Lecture Notes on Econometrics by Qiang Gao, updated at May 17, 2017
Chapter 1 Finite-Sample Properties of OLS Section 7 Application: Returns to Scale in Electricity Supply ...
How to derive the cost function (1.7.2) Assume the firms are engaged in cost minimization ,
$$
\begin{gather}
\min \quad p_{i1} x_{i1} + p_{i2} x_{i2} + p_{i3} x_{i3}
\tag{1}
\\
\text{s.t.} \qquad
A_i x_{i1}^{\alpha_1} x_{i2}^{\alpha_2} x_{i3}^{\alpha_3} = Q_i.
\tag{2}
\end{gather}
$$
Then the Lagrangian is
$$
\mathcal{L} ( x_{i1}, x_{i2}, x_{i3} ) =
p_{i1} x_{i1} + p_{i2} x_{i2} + p_{i3} x_{i3} -
\lambda ( A_i x_{i1}^{\alpha_1} x_{i2}^{\alpha_2} x_{i3}^{\alpha_3} - Q_i ),
\tag{3}
$$
where $$ \lambda > 0 $$ is the shadow price of the exogenous variable $$ Q_i $$.
The first-order conditions are
$$
\begin{align}
\frac{ \partial \mathcal{L} ( x_{i1}, x_{i2}, x_{i3} ) }
{ \partial x_{i1} } =
p_{i1} - \lambda \alpha_1 A_i x_{i1}^{\alpha_1 - 1} x_{i2}^{\alpha_2} x_{i3}^{\alpha_3} = 0,
\tag{4}
\\
\frac{ \partial \mathcal{L} ( x_{i1}, x_{i2}, x_{i3} ) }
{ \partial x_{i2} } =
p_{i2} - \lambda \alpha_2 A_i x_{i1}^{\alpha_1} x_{i2}^{\alpha_2 - 1} x_{i3}^{\alpha_3} = 0,
\tag{5}
\\
\frac{ \partial \mathcal{L} ( x_{i1}, x_{i2}, x_{i3} ) }
{ \partial x_{i3} } =
p_{i3} - \lambda \alpha_1 A_i x_{i1}^{\alpha_1} x_{i2}^{\alpha_2} x_{i3}^{\alpha_3 - 1} = 0.
\tag{6}
\end{align}
$$
Dividing (5) over (4) to eliminate the shadow price $$ \lambda $$,
$$
\begin{gather}
\frac{ p_{i2} }{ p_{i1} } =
\frac{ \alpha_2 }{ \alpha_1 }
\frac{ x_{i1} }{ x_{i2} },
\tag{7}
\\
x_{i2} = p_{i1} p_{i2}^{-1} \alpha_1^{-1} \alpha_2 x_{i1}.
\tag{8}
\end{gather}
$$
Similarly, dividing (6) over (4),
$$
\begin{gather}
\frac{ p_{i3} }{ p_{i1} } =
\frac{ \alpha_3 }{ \alpha_1 }
\frac{ x_{i1} }{ x_{i3} },
\tag{9}
\\
x_{i3} = p_{i1} p_{i3}^{-1} \alpha_1^{-1} \alpha_3 x_{i1}.
\tag{10}
\end{gather}
$$
Substituting (8) and (10) into (2),
$$
A_i x_{i1}^{ \alpha_1 + \alpha_2 + \alpha_3 }
p_{i1}^{ \alpha_2 + \alpha_3 }
p_{i2}^{ - \alpha_2 }
p_{i3}^{ - \alpha_3 }
\alpha_1^{ - \alpha_2 - \alpha_3}
\alpha_2^{ \alpha_2 }
\alpha_3^{ \alpha_3 }
= Q_i.
\tag{11}
$$
Using $$ r \equiv \alpha_1 + \alpha_2 + \alpha_3 $$, we solved from (11) that
$$
\begin{align}
x_{i1} & = A_i^{-1/r} Q_i^{1/r}
p_{i1}^{-1 + \alpha_1/r}
p_{i2}^{ \alpha_2/r }
p_{i3}^{ \alpha_3/r }
\alpha_1^{1 - \alpha_1/r }
\alpha_2^{- \alpha_2/r}
\alpha_3^{- \alpha_3/r}
\ & =
\alpha_1 \cdot ( A_i \alpha_1^{\alpha_1} \alpha_2^{\alpha_2} \alpha_3^{\alpha^3} )^{-1/r}
Q_i^{1/r}
\frac{1}{ p_{i1} }
p_{i1}^{ \alpha_1/r}
p_{i2}^{ \alpha_2/r }
p_{i3}^{ \alpha_3/r }.
\tag{12}
\end{align}
$$
Similarly, because of the symmetry of $$x_{i1}$$ , $$x_{i2}$$ , and $$ x_{i3} $$, we can also solve that
$$
\begin{align}
x_{i2} & = \alpha_2 \cdot ( A_i \alpha_1^{\alpha_1} \alpha_2^{\alpha_2} \alpha_3^{\alpha^3} )^{-1/r}
Q_i^{1/r}
\frac{1}{ p_{i2} }
p_{i1}^{ \alpha_1/r}
p_{i2}^{ \alpha_2/r }
p_{i3}^{ \alpha_3/r },
\tag{13}
\\
x_{i3} & = \alpha_3 \cdot ( A_i \alpha_1^{\alpha_1} \alpha_2^{\alpha_2} \alpha_3^{\alpha^3} )^{-1/r}
Q_i^{1/r}
\frac{1}{ p_{i3} }
p_{i1}^{ \alpha_1/r}
p_{i2}^{ \alpha_2/r }
p_{i3}^{ \alpha_3/r }.
\tag{14}
\end{align}
$$
Substituting solutions (12), (13) and (14) of the endogenous variables into the objective function (1), we get
$$
TC_i = r \cdot ( A_i \alpha_1^{ \alpha_1 } \alpha_2^{ \alpha_2 } \alpha_3^{ \alpha_3 } )^{-1/r} Q_i^{1/r} p_{i1}^{\alpha_1 / r} p_{i2}^{\alpha_2 / r} p_{i3}^{\alpha_3 / r}.
\tag{1.7.2}
$$
Copyright ©2017 by Qiang Gao