4
$\begingroup$

I'm reading "Competition in the Promised Land – Black Migrants in Northern Cities and Labor Markets", by Leah Boustan, and I'm trying to understand her computation of black and white elasticity of substitution in labor market

She uses a Cobb-Douglas function:

$$Y=AL^{\alpha}K^{1-\alpha}$$

and then defines aggregate labor as:

$$L=\left(\sum_{e=1}^{n} \left(L_{e} \theta_{e}\right)^{\frac{\delta - 1}{\delta}}\right)^{\frac{\delta}{\delta - 1}}$$

aggregate labor by educational group as:

$$L_{e}=\left(\sum_{\substack{1 \leq e \leq n\\1 \leq x \leq m}} \left(L_{ex} \theta_{ex}\right)^{\frac{\eta - 1}{\eta}}\right)^{\frac{\eta}{\eta - 1}}$$

and aggregate labor by experience and educational group by:

$$\left(\left(L_{exb} \theta_{exb}\right)^{\frac{\sigma - 1}{\sigma}} + \left(L_{exw} \theta_{exw}\right)^{\frac{\sigma - 1}{\sigma}}\right)^{\frac{\sigma}{\sigma - 1}} $$

Then she derivates the production function relative to $L_{exr}$ (where $r=w,b$) and obtains:

$$ln w_{exr}=ln(A^{\frac{1}{\alpha}}k^{\frac{(1-\alpha)}{\alpha}})+\frac{1}{\delta}\cdot ln(L)+ln\theta_{e}-(\frac{1}{\delta}-\frac{1}{\eta})\cdot ln(L_{e})+ln\theta_{ex}-(\frac{1}{\eta}-\frac{1}{\delta})\cdot ln(L_{ex})+ln\theta_{e_{exr}}-\frac{1}{\delta}\cdot ln(L_{exr})$$

I cannot figure out how to reach this expression. I thought what I need is just 1) to substitute the L terms backwardly, 2) take the derivative, and 3) apply log. The substitution gives:

$$A K^{1 - \alpha} \left(\left(\sum_{e=1}^{n} \left(\theta_{e} \left(\sum_{\substack{1 \leq e \leq n\\1 \leq x \leq m}} \left(\theta_{ex} \left(\left(L_{exb} \theta_{exb}\right)^{\frac{\sigma - 1}{\sigma}} + \left(L_{exw} \theta_{exw}\right)^{\frac{\sigma - 1}{\sigma}}\right)^{\frac{\sigma}{\sigma - 1}}\right)^{\frac{\eta - 1}{\eta}}\right)^{\frac{\eta}{\eta - 1}}\right)^{\frac{\delta - 1}{\delta}}\right)^{\frac{\delta}{\delta - 1}}\right)^{\alpha} $$

But the derivative of this expression relative, for example, to $L_{exb}$ is a equation very different from the expression showed by the author (even after the log).

I must be making a very silly mistake here, but how can I get the desired expression from the equations of L? Is there a textbook where I can find a discusson about elasticity of substitution between demographic groups?

EDIT: paper version of this part of the book can be found here. In my research I noted that Boustan's work is based on Ottaviano (2006), and Card and Lemieux (2001).

Here is a jupyter notebook with the equations

$\endgroup$
2
  • 1
    $\begingroup$ Are you sure that the two first inner powers are $(\sigma-1)/\sigma$ and $1/\sigma$? In most cases aggregate labour inputs are homogeneous of degree one in the elementary labour inputs. $\endgroup$
    – Bertrand
    Feb 27, 2021 at 17:30
  • $\begingroup$ You are right. I will edit the question to correct theses expoents. $\endgroup$
    – Lucas
    Feb 27, 2021 at 17:41

1 Answer 1

3
+50
$\begingroup$

I dont want to be rude but the only equation you copied correctly is the productivity augmented Cobb-Douglas production function.

Equation 2 is equation 2 from Ottaviano, Peri (2008) (on page 8) it says:

$$L=\left(\sum_{e=1}^{n} \theta_{e}L_{e}^{\frac{\delta - 1}{\delta}}\right)^{\frac{\delta}{\delta - 1}}$$.

In the remaining equations $\theta$ is not under the exponent as well:

$$L_{e}=\left(\sum_{\substack{1 \leq e \leq n\\1 \leq x \leq m}} \theta_{ex} L_{ex} ^{\frac{\eta - 1}{\eta}}\right)^{\frac{\eta}{\eta - 1}}$$

Is the equivalent of Ottaviano, Peri (2008) equation 3 on page 9 and

$$\left(\theta_{exb} L_{exb}^{\frac{\sigma - 1}{\sigma}} + \theta_{exw} L_{exw} ^{\frac{\sigma - 1}{\sigma}}\right)^{\frac{\sigma}{\sigma - 1}} $$

Is the equivalent of Ottaviano, Peri (2008) equation 4 on page 9 and

the expression for the logarithm of the wage is equation 2 on page 11 of Boustan(2008):

$$ln w_{exr}=ln(A^{\frac{1}{\alpha}}k^{\frac{(1-\alpha)}{\alpha}})+\frac{1}{\delta}\cdot ln(L)+ln\theta_{e}-(\frac{1}{\delta}-\frac{1}{\eta})\cdot ln(L_{e})+ln\theta_{ex}-(\frac{1}{\eta}-\frac{1}{\sigma})\cdot ln(L_{ex})+ln\theta_{e_{exr}}-\frac{1}{\sigma}\cdot ln(L_{exr})$$

Whith the correct expressions its all boring algebra:

Fn. 29 on page 11 says:

Following Ottaviano and Peri (2006), I first express output as a function of the capital-output ratio (κ = K/Y)

Thus we have:

$Y=AK^{1-\alpha}L^{\alpha} \Leftrightarrow Y=A\big(\frac{K}{Y}\big)^{1-\alpha}L^{\alpha}Y^{1-\alpha} \Leftrightarrow \\Y^{\alpha}=Ak^{1-\alpha}L^{\alpha}\Leftrightarrow Y=A^{\frac{1}{\alpha}}k^{\frac{1-\alpha}{\alpha}}L$

Observe that:

$$W = x^{\frac{\zeta}{\zeta-1}} \Leftrightarrow \frac{\partial W}{\partial L_{exr}} = \frac{\zeta}{\zeta-1}x^{\frac{1}{\zeta-1}}\frac{\partial x}{\partial L_{exr}} =\frac{\zeta}{\zeta-1}W^{\frac{1}{\zeta}}\frac{\partial x}{\partial L_{exr}} (*)$$

$\omega_{exr} = \frac{\partial Y}{\partial L_{exr}} = A^{\frac{1}{\alpha}}k^{\frac{1-\alpha}{\alpha}}\big(\sum_{e=1}^n\theta_eL_e^{\frac{\delta-1}{\delta}}\big)^{\frac{1}{\delta-1}}\theta_eL_e^{\frac{-1}{\delta}}\frac{\partial L_e}{\partial L_{exr}} \overset{(*)}= A^{\frac{1}{\alpha}}k^{\frac{1-\alpha}{\alpha}}\theta_e\frac{\partial L_e}{\partial L_{exr}}L^{\frac{1}{\delta}}L_e^{\frac{-1}{\delta}}(1)$

$\frac{\partial L_e}{\partial L_{exr}} \overset{(*)}= \theta_{ex}\frac{\partial L_{ex}}{\partial L_{exr}}L_e^{\frac{1}{\eta}}L_{ex}^{\frac{-1}{\eta}} (2)$

$\frac{\partial L_{ex}}{\partial L_{exr}} \overset{(*)}= \theta_{exr}L_{ex}^{\frac{1}{\sigma}}L_{exr}^{\frac{-1}{\sigma}} (3)$

Plug back (2) and (3) into (1), take logs and you will obtain the result.

NOTE I dont quite understand why the derivative of $k = \frac{K}{Y}$ wrt L is not taken here.

$\endgroup$
2
  • 1
    $\begingroup$ The first equation in the solution shouldn't be $$\frac{\partial Y}{\partial L_{exr}} = A^{\frac{1}{\alpha}}k^{\frac{1-\alpha}{\alpha}}\big(\sum_{e=1}^n\theta_eL_e^{\frac{\delta-1}{\delta}}\big)^{\frac{1}{\delta-1}}\theta_eL_e^{\frac{-1}{\delta}}\frac{\partial L_e}{\partial L_{exr}} \overset{(*)}=A^{\frac{1}{\alpha}}k^{\frac{1-\alpha}{\alpha}} \theta_e\frac{\partial L_e}{\partial L_{exr}}L^{\frac{1}{\delta}}L_e^{\frac{-1}{\delta}}$$? $\endgroup$
    – Lucas
    Feb 28, 2021 at 16:45
  • $\begingroup$ @Lucas ah yes, copy-paste $\endgroup$ Feb 28, 2021 at 16:48

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.