Education, social mobility, and the mismatch of talents

Abstract

This study presents a two-class, overlapping-generation model featuring social mobility inhibited by the mismatch of talents. Mobility decreases as the private education gap between the two classes widens, whereas it increases with increased public education spending. Within this framework, the study considers voting on public education and shows that when the political power of the rich is strong, the government implements low redistributive expenditures, which in turn induces a cyclical motion of social mobility across generations.

Appendix

1.1 Derivation of Eq. (2)

As shown in Table 1, workers in period $t+1$ can be classified into four types according to the transmission of innate ability from their parents to them and their parents’ social classes. Given Assumptions (A1) and (A2), type A and type D workers are allocated to the correct social class with a probability 1. Type B and type C workers are allocated to the correct social class with a probability $\alpha _{t+1}$. By using this classification, Eq. (2) is obtained as follows:

$$\begin{aligned} m_{t+1}= & {} \underbrace{m_{t}q}_{\text {type A}}+\underbrace{(1-m_{t} )q\alpha _{t+1}}_{\text {type B}}+\underbrace{m_{t}(1-q)\alpha _{t+1} }_{\text {type C}}+\underbrace{(1-m_{t})(1-q)}_{\text {type D}}\\= & {} m_{t}q+(1-m_{t})(1-q)+\left( (1-m_{t})q+m_{t}(1-q)\right) \alpha _{t+1}\\= & {} 2-\left( c-d\cdot \left( e_{t}-\ln \left( s_{t}^{\mathrm{R}}/s_{t}^{\mathrm{P}}\right) \right) \right) -q+m_{t}(2q-1). \end{aligned}$$

Table 1 Classification of workers in period $t+1$ per transmission of their parents’ innate ability and social classes

1.2 Derivation of Eq. (3)

Given the classification in Table 1 and Assumptions (A1) and (A2), type A and type D workers are allocated to the same social class as that of their parents with a probability 1. Type B and type C workers are allocated to the same social class as that of their parents with a probability $1-\alpha _{t+1}$. By using this classification, Eq. (3) is obtained as follows:

$$\begin{aligned} {\tilde{q}}_{t+1}= & {} \underbrace{m_{t}q}_{\text {type A}}+\underbrace{m_{t}(1-q)(1-\alpha _{t+1})}_{\text {type B}}+\underbrace{(1-m_{t} )q(1-\alpha _{t+1})}_{\text {type C}}+\underbrace{(1-m_{t})(1-q)}_{\text {type D}}\\= & {} 1-\left( (1-m_{t})q+m_{t}(1-q)\right) \alpha _{t+1}\\= & {} c-d\cdot \left( e_{t}-\ln \left( s_{t}^{\mathrm{R}}\Big /s_{t}^{\mathrm{P}}\right) \right) . \end{aligned}$$

1.3 Proof of Proposition 1

The procedure for the proof of Proposition 1 is identical to that in Bernasconi and Profeta (2007, 2012). The first- and second-order derivatives of $\tau _{t}$ and $\gamma _{t}$ are obtained as follows:

$$\begin{aligned} \frac{\partial W}{\partial \tau _{t}}= & {} \left( \omega (1+\eta +B)-B\right) \frac{-y_{t}^{\mathrm{P}}+\gamma _{t}{\overline{y}}_{t}}{(1-\tau _{t} )y_{t}^{\mathrm{P}}+\gamma _{t}\tau _{t}{\overline{y}}_{t}}+((1+\eta +B)-\omega B)\nonumber \\&\frac{-y_{t} ^{\mathrm{R}}+\gamma _{t}{\overline{y}}_{t}}{(1-\tau _{t})y_{t}^{\mathrm{R}}+\gamma _{t}\tau _{t}{\overline{y}}_{t}} +\frac{1+\omega }{\tau _{t}}\xi -B(1-\omega )(1-\gamma _{t}){\overline{y}}_{t}, \end{aligned}$$

(22)

$$\begin{aligned} \frac{\partial W}{\partial \gamma _{t}}= & {} \left( \omega (1+\eta +B)-B\right) \frac{\tau _{t}{\overline{y}}_{t}}{(1-\tau _{t})y_{t}^{\mathrm{P}}+\gamma _{t}\tau _{t}{\overline{y}}_{t}}+((1+\eta +B)-\omega B) \nonumber \\&\frac{\tau _{t}{\overline{y}}_{t} }{(1-\tau _{t})y_{t}^{\mathrm{R}}+\gamma _{t}\tau _{t}{\overline{y}}_{t}}-\frac{1+\omega }{1-\gamma _{t}}\xi +B(1-\omega )\tau _{t}{\overline{y}}_{t}, \\ \frac{\partial ^{2}W}{\partial \tau _{t}^{2}}= & {} -\left( \omega (1+\eta +B)-B\right) \left( \frac{-y_{t}^{\mathrm{P}}+\gamma _{t}{\overline{y}}_{t}}{(1-\tau _{t})y_{t}^{\mathrm{P}}+\gamma _{t}\tau _{t}{\overline{y}}_{t}}\right) ^{2}-((1+\eta +B)-\omega B) \nonumber \\&\left( \frac{-y_{t}^{\mathrm{R}}+\gamma _{t}{\overline{y}}_{t}}{(1-\tau _{t})y_{t}^{\mathrm{R}}+\gamma _{t}\tau _{t}{\overline{y}}_{t}}\right) ^{2}-\frac{1+\omega }{\tau _{t}^{2}}\xi , \nonumber \\ \frac{\partial ^{2}W}{\partial \gamma _{t}^{2}}= & {} -\left( \omega (1+\eta +B)-B\right) \left( \frac{\tau _{t}{\overline{y}}_{t}}{(1-\tau _{t})y_{t}^{\mathrm{P}}+\gamma _{t}\tau _{t}{\overline{y}}_{t}}\right) ^{2}-((1+\eta +B)-\omega B) \nonumber \\&\left( \frac{\tau _{t}{\overline{y}}_{t}}{(1-\tau _{t})y_{t}^{\mathrm{R}}+\gamma _{t}\tau _{t}{\overline{y}}_{t}}\right) ^{2}-\frac{1+\omega }{(1-\gamma _{t})^{2}}\xi . \nonumber \end{aligned}$$

(23)

Given $\frac{B}{1+\eta +B}<\omega <\frac{1+\eta +B}{B}$ from Assumption 1, $\frac{\partial ^{2}W}{\partial \tau _{t}^{2}}<0$ and $\frac{\partial ^{2}W}{\partial \gamma _{t}^{2}}<0$ hold. Note that $\lim _{\tau _{t}\rightarrow 0}\frac{\partial W}{\partial \tau _{t}}=+\infty$, $\lim _{\gamma _{t}\rightarrow 1}\frac{\partial W}{\partial \gamma _{t}}=-\infty$, $\lim _{\tau _{t}\rightarrow 1}\frac{\partial W}{\partial \tau _{t}}\big |_{\gamma _{t}=0}=-\infty$, and $\lim \nolimits _{\gamma _{t}\rightarrow 0}\frac{\partial W}{\partial \gamma _{t}}\big |_{\tau _{t}=1}=+\infty$. Therefore, the optimal pair of policies, $(\tau _{t}^{*},\gamma _{t}^{*})$, can be classified into the following three types: $\tau _{t}\in (0,1)$ and $\gamma _{t}=0$, $\tau _{t}=1$ and $\gamma _{t}\in (0,1)$, and $\tau _{t}\in (0,1)$ and $\gamma _{t}\in (0,1)$. All variables are indexed by t and thus we omit the index t from the expressions.

• Case $1\le \omega <\frac{1+\eta +B}{B}$

Suppose that $1\le \omega <\frac{1+\eta +B}{B}$. The following inequality holds:

$$\begin{aligned} (\omega -1)((1+\eta +B)+B)\ge 0 \Leftrightarrow \omega (1+\eta +B)-B\ge (1+\eta +B)-\omega B. \qquad \end{aligned}$$

(24)

To simplify the notations, we define the functions as follows:

$$\begin{aligned} C(\tau ,\gamma ,\omega )\equiv & {} (\omega (1+\eta +B)-B)\frac{-y^{\mathrm{P}}+\gamma {\overline{y}}}{(1-\tau )y^{\mathrm{P}}+\gamma \tau {\overline{y}}}\\&+\,((1+\eta +B)-\omega B)\frac{-y^{\mathrm{R}}+\gamma {\overline{y}}}{(1-\tau )y^{\mathrm{R}}+\gamma \tau {\overline{y}}},\\ D(\tau ,\gamma ,\omega )\equiv & {} \frac{1+\omega }{\tau }\xi -B(1-\omega )(1-\gamma ){\overline{y}}. \end{aligned}$$

Given these definitions, we can rewrite (22) as $\partial W/\partial \tau =C(\tau ,\gamma ,\omega )+D(\tau ,\gamma ,\omega )$.

We obtain the first derivatives of $C(\tau ,\gamma ,\omega )$ with respect to $\tau$ as follows:

$$\begin{aligned} \frac{\partial C(\tau ,\gamma ,\omega )}{\partial \tau }= & {} -\,(\omega (1+\eta +B)-B)\left( \frac{-y^{\mathrm{P}}+\gamma {\overline{y}}}{(1-\tau )y^{\mathrm{P}}+\gamma \tau {\overline{y}}}\right) ^{2}\nonumber \\&-\,((1+\eta +B)-\omega B)\left( \frac{-y^{\mathrm{R}}+\gamma {\overline{y}}}{(1-\tau )y^{\mathrm{R}}+\gamma \tau {\overline{y}}}\right) ^{2}<0. \end{aligned}$$

(25)

To compute the optimal pair of policies, we perform four steps. In step 1, we show that $C(1,1,1)=0$. In step 2, we show that $C(\tau , \gamma ^{*}, \omega )+D(\tau , \gamma ^{*}, \omega )>C(\tau ,1,1)$ for $\gamma ^{*}$, such that $\frac{\partial W}{\partial \gamma }|_{\gamma =\gamma ^{*}}=0$. In step 3, we show that if the optimal solution exists, it is a pair of policies, ($\tau ^{*}=1$, $\gamma ^{*}\in (0,1)$). In step 4, we show that such an equilibrium uniquely exists and compute the value of $\gamma ^{*}\in (0,1)$.

Step 1

Given the definition of $C(\tau ,\gamma ,\omega )$, we have

$$\begin{aligned} C(\tau ,1,1)= & {} (1+\eta )\frac{-y^{\mathrm{P}}+{\overline{y}}}{(1-\tau )y^{\mathrm{P}}+\tau {\overline{y}}}+(1+\eta )\frac{-y^{\mathrm{R}}+{\overline{y}}}{(1-\tau )y^{\mathrm{R}}+\tau {\overline{y}}} \nonumber \\&=\frac{1}{2}\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) (1+\eta )\left( \frac{1}{(1-\tau )y^{\mathrm{P}}+\tau {\overline{y}}}-\frac{1}{(1-\tau )y^{\mathrm{R}}+\tau {\overline{y}}}\right) .\nonumber \\ \end{aligned}$$

(26)

Hence, we obtain

$$\begin{aligned} C(1,1,1)=0. \end{aligned}$$

(27)

Step 2

Let $\gamma ^{*}$ satisfy $\frac{\partial W}{\partial \gamma }|_{\gamma =\gamma ^{*}}=0$. Then, we can rewrite (23) as follows:

$$\begin{aligned}&(\omega (1+\eta +B)-B)\frac{\tau {\overline{y}}}{(1-\tau )y^{\mathrm{P}}+\gamma ^{*}\tau {\overline{y}}}+((1+\eta +B)-\omega B)\nonumber \\&\qquad \frac{\tau {\overline{y}}}{(1-\tau )y^{\mathrm{R}}+\gamma ^{*}\tau {\overline{y}}} =\frac{1+\omega }{1-\gamma ^{*}}\xi -B(1-\omega )\tau {\overline{y}} \nonumber \\&\quad \Leftrightarrow (\omega (1+\eta +B)-B)\frac{(1-\gamma ^{*}){\overline{y}}}{(1-\tau )y^{\mathrm{P}}+\gamma ^{*}\tau {\overline{y}}}+((1+\eta +B)-\omega B)\nonumber \\&\qquad \frac{(1-\gamma ^{*}){\overline{y}}}{(1-\tau )y^{\mathrm{R}}+\gamma ^{*}\tau {\overline{y}}} \nonumber \\&\quad =\frac{1+\omega }{\tau }\xi -B(1-\omega )(1-\gamma ^{*}){\overline{y}} \nonumber \\&\quad \Leftrightarrow (\omega (1+\eta +B)-B)\frac{(1-\gamma ^{*}){\overline{y}}}{(1-\tau )y^{\mathrm{P}}+\gamma ^{*}\tau {\overline{y}}}+((1+\eta +B)-\omega B)\nonumber \\&\qquad \frac{(1-\gamma ^{*}){\overline{y}}}{(1-\tau )y^{\mathrm{R}}+\gamma ^{*}\tau {\overline{y}}}=D(\tau ,\gamma ^{*},\omega ). \end{aligned}$$

(28)

Suppose that $\gamma ^{*}\in (0,1)$ and $y^{\mathrm{R}}>y^{\mathrm{P}}$. Given (24), we have

$$\begin{aligned}&\left( \frac{\omega (1+\eta +B)-B}{(1-\tau )y^{\mathrm{P}}+\gamma ^{*}\tau {\overline{y}}}-\frac{(1+\eta +B)-\omega B}{(1-\tau )y^{\mathrm{R}}+\gamma ^{*}\tau {\overline{y}}}\right) \nonumber \\&\qquad -\left( \frac{\omega (1+\eta +B)-B}{(1-\tau )y^{\mathrm{P}}+\tau {\overline{y}}}-\frac{(1+\eta +B)-\omega B}{(1-\tau )y^{\mathrm{R}}+\tau {\overline{y}}}\right) \nonumber \\&\quad =\left( 1-\gamma ^{*}\right) \tau {\overline{y}}\left( \frac{\omega (1+\eta +B)-B}{\left( (1-\tau )y^{\mathrm{P}}+\gamma ^{*}\tau {\overline{y}}\right) \left( (1-\tau )y^{\mathrm{P}}+\tau {\overline{y}}\right) }\right. \nonumber \\&\qquad \left. -\,\frac{(1+\eta +B)-\omega B}{\left( (1-\tau )y^{\mathrm{R}}+\gamma ^{*}\tau {\overline{y}}\right) \left( (1-\tau )y^{\mathrm{R}}+\tau {\overline{y}}\right) }\right) \nonumber \\&\quad >0. \end{aligned}$$

(29)

Given $1\le \omega <\frac{1+\eta +B}{B}$, $y^{\mathrm{R}}>y^{\mathrm{P}}$, (26), (28), and (29), we obtain

$$\begin{aligned}&C\left( \tau ,\gamma ^{*},\omega \right) +D\left( \tau ,\gamma ^{*},\omega \right) \nonumber \\&\quad = (\omega (1+\eta +B)-B)\frac{-y^{\mathrm{P}}+\gamma ^{*}{\overline{y}}}{(1-\tau )y^{\mathrm{P}}+\gamma ^{*}\tau {\overline{y}}} +((1+\eta +B)-\omega B)\nonumber \\&\qquad \times \, \frac{-y^{\mathrm{R}}+\gamma ^{*}{\overline{y}}}{(1-\tau )y^{\mathrm{R}}+\gamma ^{*}\tau {\overline{y}}} +(\omega (1+\eta +B)-B)\frac{(1-\gamma ^{*}){\overline{y}}}{(1-\tau )y^{\mathrm{P}}+\gamma ^{*}\tau {\overline{y}}} \nonumber \\&\qquad +\,((1+\eta +B)-\omega B)\frac{\left( 1-\gamma ^{*}\right) {\overline{y}}}{(1-\tau )y^{\mathrm{R}}+\gamma ^{*}\tau {\overline{y}}}\nonumber \\&\quad = \frac{1}{2}\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \left( \frac{\omega (1+\eta +B)-B}{(1-\tau )y^{\mathrm{P}}+\gamma ^{*}\tau {\overline{y}}}-\frac{(1+\eta +B)-\omega B}{(1-\tau )y^{\mathrm{R}}+\gamma ^{*}\tau {\overline{y}}}\right) \nonumber \\&\quad > \frac{1}{2}\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \left( \frac{\omega (1+\eta +B)-B}{(1-\tau )y^{\mathrm{P}}+\tau {\overline{y}}}-\frac{(1+\eta +B)-\omega B}{(1-\tau )y^{\mathrm{R}}+\tau {\overline{y}}}\right) \nonumber \\&\quad \ge \frac{1}{2}\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) (1+\eta )\left( \frac{1}{(1-\tau )y^{\mathrm{P}}+\tau {\overline{y}}}-\frac{1}{(1-\tau )y^{\mathrm{R}}+\tau {\overline{y}}}\right) \nonumber \\&\quad = C(\tau ,1,1). \end{aligned}$$

(30)

Step 3

Given (25), (27), and (30), the following inequality holds:

$$\begin{aligned} \left. \frac{\partial W}{\partial \tau }\right| _{\gamma =\gamma ^{*}}= & {} C\left( \tau ,\gamma ^{*},\omega \right) +D\left( \tau ,\gamma ^{*},\omega \right) >C(\tau ,1,1)\ge C(1,1,1) =0. \qquad \end{aligned}$$

(31)

Given $\partial ^{2}W/\partial \tau ^{2}<0$ and (31), we find that $\tau ^{*}=1$ is optimal when the optimal fraction $\gamma ^{*}$ is internal.

Step 4

First, we show the existence and uniqueness of the $\gamma ^{*}$ considered in step 3 in the interval (0, 1). Given $\lim \nolimits _{\gamma \rightarrow 1}\frac{\partial W}{\partial \gamma }=-\infty$, $\lim \nolimits _{\gamma \rightarrow 0}\frac{\partial W}{\partial \gamma }\big |_{\tau =1}=+\infty$, and $\partial ^{2}W/\partial \gamma ^{2}<0$, we find that $\gamma ^{*}$ uniquely exists in the interval (0, 1).

Next, we show that a pair of policies ($\tau \in (0,1), \gamma =0$), is not optimal. Suppose that $\tau \in (0,1)$ and $\gamma =0$. Given (22), we obtain the following equation:

$$\begin{aligned} \left. \frac{\partial W}{\partial \tau }\right| _{\tau \in (0,1), \gamma =0}=0 \Leftrightarrow \frac{\tau }{1-\tau }(\omega +1)(1+\eta )=(1+\omega )\xi -B(1-\omega )\tau {\overline{y}}.\qquad \end{aligned}$$

(32)

Substituting (32) into (23) leads to

$$\begin{aligned}&\left. \frac{\partial W}{\partial \gamma }\right| _{\tau \in (0,1), \gamma =0}=(\omega (1+\eta +B)-B)\frac{\tau {\overline{y}}}{(1-\tau )y^{\mathrm{P}}} +((1+\eta +B)-\omega B)\frac{\tau {\overline{y}}}{(1-\tau )y^{\mathrm{R}}}\\&\qquad -\frac{\tau }{1-\tau }(1+\omega )(1+\eta ) \\&\quad =\frac{\tau }{1-\tau }\left( (\omega (1+\eta +B)-B)\frac{{\overline{y}}}{y^{\mathrm{P}}}+((1+\eta +B)-\omega B)\frac{{\overline{y}}}{y^{\mathrm{R}}}-(1+\omega )(1+\eta )\right) \\&\quad =\frac{\tau }{2(1-\tau )}\left( (\omega (1+\eta +B)-B)\frac{y^{\mathrm{R}}}{y^{\mathrm{P}}}+((1+\eta +B)-\omega B)\frac{y^{\mathrm{P}}}{y^{\mathrm{R}}}-(1+\omega )(1+\eta )\right) . \end{aligned}$$

Therefore, the following property holds:

$$\begin{aligned}&\frac{\partial W}{\partial \gamma }\bigg |_{\tau \in (0,1), \gamma =0}\nonumber \\&\quad \gtreqless 0\Leftrightarrow (\omega (1+\eta +B)-B)\frac{y^{\mathrm{R}}}{y^{\mathrm{P}}}+((1+\eta +B)-\omega B)\frac{y^{\mathrm{P}}}{y^{\mathrm{R}}}\nonumber \\&\qquad -\,(1+\omega )(1+\eta )\gtreqless 0\nonumber \\&\quad \Leftrightarrow \left( \frac{y^{\mathrm{R}}}{y^{\mathrm{P}}}\right) ^{2}+\frac{(1+\eta +B)-\omega B}{\omega (1+\eta +B)-B}-\frac{(1+\omega )(1+\eta )}{\omega (1+\eta +B)-B}\cdot \frac{y^{\mathrm{R}}}{y^{\mathrm{P}}}\gtreqless 0\nonumber \\&\quad \Leftrightarrow \left( \frac{y^{\mathrm{R}}}{y^{\mathrm{P}}}-\frac{(1+\eta +B)-\omega B}{\omega (1+\eta +B)-B}\right) \left( \frac{y^{\mathrm{R}}}{y^{\mathrm{P}}}-1\right) \gtreqless 0. \end{aligned}$$

(33)

Given $y^{\mathrm{R}}>y^{\mathrm{P}}$ and (24), $\frac{\partial W}{\partial \gamma }\Big |_{\tau \in (0,1), \gamma =0}>0$ holds, which implies that a pair of policies ($\tau \in (0,1), \gamma =0$), cannot be optimal.

We now compute the value of $\gamma ^{*}$, which solves the following equation:

$$\begin{aligned} \frac{\partial W}{\partial \gamma }\bigg |_{\tau =1}=0\Leftrightarrow f(\gamma )=0, \end{aligned}$$

where $f(\gamma )$ is defined as follows:

$$\begin{aligned} f(\gamma )\equiv & {} B(\omega -1){\overline{y}}\gamma ^{2}-\left( (1+\omega )(1+\eta )+(1+\omega )\xi +B(\omega -1){\overline{y}}\right) \gamma \\&+\,(1+\omega )(1+\eta ). \end{aligned}$$

Note that the coefficient of $\gamma ^{2}$ is positive and $f(\gamma )$ has the following properties:

$$\begin{aligned} f(0)= & {} (1+\eta )(1+\omega )>0, \\ f(1)= & {} -(1+\omega )\xi <0. \end{aligned}$$

Therefore, we obtain the value of $\gamma ^{*}$ as follows:

$$\begin{aligned} \gamma ^{*}=\frac{E-\sqrt{E^{2}-4B(\omega -1){\overline{y}}(1+\omega )(1+\eta )} }{2B(\omega -1){\overline{y}}}, \end{aligned}$$

where E is defined by $E\equiv (1+\omega )(1+\eta )+(1+\omega )\xi +B(\omega -1){\overline{y}}$.

• Case $\frac{B}{1+\eta +B}<\omega <1$

First, we show that $\tau =1$ cannot be optimal when $\frac{B}{1+\eta +B}<\omega <1$. Recall that the optimal pair of policies can be classified into the following three types: $\tau \in (0,1)$ and $\gamma =0$, $\tau =1$ and $\gamma \in (0,1)$, and $\tau \in (0,1)$ and $\gamma \in (0,1)$. Therefore, if $\tau =1$ is optimal, the optimal $\gamma$ is ${\hat{\gamma }}\in (0,1)$, which satisfies the following equation:

$$\begin{aligned} \frac{\partial W}{\partial \gamma }\bigg |_{\tau =1, \gamma ={\hat{\gamma }}}=0\Leftrightarrow \frac{1-{\hat{\gamma }}}{{\hat{\gamma }}}(1+\omega )(1+\eta )=(1+\omega )\xi -B(1-\omega )(1-{\hat{\gamma }}){\overline{y}}.\nonumber \\ \end{aligned}$$

(34)

Suppose that $\tau =1$, $\gamma ={\hat{\gamma }}$, $\frac{B}{1+\eta +B}<\omega <1$, and $y^{\mathrm{R}}>y^{\mathrm{P}}$. We obtain

$$\begin{aligned} \frac{\partial W}{\partial \tau }\bigg |_{\tau =1,\gamma ={\hat{\gamma }}}= & {} (\omega (1+\eta +B)-B)\frac{-y^{\mathrm{P}}+{\hat{\gamma }}{\overline{y}}}{{\hat{\gamma }}{\overline{y}}}+((1+\eta +B)-\omega B)\frac{-y^{\mathrm{R}}+{\hat{\gamma }}{\overline{y}}}{{\hat{\gamma }}{\overline{y}}}\nonumber \\&+\,(1+\omega )\xi -B(1-\omega )(1-{\hat{\gamma }}){\overline{y}}\nonumber \\= & {} (\omega (1+\eta +B)-B)\frac{-y^{\mathrm{P}}+{\hat{\gamma }}{\overline{y}}}{{\hat{\gamma }}{\overline{y}}}+((1+\eta +B)-\omega B)\nonumber \\&\times \,\frac{-y^{\mathrm{R}}+{\hat{\gamma }}{\overline{y}}}{{\hat{\gamma }}{\overline{y}}}+\frac{1-{\hat{\gamma }}}{{\hat{\gamma }}}(1+\omega )(1+\eta )\nonumber \\= & {} \frac{1}{2{\hat{\gamma }}{\overline{y}}}\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) (\omega -1)((1+\eta +B)+B)< 0, \end{aligned}$$

(35)

where the second line uses (34). Given $\partial ^{2}W/\partial \tau ^{2}<0, \lim _{\tau \rightarrow 0}\frac{\partial W}{\partial \tau }=+\infty$, and (35), we find that a pair of policies ($\tau =1, {\hat{\gamma }}\in (0,1)$), cannot be optimal.

Hence, the optimal pair of policies can be of the following two types: $\tau \in (0,1)$ and $\gamma =0$, and $\tau \in (0,1)$ and $\gamma \in (0,1)$. Given $\partial ^{2}W/\partial \gamma ^{2}<0$ and $\lim _{\gamma \rightarrow 1}\frac{\partial W}{\partial \gamma }=-\infty$, (i) a pair of policies, ($\tau \in (0,1), \gamma =0$), is optimal if $\frac{\partial W}{\partial \gamma }\big |_{\tau \in (0,1),\gamma =0}\le 0$ holds and (ii) a pair of policies, ($\tau \in (0,1), \gamma \in (0,1)$), is optimal if $\frac{\partial W}{\partial \gamma }\big |_{\tau \in (0,1),\gamma =0}>0$ holds. Given (33), we can rewrite these conditions as follows:

$$\begin{aligned} \frac{\partial W}{\partial \gamma }\bigg |_{\tau \in (0,1),\gamma =0} \gtreqless 0\Leftrightarrow \frac{y^{\mathrm{R}}}{y^{\mathrm{P}}} \gtreqless \frac{(1+\eta +B)-\omega B}{\omega (1+\eta +B)-B}\Leftrightarrow \omega \gtreqless \frac{By^{\mathrm{R}}+(1+\eta +B)y^{\mathrm{P}}}{(1+\eta +B)y^{\mathrm{R}}+By^{\mathrm{P}}}. \end{aligned}$$

• Sub-case $\frac{B}{1+\eta +B}<\omega \le \frac{By^{\mathrm{R}}+(1+\eta +B)y^{\mathrm{P}}}{(1+\eta +B)y^{\mathrm{R}}+By^{\mathrm{P}}}$

In this case, a pair of policies ($\tau ^{*}\in (0,1), \gamma ^{*}=0$), is optimal. Therefore, $\tau ^{*}$ solves the following equation:

$$\begin{aligned} \frac{\partial W}{\partial \tau }\bigg |_{\gamma =0}=0\Leftrightarrow g(\tau )=0, \end{aligned}$$

where $g(\tau )$ is defined as follows:

$$\begin{aligned} g(\tau )\equiv B(1-\omega ){\overline{y}}\tau ^{2}-((1+\omega )(1+\eta )+(1+\omega )\xi +B(1-\omega ){\overline{y}})\tau +(1+\omega )\xi . \end{aligned}$$

Note that the coefficient of $\tau ^{2}$ is positive, and $g(\tau )$ has the following properties:

$$\begin{aligned} g(0)= & {} (1+\omega )\xi >0, \\ g(1)= & {} -(1+\omega )(1+\eta )<0. \end{aligned}$$

Therefore, we obtain the value of $\tau ^{*}$ as follows:

$$\begin{aligned} \tau ^{*}=\frac{F-\sqrt{F^{2}-4B(1-\omega )(1+\omega ){\overline{y}}\xi } }{2B(1-\omega ){\overline{y}}}, \end{aligned}$$

where F is defined by $F\equiv (1+\omega )(1+\eta )+(1+\omega )\xi +B(1-\omega ){\overline{y}}$.

• Sub-case $\frac{By^{\mathrm{R}}+(1+\eta +B)y^{\mathrm{P}}}{(1+\eta +B)y^{\mathrm{R}}+By^{\mathrm{P}}}<\omega <1$

In this case, a pair of policies ($\tau ^{*}\in (0,1), \gamma ^{*}\in (0,1)$), is optimal. Therefore, $\tau ^{*}$ and $\gamma ^{*}$ solve the following equations:

$$\begin{aligned} \frac{\partial W}{\partial \tau }= & {} 0 \nonumber \\\Leftrightarrow & {} -(\omega (1+\eta +B)-B)\frac{\tau \left( -y^{\mathrm{P}}+\gamma {\overline{y}}\right) }{(1-\tau )y^{\mathrm{P}}+\gamma \tau {\overline{y}}}\nonumber \\&-\,((1+\eta +B)-\omega B)\frac{\tau \left( -y^{\mathrm{R}}+\gamma {\overline{y}}\right) }{(1-\tau )y^{\mathrm{R}}+\gamma \tau {\overline{y}}} \nonumber \\= & {} (1+\omega )\xi -B(1-\omega )(1-\gamma ){\overline{y}}\tau , \end{aligned}$$

(36)

$$\begin{aligned} \frac{\partial W}{\partial \gamma }= & {} 0 \nonumber \\\Leftrightarrow & {} (\omega (1+\eta +B)-B)\frac{\tau {\overline{y}}(1-\gamma )}{(1-\tau )y^{\mathrm{P}}+\gamma \tau {\overline{y}}}\nonumber \\&+\,((1+\eta +B)-\omega B)\frac{\tau {\overline{y}}(1-\gamma )}{(1-\tau )y^{\mathrm{R}}+\gamma \tau {\overline{y}}} \nonumber \\= & {} (1+\omega )\xi -B(1-\omega )(1-\gamma ){\overline{y}}\tau . \end{aligned}$$

(37)

By substituting (36) into (37) and rearranging the terms, we obtain

$$\begin{aligned} \frac{\omega (1+\eta +B)-B}{(1-\tau )y^{\mathrm{P}}+\gamma \tau {\overline{y}}}= & {} \frac{(1+\eta +B)-\omega B}{(1-\tau )y^{\mathrm{R}}+\gamma \tau {\overline{y}}} \end{aligned}$$

(38)

$$\begin{aligned} \Leftrightarrow \gamma \tau= & {} \frac{(1-\tau )\left( (\omega (1+\eta +B)-B)y^{\mathrm{R}}-((1+\eta +B)-\omega B)y^{\mathrm{P}}\right) }{(1-\omega )((1+\eta +B)+B){\overline{y}}}.\nonumber \\ \end{aligned}$$

(39)

Substituting (38) into (37) results in

$$\begin{aligned} 2((1+\eta +B)-\omega B)\frac{(1-\gamma )\tau {\overline{y}}}{(1-\tau )y^{\mathrm{R}}+\gamma \tau {\overline{y}}}= & {} (1+\omega )\xi -B(1-\omega )(1-\gamma )\tau {\overline{y}} \nonumber \\ \Leftrightarrow (1-\gamma )\tau= & {} \frac{(1+\omega )\xi (1-\tau )\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) }{(1-\omega )\left( 2((1+\eta +B)+B)+B(1-\tau )\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \right) {\overline{y}}}.\nonumber \\ \end{aligned}$$

(40)

To eliminate $\gamma$ from (40), substituting (39) into (40) gives us

$$\begin{aligned}&\tau - \frac{(1-\tau )\left( (\omega (1+\eta +B)-B)y^{\mathrm{R}}-((1+\eta +B)-\omega B)y^{\mathrm{P}}\right) }{(1-\omega )((1+\eta +B)+B){\overline{y}}} \\&\qquad =\frac{(1+\omega )\xi (1-\tau )(y^{\mathrm{R}}-y^{\mathrm{P}})}{(1-\omega )(2((1+\eta +B)+B)+B(1-\tau )(y^{\mathrm{R}}-y^{\mathrm{P}})){\overline{y}}} \\&\qquad \Leftrightarrow h(\tau )=0, \end{aligned}$$

where $h(\tau )$ is defined as follows:

$$\begin{aligned} h( \tau )= & {} \frac{1}{2}B(1+\eta )(1+\omega )\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) ^{2}\tau ^{2} \\&-\,\left\{ {\overline{y}}(1-\omega )((1+\eta +B)+B)\left[ 2((1+\eta +B)+B)+B\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \right] \right. \\&+\,2\left[ (\omega (1+\eta +B)-B)y^{\mathrm{R}}-((1+\eta +B)-\omega B)y^{\mathrm{P}}\right] \\&\left[ ((1+\eta +B)+B)+B\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \right] \\&\left. +\,((1+\eta +B)+B)(1+\omega )\xi \left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \right\} \tau \\&+\,\left[ 2((1+\eta +B)+B)+B\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \right] \\&\left[ (\omega (1+\eta +B)-B)y^{\mathrm{R}}-((1+\eta +B)-\omega B)y^{\mathrm{P}}\right] \\&+\,(1+\omega )((1+\eta +B)+B)\xi \left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) . \end{aligned}$$

Note that the coefficient of $\tau ^{2}$ is positive and $h(\tau )$ has the following properties:

$$\begin{aligned} h(0)= & {} \left[ 2((1+\eta +B)+B)+B\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \right] \\&\left[ (\omega (1+\eta +B)-B)y^{\mathrm{R}}-((1+\eta +B)-\omega B)y^{\mathrm{P}}\right] \\&+\,(1+\omega )((1+\eta +B)+B)\xi \left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) >0, \\ h(1)= & {} -2{\overline{y}}(1-\omega )((1+\eta +B)+B)^{2}<0. \end{aligned}$$

Therefore, we obtain the value of $\tau ^{*}$ as follows:

$$\begin{aligned} \tau ^{*}=\frac{G-\sqrt{G^{2}-2B(1+\eta )(1+\omega )\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) ^{2}I}}{B(1+\eta )(1+\omega )\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) ^{2}}, \end{aligned}$$

(41)

where G and I are defined as follows:

$$\begin{aligned} G\equiv & {} {\overline{y}}(1-\omega )((1+\eta +B)+B)\left[ 2((1+\eta +B)+B)+B\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \right] \\&+\,2\left[ (\omega (1+\eta +B)-B)y^{\mathrm{R}}-((1+\eta +B)-\omega B)y^{\mathrm{P}}\right] \\&\left[ ((1+\eta +B)+B)+B\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \right] \\&+\,((1+\eta +B)+B)(1+\omega )\xi \left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) ,\\ I\equiv & {} \left[ 2((1+\eta +B)+B)+B\left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) \right] \\&\left[ (\omega (1+\eta +B)-B)y^{\mathrm{R}}-((1+\eta +B)-\omega B)y^{\mathrm{P}}\right] \\&+\,(1+\omega )((1+\eta +B)+B)\xi \left( y^{\mathrm{R}}-y^{\mathrm{P}}\right) . \end{aligned}$$

By substituting (41) into (39), the value of $\gamma ^{*}$ is obtained as follows:

$$\begin{aligned} \gamma ^{*}=\frac{(1-\tau ^{*})\left( (\omega (1+\eta +B)-B)y^{\mathrm{R}}-((1+\eta +B)-\omega B)y^{\mathrm{P}}\right) }{(1-\omega )((1+\eta +B)+B)\tau ^{*}{\overline{y}}}. \end{aligned}$$

1.4 Derivation of Eqs. (17) and (18)

Table 2 summarizes the information on the allocation of workers and output levels in a low-income occupation. From the result presented in Table 2, we can compute $y_{t+1}^{\mathrm{P}}$ as follows:

$$\begin{aligned} y_{t+1}^{\mathrm{P}}= & {} \left( \epsilon _{t+1}^{2}+2\epsilon _{t+1}\theta _{t+1}+2\epsilon _{t+1}(1-\epsilon _{t+1}-\theta _{t+1})\right) h_{t+1}^{\mathrm{P,L}}\\&+\, \left( \theta _{t+1}^{2}+2\theta _{t+1}(1-\epsilon _{t+1}-\theta _{t+1})\right) h_{t+1}^{\mathrm{R,L}} + \left( 1-\epsilon _{t+1}-\theta _{t+1}\right) ^{2}h_{t+1} ^{\mathrm{P,H}}\\= & {} \epsilon _{t+1}(2-\epsilon _{t+1})h_{t+1}^{\mathrm{P,L}}+\left( -\theta _{t+1}^{2} -2\theta _{t+1}\epsilon _{t+1}+2\theta _{t+1}\right) h_{t+1}^{\mathrm{R,L}}\\&+\,(1-\theta _{t+1}-\epsilon _{t+1})^{2}h_{t+1}^{\mathrm{P,H}}. \end{aligned}$$

Table 2 Pairs of workers and their fractions and output levels in a low-income occupation in period $t+1$

Table 3 summarizes the information on the allocation of workers and output levels in a high-income occupation. From the result presented in Table 3, we can compute $y_{t+1}^{\mathrm{R}}$ as follows:

$$\begin{aligned} y_{t+1}^{\mathrm{R}}= & {} \left( (1-\epsilon _{t+1}-\theta _{t+1})^{2} + 2(1-\epsilon _{t+1}-\theta _{t+1})\theta _{t+1} + 2\epsilon _{t+1}(1-\epsilon _{t+1} -\theta _{t+1})\right) h_{t+1}^{\mathrm{R,L}}\\&+\,(\theta _{t+1}^{2}+2\epsilon _{t+1}\theta _{t+1})h_{t+1}^{\mathrm{P,H}} + \epsilon _{t+1}^{2}h_{t+1}^{\mathrm{R,H}}\\= & {} \left( 1-(\theta _{t+1}+\epsilon _{t+1})^{2}\right) h_{t+1}^{\mathrm{R,L}}+\theta _{t+1}(\theta _{t+1}+2\epsilon _{t+1})h_{t+1}^{\mathrm{P,H}}+\epsilon _{t+1}^{2} h_{t+1}^{\mathrm{R,H}}. \end{aligned}$$

Table 3 Pairs of workers and their fractions and output levels in a high-income occupation in period $t+1$

1.5 Mismatch of talents, GDP, and pre-tax inequality

By using Eqs. (17) and (18), the following equations are obtained as follows:

$$\begin{aligned} \frac{\partial y_{t+1}^{\mathrm{P}}}{\partial \theta _{t+1}}= & {} 2(1-m_{t+1} )\left( h_{t+1}^{\mathrm{R,L}}-h_{t+1}^{\mathrm{P,H}}\right) , \end{aligned}$$

(42)

$$\begin{aligned} \frac{\partial y_{t+1}^{\mathrm{R}}}{\partial \theta _{t+1}}= & {} 2m_{t+1}\left( h_{t+1} ^{\mathrm{P,H}}-h_{t+1}^{\mathrm{R,L}}\right) , \end{aligned}$$

(43)

If $h_{t+1}^{\mathrm{P,H}}\ge h_{t+1}^{\mathrm{R,L}}$ is satisfied, $\frac{\partial y_{t+1} ^{\mathrm{P}}}{\partial \theta _{t+1}}\le 0$, and $\frac{\partial y_{t+1}^{\mathrm{R}}}{\partial \theta _{t+1}}\ge 0$ hold. This implies that a higher level of social mobility results in a larger level of pre-tax inequality between the rich and poor.

From Eqs. (16), (42), and (43), the following equations are obtained as follows:

$$\begin{aligned} \frac{\partial {\overline{y}}_{t+1}}{\partial \theta _{t+1}}= & {} \frac{1}{2} \frac{\partial y_{t+1}^{\mathrm{P}}}{\partial \theta _{t+1}}+\frac{1}{2}\frac{\partial y_{t+1}^{\mathrm{R}}}{\partial \theta _{t+1}}\\= & {} (1-m_{t+1})\left( h_{t+1}^{\mathrm{R,L}}-h_{t+1}^{\mathrm{P,H}}\right) +m_{t+1}\left( h_{t+1}^{\mathrm{P,H}} -h_{t+1}^{\mathrm{R,L}}\right) \\= & {} (1-2m_{t+1})\left( h_{t+1}^{\mathrm{R,L}}-h_{t+1}^{\mathrm{P,H}}\right) . \end{aligned}$$

If $h_{t+1}^{\mathrm{P,H}}\ge h_{t+1}^{\mathrm{R,L}}$ and $m_{t+1}\ge 1/2$ are satisfied, $\frac{\partial {\overline{y}}_{t+1}}{\partial \theta _{t+1}}\ge 0$ holds.

We now turn to verify whether $h_{t+1}^{\mathrm{P,H}}\ge h_{t+1}^{\mathrm{R,L}}$ and $m_{t+1}\ge 1/2$ for all t are satisfied.

Fig. 10

a, c Plot the dynamics of individual human capital. b, d Plot the dynamics of the fraction of workers allocated to the correct social class. We set $\omega =1.0$ in a, b and $\omega =0.3$ in c, d. In a and c, the solid curve presents the human capital of workers with high innate ability whose parents are poor and the dotted curve presents the human capital of workers with low innate ability whose parents are rich

As shown in Fig. 10, $h_{t+1}^{\mathrm{P,H}}\ge h_{t+1}^{\mathrm{R,L}}$ and $m_{t+1}\ge 1/2$ for all t are satisfied in the present numerical analysis. Therefore, $\frac{\partial y_{t+1}^{\mathrm{P}}}{\partial \theta _{t+1}}\le 0$, $\frac{\partial y_{t+1}^{\mathrm{R}}}{\partial \theta _{t+1}}\ge 0$, and $\frac{\partial \overline{y}_{t+1}}{\partial \theta _{t+1}}\ge 0$ hold for all t.

1.6 Numerical algorithm

We describe the numerical procedure to simulate the dynamics of GDP, aggregate human capital, social mobility, and the fraction of workers who have the same ability as their parents and belong to the occupation reflecting their innate ability. It is sufficient to use GDP to check economic convergence because it summarizes all the information on the economy. The procedure is as follows.

Step 1

We set the initial values $y_{1}^{\mathrm{P}}, y_{1}^{\mathrm{R}}, {\overline{y}}_{1}, \overline{H}_{1}$, and $m_{1}$.

Step 2

Given $y_{t}^{\mathrm{P}}, y_{t}^{\mathrm{R}}, {\overline{y}}_{t}, {\overline{H}}_{t}$, and $m_{t}$ for any t, we obtain $y_{t+1}^{\mathrm{P}}, y_{t+1}^{\mathrm{R}}, {\overline{y}}_{t+1}, {\overline{H}}_{t+1}, m_{t+1}$ by using the following equations:

• 1. $\tau _{t}=\left\{ \begin{array}{ll} \frac{F-\sqrt{F^{2}-4B(1-\omega )(1+\omega ){\overline{y}}_{t}\xi } }{2B(1-\omega ){\overline{y}}_{t}}&{}\quad \text {if}\,\frac{B}{1+\eta +B}<\omega \le {\tilde{\omega }}_{t},\\ \frac{G-\sqrt{G^{2}-2B(1+\eta )(1+\omega )\left( y_{t}^{\mathrm{R}}-y_{t}^{\mathrm{P}}\right) ^{2}I} }{B(1+\eta )(1+\omega )\left( y_{t}^{\mathrm{R}}-y_{t}^{\mathrm{P}}\right) ^{2}} &{}\quad \text {if}\,\tilde{\omega }_{t}<\omega<1,\\ 1&{}\quad \text {if}\, 1\le \omega <\frac{1+\eta +B}{B}, \end{array} \right.$

  where

  $$\begin{aligned} F= & {} (1+\omega )(1+\eta )+(1+\omega )\xi +B(1-\omega ){\overline{y}}_{t},\\ G= & {} {\overline{y}}_{t}(1-\omega )((1+\eta +B)+B)\left[ 2((1+\eta +B)+B)+B\left( y_{t}^{\mathrm{R}}-y_{t}^{\mathrm{P}}\right) \right] \\&+\,2\left[ (\omega (1+\eta +B)-B)y_{t}^{\mathrm{R}}-((1+\eta +B)-\omega B)y_{t}^{\mathrm{P}}\right] \\&\left[ ((1+\eta +B)+B)+B\left( y_{t}^{\mathrm{R}}-y_{t}^{\mathrm{P}}\right) \right] \\&+\,((1+\eta +B)+B)(1+\omega )\xi \left( y_{t}^{\mathrm{R}}-y_{t}^{\mathrm{P}}\right) ,\\ I= & {} \left[ 2((1+\eta +B)+B)+B\left( y_{t}^{\mathrm{R}}-y_{t}^{\mathrm{P}}\right) \right] \\&\left[ (\omega (1+\eta +B)-B)y_{t}^{\mathrm{R}}-((1+\eta +B)-\omega B)y_{t}^{\mathrm{P}}\right] \\&+\,(1+\omega )((1+\eta +B)+B)\xi \left( y_{t}^{\mathrm{R}}-y_{t}^{\mathrm{P}}\right) , \end{aligned}$$

• 2. $\gamma _{t}=\left\{ \begin{array}{ll} 0 &{}\quad \text {if}\, \frac{B}{1+\eta +B}<\omega \le {\tilde{\omega }}_{t},\\ \frac{(1-\tau _{t})\left( (\omega (1+\eta +B)-B)y_{t}^{\mathrm{R}}-((1+\eta +B)-\omega B)y_{t}^{\mathrm{P}} \right) }{(1-\omega )((1+\eta +B)+B)\tau _{t}{\overline{y}}_{t}}&{} \quad \text {if}\,{\tilde{\omega }}_{t}<\omega<1,\\ \frac{E-\sqrt{E^{2}-4B(\omega -1){\overline{y}}_{t}(1+\omega )(1+\eta )} }{2B(\omega -1){\overline{y}}_{t}}&{}\quad \text {if}\,1\le \omega <\frac{1+\eta +B}{B}, \end{array} \right.$

  where $E=(1+\omega )(1+\eta )+(1+\omega )\xi -B(1-\omega ){\overline{y}} _{t},$

• 3. $b_{t}=\gamma _{t}\tau _{t}{\overline{y}}_{t},$

• 4. $e_{t}=(1-\gamma _{t})\tau _{t}{\overline{y}}_{t},$

• 5. $s_{t}^{i}=\frac{\eta +B}{1+\eta +B}((1-\tau _{t})y_{t}^{i}+b_{t}),\quad i=P,R,$

• 6. $h_{t+1}^{i,j}=e_{t}^{\xi }(s_{t}^{i})^{\eta }{\overline{H}}_{t}^{\delta }A^{j}, \quad i=P,R, \quad j=L,H,$

• 7. $\theta _{t+1}=1-c+d \cdot (e_{t}-\ln (s_{t}^{\mathrm{R}}/s_{t}^{\mathrm{P}})),$

• 8. $\epsilon _{t+1}=m_{t}q+(1-m_{t})(1-q),$

• 9. $m_{t+1}=\epsilon _{t+1}+\theta _{t+1},$

• 10. ${\overline{H}}_{t+1}=\frac{1}{2}\epsilon _{t+1}h_{t+1}^{\mathrm{P,L}}+\frac{1}{2}(1-\epsilon _{t+1})h_{t+1}^{\mathrm{R,L}}+\frac{1}{2}(1-\epsilon _{t+1})h_{t+1} ^{\mathrm{P,H}}+\frac{1}{2}\epsilon _{t+1}h_{t+1}^{\mathrm{R,H}},$

• 11. $y_{t+1}^{\mathrm{P}}=\epsilon _{t+1}(2-\epsilon _{t+1})h_{t+1}^{\mathrm{P,L}}+\theta _{t+1}(2-2\epsilon _{t+1}-\theta _{t+1})h_{t+1}^{\mathrm{R,L}}+(1-\epsilon _{t+1} -\theta _{t+1})^{2}h_{t+1}^{\mathrm{P,H}},$

• 12. $y_{t+1}^{\mathrm{R}}=\left( 1-(\epsilon _{t+1}+\theta _{t+1})^{2}\right) h_{t+1} ^{\mathrm{R,L}}+\theta _{t+1}(\theta _{t+1}+2\epsilon _{t+1})h_{t+1}^{\mathrm{P,H}}+\epsilon _{t+1}^{2}h_{t+1}^{\mathrm{R,H}},$

• 13. ${\overline{y}}_{t+1}=\frac{1}{2}y_{t+1}^{\mathrm{P}}+\frac{1}{2}y_{t+1}^{\mathrm{R}}.$

Step 3

If $\big |\frac{{\overline{y}}_{t+1}}{{\overline{y}}_{t}}-1\big |$ is sufficiently small, we stop the iterative calculation. If not, we return to step 2.