2
$\begingroup$

I am reading a paper Okuguchi,1981 with a Ramsey growth model. There is a constant population growth $n$ such as $L(t)=e^{nt}$. The way he writes the Hamiltonian is quite interesting (equation 5 on page 658.)

$$H=e^{-(\rho+n\left(1-\sigma\right))t}\left[\frac{C^{1-\sigma}}{1-\sigma}+\psi AG+\phi\left(Q-C\right)\right]+\lambda R$$

where capital letters represent the aggregate variables and not per capita variables. $\lambda$ is constant in the model.

In fact, the author does not deal with per capita variables but aggregate variables. Of course, he takes into account the population growth since $n$ appears in co-state dynamics. However, I can not understand how he puts the exponential term $e^{-(\rho+n\left(1-\sigma\right))t}$ in the Hamiltonian in this way. I would really appreciate if you give me some hints or suggestions.

In the same fashion, is it correct to write something like

$$H=e^{-(\rho-n)t}\left[U\left(C\right)+\lambda\left(AK-C\right)+\mu\left(\left(1-S\right)S-\gamma AK\right)\right]$$

$\endgroup$
1
$\begingroup$

For per capita $c=C/L$, I guess it comes from

$$\int_0^{\infty}e^{-\rho t}\frac{c(t)^{1-\sigma}}{1-\sigma} dt = \int_0^{\infty}e^{-\rho t}\frac{(C(t)/L(t))^{1-\sigma}}{1-\sigma} dt$$

$$= \int_0^{\infty}L(t)^{-(1-\sigma)}e^{-\rho t}\frac{C^{1-\sigma}}{1-\sigma} dt$$

Normalizing initial population to $L_0=1$ we have $L(t) = e^{nt}$ so

$$...= \int_0^{\infty}(e^{nt})^{-(1-\sigma)}e^{-\rho t}\frac{C(t)^{1-\sigma}}{1-\sigma} dt = \int_0^{\infty}e^{-(1-\sigma)nt}e^{-\rho t}\frac{C(t)^{1-\sigma}}{1-\sigma} dt$$

$$=\int_0^{\infty}e^{-[\rho+(1-\sigma)n]t}\frac{C(t)^{1-\sigma}}{1-\sigma} dt$$

Regarding the application of the discounte factor throughout the whole Hamiltonian, it just implies that the multipliers $\lambda$ and $\mu$ are to be treated as present-value multipliers, and not current-value multipliers. In the textbook formulation and in per capita magnitudes, we usually use current (running) value mutipliers and we totally ignore the discount factor. This affects the intertemporal first-order condition. In generic notation, denote $B$ the state variable, $q$ the mutliplier and $r$ the discount factor. Then, with present-value mutlipliers, we have

$$\frac{\partial H}{\partial B} = -\dot q$$

while if we interpret $q$ as a current-value multiplier we have

$$\frac{\partial H}{\partial B} = r q-\dot q $$

$\endgroup$
  • $\begingroup$ Thanks Alecos for your explanation. However, I do not understand how he does apply this exponential to all the Hamiltonian. $\endgroup$ – optimal control Aug 9 '18 at 10:02
  • $\begingroup$ Thanks a lot Alecos! Now I understood it. Just a last question. I imagine that all variables which are aggregate and not per capita will always grow at the steady-state no? It is just the variables per capita which admit constant steady states values. $\endgroup$ – optimal control Aug 9 '18 at 10:33
  • $\begingroup$ @optimalcontrol Indeed. $\endgroup$ – Alecos Papadopoulos Aug 9 '18 at 10:35

Your Answer

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

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