<p>The idea is indeed to Taylor expand the production function. To justify it, you can start with the constant elasticity of substitution function, which in the two-factor case can be written <a href="https://en.wikipedia.org/wiki/Cobb%E2%80%93Douglas_production_function#Relationship_to_the_CES_production_function" rel="noreferrer">as</a></p>

<p><span class="math-container">$$
Y = A[\alpha K^\gamma + (1 - \alpha)L^\gamma]^{1/\gamma} \tag{1}
$$</span></p>

<p>in this case <span class="math-container">$X_1 = K$</span>, <span class="math-container">$X_2 = L$</span>. Now we expand <span class="math-container">$\ln Y$</span> around <span class="math-container">$\gamma = 0$</span> (recall the CES approximates to a cobb-douglas production function when <span class="math-container">$\gamma \approx0).$</span> </p>

<blockquote>
  <p><span class="math-container">$\gamma^0$</span> term</p>
</blockquote>

<p><span class="math-container">$$
\lim_{\gamma \to 0} \ln Y = \ln (A K^\alpha L^{1 - \alpha}) \tag{2}
$$</span></p>

<blockquote>
  <p><span class="math-container">$\gamma^1$</span> term</p>
</blockquote>

<p><span class="math-container">\begin{eqnarray}
\lim_{\gamma \to 0} \frac{\partial \ln Y}{\partial \gamma} &amp;=&amp; \lim_{\gamma \to 0}\frac{\alpha  K^{\gamma } \ln (L)+(1-\alpha ) l^{\gamma } \log (L)}{\gamma 
   \left(\alpha  K^{\gamma }+(1-\alpha ) L^{\gamma }\right)}-\frac{\ln \left(\alpha 
   K^{\gamma }+(1-\alpha ) L^{\gamma }\right)}{\gamma ^2}\\
&amp;=&amp; \frac{1}{2} (1 - \alpha) \alpha  (\ln (K)-\ln (L))^2 \tag{3}
\end{eqnarray}</span></p>

<p>Up to first order we have then</p>

<p><span class="math-container">\begin{eqnarray}
\ln Y &amp;\approx&amp; \color{blue}{(\ln Y)_{\gamma = 0}} + \color{red}{\left(\frac{\partial \ln Y}{\partial \gamma}\right)_{\gamma = 0} \gamma} \\
&amp;\stackrel{(2),(3)}{=}&amp; \color{blue}{\ln A + \alpha \ln K + (1 - \alpha) \ln L} + \color{red}{\frac{1}{2}\alpha(1 - \alpha)\gamma [\ln K - \ln L]^2} \\
&amp;=&amp; \ln A + \alpha \ln X_1 + (1 - \alpha) \ln X_2 + \frac{\alpha \gamma (1 -\alpha)}{2}\left[\ln^2 X_1 -2\ln X_1\ln X_2 + \ln^2 X_2 \right] \\
&amp;=&amp; \alpha_0 + \sum_{i = 1}^2 \alpha_i \ln X_i + \frac{1}{2}\sum_{i, j = 1}^2 \beta_{ij}\ln X_i \ln X_j \tag{3}
\end{eqnarray}</span></p>

<p>This is naturally extended to <span class="math-container">$n &gt; 2$</span> as </p>

<p><span class="math-container">$$
\ln Y = \alpha_0 + \sum_{i = 1}^n \alpha_i \ln X_i + \frac{1}{2}\sum_{i, j = 1}^n \beta_{ij}\ln X_i \ln X_j
$$</span></p>

CC BY-SA 4.0
2018-12-25T23:55:34.190Z
2019-01-18T17:44:13.230Z
2019-01-18T17:44:13.230Z
caverac
<p>The translog is a second order Taylor development of <span class="math-container">$$\log y = f(\log x) $$</span> in <span class="math-container">$\log x$</span> around an arbitrary point <span class="math-container">$x_0 \ne 0$</span>:</p>

<p><span class="math-container">$$\log y = f(\log x_0) + \frac{\partial f}{\partial \log x^T}(\log x_0)\cdot(\log x-\log x_0) + \frac{1}{2}(\log x-\log x_0)^T\frac{\partial^2 f}{\partial \log x\log x^T}(\log x_0)\cdot(\log x-\log x_0) $$</span>
<span class="math-container">$$=\alpha_0+\alpha_x^T \log x+\frac{1}{2}\log x^TB\log x $$</span>
where this last reparameterization is obtained when defining:
<span class="math-container">$$\alpha_0=f(\log x_0) - \frac{\partial f}{\partial \log x^T}(\log x_0)\cdot\log x_0 + \frac{1}{2}\log x_0^T\frac{\partial^2 f}{\partial \log x\log x^T}(\log x_0)\cdot\log x_0$$</span>
<span class="math-container">$$\alpha_x=\frac{\partial f}{\partial \log x}(\log x_0) - \frac{\partial^2 f}{\partial \log x\log x^T}(\log x_0)\log x_0$$</span>
<span class="math-container">$$B=\frac{\partial^2 f}{\partial \log x\log x^T}(\log x_0). $$</span></p>

CC BY-SA 4.0
2019-01-18T21:15:47.837Z
2019-01-18T22:20:21.173Z
2019-01-18T22:20:21.173Z
Bertrand
<p>Ive been having difficulty deriving the translog production function defined as:</p>

<p><span class="math-container">$$\ln y=\alpha_0+\sum_{i=1}^n\alpha_i \ln x_i+\frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n\ \beta_{ij}\ln x_i\ln x_j $$</span></p>

<p>I know we start with a log-log production function. 
<span class="math-container">$$\ln y=\alpha_0+\sum_{i=1}^n\alpha_i\ln x_i$$</span></p>

<p>the next step from what i recall is to take the taylor series of this function around the point <span class="math-container">$x_i=0$</span>. the reason why this is an issue is because <span class="math-container">$\ln(0)$</span> is undefined.</p>

<p>How exactly is this function derived? </p>

<microeconomics><econometrics><production-function>
Deriving the translog production function
EconJohn
# Deriving the translog production function

(Asked by: username_0 on 2018-12-25)

Ive been having difficulty deriving the translog production function defined as:

$$\ln y=\alpha\_0+\sum\_{i=1}^n\alpha\_i \ln x\_i+\frac{1}{2}\sum\_{i=1}^n\sum\_{j=1}^n\ \beta\_{ij}\ln x\_i\ln x\_j $$

I know we start with a log-log production function.
$$\ln y=\alpha\_0+\sum\_{i=1}^n\alpha\_i\ln x\_i$$

the next step from what i recall is to take the taylor series of this function around the point $x\_i=0$. the reason why this is an issue is because $\ln(0)$ is undefined.

How exactly is this function derived?

**Tags:** microeconomics, econometrics, production-function

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## 2 answers

### Answer by username_2 on 2019-01-18. Score: 3 | Accepted answer

The translog is a second order Taylor development of $$\log y = f(\log x) $$ in $\log x$ around an arbitrary point $x\_0 \ne 0$:

$$\log y = f(\log x\_0) + \frac{\partial f}{\partial \log x^T}(\log x\_0)\cdot(\log x-\log x\_0) + \frac{1}{2}(\log x-\log x\_0)^T\frac{\partial^2 f}{\partial \log x\log x^T}(\log x\_0)\cdot(\log x-\log x\_0) $$
$$=\alpha\_0+\alpha\_x^T \log x+\frac{1}{2}\log x^TB\log x $$
where this last reparameterization is obtained when defining:
$$\alpha\_0=f(\log x\_0) - \frac{\partial f}{\partial \log x^T}(\log x\_0)\cdot\log x\_0 + \frac{1}{2}\log x\_0^T\frac{\partial^2 f}{\partial \log x\log x^T}(\log x\_0)\cdot\log x\_0$$
$$\alpha\_x=\frac{\partial f}{\partial \log x}(\log x\_0) - \frac{\partial^2 f}{\partial \log x\log x^T}(\log x\_0)\log x\_0$$
$$B=\frac{\partial^2 f}{\partial \log x\log x^T}(\log x\_0). $$


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### Answer by username_1 on 2018-12-25. Score: 7

The idea is indeed to Taylor expand the production function. To justify it, you can start with the constant elasticity of substitution function, which in the two-factor case can be written [as](https://en.wikipedia.org/wiki/Cobb%E2%80%93Douglas_production_function#Relationship_to_the_CES_production_function)

$$
Y = A[\alpha K^\gamma + (1 - \alpha)L^\gamma]^{1/\gamma} \tag{1}
$$

in this case $X\_1 = K$, $X\_2 = L$. Now we expand $\ln Y$ around $\gamma = 0$ (recall the CES approximates to a cobb-douglas production function when $\gamma \approx0).$

> $\gamma^0$ term

$$
\lim\_{\gamma \to 0} \ln Y = \ln (A K^\alpha L^{1 - \alpha}) \tag{2}
$$

> $\gamma^1$ term

\begin{eqnarray}
\lim\_{\gamma \to 0} \frac{\partial \ln Y}{\partial \gamma} &=& \lim\_{\gamma \to 0}\frac{\alpha K^{\gamma } \ln (L)+(1-\alpha ) l^{\gamma } \log (L)}{\gamma
\left(\alpha K^{\gamma }+(1-\alpha ) L^{\gamma }\right)}-\frac{\ln \left(\alpha
K^{\gamma }+(1-\alpha ) L^{\gamma }\right)}{\gamma ^2}\\
&=& \frac{1}{2} (1 - \alpha) \alpha (\ln (K)-\ln (L))^2 \tag{3}
\end{eqnarray}

Up to first order we have then

\begin{eqnarray}
\ln Y &\approx& \color{blue}{(\ln Y)\_{\gamma = 0}} + \color{red}{\left(\frac{\partial \ln Y}{\partial \gamma}\right)\_{\gamma = 0} \gamma} \\
&\stackrel{(2),(3)}{=}& \color{blue}{\ln A + \alpha \ln K + (1 - \alpha) \ln L} + \color{red}{\frac{1}{2}\alpha(1 - \alpha)\gamma [\ln K - \ln L]^2} \\
&=& \ln A + \alpha \ln X\_1 + (1 - \alpha) \ln X\_2 + \frac{\alpha \gamma (1 -\alpha)}{2}\left[\ln^2 X\_1 -2\ln X\_1\ln X\_2 + \ln^2 X\_2 \right] \\
&=& \alpha\_0 + \sum\_{i = 1}^2 \alpha\_i \ln X\_i + \frac{1}{2}\sum\_{i, j = 1}^2 \beta\_{ij}\ln X\_i \ln X\_j \tag{3}
\end{eqnarray}

This is naturally extended to $n > 2$ as

$$
\ln Y = \alpha\_0 + \sum\_{i = 1}^n \alpha\_i \ln X\_i + \frac{1}{2}\sum\_{i, j = 1}^n \beta\_{ij}\ln X\_i \ln X\_j
$$


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