Endogenous cycles and human capital

Abstract

Existing evidence shows that human capital investment is countercyclical. In a dynamic model, I show that countercyclical investment in human capital arises as a result of a parametric combination relating to preferences and technologies. This countercyclical reaction is responsible for complex dynamics in the evolution of human capital, thus initiating a self-sustained sequence of events that generate endogenous cycles.

Appendix

1.1 Proof of Proposition 2

For the case where $\sigma =1$, we can substitute (6) and (7) in (9) to get

$$\begin{aligned} h_{t+1} =\frac{\varphi \beta }{1+\beta }h_t^\psi =F(h_t ). \end{aligned}$$

(15)

We can see that there are two possible steady state equilibria. One is $h_{t+1} =h_t =0$ and the other is $h_{t+1} =h_t =\hat{h}=[\varphi \beta /(1+\beta )]^{1/(1-\psi )}$. Given (15), the derivative of $F(h_t )$ is

$$\begin{aligned} {F}'(h_t )=\frac{\psi \varphi \beta }{1+\beta }h_t^{\psi -1} . \end{aligned}$$

(16)

It is straightforward to check that ${F}'(0)=\infty$ and ${F}'(\hat{h})=\psi \in (0,1)$. Thus, the only asymptotically stable equilibrium is $\hat{h}$.

Now, let us consider what happens when $\sigma >1$. Note that in this case $\delta =\frac{\sigma -1}{\sigma }(1-\psi )>0$ and $\delta +\psi =\left[ {\frac{\sigma -1}{\sigma }(1-\psi )+\psi } \right] \in (0,1)$. Given these, it is $F(0)=0$ and $F(\infty )=\infty$. Furthermore, the derivative

$$\begin{aligned} {F}'(h_t )=\frac{\varphi \omega h_t^{\delta +\psi -1} }{1+\omega h_t^\delta }\left[ {\psi +\delta ( {1-\frac{\omega h_t^\delta }{1+\omega h_t^\delta }})} \right] , \end{aligned}$$

(17)

is positive, while (17) reveals that ${F}'(0)=\infty$, meaning that $h_{t+1} =h_t =0$ is an unstable solution. Next, we can define the function

$$\begin{aligned} M(h_t )=\frac{F(h_t )}{h_t }=\frac{\varphi \omega h_t^{\delta +\psi -1} }{1+\omega h_t^\delta }, \end{aligned}$$

(18)

That satisfies $M(0) \text{= }\infty$ and $M(\infty )=0$ for $0<\delta +\psi <1$. The derivative ${M}'(h_t )$ is equal to

$$\begin{aligned} {M}'(h_t )=\frac{\varphi \omega h_t^{\delta +\psi -2} }{1+\omega h_t^\delta }\left[ {\delta ( {1-\frac{\omega h_t^\delta }{1+\omega h_t^\delta }})-(1-\psi )} \right] , \end{aligned}$$

(19)

and it is negative by virtue of $0<\delta +\psi <1$. We conclude that there is a unique interior $\hat{h}$ such that $h_{t+1} =h_t =\hat{h}$ and, therefore, $M(\hat{h})=1.$ Furthermore, this solution satisfies ${M}'(\hat{h})<0\Leftrightarrow {F}'(\hat{h})<1$. Combined with ${F}'(h_t )>0$, this analysis reveals that $\hat{h}$ is asymptotically stable because $0<{F}'(\hat{h})<1$. $\square$

1.2 Proof of Proposition 3

With $\sigma \in (0,1)$ and $\sigma +\psi >1$, it is $\delta <0$ and $\delta +\psi >0$. From (9), these imply that $F(0)=0$ and $F(\infty )=\infty$. Now let us consider the derivative in (17). Since $\delta <0$ and $\delta +\psi >0$, it is obvious that $0<\delta +\psi <1$ and that ${F}'(h_t )>0$. Furthermore, it is ${F}'(0)=\infty$, meaning that $h_{t+1} =h_t =0$ is an unstable solution. As for the function $M(h_t )$ in (18), the same conditions reveal that $M(0) \text{= }\infty$, $M(\infty )=0$ and ${M}'(h_t )<0$ still hold. Consequently, there is a unique interior $\hat{h}$ such that $h_{t+1} =h_t =\hat{h}$ and, therefore, $M(\hat{h})=1.$ Furthermore, this solution satisfies ${M}'(\hat{h})<0\Leftrightarrow {F}'(\hat{h})<1$. Combined with ${F}'(h_t )>0$, it is clear that $\hat{h}$ is asymptotically stable because $0<{F}'(\hat{h})<1$. $\square$

1.3 Proof of Proposition 4

When $\sigma \in (0,1)$ and $\sigma +\psi >1$, it is $\delta <0$ and $\delta +\psi <0$. Under these conditions, Eq. (9) reveals that $F(0)=F(\infty )=0$. Furthermore, we can use Eq. (17) to establish that

$$\begin{aligned} {F}'(h_t ) \left\{ {{\begin{array}{lll} {>0} &{}\quad {\text{ if }} &{} {h_t<\xi } \\ &{} &{} \\ {<0} &{}\quad {\text{ if }} &{} {h_t >\xi } \\ \end{array} }} \right. , \end{aligned}$$

(20)

where $\xi >0$ is defined in Eq. (10) of the main text. Additionally, we can check that (17) reveals ${F}'(0)=\infty$, therefore $h_{t+1} =h_t =0$ is an unstable solution.

Now, let us turn our attention to Eq. (18) and (19). We can establish that $M(0) \text{= }\infty$, $M(\infty )=0$ and ${M}'(h_t )<0$ still hold. These imply that there is a unique $\hat{h}$ such that $h_{t+1} =h_t =\hat{h}$ and, therefore, $M(\hat{h})=1.$ Furthermore, it is ${M}'(\hat{h})<0$ or, alternatively, ${F}'(\hat{h})<1$. In this case, however, we cannot make any definite conclusions concerning the stability of this equilibrium as we do not yet know whether $\hat{h}$ lies on the downward sloping part of $F(h_t )$. For this reason, we have to examine two different scenarios.

Given the properties of the function $F(h_t )$, as long as $F(\xi )<\xi$ then $\hat{h}$ will lie on the upward sloping part of $F(h_t )$ and it will satisfy $\hat{h}<\xi$ and $0<{F}'(\hat{h})<1$. In this case, $\hat{h}$ is asymptotically stable. When $F(\xi )>\xi$, $\hat{h}$ will lie on the downward sloping part of $F(h_t )$ and it will satisfy $\hat{h}>\xi$ and ${F}'(\hat{h})<0$. This means that dynamics are oscillatory, rather than monotonic, and that there may be periodic solutions for which oscillations are permanent. This can happen because the function $F(h_t )$ satisfies all the conditions of Theorem 8.2 in Azariadis (1993). Therefore, given that Theorem, ${F}'(\hat{h})<-1$ is a sufficient condition for the existence of a period-2 cycle $\{{\mathop h\limits ^{ \smile }}_1 ,{\mathop h\limits ^{ \smile }}_2 \}$ such that ${\mathop h\limits ^{ \smile }}_1 =F({\mathop h\limits ^{ \smile }}_2 )$, ${\mathop h\limits ^{ \smile }}_2 =F({\mathop h\limits ^{ \smile }}_1 )$ and ${\mathop h\limits ^{ \smile }}_1<\hat{h}<{\mathop h\limits ^{ \smile }}_2$. $\square$