<p>In this static model with no savings/intertemporal aspect, labor supply does not depend on the wage. Using the equality relations, we can maximize over $L^s$ without constraints</p>

<p>$$\max_{L^s} U = \left(\frac{L^s\cdot W}{P}\right)^a\cdot (T-L^s)^{1-a}$$</p>

<p>$$\frac {\partial U}{\partial L^s} = \frac a{L^s}\cdot U - \frac {1-a}{T-L^s}\cdot U$$</p>

<p>and setting this equal to zero we get the optimal $L^s$</p>

<p>$$\frac a{L^s}= \frac {1-a}{T-L^s} \implies L^s = aT$$</p>

<p>One can verify that, at the optimum, the second derivative of the Utility function is negative so we do have a maximum.</p>

CC BY-SA 3.0
2017-06-18T20:21:15.713Z
2017-08-18T09:04:45.303Z
2017-08-18T09:04:45.303Z
Alecos Papadopoulos
<p>Let utility curve an individual given as $U(C,R) = C^aR^{1-a}$ where $(0\lt a  \lt 1)$ and $C$ denotes consumption commodity and $R$ denotes its leisure, and price of $C$ is given as $P$, and the nominal wage for a unit of labor given as $W$. Total amount time available for the individual is $T$.  </p>

<p>Now I would like to derive the change of labor supply when nominal wage get increased with two different effect decomposed - income effect/substitution effect.</p>

<p>What I have got is $L^s$ only, which is equal to $T-R$, following below process:</p>

<p>first, $P\cdot C = L^s\cdot W $ </p>

<p>second, $T-R = L^s$</p>

<p>by, first and second, $P\cdot C= (T-R)\cdot W$ (*)</p>

<p>and the from the utility function we can derive the condition of maxmizing its utility, $MU_C = MU_R$, which is eqaul to $\frac{a}{1-a}=\frac{C}{R}$(**)</p>

<p>Now, since the (*), (**) must hold at the same time, we can derive equation between $W$ and $R$ without $C$ as below:</p>

<p>$P\cdot \frac{a}{1-a}\cdot R = (T-R)\cdot W$</p>

<p>Then we get $R = \frac{TW}{P(\frac{a}{1-a})+W}$ and $L^s =T\frac{P(\frac{a}{1-a})}{P(\frac{a}{1-a})+W}$ </p>

<p>but my derivation of $L^s$ looks only decreasing while W is increasing.</p>

<p>Any points did I do wrong? I want to did some decomposition of Slutsky, but lack of skills to deal with partial derivatives, it makes me hard to do also.</p>

<labor-economics>
Slutsky Decomposition of Given Labor Supply Model
Beverlie
# Slutsky Decomposition of Given Labor Supply Model

(Asked by: username_0 on 2017-06-17)

Let utility curve an individual given as $U(C,R) = C^aR^{1-a}$ where $(0\lt a \lt 1)$ and $C$ denotes consumption commodity and $R$ denotes its leisure, and price of $C$ is given as $P$, and the nominal wage for a unit of labor given as $W$. Total amount time available for the individual is $T$.

Now I would like to derive the change of labor supply when nominal wage get increased with two different effect decomposed - income effect/substitution effect.

What I have got is $L^s$ only, which is equal to $T-R$, following below process:

first, $P\cdot C = L^s\cdot W $

second, $T-R = L^s$

by, first and second, $P\cdot C= (T-R)\cdot W$ (\*)

and the from the utility function we can derive the condition of maxmizing its utility, $MU\_C = MU\_R$, which is eqaul to $\frac{a}{1-a}=\frac{C}{R}$(\*\*)

Now, since the (\*), (\*\*) must hold at the same time, we can derive equation between $W$ and $R$ without $C$ as below:

$P\cdot \frac{a}{1-a}\cdot R = (T-R)\cdot W$

Then we get $R = \frac{TW}{P(\frac{a}{1-a})+W}$ and $L^s =T\frac{P(\frac{a}{1-a})}{P(\frac{a}{1-a})+W}$

but my derivation of $L^s$ looks only decreasing while W is increasing.

Any points did I do wrong? I want to did some decomposition of Slutsky, but lack of skills to deal with partial derivatives, it makes me hard to do also.

**Tags:** labor-economics

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## 1 answers

### Answer by username_1 on 2017-06-18. Score: 2

In this static model with no savings/intertemporal aspect, labor supply does not depend on the wage. Using the equality relations, we can maximize over $L^s$ without constraints

$$\max\_{L^s} U = \left(\frac{L^s\cdot W}{P}\right)^a\cdot (T-L^s)^{1-a}$$

$$\frac {\partial U}{\partial L^s} = \frac a{L^s}\cdot U - \frac {1-a}{T-L^s}\cdot U$$

and setting this equal to zero we get the optimal $L^s$

$$\frac a{L^s}= \frac {1-a}{T-L^s} \implies L^s = aT$$

One can verify that, at the optimum, the second derivative of the Utility function is negative so we do have a maximum.


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