I was discussing today with a classmate about the relationship between prodution functions, and we tried to prove that the CES is a special case of the translog production function, but we failed. We have related the CES, the cobb-douglas and the leontief already, but we are missing the link with the translog. Is it possible to show the assumption in terms of CES parameters that yield a CES? Any help is more than welcome.

Thanks! Jonas.


1 Answer 1


the translog is an approximation of the CES function. Let consider a 2 factors production function $A(\alpha K^\gamma+(1-\alpha)L^\gamma)^{1/\gamma}$. First, take the log of this production function, and then approximate this to the first-order using the McLaurin series for $\gamma$.

Using Mathematica (laziness, sorry):

Series[Log[A(\[Alpha] K^\[Gamma]+(1-\[Alpha])L^\[Gamma])^(1/\[Gamma])], {\[Gamma], 0, 1}] // FullSimplify

You reach $%\log \left(A K^{\alpha } L^{1-\alpha }\right)+\frac{1}{2} \gamma(1-\alpha) \alpha \left((\log (K)-\log (L))^2\right)+O\left(\gamma ^2\right)$

$\ln Y=\ln(A)+\alpha\ln K+(1-\alpha)\ln L+\gamma \alpha(1-\alpha)\left(\ln K+\ln L -\ln K\ln L \right)+O\left(\gamma ^2\right)$

This corresponds to the usual translog functional form used in econometric estimation.

  • $\begingroup$ Thanks. Is the expansion around a certain number? $\endgroup$
    – JonasanoJ
    Commented Nov 29, 2018 at 18:43
  • $\begingroup$ around $\gamma=0$ (McLaurin is the expansion about 0 while Taylor is about any number) $\endgroup$
    – Yann
    Commented Nov 29, 2018 at 18:59
  • $\begingroup$ so then the translog assumes always that gamma is zero (cobb-douglas)? Why is gamma in the equation then? $\endgroup$
    – JonasanoJ
    Commented Nov 29, 2018 at 20:39
  • $\begingroup$ No, we do not assume that $\gamma=0$ but the approximation is better the closer we are to 0 (maybe the animation may help you to visualize the focus of these series: en.wikipedia.org/wiki/…) $\endgroup$
    – Yann
    Commented Nov 30, 2018 at 14:31

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.