{\theta }-\frac{\eta -\beta (1-\alpha )\varepsilon ^{*}}{\eta }g^{*}\right] . \end{aligned}$$

Therefore, the following statements are true.

• 1. If $\eta -\beta (1-\alpha )\varepsilon ^{*}\le 0$,

  $$\begin{aligned} \frac{\mathrm{d}n^{*}}{\mathrm{d}\tau }|_{\tau =0}>0. \end{aligned}$$

• 2. If $\eta -\beta (1-\alpha )\varepsilon ^{*}>0$,

  $$\begin{aligned} g^{*}<\frac{\eta (1-\alpha )\varepsilon ^{*}}{\theta [\eta -\beta (1-\alpha )\varepsilon ^{*}]}\Rightarrow \frac{\mathrm{d}n^{*}}{\mathrm{d}\tau } |_{\tau =0}>0 \\ g^{*} >\frac{\eta (1-\alpha )\varepsilon ^{*}}{\theta [\eta -\beta (1-\alpha )\varepsilon ^{*}]}\Rightarrow \frac{\mathrm{d}n^{*}}{\mathrm{d}\tau } |_{\tau =0}<0 \end{aligned}$$

  which completes the proof of Proposition 7. $\square$

Appendix 5

[Parameters and simulation results]

• 1. Time allocation

We use time data in Rogerson and Wallenius (2016), which indicate the time use by age compiled from the American Time Use Survey (ATUS) from 2003 to 2011. Market work at age 60 is 26.17 hours per week, which decreases to 11.20 at age 65. By normalizing the former market work to unity, the elderly labor in our model is given by $l_{0}=0.43$, where a subscript 0 indicates variables when the contribution rate of public pensions is zero. The remaining time is allocated between leisure (0.31) and home production, etc. (0.26). Therefore, we set time endowment in old age by ${\bar{l}}=0.69$, and time input in home production by $l_{0}^{H}=0.26$, respectively.

We normalize the fertility rate to unity when the contribution rate is zero ($n_{0}=1$). Since $l_{0}^{H}=n_{0}g_{0}$, grandparenting per child is given by $g_{0}=0.26$.

• 2. Preference parameters

We assume that one period in the model economy is 30 years. Assuming that annual rate of time preference is 1%, the private discount factor is given by $\delta =1.01^{-30}=0.742$.

In our model setting, $\delta =\beta /(1-\rho -\beta -\eta )$. Therefore, we obtain

$$\begin{aligned} \beta =\frac{\delta }{1+\delta }(1-\rho -\eta ). \end{aligned}$$

(43)

In a benchmark case, we assume that the productivity of old workers is the same as young workers ($\theta =1$). Then, Eq. (24) with $\tau =0$ yields

$$\begin{aligned} l_{0}=\frac{{\bar{l}}\beta (1-\alpha )-\eta \alpha n_{0}}{\eta +\beta (1-\alpha )}. \end{aligned}$$

(44)

(43) and (44) solve for $\beta$ and $\eta$ as follows:

$$\begin{aligned} \beta&=\frac{1-\rho }{\frac{1+\delta }{\delta }+\frac{(1-\alpha )({\bar{l}} -l_{0})}{l_{0}+\alpha n_{0}}} \\ \eta&=\frac{1-\rho }{1+\frac{1+\delta }{\delta }\frac{l_{0}+\alpha n_{0}}{ (1-\alpha )({\bar{l}}-l_{0})}}. \end{aligned}$$

We assume the capital share is $\alpha =0.33$, and the preference for the number of children is $\rho =0.1$. Using $\delta =0.742$, ${\bar{l}}=0.69$, $l_{0}=0.43$, $n_{0}=1$, we obtain $\beta =0.349$, and $\eta =0.08$.

• 3. $\phi (g)$

Since analytical results in our model are represented in terms of elasticity of grandparenting, we use a power function, $\phi (g)=\phi g^{\varepsilon }$ , where $0<\varepsilon <1$ and $\phi >0$. The stability condition in Proposition 5 is given by $\varepsilon \le 1-\eta -\beta (1-\alpha )=0.686$ , and the condition for Proposition 7 (i) is $\varepsilon \ge \eta /[\beta (1-\alpha )]=0.342$. We use $\varepsilon =0.4$ in a benchmark case of efficient grandparenting. Alternatively, we use $\varepsilon =0.2$ when grandparenting is inefficient.

Equation (27) with $\tau =0$ and $\theta =1$ yields

$$\begin{aligned} \phi (g_{0})^{\varepsilon }=\frac{n_{0}}{\rho }\left[ 1-\frac{{\bar{l}}(\eta +\beta (1-\alpha ))}{\alpha n_{0}+{\bar{l}}}\right] . \end{aligned}$$

Using parameter values derived above, we obtain $\phi =13.501$ when $\varepsilon =0.4$, and $\phi =10.313$ when $\varepsilon =0.2$.

• 4. Growth rate

Assuming that annual growth rate of per capita income is 2% without public pensions, $\gamma _{0}=1.02^{30}=1.8114$. From (23) with $\theta =1$,

$$\begin{aligned} \gamma _{0}=\frac{[\eta +\beta (1-\alpha )]\alpha A}{\alpha n_{0}+{\bar{l}} [1-\eta -\beta (1-\alpha )]} \end{aligned}$$

which yields $A=14.053$. The interest factor is $R=\alpha A=4.6375$, which implies that annual interest rate is 5.25%.

• 5. Simulation results

We calculated the steady-state values of fertility, grandparenting, elderly labor, and annual growth rate. For the interior solution, we use (19), (25), and (23) in the main body. For the corner solution, we use (45)–(47) in Appendix 6. Table 1 shows a benchmark case in which grandparenting is assumed to be efficient ($\varepsilon =0.4$). Tables 2 and 3 show a case of inefficient grandparenting ($\varepsilon =0.2$), and low productivity of elderly labor ($\theta =0.8$), respectively.

Table 1 Public pensions, fertility, grandparenting, elderly labor, and growth

Table 2 A case of inefficient grandparenting

Table 3 A case of low productivity of elderly labor

Appendix 6

[Corner solution]

As the contribution rate increases, the elderly labor could decrease to reach zero. The time constraint at the corner solution is given by

$$\begin{aligned} {\bar{l}}=n_{t}g_{t+1}. \end{aligned}$$

(45)

The household maximization problem is modified as

$$\begin{aligned} \max \,u_{t}=(1-\rho -\beta -\eta )\ln c_{1t}+\rho \ln n_{t}+\beta \ln c_{2t+1}+\eta \ln {\bar{l}} \end{aligned}$$

subject to

$$\begin{aligned} (1-\tau )w_{t}+\frac{P_{t+1}{\bar{l}}}{R_{t+1}}=c_{1t}+\frac{w_{t}}{\phi (g_{t})}n_{t}+\frac{c_{2t+1}}{R_{t+1}}. \end{aligned}$$

Solving the problem yields the fertility rate and the saving function:

$$\begin{aligned} n_{t}& {}= \frac{\rho }{1-\eta }\frac{\phi (g_{t})}{w_{t}}I_{t} \\ s_{t}& {}= \frac{\beta }{1-\eta }(1-\tau )w_{t}-\left( 1-\frac{\beta }{1-\eta } \right) \frac{P_{t+1}{\bar{l}}}{R_{t+1}}, \end{aligned}$$

where $I_{t}\equiv (1-\tau )w_{t}+P_{t+1}{\bar{l}}/R_{t+1}$ represents lifetime income.

The government budget constraint and the labor market-clearing condition are modified as follows:

$$\begin{aligned} N_{t}\tau w_{t}& {}= N_{t-1}P_{t}{\bar{l}} \\ L_{t}& {}= N_{t}. \end{aligned}$$

Substituting the saving function into the capital market-clearing condition, $k_{t+1}=s_{t}/n_{t}$, and using the government budget constraint, $P_{t+1} {\bar{l}}=\tau w_{t+1}n_{t}$, the growth rate of the capital–labor ratio is given by

$$\begin{aligned} \gamma _{t}=\frac{k_{t+1}}{k_{t}}=\frac{\beta (1-\tau )(1-\alpha )A}{n_{t} \left[ 1-\eta +\left( 1-\eta -\beta \right) \frac{1-\alpha }{\alpha }\tau \right] }. \end{aligned}$$

(46)

Substituting (46) into the present value of pension benefits, $P_{t+1}{\bar{l}} /R_{t+1}=(1-\alpha )\tau n_{t}k_{t+1}/\alpha$, we obtain

$$\begin{aligned} I_{t}=\frac{\left( 1+\frac{1-\alpha }{\alpha }\tau \right) (1-\tau )w_{t}}{ 1+\left( 1-\frac{\beta }{1-\eta }\right) \frac{1-\alpha }{\alpha }\tau } \end{aligned}$$

which yields the fertility equation,

$$\begin{aligned} n_{t}=\frac{\rho \phi (g_{t})(1-\tau )\left( 1+\frac{1-\alpha }{\alpha }\tau \right) }{1-\eta +\left( 1-\eta -\beta \right) \frac{1-\alpha }{\alpha }\tau }. \end{aligned}$$

(47)

Given $g_{0}$, Eqs. (45) and (47) determine the dynamics of $(n_{t},g_{t})$.