# Ramsey growth model - per capita vs aggregate variables

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

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

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

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

• Thanks Alecos for your explanation. However, I do not understand how he does apply this exponential to all the Hamiltonian. – optimal control Aug 9 '18 at 10:02
• 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. – optimal control Aug 9 '18 at 10:33
• @optimalcontrol Indeed. – Alecos Papadopoulos Aug 9 '18 at 10:35