Human Capital Externalities, Migration, and Wage Inequality

Jiancai Pi; Shumin Yu; Ping Xu; Yanwei Fan

doi:10.1515/snde-2025-0010

Article

Human Capital Externalities, Migration, and Wage Inequality

Jiancai Pi , Shumin Yu , Ping Xu and Yanwei Fan

Published/Copyright: August 22, 2025

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From the journal Studies in Nonlinear Dynamics & Econometrics

Abstract

This paper explores the impacts of human capital externalities on rural-urban migration and skilled-unskilled wage inequality. In the basic model with full employment, an increase in the intensity of human capital externalities will increase migration, and reduce wage inequality if the skilled sector is more capital intensive than the unskilled sector. The results of the basic model remain robust under various conditions, which include the incorporation of human capital externalities in the rural sector into the model, the use of the average human capital level to represent human capital externalities in the urban area. When there is an urban minimum wage in the economy (irrespective of whether an informal sector exists or not), wage inequality will be conditionally decreased, but the impact on migration remains unchanged. This also applies when human capital externalities affect skilled and unskilled sectors differently. In the case of a partially open capital market in the economy, migration will still be encouraged, yet wage inequality will be widened.

Keywords: human capital externalities; rural-urban migration; skilled-unskilled wage inequality

JEL Classification: J24; J31; J61

Corresponding author: Jiancai Pi, Department of Economics, School of Business, Nanjing University, 22, Hankou Road, Nanjing, 210093, China, E-mail: pi2008@nju.edu.cn

Appendix A: Dynamic Adjustment Process of Equation (9)

Following Beladi et al. (2008), we build the excess demand functions through the dynamic adjustment process. Then, the differential equations can be expressed as follows:

(A1) X ˙ = d 1 p X − a S X w S − a K X r ,

(A2) Y ˙ = d 2 p Y − a U Y w U − a K Y r ,

(A3) Z ˙ = d 3 1 − a U Z w U − a T Z τ ,

(A4) w S ˙ = d 4 a S X X − L ̄ S ,

(A5) w U ˙ = d 5 a U Y Y + a U Z Z − L ̄ U ,

(A6) r ˙ = d 6 a K X X + a K Y Y − K ̄ ,

(A7) τ ˙ = d 7 a T Z Z − T ̄ ,

(A8) H ˙ = d 8 H ̄ − L U Z H U − H ,

where d _i > 0 (i = 1, 2, …, 8) represents the speed of adjustment. The notation “·” represents the differentiation with respect to time . The determinant of the Jacobi matrix of Equations (A1)–(A8) are given by:

J = Q 0 0 0 θ S X 0 θ K X 0 − α 0 0 0 0 θ U Y θ K Y 0 − α 0 0 0 0 θ U Z 0 θ T Z 0 1 0 0 − θ K X σ X 0 θ K X σ X 0 − α 0 λ U Y λ U Z 0 − λ U Y σ Y θ K Y − λ U Z σ Z θ T Z λ U Y σ Y θ K Y λ U Z σ Z θ T Z − α λ U Y λ K X λ K Y 0 λ K X σ X θ S X λ K Y σ Y θ U Y − λ K Y σ Y θ U Y − λ K X σ X θ S X 0 − α 0 0 1 0 θ U Z σ Z 0 − θ U Z σ Z 0 0 0 λ Z H 0 − λ Z H σ Z θ T Z 0 λ Z H σ Z θ T Z 1 − λ Z H = Q Δ ̃ 1 ,

where Q = ∏ i = 1 8 d i p X p Y L ¯ S L ¯ U K ¯ T ¯ H ¯ X Y Z H w S w U r τ > 0 , and

Δ ̃ 1 = 0 0 0 θ S X 0 θ K X 0 − α 0 0 0 0 θ U Y θ K Y 0 − α 0 0 0 0 θ U Z 0 θ T Z 0 1 0 0 − θ K X σ X 0 θ K X σ X 0 − α 0 λ U Y λ U Z 0 − λ U Y σ Y θ K Y − λ U Z σ Z θ T Z λ U Y σ Y θ K Y λ U Z σ Z θ T Z − α λ U Y λ K X λ K Y 0 λ K X σ X θ S X λ K Y σ Y θ U Y − λ K Y σ Y θ U Y − λ K X σ X θ S X 0 − α 0 0 1 0 θ U Z σ Z 0 − θ U Z σ Z 0 0 0 λ Z H 0 − λ Z H σ Z θ T Z 0 λ Z H σ Z θ T Z 1 − λ Z H .

We assume the economic system is stable. Then, according to the Routh-Hurwitz theorem, ensuring the stability of the economic system require the Jacobi matrix of the adjustment system to be negative definite. Based on the judgement of a negative definite matrix (its odd order principal sub-equations are negative and even order principal sub-equations are positive), a necessary condition for the Jacobian matrix to be negative definite is that the sign of |J| is (−1)ⁿ, where n describes the row of the Jacobian matrix. In our situation, n = 8, hence, |J| > 0. Since Q > 0, we obtain that the sign of Δ ̃ 1 is positive. And Δ₁ can be obtained from Δ ̃ 1 by a series of primary transformations. The specific transformation process is as follows:

Making the sixth line subtract λ _KX times the fourth line in Δ ̃ 1 , we have:

Δ ̃ 1 = 0 0 0 θ S X 0 θ K X 0 − α 0 0 0 0 θ U Y θ K Y 0 − α 0 0 0 0 θ U Z 0 θ T Z 0 1 0 0 − θ K X ⁢ σ X 0 θ K X ⁢ σ X 0 − α 0 λ U Y λ U Z 0 − λ U Y ⁢ σ Y ⁢ θ K Y − λ U Z ⁢ σ Z ⁢ θ T Z λ U Y ⁢ σ Y ⁢ θ K Y λ U Z ⁢ σ Z ⁢ θ T Z − α λ U Y 0 λ K Y 0 λ K X ⁢ σ X λ K Y ⁢ σ Y ⁢ θ U Y − λ K Y ⁢ σ Y ⁢ θ U Y − λ K X ⁢ σ X 0 − α λ K Y 0 0 1 0 θ U Z ⁢ σ Z 0 − θ U Z ⁢ σ Z 0 0 0 λ Z H 0 − λ Z H ⁢ σ Z ⁢ θ T Z 0 λ Z H ⁢ σ Z ⁢ θ T Z 1 − λ Z H = 1 × − 1 5 × 0 0 θ S X 0 θ K X 0 − α 0 0 0 θ U Y θ K Y 0 − α 0 0 0 θ U Z 0 θ T Z 0 λ U Y λ U Z 0 − λ U Y ⁢ σ Y ⁢ θ K Y − λ U Z ⁢ σ Z ⁢ θ T Z λ U Y ⁢ σ Y ⁢ θ K Y λ U Z ⁢ σ Z ⁢ θ T Z − α λ U Y λ K Y 0 λ K X ⁢ σ X λ K Y ⁢ σ Y ⁢ θ U Y − λ K Y ⁢ σ Y ⁢ θ U Y − λ K X ⁢ σ X 0 − α λ K Y 0 1 0 θ U Z ⁢ σ Z 0 − θ U Z ⁢ σ Z 0 0 λ Z H 0 − λ Z H ⁢ σ Z ⁢ θ T Z 0 λ Z H ⁢ σ Z ⁢ θ T Z 1 − λ Z H .

Then, let the fourth line subtract λ _UZ times the sixth line, and the seventh line subtract λ _ZH times the sixth line, and we obtain:

Δ ̃ 1 = 1 × − 1 5 × 0 0 θ S X 0 θ K X 0 − α 0 0 0 θ U Y θ K Y 0 − α 0 0 0 θ U Z 0 θ T Z 0 λ U Y 0 0 − λ U Y ⁢ σ Y ⁢ θ K Y − λ U Z ⁢ σ Z λ U Y ⁢ σ Y ⁢ θ K Y λ U Z ⁢ σ Z − α λ U Y λ K Y 0 λ K X ⁢ σ X λ K Y ⁢ σ Y ⁢ θ U Y − λ K Y ⁢ σ Y ⁢ θ U Y − λ K X ⁢ σ X 0 − α λ K Y 0 1 0 θ U Z ⁢ σ Z 0 − θ U Z ⁢ σ Z 0 0 0 0 − λ Z H ⁢ σ Z 0 λ Z H ⁢ σ Z 1 − λ Z H = 1 × − 1 5 × − 1 8 × 0 θ S X 0 θ K X 0 − α 0 0 θ U Y θ K Y 0 − α 0 0 θ U Z 0 θ T Z 0 λ U Y 0 − λ U Y ⁢ σ Y ⁢ θ K Y − λ U Z ⁢ σ Z λ U Y ⁢ σ Y ⁢ θ K Y λ U Z ⁢ σ Z − α λ U Y λ K Y λ K X ⁢ σ X λ K Y ⁢ σ Y ⁢ θ U Y − λ K Y ⁢ σ Y ⁢ θ U Y − λ K X ⁢ σ X 0 − α λ K Y 0 0 − λ Z H ⁢ σ Z 0 λ Z H ⁢ σ Z 1 − λ Z H .

After that, we carry out a series of exchange for the columns in the gained determinant, and then we have:

Δ ̃ 1 = − θ S X 0 θ K X 0 0 − α 0 θ U Y θ K Y 0 0 − α 0 θ U Z 0 θ T Z 0 0 0 − λ U Y σ Y θ K Y − λ U Z σ Z λ U Y σ Y θ K Y λ U Z σ Z λ U Y − α λ U Y λ K X σ X λ K Y σ Y θ U Y B 0 λ K Y − α λ K Y 0 − λ Z H σ Z 0 λ Z H σ Z 0 1 − λ Z H = − Δ 1 .

We have known that Δ ̃ 1 > 0 . Hence, it is easy to obtain Δ₁ < 0.

Appendix B: Dynamic Adjustment Process of Equation (13)

Using the above method, we replace Equations (A2) and (A5) with the following equations:

(B1) Y ˙ = d 2 p Y − a U Y w ̄ U − a K Y r ,

(B2) w ˙ U = d 5 w U ̄ w U a U Y Y + a U Z Z − L ̄ U .

The determinant of the Jacobian matrix of Equations (A1), (A3), (A4), (A6)–(A8), (B1), and (B2) is given by:

J = Q ⁡ 0 0 0 θ S X 0 θ K X 0 − α 0 0 0 0 0 θ K Y 0 − α 0 0 0 0 θ U Z 0 θ T Z 0 1 0 0 − θ K X ⁢ σ X 0 θ K X ⁢ σ X 0 − α 0 1 + λ ⁢ λ U Y λ U Z 0 − 1 + λ ⁢ λ U Y − λ U Z ⁢ σ Z ⁢ θ T Z 1 + λ ⁢ λ U Y ⁢ σ Y ⁢ θ K Y λ U Z ⁢ σ Z ⁢ θ T Z − α ⁡ 1 + λ λ U Y λ K X λ K Y 0 λ K X ⁢ σ X ⁢ θ S X 0 − λ K Y ⁢ σ Y ⁢ θ U Y − λ K X ⁢ σ X ⁢ θ S X 0 − α 0 0 1 0 θ U Z ⁢ σ Z 0 − θ U Z ⁢ σ Z 0 0 0 λ Z H 0 − λ Z H ⁢ σ Z ⁢ θ T Z 0 λ Z H ⁢ σ Z ⁢ θ T Z 1 − λ Z H = Q Δ ̃ 2 ,

where Δ ̃ 2 = 0 0 0 θ S X 0 θ K X 0 − α 0 0 0 0 0 θ K Y 0 − α 0 0 0 0 θ U Z 0 θ T Z 0 1 0 0 − θ K X σ X 0 θ K X σ X 0 − α 0 1 + λ λ U Y λ U Z 0 − 1 + λ λ U Y − λ U Z σ Z θ T Z 1 + λ λ U Y σ Y θ K Y λ U Z σ Z θ T Z − α 1 + λ λ U Y λ K X λ K Y 0 λ K X σ X θ S X 0 − λ K Y σ Y θ U Y − λ K X σ X θ S X 0 − α 0 0 1 0 θ U Z σ Z 0 − θ U Z σ Z 0 0 0 λ Z H 0 − λ Z H σ Z θ T Z 0 λ Z H σ Z θ T Z 1 − λ Z H .

After the same transformation process as in Appendix A, we have:

Δ ̃ 2 = − θ S X 0 θ K X 0 0 − α 0 0 θ K Y 0 0 − α 0 θ U Z 0 θ T Z 0 0 0 C 1 + λ λ U Y σ Y θ K Y λ U Z σ Z 1 + λ λ U Y − α 1 + λ λ U Y λ K X σ X 0 B 0 λ K Y − α λ K Y 0 − λ Z H σ Z 0 λ Z H σ Z 0 1 − λ Z H = − Δ 2 .

Similar to Appendix A, the sign of |J| is the same as that of (−1)⁸ according to the Routh-Hurwitz theorem, which means |J| = −QΔ₂ > 0. Thus, we have Δ₂ < 0.

Appendix C: Dynamic Adjustment Process of Equation (17)

Using the above method, we replace Equations (A1) and (A6) with the following equations:

(C1) X ˙ = d 1 p X − a S X w S − a K X r ̄ ,

(C2) r ˙ = d 6 a K Y Y − K ¯ Y ,

(C3) K ˙ X = d 9 a K X X − K X .

The determinant of the Jacobian matrix of Equations (A2)–(A5), (A7), (A8), and (C1)–(C3) is given by:

J = N 0 0 0 θ S X 0 0 0 0 − α 0 0 0 0 θ U Y 0 θ K Y 0 − α 0 0 0 0 θ U Z 0 0 θ T Z 0 1 0 0 − θ K X σ X 0 0 θ K X σ X 0 − α 0 λ U Y λ U Z 0 − λ U Y σ Y θ K Y − λ U Z σ Z θ T Z 0 λ U Y σ Y θ K Y λ U Z σ Z θ T Z − α λ U Y 1 0 0 θ S X σ X 0 − 1 0 0 − α 0 1 0 0 σ Y θ U Y 0 − σ Y θ U Y 0 − α 0 0 1 0 θ U Z σ Z 0 0 − θ U Z σ Z 0 0 0 λ Z H 0 − λ Z H σ Z θ T Z 0 0 λ Z H σ Z θ T Z 1 − λ Z H = N Δ ̃ 3 ,

where N = ∏ i = 1 9 d i p X p Y L ̄ S L ̄ U K ̄ Y T ̄ H ̄ X Y Z H w S w U r τ > 0 , and

Δ ̃ 3 = 0 0 0 θ S X 0 0 0 0 − α 0 0 0 0 θ U Y 0 θ K Y 0 − α 0 0 0 0 θ U Z 0 0 θ T Z 0 1 0 0 − θ K X σ X 0 0 θ K X σ X 0 − α 0 λ U Y λ U Z 0 − λ U Y σ Y θ K Y − λ U Z σ Z θ T Z 0 λ U Y σ Y θ K Y λ U Z σ Z θ T Z − α λ U Y 1 0 0 θ S X σ X 0 − 1 0 0 − α 0 1 0 0 σ Y θ U Y 0 − σ Y θ U Y 0 − α 0 0 1 0 θ U Z σ Z 0 0 − θ U Z σ Z 0 0 0 λ Z H 0 − λ Z H σ Z θ T Z 0 0 λ Z H σ Z θ T Z 1 − λ Z H .

After a series of transformation steps, we obtain:

Δ ̃ 3 = θ S X 0 0 0 0 − α 0 θ U Y θ K Y 0 0 − α 0 θ U Z 0 θ T Z 0 0 0 − λ U Y σ Y θ K Y − λ U Z σ Z λ U Y σ Y θ K Y λ U Z σ Z λ U Y − α λ U Y 0 σ Y θ U Y − σ Y θ U Y 0 1 − α 0 − λ Z H σ Z 0 λ Z H σ Z 0 1 − λ Z H = Δ 3 .

The sign of |J| is the same as that of (−1)⁹ according to the Routh-Hurwitz theorem, which means |J| = NΔ₃ < 0. Thus, we have Δ₃ < 0.

Appendix D: Dynamic Adjustment Process of Equation (21)

Using the above method, we replace Equation (A8) with the following equation:

(D1) H ˙ a = d 8 H ¯ − L ¯ − L U Z H a − L U Z H U .

The determinant of the Jacobian matrix of Equations (A1)–(A7), and (D1) is given by:

Following the same transformation process outlined in Appendix A, we obtain:

J = − Q θ S X 0 θ K X 0 0 − α 0 θ U Y θ K Y 0 0 − α 0 θ U Z 0 θ T Z 0 0 0 A λ U Y σ Y θ K Y λ U Z σ Z λ U Y − α λ U Y λ K X σ X λ K Y σ Y θ U Y B 0 λ K Y − α λ K Y 0 − F σ Z 0 F σ Z 0 D = − Q Δ 5 .

Similar to Appendix A, the sign of |J| is the same as that of (−1)⁸ according to the Routh-Hurwitz theorem, which means |J| = −QΔ₅ > 0. Thus, we have Δ₅ < 0.

Appendix E: Dynamic Adjustment Process of Equation (25)

Using the above method, the dynamic adjustment functions for the informal sector are given by:

(E1) R ˙ = d 9 p R − a U R w U − a K R r ,

(E2) w ˙ U = d 5 w ̄ U w U a U Y Y + a U R R + a U Z Z − L ̄ U .

We replace Equation (A6) with the following equation:

(E3) r ˙ = d 6 a K X X + a K Y Y + a K R R − K ̄ .

The determinant of the Jacobian matrix of Equations (A1), (A3), (A4), (A7), (A8), (B1), and (E1)–(E3) is given by:

J = S ⁡ 0 0 0 0 θ S X 0 θ K X 0 − α 0 0 0 0 0 0 θ K Y 0 − α 0 0 0 0 0 θ U R θ K R 0 − α 0 0 0 0 0 θ U Z 0 θ T Z 0 1 0 0 0 − θ K X ⁢ σ X 0 θ K X ⁢ σ X 0 − α 0 1 + λ ⁢ λ U Y λ U R λ U Z 0 − 1 + λ ⁢ λ U Y − λ U Z ⁢ σ Z ⁢ θ T Z 1 + λ ⁢ λ U Y ⁢ σ Y ⁢ θ K Y λ U Z ⁢ σ Z ⁢ θ T Z − α ⁡ 1 + λ λ U Y − λ U R ⁢ σ R ⁢ θ K R + λ U R ⁢ σ R ⁢ θ K R − α λ U R λ K X λ K Y λ K R 0 λ K X ⁢ σ X ⁢ θ S X λ K R ⁢ σ R ⁢ θ U R − λ K Y ⁢ σ Y ⁢ θ U Y − λ K R ⁢ σ R ⁢ θ U R 0 − α − λ K X ⁢ σ X ⁢ θ S X 0 0 0 1 0 θ U Z ⁢ σ Z 0 − θ U Z ⁢ σ Z 0 0 0 0 λ Z H 0 − λ Z H ⁢ σ Z ⁢ θ T Z 0 λ Z H ⁢ σ Z ⁢ θ T Z 1 − λ Z H = S Δ ̃ 6 ,

where S = − ∏ i = 1 9 d i p X p Y p R L ̄ S L ̄ U K ̄ T ̄ H ̄ X Y Z R H w S w U r τ < 0 , and

Δ ̃ 6 = 0 0 0 0 θ S X 0 θ K X 0 − α 0 0 0 0 0 0 θ K Y 0 − α 0 0 0 0 0 θ U R θ K R 0 − α 0 0 0 0 0 θ U Z 0 θ T Z 0 1 0 0 0 − θ K X ⁢ σ X 0 θ K X ⁢ σ X 0 − α 0 1 + λ ⁢ λ U Y λ U R λ U Z 0 − 1 + λ ⁢ λ U Y − λ U Z ⁢ σ Z ⁢ θ T Z 1 + λ ⁢ λ U Y ⁢ σ Y ⁢ θ K Y λ U Z ⁢ σ Z ⁢ θ T Z − α ⁡ 1 + λ λ U Y − λ U R ⁢ σ R ⁢ θ K R + λ U R ⁢ σ R ⁢ θ K R − α λ U R λ K X λ K Y λ K R 0 λ K X ⁢ σ X ⁢ θ S X λ K R ⁢ σ R ⁢ θ U R − λ K Y ⁢ σ Y ⁢ θ U Y − λ K R ⁢ σ R ⁢ θ U R 0 − α − λ K X ⁢ σ X ⁢ θ S X 0 0 0 1 0 θ U Z ⁢ σ Z 0 − θ U Z ⁢ σ Z 0 0 0 0 λ Z H 0 − λ Z H ⁢ σ Z ⁢ θ T Z 0 λ Z H ⁢ σ Z ⁢ θ T Z 1 − λ Z H .

After a series of transformation steps, we obtain:

Δ ̃ 6 = θ S X 0 θ K X 0 0 0 − α 0 0 θ K Y 0 0 0 − α 0 θ U R θ K R 0 0 0 − α 0 θ U Z 0 θ T Z 0 0 0 0 L W λ U Z σ Z 1 + λ λ U Y λ U R − α 1 + λ λ U Y − α λ U R λ K X σ X λ K R σ R θ U R T 0 λ K Y λ K R − α 1 − λ K X 0 − λ Z H σ Z 0 λ Z H σ Z 0 0 1 − λ Z H = Δ 6 .

Similar to Appendix A, the sign of |J| is the same as that of (−1)⁹ according to the Routh-Hurwitz theorem, which means |J| = SΔ₆ < 0. Thus, we have Δ₆ > 0.

Appendix F: Dynamic Adjustment Process of Equation (26)

After the production function of the urban unskilled sector is changed to Y = H ^δα F ²(L _UY, K _Y), the determinant of the Jacobian matrix of Equations (A1)–(A8) is given by:

J = Q ⁡ 0 0 0 θ S X 0 θ K X 0 − α 0 0 0 0 θ U Y θ K Y 0 − α δ 0 0 0 0 θ U Z 0 θ T Z 0 1 0 0 − θ K X ⁢ σ X 0 θ K X ⁢ σ X 0 − α 0 λ U Y λ U Z 0 − λ U Y ⁢ σ Y ⁢ θ K Y − λ U Z ⁢ σ Z ⁢ θ T Z λ U Y ⁢ σ Y ⁢ θ K Y λ U Z ⁢ σ Z ⁢ θ T Z − α δ λ U Y λ K X λ K Y 0 λ K X ⁢ σ X ⁢ θ S X λ K Y ⁢ σ Y ⁢ θ U Y − λ K Y ⁢ σ Y ⁢ θ U Y − λ K X ⁢ σ X ⁢ θ S X 0 − α ⁡ δ λ K Y + λ K X 0 0 1 0 θ U Z ⁢ σ Z 0 − θ U Z ⁢ σ Z 0 0 0 λ Z H 0 − λ Z H ⁢ σ Z ⁢ θ T Z 0 λ Z H ⁢ σ Z ⁢ θ T Z 1 − λ Z H = Q Δ ̃ 7 ,

where Q = ∏ i = 1 8 d i p X p Y L ̄ S L ̄ U K ̄ T ̄ H ̄ X Y Z H w S w U r τ > 0 , and

Δ ̃ 7 = 0 0 0 θ S X 0 θ K X 0 − α 0 0 0 0 θ U Y θ K Y 0 − α δ 0 0 0 0 θ U Z 0 θ T Z 0 1 0 0 − θ K X σ X 0 θ K X σ X 0 − α 0 λ U Y λ U Z 0 − λ U Y σ Y θ K Y − λ U Z σ Z θ T Z λ U Y σ Y θ K Y λ U Z σ Z θ T Z − α δ λ U Y λ K X λ K Y 0 λ K X σ X θ S X λ K Y σ Y θ U Y − λ K Y σ Y θ U Y − λ K X σ X θ S X 0 − α δ λ K Y + λ K X 0 0 1 0 θ U Z σ Z 0 − θ U Z σ Z 0 0 0 λ Z H 0 − λ Z H σ Z θ T Z 0 λ Z H σ Z θ T Z 1 − λ Z H .

After the same transformation process as in Appendix A, we have:

Δ ̃ 7 = − θ S X 0 θ K X 0 0 − α 0 θ U Y θ K Y 0 0 − α δ 0 θ U Z 0 θ T Z 0 0 0 − λ U Y σ Y θ K Y − λ U Z σ Z λ U Y σ Y θ K Y λ U Z σ Z λ U Y − α δ λ U Y λ K X σ X λ K Y σ Y θ U Y − λ K Y σ Y θ U Y − λ K X σ X 0 λ K Y − α δ λ K Y 0 − λ Z H σ Z 0 λ Z H σ Z 0 1 − λ Z H = − Δ 7 .

Similar to Appendix A, the sign of |J| is the same as that of (−1)⁸ according to the Routh-Hurwitz theorem, which means |J| = −QΔ₇ > 0. Thus, we have Δ₇ < 0.

Appendix G: Sensitivity Test for Numerical Simulation

According to the 2020 China Family Panel Studies (CFPS), the estimated average annual wage of skilled labor is about two times that of unskilled labor. Hence, we adjust the ratio of human capital per unit of skilled labor to unskilled labor to 2. Specifically, we reset parameter H _S to 2 and keep the values of other parameters unchanged. Based on the new parameter configuration, a single numerical simulation is performed using the same methodology and the results obtained are shown in Figure A1.

Figure A1:

The sensitivity test for numerical simulation when H _S = 2.

Then, we also change the value of the parameter of the land endowment and then carry out the sensitivity test. The results are shown in Figure A2.

$Figure A2: The sensitivity test for numerical simulation when T ̄ = 300 $\bar{T}=300$ .$

Figure A2:

The sensitivity test for numerical simulation when T ̄ = 300 .

At last, we change the product prices in the urban skilled and unskilled sectors relative to the agricultural sector. Specifically, we reset p _X and p _Y as 2 and 1, respectively. Other parameters’ values remain unchanged. Then, we conduct numerical simulation with the same methods and the results are shown in Figure A3.

Figure A3:

The sensitivity test for numerical simulation when p _X = 2 and p _Y = 1.

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Received: 2025-02-03

Accepted: 2025-08-04

Published Online: 2025-08-22

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https://doi.org/10.1515/snde-2025-0010

Keywords for this article

human capital externalities; rural-urban migration; skilled-unskilled wage inequality