Suppose an agent wants to maximize:

$Max \sum_{t=0}^\infty \beta^t \ln(C_t) $


$C_t + I_t = Y_t$

$Y_t = K_t$

$K_{t+1} = K_t^\alpha I_t^\gamma$

where $\alpha, \gamma >0$ and $\alpha + \gamma < 1$.

One can combine the three constraints in: $C_t = K_t - K_{t+1}^{\frac{1}{\gamma}}K_{t}^{\frac{-\alpha}{\gamma}}$

Write the lagrangean as:

${\cal L} = \sum_{t=0}^\infty \beta^t \ln(C_t) - \lambda_t(C_t - K_t +K_{t+1}^{\frac{1}{\gamma}}K_{t}^{\frac{-\alpha}{\gamma}})$

I think the FOC yields:

$\frac{1}{\gamma C_t} K_{t+1}^{\frac{1}{\gamma}-1} K_t^{-\frac{\alpha}{\gamma}} = \frac{1}{\beta C_{t+1}}\left(1 + \frac{\alpha}{\gamma} K_{t+2}^{\frac{1}{\gamma}}K_{t+1}^{-\frac{\alpha+\gamma}{\gamma}}\right)$

If the FOC is correct, how to compute the steady state from here?

  • $\begingroup$ Steady state implies that variables do not change, so $K_t = K_{t+1}$, etc. $\endgroup$ – Giskard Dec 8 '15 at 10:52
  • $\begingroup$ Not really. I think this model might imply endogenous growth. So in fact: $\frac{c_{t+1}}{c_t} = constant$, which is different from 0. $\endgroup$ – phdstudent Dec 8 '15 at 11:03
  • $\begingroup$ I would call that the steady growth path, not the steady state, but I am unfamiliar with the naming conventions in this area. Whether growth is optimal is determined by the variables $\alpha, \gamma$. Are there any specific limits on those? $\endgroup$ – Giskard Dec 8 '15 at 11:41
  • 1
    $\begingroup$ If $\alpha + \gamma < 1$ I am pretty sure you will have a steady state without growth. $\endgroup$ – Giskard Dec 8 '15 at 11:50
  • $\begingroup$ Just edited. All $C_t$ should be capitalized. Why if $\gamma + \alpha <1 $ there is no growth? $\endgroup$ – phdstudent Dec 8 '15 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.