<p>The typical translog production function <span class="math-container">$q \left(K, L\right)$</span> satisfies
<span class="math-container">\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}</span>
with parameters <span class="math-container">$A &gt; 0$</span>, <span class="math-container">$\alpha \in \left( 0, 1 \right)$</span>, <span class="math-container">$\gamma$</span> &quot;close&quot; to zero (more on that later).</p>
<p>To get the production function, apply the exponential:
<span class="math-container">\begin{align}
q \left(K, L \right) &amp;= e^{\ln \left( q \left(K, L \right) \right)} \\
&amp;= 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} \\
&amp;= 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} \\
&amp;= 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}</span></p>
<p>There are 2 elements:</p>
<ol>
<li>A Cobb-Douglas function: <span class="math-container">$A K^\alpha L^{1-\alpha}$</span></li>
<li>Another term: <span class="math-container">$e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2}$</span></li>
</ol>
<p>Let <span class="math-container">$\lambda &gt; 1$</span>. The objective is to analyze
<span class="math-container">\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}</span></p>
<p>Note that
<span class="math-container">\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}</span>
and therefore
<span class="math-container">\begin{align}
\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( \lambda K \right) - \ln \left( \lambda L \right) \right)^2 &amp;= \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 \\
&amp;= \frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2
\end{align}</span></p>
<p>Consequently,
<span class="math-container">\begin{align}
q \left(\lambda K, \lambda L \right) &amp;= \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} \\
&amp;= \lambda q \left(K, L \right)
\end{align}</span></p>
<p>The production function has constant returns to scale.</p>
<p>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 <span class="math-container">$\left( \delta K^\gamma + \left( 1 - \delta \right) L^\gamma \right)^{\frac{1}{\gamma}}$</span> when <span class="math-container">$\gamma$</span> is close to zero. And what is a CES when <span class="math-container">$\gamma = 0$</span>? It's a Cobb-Douglas! So that extra term in the translog (<span class="math-container">$e^{\frac{1}{2} \gamma \alpha \left( 1 - \alpha \right) \left( \ln \left( K \right) - \ln \left( L \right) \right)^2}$</span>) is an error term for approximating that CES function.</p>

CC BY-SA 4.0
2022-02-24T10:03:42.847Z
2022-02-24T10:03:42.847Z
Thomas Favory
<p>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.</p>
<p>How do I calculate the degree of Returns to Scale (whether it is increasing, decreasing, or constant) in this case?</p>
<p>Should I simply add the coefficients? Just as is the case with the Cobb-Douglas production function. Or is there some other way out?</p>
<p>Lastly, how do I calculate the elasticity of scale in this case?</p>
<p>The production function has been specified in the following form:</p>
<p>In(Q)= a_0 + a_L(InL)+ a_KK(InK)+ a_LL(InL)^2 + a_KK(InK)^2 + a_LK(InL)*(InK)</p>
<p>Wherein, Q = Output, L = Labour input, K = Capital Input.</p>
<p>I hope this helps.</p>
<p>Thank You.</p>

CC BY-SA 4.0
2022-02-18T17:05:08.217Z
Have you tried applying the definition of ... Returns to Scale?
Giskard
CC BY-SA 4.0
2022-02-18T17:06:27.000Z
Can you please help me by further elaborating on your question? Thank You.
Bhagirath Baria
CC BY-SA 4.0
2022-02-18T18:38:39.997Z
@BhagirathBaria Why don't you write out the functional form and apply the *definition*? What have you tried?
Golden_Ratio
CC BY-SA 4.0
2022-02-21T05:57:40.343Z
Please write the production function explicitly in your question for more clarity.
Dayne
<microeconomics><production-function><economies-of-scale>
How to calculate Returns to Scale for Translog production function with two inputs?
Bhagirath Baria
# How to calculate Returns to Scale for Translog production function with two inputs?

(Asked by: username_0 on 2022-02-18)

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.

**Tags:** microeconomics, production-function, economies-of-scale
> **Comments:**
> **username_1**: Have you tried applying the definition of ... Returns to Scale?  
> **username_0**: Can you please help me by further elaborating on your question? Thank You.  
> **username_2**: @BhagirathBaria Why don't you write out the functional form and apply the *definition*? What have you tried?  
> **username_3**: Please write the production function explicitly in your question for more clarity.  

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## 1 answers

### Answer by username_4 on 2022-02-24. Score: 2

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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