5
$\begingroup$

Following the work of Raurich et al. (2012) I got stuck trying to derive the average productivity starting from the following CES production function:

$$Y=A\left [ \alpha K^{\frac{\sigma -1}{\sigma }}+(1-\alpha)(AL)^{\frac{\sigma -1}{\sigma }} \right ]^{\frac{\sigma}{\sigma-1 }}$$

The result they get is: $$\left (\frac{Y}{AL} \right )^{\frac{1-\sigma }{\sigma }}=\frac{1}{\alpha \left ( \frac{K}{AL} \right )^{\frac{\sigma-1 }{\sigma }}+(1-\alpha )}$$

Is there anyone who can help me with this?

$\endgroup$
1
  • 1
    $\begingroup$ I checked just to be sure. But actually the exponent of the LHS is right. $\endgroup$
    – Alessandro
    Sep 29, 2018 at 23:52

1 Answer 1

2
$\begingroup$

Solution

This is the solution to your difficulty. By the way, which text are you following?

$\endgroup$
4
  • $\begingroup$ I think OP's confusion is over why average productivity is defined in this manner. To me, average labor productivity (in efficiency units) would just be $\frac{Y}{AL}$. The paper in question is here (see page 6 of the paper). Maybe it's just a language issue -- it is a working paper. $\endgroup$ Sep 30, 2018 at 16:35
  • $\begingroup$ Yes, I'm also trying to understand this. Anyway, it is not a working paper, it has been published on Journal of Macroeconomics (Volume 34, Issue 1, March 2012, Pages 181-198). $\endgroup$
    – Alessandro
    Oct 1, 2018 at 11:42
  • 1
    $\begingroup$ I think this is a language issue. I see no relevance in raising average productivity to a function of elasticity of substitution and then calling it average productivity. If you find an answer, please do let me know. $\endgroup$ Oct 1, 2018 at 17:53
  • $\begingroup$ Agreed that this is likely just semantics. The language preceding the expression given for average productivity is indicative to me. $\endgroup$ Oct 1, 2018 at 21:02

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.