2
$\begingroup$

Let's say that we have the sum of the utility of a social planner

$$\int_{0}^{\infty}U\left(C\right)e^{-\rho t}dt$$

where $C$ is the total consumption. If we want to write this by a per capita variable $c=\frac{C}{L}$ where $L=e^{nt}$ is the total number of population that grows exogenously at rate $n$. We can reformulate this such as

$$\int_{0}^{\infty}U\left(c\right)e^{nt}e^{-\rho t}dt$$

Until now, everything is straightforward. However, if we have started by a specified functional form such as a CRRA utility

$$\int_{0}^{\infty}\frac{C^{1-\sigma}}{1-\sigma}e^{-\rho t}dt$$

We would not have reached the result above and should have some exponent term with $\sigma$. What is the logic behind? There is something I miss.

$\endgroup$

1 Answer 1

2
$\begingroup$

The reason why the $e^{nt}$ term is there is because you want to multiply the whole utility by the number of people. You are actually not substituting consumption into the utility but multiplying the utility of an individual by the number of individuals. So the problem would look like:

$$\int_{0}^{\infty}\frac{c^{1-\sigma}}{1-\sigma}e^{nt} e^{-\rho t}dt$$

At least this is how it is explained by Barro & Sala-i-Martin in their textbook Economic Growth 2nd ed. pp 87 where they say [emphasis mine]:

The multiplication of $u(c)$ in equation ... by family size, $L = e^{nt}$, represents the adding up of utils for all family members alive at time $t$.

So the way how I understand it you are multiplying all the individual utilities by the population.

In fact I am quite sure of it since in Romer Advanced Economics 4th ed in chapter 2 pp 54 there is Ramsey model where the objective function is given as:

$$\int_{t=0}^{\infty} e^{-\rho t}\frac{c^{1-\sigma}}{1-\sigma} \frac{L(t)}{H} dt$$

which is essentially exactly the case you are looking for where the only difference is that Romer also adds the $H$ parameter which is the number of households so $L(t)/H$ would be population per household. Romer also later substitutes $L(t)=L(0)e^{nt}$ which for $L(0)$ normalized to 1 gives almost the same function (with that extra $H$ parameter - but for $H=1$ it would give you exactly the same result).

$\endgroup$

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.