Investment in children, social security, and intragenerational risk sharing

Extend your filial piety for your aged parents to all the aged, and extend your care for your own children to all children.

— Mencius (372 BC – 289 BC, the “Second Sage” of Confucianism)

Abstract

We analyze the role of pay-as-you-go social security in intragenerational risk sharing in an overlapping-generations model with individual heterogeneity. Parents invest in their children’s education in state schools in exchange for old-age financial support. Due to random factors such as luck in the job market, children may have different earning capacities despite that they receive the same education. Without social security, a parent gets a transfer payment from her own child, so the received amount is uncertain as it depends on the child’s earnings. The social security scheme, which essentially serves to pool transfer contributions from all children and then redistribute them equally to each parent, insures parents against the risk of educational investments. Our model shows that social security stimulates educational spending, enhances labor earnings, and increases ex ante individual utility. However, it may worsen ex post intragenerational inequality of lifetime income.

Appendices

Appendix 1: Mathematical proofs

1.1 Proof of Lemma 1

Substituting (7) into (6) yields

$$\begin{aligned} U_{it}=u[(1-\delta )(1+\varepsilon _i)H(e_{t-1})-s_{it}-e_t)]+v[(1+r)s_{it}-f_t], \end{aligned}$$

(27)

where $f_t$ is independent of $s_{it}$ by (8). The maximization problem of an i-type member implies

$$\begin{aligned} u^{\prime}[(1-\delta )(1+\varepsilon _i)H(e_{t-1})-s_{it}-e_t)]=(1+r)v^{\prime}[(1+r)s_{it}-f_t]. \end{aligned}$$

(28)

Suppose that $s_{lt}\geqslant s_{ht}$. It follows directly from (28) that

$$\begin{aligned}&u^{\prime}[(1-\delta )(1+\varepsilon _h)H(e_{t-1})-s_{ht}-e_t)] \\&\quad =(1+r)v^{\prime}[(1+r)s_{ht}-f_t] \\&\quad \geqslant (1+r)v^{\prime}[(1+r)s_{lt}-f_t] \\&\quad = u^{\prime}[(1-\delta )(1+\varepsilon _l)H(e_{t-1})-s_{lt}-e_t)] \\&\quad > u^{\prime}[(1-\delta )(1+\varepsilon _h)H(e_{t-1})-s_{ht}-e_t)], \end{aligned}$$

(29)

where the first inequality results from $v^{\prime\prime}<0$, and the second one results from $u^{\prime\prime}<0$ and $\varepsilon _l<\varepsilon _h$. Because (29) is a self-contradiction, the presumption $s_{lt}\geqslant s_{ht}$ violates, which means $s_{lt}<s_{ht}$.

1.2 Proof of Proposition 1

Expanding the first-order condition (12) obtains

$$\begin{aligned} \frac{dU^n_t}{de^n_t}& = pu^{\prime}[(1-\delta )(1+\varepsilon _h)H(e^n_{t-1}) -s^n_{ht}-e^n_{t}]\left( -\frac{ds^n_{ht}}{de^n_t}-1\right) \\&\quad+p^2v^{\prime}[(1+r)s^n_{ht}+\delta (1+\varepsilon _h)H(e^n_{t})] \\&\qquad\left[ (1+r) \frac{ds^n_{ht}}{de^n_t}+\delta (1+\varepsilon _h)H^{\prime }(e^n_{t})\right] \\&\quad+p(1-p)v^{\prime}[(1+r)s^n_{ht}+\delta (1+\varepsilon _l)H(e^n_{t})] \\&\qquad\left[ (1+r) \frac{ds^n_{ht}}{de^n_t}+\delta (1+\varepsilon _l)H^{\prime }(e^n_{t})\right] \\&\quad+(1-p)u^{\prime}[(1-\delta )(1+\varepsilon _l)H(e^n_{t-1})-s^n_{lt}-e^n_{t}] \left( -\frac{ds^n_{lt}}{de^n_t}-1\right) \\&\quad+p(1-p)v^{\prime}[(1+r)s^n_{lt}+\delta (1+\varepsilon _h)H(e^n_{t})] \\&\qquad\left[ (1+r)\frac{ds^n_{lt}}{de^n_t}+\delta (1+\varepsilon _h) H^{\prime }(e^n_{t})\right] \\&\quad+(1-p)^2v^{\prime}[(1+r)s^n_{lt}+\delta (1+\varepsilon _l)H(e^n_{t})] \\&\qquad\left[ (1+r)\frac{ds^n_{lt}}{de^n_t}+\delta (1+\varepsilon _l) H^{\prime }(e^n_{t})\right] =0 \\\Leftrightarrow & {} p^2v^{\prime}[(1+r)s^n_{ht}+\delta (1+\varepsilon _h) H(e^n_{t})]\left[ \delta (1+\varepsilon _h)H^{\prime }(e^n_{t})-(1+r)\right] \\&\quad+p(1-p)v^{\prime}[(1+r)s^n_{ht}+\delta (1+\varepsilon _l)H(e^n_{t})] \\&\qquad\left[ \delta (1+\varepsilon _l)H^{\prime }(e^n_{t})-(1+r)\right] \\&\quad+p(1-p)v^{\prime}[(1+r)s^n_{lt}+\delta (1+\varepsilon _h)H(e^n_{t})] \\&\qquad\left[ \delta (1+\varepsilon _h)H^{\prime }(e^n_{t})-(1+r)\right] \\&\quad+(1-p)^2v^{\prime}[(1+r)s^n_{lt}+\delta (1+\varepsilon _l)H(e^n_{t})] \\&\qquad\left[ \delta (1+\varepsilon _l)H^{\prime }(e^n_{t})-(1+r)\right] =0 \\\Leftrightarrow & {} \Phi _t p\left[ (1+\varepsilon _h)H^{\prime }(e^n_{t}) -\frac{1+r}{\delta }\right] \\&\quad+\Omega _t(1-p)\left[ (1+\varepsilon _l) H^{\prime }(e^n_{t})-\frac{1+r}{\delta }\right] =0, \end{aligned}$$

(30)

where two notations are introduced:

$$\begin{aligned} \Phi _t&:=pv^{\prime}[(1+r)s^n_{ht}+\delta (1+\varepsilon _h)H(e^n_{t})] \\&\quad +(1-p)v^{\prime} [(1+r)s^n_{lt}+\delta (1+\varepsilon _h)H(e^n_{t})]>0, \end{aligned}$$

(31)

$$\begin{aligned} \Omega _t&:=pv^{\prime}[(1+r)s^n_{ht}+\delta (1+\varepsilon _l)H(e^n_{t})] \\&\quad +(1-p)v^{\prime} [(1+r)s^n_{lt}+\delta (1+\varepsilon _l)H(e^n_{t})]>0. \end{aligned}$$

(32)

By $\varepsilon _h>\varepsilon _l$ and $H^{\prime }>0$, we can infer from (30)–(32) that

$$\begin{aligned} \frac{1+r}{\delta (1+\varepsilon _h)}<H^{\prime }(e^n_t)<\frac{1+r}{\delta (1+\varepsilon _l)}. \end{aligned}$$

(33)

Since (10) presents $s^n_{it}$ as a function of $(e^n_{t-1}, e^n_t)$, (30) implies that $e^n_t$ is time-varying for some t. Because $v^{\prime\prime}<0$ and $\varepsilon _h>\varepsilon _l$, we have $\Phi _t<\Omega _t$. Rewrite (30) and then use (4) to obtain

$$\begin{aligned}&[\Phi _tp(1+\varepsilon _h)+\Omega _t(1-p)(1+\varepsilon _l)]H^{\prime }(e^n_t) =\frac{1+r}{\delta }[\Phi _tp+\Omega _t(1-p)] \\&\quad \Leftrightarrow \frac{1+r}{\delta H^{\prime }(e_t)}-1= \frac{\Phi _tp\varepsilon _h+\Omega _t(1-p)\varepsilon _l}{\Phi _tp+\Omega _t(1-p)}<\frac{\Omega _t[p\varepsilon _h+(1-p) \varepsilon _l]}{\Phi _tp+\Omega _t(1-p)}=0 \\&\quad \Leftrightarrow H^{\prime }(e^n_t)>\frac{1+r}{\delta }. \end{aligned}$$

(34)

Combining (33) and (34) derives Eq. (13).

1.3 Proof of Proposition 2

Taking the first-order condition of (16) with respect to $e^k_t$ and then inserting (15) obtains

$$\begin{aligned} \frac{dU^k_t}{de^k_t}& = p\bigg \{-u^{\prime}[(1-\delta )(1+\varepsilon _h)H(e^k_{t-1})-s^k_{ht}-e^k_t]\bigg (1+\frac{\partial s^k_{ht}}{\partial e^k_t}\bigg ) \\&\quad +v^{\prime}[(1+r)s^k_{ht}+\delta H(e^k_t)]\bigg [(1+r)\frac{\partial s^k_{ht}}{\partial e^k_t}+\delta H^{\prime }(e^k_t)\bigg ]\bigg \} +(1-p) \\&\qquad\bigg \{-u^{\prime}[(1-\delta )(1+\varepsilon _l)H(e^k_{t-1})-s^k_{lt}-e^k_t]\bigg (1+\frac{\partial s^k_{lt}}{\partial e^k_t}\bigg ) \\&\quad +v^{\prime}\big [(1+r)s^k_{lt}+\delta H(e^k_t)\big ]\bigg [(1+r)\frac{\partial s^k_{lt}}{\partial e^k_t}+\delta H^{\prime }(e^k_t)\bigg ]\bigg \}=0 \\\Leftrightarrow & {} pu^{\prime}[(1-\delta )(1+\varepsilon _h)H(e^k_{t-1})-s^k_{ht}-e^k_t] \\&\quad+(1-p)u^{\prime}[(1-\delta )(1+\varepsilon _l)H(e^k_{t-1})-s^k_{lt}-e^k_t] \\&\qquad\left\{ pv^{\prime}[(1+r)s^k_{ht}+\delta H(e^k_t)]+(1-p)v^{\prime}[(1+r)s^k_{lt} +\delta H(e^k_t)]\right\} \\&\qquad\delta H^{\prime }(e^k_t)=0 \\\Leftrightarrow & {} \left[ H^{\prime }(e^k_t)-\frac{1+r}{\delta }\right] \\&\qquad\bigg \{\frac{p v^{\prime}[(1+r)s^k_{ht}+\delta H(e^k_t)]}{1-p}+v^{\prime}[(1+r)s^k_{lt}+\delta H(e^k_t)]\bigg \}=0. \end{aligned}$$

(35)

Since $p\in (0,1)$ and $v^{\prime}>0$, it is straightforward to infer from (35) that (17) holds.

1.4 Proof of Proposition 3

The coefficient of variation of $f_t$, as denoted by $\varphi (f_t)$, equals the standard deviation of $f_t$ to the mean of $f_t$, namely,

$$\begin{aligned} \varphi (f_t):=\frac{SD(f_t)}{E(f_t)}=\frac{\sqrt{E(f_t^2)-[E(f_t)]^2}}{E(f_t)}, \end{aligned}$$

(36)

Without social security, we can use (4) and (8) to rewrite (36) as

$$\begin{aligned} \varphi (f^n_t)=\frac{\sqrt{[\delta H(e_t)]^2[p\varepsilon _h^2+(1-p)\varepsilon _l^2]}}{\delta H(e_t)}=\sqrt{\frac{\varepsilon _l\varepsilon _h^2}{\varepsilon _l-\varepsilon _h}-\frac{\varepsilon _l^2\varepsilon _h}{\varepsilon _l-\varepsilon _h}}=\sqrt{-\varepsilon _l\varepsilon _h}. \end{aligned}$$

(37)

With social security, we can use (8) to rewrite (36) as

$$\begin{aligned} \varphi (f^k_t)=\frac{\sqrt{[\delta H(e_t)]^2-[\delta H(e_t)]^2}}{\delta H(e_t)}=0. \end{aligned}$$

(38)

1.5 Proof of Proposition 4

Propositions 1 and 2 imply that $H^{\prime }(e^n)>H^{\prime }(e^k)$. Therefore, $e^n<e^k$ under the assumption of $H^{\prime \prime }<0$.

1.6 Proof of Proposition 5

We can infer from (17) and (19) that $H^{\prime }(e^*)<H^{\prime }(e^k)$ for $\delta \in (0,1)$. By the assumption $H^{\prime \prime }<0$ and Proposition 4, we have $e^*>e^k_t>e^n_t$.

1.7 Proof of Proposition 6

Given that $v^{\prime\prime}<0$, for any $w_{it}$, $s_{it}$, and $e_t$ where $i\in \{h,l\}$, we can infer from Eq. (11):

$$\begin{aligned} U^n(e_t)& = p\big \{u[(1-\delta )w_{ht}-s_{ht}-e_t]+Ev[(1+r)s_{ht}+\delta (1+\varepsilon )H(e_t)]\big \} \\&\quad+(1-p)\big \{u[(1-\delta )w_{lt}-s_{lt}-e_t]+Ev[(1+r)s_{lt}+\delta (1+\varepsilon )H(e_t)]\big \} \\< & {} p\big \{u[(1-\delta )w_{ht}-s_{ht}-e_t]+v[(1+r)s_{ht}+\delta E(1+\varepsilon )H(e_t)]\big \} \\&\quad+(1-p)\big \{u[(1-\delta )w_{lt}-s_{lt}-e_t]+v[(1+r)s_{lt}+\delta E(1+\varepsilon )H(e_t)]\big \} \\& = p\big \{u[(1-\delta )w_{ht}-s_{ht}-e_t]+v[(1+r)s_{ht}+\delta H(e_t)]\big \} \\&\quad+(1-p)\big \{u[(1-\delta )w_{lt}-s_{lt}-e_t]+v[(1+r)s_{lt}+\delta H(e_t)]\big \}=U^k(e_t). \end{aligned}$$

(39)

It follows that for any $e^n_t$, $\max \{U^k_t\}\geqslant U^k(e^{n}_t)>U^{n}(e^{n}_t)=\max \{U^{n}_t\}.$

1.8 Proof of Proposition 7

We first examine the steady-state Gini coefficient without social security. By (21), the total income of each generation amounts to $H(e^n)$. Referring to Figure 1(a), we obtain $\lambda _1$ and $\lambda _3$ as follows:

$$\begin{aligned} \lambda _1& = \frac{(1-p)^2[(1+\varepsilon _l)H(e^n)]}{H(e^n)}= (1-p)^2(1+\varepsilon _l), \end{aligned}$$

(40)

$$\begin{aligned} \lambda _3& = \lambda _1+\frac{p(1-p)[1+(1-\delta ) \varepsilon _l+\delta \varepsilon _h]H(e^n)+p(1-p) [1+(1-\delta )\varepsilon _h+\delta \varepsilon _l]H(e^n)}{H(e^n)} \\& = (1-p)[1+(1+\varepsilon _l)p]. \end{aligned}$$

(41)

However, the expression of $\lambda _2$ depends on the income ranking. In case that $\delta <\frac{1}{2}$, the ranking is $I_{ll}<I_{lh}<I_{hl}<I_{hh}$. It follows that

$$\begin{aligned} \lambda _2=\lambda _1+\frac{p(1-p)[1+(1-\delta )\varepsilon _l+\delta \varepsilon _h]H(e^n)}{H(e^n)} =(1-p)[1+(1-\delta )\varepsilon _l]. \end{aligned}$$

(42)

In case that $\delta \geqslant \frac{1}{2}$, the income ranking is $I_{ll}<I_{hl}\leqslant I_{lh}<I_{hh}$. It follows that

$$\begin{aligned} \lambda _2=\lambda _1+\frac{(1-p)p[1+(1-\delta )\varepsilon _h+\delta \varepsilon _l]H(e^n)}{H(e^n)} =(1-p)(1+\delta \varepsilon _l). \end{aligned}$$

(43)

Plugging (40)–(43) into (22) and rearranging obtains $G^n$ as

$$\begin{aligned} G^n=\left\{ \begin{array}{ll} (1-p)[2p(1-p)\delta \varepsilon _l-\varepsilon _l]&{}\hbox { if}\ \delta <\frac{1}{2}\\ (1-p)[2p(1-p)(1-\delta )\varepsilon _l-\varepsilon _l]&{}\text {otherwise} \end{array} \right. , \end{aligned}$$

(44)

We then examine the Gini coefficient with social security. The share of total income earned by all l-type members can be expressed by

$$\begin{aligned} \lambda _4=\frac{(1-p)(1+\varepsilon _l)H(e^k)}{(1-p)(1+\varepsilon _l)H(e^k)+p(1+\varepsilon _h)H(e^k)} =(1-p)(1+\varepsilon _l). \end{aligned}$$

(45)

Inserting (45) into (23) yields $G^k=-(1-p)\varepsilon _l$. By this equation and (44), we show that

$$\begin{aligned} G^n-G^k=\left\{ \begin{array}{ll} 2p(1-p)^2\delta \varepsilon _l<0&{}\hbox { if}\ \delta<\frac{1}{2}\\ 2p(1-p)^2(1-\delta )\varepsilon _l<0&{}\text {otherwise} \end{array} \right. . \end{aligned}$$

(46)

1.9 Proof of Proposition 8

Given the specific functions in (24), the optimal educational spending per child can be written as

$$\begin{aligned} e^k_t=\frac{\beta \delta }{1+r}. \end{aligned}$$

(47)

Plugging (47) into (15) solves the i-type individual’s optimal (interior) saving level as

$$\begin{aligned}&\frac{1}{\beta (1-\delta )(1+\varepsilon _i)\ln e^k_{t-1}-s^k_{it}-e_{t}}=\frac{\alpha (1+r)}{(1+r)s^k_{it}+\beta \delta \ln e^k_t} \\&\quad \Leftrightarrow (1+\alpha )s^k_{it}=\alpha \left[ \beta (1-\delta )(1+\varepsilon _i)\ln \frac{\beta \delta }{1+r}-\frac{\beta \delta }{1+r}\right] -\frac{\beta \delta }{1+r}\ln \frac{\beta \delta }{1+r}, \\&\quad \Leftrightarrow s^k_{it}=\frac{\beta }{1+\alpha }\left[ \alpha (1-\delta )(1+\varepsilon _i)-\frac{\delta }{1+r}\right] \ln \frac{\beta \delta }{1+r}-\frac{\alpha \beta \delta }{(1+\alpha )(1+r)} \end{aligned}$$

(48)

Substituting (47) and (48) into (16) derives the steady-state social welfare under the social security system as

$$\begin{aligned} U^k& = p\ln \\&\left[ \frac{\beta (1-\delta )(1+\varepsilon _h)}{1+\alpha }\ln \frac{\beta \delta }{1+r}+\frac{\beta \delta }{(1+\alpha )(1+r)}\ln \frac{\beta \delta }{1+r}-\frac{\beta \delta }{(1+\alpha )(1+r)}\right] \\&\quad+\alpha p\ln \left[ \frac{\alpha \beta (1+\delta )(1+\varepsilon _h)(1+r)}{1+\alpha }\ln \frac{\beta \delta }{1+r}+\frac{\alpha \beta \delta }{1+\alpha }\ln \frac{\beta \delta }{1+r}-\frac{\alpha \beta \delta }{1+r}\right] \\&\quad+(1-p)\ln \left[ \frac{\beta (1-\delta )(1+\varepsilon _l)}{1+\alpha }\ln \frac{\beta \delta }{1+r}+\frac{\beta \delta }{(1+\alpha )(1+r)}\ln \frac{\beta \delta }{1+r}-\frac{\beta \delta }{(1+\alpha )(1+r)}\right] \\&\quad+\alpha (1-p)\ln \left[ \frac{\alpha \beta (1+\delta )(1+\varepsilon _l)(1+r)}{1+\alpha }\ln \frac{\beta \delta }{1+r}+\frac{\alpha \beta \delta }{1+\alpha }\ln \frac{\beta \delta }{1+r}-\frac{\alpha \beta \delta }{1+r}\right] \\& = \ln \bigg [\frac{\alpha ^\alpha }{1+r}\left( \frac{\beta }{1+\alpha }\right) ^{1+\alpha } \bigg ] \\&\quad+p(1+\alpha )\ln \left\{ [(1+r)(1+\varepsilon _h)(1-\delta )+\delta ]\ln \frac{\beta \delta }{1+r}-\delta \right\} \\&\quad+(1-p)(1+\alpha )\ln \left\{ [(1+r)(1+\varepsilon _l)(1-\delta )+\delta ]\ln \frac{\beta \delta }{1+r}-\delta \right\} . \end{aligned}$$

(49)

Differentiating (49) with respect to $\delta$ and setting it to zero yields the interior solution to $\delta$:

$$\begin{aligned} \frac{\partial U^k_t}{\partial \delta }& = p(1+\alpha ) \left\{ \frac{-(r+\varepsilon _h+r\varepsilon _h)\ln \frac{\beta \delta }{1+r}+[(1+r)(1+\varepsilon _h)(1-\delta )+\delta ]/\delta -1}{[(1+r)(1+\varepsilon _h)(1-\delta )+\delta ]\ln \frac{\beta \delta }{1+r}-\delta }\right\} \\&\quad +(1-p)(1+\alpha ) \left\{ \frac{-(r+\varepsilon _l+r\varepsilon _l)\ln \frac{\beta \delta }{1+r}+[(1+r)(1+\varepsilon _l)(1-\delta )+\delta ]/\delta -1}{[(1+r)(1+\varepsilon _l)(1-\delta )+\delta ]\ln \frac{\beta \delta }{1+r}-\delta }\right\} =0\\\Leftrightarrow & {} \frac{1-p}{p}\left\{ \frac{(1+r+\varepsilon _l+r\varepsilon _l)(1-\delta )-\delta (r+\varepsilon _l+r\varepsilon _l)\ln \frac{\beta \delta }{1+r}}{[(1+r+\varepsilon _l+r\varepsilon _l)(1-\delta )+\delta ]\ln \frac{\beta \delta }{1+r}-\delta }\right\} \\&\quad+ \frac{(1+r+\varepsilon _h+r\varepsilon _h)(1-\delta )-\delta (r+\varepsilon _h+r\varepsilon _h)\ln \frac{\beta \delta }{1+r}}{[(1+r+\varepsilon _h+r\varepsilon _h)(1-\delta )+\delta ]\ln \frac{\beta \delta }{1+r}-\delta }=0\\\Leftrightarrow & {} \frac{\varepsilon _h}{\varepsilon _l} \frac{(r+\varepsilon _l+r\varepsilon _l)\left( 1-\delta - \delta \ln \frac{\beta \delta }{1+r}\right) +1-\delta }{[1+(r+\varepsilon _l+r\varepsilon _l)(1-\delta )] \ln \frac{\beta \delta }{1+r}-\delta } \\& = \frac{(r+\varepsilon _h+r\varepsilon _h)\left( 1-\delta -\delta \ln \frac{\beta \delta }{1+r}\right) +1-\delta }{[1+(r+\varepsilon _h+r\varepsilon _h)(1-\delta )]\ln \frac{\beta \delta }{1+r}-\delta }, \end{aligned}$$

which is equivalent to Eq. (25).

1.10 Proof of Proposition 9

It is clear from (25) that ${\widetilde{\delta }}^k$ is not a function of $\alpha$. Besides, totally differentiating (25) with respect to $\delta$ and $\beta$ obtains:

$$\begin{aligned}&\frac{\left\{ (r+\varepsilon _h+r\varepsilon _h)\left[ -d\delta -\ln \frac{\beta \delta }{1+r}d\delta -\delta \left( \frac{d\beta }{\beta }+\frac{d\delta }{\delta }\right) \right] -d\delta \right\} \left[ (r+\varepsilon _l+r\varepsilon _l)\left( 1-\delta -\delta \ln \frac{\beta \delta }{1+r}\right) +1-\delta \right] }{\left[ (r+\varepsilon _l+r\varepsilon _l)\left( 1-\delta -\delta \ln \frac{\beta \delta }{1+r}\right) +1-\delta \right] ^2} \\&\qquad -\frac{\left\{ (r+\varepsilon _l+r\varepsilon _l)\left[ -d\delta -\ln \frac{\beta \delta }{1+r}d\delta -\delta \left( \frac{d\beta }{\beta }+\frac{d\delta }{\delta }\right) \right] -d\delta \right\} \left[ (r+\varepsilon _h+r\varepsilon _h)\left( 1-\delta -\delta \ln \frac{\beta \delta }{1+r}\right) +1-\delta \right] }{\left[ (r+\varepsilon _l+r\varepsilon _l)\left( 1-\delta -\delta \ln \frac{\beta \delta }{1+r}\right) +1-\delta \right] ^2} \\&\quad =\frac{\left\{ -(r+\varepsilon _h+r\varepsilon _h)\ln \frac{\beta \delta }{1+r}d\delta +[1+(r+\varepsilon _l+r\varepsilon _l)(1-\delta )]\left( \frac{d\beta }{\beta }+\frac{d\delta }{\delta }\right) -d\delta \right\} \left\{ [1+(r+\varepsilon _l+r\varepsilon _l)(1-\delta )]\ln \frac{\beta \delta }{1+r}-\delta \right\} }{\frac{\varepsilon _l}{\varepsilon _h}\left\{ [1+(r+\varepsilon _l+r\varepsilon _l)(1-\delta )]\ln \frac{\beta \delta }{1+r}-\delta \right\} ^2}\\&\qquad -\frac{\left\{ -(r+\varepsilon _l+r\varepsilon _l)\ln \frac{\beta \delta }{1+r}d\delta +[1+(r+\varepsilon _h+r\varepsilon _h)(1-\delta )]\left( \frac{d\beta }{\beta }+\frac{d\delta }{\delta }\right) -d\delta \right\} \left\{ [1+(r+\varepsilon _h+r\varepsilon _h)(1-\delta )]\ln \frac{\beta \delta }{1+r}-\delta \right\} }{\frac{\varepsilon _l}{\varepsilon _h}\left\{ [1+(r+\varepsilon _l+r\varepsilon _l)(1-\delta )]\ln \frac{\beta \delta }{1+r}-\delta \right\} ^2}\\&\quad \Leftrightarrow \frac{(\varepsilon _h+r\varepsilon _h-\varepsilon _l-r\varepsilon _l)(1-\delta )\left( -2d\delta -\ln \frac{\beta \delta }{1+r}d\delta -\frac{\delta }{\beta }d\beta \right) +(\varepsilon _h+r\varepsilon _h-\varepsilon _l-r\varepsilon _l)\left( 1-\delta -\delta \ln \frac{\beta \delta }{1+r}\right) d\delta }{\left[ (r+\varepsilon _l+r\varepsilon _l)\left( 1-\delta -\delta \ln \frac{\beta \delta }{1+r}\right) +1-\delta \right] ^2} \\&\quad =\frac{(\varepsilon _h+r\varepsilon _h-\varepsilon _l-r\varepsilon _l)\left\{ \left[ (1-\delta )\frac{d\beta }{\beta }+(1-\delta )\frac{d\delta }{\delta }-\ln \frac{\beta \delta }{1+r}d\delta \right] \left( \ln \frac{\beta \delta }{1+r}-\delta \right) +(1-\delta )\ln \frac{\beta \delta }{1+r}\left( d\delta -\frac{d\beta }{\beta }-\frac{d\delta }{\delta }\right) \right\} }{\frac{\varepsilon _l}{\varepsilon _h}\left\{ [1+(r+\varepsilon _l+r\varepsilon _l)(1-\delta )]\ln \frac{\beta \delta }{1+r}-\delta \right\} ^2} \\&\quad \Leftrightarrow \frac{\frac{\delta }{\beta }d\beta +\left( 1+\frac{1}{1-\delta } \ln \frac{\beta \delta }{1+r}\right) d\delta }{\left[ (r+\varepsilon _l+r\varepsilon _l) \left( 1-\delta -\delta \ln \frac{\beta \delta }{1+r}\right) +1-\delta \right] ^2} \\&\quad =\frac{\frac{\delta }{\beta }d\beta +\left( 1+\frac{1}{1-\delta } \ln \frac{\beta \delta }{1+r}\right) d\delta +\frac{1}{1-\delta } \left( \ln \frac{\beta \delta }{1+r}\right) ^2d\delta -\frac{2}{1-\delta } \ln \frac{\beta \delta }{1+r}d\delta }{\frac{\varepsilon _l}{\varepsilon _h} \left\{ [1+(r+\varepsilon _l+r\varepsilon _l)(1-\delta )]\ln \frac{\beta \delta }{1+r} -\delta \right\} ^2} \\&\quad \Leftrightarrow \frac{\varepsilon _l}{\varepsilon _h} \left\{ \frac{[1+(r+\varepsilon _l+r\varepsilon _l)(1-\delta )] \ln \frac{\beta \delta }{1+r}-\delta }{(r+\varepsilon _l+r\varepsilon _l) \left( 1-\delta -\delta \ln \frac{\beta \delta }{1+r}\right) +1-\delta }\right\} ^2-1 \\&\quad =\frac{\frac{1}{1-\delta }\ln \frac{\beta \delta }{1+r}\left( \ln \frac{\beta \delta }{1+r}-2\right) }{\left( 1+\frac{1}{1-\delta }\ln \frac{\beta \delta }{1+r}\right) +\frac{\delta }{\beta }\frac{d\beta }{d\delta }} \\&\quad \Leftrightarrow \frac{d\beta }{d\delta }=\frac{\beta }{\delta }\left\{ \frac{\frac{1}{1-\delta }\ln \frac{\beta \delta }{1+r}\left( \ln \frac{\beta \delta }{1+r}-2\right) }{\frac{\varepsilon _l}{\varepsilon _h}\left\{ \frac{[1+(r+\varepsilon _l+r\varepsilon _l)(1-\delta )]\ln \frac{\beta \delta }{1+r}-\delta }{(r+\varepsilon _l+r\varepsilon _l)\left( 1-\delta -\delta \ln \frac{\beta \delta }{1+r}\right) +1-\delta }\right\} ^2-1}-\left( 1+\frac{1}{1-\delta }\ln \frac{\beta \delta }{1+r}\right) \right\} , \end{aligned}$$

where the first term in the bracket of the right-hand-side is negative if $\ln \frac{\beta \delta }{1+r}-2>0$, while the second term is positive. Therefore, $\frac{d\delta }{d\beta }<0$ if $\ln \frac{\beta \delta }{1+r}>2$.

Appendix 2: A numerical example of Sect. 3.1

This appendix provides a simulation on the evolution of public educational expenditure, private savings, and social welfare without social security. Our numerical exercise rests on specific functional forms in (24). We first rewrite the i-type individual’s maximization problem (10) as a quadratic function of $s^n_{it}$:

$$\begin{aligned}&\frac{\alpha (1+r) p}{(1+r)s^n_{it}+\beta \delta (1+\varepsilon _h)\ln (e^n_t)}+\frac{\alpha (1+r)(1-p)}{(1+r)s^n_{it}+\beta \delta (1+\varepsilon _l)\ln (e^n_t)} \\&\quad =\frac{1}{\beta (1-\delta )(1+\varepsilon _i)\ln (e^n_{t-1})-s^n_{it}-e^n_t}, \\&\quad \Leftrightarrow \alpha \left[ s^n_{it}+\frac{\beta \delta }{1+r}(1+\varepsilon _h+\varepsilon _l)\ln (e^n_t)\right] =\frac{\left[ s^n_{it}+\frac{\beta \delta (1+\varepsilon _h)\ln (e^n_t)}{1+r}\right] \left[ s^n_{it}+\frac{\beta \delta (1+\varepsilon _l)\ln (e^n_t)}{1+r}\right] }{\beta (1-\delta )(1+\varepsilon _i)\ln (e^n_{t-1})-s^n_{it}-e^n_t} \\&\quad \Leftrightarrow (s^n_{it})^2+\left\{ \frac{\beta \delta \ln e^n_t}{1+r}\left( \varepsilon _h+\varepsilon _l+\frac{2+\alpha }{1+\alpha }\right) -\frac{\alpha [\beta (1-\delta )(1+\varepsilon _i)\ln e^n_{t-1}-e^n_t]}{1+\alpha }\right\} s^n_{it} \\&\qquad +\left( \frac{\beta \delta \ln e^n_t}{1+r}\right) ^2\frac{(1+\varepsilon _h)(1+\varepsilon _l)}{1+\alpha } \\&\qquad -\frac{\alpha \beta \delta (1+\varepsilon _h+\varepsilon _l)[\beta (1-\delta )(1+\varepsilon _i)\ln e^n_{t-1}-e^n_t]\ln e^n_t}{(1+\alpha )(1+r)}=0. \end{aligned}$$

(50)

The government’s optimal choice of $e^n_t$ characterized by (12) can be solved in the next equation:

$$\begin{aligned}&\frac{p[(1+r)(ps^n_{lt}+(1-p)s^n_{ht})+\beta \delta (1+\varepsilon _h)\ln e^n_t]\left[ (1+r)e^n_t-\beta \delta (1+\varepsilon _h)\right] }{[(1+r)s^n_{ht}+\beta \delta (1+\varepsilon _h)\ln e^n_t][(1+r)s^n_{lt}+\beta \delta (1+\varepsilon _h)\ln e^n_t]} \\&\quad =\frac{(p-1)[(1+r)(ps^n_{lt}+(1-p)s^n_{ht})+\beta \delta (1+\varepsilon _l)\ln e^n_t]\left[ (1+r)e^n_t-\beta \delta (1+\varepsilon _l)\right] }{[(1+r)s^n_{ht}+\beta \delta (1+\varepsilon _l)\ln e^n_t][(1+r)s^n_{lt}+\beta \delta (1+\varepsilon _l)\ln e^n_t]}, \end{aligned}$$

(51)

where $p=\frac{\varepsilon _l}{\varepsilon _l-\varepsilon _h}$ by (4). Combining (50) and (51) implies that $e^n_t$ is governed by $e^n_{t-1}$ and six parameters, namely $(e^n_{t-1}; \alpha , \beta , r, \varepsilon _h, \varepsilon _l, \delta )$.

To do the simulation, we rely on the values of parameters as in Table 1. In addition, we set the initial level of educational spending to 9 (i.e., $e^n_0=9$) and the fraction of earnings transferred to parents to 0.4 (i.e., $\delta =40\%$). Table 2 reports the simulated results of $e^n_t$, $s^n_{lt}$, $s^n_{ht}$, and $U^n_t$ in every period t. For example, starting from 9 in the period 0, educational spending per child stands at 17.076 in period 1, increases to 17.497 in period 2, and then converges to its steady-state level of 17.511. The other three variables exhibit a similar pattern: private savings and social welfare rise gradually to their respective steady-state levels.

Table 2 Transitional dynamics and steady state under the intrafamily transfer scheme