Appendix 1: Derivations for the Benchmark Model (Perfect Information Case)
Because in the information economics literature, a terminology “type“ contains the meaning of ability of a worker in this paper, in the following derivation in Appendix 1 and 2 I use “type” in place of “ability” appearing in the main text without any change of meaning.
Assume that types, \({\uptheta }\), are uniformly distributed as follows:
$$\begin{aligned} {\uptheta }\left( \hbox {i} \right)= & {} \left( \bar{\uptheta }- \underline{{\uptheta }}\right) \hbox {i}+\underline{\uptheta }\\ \hbox {F}\left( {\uptheta }\right)= & {} \left\{ \begin{array}{ll} 0&{}\quad {\uptheta }<\underline{{\uptheta }}\\ \frac{{\uptheta }-\underline{{\uptheta }}}{\bar{{\uptheta }}- \underline{{\uptheta }}}&{} \quad {\uptheta }\in \left[ \underline{{\uptheta }},\bar{{\uptheta }}\right) \\ 1&{}\quad {{\uptheta }}\ge \bar{{\uptheta }} \end{array} \right. \\ \hbox {f}\left( {\uptheta }\right)= & {} \left\{ \begin{array}{ll} \frac{1}{\bar{{\uptheta }}-\underline{\uptheta }}&{}\quad {\uptheta }\in \left[ \underline{{\uptheta }},\bar{{\uptheta }}\right] \\ 0&{}\quad \hbox {otherwise} \end{array} \right. \\ \hbox {E}\left[ {\uptheta }\right]= & {} \mathop \int \nolimits _0^1 \left( {\bar{{\uptheta }}-\underline{\uptheta }}\right) \hbox {idi}+\underline{\uptheta }=\frac{\bar{{\uptheta }}- \underline{\uptheta }}{2} \end{aligned}$$
Utility:
$$\begin{aligned} \hbox {u}\left( {\hbox {c}_{\mathrm{it}} ,\hbox {b}_{\mathrm{it}} } \right) =\hbox {c}_{\mathrm{it}}^{\upeta } \hbox {b}_{\mathrm{it}}^{1-{\upeta }} \end{aligned}$$
Utility Maximization Problem choice variables are c, b. It is also the decision about consumption and bestow, and made during the adulthood period.
$$\begin{aligned} \max \hbox {u}=\hbox {c}_{\mathrm{it}}^{\upeta } \hbox {b}_{\mathrm{it}}^{1-{\upeta }} \end{aligned}$$
Subject to
$$\begin{aligned} \hbox {c}_{\mathrm{it}} +\hbox {b}_{\mathrm{it}} \le \hbox {m}_{\mathrm{it}} =\hbox {w}_{\mathrm{it}} \hbox {h}_{\mathrm{it}} +\hbox {R}_{\mathrm{t}} \hbox {a}_{\mathrm{it}} \end{aligned}$$
Equilibrium conditions: \(\forall \hbox {i}\)
$$\begin{aligned} \hbox {c}_{\mathrm{it}}= & {} {\upeta } \hbox {m}_{\mathrm{it}} \end{aligned}$$
(17)
$$\begin{aligned} \hbox {b}_{\mathrm{it}}= & {} \left( {1-{\upeta }} \right) \hbox {m}_{\mathrm{it}} \end{aligned}$$
(18)
$$\begin{aligned} \hbox {m}_{\mathrm{it}}= & {} \hbox {w}_{\mathrm{it}} \hbox {h}_{\mathrm{it}} +\hbox {R}_{\mathrm{t}} \hbox {a}_{\mathrm{it}} \end{aligned}$$
(19)
Schooling investment
$$\begin{aligned} \hbox {x}_{\mathrm{it}} =\frac{\hbox {Zh}_{\mathrm{it}+1}^{1+{\upphi } } }{{\uptheta }_{\mathrm{i}} } \end{aligned}$$
Family Income Maximization Problem choice variable is h. It is also the decision about human capital investment and financial assets investment, and made during the childhood period.
$$\begin{aligned} \max \hbox {w}_{\mathrm{it}} \hbox {h}_{\mathrm{it}} +\hbox {R}_{\mathrm{t}} \hbox {a}_{\mathrm{it}} \end{aligned}$$
Subject to
$$\begin{aligned}&\hbox {a}_{\mathrm{it}} +\left( {1-{\uptau }} \right) \hbox {x}_{\mathrm{it}-1} +\hbox {T}_{\mathrm{it}-1} \le \hbox {b}_{\mathrm{it}-1} \end{aligned}$$
(20)
$$\begin{aligned}&\hbox {x}_{\mathrm{it}-1} =\frac{\hbox {Zh}_{\mathrm{it}}^{1+{\upphi } } }{{\uptheta }_{\mathrm{i}} } \end{aligned}$$
(21)
where \(\hbox {T}_{\mathrm{it}} ={\uptau } \hbox {x}_{\mathrm{it}} \).
Equilibrium conditions
$$\begin{aligned} \hbox {w}_{\mathrm{it}}= & {} \frac{\hbox {R}_{\mathrm{t}} \left( {1-{\uptau }} \right) \left( {1+{\upphi } } \right) \hbox {Zh}_{\mathrm{it}}^{\upphi } }{{\uptheta }_{\mathrm{i}} }\nonumber \\ \hbox {b}_{\mathrm{i}0}= & {} \hbox {b}_0 \end{aligned}$$
(22)
Production: \(\forall \hbox {i}\)
$$\begin{aligned} \hbox {y}_{\mathrm{it}} ={\uptheta }_{\mathrm{i}} \hbox {k}_{\mathrm{it}}^{\upalpha } \hbox {h}_{\mathrm{it}}^{1-{\upalpha }} \end{aligned}$$
(23)
Profit Maximization Problem choice variables are h, k. It is also the decision about physical capital investment and employment, made by the representative firm. \(\forall \hbox {t}\)
$$\begin{aligned} \max \mathop \int \nolimits _{\underline{{\uptheta }} }^{\bar{{\uptheta }}} \left[ {\hbox {y}\left( {{\uptheta }_{\mathrm{i}} } \right) -\hbox {w}_{\mathrm{i}} \hbox {h}_{\mathrm{i}} -\hbox {Rk}_{\mathrm{i}} } \right] \hbox {f}\left( {{\uptheta }_{\mathrm{i}} } \right) \hbox {d}{\uptheta }_{\mathrm{i}} \end{aligned}$$
Factor market clearing: \(\forall \hbox {i}\)
$$\begin{aligned} \hbox {R}_{\mathrm{t}}= & {} {\upalpha } {\uptheta }_{\mathrm{i}} \hbox {k}_{\mathrm{it}}^{{\upalpha }-1} \hbox {h}_{\mathrm{it}}^{1-{\upalpha }} \end{aligned}$$
(24)
$$\begin{aligned} \hbox {w}_{\mathrm{it}}= & {} \left( {1-{\upalpha }} \right) {\uptheta }_{\mathrm{i}} \hbox {k}_{\mathrm{it}}^{\upalpha } \hbox {h}_{\mathrm{it}}^{-{\upalpha }} \end{aligned}$$
(25)
Market clearing condition
$$\begin{aligned} \hbox {K}_{\mathrm{t}+1} =\mathop \int \nolimits _0^1 \hbox {k}_{\mathrm{it}+1} \hbox {di}=\mathop \int \nolimits _0^1 \hbox {a}_{\mathrm{it}+1} \hbox {di}=\hbox {A}_{\mathrm{t}+1} \end{aligned}$$
(26)
Ten endogenous variables are c, b, m, a, x, h, y, k, R, w. And we have ten equilibrium conditions are Eqs. (17)–(26) to figure them out.
Derivations
Equations (18)–(20) turn out to be
$$\begin{aligned} \hbox {a}_{\mathrm{it}+1} +\left( {1-{\uptau }} \right) \hbox {x}_{\mathrm{it}} +\hbox {T}_{\mathrm{it}} =\left( {1-{\upeta }} \right) \left[ {\hbox {w}_{\mathrm{it}} \hbox {h}_{\mathrm{it}} +R_t a_{it} } \right] \end{aligned}$$
Aggregate RHS of the above equations:
$$\begin{aligned} \hbox {A}_{\mathrm{t}+1} +\mathop \int \nolimits _0^1 \hbox {x}_{\mathrm{it}} \hbox {di}=\left( {1-{\upeta }} \right) \mathop \int \nolimits _0^1 \hbox {w}_{\mathrm{it}} \hbox {h}_{\mathrm{it}} \hbox {di}+\left( {1-{\upeta }} \right) \hbox {R}_{\mathrm{t}} \hbox {A}_{\mathrm{t}} \end{aligned}$$
(27)
Remember that (24) and (25) turn out to be
$$\begin{aligned} \hbox {w}_{\mathrm{it}} \hbox {h}_{\mathrm{it}} =\frac{1-{\upalpha }}{{\upalpha }}\hbox {R}_{\mathrm{t}} \hbox {k}_{\mathrm{it}} \end{aligned}$$
Hence,
$$\begin{aligned} \mathop \int \nolimits _0^1 \hbox {w}_{\mathrm{it}} \hbox {h}_{\mathrm{it}} \hbox {di}=\frac{1-{\upalpha }}{{\upalpha }}\hbox {R}_{\mathrm{t}} \mathop \int \nolimits _0^1 \hbox {k}_{\mathrm{it}} \hbox {di}=\frac{1-{\upalpha }}{{\upalpha }}\hbox {R}_{\mathrm{t}} \hbox {K}_{\mathrm{t}} \end{aligned}$$
(28)
Remember that (22) turns out to be
$$\begin{aligned} \hbox {w}_{\mathrm{it}} \hbox {h}_{\mathrm{it}} =\frac{\hbox {R}_{\mathrm{t}} \left( {1-{\uptau }} \right) \left( {1+{\upphi } } \right) \hbox {Zh}_{\mathrm{it}}^{1+{\upphi } } }{{\uptheta }_{\mathrm{i}} }=\hbox {R}_{\mathrm{t}} \left( {1-{\uptau }} \right) \left( {1+{\upphi } } \right) \hbox {x}_{\mathrm{it}-1} \end{aligned}$$
Hence, together with (28)
$$\begin{aligned} \mathop \int \nolimits _0^1 \hbox {x}_{\mathrm{it}} \hbox {di}= & {} \frac{1}{\left( {1+{\upphi } } \right) \left( {1-{\uptau }} \right) \hbox {R}_{\mathrm{t}+1} }\frac{1-{\upalpha }}{{\upalpha }}\hbox {R}_{\mathrm{t}+1} \hbox {K}_{\mathrm{t}+1} \nonumber \\= & {} \frac{1-{\upalpha }}{{\upalpha }\left( {1+{\upphi } } \right) \left( {1-{\uptau }} \right) }\hbox {K}_{\mathrm{t}+1} \end{aligned}$$
(29)
Combine Eqs. (26), (27), (28), and (29), and we have,
$$\begin{aligned} \frac{1+{\upalpha }{\upphi } -{\upalpha } {\uptau }\left( {1+{\upphi } } \right) }{{\upalpha }\left( {1+{\upphi } } \right) \left( {1-{\uptau }} \right) }\hbox {K}_{\mathrm{t}+1}= & {} \left[ {\frac{1-{\upeta }}{{\upalpha }}} \right] \hbox {R}_{\mathrm{t}} \hbox {K}_{\mathrm{t}} \Rightarrow \hbox {K}_{\mathrm{t}+1} \nonumber \\= & {} \frac{\left( {1-{\upeta }} \right) \left( {1+{\upphi } } \right) \left( {1-{\uptau }} \right) }{1+{\upalpha }{\upphi } -{\upalpha } {\uptau }\left( {1+{\upphi } } \right) }\hbox {R}_{\mathrm{t}} \hbox {K}_{\mathrm{t}} \end{aligned}$$
(30)
This is the law of motion for K (with one unknown R). Next, we should know the relation between K and R.
Equation (24) turns out to be
$$\begin{aligned} \hbox {R}_{\mathrm{t}}^{\frac{{\upalpha }}{{\upalpha }-1}} ={\upalpha }^{\frac{{\upalpha }}{{\upalpha }-1}}{\uptheta }_{\mathrm{i}}^{\frac{{\upalpha }}{{\upalpha }-1}} \hbox {k}_{\mathrm{it}}^{\upalpha } \hbox {h}_{\mathrm{it}}^{-{\upalpha }} \end{aligned}$$
Combine this equation with Eq. (25), we have
$$\begin{aligned} \hbox {w}_{\mathrm{it}} =\frac{1-{\upalpha }}{{\upalpha }^{\frac{{\upalpha }}{{\upalpha }-1}}}{\uptheta }_{\mathrm{i}}^{\frac{1}{1-{\upalpha }}} \hbox {R}_{\mathrm{t}}^{\frac{{\upalpha }}{{\upalpha }-1}} \end{aligned}$$
(31)
For any \(\hbox {i}\), and for any \(\hbox {t}=1,2,\ldots \)
Equation (22) turns out to be
$$\begin{aligned} \hbox {h}_{\mathrm{it}} =\left[ {\frac{{\uptheta }_{\mathrm{i}} \hbox {w}_{\mathrm{it}} }{\left( {1+{\upphi } } \right) \left( {1-{\uptau }} \right) \hbox {ZR}_{\mathrm{t}} }} \right] ^{\frac{1}{{\phi } }} \end{aligned}$$
(32)
Equations (32) and (21) turn out to be
$$\begin{aligned} \hbox {x}_{\mathrm{it}-1} =\left[ {\frac{{\uptheta }_{\mathrm{i}} }{\hbox {Z}}} \right] ^{\frac{1}{{\phi } }}\left[ {\frac{w_{it} }{\left( {1+{\phi } } \right) \left( {1-\tau } \right) R_t }} \right] ^{\frac{1+{\phi } }{{\phi } }} \end{aligned}$$
Substitute (31) into them, and let \(\hbox {C}=\frac{1-{\upalpha }}{\hbox {Z}\left( {1-{\uptau }} \right) \left( {1+{\upphi } } \right) }{\upalpha }^{\frac{{\upalpha }}{1-{\upalpha }}}\), we have,
$$\begin{aligned} \hbox {h}_{\mathrm{it}}= & {} \hbox {C}^{\frac{1}{{\phi } }}\times {\uptheta }_{\mathrm{i}}^{\frac{2-{\upalpha }}{\left( {1-{\upalpha }} \right) {\upphi } }} \hbox {R}_{\mathrm{t}}^{\frac{-1}{\left( {1-{\upalpha }} \right) {\upphi } }}\nonumber \\ \hbox {x}_{\mathrm{it}-1}= & {} \hbox {Z}\times \hbox {C}^{\frac{1+{\phi } }{{\phi } }}\times {\uptheta }_{\mathrm{i}}^{\frac{{\upphi } +2-{\upalpha }}{\left( {1-{\upalpha }} \right) {\upphi } }} \times \hbox {R}_{\mathrm{t}}^{\frac{-\left( {1+{\upphi } } \right) }{\left( {1-{\upalpha }} \right) {\upphi } }} \end{aligned}$$
(33)
Substitute Eq. (33) into Eq. (24), we have
$$\begin{aligned} \hbox {R}_{\mathrm{t}}= & {} {\upalpha } {\uptheta }_{\mathrm{i}} \hbox {k}_{\mathrm{it}}^{{\upalpha }-1} \left[ {\hbox {C}^{\frac{1}{{\phi } }}\times {\uptheta }_{\mathrm{i}}^{\frac{2-{\upalpha }}{\left( {1-{\upalpha }} \right) {\upphi } }} \hbox {R}_{\mathrm{t}}^{\frac{-1}{\left( {1-{\upalpha }} \right) {\upphi } }} } \right] ^{1-\alpha }\nonumber \\ \hbox {R}_{\mathrm{t}}^{1+\frac{1}{{\upphi } }} \hbox {k}_{\mathrm{it}}^{1-{\upalpha }}= & {} {\upalpha } {\uptheta }_{\mathrm{i}}^{1+\frac{2-{\upalpha }}{{\upphi } }} \hbox {C}^{\frac{1-\alpha }{{\phi } }}\nonumber \\ \hbox {R}_{\mathrm{t}}^{\frac{1+{\upphi } }{\left( {1-{\upalpha }} \right) {\upphi } }} \hbox {K}_{\mathrm{t}}= & {} {\upalpha }^{\frac{1}{1-{\upalpha }}}\hbox {C}^{\frac{1}{{\phi } }}\mathop \int \nolimits _0^1 {\uptheta }_{\mathrm{i}}^{\frac{{\upphi } +2-{\upalpha }}{\left( {1-{\upalpha }} \right) {\upphi } }} \hbox {di} \end{aligned}$$
(34)
Appendix 2: Derivations for Asymmetric Information Cases
Suppose that workers know their types but firms do not. Workers choose human capital investment, and firms observe workers’ human capital levels and pay wages. Specifications are the same as the benchmark, except for the Income Maximization Problem and the Profit Maximization Problem.
Separating equilibrium (SE)
Income Maximization Problem households choose h, according to firms’ wage schedule, w, which is a function of h.
$$\begin{aligned} \max \hbox {w}\left[ {\hbox {h}_{\mathrm{it}} } \right] +\hbox {R}_{\mathrm{t}} \hbox {a}_{\mathrm{it}} \end{aligned}$$
Subject to
$$\begin{aligned}&\hbox {a}_{\mathrm{it}} +\left( {1-{\uptau }} \right) \hbox {x}_{\mathrm{it}-1} +\hbox {T}_{\mathrm{it}-1} \le \hbox {b}_{\mathrm{it}-1} \end{aligned}$$
(35)
$$\begin{aligned}&\hbox {x}_{\mathrm{it}-1} =\frac{\hbox {Zh}_{\mathrm{it}}^{1+{\upphi } } }{{\uptheta }_{\mathrm{i}} } \end{aligned}$$
(36)
where \(\hbox {T}_{\mathrm{it}} ={\uptau } \hbox {x}_{\mathrm{it}} \).
Before showing the equilibrium condition of SE, we first derive an important implication result.
Lemma
Incentive compatibility alone requires \(\frac{\mathrm {dh}}{\mathrm {d}{\uptheta }}>0\).
Proof
\(\forall {\uptheta },\tilde{{\uptheta }} \), and \(\tilde{{\uptheta }} >\theta \)
$$\begin{aligned} \hbox {w}\left[ {\hbox {h}\left( {\uptheta } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {\uptheta } \right) ;{\uptheta }} \right]\ge & {} \hbox {w}\left[ {\hbox {h}\left( {\tilde{{\uptheta }} } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {\tilde{{\uptheta }} } \right) ;{\uptheta }} \right] \\ \hbox {w}\left[ {\hbox {h}\left( {\tilde{{\uptheta }} } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {\tilde{{\uptheta }} } \right) ;\tilde{{\uptheta }} } \right]\ge & {} \hbox {w}\left[ {\hbox {h}\left( {\uptheta } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {\uptheta } \right) ;\tilde{{\uptheta }} } \right] \end{aligned}$$
The above two equations turn out to be:
$$\begin{aligned}&\left[ {\hbox {x}\left[ {\hbox {h}\left( {\tilde{{\uptheta }} } \right) ;{\uptheta }} \right] -\hbox {x}\left[ {\hbox {h}\left( {\uptheta } \right) ;{\uptheta }} \right] } \right] -\left[ {\hbox {x}\left[ {\hbox {h}\left( {\tilde{{\uptheta }} } \right) ;\tilde{{\uptheta }} } \right] -\hbox {x}\left[ {\hbox {h}\left( {\uptheta } \right) ;\tilde{{\uptheta }} } \right] } \right] \ge 0\\&\left( {\frac{1}{{\uptheta }}-\frac{1}{\tilde{{\uptheta }} }} \right) \left( {\left[ {\hbox {h}\left( {\tilde{{\uptheta }} } \right) } \right] ^{1+{\phi } }-\left[ {\hbox {h}\left( {\uptheta } \right) } \right] ^{1+{\phi } }} \right) \ge 0 \end{aligned}$$
Thus, \(\hbox {h}\left( {\tilde{{\uptheta }} } \right) >h\left( {\uptheta } \right) \), or \(\frac{\hbox {dh}}{\hbox {d}{{\uptheta }}}\ge 0\). \(\square \)
Remember that workers can only choose their human capital levels but not their types. Then \(\forall {\uptheta }\)
$$\begin{aligned} \frac{\hbox {dw}}{\hbox {dh}}\left[ {\hbox {h}\left( {\uptheta } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \frac{\partial \hbox {x}}{\partial \hbox {h}}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] =0 \end{aligned}$$
This is the First Order Condition (FOC). And the Second order condition (SOC) is:
$$\begin{aligned} \frac{\hbox {d}^{2}\hbox {w}}{\hbox {dh}^{2}}\left[ {\hbox {h}\left( {\uptheta } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \frac{\partial ^{2}\hbox {x}}{\partial \hbox {h}^{2}}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] \le 0 \end{aligned}$$
Differentiating the FOC by \({\uptheta }\),
$$\begin{aligned} \left[ {\frac{\hbox {d}^{2}\hbox {w}}{\hbox {dh}^{2}}\left[ {\hbox {h}\left( {\uptheta } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \frac{\partial ^{2}\hbox {x}}{\partial \hbox {h}^{2}}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] } \right] \frac{\hbox {dh}}{\hbox {d}{\uptheta }}-\hbox {R}\left( {1-{\uptau }} \right) \frac{\partial ^{2}\hbox {x}}{\partial \hbox {h}\partial {\uptheta }}\left( {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right) =0 \end{aligned}$$
Then the SOC becomes
$$\begin{aligned} \frac{\partial ^{2}\hbox {x}}{\partial \hbox {h}\partial {\uptheta }}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] \le 0 \end{aligned}$$
If the above condition is a strict inequality, then it is the Spence-Mirrlees (single crossing) condition in the principal-agent literature. Economically, it simply says that a more efficient type is also more efficient at the margin.
Define worker’s indirect utility function \(\hbox {U}\left( {\uptheta } \right) =\hbox {w}-\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\). Then the FOC becomes
$$\begin{aligned} \frac{\hbox {dU}}{\hbox {d}{\uptheta }}\left[ {\uptheta } \right] =-\hbox {R}\left( {1-{\uptau }} \right) \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] \end{aligned}$$
And the relevant participation constraint (PC) is \(\forall {\uptheta }\),
$$\begin{aligned} \hbox {U}\left( {\uptheta } \right) \ge \underline{\hbox {U}} \end{aligned}$$
Because \(\frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] <0\), so\(\frac{\hbox {dU}}{\hbox {d}{\uptheta }}\left[ {\uptheta } \right] >0\), and let the reserve price be zero. PC becomes
$$\begin{aligned} \hbox {U}\left( {\underline{{\uptheta }} } \right) \ge \underline{\hbox {U}} =0 \end{aligned}$$
The FOC, SOC, and PC characterize the Income Maximization Problem.
Production function: \(\forall \hbox {i}\)
$$\begin{aligned} \hbox {y}_{\mathrm{it}} ={\uptheta }_{\mathrm{i}} \hbox {k}_{\mathrm{it}}^{\upalpha } \hbox {h}_{\mathrm{it}}^{1-{\upalpha }} \end{aligned}$$
(37)
Profit Maximization Problem \(\forall \hbox {t}\)
$$\begin{aligned} \max \mathop \int \nolimits _{\underline{{\uptheta }} }^{\bar{{\uptheta }}} \left[ {\hbox {y}\left( {{\uptheta }_{\mathrm{i}} } \right) -\hbox {U}_{\mathrm{i}} -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}-\hbox {Rk}_{\mathrm{i}} } \right] \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) \end{aligned}$$
Subject to FOC, SOC and PC:
$$\begin{aligned}&\frac{\hbox {dU}}{\hbox {d}{\uptheta }}\left[ {\uptheta } \right] =-\hbox {R}\left( {1-{\uptau }} \right) \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] \nonumber \\&\frac{\partial ^{2}\hbox {x}}{\partial \hbox {h}\partial {\uptheta }}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] \le 0 \end{aligned}$$
(38)
$$\begin{aligned}&\hbox {U}\left( {\underline{{\uptheta }} } \right) =0 \end{aligned}$$
(39)
The FOC can be rewritten as:
$$\begin{aligned} \hbox {U}\left( {\uptheta } \right) =-\hbox {R}\left( {1-{\uptau }} \right) \mathop \int \nolimits _{\underline{{\uptheta }} }^{\uptheta } \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] \hbox {d}{\uptheta } \end{aligned}$$
or
$$\begin{aligned} \hbox {w}\left[ {\hbox {h}\left( {\uptheta } \right) } \right] =\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] -\hbox {R}\left( {1-{\uptau }} \right) \mathop \int \nolimits _{\underline{{\uptheta }} }^{\uptheta } \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] \hbox {d}{\uptheta } \end{aligned}$$
(40)
Then the maximand becomes
$$\begin{aligned}&\max \mathop \int \nolimits _{\underline{{\uptheta }} }^{\bar{{\uptheta }}} \left[ \hbox {y}\left( {{\uptheta }_{\mathrm{i}} } \right) +\hbox {R}\left( {1-{\uptau }} \right) \mathop \int \nolimits _{\underline{{\uptheta }} }^{\uptheta } \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] \hbox {d}{\uptheta }_{\mathrm{i}}\right. \\&\quad \left. -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] -\hbox {Rk}_{\mathrm{i}} \right] \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) \end{aligned}$$
Factor market clearing condition w.r.t. physical capital: \(\forall \hbox {i}\)
$$\begin{aligned} \hbox {R}_{\mathrm{t}} ={\upalpha } {\uptheta }_{\mathrm{i}} \hbox {k}_{\mathrm{it}}^{{\upalpha }-1} \hbox {h}_{\mathrm{it}}^{1-{\upalpha }} \end{aligned}$$
(41)
This result is the same as that in the benchmark.
To derive the FOC w.r.t. human capital, we need simplify the integration inside the maximand.
$$\begin{aligned}&\mathop \int \nolimits _{\underline{{\uptheta }} }^{\bar{{\uptheta }}} \left( {\mathop \int \nolimits _{\underline{{\uptheta }} }^{\uptheta } \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] \hbox {d}{\uptheta }_{\mathrm{i}} } \right) \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) \\&\quad =\left( {\mathop \int \nolimits _{\underline{{\uptheta }} }^{\bar{{\uptheta }}} \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] \hbox {d}{\uptheta }_{\mathrm{i}} } \right) \hbox {F}\left( {\bar{{\uptheta }}} \right) -\mathop \int \nolimits _{\underline{{\uptheta }} }^{\bar{{\uptheta }}} \hbox {F}\left( {{\uptheta }_{\mathrm{i}} } \right) \hbox {d}\left( {\mathop \int \nolimits _{\underline{{\uptheta }} }^{\uptheta } \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] \hbox {d}{\uptheta }_{\mathrm{i}} } \right) \\&\quad =\mathop \int \nolimits _{\underline{{\uptheta }} }^{\bar{{\uptheta }}} \left[ {1-\hbox {F}\left( {{\uptheta }_{\mathrm{i}} } \right) } \right] \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] \hbox {d}{\uptheta }_{\mathrm{i}} \end{aligned}$$
The maximand becomes
$$\begin{aligned} \max \mathop \int \nolimits _{\underline{{\uptheta }} }^{\bar{{\uptheta }}} \left[ {\hbox {y}\left( {{\uptheta }_{\mathrm{i}} } \right) -\hbox {R}\left( {1-{\uptau }} \right) \left[ {\hbox {x}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] -\frac{1-\hbox {F}\left( {{\uptheta }_{\mathrm{i}} } \right) }{\hbox {f}\left( {{\uptheta }_{\mathrm{i}} } \right) }\frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] } \right] } \right] \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) \end{aligned}$$
Factor market clearing condition w.r.t. human capital: \(\forall \hbox {i}\)
$$\begin{aligned} \left( {1-{\upalpha }} \right) {\uptheta }_{\mathrm{i}} \hbox {k}_{\mathrm{it}}^{\upalpha } \hbox {h}_{\mathrm{it}}^{-{\upalpha }} =\hbox {R}_{\mathrm{t}} \left( {1-{\uptau }} \right) \left[ {\frac{\partial \hbox {x}}{\partial \hbox {h}}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] -\frac{1-\hbox {F}\left( {{\uptheta }_{\mathrm{i}} } \right) }{\hbox {f}\left( {{\uptheta }_{\mathrm{i}} } \right) }\frac{\partial ^{2}\hbox {x}}{\partial {\uptheta }\partial \hbox {h}}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] } \right] \end{aligned}$$
(42)
Market clearing condition
$$\begin{aligned} \hbox {K}_{\mathrm{it}+1} =\mathop \int \nolimits _0^1 \hbox {k}_{\mathrm{it}+1} \hbox {di}=\mathop \int \nolimits _0^1 \hbox {a}_{\mathrm{it}+1} \hbox {di}=\hbox {A}_{\mathrm{it}+1} \end{aligned}$$
(43)
Ten endogenous variables: c, b, m, a, x, h, w, R, y, k. Equations (35), (36), (37), (40)–(43) together with three conditions associated with the Utility Maximization Problem:
$$\begin{aligned} \hbox {c}_{\mathrm{it}}= & {} {\upeta } \hbox {m}_{\mathrm{it}} \end{aligned}$$
(44)
$$\begin{aligned} \hbox {b}_{\mathrm{it}}= & {} \left( {1-{\upeta }} \right) \hbox {m}_{\mathrm{it}} \end{aligned}$$
(45)
$$\begin{aligned} \hbox {m}_{\mathrm{it}}= & {} \hbox {w}_{\mathrm{it}} +\hbox {R}_{\mathrm{t}} \hbox {a}_{\mathrm{it}} \end{aligned}$$
(46)
Derivations
Similar to the benchmark, substitute (45) into (35) and aggregate both sides, we have
$$\begin{aligned} \hbox {K}_{\hbox {t}+1} +\mathop \int \nolimits _0^1 \hbox {x}_{\mathrm{it}} \hbox {di}=\left( {1-{\upeta }} \right) \mathop \int \nolimits _0^1 \hbox {w}_{\mathrm{it}} \hbox {di}+\left( {1-{\upeta }} \right) \hbox {R}_{\mathrm{t}} \hbox {K}_{\mathrm{t}} \end{aligned}$$
(47)
First, by the Cobb–Douglas production function,
$$\begin{aligned} \mathop \int \nolimits _0^1 \hbox {w}_{\mathrm{it}} \hbox {di}=\frac{1-{\upalpha }}{{\upalpha }}\mathop \int \nolimits _0^1 \hbox {R}_{\mathrm{t}} \hbox {k}_{\mathrm{it}} \hbox {di}=\frac{1-{\upalpha }}{{\upalpha }}\hbox {R}_{\mathrm{t}} \hbox {K}_{\mathrm{t}} \end{aligned}$$
(48)
Define marginal product of human capital\(\left[ {\hbox {MP}_{\mathrm{h}} } \right] _{\mathrm{it}} =\left( {1-{\upalpha }} \right) {\uptheta }_{\mathrm{i}} \hbox {k}_{\mathrm{it}}^{\upalpha } \hbox {h}_{\mathrm{it}}^{-{\upalpha }} \). Then by (42)
$$\begin{aligned} \left[ {\hbox {MP}_{\mathrm{h}} } \right] _{\mathrm{it}} =\hbox {R}_{\mathrm{t}} \left( {1-{\uptau }} \right) \left( {1+{\upphi } } \right) \hbox {Z}\frac{\bar{{\uptheta }}}{{\uptheta }_{\mathrm{i}}^2 }\hbox {h}_{\mathrm{it}}^{\upphi } \end{aligned}$$
From Eq. (41), we have
$$\begin{aligned} \hbox {R}_{\mathrm{t}}^{\frac{{\upalpha }}{{\upalpha }-1}} ={\upalpha }^{\frac{{\upalpha }}{{\upalpha }-1}}{\uptheta }_{\mathrm{i}}^{\frac{{\upalpha }}{{\upalpha }-1}} \hbox {k}_{\mathrm{it}}^{\upalpha } \hbox {h}_{\mathrm{it}}^{-{\upalpha }} \end{aligned}$$
Substitute the above equation into the definition of marginal product of human capital, we have
$$\begin{aligned} \left[ {\hbox {MP}_{\mathrm{h}} } \right] _{\mathrm{it}} =\frac{1-{\upalpha }}{{\upalpha }^{\frac{{\upalpha }}{{\upalpha }-1}}}{\uptheta }_{\mathrm{i}}^{\frac{1}{1-{\upalpha }}} \hbox {R}_{\mathrm{t}}^{\frac{{\upalpha }}{{\upalpha }-1}} \end{aligned}$$
So combine the above two expressions for MPH, and we have
$$\begin{aligned} \hbox {h}_{\mathrm{it}} =\hbox {C}^{\frac{1}{{\phi } }}\times \bar{{\uptheta }} ^{\frac{-1}{{\phi } }}{\uptheta }_{\mathrm{i}}^{\frac{3-2{\upalpha }}{\left( {1-{\upalpha }} \right) {\upphi } }} \hbox {R}_{\mathrm{t}}^{\frac{-1}{\left( {1-{\upalpha }} \right) {\upphi } }} \end{aligned}$$
where \(\hbox {C}=\frac{1-{\upalpha }}{\hbox {Z}\left( {1-{\uptau }} \right) \left( {1+{\upphi } } \right) }{\upalpha }^{\frac{{\upalpha }}{1-{\upalpha }}}\). And
$$\begin{aligned} \hbox {x}_{\mathrm{it}} =\frac{\hbox {Zh}_{\mathrm{it}+1}^{1+{\upphi } } }{{\uptheta }_{\mathrm{i}} }=\hbox {Z}\times \hbox {C}^{\frac{1+{\phi } }{{\phi } }}\times \bar{{\uptheta }} ^{\frac{-\left( {1+{\phi } } \right) }{{\phi } }}{\uptheta }_{\mathrm{i}}^{\frac{\left( {2-{\upalpha }} \right) {\upphi } +3-2{\upalpha }}{\left( {1-{\upalpha }} \right) {\upphi } }} \times \hbox {R}_{\mathrm{t}+1}^{\frac{-\left( {1+{\upphi } } \right) }{\left( {1-{\upalpha }} \right) {\upphi } }} \end{aligned}$$
Thus
$$\begin{aligned} \mathop \int \nolimits _0^1 \hbox {x}_{\mathrm{it}} \hbox {di}=\hbox {Z}\times \hbox {C}^{\frac{1+{\phi } }{{\phi } }}\times \bar{{\uptheta }} ^{\frac{-\left( {1+{\phi } } \right) }{{\phi } }}\times \hbox {R}_{\mathrm{t}+1}^{\frac{-\left( {1+{\upphi } } \right) }{\left( {1-{\upalpha }} \right) {\upphi } }} \times \mathop \int \nolimits _0^1 {\uptheta }_{\mathrm{i}}^{\frac{\left( {2-{\upalpha }} \right) {\upphi } +3-2{\upalpha }}{\left( {1-{\upalpha }} \right) {\upphi } }} di \end{aligned}$$
(49)
Substitute (48) and (49) into (47)
$$\begin{aligned} \hbox {K}_{\mathrm{t}+1} +\hbox {Z}\times \hbox {C}^{\frac{1+{\phi } }{{\phi } }}\times \bar{{\uptheta }} ^{\frac{-\left( {1+{\phi } } \right) }{{\phi } }}\times \hbox {R}_{\mathrm{t}+1}^{\frac{-\left( {1+{\upphi } } \right) }{\left( {1-{\upalpha }} \right) {\upphi } }} \times \mathop \int \nolimits _0^1 {\uptheta }_{\mathrm{i}}^{\frac{\left( {2-{\upalpha }} \right) {\upphi } +3-2{\upalpha }}{\left( {1-{\upalpha }} \right) {\upphi } }} di=\frac{1-{\upeta }}{{\upalpha }}\hbox {R}_{\mathrm{t}} \hbox {K}_{\mathrm{t}} \end{aligned}$$
Next, we derive the relation between R and K.
$$\begin{aligned} \hbox {R}_{\mathrm{t}}= & {} {\upalpha } {\uptheta }_{\mathrm{i}} \hbox {k}_{\mathrm{it}}^{{\upalpha }-1} \left[ {\hbox {C}^{\frac{1}{{\phi } }}\times \bar{{\uptheta }} ^{\frac{-1}{{\phi } }}{\uptheta }_{\mathrm{i}}^{\frac{3-2{\upalpha }}{\left( {1-{\upalpha }} \right) {\upphi } }} \hbox {R}_{\mathrm{t}}^{\frac{-1}{\left( {1-{\upalpha }} \right) {\upphi } }} } \right] ^{1-\alpha }\nonumber \\ \hbox {R}_{\mathrm{t}}^{1+\frac{1}{{\upphi } }} \hbox {k}_{\mathrm{it}}^{1-{\upalpha }}= & {} {\upalpha }\bar{{\uptheta }} ^{\frac{-\left( {1-\alpha } \right) }{{\phi } }}{\uptheta }_{\mathrm{i}}^{1+\frac{3-2{\upalpha }}{{\upphi } }} \hbox {C}^{\frac{1-\alpha }{{\phi } }}\nonumber \\ \hbox {R}_{\mathrm{t}}^{\frac{1+{\upphi } }{\left( {1-{\upalpha }} \right) {\upphi } }} \hbox {K}_{\mathrm{t}}= & {} {\upalpha }^{\frac{1}{1-{\upalpha }}}\bar{{\uptheta }} ^{\frac{-1}{{\phi } }} \hbox {C}^{\frac{1}{{\phi } }}\mathop \int \nolimits _0^1 {\uptheta }_{\mathrm{i}}^{\frac{{\upphi } +3-2{\upalpha }}{\left( {1-{\upalpha }} \right) {\upphi } }} \hbox {di} \end{aligned}$$
(50)
Pooling Equilibrium (PE)
Suppose that all workers choose the same human capital level \(\hbox {h}^{\mathrm{p}}\). Then for the lowest type, PC is binding:
$$\begin{aligned} \hbox {w}_{\mathrm{t}}^{\mathrm{p}} -\hbox {R}_{\mathrm{t}} \left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} , \underline{\uptheta }} \right] =0 \end{aligned}$$
(51)
And the Profit Maximization Problem becomes
$$\begin{aligned} \max \mathop \int \nolimits _{\underline{\uptheta }}^{\bar{{\uptheta }}} \left[ {\uptheta }\left[ {\hbox {k}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{{\upalpha }} \left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{1-\alpha }- \hbox {w}\left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} } \right] -\hbox {R}_{\mathrm{t}} \hbox {k}_{\mathrm{t}}^{\mathrm{p}} \right] \hbox {dF}\left( {\uptheta } \right) \end{aligned}$$
Subject to
$$\begin{aligned} \hbox {w}\left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} } \right] =\hbox {R}_{\mathrm{t}} \left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} ,\underline{\uptheta }} \right] \end{aligned}$$
FOC
$$\begin{aligned}&\hbox {E}\left[ {\uptheta } \right] {\upalpha }\left[ {\hbox {k}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{{\upalpha }-1}\left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{1-\alpha }=\hbox {R}_{\mathrm{t}}^{\mathrm{p}} \end{aligned}$$
(52)
$$\begin{aligned}&\hbox {E}\left[ {\uptheta } \right] \left( {1-{\upalpha }} \right) \left[ {\hbox {k}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{{\upalpha }}\left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{-\alpha }=\left[ {\hbox {MP}_{\mathrm{h}} } \right] _{\mathrm{t}} =\hbox {R}_{\mathrm{t}}^{\mathrm{p}} \left( {1-{\uptau }} \right) \left( {1+{\upphi } } \right) \frac{\hbox {Z}\left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{{\upphi } }}{\underline{\uptheta }} \end{aligned}$$
(53)
And the law of motion for K is
$$\begin{aligned} \hbox {K}_{\mathrm{t}+1} +\mathop \int \nolimits _0^1 \hbox {x}_{\mathrm{it}} \hbox {di}=\left( {1-{\upeta }} \right) \mathop \int \nolimits _0^1 \hbox {w}_{\mathrm{it}} \hbox {di}+\left( {1-{\upeta }} \right) \hbox {R}_{\mathrm{t}} \hbox {K}_{\mathrm{t}} \end{aligned}$$
Since
$$\begin{aligned} \mathop \int \nolimits _0^1 \hbox {w}_{\mathrm{it}} \hbox {di}= & {} \frac{1-{\upalpha }}{{\upalpha }} \hbox {R}_{\mathrm{t}} \hbox {K}_{\mathrm{t}}\\ \mathop \int \nolimits _0^1 \hbox {x}_{\mathrm{it}} \hbox {di}= & {} \mathop \int \nolimits _{\underline{\uptheta }}^{\bar{{\uptheta }}} \frac{\hbox {Z}\left[ {\hbox {h}_{\mathrm{t}+1}^{\mathrm{p}} } \right] ^{1+{\phi } }}{{\uptheta }_{\mathrm{i}} }\hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) \\= & {} \frac{\hbox {Z}\left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{1+{\phi } }}{\underline{\uptheta }}\frac{{\underline{{\uptheta }}}}{{{\bar{\uptheta }} - \underline{{\uptheta }}}}\ln \frac{{{\bar{\uptheta }}}}{{\underline{{\uptheta }}}} \\= & {} \frac{{{\mathrm{w}}\left[ {{\mathrm{h}}_{{\mathrm{t}} + 1}^{\mathrm{P}}} \right] }}{{{{\mathrm{R}}_{{\mathrm{t}} + 1}}\left( {1 - {{\uptau }}} \right) }}\frac{{\underline{{\uptheta }}}}{{{\bar{\uptheta }} - \underline{{\uptheta }} }}\ln \frac{{{\bar{\uptheta }}}}{{\underline{{\uptheta }}}} \\= & {} \frac{{1 - {{\upalpha }}}}{{{\upalpha }}}\frac{{{\mathrm{k}}_{{\mathrm{t}} + 1}^{\mathrm{P}}}}{{1 - {{\uptau }}}}\frac{{\underline{{\uptheta }}}}{{{\bar{\uptheta }} - \underline{{\uptheta }} }}\ln \frac{{{\bar{\uptheta }}}}{{\underline{{\uptheta }}}} = \frac{{1 - {{\upalpha }}}}{{{\upalpha }}}\frac{{{{\mathrm{K}}_{{\mathrm{t}} + 1}}}}{{1 - {{\uptau }}}}\frac{{\underline{{\uptheta }}}}{{{\bar{\uptheta }} - \underline{{\uptheta }} }}\ln \frac{{{\bar{\uptheta }}}}{{\underline{{\uptheta }}}} \end{aligned}$$
Let \(\hbox {b}=\frac{{\underline{{\uptheta }}}}{{{\bar{\uptheta }} - \underline{{\uptheta }}}}\ln \frac{{{\bar{\uptheta }}}}{{\underline{{\uptheta }}}}\). Then
$$\begin{aligned} \hbox {K}_{\mathrm{t}+1} =\frac{\left( {1-{\upeta }} \right) \left( {1-{\uptau }} \right) }{{\upalpha }\left( {1-{\uptau }} \right) +\left( {1-{\upalpha }} \right) \hbox {b}}\hbox {R}_{\mathrm{t}} \hbox {K}_{\mathrm{t}} \end{aligned}$$
(54)
Next we derive the relation between R and K. Remember (52) can be rewritten as:
$$\begin{aligned} \left[ {\hbox {E}\left[ {\uptheta } \right] {\upalpha }} \right] ^{\frac{\alpha }{\alpha -1}}\left[ {\hbox {k}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{{\upalpha }}\left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{-\alpha }=\left[ {\hbox {R}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{\frac{\alpha }{\alpha -1}} \end{aligned}$$
Combine it with (53), we have
$$\begin{aligned} \left[ {\hbox {MP}_{\mathrm{h}} } \right] _{\mathrm{t}} =\hbox {E}\left[ {\uptheta } \right] ^{\frac{1}{1-\alpha }}\frac{1-\alpha }{\alpha ^{\frac{\alpha }{\alpha -1}}}\left[ {\hbox {R}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{\frac{\alpha }{\alpha -1}} \end{aligned}$$
And
$$\begin{aligned} \hbox {R}_{\mathrm{t}}^{\mathrm{p}} \left( {1-{\uptau }} \right) \left( {1+{\upphi } } \right) \frac{\hbox {Z}\left[ {\hbox {h}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{{\upphi } }}{\underline{\uptheta }} =\left[ {\hbox {R}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{\frac{\alpha }{\alpha -1}}\hbox {E}\left[ {\uptheta } \right] ^{\frac{1}{1-\alpha }}\frac{1-\alpha }{\alpha ^{\frac{\alpha }{\alpha -1}}} \end{aligned}$$
Let \(\hbox {C}=\frac{1-{\upalpha }}{\hbox {Z}\left( {1-{\uptau }} \right) \left( {1+{\upphi } } \right) }{\upalpha }^{\frac{{\upalpha }}{1-{\upalpha }}}\)
Then
$$\begin{aligned} \hbox {h}_{\mathrm{t}}^{\mathrm{P}} =\hbox {C}^{\frac{1}{{\upphi } }}\times \underline{\uptheta } ^{\frac{1}{{\phi } }}\hbox {E}\left[ {\uptheta } \right] ^{\frac{1}{\left( {1-\alpha } \right) {\phi } }}\times \left[ {\hbox {R}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{\frac{1}{\left( {{\upalpha }-1} \right) {\upphi }}} \end{aligned}$$
(55)
Substitute (55) into (52)
$$\begin{aligned} \hbox {E}\left[ {\uptheta } \right] {\upalpha }\left[ {\hbox {k}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{{\upalpha }-1}\left[ {\hbox {C}^{\frac{1}{{\upphi } }}\times \underline{\uptheta }} ^{\frac{1}{{\phi } }}\hbox {E}\left[ {\uptheta } \right] ^{\frac{1}{\left( {1-\alpha } \right) {\phi } }}\times \left[ {\hbox {R}_{\mathrm{t}}^{\mathrm{p}} } \right] ^{\frac{1}{\left( {{\upalpha }-1} \right) {\upphi } }} \right] ^{1-\alpha }=\hbox {R}_{\mathrm{t}}^{\mathrm{p}} \end{aligned}$$
or
$$\begin{aligned} \left[ {\hbox {R}_{\mathrm{t}}^{\mathrm{P}} } \right] ^{\frac{1+{\upphi } }{\left( {1-{\upalpha }} \right) {\upphi } }}\hbox {K}_{\mathrm{t}}^{\mathrm{P}} ={\upalpha }^{\frac{1}{1-{\upalpha }}}\hbox {C}^{\frac{1}{{\phi } }}{\underline{\uptheta }}^{\frac{1}{{\phi } }}\hbox {E}\left[ {\uptheta } \right] ^{\frac{1+{\upphi } }{\left( {1-{\upalpha }} \right) {\upphi } }} \end{aligned}$$
(56)
Hybrid Equilibrium (HE)
Suppose workers are split into two groups: some workers choose a common human capital level\(\hbox {h}^{1}\), the others choose\(\hbox {h}^{2}>\hbox {h}^{1}\). And the associated type threshold is \({\uptheta }^{{*}}\).
Profit Maximization Problem is \(\forall \hbox {t}\)
$$\begin{aligned}&\max \mathop \int \nolimits _{\underline{\uptheta }}^{{\uptheta }^{{*}}} \left[ {\uptheta }_{\mathrm{i}} \left[ {\hbox {k}^{1}} \right] ^{{\upalpha }}\left[ {\hbox {h}^{1}} \right] ^{1-\alpha }-\hbox {w}\left[ {\hbox {h}^{1}} \right] -\hbox {Rk}^{1} \right] \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) \\&\quad +\mathop \int \nolimits _{{\uptheta }^{{*}}}^{\bar{{\uptheta }}} \left[ {{\uptheta }_{\mathrm{i}} \left[ {\hbox {k}^{2}} \right] ^{{\upalpha }}\left[ {\hbox {h}^{2}} \right] ^{1-\alpha }-\hbox {w}\left[ {\hbox {h}^{2}} \right] -\hbox {Rk}^{2}} \right] \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) \end{aligned}$$
FOC with respect to physical capital:
$$\begin{aligned} \hbox {E}\left[ {{\uptheta }^{1}} \right] {\upalpha }\left[ {\hbox {k}_{\mathrm{t}}^1 } \right] ^{\alpha -1}\left[ {\hbox {h}_{\mathrm{t}}^1 } \right] ^{1-{\upalpha }}= & {} \hbox {R}_{\mathrm{t}} \end{aligned}$$
(57)
$$\begin{aligned} \hbox {E}\left[ {{\uptheta }^{2}} \right] {\upalpha }\left[ {\hbox {k}_{\mathrm{t}}^2 } \right] ^{\alpha -1}\left[ {\hbox {h}_{\mathrm{t}}^2 } \right] ^{1-{\upalpha }}= & {} \hbox {R}_{\mathrm{t}} \end{aligned}$$
(58)
For \({\uptheta }\in \left[ {\underline{\uptheta }} ,{\uptheta }^{{*}} \right] \), the participation constraint requires
$$\begin{aligned} \hbox {w}\left[ {\hbox {h}^{1}} \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}^{1},\underline{\uptheta }} \right] \ge 0 \end{aligned}$$
In equilibrium, it is binding only for the lowest type:
$$\begin{aligned} \hbox {w}\left[ {\hbox {h}^{1}} \right] =\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}^{1},\underline{\uptheta }} \right] \end{aligned}$$
(59)
For \({\uptheta }\in \left[ {{\uptheta }^{{*}},\bar{\uptheta }} \right] \), the incentive compatibility constraint requires
$$\begin{aligned} \hbox {w}\left[ {\hbox {h}^{2}} \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}^{2},{\uptheta }^{{*}}} \right]\ge & {} \hbox {w}\left[ {\hbox {h}^{1}} \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}^{1},{\uptheta }^{{*}}} \right] \\= & {} \hbox {R}\left( {1-{\uptau }} \right) \left[ {\hbox {x}\left[ {\hbox {h}^{1},\underline{\uptheta }} \right] -\hbox {x}\left[ {\hbox {h}^{1},{\uptheta }^{{*}}} \right] } \right] \end{aligned}$$
In equilibrium, it is binding only for the threshold type:
$$\begin{aligned} \hbox {w}\left[ {\hbox {h}^{2}} \right] =\hbox {R}\left( {1-{\uptau }} \right) \left[ {\hbox {x}\left[ {\hbox {h}^{2},{\uptheta }^{{*}}} \right] +\hbox {x}\left[ {\hbox {h}^{1},\underline{\uptheta }} \right] -\hbox {x}\left[ {\hbox {h}^{1},{\uptheta }^{{*}}} \right] } \right] \end{aligned}$$
(60)
Now the maximand turns out to be
$$\begin{aligned}&\max \hbox {E}\left[ {{\uptheta }^{1}} \right] \left[ {\hbox {k}^{1}} \right] ^{{\upalpha }}\left[ {\hbox {h}^{1}} \right] ^{1-\alpha }-\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}^{1},\underline{\uptheta }} \right] \hbox {F}\left( {{\uptheta }^{{*}}} \right) +\hbox {E}\left[ {{\uptheta }^{2}} \right] \left[ {\hbox {k}^{2}} \right] ^{{\upalpha }}\left[ {\hbox {h}^{2}} \right] ^{1-\alpha }\\&\quad -\hbox {R}\left( {1-{\uptau }} \right) \left[ {\hbox {x}\left[ {\hbox {h}^{2},{\uptheta }^{{*}}} \right] +\hbox {x}\left[ {\hbox {h}^{1},\underline{\uptheta }} \right] -\hbox {x}\left[ {\hbox {h}^{1},{\uptheta }^{{*}}} \right] } \right] \left[ {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right] \end{aligned}$$
FOC with respect to human capital:
$$\begin{aligned}&\hbox {E}\left[ {{\uptheta }^{1}} \right] \left( {1-{\upalpha }} \right) \left[ {\hbox {k}^{1}} \right] ^{{\upalpha }}\left[ {\hbox {h}^{1}} \right] ^{-\alpha }\nonumber \\&\quad =\hbox {R}\left( {1-{\uptau }} \right) \left[ {\frac{\partial \hbox {x}}{\partial \hbox {h}^{1}}\left[ {\hbox {h}^{1},\underline{\uptheta }} \right] -\frac{\partial \hbox {x}}{\partial \hbox {h}^{1}}\left[ {\hbox {h}^{1},{\uptheta }^{{*}}} \right] \left[ {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right] } \right] \end{aligned}$$
(61)
$$\begin{aligned}&\hbox {E}\left[ {{\uptheta }^{2}} \right] \left( {1-{\upalpha }} \right) \left[ {\hbox {k}^{2}} \right] ^{{\upalpha }}\left[ {\hbox {h}^{2}} \right] ^{-\alpha }\nonumber \\&\quad =\hbox {R}\left( {1-{\uptau }} \right) \frac{\partial \hbox {x}}{\partial \hbox {h}^{2}}\left[ {\hbox {h}^{2},{\uptheta }^{{*}}} \right] \left[ {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right] \end{aligned}$$
(62)
Derivations
Similar to the benchmark, we have
$$\begin{aligned} \mathop \int \nolimits _0^1 \left[ {\hbox {a}_{\mathrm{it}+1} +\left( {1-{\uptau }} \right) \hbox {x}_{\mathrm{it}} +\hbox {T}_{\mathrm{it}} } \right] \hbox {di}=\left( {1-{\upeta }} \right) \mathop \int \nolimits _0^1 \left( {\hbox {w}\left[ {\hbox {h}_{\mathrm{it}} } \right] +\hbox {R}_{\mathrm{t}} \hbox {a}_{\mathrm{it}} } \right) \hbox {di}\nonumber \\ \hbox {K}_{\mathrm{t}+1} +\mathop \int \nolimits _{\underline{\uptheta }} ^{{\uptheta }^{{*}}} \hbox {x}\left[ {\hbox {h}_{\mathrm{t}+1}^1 ,{\uptheta }} \right] \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) +\mathop \int \nolimits _{{\uptheta }^{{*}}}^{\bar{{\uptheta }}} \hbox {x}\left[ {\hbox {h}_{\mathrm{t}+1}^2 ,{\uptheta }} \right] \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) =\frac{1-{\upeta }}{{\upalpha }}\hbox {R}_{\mathrm{t}} \hbox {K}_{\mathrm{t}} \end{aligned}$$
(63)
And we are able to integrate two integrals in (63).
$$\begin{aligned} \hbox {I}_1= & {} \mathop \int \nolimits _{\underline{\uptheta }} ^{{\uptheta }^{{*}}} \hbox {x}\left[ {\hbox {h}_{\mathrm{t}+1}^1 ,{\uptheta }} \right] \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) =\mathop \int \nolimits _{\underline{\uptheta }} ^{{\uptheta }^{{*}}} \frac{\hbox {Z}\left[ {\hbox {h}_{\mathrm{t}+1}^1 } \right] ^{1+{\phi } }}{{\uptheta }_{\mathrm{i}} }\hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) =\frac{\hbox {Z}\left[ {\hbox {h}_{\mathrm{t}+1}^1 } \right] ^{1+{\phi } }}{\underline{\uptheta }} \frac{\underline{\uptheta }}{{\bar{{\uptheta }}-\underline{\uptheta }}}\hbox {ln}\frac{\uptheta ^{*}}{\underline{\uptheta }}\\= & {} \frac{{\hbox {w}}{[\hbox {h}_{\mathrm{t}+1}^{1}]}}{\hbox {R}_{\hbox {t}+1}(1-\uptau )} \frac{\underline{\uptheta }}{\bar{\uptheta }-\underline{\uptheta }}\hbox {ln}\frac{\uptheta ^{*}}{\underline{\uptheta }}= {\frac{1-\upalpha }{\upalpha }}\frac{\hbox {k}_{\mathrm{t}+1}^{1}}{1-\uptau } \frac{\underline{\uptheta }}{\bar{\uptheta }-\underline{\uptheta }} \hbox {ln}\frac{\uptheta ^{*}}{\underline{\uptheta }} \end{aligned}$$
The fourth equality comes from (59), and the fifth equality comes from (57).
$$\begin{aligned} {{\mathrm{I}}_2}= & {} \mathop \int \nolimits _{{{{\uptheta }}^{{*}}}}^{{\bar{\uptheta }}} {\mathrm{x}}\left[ {{{\mathrm{h}}^2},{{\uptheta }}} \right] {\mathrm{dF}}\left( {{{{\uptheta }}_{\mathrm{i}}}} \right) = \mathop \int \nolimits _{{{{\uptheta }}^{{*}}}}^{{\bar{\uptheta }}} \frac{{{\mathrm{Z}}{{\left[ {{{\mathrm{h}}^2}} \right] }^{1 + \phi }}}}{{{{{\uptheta }}_{\mathrm{i}}}}}{\mathrm{dF}}\left( {{{{\uptheta }}_{\mathrm{i}}}} \right) = \frac{{{\mathrm{Z}}{{\left[ {{{\mathrm{h}}^2}} \right] }^{1 + \phi }}}}{{{\bar{\uptheta }} - \underline{{\uptheta }} }}\ln \frac{{{\bar{\uptheta }}}}{{{{{\uptheta }}^{{*}}}}}\\= & {} \left[ {\frac{{{\mathrm{w}}\left[ {{\mathrm{h}}_{{\mathrm{t}} + 1}^2} \right] }}{{{{\mathrm{R}}_{{\mathrm{t}} + 1}}\left( {1 - {{\uptau }}} \right) }} - \frac{{{\mathrm{w}}\left[ {{\mathrm{h}}_{{\mathrm{t}} + 1}^1} \right] }}{{{{\mathrm{R}}_{{\mathrm{t}} + 1}}\left( {1 - {{\uptau }}} \right) }}\frac{{{{{\uptheta }}^{{*}}} - \underline{{\uptheta }} }}{{{{{\uptheta }}^{{*}}}}}} \right] \frac{{{{{\uptheta }}^{{*}}}}}{{{\bar{\uptheta }} - \underline{{\uptheta }} }}\ln \frac{{{\bar{\uptheta }}}}{{{{{\uptheta }}^{{*}}}}} \\= & {} \frac{{1 - {{\upalpha }}}}{{{{\upalpha }}\left( {1 - {{\uptau }}} \right) }}\left[ {{\mathrm{k}}_{{\mathrm{t}} + 1}^2 - {\mathrm{k}}_{{\mathrm{t}} + 1}^1\frac{{{{{\uptheta }}^{{*}}} - \underline{{\uptheta }} }}{{{{{\uptheta }}^{{*}}}}}} \right] \frac{{{{{\uptheta }}^{{*}}}}}{{{\bar{\uptheta }} - \underline{{\uptheta }} }}\ln \frac{{{\bar{\uptheta }}}}{{{{{\uptheta }}^{{*}}}}} \end{aligned}$$
The fourth equality comes from (59) and (60), and the fifth equality comes from (57) and (58).
Now we derive the relation between K1 and K2. By Eqs. (57) and (58)
$$\begin{aligned} \hbox {k}^{2}=\hbox {k}^{1}\frac{\hbox {h}^{2}}{\hbox {h}^{1}}\left( {\frac{\hbox {E}\left[ {{\uptheta }^{2}} \right] }{\hbox {E}\left[ {{\uptheta }^{1}} \right] }} \right) ^{\frac{1}{1-\alpha }} \end{aligned}$$
Define marginal product of human capital. Then Eqs. (61) and (62) become
$$\begin{aligned} \hbox {MP}_{\mathrm{h}}^1= & {} \hbox {R}\left( {1-{\uptau }} \right) \hbox {Z}\left( {1+{\upphi } } \right) \left[ {\hbox {h}^{1}} \right] ^{{\upphi } }\left[ {\frac{1}{\underline{\uptheta }} }-\frac{1}{{\uptheta }^{{*}}}\left[ {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right] \right] \end{aligned}$$
(64)
$$\begin{aligned} \hbox {MP}_{\mathrm{h}}^2= & {} \hbox {R}\left( {1-{\uptau }} \right) \hbox {Z}\left( {1+{\upphi } } \right) \left[ {\hbox {h}^{2}} \right] ^{{\upphi } }\frac{\left[ {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right] }{{\uptheta }^{{*}}} \end{aligned}$$
(65)
Use the definitions of MPK and MPH and some algebraic manipulation, we have
$$\begin{aligned} \left[ {\hbox {MP}_{\mathrm{h}}^1 } \right] _{\mathrm{t}}= & {} \hbox {E}\left[ {{\uptheta }^{1}} \right] ^{\frac{1}{1-\alpha }}\frac{1-{\upalpha }}{{\upalpha }^{\frac{{\upalpha }}{{\upalpha }-1}}}\hbox {R}_{\mathrm{t}}^{\frac{{\upalpha }}{{\upalpha }-1}} \end{aligned}$$
(66)
$$\begin{aligned} \left[ {\hbox {MP}_{\mathrm{h}}^2 } \right] _{\mathrm{t}}= & {} \hbox {E}\left[ {{\uptheta }^{2}} \right] ^{\frac{1}{1-\alpha }}\frac{1-{\upalpha }}{{\upalpha }^{\frac{{\upalpha }}{{\upalpha }-1}}}\hbox {R}_{\mathrm{t}}^{\frac{{\upalpha }}{{\upalpha }-1}} \end{aligned}$$
(67)
Then (64), (65), (66) and (67) give
$$\begin{aligned} \frac{\hbox {h}^{2}}{\hbox {h}^{1}}=\left[ {\frac{\hbox {E}\left[ {{\uptheta }^{2}} \right] }{\hbox {E}\left[ {{\uptheta }^{1}} \right] }} \right] ^{\frac{1}{\left( {1-\alpha } \right) {\phi } }}\left[ {\frac{\theta ^{*}}{\underline{\theta } \left[ {1-F\left( {\theta ^{*}} \right) } \right] }-1} \right] ^{\frac{1}{{\phi } }} \end{aligned}$$
so
$$\begin{aligned} \hbox {k}^{2}=\hbox {k}^{1}\left( {\frac{\hbox {E}\left[ {{\uptheta }^{2}} \right] }{\hbox {E}\left[ {{\uptheta }^{1}} \right] }} \right) ^{\frac{1}{1-\alpha }\left( {1+\frac{1}{{\phi } }} \right) }\left[ {\frac{\theta ^{*}}{\underline{\theta } \left[ {1-F\left( {\theta ^{*}} \right) } \right] }-1} \right] ^{1/{\phi } }=\hbox {CC}\times \hbox {k}^{1} \end{aligned}$$
(68)
where \(\hbox {CC}=\left( {\frac{\hbox {E}\left[ {{\uptheta }^{2}} \right] }{\hbox {E}\left[ {{\uptheta }^{1}} \right] }} \right) ^{\frac{1}{1-\alpha }\left( {1+\frac{1}{{\phi } }} \right) }\left[ {\frac{\theta ^{*}}{\underline{\theta } \left[ {1-F\left( {\theta ^{*}} \right) } \right] }-1} \right] ^{1/{\phi } }\)
By definition, \(\hbox {F}\left( {{\uptheta }^{{*}}} \right) \hbox {k}^{1}+\left[ {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right] \hbox {k}^{2}=\hbox {k}=\hbox {K}.\) So
$$\begin{aligned} \hbox {k}^{1}= & {} \frac{\hbox {K}}{\hbox {F}\left( {{\uptheta }^{{*}}} \right) +\hbox {CC}\left( {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right) } \end{aligned}$$
(69)
$$\begin{aligned} \hbox {k}^{2}= & {} \frac{\hbox {CC}\times \hbox {K}}{\hbox {F}\left( {{\uptheta }^{{*}}} \right) +\hbox {CC}\left( {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right) } \end{aligned}$$
(70)
Now substitute (69) and (70) into the two integrals in (63). It is the law of motion between K (t+1), K (t) and R (t).
Finally, we derive the relation between R(t) and K(t). Remember that R (t) equals MPK (t).
$$\begin{aligned} \hbox {E}\left[ {{\uptheta }^{1}} \right] {\upalpha }\left[ {\hbox {k}_{\mathrm{t}}^1 } \right] ^{\alpha -1}\left[ {\hbox {h}_{\mathrm{t}}^1 } \right] ^{1-{\upalpha }}=\hbox {R}_{\mathrm{t}} \end{aligned}$$
And MPH1 connects Eqs. (64) and (66)
$$\begin{aligned} \frac{1-{\upalpha }}{{\upalpha }^{\frac{{\upalpha }}{{\upalpha }-1}}}\hbox {E}\left[ {{\uptheta }^{1}} \right] ^{\frac{1}{1-\alpha }}\hbox {R}^{\frac{{\upalpha }}{{\upalpha }-1}}=\hbox {R}\left( {1-{\uptau }} \right) \hbox {Z}\left( {1+{\upphi } } \right) \left[ {\hbox {h}^{1}} \right] ^{{\upphi } }\left[ {\frac{1}{\underline{\uptheta }} }-\frac{1}{{\uptheta }^{{*}}}\left[ {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right] \right] \end{aligned}$$
Let \(\hbox {C}=\frac{1-{\upalpha }}{\hbox {Z}\left( {1-{\uptau }} \right) \left( {1+{\upphi } } \right) }{\upalpha }^{\frac{{\upalpha }}{1-{\upalpha }}}\). Then
$$\begin{aligned} \hbox {h}_{\mathrm{t}}^1 =\hbox {C}^{\frac{1}{{\upphi } }}\times \left[ {\frac{1}{\underline{\uptheta }} }-\frac{1}{{\uptheta }^{{*}}}\left[ {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right] \right] ^{\frac{-1}{{\phi } }}\hbox {E}\left[ {{\uptheta }^{1}} \right] ^{\frac{1}{\left( {1-\alpha } \right) {\upphi } }}\times \left[ {\hbox {R}_{\mathrm{t}} } \right] ^{\frac{1}{\left( {{\upalpha }-1} \right) {\upphi } }} \end{aligned}$$
(71)
Similarly,
$$\begin{aligned} \hbox {h}_{\mathrm{t}}^2 =\hbox {C}^{\frac{1}{{\upphi } }}\times \left[ {\frac{{\uptheta }^{{*}}}{1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) }} \right] ^{\frac{1}{{\phi } }}\hbox {E}\left[ {{\uptheta }^{2}} \right] ^{\frac{1}{\left( {1-\alpha } \right) {\phi } }}\times \left[ {\hbox {R}_{\mathrm{t}} } \right] ^{\frac{1}{\left( {{\upalpha }-1} \right) {\upphi } }} \end{aligned}$$
Substitute (71) into (57)
$$\begin{aligned} \hbox {E}\left[ {{\uptheta }^{1}} \right] {\upalpha }\left[ {\hbox {k}_{\mathrm{t}}^1 } \right] ^{\alpha -1}\left( {\hbox {C}^{\frac{1}{{\upphi } }}\times \left[ {\frac{1}{\underline{\uptheta }} }-\frac{1}{{\uptheta }^{{*}}}\left[ {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right] \right] ^{\frac{-1}{{\phi } }}\hbox {E}\left[ {{\uptheta }^{1}} \right] ^{\frac{1}{\left( {1-\alpha } \right) {\phi } }}\times \left[ {\hbox {R}_{\mathrm{t}} } \right] ^{\frac{1}{\left( {{\upalpha }-1} \right) {\upphi } }}} \right) ^{1-{\upalpha }}=\hbox {R}_{\mathrm{t}} \end{aligned}$$
Then
$$\begin{aligned} \left[ {\hbox {R}_{\mathrm{t}} } \right] ^{\frac{1+{\phi } }{\left( {1-\alpha } \right) {\phi } }}\hbox {K}_{\mathrm{t}}= & {} \left[ {\hbox {F}\left( {{\uptheta }^{{*}}} \right) +\hbox {CC}\left( {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right) } \right] {\upalpha }^{\frac{1}{1-{\upalpha }}}C^{\frac{1}{{\phi } }}\nonumber \\&\quad \times \left[ {\frac{1}{\underline{\uptheta }} }-\frac{1}{{\uptheta }^{{*}}}\left[ {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right] \right] ^{\frac{-1}{{\phi } }}\hbox {E}\left[ {{\uptheta }^{1}} \right] ^{\frac{1+{\phi } }{\left( {1-\alpha } \right) {\phi } }} \end{aligned}$$
(72)
or
$$\begin{aligned} \left[ {\hbox {R}_{\mathrm{t}} } \right] ^{\frac{1+{\phi } }{\left( {1-\alpha } \right) {\phi } }}\hbox {K}_{\mathrm{t}} =\frac{\left[ {\hbox {F}\left( {{\uptheta }^{{*}}} \right) +\hbox {CC}\left( {1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) } \right) } \right] }{\hbox {CC}}{\upalpha }^{\frac{1}{1-{\upalpha }}}\left[ {\hbox {C}\times \frac{{\uptheta }^{{*}}}{1-\hbox {F}\left( {{\uptheta }^{{*}}} \right) }} \right] ^{\frac{1}{{\phi } }}\hbox {E}\left[ {{\uptheta }^{2}} \right] ^{\frac{1+{\phi } }{\left( {1-\alpha } \right) {\phi } }} \end{aligned}$$
Appendix: Proof of the Lemma
Lemma
Incentive compatibility alone requires \(\frac{\mathrm {dh}}{\mathrm {d}{\uptheta }}>0\).
Proof
\(\forall {\uptheta },\tilde{\uptheta } ,\) and \(\tilde{\uptheta } >\theta \), incentive compatibility implies,
$$\begin{aligned} \hbox {w}\left[ {\hbox {h}\left( {\uptheta } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {\uptheta } \right) ;{\uptheta }} \right]\ge & {} \hbox {w}\left[ {\hbox {h}\left( {\tilde{\uptheta } } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {\tilde{\uptheta } } \right) ;{\uptheta }} \right] \\ \hbox {w}\left[ {\hbox {h}\left( {\tilde{\uptheta } } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {\tilde{\uptheta } } \right) ;\tilde{\uptheta } } \right]\ge & {} \hbox {w}\left[ {\hbox {h}\left( {\uptheta } \right) } \right] -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {\uptheta } \right) ;\tilde{\uptheta } } \right] \end{aligned}$$
The above two equations turn out to be,
$$\begin{aligned}&\left[ {\hbox {x}\left[ {\hbox {h}\left( {\tilde{\uptheta } } \right) ;{\uptheta }} \right] -\hbox {x}\left[ {\hbox {h}\left( {\uptheta } \right) ;{\uptheta }} \right] } \right] -\left[ {\hbox {x}\left[ {\hbox {h}\left( {\tilde{\uptheta } } \right) ;\tilde{\uptheta } } \right] -\hbox {x}\left[ {\hbox {h}\left( {\uptheta } \right) ;\tilde{\uptheta } } \right] } \right] \ge 0\\&\left( {\frac{1}{{\uptheta }}-\frac{1}{\tilde{\uptheta } }} \right) \left( {\left[ {\hbox {h}\left( {\tilde{\uptheta } } \right) } \right] ^{1+{\phi } }-\left[ {\hbox {h}\left( {\uptheta } \right) } \right] ^{1+{\phi } }} \right) \ge 0 \end{aligned}$$
Thus, \(\hbox {h}\left( {\tilde{\uptheta } } \right) >h\left( {\uptheta } \right) \), or \(\frac{\hbox {dh}}{\hbox {d}{\uptheta }}\ge 0\) . \(\square \)
Derivation of Eq. (67)
First rewrite the first order condition as,
$$\begin{aligned} \frac{\hbox {dU}}{\hbox {d}{\uptheta }}\left[ {\uptheta } \right] =-\hbox {R}\left( {1-{\uptau }} \right) \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] \end{aligned}$$
In addition, the associated participation constraint is \(\forall {\uptheta }\),
$$\begin{aligned} \hbox {U}\left( {\uptheta } \right) \ge \underline{\hbox {U}} \end{aligned}$$
where \(\underline{\hbox {U}} \) is the reserve utility. Because \(\frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] <0\), so \(\frac{\hbox {dU}}{\hbox {d}{\uptheta }}\left[ {\uptheta } \right] >0\). Let the reserve utility be zero. Then the participation condition becomes,
$$\begin{aligned} \hbox {U}\left( {\underline{\uptheta }} \right) \ge \underline{\hbox {U}} =0 \end{aligned}$$
The first order condition, second order condition, and participation condition fully characterize the income maximization problem.
Then integrate both sides of the first order condition and take advantage of that \(\hbox {U}\left( {\underline{\uptheta }} \right) =0\),
$$\begin{aligned} \hbox {U}\left( {\uptheta } \right) =-\hbox {R}\left( {1-{\uptau }} \right) \mathop \int \nolimits _{\underline{\uptheta }} ^{\uptheta } \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {\uptheta } \right) ,{\uptheta }} \right] \hbox {d}{\uptheta } \end{aligned}$$
Substitute the above expression into the maximand of firms’ problem,
$$\begin{aligned}&\max \mathop \int \nolimits _{\underline{\uptheta }} ^{\bar{{\uptheta }}} \left[ {\hbox {y}\left( {{\uptheta }_{\mathrm{i}} } \right) +\hbox {R}\left( {1-{\uptau }} \right) \mathop \int \nolimits _{\underline{\uptheta }}}^{\uptheta } \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] \hbox {d}{\uptheta }_{\mathrm{i}}\right. \\&\quad \left. -\hbox {R}\left( {1-{\uptau }} \right) \hbox {x}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] -\hbox {Rk}_{\mathrm{i}} \right] \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) \end{aligned}$$
The first order condition with respect to physical capital is, \(\forall \hbox {i}\)
$$\begin{aligned} \hbox {R}_{\mathrm{t}} ={\upalpha } {\uptheta }_{\mathrm{i}} \hbox {k}_{\mathrm{it}}^{{\upalpha }-1} \hbox {h}_{\mathrm{it}}^{1-{\upalpha }} \end{aligned}$$
which is the same as Eq. (65).
To derive the first order condition with respect to human capital, a few more algebraic manipulations are needed. Consider the integral inside the maximand, and integrate it by parts,
$$\begin{aligned} \mathop \int \nolimits _{\underline{\uptheta }} ^{\bar{{\uptheta }}} \left( {\mathop \int \nolimits _{\underline{\uptheta }} }^{\uptheta } \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] \hbox {d}{\uptheta }_{\mathrm{i}} \right) \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right)= & {} \left( {\mathop \int \nolimits _{\underline{\uptheta }} }^{\bar{{\uptheta }}} \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] \hbox {d}{\uptheta }_{\mathrm{i}} \right) \hbox {F}\left( {\bar{{\uptheta }}} \right) \\&-\mathop \int \nolimits _{\underline{\uptheta }} ^{\bar{{\uptheta }}} \hbox {F}\left( {{\uptheta }_{\mathrm{i}} } \right) \hbox {d}\left( {\mathop \int \nolimits _{\underline{\uptheta }} }^{\uptheta } \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] \hbox {d}{\uptheta }_{\mathrm{i}} \right) \\= & {} \mathop \int \nolimits _{\underline{\uptheta }} ^{\bar{{\uptheta }}} \left[ {1-\hbox {F}\left( {{\uptheta }_{\mathrm{i}} } \right) } \right] \frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] \hbox {d}{\uptheta }_{\mathrm{i}} \end{aligned}$$
Substitute it into the maximand,
$$\begin{aligned} \max \mathop \int \nolimits _{\underline{\uptheta }} ^{\bar{{\uptheta }}} \left[ {\hbox {y}\left( {{\uptheta }_{\mathrm{i}} } \right) -\hbox {R}\left( {1-{\uptau }} \right) \left[ {\hbox {x}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] -\frac{1-\hbox {F}\left( {{\uptheta }_{\mathrm{i}} } \right) }{\hbox {f}\left( {{\uptheta }_{\mathrm{i}} } \right) }\frac{\partial \hbox {x}}{\partial {\uptheta }}\left[ {\hbox {h}\left( {{\uptheta }_{\mathrm{i}} } \right) ,{\uptheta }_{\mathrm{i}} } \right] } \right] } \right] \hbox {dF}\left( {{\uptheta }_{\mathrm{i}} } \right) \end{aligned}$$
And we get Eq. (67) in the text.
