Human capital and knowledge

NOTE: The present notebook is coded in R. It relies heavily on the tidyverse ecosystem of packages. We load the tidyverse below as a prerequisite for the rest of the notebook - along with a few other libraries.

\(\rightarrow\) Don’t forget that code flows sequentially. A random chunk may not work if the previous have have not been executed.

library(tidyverse)   # Package for data wrangling
library(readxl)      # Package to import MS Excel files
library(latex2exp)   # Package for LaTeX expressions
library(quantmod)    # Package for stock data extraction
library(highcharter) # Package for reactive plots

The content of the notebook is heavily inspired from the book Advanced Macro-economics - An Easy Guide.

Context

The models we have covered so far do not allow for strictly positive growth in equilibrium. There are several ways in which they can be changed in order to try to foster endogenous growth.

The model

Production

In this section, the change comes from the production function:

\[Y = K^\alpha H^\beta (AL)^\gamma,\]

where \(H\) stands for human capital and returns to scale are constant, increasing or decreasing if \(\alpha+\beta+\gamma\) is equal to one, strictly above, or strictly below. The productivity (or technological) factor \(A\) is associated to labor in this case. Dividing by \(L\), we get the per-capita version:

\[y = A^\gamma k^\alpha h^\beta , \tag{1}\]

Evolution through saving & depreciation

As in the simpler models, we assume savings rates for physical and human capital \(s_k\) and \(s_h\); as well as depreciation rates, too: \(\delta_k\) and \(\delta_h\). Then, time-variations are given by

\[\dot{K} = s_kY-\delta_k K \quad \Leftrightarrow \quad \frac{\dot{K}}{L}=s_ky-\delta_kk\] \[\dot{H} = s_hY-\delta_h H \quad \Leftrightarrow \quad \frac{\dot{H}}{L}=s_hy-\delta_hh. \] Moreover, by definition, \(\dot{k}=\dot{K}/L-nk\) and \(\dot{h}=\dot{H}/L-nh\), where \(n\) is the constant growth rate of the population \((L)\). Plugging this into the above expressions, we get \[\dot{k}=s_k A^\gamma k^\alpha h^\beta-(\delta_k+n)k, \quad \text{and} \quad \dot{h}=s_h A^\gamma k^\alpha h^\beta-(\delta_h+n)h,\] and the corresponding relative growth rates are: \[\gamma_k=\frac{\dot{k}}{k}=s_k A^\gamma k^{\alpha-1} h^\beta-(\delta_k+n)\] \[\gamma_h=\frac{\dot{h}}{h}=s_h A^\gamma k^{\alpha} h^{\beta-1}-(\delta_h+n).\]

Steady states

The balanced growth path requires that \(\gamma_k\) and \(\gamma_h\) be constant. Since \(\delta_k+n\) and \(\delta_h+n\) are already constant, we need that \(s_k A^\gamma k^{\alpha-1} h^\beta\) and \(s_h A^\gamma k^{\alpha} h^{\beta-1}\) be constant. To simplify these quantities, we take their log (they, too, have to be constant), and we differentiate with respect to time (log(\(s_k\)), log(\(A^\gamma\)), and log(\(s_h\)) are constant so they vanish): \[\frac{\partial}{\partial t} \left((\alpha-1) \log(k)+\beta \log(h) \right)=(\alpha-1) \frac{\dot{k}}{k}+\beta \frac{\dot{h}}{h}=(\alpha-1)\gamma_k+\beta \gamma_h=0\] \[\frac{\partial}{\partial t} \left(\alpha \log(k)+(\beta-1) \log(h) \right)=\alpha \frac{\dot{k}}{k}+(\beta-1) \frac{\dot{h}}{h}=\alpha\gamma_k+(\beta-1) \gamma_h=0.\] Both l.h.s. expressions have to be equal to zero, i.e, from the second equation, \(\gamma_h=\frac{\alpha}{1-\beta}\gamma_k\) and substituting this into the first equation, \[\left((\alpha-1)+\frac{\beta \alpha}{1-\beta}\right) \gamma_k=\frac{(\alpha-1)(1-\beta)+\alpha \beta}{1-\beta}\gamma_k=\frac{\alpha+\beta-1}{1-\beta} \gamma_k=0.\] If we assume constant returns to scale, \(\alpha+\beta \neq 1\) because else \(\gamma=0\). And obviously, \(\beta \neq 0\) too, hence this imposes \(\gamma_k=0\).

Important

Again, this means zero growth in equilibrium!

Another model

From Barro and Sala-ì-Martin, Section 5.1. The production is of Cobb-Douglas type:

\[Y=AK^aH^{1-a}, \quad a \in (0,1),\] with \(H\) being human capital. The resource constraint is: \[Y=C+I_K+I_H,\] where \(I_K\) and \(I_H\) are the investments in capital and the workforce. Moreover, \[\dot{K}=I_K-\delta_KK, \quad \dot{H}=I_H-\delta_HH.\]

The population seeks to maximize utility across generations: \(\int_0^\infty e^{-\rho t}u(C_t)dt\). Given the 3 constraints listed above, the Hamiltonian is \[J=e^{-\rho t}u(C)+\nu(I_K-\delta_KK)+\mu(I_H-\delta_HH)+\omega(AK^aH^{1-a}-C-I_K-I_H).\]

The model then assumes a CRRA utility function, written as \(u(x)=x^{1-\theta}/(1-\theta)\). We have

\[\frac{\partial J}{\partial C}=e^{-\rho t}C^{-\theta} -\omega, \quad \frac{\partial J}{\partial I_K}=\nu -\omega, \quad \frac{\partial J}{\partial I_H}=\mu -\omega \] we get \[\omega = \nu = \mu = e^{-\rho t}C^{-\theta}\] Hence, differentiating with respect to t, we obtain (equating with \(\partial J/\partial K\) and \(\partial J/\partial H\)) \[\dot{\nu}=-\rho e^{-\rho t}C^{-\theta}-\theta e^{-\rho t}C^{-\theta-1} \dot{C}=\frac{\partial J}{\partial K}=\nu \delta_K-\omega a A K^{a-1}H^{1-a}\]

\[\dot{\mu}=-\rho e^{-\rho t}C^{-\theta}-\theta e^{-\rho t}C^{-\theta-1} \dot{C}=\frac{\partial J}{\partial H}=-\mu \delta_H-\omega (1-a) A K^{a}H^{-a}\]

And since \(\omega = \nu=\mu\), \[\frac{\dot{\nu}}{\nu}=\delta_K-aAK^{a-1}H^{1-a}, \quad \frac{\dot{\mu}}{\mu}=-\delta_H+(1-a)AK^{a}H^{-a}\] At the same time, from the expressions of \(\dot{\nu}\) and \(\dot{\mu}\), \[\frac{\dot{\nu}}{\nu}=-\rho -\theta \frac{\dot{C}}{C}, \quad \frac{\dot{\mu}}{\mu}=-\rho -\theta \frac{\dot{C}}{C}\]

Hence,

\[\frac{\dot{C}}{C}=\theta^{-1}(aA(K/H)^{a-1}-\delta_K-\rho)=\theta^{-1}((a-1)A(K/H)^{a}-\delta_H-\rho) \tag{2}\]

If we set \(\delta_K=\delta_H\), it must hold that \(aA(K/H)^{a-1}=(a-1)A(K/H)^{a}\), i.e., \[K/H=a/(1-a). \tag{3}\] We can then plug this in the above equation to get:

\[\frac{\dot{C}}{C}=\theta^{-1}(Aa^a(1-a)^{1-a}-\delta-\rho)\]

If \(Y=AK^aH^{1-a}= AK (K/H)^{a-1}\) and Equation 3 holds, we have \[Y=AK\left(\frac{a}{1-a} \right)^{a-1}\] \(\rightarrow\) The AK model! More on that soon.

Two imperfect avenues towards growth

Returns to scale

One reason why growth is not possible in the above model is the assumption of constant returns to scale. Indeed, if we allow \(\alpha+\beta=1\) (and \(\gamma >0\), so that returns to scale are increasing), then \(\gamma_k \neq 0\).

Exogenous growth

If \(\gamma_k= s_k y/k-(\delta_k+n)\) and \(\gamma_h=s_h y/h-(\delta_h+n)\) are constant, then \(k/y\) and \(h/y\) are constant and all variables grow at the same pace. Let’s take log-output:

\[\log(y)=\gamma \log(A) + \alpha \log(k) +\beta \log(h), \]

hence, differentiating with respect to time (recalling a Cobb-Douglas function \(y=Ak^\alpha h^\beta\)), \[\frac{\dot{y}}{y} = \gamma \frac{\dot{A}}{A}+\alpha \frac{\dot{k}}{k} + \beta \frac{\dot{h}}{h}\]

Since \(x = \frac{\dot{y}}{y} =\frac{\dot{k}}{k} = \frac{\dot{h}}{h}\), if we write \(g\) for the relative growth rate of \(A\) we get \[x = \gamma g + \alpha x + \beta x,\] and assuming constant returns to scale \(1-\alpha-\beta=\gamma\), we get \(x=g\), i.e., all growth rates are equal to \(g\), the rate of increase of the productivity factor \(A\). But this is not endogenous growth…

The AK model

Assumptions and BGP

This is a special model because it assumes

\[y=Ak,\] hence its name. As in the Ramsey model, if we assume \(n=0\), capital evolves as: \[\dot{k}_t=y_t-c_t=Ak_t-c_t \tag{4}\]

We then set the utility function \[u(c_t)=\left(\frac{\sigma}{\sigma-1} \right)c_t^{(\sigma-1)/\sigma}\] which implies \(u'(c_t)=c_t^{-1/\sigma}\) and \(u''(c_t)=-\sigma^{-1}c_t^{-1-1/\sigma}\) so that indeed \(-\frac{u'(c)}{u''(c)c}=\sigma\).

As before, the representative agent seeks to maximize discounted utility over an infinite horizon: \(\int_0^\infty u(c_t)e^{-\rho t}dt\).

The corresponding Hamiltonian is \[H=\left( \frac{\sigma}{\sigma - 1} \right)c_t^{(\sigma-1)/\sigma}+\lambda_t (A k_t - c_t),\] leading to

\[\frac{\partial H}{\partial c_t}=c_t^{-1/\sigma}-\lambda_t=0\] and \[\dot{\lambda}_t=-\frac{\partial H}{\partial k_t}+\rho \lambda_t=(\rho-A)\lambda_t\]

The first FOC implies \(\lambda_t=c_t^{-1/\sigma}\), hence \(\dot{\lambda}_t=-\sigma^{-1}c_t^{-1-1/\sigma} \dot{c}_t\). If we divide this by the FOC, we get

\[\frac{\dot{\lambda}_t}{\lambda_t}=-\sigma^{-1} \frac{\dot{c}_t}{c_t}\]

From the second FOC, this implies \[\frac{\dot{c}_t}{c_t}=\sigma(A-\rho). \tag{5}\]

For the BGP, the variables need to grow at constant speed. This is already the case for consumption, but what about capital? From Equation 4, we have \[\frac{\dot{k}_t}{k_t}=A-c_t/k_t, \tag{6}\] hence we need the same growth rate for \(c_t\) and \(k_t\). In the model, the BGP is such that all variables, including output \(y_t\) grow at the same pace, \(\sigma(A-\rho)\). In addition, from Equation 6, we have \[\sigma(A-\rho)=A-c_t/k_t \Leftrightarrow k_t = c_t (\sigma(A-\rho)-A)\]

Hence, consumption is exactly proportional to capital (and output).

The transversality condition

In the model, the TVC \(e^{-\rho t}k_t \lambda_t \rightarrow 0\) is \(e^{-\rho t}k_t c^{-\sigma^{-1}_t} \rightarrow 0\). The differential Equation 5 has simple solution \(c_t=c_0e^{(\sigma(A-\rho))t}\). But remember that this is also true for \(k\): \(k_t=k_0 e^{(\sigma(A-\rho))t}\), hence the condition becomes \(k_0c_0^{-\sigma^{-1}}e^{(\sigma(A-\rho)-A)t}\rightarrow 0\), which requires \(\sigma(A-\rho)<A\).

Knowledge & learning by doing

To get growth, one way may be to rely on some form productivity. The idea here is to assume that experience gets us better at production. We follow the survey Learning by doing, Section 5.

\[Y=AK^aL^{1-a},\] with \(A\) being knowledge. Labor increases exogenously at rate \(n\) and \(K\) satisfies \(\dot{K}=sY\), i.e., via savings and without depreciation. Knowledge is linked to cumulative investment (e.g., R&D) via \(A=bK^\lambda\).

We simplify to

\[y=Ak^a \tag{7}\]

and laws of motion are (same as in the Solow model below):

\[\frac{\dot{k}}{k}=sAk^{a-1}-n \tag{8}\]

and \[\frac{\dot{A}}{A}=\lambda n+\lambda \frac{\dot{k}}{k}. \tag{9}\]

Indeed, \(\dot{A}=b\lambda K^{\lambda-1}\dot{K}=\lambda A \frac{\dot{K}}{K}=\lambda A(\dot{k}/k+n)\) where the final equality comes from \(\dot{K}/K=sY/K=sA(L/K)^{1-a}=sAk^{a-1}=\dot{k}/k+n\) (from Equation 8).

This implies

\[\frac{\dot{y}}{y}=\frac{\dot{A}k^a+Aa k^{a-1}\dot{k}}{Ak^a}=\frac{\dot{A}}{A}+a\frac{\dot{k}}{k}=\lambda n+(\lambda +a)(sAk^{a-1}-n)\] In short, we have \(\frac{\dot{y}}{y}=\lambda n+ (\lambda+a)\frac{\dot{k}}{k}\): output growth is an affine function of capital growth. If we assume \(\frac{\dot{k}}{k}=0\), then growth comes from learning only. We can find a situation where all growths are positive. It corresponds to \(\frac{\dot{y}}{y}=\frac{\dot{k}}{k}\) and \(\frac{\dot{A}}{A}=(1-a)\frac{\dot{k}}{k}\). This yields \[\frac{\dot{k}}{k}=\frac{\dot{y}}{y}=\frac{\lambda n}{1-a-\lambda}\] Note that \(n=0\) implies \(a+\lambda=1\) (constant returns to scale). If population grows, then it must hold that \(a+\lambda<1\) (DRS).

NOTE: a simple alternative model can be found in Economic Growth, by Barro and Sala-i-Martin, Section 4.3.