# Is special case of utility function in “Intertemporal Substitution in Macroeconomics” also a special case of a Ramsey growth model utility function?

So in RBC and Ramsey-derived utility function, the following is usually the form of utility:

$$u(c,l) = c^{1-\sigma}(1 + \omega(l))$$

where $\omega(l)$ is arbitrary function of $l$, labor, that satisfies $u_c>0$, $u_l <0$, $u_{ll} \leq 0$ and $u_{cc} < 0$. $c$ is consumption.

In Mankiw/Rotemberg/Summers paper Intertemporal Substitution in Macroeconomics (link: http://scholar.harvard.edu/files/mankiw/files/intertemporal_substitution.pdf), utility function of following is used to test RBC model:

$$u(c,l) = \frac{1}{1-\gamma}\left[\frac{c^{1-\alpha} - 1}{1-\alpha} + d\frac{l^{1-\beta} - 1}{1-\beta}\right]^{1-\gamma}$$

As a special case, not considering multiplicative and additive constants, a special case of $$u(c,l) = \frac{c^{1-\alpha}}{1-\alpha} - \frac{l^{1+\beta}}{1+\beta}$$ can be considered.

Now for the third utility, it seems that third utility must be special case of the first utility functional form, but I cannot see how this is possible.

• For completeness, can you please provide a reference where the "first form" of the utility function is used? – Alecos Papadopoulos Jan 3 '15 at 21:44

Let's call your first utility function [1], and the third utility function [3].

If [3] is a special case of [1], then for some [1] we can find parameter values such that the first derivatives of [1] and [3] will be equal for all values of $c$ and $l$.

So the following must hold:

$$c^{1-\sigma} = -(1+\beta)l^{\beta}/\omega_l(l)$$

$$c^{\sigma -\alpha} = (1-\sigma)(1+\omega(l))$$

Solving both for $c$, and the setting the equations equal to one another, you'll find a functional equation which we need to solve for $\omega$ given its restrictions.

$$(-(1+\beta)l^{\beta}/\omega_l(l))^{1/(1-\sigma)} = ((1-\sigma)(1+\omega(l)))^{1/(\sigma-\alpha)}$$

Now check this equation to see if you can satisfy the sign restrictions on $\omega$'s derivatives, and you fill find that you cannot for any meaningful specification.