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Economics
Johns Hopkins University

Endogenous Growth: AK, Romer, and Lucas Models

This notebook is a computational companion to the Growth Theory II lecture (Module 12). It covers three canonical endogenous growth models: the Rebelo Rebelo (1991) AK model, the Romer Romer (1986) model of knowledge externalities, and the Lucas Lucas (1988) model of human capital accumulation.

The notebook proceeds in four parts. Part I derives Rebelo’s taxonomy of steady-state growth possibilities. Part II solves the AK model and contrasts its dynamics with the RCK model. Part III examines the private-vs-social growth wedge in the Romer model. Part IV analyzes the Lucas model with human capital externalities.

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from scipy import optimize
from sympy import (
    symbols, Rational, simplify, solve, diff, factor, latex, Eq,
    init_printing, oo, limit,
)

init_printing()

# Theme colors
ACCENT = '#107895'
RED    = '#9a2515'

Model Parameters

We define three namedtuple types, one per model.

from collections import namedtuple

AKModel    = namedtuple('AKModel',    ['A', 'delta_t', 'rho'])
RomerModel = namedtuple('RomerModel', ['alpha', 'eta', 'delta_t', 'rho'])
LucasModel = namedtuple('LucasModel', ['alpha', 'phi', 'delta_t', 'rho', 'psi'])

ak = AKModel(A=0.15, delta_t=0.04, rho=2.0)
rm = RomerModel(alpha=0.33, eta=0.67, delta_t=0.04, rho=2.0)
lm = LucasModel(alpha=0.33, phi=0.10, delta_t=0.04, rho=2.0, psi=0.2)

Part I: Rebelo’s Taxonomy

The Growth Equation

Rebelo Rebelo (1991) considers a general Cobb-Douglas production function Y=AKαLβY = A K^\alpha L^\beta where α+β\alpha + \beta need not equal one. With a constant saving rate ss and population growth nn, steady-state growth at rate γ\gamma requires

0=(α−1)γ+n(α+β−1).0 = (\alpha - 1)\gamma + n(\alpha + \beta - 1).
(1)

We derive this condition symbolically and solve for γ\gamma in each case.

alpha_s, beta_s, gamma_s, n_s = symbols('alpha beta gamma n', real=True)

# The steady-state growth condition
rebelo_eq = Eq((alpha_s - 1) * gamma_s + n_s * (alpha_s + beta_s - 1), 0)

# Solve for gamma
gamma_sol = solve(rebelo_eq, gamma_s)[0]
print(f"General solution: gamma = {gamma_sol}")

Catalog of Growth Possibilities

cases = []

# Case 1a: CRS with alpha < 1 (Solow)
g1a = gamma_sol.subs(beta_s, 1 - alpha_s)  # alpha + beta = 1
cases.append({
    'Case': '1a: CRS, α < 1',
    'Condition': r'$\alpha + \beta = 1$, $\alpha < 1$',
    'gamma': f'{simplify(g1a)}',
    'Name': 'Solow',
})

# Case 1b: CRS with alpha = 1 (AK)
cases.append({
    'Case': '1b: CRS, α = 1',
    'Condition': r'$\alpha = 1$, $\beta = 0$',
    'gamma': 'any (from Euler eq.)',
    'Name': 'Rebelo AK',
})

# Case 2a.i: IRS with n > 0
g2ai = gamma_sol  # general form
cases.append({
    'Case': '2a.i: IRS, n > 0',
    'Condition': r'$\alpha + \beta > 1$, $\alpha < 1$, $n > 0$',
    'gamma': rf'$\frac{{(\alpha+\beta-1)}}{{(1-\alpha)}} n$',
    'Name': 'Scale effects',
})

# Case 2a.ii: IRS with n = 0
g2aii = gamma_sol.subs(n_s, 0)
cases.append({
    'Case': '2a.ii: IRS, n = 0',
    'Condition': r'$\alpha + \beta > 1$, $\alpha < 1$, $n = 0$',
    'gamma': f'{g2aii}',
    'Name': 'No growth',
})

# Case 2b.i: IRS with alpha = 1, beta > 0, n > 0
cases.append({
    'Case': '2b.i: IRS, α = 1, n > 0',
    'Condition': r'$\alpha = 1$, $\beta > 0$, $n > 0$',
    'gamma': 'No steady state',
    'Name': '(inconsistent)',
})

# Case 2b.ii: IRS with alpha = 1, beta > 0, n = 0
cases.append({
    'Case': '2b.ii: IRS, α = 1, n = 0',
    'Condition': r'$\alpha = 1$, $\beta > 0$, $n = 0$',
    'gamma': 'any (from Euler eq.)',
    'Name': 'Second endogenous config.',
})

df_cases = pd.DataFrame(cases)
df_cases

Numerical Examples

# Compute growth rates for specific parameterizations
examples = []
for alpha_val, beta_val, n_val, label in [
    (0.33, 0.67, 0.01, 'Solow (CRS)'),
    (1.0,  0.0,  0.01, 'AK'),
    (0.4,  0.8,  0.02, 'IRS, n=0.02'),
    (0.4,  0.8,  0.0,  'IRS, n=0'),
]:
    if alpha_val == 1.0:
        g_val = 'N/A (no transition)'
    elif alpha_val + beta_val == 1.0 and alpha_val < 1.0:
        g_val = 0.0
    else:
        g_val = float(gamma_sol.subs({alpha_s: alpha_val, beta_s: beta_val, n_s: n_val}))
    examples.append({
        'Model': label, 'alpha': alpha_val, 'beta': beta_val,
        'n': n_val, 'gamma': g_val,
    })

pd.DataFrame(examples)

Part II: The AK Model

Growth Rate and Saving

In the AK model (Y=AKY = AK, no depreciation or population growth), the Euler equation gives a constant consumption growth rate:

C˙C=1ρ(A−δt).\frac{\dot{C}}{C} = \frac{1}{\rho}(A - \delta_t).
(2)

The saving rate is constant and proportional to growth:

s=γ/A,where γ=A−δtρ.s = \gamma / A, \quad \text{where } \gamma = \frac{A - \delta_t}{\rho}.
(3)
A_s, delta_t_s, rho_s = symbols('A delta_t rho', positive=True)

gamma_ak = (A_s - delta_t_s) / rho_s
s_ak = gamma_ak / A_s

print(f"AK growth rate: {gamma_ak}")
print(f"AK saving rate: {simplify(s_ak)}")

# Numerical
gamma_num = (ak.A - ak.delta_t) / ak.rho
s_num = gamma_num / ak.A
print(f"\nBaseline: gamma = {gamma_num:.4f} ({gamma_num*100:.2f}%)")
print(f"          s     = {s_num:.4f} ({s_num*100:.2f}%)")

Impatience Condition

The AK model has a well-defined solution only when γ<A\gamma < A. For a given AA, the minimum feasible ρ\rho satisfies ρ−1(A−δt)<A\rho^{-1}(A - \delta_t) < A. We find the critical ρ\rho numerically.

# Find the rho at which the impatience condition binds: gamma = A
# gamma = (A - delta_t) / rho = A  =>  rho = (A - delta_t) / A
rho_crit_exact = (ak.A - ak.delta_t) / ak.A

# Verify numerically with brentq
def impatience_residual(rho_val):
    return (ak.A - ak.delta_t) / rho_val - ak.A

rho_crit = optimize.brentq(impatience_residual, 0.01, 10.0)
print(f"Critical rho (analytic): {rho_crit_exact:.4f}")
print(f"Critical rho (brentq):   {rho_crit:.4f}")
print(f"Baseline rho = {ak.rho} is {'feasible' if ak.rho > rho_crit else 'infeasible'}")

AK vs. RCK Dynamics

The AK model has no transitional dynamics: all economies grow at the same rate regardless of initial capital. In contrast, the RCK model exhibits convergence to a common steady state.

T = 80
t = np.linspace(0, T, 500)

fig, axes = plt.subplots(1, 2, figsize=(10, 4))

# AK: parallel growth paths
for K0, label in [(1.0, 'Rich country'), (0.3, 'Poor country')]:
    Y = ak.A * K0 * np.exp(gamma_num * t)
    axes[0].plot(t, Y / (ak.A * 1.0), lw=2, label=label)
axes[0].set_xlabel('Time')
axes[0].set_ylabel('Output (relative to rich at $t=0$)')
axes[0].set_title('AK: no convergence')
axes[0].legend(fontsize=9, frameon=False)
axes[0].grid(True, alpha=0.3)

# RCK: convergence (stylized linearization)
alpha_rck, delta_rck, delta_t_rck = 0.33, 0.08, 0.04
n_rck, g_rck, rho_rck = 0.01, 0.02, 2.0
k_ss_rck = (alpha_rck / (delta_rck + delta_t_rck + rho_rck * g_rck)
            ) ** (1 / (1 - alpha_rck))
lam_approx = -0.04

for frac, label in [(0.3, 'Poor country'), (1.5, 'Rich country')]:
    k0 = frac * k_ss_rck
    k_path = k_ss_rck + (k0 - k_ss_rck) * np.exp(lam_approx * t)
    axes[1].plot(t, k_path**alpha_rck, lw=2, label=label)
axes[1].axhline(k_ss_rck**alpha_rck, color='gray', ls=':', lw=1)
axes[1].set_xlabel('Time')
axes[1].set_ylabel('Output per efficiency unit')
axes[1].set_title('RCK: convergence to steady state')
axes[1].legend(fontsize=9, frameon=False)
axes[1].grid(True, alpha=0.3)

plt.tight_layout()
plt.show()

Sensitivity to Technology Parameter

A_grid = np.linspace(0.05, 0.30, 200)

fig, axes = plt.subplots(1, 2, figsize=(10, 4))

for rho_val in [1.0, 2.0, 4.0]:
    gamma_grid = (A_grid - ak.delta_t) / rho_val
    s_grid = gamma_grid / A_grid
    axes[0].plot(A_grid, gamma_grid * 100, lw=2, label=rf'$\rho = {rho_val}$')
    axes[1].plot(A_grid, s_grid * 100, lw=2, label=rf'$\rho = {rho_val}$')

axes[0].set_xlabel(r'Technology parameter $A$')
axes[0].set_ylabel('Growth rate (%)')
axes[0].set_title('AK growth rate increases linearly in $A$')
axes[0].legend(fontsize=9, frameon=False)
axes[0].grid(True, alpha=0.3)

axes[1].set_xlabel(r'Technology parameter $A$')
axes[1].set_ylabel('Saving rate (%)')
axes[1].set_title('AK saving rate increases with $A$')
axes[1].legend(fontsize=9, frameon=False)
axes[1].grid(True, alpha=0.3)

plt.tight_layout()
plt.show()

Part III: Romer (1986) — Knowledge Externalities

Private vs. Social Returns

In the Romer model, firm-level production is CRS in (k,ℓ)(k, \ell):

yj=kjα ℓj1−α Ξηy_j = k_j^\alpha \, \ell_j^{1-\alpha} \, \Xi^\eta
(4)

where Ξ=K\Xi = K is aggregate knowledge (a byproduct of investment). On a balanced growth path with α+η=1\alpha + \eta = 1, the private and social growth rates are

γprivate=1ρ(α−δt)\gamma_{\text{private}} = \frac{1}{\rho}(\alpha - \delta_t)
(5)
γsocial=1ρ(α+η−δt)=1ρ(1−δt).\gamma_{\text{social}} = \frac{1}{\rho}(\alpha + \eta - \delta_t) = \frac{1}{\rho}(1 - \delta_t).
(6)
alpha_s, eta_s, delta_t_s, rho_s = symbols('alpha eta delta_t rho', positive=True)

gamma_priv = (alpha_s - delta_t_s) / rho_s
gamma_soc  = (alpha_s + eta_s - delta_t_s) / rho_s
wedge      = simplify(gamma_soc - gamma_priv)

print(f"Private growth rate: {gamma_priv}")
print(f"Social growth rate:  {gamma_soc}")
print(f"Externality wedge:   {wedge}")

Optimal Subsidy

The optimal Pigouvian subsidy s∗s^* on capital equates private after-subsidy MPK to social MPK:

(1+s∗)α=α+η⟹s∗=η/α.(1 + s^*)\alpha = \alpha + \eta \quad \Longrightarrow \quad s^* = \eta / \alpha.
(7)
s_star = eta_s / alpha_s
print(f"Optimal subsidy rate: s* = {s_star}")
print(f"Baseline: s* = {rm.eta / rm.alpha:.4f} ({rm.eta / rm.alpha * 100:.1f}%)")

Private vs. Social Growth Rates

alpha_grid = np.linspace(0.15, 0.85, 200)
eta_grid = 1 - alpha_grid  # balanced growth path condition

g_priv = (alpha_grid - rm.delta_t) / rm.rho
g_soc  = (1 - rm.delta_t) / rm.rho * np.ones_like(alpha_grid)

fig, ax = plt.subplots(figsize=(8, 5))
ax.plot(alpha_grid, g_priv * 100, lw=2, color=ACCENT, label='Private (decentralized)')
ax.plot(alpha_grid, g_soc * 100, lw=2, ls='--', color=RED, label='Social (planner)')
ax.fill_between(alpha_grid, g_priv * 100, g_soc * 100, alpha=0.12, color=RED)

mid_y = (g_priv[100] + g_soc[100]) / 2 * 100
ax.annotate(r'Wedge $= \eta / \rho$', xy=(0.5, mid_y), fontsize=11,
            ha='center', color=RED)

ax.set_xlabel(r'Capital share $\alpha$ (with $\eta = 1 - \alpha$)', fontsize=12)
ax.set_ylabel('Growth rate (%)', fontsize=12)
ax.set_title('Romer (1986): knowledge externalities create underinvestment', fontsize=13)
ax.legend(fontsize=10, frameon=False)
ax.grid(True, alpha=0.3)
plt.tight_layout()
plt.show()

Subsidy Rates by Capital Share

subsidy_grid = (1 - alpha_grid) / alpha_grid

fig, ax = plt.subplots(figsize=(8, 4))
ax.plot(alpha_grid, subsidy_grid * 100, lw=2, color=ACCENT)
ax.set_xlabel(r'Capital share $\alpha$', fontsize=12)
ax.set_ylabel(r'Optimal subsidy rate $s^* = \eta/\alpha$ (%)', fontsize=12)
ax.set_title('Optimal investment subsidy falls as private capital share rises', fontsize=12)
ax.grid(True, alpha=0.3)
plt.tight_layout()
plt.show()

Part IV: Lucas (1988) — Human Capital

The Planner’s Growth Rate

Lucas Lucas (1988) introduces a time allocation between work (uu) and study (1−u1-u). Human capital hh accumulates at rate h˙/h=ϕ(1−u)\dot{h}/h = \phi(1-u), with no diminishing returns.

The social planner’s optimal growth rate is

γplanner=1ρ(ϕ−δt).\gamma_{\text{planner}} = \frac{1}{\rho}(\phi - \delta_t).
(8)

Decentralized Growth with Externalities

When individual productivity depends on average human capital hˉψ\bar{h}^\psi, the decentralized growth rate is

γh=ρ−1(ϕ−δt)1+ψ(1−1/ρ)/(1−α).\gamma_h = \frac{\rho^{-1}(\phi - \delta_t)}{1 + \psi(1 - 1/\rho)/(1-\alpha)}.
(9)

The direction of the externality’s effect depends on whether ρ≷1\rho \gtrless 1.

phi_s, psi_s = symbols('phi psi', positive=True)

gamma_planner = (phi_s - delta_t_s) / rho_s
gamma_decent  = gamma_planner / (1 + psi_s * (1 - 1/rho_s) / (1 - alpha_s))

print(f"Planner growth rate:       {gamma_planner}")
print(f"Decentralized growth rate: {simplify(gamma_decent)}")
# Numerical comparison
g_plan = (lm.phi - lm.delta_t) / lm.rho
g_dec  = g_plan / (1 + lm.psi * (1 - 1/lm.rho) / (1 - lm.alpha))

print(f"Planner:       {g_plan*100:.2f}%")
print(f"Decentralized: {g_dec*100:.2f}%")
print(f"Ratio:         {g_dec/g_plan:.4f}")

Externality and Risk Aversion

psi_grid = np.linspace(0, 0.5, 200)

fig, ax = plt.subplots(figsize=(8, 5))
for rho_val in [0.5, 1.0, 2.0, 4.0]:
    g_plan_val = (lm.phi - lm.delta_t) / rho_val
    g_dec_val = g_plan_val / (1 + psi_grid * (1 - 1/rho_val) / (1 - lm.alpha))
    ax.plot(psi_grid, g_dec_val * 100, lw=2, label=rf'$\rho = {rho_val}$')

ax.set_xlabel(r'Externality parameter $\psi$', fontsize=12)
ax.set_ylabel('Decentralized growth rate (%)', fontsize=12)
ax.set_title(r'Lucas externality: direction depends on $\rho$ relative to 1', fontsize=13)
ax.legend(fontsize=10, frameon=False)
ax.grid(True, alpha=0.3)
plt.tight_layout()
plt.show()

Comparing All Three Models

comparison = pd.DataFrame([
    {
        'Model': 'Rebelo AK',
        'Engine': r'$Y = AK$',
        'Growth rate': rf'$({ak.A} - {ak.delta_t}) / {ak.rho} = {gamma_num:.3f}$',
        'Saving rate (%)': round(s_num * 100, 1),
        'Transitional dynamics': 'None',
        'Market failure': 'None',
    },
    {
        'Model': 'Romer (1986)',
        'Engine': r'Knowledge spillovers ($\Xi = K$)',
        'Growth rate': rf'$({rm.alpha} - {rm.delta_t}) / {rm.rho} = {(rm.alpha - rm.delta_t)/rm.rho:.3f}$ (private)',
        'Saving rate (%)': 'N/A (no closed form)',
        'Transitional dynamics': 'None (on BGP)',
        'Market failure': 'Externality',
    },
    {
        'Model': 'Lucas (1988)',
        'Engine': r'Human capital ($\dot{h}/h = \phi(1-u)$)',
        'Growth rate': rf'$({lm.phi} - {lm.delta_t}) / {lm.rho} = {g_plan:.3f}$ (planner)',
        'Saving rate (%)': 'N/A',
        'Transitional dynamics': 'None (on BGP)',
        'Market failure': rf'Externality ($\psi = {lm.psi}$)',
    },
])
comparison

Exercises

Solution to Exercise 1
# (a)
target_gamma = 0.02
rho_cal = 2.0
delta_t_cal = 0.04
A_cal = target_gamma * rho_cal + delta_t_cal
print(f"(a) A = gamma * rho + delta_t = {A_cal:.2f}")

# (b)
s_cal = target_gamma / A_cal
print(f"(b) Implied saving rate: {s_cal * 100:.1f}%")

# (c)
target_s = 0.03
def residual(rho_val):
    gamma_val = (A_cal - delta_t_cal) / rho_val
    return gamma_val / A_cal - target_s

rho_star = optimize.brentq(residual, 0.1, 50.0)
print(f"(c) rho for s = 3%: {rho_star:.4f}")
print(f"    Implied growth rate: {(A_cal - delta_t_cal) / rho_star * 100:.2f}%")

With A=0.08A = 0.08, the marginal product of capital is 8%, consistent with standard calibrations. The saving rate of 25% is reasonable. But part (c) reveals a tension: matching a 3% saving rate requires ρ≈16.7\rho \approx 16.7, far above the conventional range of 1 to 4.

Solution to Exercise 2
rows = []
for alpha_val in [0.2, 0.33, 0.5, 0.7]:
    eta_val = 1 - alpha_val
    s_star = eta_val / alpha_val
    g_priv = (alpha_val - rm.delta_t) / rm.rho
    g_soc  = (1 - rm.delta_t) / rm.rho
    rows.append({
        'alpha': alpha_val,
        'eta': round(eta_val, 2),
        's* (%)': round(s_star * 100, 1),
        'gamma_priv (%)': round(g_priv * 100, 2),
        'gamma_soc (%)': round(g_soc * 100, 2),
        'wedge (pp)': round((g_soc - g_priv) * 100, 2),
    })

pd.DataFrame(rows)

When α\alpha is small (capital contributes little to private output), the externality is large and the optimal subsidy exceeds 100%. As α\alpha rises, private returns capture more of the social benefit and the required subsidy falls.

Solution to Exercise 3
# On the BGP, gamma = phi*(1-u). The planner's optimal growth rate is
# gamma = (phi - delta_t)/rho. Substituting and solving for u*:
#   u* = 1 - gamma/phi = 1 - (phi - delta_t) / (rho * phi)

phi_grid = np.linspace(0.05, 0.20, 200)

u_star = 1 - (phi_grid - lm.delta_t) / (lm.rho * phi_grid)
gamma_vals = phi_grid * (1 - u_star)

fig, axes = plt.subplots(1, 2, figsize=(10, 4))

axes[0].plot(phi_grid, u_star, lw=2, color=ACCENT)
axes[0].set_xlabel(r'Human capital efficiency $\phi$')
axes[0].set_ylabel(r'Optimal working fraction $u^*$')
axes[0].set_title(r'Higher $\phi$ lowers time spent working')
axes[0].grid(True, alpha=0.3)

axes[1].plot(phi_grid, gamma_vals * 100, lw=2, color=RED)
axes[1].set_xlabel(r'Human capital efficiency $\phi$')
axes[1].set_ylabel('Growth rate (%)')
axes[1].set_title('Growth accelerates with human capital efficiency')
axes[1].grid(True, alpha=0.3)

plt.tight_layout()
plt.show()

# (a) Baseline
u_baseline = 1 - (lm.phi - lm.delta_t) / (lm.rho * lm.phi)
print(f"(a) u* = {u_baseline:.4f} (spend {(1-u_baseline)*100:.1f}% of time studying)")
print(f"    Growth rate: {lm.phi * (1 - u_baseline) * 100:.2f}%")

As ϕ\phi rises, the return to studying increases, so the planner allocates more time to human capital accumulation (u∗u^* falls). The growth rate rises because both channels reinforce: more efficient studying and more time spent studying.

References

References
  1. Rebelo, S. (1991). Long-Run Policy Analysis and Long-Run Growth. Journal of Political Economy, 99(3), 500–521. 10.1086/261764
  2. Romer, P. M. (1986). Increasing Returns and Long-Run Growth. Journal of Political Economy, 94(5), 1002–1037. 10.1086/261420
  3. Lucas, Jr., Robert E. (1988). On the Mechanics of Economic Development. Journal of Monetary Economics, 22(1), 3–42. 10.1016/0304-3932(88)90168-7
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