<p>Suppose that the total amount of land (<span class="math-container">$\Lambda$</span>) is fixed but the population measure increases exogenously (<span class="math-container">$\dot{L}&gt;0$</span>).</p>

<p>The only way the <em>per capita</em> product (<span class="math-container">$y(t)$</span>) remains <strong>constant over time</strong> is when you can fully substitute capital against land. In formulas, you need to set <span class="math-container">$\psi=1$</span> to get:
<span class="math-container">$$ y(t) = A[a k(t)^\alpha + (1-a) \lambda(t)] $$</span>
and, as you can see, you can always trade-off declining land per capita (<span class="math-container">$\lambda(t) =\frac{\Lambda}{L(t)}$</span>, then <span class="math-container">$\dot{\lambda}&lt;0$</span>) with more investment in capital.</p>

<p>On the other hand, as soon as <span class="math-container">$\psi \neq 1$</span>, then <span class="math-container">$Y/L$</span> will <strong>steadily decrease over time</strong> as a consequence of population growth in conjunction with decrease land per capita not being fully substituted by increases in capital.</p>

<p>This would be my intuition, hope it helps, I'm sure one can make the argument more rigorous, especially for the case of <span class="math-container">$\psi\neq 1$</span>.</p>

CC BY-SA 4.0
2019-05-08T21:54:20.513Z
I think this logic is sound. Perhaps to do it formally you should solve the steady state of the model for an arbitrary $\psi$, and see directly how the growth rate of $y(t)$ depends on $\psi$.
Regio
CC BY-SA 4.0
2019-05-09T04:24:04.323Z
thanks for the input! I will think more thoroughly about the condition when ≠1 .
June
CC BY-SA 4.0
2019-05-08T11:25:31.673Z
2019-05-08T11:25:31.673Z
GabMac
<p>I am an undergrad student trying to study grad macro. I have been trying to solve exercise problem in Barro's economic growth.</p>

<p>The model is about adding land in the Ramsey Model. The model goes as follows:
<span class="math-container">$$Y=A\Bigl[a(K^\alpha L^{1-\alpha})^\psi+(1-a)\Lambda^\psi\Bigr]^{1/\psi}$$</span></p>

<p>Now, the questions asks what condition of <span class="math-container">$\psi$</span> can ensure that <span class="math-container">$Y/L$</span> is constant over time, and what condition will make <span class="math-container">$Y/L$</span> steadily decrease over time.</p>

<p>I have derived the per capita intensive form as follows:
<span class="math-container">$$y(t)=A\Bigl[a(k(t)^\alpha)^\psi+(1-a)\lambda^\psi\Bigr]^{1/\psi}$$</span></p>

<p>However, I do not grasp the meaning of how different value of <span class="math-container">$\psi$</span> can affect the outcome. </p>

<macroeconomics><economic-growth>
[Graduate Macro]Land in the Economic Growth Model: CES Function
June
# [Graduate Macro]Land in the Economic Growth Model: CES Function

(Asked by: username_0 on 2019-05-07)

I am an undergrad student trying to study grad macro. I have been trying to solve exercise problem in Barro's economic growth.

The model is about adding land in the Ramsey Model. The model goes as follows:
$$Y=A\Bigl[a(K^\alpha L^{1-\alpha})^\psi+(1-a)\Lambda^\psi\Bigr]^{1/\psi}$$

Now, the questions asks what condition of $\psi$ can ensure that $Y/L$ is constant over time, and what condition will make $Y/L$ steadily decrease over time.

I have derived the per capita intensive form as follows:
$$y(t)=A\Bigl[a(k(t)^\alpha)^\psi+(1-a)\lambda^\psi\Bigr]^{1/\psi}$$

However, I do not grasp the meaning of how different value of $\psi$ can affect the outcome.

**Tags:** macroeconomics, economic-growth


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## 1 answers

### Answer by username_1 on 2019-05-08. Score: 3 | Accepted answer

Suppose that the total amount of land ($\Lambda$) is fixed but the population measure increases exogenously ($\dot{L}>0$).

The only way the *per capita* product ($y(t)$) remains **constant over time** is when you can fully substitute capital against land. In formulas, you need to set $\psi=1$ to get:
$$ y(t) = A[a k(t)^\alpha + (1-a) \lambda(t)] $$
and, as you can see, you can always trade-off declining land per capita ($\lambda(t) =\frac{\Lambda}{L(t)}$, then $\dot{\lambda}<0$) with more investment in capital.

On the other hand, as soon as $\psi \neq 1$, then $Y/L$ will **steadily decrease over time** as a consequence of population growth in conjunction with decrease land per capita not being fully substituted by increases in capital.

This would be my intuition, hope it helps, I'm sure one can make the argument more rigorous, especially for the case of $\psi\neq 1$.

> **Comments:**
> **username_2**: I think this logic is sound. Perhaps to do it formally you should solve the steady state of the model for an arbitrary $\psi$, and see directly how the growth rate of $y(t)$ depends on $\psi$.  
> **username_0**: thanks for the input! I will think more thoroughly about the condition when ≠1 .  

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