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.


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.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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