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

<p>Using Mathematica (laziness, sorry): </p>

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

<p>You reach 
<span class="math-container">$%\log \left(A K^{\alpha } L^{1-\alpha }\right)+\frac{1}{2} \gamma(1-\alpha) \alpha  \left((\log (K)-\log (L))^2\right)+O\left(\gamma ^2\right)$</span></p>

<p><span class="math-container">$\ln Y=\ln(A)+\alpha\ln K+(1-\alpha)\ln L+\gamma \alpha(1-\alpha)\left(\ln K+\ln L -\ln K\ln L \right)+O\left(\gamma ^2\right)$</span></p>

<p>This corresponds to the usual translog functional form used in econometric estimation.</p>

CC BY-SA 4.0
2018-11-29T18:43:46.830Z
Thanks. Is the expansion around a certain number?
JonasanoJ
CC BY-SA 4.0
2018-11-29T18:59:33.477Z
around $\gamma=0$ (McLaurin is the expansion  about 0 while Taylor is about any number)
Yann
CC BY-SA 4.0
2018-11-29T20:39:56.157Z
so then the translog assumes always that gamma is zero (cobb-douglas)? Why is gamma in the equation then?
JonasanoJ
CC BY-SA 4.0
2018-11-30T14:31:44.843Z
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: https://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions)
Yann
CC BY-SA 4.0
2018-11-29T11:14:06.503Z
2018-11-29T11:14:06.503Z
Yann
<p>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. </p>

<p>Thanks! Jonas.</p>

<production-function>
Is the CES a special case of the translog production function?
JonasanoJ
# Is the CES a special case of the translog production function?

(Asked by: username_0 on 2018-11-28)

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.

**Tags:** production-function

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## 1 answers

### Answer by username_1 on 2018-11-29. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: Thanks. Is the expansion around a certain number?  
> **username_1**: around $\gamma=0$ (McLaurin is the expansion  about 0 while Taylor is about any number)  
> **username_0**: so then the translog assumes always that gamma is zero (cobb-douglas)? Why is gamma in the equation then?  
> **username_1**: 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: https://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions)  

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