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?


1 Answer 1


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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