# Translog Production and Cost Functions

Teaching Notes
Research Notes
Author
Affiliation

Macquarie University

Published

October 13, 2023

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.

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:

$$$\label{eq:ces-production} Q = A \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}}$$ \tag{1}$

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:

$$$\label{eq:log-form-ces-production} \ln Q = \ln A + \frac{1}{\rho} \ln \left[ \alpha X_1^{\rho} + (1-\alpha) X_2^{\rho} \right]$$ \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]

$$$\label{eq:taylor-expansion-of-ces} \alpha \ln X_1 + (1-\alpha) \ln X_2-\frac{1}{2} \rho \left[(\alpha -1) \alpha (\ln X_1-\ln X_2)^2\right]+O\left(\rho ^2\right)$$ \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.

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

$Q = A \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}}$

The firm’s cost function is

$$$\label{eq:cost-function} C = w_1 X_1 + w_2 X_2$$ \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:

$$$\min_{X_1, X_2} \quad C=w_1 X_1 + w_2 X_2$$$

subject to:

$$$A \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}} = Q$$$

### Solving the problem

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

The Lagrangian for this problem is:

$$$\mathcal{L} = w_1 X_1 + w_2 X_2 + \lambda \left[ Q - A \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}} \right]$$$

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 $$\lambda$$:

$$$\label{eq:lambda} \lambda = \frac{w_1}{A \alpha_1 \rho X_1^{\rho-1} \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}-1}} = \frac{w_2}{A \alpha_2 \rho X_2^{\rho-1} \left( \alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho} \right)^{\frac{1}{\rho}-1}}$$ \tag{6}$

Simplifying Equation 6, we get:

$$$\label{eq:lambda-equal} w_1 X_2^{\rho-1} \alpha_2 = w_2 X_1^{\rho-1} \alpha_1$$ \tag{7}$

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{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

$$$\label{eq:x2_before_simplification} \left(w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}} \right)^{\frac{1}{\rho}} \alpha_2^{\frac{1}{\rho}} X_2 = \left(w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right)^{\frac{1}{\rho}} \left(\alpha_1 X_1^{\rho} + \alpha_2 X_2^{\rho}\right)^{\frac{1}{\rho}}$$ \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

$$$K \alpha_2^{\frac{1}{\rho}} X_2 = \left(w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right)^{\frac{1}{\rho}} \frac{Q}{A}$$$

Therefore, $$X_2$$ is given by

$$$\label{eq:x2} X_2 = K^{-1} w_2^{\frac{1}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}} \frac{Q}{A}$$ \tag{9}$

We can similarly get $$X_1$$

$$$\label{eq:x1} X_1 = K^{-1} w_1^{\frac{1}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} \frac{Q}{A}$$ \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:

Cost function derived from CES production function

$$$\label{eq:derived-cost-function} C = \frac{Q}{A} \left(w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}}\right)^{\frac{\rho-1}{\rho}}$$ \tag{11}$

## Translog Cost Function

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

$$$\label{eq:log-cost-function} \ln(C) = \ln \left( \frac{Q}{A} \right) + \frac{\rho-1}{\rho} \ln \left( w_1^{\frac{\rho}{\rho-1}} \alpha_1^{\frac{-1}{\rho-1}} + w_2^{\frac{\rho}{\rho-1}} \alpha_2^{\frac{-1}{\rho-1}} \right)$$ \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}

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{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:

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

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,

$$$d\ln C = \sum_{i=1}^{n} \gamma_i d\ln W_i+ \frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n} \theta_{ij} \ln W_j d\ln W_i + \sum_{i=1}^{n}\phi_i \ln Q d\ln W_i$$$

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,

$$$d\ln C = d\ln \bar{W} \sum_{i=1}^{n} \gamma_i + d\ln \bar{W}^2 \frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n} \theta_{ij} + d\ln \bar{W} \ln Q \sum_{i=1}^{n}\phi_i$$$

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}

Tip

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