Translog Production and Cost Functions
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.
flowchart TB
subgraph Production
A[Production Function] -. approximation .-> D(Translog Production Function)
end
subgraph Cost
B[Cost Function] -. approximation .-> C(Translog Cost Function)
end
A == Conversion via Duality ==> B
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:
\[ \begin{equation} \label{eq:ces-production} Q = A \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}} \end{equation} \tag{1}\]
where \(\alpha_1+\alpha_2=1\), \(A\) is a scale parameter, \(\alpha\) is the distribution parameter, and \(\rho\) is the substitution parameter.
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:
\[ \begin{equation} \label{eq:log-form-ces-production} \ln Q = \ln A + \frac{1}{\rho} \ln \left[ \alpha X_1^{\rho} + (1-\alpha) X_2^{\rho} \right] \end{equation} \tag{2}\]
The Taylor expansion of \(\frac{1}{\rho} \ln \left[ \alpha X_1^{\rho} + (1-\alpha) X_2^{\rho} \right]\) 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]\[ \begin{equation} \label{eq:taylor-expansion-of-ces} \alpha \ln X_1 + (1-\alpha) \ln X_2-\frac{1}{2} \rho \left[(\alpha -1) \alpha (\ln X_1-\ln X_2)^2\right]+O\left(\rho ^2\right) \end{equation} \tag{3}\]
Omitting \(O\left(\rho ^2\right)\) and substituting the Taylor expansion into Equation Equation 2, we have
\[ \begin{align} \label{eq:log-production-with-taylor} \ln Q &= \ln A \\ &+ \alpha \ln X_1 + (1-\alpha) \ln X_2-\frac{1}{2} \rho \left[(\alpha -1) \alpha (\ln X_1-\ln X_2)^2\right] \nonumber \end{align} \tag{4}\]
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:
\[ \begin{align} \ln Q &= a_0 + a_1 \ln X_1 + a_2 \ln X_2 \\ &+ b_{11} (\ln X_1)^2 + b_{22} (\ln X_2)^2 + b_{12} \ln X_1 \ln X_2 \nonumber \end{align} \]
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.
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.
- The production function describes the maximum output \(Q\) that can be produced given the input factors.
- 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
\[ Q = A \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}} \]
The firm’s cost function is
\[ \begin{equation} \label{eq:cost-function} C = w_1 X_1 + w_2 X_2 \end{equation} \tag{5}\]
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:
\[ \begin{equation} \min_{X_1, X_2} \quad C=w_1 X_1 + w_2 X_2 \end{equation} \]
subject to:
\[ \begin{equation} A \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}} = Q \end{equation} \]
Solving the problem
This part is math-heavy. The derived cost function is given by Equation Equation 11.
The Lagrangian for this problem is:
\[ \begin{equation} \mathcal{L} = w_1 X_1 + w_2 X_2 + \lambda \left[ Q - A \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}} \right] \end{equation} \]
Take the first-order conditions:
\[ \begin{align} \frac{\partial \mathcal{L}}{\partial X_1} &= w_1 - \lambda A \alpha_1 \rho X_1^{\rho-1} \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}-1} = 0 \\ \frac{\partial \mathcal{L}}{\partial X_2} &= w_2 - \lambda A \alpha_2 \rho X_2^{\rho-1} \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}-1} = 0 \\ \frac{\partial \mathcal{L}}{\partial \lambda} &= Q - A \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}} = 0 \end{align} \]
Solve the first two equations for \(\lambda\):
\[ \begin{equation} \label{eq:lambda} \lambda = \frac{w_1}{A \alpha_1 \rho X_1^{\rho-1} \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}-1}} = \frac{w_2}{A \alpha_2 \rho X_2^{\rho-1} \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}-1}} \end{equation} \tag{6}\]
Simplifying Equation 6, we get:
\[ \begin{equation} \label{eq:lambda-equal} w_1 X_2^{\rho-1} \alpha_2 = w_2 X_1^{\rho-1} \alpha_1 \end{equation} \tag{7}\]
Manipulating Equation 7, we have:
\[ \begin{equation} \frac{X_1}{X_2} = \left(\frac{\alpha_2 w_1}{\alpha_1 w_2}\right)^{\frac{1}{\rho-1}} \end{equation} \]
so that
\[ \begin{align} (\alpha_2 w_1)^{\frac{\rho}{\rho-1}} X_2^{\rho} &= (\alpha_1 w_2)^{\frac{\rho}{\rho-1}} X_1^{\rho} \\ \left(w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}}\right) \alpha_2 X_2^{\rho} &= \left(w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right) \alpha_1 X_1^{\rho} \end{align} \]
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
\[ \begin{align} \left(w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}}\right) \alpha_2 X_2^{\rho} + \left(w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right) \alpha_2 X_2^{\rho} &= \left(w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right) \alpha_1 X_1^{\rho} + \left(w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right) \alpha_2 X_2^{\rho} \nonumber \\ \left(w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right) \alpha_2 X_2^{\rho} &= \left(w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right) \left(\alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho}\right) \end{align} \]
Raise both sides to the power of \(\frac{1}{\rho}\), we have
\[ \begin{equation} \label{eq:x2_before_simplification} \left(w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}} \right)^{\frac{1}{\rho}} \alpha_2^{\frac{1}{\rho}} X_2 = \left(w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right)^{\frac{1}{\rho}} \left(\alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho}\right)^{\frac{1}{\rho}} \end{equation} \tag{8}\]
Let \(K = \left(w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}} \right)^{\frac{1}{\rho}}\), observe that \(\frac{Q}{A} = \left(\alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho}\right)^{\frac{1}{\rho}}\), we can simplify Equation 8 to
\[ \begin{equation} K \alpha_2^{\frac{1}{\rho}} X_2 = \left(w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right)^{\frac{1}{\rho}} \frac{Q}{A} \end{equation} \]
Therefore, \(X_2\) is given by
\[ \begin{equation} \label{eq:x2} X_2 = K^{-1} w_2^{\frac{1}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}} \frac{Q}{A} \end{equation} \tag{9}\]
We can similarly get \(X_1\)
\[ \begin{equation} \label{eq:x1} X_1 = K^{-1} w_1^{\frac{1}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} \frac{Q}{A} \end{equation} \tag{10}\]
Substituting Equation 10 and Equation 9 into the cost function Equation 5, we have
\[ \begin{align} C &= w_1 X_1 + w_2 X_2 \nonumber \\ &= K^{-1} w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} \frac{Q}{A} + K^{-1} w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}} \frac{Q}{A} \\ &= \frac{Q}{A} K^{-1} \left(w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right) \end{align} \]
Since \(K = \left(w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}} \right)^{\frac{1}{\rho}}\), we have the derived cost function:
\[ \begin{equation} \label{eq:derived-cost-function} C = \frac{Q}{A} \left(w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right)^{\frac{\rho-1}{\rho}} \end{equation} \tag{11}\]
Translog Cost Function
Taking the natural logarithm of both sides of Equation Equation 11, we get:
\[ \begin{equation} \label{eq:log-cost-function} \ln(C) = \ln \left( \frac{Q}{A} \right) + \frac{\rho-1}{\rho} \ln \left( w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}} \right) \end{equation} \tag{12}\]
The Taylor expansion of \(\frac{\rho-1}{\rho} \ln \left( w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}} \right)\) around \(\rho=0\) is 2
2 This is computed in Mathematica, too.
\[ \begin{align} \label{eq:taylor-expansion-of-log-cost} &((\alpha_2-1)\ln\alpha_1-\alpha_2(\ln\alpha_2+\ln w_1-\ln w_2)+\ln w_1)\nonumber\\ &+\frac{1}{2}(\alpha_2-1)\alpha_2\rho(\ln \alpha_1-\ln \alpha_2-\ln w_1+\ln w_2)^2+O\left(\rho^2\right) \end{align} \tag{13}\]
Omitting \(O\left(\rho ^2\right)\) and substituting the Taylor expansion into Equation Equation 12, we have
\[ \begin{align} \ln C &= -\ln A + \ln Q \nonumber\\ &+((\alpha_2-1)\ln\alpha_1-\alpha_2(\ln\alpha_2+\ln w_1-\ln w_2)+\ln w_1)\nonumber\\ &+\frac{1}{2}(\alpha_2-1)\alpha_2\rho(\ln \alpha_1-\ln \alpha_2-\ln w_1+\ln w_2)^2 \label{eq:log-cost-taylor} \end{align} \tag{14}\]
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:
\[ \begin{align} \label{eq:translog-cost} \ln C &= a_0 + a_1 \ln Q \\ &+ b_{11} \ln w_1 + b_{22} \ln w_2 + b_{12} \ln w_1 \ln w_2 \nonumber \end{align} \tag{15}\]
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
\[ \begin{align} \ln C(Q, W) &= \beta_0 + \beta_1 \ln Q + \frac{1}{2} \beta_2 (\ln Q)^2 \\ &+ \sum_{i=1}^{n} \gamma_i \ln W_i + \frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n} \theta_{ij} \ln W_i \ln W_j \nonumber \\ &+ \sum_{i=1}^{n} \phi_i \ln Q \ln W_i \nonumber \label{eq:translog-cost-general} \end{align} \tag{16}\]
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:
\[ \begin{equation} C(Q, \lambda W) = \lambda C(Q, W) \end{equation} \]
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,
\[ \begin{equation} d\ln C = \sum_{i=1}^{n} \gamma_i d\ln W_i+ \frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n} \theta_{ij} \ln W_j d\ln W_i + \sum_{i=1}^{n}\phi_i \ln Q d\ln W_i \end{equation} \]
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,
\[ \begin{equation} d\ln C = d\ln \bar{W} \sum_{i=1}^{n} \gamma_i + d\ln \bar{W}^2 \frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n} \theta_{ij} + d\ln \bar{W} \ln Q \sum_{i=1}^{n}\phi_i \end{equation} \]
To ensure \(\frac{d\ln C}{d\ln \bar{W}}=1\) hence linear homogeneity in the translog cost function, the following conditions must be met:
\[ \begin{align} \sum_{i=1}^{n} \gamma_i &= 1 \\ \sum_{j=1}^{n} \theta_{ij} &= 0 \quad \text{for all } i \\ \sum_{i=1}^{n} \phi_{i} &= 0 \end{align} \]
See Translog Cost Function Estimation for estimation notes and code example.