<p>Continuing on my comment, if <em>both</em> parameters are set strictly higher than unity, then we can get all the range, indicatively:
<a href="https://i.sstatic.net/qyLHD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/qyLHD.png" alt="enter image description here"></a></p>

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

<p>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.</p>

CC BY-SA 3.0
2016-08-12T13:42:41.817Z
Did you chose $T = 1$?
FooBar
CC BY-SA 3.0
2016-08-12T13:44:30.487Z
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.
FooBar
CC BY-SA 3.0
2016-08-12T13:45:32.750Z
Τhe table shows n/T
Alecos Papadopoulos
CC BY-SA 3.0
2016-08-12T13:34:35.493Z
2016-08-12T13:34:35.493Z
Alecos Papadopoulos
<p>KPR preferences are generally given by</p>

<p>$$ \frac{c^{1-\sigma}}{1-\sigma} \cdot v(l)$$</p>

<p>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 </p>

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

<p>An interior solution requires</p>

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

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

<p>$$\frac{T-n}{1-\gamma} = \frac{n}{1-\sigma}$$</p>

<p>or</p>

<p>$$ n = (1-\sigma)\frac{T}{1 - \gamma + 1 - \sigma} $$</p>

<p>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</p>

<p>$$ n = \frac{T}{2 - \gamma} \tag 2$$</p>

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

<p>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 $U = \log c + g(v(l))$.</p>

CC BY-SA 3.0
2016-08-12T12:12:23.730Z
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?
Alecos Papadopoulos
CC BY-SA 3.0
2016-08-12T13:22:31.367Z
@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).
FooBar
<macroeconomics><labor-economics><preferences>
Controling the level of hours worked in the multiplicative formulation of KPR
FooBar
# Controling the level of hours worked in the multiplicative formulation of KPR

(Asked by: username_0 on 2016-08-11)

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

**Tags:** macroeconomics, labor-economics, preferences
> **Comments:**
> **username_1**: 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?  
> **username_0**: @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).  

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## 1 answers

### Answer by username_1 on 2016-08-12. Score: 1

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](https://i.sstatic.net/qyLHD.png)](https://i.sstatic.net/qyLHD.png)

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.

> **Comments:**
> **username_0**: Did you chose $T = 1$?  
> **username_0**: 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.  
> **username_1**: Τhe table shows n/T  

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