# Controling the level of hours worked in the multiplicative formulation of KPR

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 $U = \log c + g(v(l))$.

• 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? Aug 12 '16 at 12:12
• @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). Aug 12 '16 at 13:22

Continuing on my comment, if both parameters are set strictly higher than unity, then we can get all the range, indicatively: This is why I asked how important for the work at hand is to constrain them in the $(0,1)$ interval instead.
• Did you chose $T = 1$? Aug 12 '16 at 13:42
• 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. Aug 12 '16 at 13:44