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I have a double-log (both inputs and output in logarithmic form) translog production function with 2 inputs [with Labour and Capital]. There are two squared terms, one for each of the inputs and there is their interaction term as well.

How do I calculate the degree of Returns to Scale (whether it is increasing, decreasing, or constant) in this case?

Should I simply add the coefficients? Just as is the case with the Cobb-Douglas production function. Or is there some other way out?

Lastly, how do I calculate the elasticity of scale in this case?

The production function has been specified in the following form:

In(Q)= a_0 + a_L(InL)+ a_KK(InK)+ a_LL(InL)^2 + a_KK(InK)^2 + a_LK(InL)*(InK)

Wherein, Q = Output, L = Labour input, K = Capital Input.

I hope this helps.

Thank You.

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    $\begingroup$ Have you tried applying the definition of ... Returns to Scale? $\endgroup$
    – Giskard
    Feb 18 at 17:05
  • $\begingroup$ Can you please help me by further elaborating on your question? Thank You. $\endgroup$ Feb 18 at 17:06
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    $\begingroup$ @BhagirathBaria Why don't you write out the functional form and apply the definition? What have you tried? $\endgroup$ Feb 18 at 18:38
  • $\begingroup$ Please write the production function explicitly in your question for more clarity. $\endgroup$
    – Dayne
    Feb 21 at 5:57

1 Answer 1

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The typical translog production function $q \left(K, L\right)$ satisfies \begin{equation} \ln \left( q \left(K, L \right) \right) = \ln \left( A \right) + \alpha \ln \left( K \right) + \left( 1 - \alpha \right) \ln \left( L \right) + \frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2 \end{equation} with parameters $A > 0$, $\alpha \in \left( 0, 1 \right)$, $\gamma$ "close" to zero (more on that later).

To get the production function, apply the exponential: \begin{align} q \left(K, L \right) &= e^{\ln \left( q \left(K, L \right) \right)} \\ &= e^{\ln \left( A \right) + \alpha \ln \left( K \right) + \left( 1 - \alpha \right) \ln \left( L \right) + \frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2} \\ &= e^{\ln \left( A \right)} e^{\alpha \ln \left( K \right)} e^{\left( 1 - \alpha \right) \ln \left( L \right)} e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2} \\ &= A K^\alpha L^{1-\alpha} e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2} \end{align}

There are 2 elements:

  1. A Cobb-Douglas function: $A K^\alpha L^{1-\alpha}$
  2. Another term: $e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2}$

Let $\lambda > 1$. The objective is to analyze \begin{equation} q \left(\lambda K, \lambda L \right) = A \left( \lambda K \right)^\alpha \left( \lambda L \right)^{1 - \alpha} e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( \lambda K \right) - \ln \left( \lambda L \right) \right)^2} \end{equation}

Note that \begin{align} \ln \left( \lambda K \right) = \ln \left( \lambda \right) + \ln \left( K \right) \\ \ln \left( \lambda L \right) = \ln \left( \lambda \right) + \ln \left( L \right) \end{align} and therefore \begin{align} \frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( \lambda K \right) - \ln \left( \lambda L \right) \right)^2 &= \frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( \lambda \right) + \ln \left( K \right) - \ln \left( \lambda \right) - \ln \left( L \right) \right)^2 \\ &= \frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2 \end{align}

Consequently, \begin{align} q \left(\lambda K, \lambda L \right) &= \lambda A \left( K \right)^\alpha \left( L \right)^{1 - \alpha} e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( \lambda K \right) - \ln \left( \lambda L \right) \right)^2} \\ &= \lambda q \left(K, L \right) \end{align}

The production function has constant returns to scale.

Note: That should not come as a surprise. The translog utility function is defined as a Taylor approximation of a CES function of the type $\left( \delta K^\gamma + \left( 1 - \delta \right) L^\gamma \right)^{\frac{1}{\gamma}}$ when $\gamma$ is close to zero. And what is a CES when $\gamma = 0$? It's a Cobb-Douglas! So that extra term in the translog ($e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2}$) is an error term for approximating that CES function.

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