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?


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

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

in this case $X_1 = K$, $X_2 = L$. Now we expand $\ln Y$ around $\gamma = 0$

$\gamma^0$ 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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