Translog Production and Cost Functions

Author

Affiliation

Mingze Gao, PhD

Macquarie University

In this post, I’ll carefully explain the derivation of cost function from a CES production function, as well as the derivation of translog (transcendental logarithmic) production and cost functions.

Figure 1: A diagram of the relationship between production functions and cost functions.

Before I start, the graph above illustrate the relations. Specifically, we can derive the cost function from a CES production function via the duality theorem. Translog production and translog cost functions are approximations to the production and corresponding cost function, respectively, via Taylor expansion.

CES Production Function

Let’s start from the a general production function, CES (Constant Elasticity of Substitution).

The standard CES production function with two factors $X_1$ and $X_2$ is given by:

$$\begin{array}{}\text{(1)}& Q=A{({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }}\end{array}$$

where ${\alpha }_1+{\alpha }_2=1$, $A$ is a scale parameter, $\alpha$ is the distribution parameter, and $\rho$ is the substitution parameter.

Note

The Cobb-Douglas production function is a special case of the CES function when $\rho \to 0$:

$$Q=A{X}_1^{\alpha }{X}_2^{1-\alpha }$$

Translog Production Function

Taking the natural logarithm of both sides of Equation Equation 1, we get:

$$\begin{array}{}\text{(2)}& \ln Q=\ln A+\frac{1}{\rho }\ln [\alpha X_1^{\rho }+(1-\alpha )X_2^{\rho }]\end{array}$$

The Taylor expansion of $\frac{1}{\rho }\ln [\alpha X_1^{\rho }+(1-\alpha )X_2^{\rho }]$ around $\rho =0$ is ¹

¹ This is computed in Mathematica:

expr = Series[1/rho * Log[alpha*X1^rho + (1 - alpha)*X2^rho], {rho, 0, 1}];
simplifiedExpr = FullSimplify[expr];
TeXForm[simplifiedExpr]

$$\begin{array}{}\text{(3)}& \alpha \ln X_1+(1-\alpha )\ln X_2-\frac{1}{2}\rho [(\alpha -1)\alpha (\ln X_1-\ln X_2{)}^2]+O\left({\rho }^2\right)\end{array}$$

Omitting $O\left({\rho }^2\right)$ and substituting the Taylor expansion into Equation Equation 2, we have

$$\begin{array}{}\text{(4)}& \begin{array}{rl}\ln Q& =\ln A\\ & +\alpha \ln X_1+(1-\alpha )\ln X_2-\frac{1}{2}\rho [(\alpha -1)\alpha (\ln X_1-\ln X_2{)}^2]\end{array}\end{array}$$

which clearly is a function of $\ln X_1$, $\ln X_2$ and their interaction terms.

We can therefore reparameterize Equation Equation 4 and get the Translog production function:

$$\begin{array}{rl}\ln Q& =a_0+a_1\ln X_1+a_2\ln X_2\\ & +b_{11}(\ln X_1{)}^2+b_{22}(\ln X_2{)}^2+b_{12}\ln X_1\ln X_2\end{array}$$

Here, $a_1$ and $a_2$ are coefficients that capture the first-order effects, and $b_{11}$, $b_{22}$, and $b_{12}$ are coefficients that capture the second-order effects.

Note

If we use fist-order Taylor expansion in Equation Equation 3 instead, we will end up with a log-linear production function.

Derive Cost Function From Production Function

Given the CES production function Equation 1, we can derive the cost function via the duality theorem.

Duality in a nutshell

1. The production function describes the maximum output $Q$ that can be produced given the input factors.
2. Given a production function and input prices, the firm aims to minimize its costs subject to the constraint of producing a given output level $Q$. This leads to a cost minimization problem.

Cost minimization and the production maximization are essentially “dual” to each other. The conditions that solve one problem can be used to solve the other. This is a manifestation of the more general concept of duality in optimization theory.

Recall that the CES production function is

$$Q=A{({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }}$$

The firm’s cost function is

$$\begin{array}{}\text{(5)}& C=w_1{X}_1+w_2{X}_2\end{array}$$

where $w_1$ and $w_2$ are the factor prices.

Cost minimization problem

To derive the cost function from the given CES production function, we need to find the minimum cost of producing a given level of output $Q$ given input prices $w_1$ and $w_2$.

The cost minimization problem is:

$$\underset{X_1,X_2}{min}C=w_1{X}_1+w_2{X}_2$$

subject to:

$$A{({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }}=Q$$

Solving the problem

This part is math-heavy. The derived cost function is given by Equation Equation 11.

The Lagrangian for this problem is:

$$\mathcal{L}=w_1{X}_1+w_2{X}_2+\lambda [Q-A{({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }}]$$

Take the first-order conditions:

$$\begin{array}{rl}\frac{\partial \mathcal{L}}{\partial X_1}& =w_1-\lambda A{\alpha }_1\rho X_1^{\rho -1}{({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }-1}=0\\ \frac{\partial \mathcal{L}}{\partial X_2}& =w_2-\lambda A{\alpha }_2\rho X_2^{\rho -1}{({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }-1}=0\\ \frac{\partial \mathcal{L}}{\partial \lambda }& =Q-A{({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }}=0\end{array}$$

Solve the first two equations for $\lambda$:

$$\begin{array}{}\text{(6)}& \lambda =\frac{w_1}{A{\alpha }_1\rho X_1^{\rho -1}{({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }-1}}=\frac{w_2}{A{\alpha }_2\rho X_2^{\rho -1}{({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }-1}}\end{array}$$

Simplifying Equation 6, we get:

$$\begin{array}{}\text{(7)}& w_1{X}_2^{\rho -1}{\alpha }_2=w_2{X}_1^{\rho -1}{\alpha }_1\end{array}$$

Manipulating Equation 7, we have:

$$\frac{X_1}{X_2}={\left(\frac{{\alpha }_2{w}_1}{{\alpha }_1{w}_2}\right)}^{\frac{1}{\rho -1}}$$

so that

$$\begin{array}{rl}({\alpha }_2{w}_1{)}^{\frac{\rho }{\rho -1}}{X}_2^{\rho }& =({\alpha }_1{w}_2{)}^{\frac{\rho }{\rho -1}}{X}_1^{\rho }\\ \left(w_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}\right){\alpha }_2{X}_2^{\rho }& =\left(w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}\right){\alpha }_1{X}_1^{\rho }\end{array}$$

Adding $\left(w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}\right){\alpha }_2{X}_2^{\rho }$ to both sides, we have

$$\begin{array}{rl}\left(w_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}\right){\alpha }_2{X}_2^{\rho }+\left(w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}\right){\alpha }_2{X}_2^{\rho }& =\left(w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}\right){\alpha }_1{X}_1^{\rho }+\left(w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}\right){\alpha }_2{X}_2^{\rho }\\ (w_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}+w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}){\alpha }_2{X}_2^{\rho }& =\left(w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}\right)({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })\end{array}$$

Raise both sides to the power of $\frac{1}{\rho }$, we have

$$\begin{array}{}\text{(8)}& {(w_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}+w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}})}^{\frac{1}{\rho }}{\alpha }_2^{\frac{1}{\rho }}{X}_2={\left(w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}\right)}^{\frac{1}{\rho }}{({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }}\end{array}$$

Let $K={(w_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}+w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}})}^{\frac{1}{\rho }}$, observe that $\frac{Q}{A}={({\alpha }_1{X}_1^{\rho }+{\alpha }_2{X}_2^{\rho })}^{\frac{1}{\rho }}$, we can simplify Equation 8 to

$$K{\alpha }_2^{\frac{1}{\rho }}{X}_2={\left(w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}\right)}^{\frac{1}{\rho }}\frac{Q}{A}$$

Therefore, $X_2$ is given by

$$\begin{array}{}\text{(9)}& X_2=K^{-1}{w}_2^{\frac{1}{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}\frac{Q}{A}\end{array}$$

We can similarly get $X_1$

$$\begin{array}{}\text{(10)}& X_1=K^{-1}{w}_1^{\frac{1}{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}\frac{Q}{A}\end{array}$$

Substituting Equation 10 and Equation 9 into the cost function Equation 5, we have

$$\begin{array}{rl}C& =w_1{X}_1+w_2{X}_2\\ & =K^{-1}{w}_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}\frac{Q}{A}+K^{-1}{w}_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}}\frac{Q}{A}\\ & =\frac{Q}{A}{K}^{-1}(w_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}+w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}})\end{array}$$

Since $K={(w_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}+w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}})}^{\frac{1}{\rho }}$, we have the derived cost function:

Cost function derived from CES production function

$$\begin{array}{}\text{(11)}& C=\frac{Q}{A}{(w_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}+w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}})}^{\frac{\rho -1}{\rho }}\end{array}$$

Translog Cost Function

Taking the natural logarithm of both sides of Equation Equation 11, we get:

$$\begin{array}{}\text{(12)}& \ln (C)=\ln \left(\frac{Q}{A}\right)+\frac{\rho -1}{\rho }\ln (w_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}+w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}})\end{array}$$

The Taylor expansion of $\frac{\rho -1}{\rho }\ln (w_1^{\frac{\rho }{\rho -1}}{\alpha }_1^{\frac{-1}{\rho -1}}+w_2^{\frac{\rho }{\rho -1}}{\alpha }_2^{\frac{-1}{\rho -1}})$ around $\rho =0$ is ²

² This is computed in Mathematica, too.

$$\begin{array}{}\text{(13)}& \begin{array}{rl}& (({\alpha }_2-1)\ln {\alpha }_1-{\alpha }_2(\ln {\alpha }_2+\ln w_1-\ln w_2)+\ln w_1)\\ & +\frac{1}{2}({\alpha }_2-1){\alpha }_2\rho (\ln {\alpha }_1-\ln {\alpha }_2-\ln w_1+\ln w_2{)}^2+O\left({\rho }^2\right)\end{array}\end{array}$$

Omitting $O\left({\rho }^2\right)$ and substituting the Taylor expansion into Equation Equation 12, we have

$$\begin{array}{}\text{(14)}& \begin{array}{rl}\ln C& =-\ln A+\ln Q\\ & +(({\alpha }_2-1)\ln {\alpha }_1-{\alpha }_2(\ln {\alpha }_2+\ln w_1-\ln w_2)+\ln w_1)\\ & +\frac{1}{2}({\alpha }_2-1){\alpha }_2\rho (\ln {\alpha }_1-\ln {\alpha }_2-\ln w_1+\ln w_2{)}^2\end{array}\end{array}$$

which clearly is a function of $\ln Q$; $\ln w_1$, $\ln w_2$ and their interaction terms.

We can therefore reparameterize Equation Equation 14 and get the Translog cost function:

$$\begin{array}{}\text{(15)}& \begin{array}{rl}\ln C& =a_0+a_1\ln Q\\ & +b_{11}\ln w_1+b_{22}\ln w_2+b_{12}\ln w_1\ln w_2\end{array}\end{array}$$

Why there is no interaction between $\ln Q$ and $\ln w$?

This is NOT an error! It is because we started from a standard CES production function, which doesn’t include interaction terms.

A more general form of translog cost function includes interaction terms $\ln Q\ln w$ because the underlying production function is even more flexible than the standard CES production function. This is the beauty of translog.

In a general form, the translog cost function $\ln C(Q,W)$ as a function of output $Q$ and a vector of $n$ input prices $W$ is represented as

$$\begin{array}{}\text{(16)}& \begin{array}{rl}\ln C(Q,W)& ={\beta }_0+{\beta }_1\ln Q+\frac{1}{2}{\beta }_2(\ln Q{)}^2\\ & +\sum _{i=1}^n{\gamma }_i\ln W_i+\frac{1}{2}\sum _{i=1}^n\sum _{j=1}^n{\theta }_{ij}\ln W_i\ln W_j\\ & +\sum _{i=1}^n{\varphi }_i\ln Q\ln W_i\end{array}\end{array}$$

Note that here it includes a quadratic term for $\ln Q$ and interactions between $\ln Q$ and $\ln W$. As a result, it can approximate a wide range of very complex cost functions (hence complex underlying production function, via duality).

Linear Homogeneity Constraint

In economic theory, a cost function is often assumed to be linearly homogeneous in input prices. This means that if all input prices $W_i$ are scaled by a constant $\lambda >0$, the total cost $C$ should also scale by the same constant $\lambda$. Mathematically, this is expressed as:

$$C(Q,\lambda W)=\lambda C(Q,W)$$

Linear homogeneity is an important property because it ensures that the cost function is consistent with the idea of constant returns to scale in prices.

Implications for parameters

If we take the total differential of the log cost, holding output constant, we have,

$$d\ln C=\sum _{i=1}^n{\gamma }_{i}d\ln W_i+\frac{1}{2}\sum _{i=1}^n\sum _{j=1}^n{\theta }_{ij}\ln W_{j}d\ln W_i+\sum _{i=1}^n{\varphi }_i\ln Qd\ln W_i$$

By assumption, all input prices scale by the same factor $\lambda$ so that $d\ln W_i$ is the same across all $n$ inputs. Therefore, we can factor it out, which gives,

$$d\ln C=d\ln \overline{W}\sum _{i=1}^n{\gamma }_i+d\ln {\overline{W}}^2\frac{1}{2}\sum _{i=1}^n\sum _{j=1}^n{\theta }_{ij}+d\ln \overline{W}\ln Q\sum _{i=1}^n{\varphi }_i$$

To ensure $\frac{d\ln C}{d\ln \overline{W}}=1$ hence linear homogeneity in the translog cost function, the following conditions must be met:

$$\begin{array}{rl}\sum _{i=1}^n{\gamma }_i& =1\\ \sum _{j=1}^n{\theta }_{ij}& =0\text{for all }i\\ \sum _{i=1}^n{\varphi }_i& =0\end{array}$$

Tip

See Translog Cost Function Estimation for estimation notes and code example.