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.


1 Answer 1


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 $\psi\neq 1$.

  • $\begingroup$ 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 $\psi$. $\endgroup$
    – Regio
    Commented May 8, 2019 at 21:54
  • $\begingroup$ thanks for the input! I will think more thoroughly about the condition when 𝜓≠1 . $\endgroup$
    – June
    Commented May 9, 2019 at 4:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.