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.


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
    Nov 29 '18 at 18:43
  • $\begingroup$ around $\gamma=0$ (McLaurin is the expansion about 0 while Taylor is about any number) $\endgroup$
    – Yann
    Nov 29 '18 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
    Nov 29 '18 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
    Nov 30 '18 at 14:31

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