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

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

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

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

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

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

<p>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 <em>present-value</em> multipliers, and not <em>current-value</em> 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</p>

<p>$$\frac{\partial H}{\partial B} = -\dot q$$</p>

<p>while if we interpret $q$ as a current-value multiplier we have </p>

<p>$$\frac{\partial H}{\partial B} = r q-\dot q
$$</p>

CC BY-SA 4.0
2018-08-09T10:02:19.413Z
Thanks Alecos for your explanation. However, I do not understand how he does apply this exponential to all the Hamiltonian.
optimal control
CC BY-SA 4.0
2018-08-09T10:33:37.613Z
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
CC BY-SA 4.0
2018-08-09T10:35:35.330Z
@optimalcontrol Indeed.
Alecos Papadopoulos
CC BY-SA 4.0
2018-08-09T07:38:45.177Z
2018-08-09T10:16:04.283Z
2018-08-09T10:16:04.283Z
Alecos Papadopoulos
<p>I am reading a paper <a href="https://www.dropbox.com/s/sodwp5j6a9i9xhf/Okuguchi%20-%201981%20-%20Population%20Growth%2C%20Costly%20Innovation%20and%20Modified%20Hartwick%27s%20Rule.pdf?dl=0" rel="nofollow noreferrer">Okuguchi,1981</a> 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.)</p>

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

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

<p>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. </p>

<p>In the same fashion, is it correct to write something like</p>

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

<macroeconomics><economic-growth>
Ramsey growth model - per capita vs aggregate variables
optimal control
# Ramsey growth model - per capita vs aggregate variables

(Asked by: username_0 on 2018-08-08)

I am reading a paper [Okuguchi,1981](https://www.dropbox.com/s/sodwp5j6a9i9xhf/Okuguchi%20-%201981%20-%20Population%20Growth%2C%20Costly%20Innovation%20and%20Modified%20Hartwick%27s%20Rule.pdf?dl=0) 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

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

**Tags:** macroeconomics, economic-growth

---

## 1 answers

### Answer by username_1 on 2018-08-09. Score: 1 | Accepted answer

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

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

> **Comments:**
> **username_0**: Thanks Alecos for your explanation. However, I do not understand how he does apply this exponential to all the Hamiltonian.  
> **username_0**: 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.  
> **username_1**: @optimalcontrol Indeed.  

---

