1
$\begingroup$

KPR preferences are generally given by

$$ \frac{c^{1-\sigma}}{1-\sigma} \cdot v(l)$$

with $l$ being leisure. Let's focus on the case where $\sigma \in (0, 1)$, where we know that $v(l)$ must be increasing and concave. Let total time be $T$, and denote $n$ by working hours. The standard problem then is

$$ \max_{n\in[0, T]} \frac{(wn)^{1-\sigma}}{1-\sigma} \cdot v(T-n) $$

An interior solution requires

$$ v(T-n) = v'(T-n)\frac{n}{1-\sigma} \tag 1$$

One increasing and concave function would be $v(x) = \frac{x^{1-\gamma}}{1-\gamma}$, $\gamma \in (0, 1)$. This yields

$$\frac{T-n}{1-\gamma} = \frac{n}{1-\sigma}$$

or

$$ n = (1-\sigma)\frac{T}{1 - \gamma + 1 - \sigma} $$

Now, generally, we should be able to pick the level of working hours - for each level of IES ($\sigma$). So let's fix $\sigma = 0$, the case of risk-neutrality, and we yield

$$ n = \frac{T}{2 - \gamma} \tag 2$$

As $\gamma \in (0, 1)$, that means that we have no fully control about the level of working hours: We cannot get that $n<0.5$ is an optimal choice with these preferences.

Note that a constant factor in $v(l)$ wouldn't help either, that would just drop out in (1). What am I missing - how can I control the level of working hours better in the multiplicative setup of KPR? That is, where $U = U(c, v(l))$ and not $U = \log c + g(v(l))$.

$\endgroup$
2
  • $\begingroup$ Constraining the $\sigma$ and $\gamma$ parameters in the $(0,1)$ is what creates the problem I think. Considering that usually these parameters are treated as higher than unity, do you need to use this restriction and be able to obtain all ranges of time worked? $\endgroup$ Aug 12, 2016 at 12:12
  • $\begingroup$ @AlecosPapadopoulos I don't think the $\sigma$ matters for this. One should be able to control the hours worked independently of the IES. Regarding $\gamma$, $0$ is definitely a lower bound. If you look at (2), you see that to get $n < T/2$, you require $\gamma < 0$, not $\gamma > 1$. So allowing a larger $\gamma$ doesnt help. Moreover, it yields weird results in (2). $\endgroup$
    – FooBar
    Aug 12, 2016 at 13:22

1 Answer 1

1
$\begingroup$

Continuing on my comment, if both parameters are set strictly higher than unity, then we can get all the range, indicatively: enter image description here

This is why I asked how important for the work at hand is to constrain them in the $(0,1)$ interval instead.

This won't be the first time when a mathematical functional form has some inherent restrictions that do not allow it to reflect the whole spectrum of possible human behavior.

$\endgroup$
3
  • $\begingroup$ Did you chose $T = 1$? $\endgroup$
    – FooBar
    Aug 12, 2016 at 13:42
  • $\begingroup$ Also, my point was not necessarily that "some mathematical form" has some inherent restriction. Say you fix $\sigma$ for whatever reason. I failed to come up with any functional form $v(l)$ that satisfies the requirements such that I could choose $T$. This being said, the $log$ transformation of the preferences yield such calibration straight-forwardly. $\endgroup$
    – FooBar
    Aug 12, 2016 at 13:44
  • $\begingroup$ Τhe table shows n/T $\endgroup$ Aug 12, 2016 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.