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

<p>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 $l$.</p>

<p>So the following must hold:</p>

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

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

<p>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.</p>

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

<p>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.</p>

CC BY-SA 3.0
2015-01-02T16:49:27.983Z
2015-01-02T20:08:17.760Z
2015-01-02T20:08:17.760Z
Bryce
<p>So in RBC and Ramsey-derived utility function, the following is usually the form of utility:</p>

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

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

<p>In Mankiw/Rotemberg/Summers paper Intertemporal Substitution in Macroeconomics (link: <a href="http://scholar.harvard.edu/files/mankiw/files/intertemporal_substitution.pdf" rel="nofollow">http://scholar.harvard.edu/files/mankiw/files/intertemporal_substitution.pdf</a>), utility function of following is used to test RBC model:</p>

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

<p>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. </p>

<p>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. </p>

CC BY-SA 3.0
2015-01-03T21:44:16.587Z
For completeness, can you please provide a reference where the "first form" of the utility function is used?
Alecos Papadopoulos
<macroeconomics><microeconomics><utility>
Is special case of utility function in "Intertemporal Substitution in Macroeconomics" also a special case of a Ramsey growth model utility function?
Kompani
# Is special case of utility function in "Intertemporal Substitution in Macroeconomics" also a special case of a Ramsey growth model utility function?

(Asked by: username_0 on 2015-01-02)

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.

**Tags:** macroeconomics, microeconomics, utility
> **Comments:**
> **username_1**: For completeness, can you please provide a reference where the "first form" of the utility function is used?  

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## 1 answers

### Answer by username_2 on 2015-01-02. Score: 5

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.


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