Kenji Sato
2016-12-27
To add decorations to the Solow model.
\[ Y = T^\beta L^{(1-\beta)}, \]
\[ F(K, AL) = K^\alpha (AL)^{1 - \alpha} \]
Recall that \[ \dot K = s Y - \delta K \] we have \[ \frac{\dot K}{K} = s \frac{Y}{K} - \delta \] If \( \dot K / K \) is constant, \( Y/K \) must be constant. Thus, \( K \) and \( Y \) share the same growth rate.
\[ g_Y = g_K \]
From \[ Y = K^\alpha (AL)^{1-\alpha} \]
we obtain \[ g_Y = \alpha g_K + (1-\alpha)(g_A + g_L) \] Since \( g_Y = g_K \), \[ \begin{aligned} g_Y &= g_K = g_A + g_L\\ &= g + n. \end{aligned} \]
Growth rate for \( Y/L \) is \( g \).
\[ \dot T = 0. \]
\[ \dot R = - E \]
Assume that a constant fraction of the stock of resource is extracted each moment.
\[ E = b R, \qquad b > 0 \]
If \( b \) is large, the resource depletes very quickly.
As before…
\[ \dot K = s Y - \delta K. \]
If \( Y \) and \( K \) grow at constant rates, the growth rates coincide (balanced growth).
Assume Cobb-Douglas:
\[ Y = K^\alpha E^\beta T^\gamma (AL)^{1-\alpha-\beta-\gamma} \]
with
\[ \begin{aligned} \alpha, \beta, \gamma &> 0, \\ \alpha + \beta + \gamma &< 1 \end{aligned} \]
Let's compute growth rates along the BGP (\( g_Y = g_K \))
From \[ Y = K^\alpha E^\beta T^\gamma (AL)^{1-\alpha-\beta-\gamma} \]
we have
\[ \begin{aligned} g_Y &= \alpha g_K + \beta g_E + \gamma g_T + (1-\alpha -\beta-\gamma) (g_A + g_L)\\ &= \alpha g_K + \beta g_E + (1-\alpha-\beta-\gamma)(g + n) \end{aligned} \]
Noticing that \( g_E = -b \)
\[ g_Y = \frac{ (1-\alpha-\beta-\gamma)(g+n) - \beta b }{ 1-\alpha } \]
Per-capita growth rate:
\[ g_{Y/L} = g_Y - n = \frac{ (1-\alpha-\beta-\gamma)g - \beta b - (\beta+\gamma)n }{ 1-\alpha } \]
\[ g_{Y/L} = g_Y - n = \frac{ (1-\alpha-\beta-\gamma)g - \beta b - (\beta+\gamma)n }{ 1-\alpha } \]
Growth rate of \( Y/L \) along the BGP (Verify this)
\[ \tilde g_{Y/L} = \frac{(1-\alpha-\beta-\gamma)g}{1-\alpha} \]
The drag is thus
\[
\tilde g_{Y/L} - g_{Y/L}
=
\frac{\beta b + (\beta+\gamma)n}{1-\alpha}
\]
Calibration according to Nordhaus (1992):
beta = 0.1; gamma = 0.1;
alpha = 0.2; n = 0.01; b = 0.005
drag = (beta * b + (beta + gamma) * n) / (1 - alpha)
drag
[1] 0.003125
The limited land and decreasing nonrenewable resources reduce per-capita growth rate by 0.3 percentage point. A more elaborate model of Nordhaus's estimates smaller drag.
The drag is not trivial but not large.
We analyzed the BGP for the baseline Solow model because it has global stability. Start from anywhere, you eventually reach that path (in very long run).
As for the environmental Solow model, we bypassed the analysis of transition path and got directly to the BGP. We still don't know whether or not this BGP will is reached in the long run.
Some macroeconomic analyses are based on unstable equilibrium paths …. but that is a bad practice.
I would strongly recommend to check stability somehow. We do this by a computer simulation. It's not rigorous mathematical proof but we can be sure that the above BGP analysis is relevant.
library(dplyr)
s = 0.3; alpha = 0.2; beta = 0.1
gamma = 0.1; delta = 0.05; b = 0.005
g = 0.02; n = 0.01; dt = 0.05;
t = seq(0.0, 250.0, by=dt)
df = data.frame(t)
df = df %>%
mutate(A = exp(g * t)) %>%
mutate(L = exp(n * t)) %>%
mutate(T = exp(0 * t)) %>%
mutate(E = b * exp(-b * df$t))
head(df)
t A L T E
1 0.00 1.000000 1.000000 1 0.005000000
2 0.05 1.001001 1.000500 1 0.004998750
3 0.10 1.002002 1.001001 1 0.004997501
4 0.15 1.003005 1.001501 1 0.004996251
5 0.20 1.004008 1.002002 1 0.004995002
6 0.25 1.005013 1.002503 1 0.004993754
F = function(K, A, L, T, E){
Y = (K^alpha) * (E^beta) * (T^gamma) * (A*L)^(1-alpha-beta-gamma)
return(Y)
}
K = 0.05
for (i in 1:nrow(df)){
A = df[i, "A"]
L = df[i, "L"]
T = df[i, "T"]
E = df[i, "E"]
Y = F(K, A, L, T, E)
df[i, "K"] = K
df[i, "Y"] = Y
K = K + dt * (s * Y - delta * K)
}
head(df)
t A L T E K Y
1 0.00 1.000000 1.000000 1 0.005000000 0.05000000 0.3233635
2 0.05 1.001001 1.000500 1 0.004998750 0.05472545 0.3295451
3 0.10 1.002002 1.001001 1 0.004997501 0.05953182 0.3354338
4 0.15 1.003005 1.001501 1 0.004996251 0.06441449 0.3410623
5 0.20 1.004008 1.002002 1 0.004995002 0.06936939 0.3464580
6 0.25 1.005013 1.002503 1 0.004993754 0.07439284 0.3516440
library(ggplot2)
ggplot(data=df, aes(x=t, y=Y)) + geom_line() + scale_y_log10() + theme(text=element_text(size=30))
It seems like the equilibrium path is converging to the BGP.
:small-code
growth.rate = (df[2:nrow(df), "Y"] - df[1:(nrow(df)-1), "Y"]) / df[1:(nrow(df)-1), "Y"] / dt
qplot(x=df[1:(nrow(df)-1),"t"], y=growth.rate, geom='line') +
geom_hline(aes(yintercept=((1-alpha-beta-gamma) * (g + n) - beta * b )/ (1 - alpha)), color="red") +
labs(y='Growth rate of Y', x='time') + theme(text=element_text(size=30))
In the Solow model, the saving function is constant.
In the Ramsey-Cass-Koopmans model (or more simply Ramsey model),
The Ramsey model predicts
There is a subtle difference concerning policy change.
This difference comes from the fact that the agents in the Ramsey model behave forward-lookingly and take there future income as given.
A permanent policy change alters the total income but doesn't raise substitution between saving and consumption.
Before tackling the model in Romer 4e, we analyze the one-sector optimal growth model:
\[ \begin{aligned} \max \int_0^\infty e^{-\rho t} u(c(t)) dt \end{aligned} \]
subject to
\[ \begin{aligned} \dot k(t) &= f(k(t)) - \delta k(t) - c(t)\\ k(t) &\ge 0\\ c(t) &\ge 0 \end{aligned} \]
Maximize discounted sum of utility from consumption stream subject to the capital accumulation constraint
We will study