<p>Regarding the presence of $l$ in the Frisch elasticity, your result is correct. It is my impression that a Frisch elasticity free of $l$ requires very specific functional forms.</p>

<p>For example we know that the following utility function </p>

<p>$$U(c,l) = ln(c) +\alpha \frac {(1-l)^{1+1/v}}{1+1/v} \tag{1}$$
gives a Frisch elasticity independent of $l$ (and then $\eta=v$).</p>

<p>We could compare this with your utility specification and say "ah, so we must have additive separability to obtain a Frisch elasticity independent of $l$"...<br>
No, this is not enough. Note that essentially $(1)$ does not express utility as a positive non-linear function of leisure but as a negative non-linear function of labor, and so the term $1-l$ appears. This is critical to eliminate $l$ from the final expression.  </p>

<p>Because, if  we assume additive separability in general, then the cross-partial derivative is zero and we would be left with</p>

<p>$$\eta = \frac{u_l}{- u_{ll}}\frac{1}{1-l} \tag{2}$$</p>

<p>If the additively separable utility function is expressed as a non-linear positive function of $l$, <em>and not as a negative non-linear function of $1-l$</em>, $(2)$ would also be dependent on $l$.</p>

<p>--</p>

<p>Also, with your utility function specification, I do not find that $\sigma$ is out of the final picture. Instead I find</p>

<p>$$ \eta = \frac{\gamma(1-\sigma)-1}{(\omega+\gamma)(1-\sigma)-1}\frac{l}{1-l}$$</p>

<p>If $\omega = 1-\gamma$ this simplifies to</p>

<p>$$ \eta = \frac{1-\gamma(1-\sigma)}{\sigma}\frac{l}{1-l}$$</p>

<p>But these are treacherous algebraic calculations - you should recheck.</p>

CC BY-SA 3.0
2015-05-20T02:40:31.380Z
2015-05-22T09:29:09.337Z
2015-05-22T09:29:09.337Z
Alecos Papadopoulos
<p>Consider the following version of KPR preferences (with $l$ being leisure):</p>

<p>$$ U(c,l) = \left(\left(c\right)^\gamma l^\omega\right)^{1-\sigma}$$</p>

<p>I'm after the Frisch elasticity:</p>

<p>$$ \frac{\partial(1-l)}{\partial w} \frac{w}{1-l}$$</p>

<p>Note that $\frac{\partial(1-l)}{\partial w} = -\frac{\partial(l)}{\partial w}$. </p>

<p>In general, one can compute the first component as $$  w\frac{\partial l}{\partial w} = \frac{u_l}{u_{ll} - \frac{u^2_{lc}}{u_cc}}$$</p>

<p>Hence, the Frisch elasticity is given by</p>

<p>$$  \eta = \frac{u_l}{\frac{u^2_{lc}}{u_cc} - u_{ll}}\frac{1}{1-l}$$</p>

<p>With the given preferences, I have that</p>

<p>$$ u_l = \omega l^{\omega-1} c^\gamma K\\
u_c = \gamma c^{\gamma-1}l^\omega K\\
u_{cc} = (2\gamma-1)\gamma c^{\gamma-2}l^\omega K\\
u_{ll} = (2\omega-1)\omega l^{\omega-2}c^\gamma K \\
u_{cl} = 2\gamma c^{\gamma-1} \omega l^{\omega-1}K
$$</p>

<p>and then the Frisch elasticity boils down to</p>

<p>$$ \eta = \frac{1}{\frac{4\omega}{2\gamma-1} + 1 - 2\omega}\frac{l}{1-l}$$</p>

<p>It makes sense to some degree: as $\sigma$ is the inter temporal elasticity of substitution, it does not appear here. However, the relative curvatures on $c$ and $l$ appear, which is fine. However, the elasticity is not independent of $l$. As I'm somewhat in the KPR environment (despite adding curvature to $c$), I didn't expect this.</p>

<p>Are my result correct? Does anyone have more insights on this matter?</p>

<utility><labor-economics><elasticity><preferences><frisch-elasticity>
Generalized KPR: Frisch Elasticity
FooBar
# Generalized KPR: Frisch Elasticity

(Asked by: username_0 on 2015-05-19)

Consider the following version of KPR preferences (with $l$ being leisure):

$$ U(c,l) = \left(\left(c\right)^\gamma l^\omega\right)^{1-\sigma}$$

I'm after the Frisch elasticity:

$$ \frac{\partial(1-l)}{\partial w} \frac{w}{1-l}$$

Note that $\frac{\partial(1-l)}{\partial w} = -\frac{\partial(l)}{\partial w}$.

In general, one can compute the first component as $$ w\frac{\partial l}{\partial w} = \frac{u\_l}{u\_{ll} - \frac{u^2\_{lc}}{u\_cc}}$$

Hence, the Frisch elasticity is given by

$$ \eta = \frac{u\_l}{\frac{u^2\_{lc}}{u\_cc} - u\_{ll}}\frac{1}{1-l}$$

With the given preferences, I have that

$$ u\_l = \omega l^{\omega-1} c^\gamma K\\
u\_c = \gamma c^{\gamma-1}l^\omega K\\
u\_{cc} = (2\gamma-1)\gamma c^{\gamma-2}l^\omega K\\
u\_{ll} = (2\omega-1)\omega l^{\omega-2}c^\gamma K \\
u\_{cl} = 2\gamma c^{\gamma-1} \omega l^{\omega-1}K
$$

and then the Frisch elasticity boils down to

$$ \eta = \frac{1}{\frac{4\omega}{2\gamma-1} + 1 - 2\omega}\frac{l}{1-l}$$

It makes sense to some degree: as $\sigma$ is the inter temporal elasticity of substitution, it does not appear here. However, the relative curvatures on $c$ and $l$ appear, which is fine. However, the elasticity is not independent of $l$. As I'm somewhat in the KPR environment (despite adding curvature to $c$), I didn't expect this.

Are my result correct? Does anyone have more insights on this matter?

**Tags:** utility, labor-economics, elasticity, preferences, frisch-elasticity

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## 1 answers

### Answer by username_1 on 2015-05-20. Score: 3 | Accepted answer

Regarding the presence of $l$ in the Frisch elasticity, your result is correct. It is my impression that a Frisch elasticity free of $l$ requires very specific functional forms.

For example we know that the following utility function

$$U(c,l) = ln(c) +\alpha \frac {(1-l)^{1+1/v}}{1+1/v} \tag{1}$$
gives a Frisch elasticity independent of $l$ (and then $\eta=v$).

We could compare this with your utility specification and say "ah, so we must have additive separability to obtain a Frisch elasticity independent of $l$"...  
No, this is not enough. Note that essentially $(1)$ does not express utility as a positive non-linear function of leisure but as a negative non-linear function of labor, and so the term $1-l$ appears. This is critical to eliminate $l$ from the final expression.

Because, if we assume additive separability in general, then the cross-partial derivative is zero and we would be left with

$$\eta = \frac{u\_l}{- u\_{ll}}\frac{1}{1-l} \tag{2}$$

If the additively separable utility function is expressed as a non-linear positive function of $l$, *and not as a negative non-linear function of $1-l$*, $(2)$ would also be dependent on $l$.

--

Also, with your utility function specification, I do not find that $\sigma$ is out of the final picture. Instead I find

$$ \eta = \frac{\gamma(1-\sigma)-1}{(\omega+\gamma)(1-\sigma)-1}\frac{l}{1-l}$$

If $\omega = 1-\gamma$ this simplifies to

$$ \eta = \frac{1-\gamma(1-\sigma)}{\sigma}\frac{l}{1-l}$$

But these are treacherous algebraic calculations - you should recheck.


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