Foreign-owned firms and partial privatization of state holding corporations

Abstract

We consider a state holding corporation with two plants producing heterogeneous goods. In the partially foreign-owned private sector, firms may be organized as uniplant or multiplant. We find that the stake that the government retains in the state corporation depends on whether goods are substitutes or complements, whether private firms are uniplant or multiplant and the percentage of foreign ownership in private firms. The main result is that when foreign ownership is high, there is more privatization with uniplant (multiplant) private firms if goods are substitute (complements). However, when foreign ownership is low, the opposite result is obtained.

Appendices

Appendix 1: Uniplant private firms

From Lemma 1, we obtain the following:

$\frac{{\partial \beta^{SU} }}{\partial \alpha } = - \frac{{6 - b - b^{2} }}{{\left( {11 - 6\alpha + b\left( {5 - 4\alpha } \right)} \right)^{2} }}$ < 0, $\frac{{\partial \beta^{SU} }}{\partial b} = - \frac{{6 - 13\alpha + 6\alpha^{2} }}{{\left( {11 - 6\alpha + b\left( {5 - 4\alpha } \right)} \right)^{2} }} > 0$ if and only if $\alpha > 2/3$,

$$ \begin{aligned} \begin{aligned} & \frac{{\partial q_{iA}^{SU} }}{\partial \alpha } = - \frac{{3\left( {1 + b} \right)\left( {3 + b} \right)^{2} }}{{\left( {17 + 18b + 5b^{2} - 3\left( {1 + b} \right)^{2} \alpha } \right)^{2} }} < 0 ,\,\,\,\,\\&\frac{{\partial q_{iB}^{SU} }}{\partial \alpha } = \frac{{3\left( {1 + b} \right)^{2} \left( {3 + b} \right)}}{{\left( {17 + 18b + 5b^{2} - 3\left( {1 + b} \right)^{2} \alpha } \right)^{2} }} > 0, \\ & \frac{{\partial (q_{iA}^{SU} + q_{iB}^{SU} )}}{\partial \alpha } = - \frac{{6\left( {3 + 4b + b^{2} } \right)}}{{\left( {17 + 18b + 5b^{2} - 3\left( {1 + b} \right)^{2} \alpha } \right)^{2} }} < 0, \\ & q_{iA}^{SU} - q_{iB}^{SU} = \frac{{5 + 3b\left( {1 - \alpha } \right) - 3\alpha }}{{17 + 18b + 5b^{2} - 3\left( {1 + b} \right)^{2} \alpha }} > 0, \\ & \pi_{iA}^{SU} - \pi_{iB}^{SU} = \frac{{\left( {5 + 3b\left( {1 - \alpha } \right) - 3\alpha } \right)\left( {1 - b + 3\left( {1 + b} \right)\alpha } \right)}}{{2\left( {17 + 18b + 5b^{2} - 3\left( {1 + b} \right)^{2} \alpha } \right)^{2} }} > 0. \\ \end{aligned} \end{aligned} $$

Appendix 2: Multiplant private firm

From Lemma 2, it can be shown the following:

$\frac{{\partial \beta^{M} }}{\partial b} = \frac{{1 + b\left( {2 + b} \right)\left( {3 - 2\alpha } \right) - \alpha }}{{\left( {11 - 6\alpha + b^{2} \left( {3 - 2\alpha } \right) + b\left( {12 - 7\alpha } \right)} \right)^{2} }}$ > 0 if b > 0, and negative for b < 0 if $\alpha > \frac{{1 + 6b + 3b^{2} }}{{1 + 4b + 2b^{2} }}$;

$$ \begin{aligned} & \frac{{\partial \beta^{M} }}{\partial \alpha } = - \frac{{\left( {1 + b} \right)\left( {2 + b} \right)\left( {3 + 2b} \right)}}{{\left( {11 - 6\alpha + b^{2} \left( {3 - 2\alpha } \right) + b\left( {12 - 7\alpha } \right)} \right)^{2} }} < 0;\,\,\,\frac{{\partial q_{iA}^{M} }}{\partial \alpha } = - \frac{{\left( {1 + b} \right)\left( {3 + 2b} \right)^{3} }}{{H_{2}^{2} }} < 0; \\ & \frac{{\partial q_{iB}^{M} }}{\partial \alpha } = \frac{{\left( {1 + b} \right)^{2} \left( {3 + 2b} \right)^{2} }}{{H_{2}^{2} }} > 0;\,\,\,q_{iA}^{M} - q_{iB}^{M} = \frac{{\left( {1 + b} \right)\left( {5 - 3\alpha + b\left( {3 - 2\alpha } \right)} \right)}}{{H_{2} }} > 0; \\ & \pi_{iA}^{M} - \pi_{iB}^{M} = \frac{{\left( {1 + b} \right)^{2} \left( {1 + b + 3\alpha + 2b\alpha } \right)\left( {5 - 3\alpha + b\left( {3 - 2\alpha } \right)} \right)}}{{2H_{2}^{2} }} > 0.\,\, \\ \end{aligned} $$

Appendix 3: Proof of Proposition 1

$\beta^{M} - \beta^{U}$ = $\frac{{b\left( {7 - 14\alpha + 6\alpha^{2} + b^{2} \left( {3 - 5\alpha + 2\alpha^{2} } \right) + b\left( {10 - 17\alpha + 7\alpha^{2} } \right)} \right)}}{{\left( {11 - 6\alpha + b\left( {5 - 4\alpha } \right)} \right)\left( {11 - 6\alpha + b^{2} \left( {3 - 2\alpha } \right) + b\left( {12 - 7\alpha } \right)} \right)}}$ that equals to zero for $\alpha$ = $\alpha^{*}$, with $\alpha^{*} = \frac{{14 + 17b + 5b^{2} - \sqrt {28 + 40b + 21b^{2} + 6b^{3} + b^{4} } }}{{12 + 14b + 4b^{2} }}$. Moreover, $\beta^{M} > \beta^{U}$ for $\alpha$ > $\alpha^{*}$ if b < 0 and for $\alpha$ < $\alpha^{*}$ if b > 0; $\beta^{M} = \beta^{U}$ if b = 0; $\beta^{M} < \beta^{U}$ for $\alpha$ > $\alpha^{*}$ if b > 0 and for $\alpha$ < $\alpha^{*}$ if b < 0.

Appendix 4: Proof of the results shown in Sect. 6

The producer surplus is given by PS = $\left( {\beta + \gamma \left( {1 - \beta } \right)} \right)(\pi_{1A} + \pi_{2A} ) + \alpha (\pi_{1B} + \pi_{2B} )$. Foreign investors own (1 − β)(1 − γ) percent of the shares in the semipublic firm. Following Lin and Matsumura (2012), the government obtains M(α, β, γ) = (1 − γ)$\left( {\pi_{1A}^{e} + \pi_{2A}^{e} } \right)$ from selling part of the public firm to foreign investors. Therefore, social welfare in the second stage of the game is given by: W = CS + $\left( {\beta + \gamma \left( {1 - \beta } \right)} \right)(\pi_{1A} + \pi_{2A} ) + \alpha (\pi_{1B} + \pi_{2B} ) + \left( {1 - \gamma } \right)M(\alpha ,\beta ,\gamma )$.

Let superscript S denote the equilibrium obtained in the second stage of the game. In the first stage of the game $\pi_{1A}^{e} + \pi_{2A}^{e}$ = $\pi_{1A}^{\text{SU}} + \pi_{2A}^{\text{SU}}$ for uniplant private firms, and $\pi_{{1{\text{A}}}}^{e} + \pi_{{2{\text{A}}}}^{e}$ = $\pi_{{1{\text{A}}}}^{\text{SM}} + \pi_{{2{\text{A}}}}^{\text{SM}}$ for a multiplant private firm. Thus, (1 − β) M(α, β, γ) = (1 − β) (1 − γ)($\pi_{1A}^{\text{Sk}} + \pi_{2A}^{\text{Sk}}$), k = U, M.

Let $G_{1} = \left( {4\left( {10 - 6\alpha - b\left( {4 - 3\alpha } \right)} \right)^{2} \left( {\gamma - 1} \right) + \left( {21 - 6\alpha \left( {2 - \gamma } \right) - 10\gamma + b\left( {9 - 4\gamma - \alpha \left( {7 - 3\gamma } \right)} \right)} \right)^{2} } \right)^{1/2}$ and $G_{2} = \left( {4\left( {2 + b} \right)^{2} \left( {5 - 3\alpha + b\left( {3 - 2\alpha } \right)} \right)^{2} \left( {\gamma - 1} \right) + \left( {\left( {3 + 2b} \right)\left( {2\left( {2 + b} \right)\alpha - 7 - 3b} \right) - \left( {2 + b} \right)\left( {5 - 3\alpha + b\left( {3 - 2\alpha } \right)} \right)\gamma } \right)^{2} } \right)^{1/2}$. Solving the game with both uniproduct and multiproduct private firms, we obtain:

$$ \begin{aligned} \beta^{U} & = \frac{{21 + 9b - 12\alpha - 7b\alpha - 10\gamma - 4b\gamma + 6\alpha \gamma + 3b\alpha \gamma - G_{1} }}{{2\left( {10 - 6\alpha + b\left( {4 - 3\alpha } \right)} \right)\left( {1 - \gamma } \right)}}, \\ \beta^{M} & = \frac{{\left( {3 + 2b} \right)\left( {7 + 3b - 2\left( {2 + b} \right)\alpha } \right) - \left( {2 + b} \right)\left( {5 - 3\alpha + b\left( {3 - 2\alpha } \right)} \right)\gamma - G_{2} }}{{2\left( {2 + b} \right)\left( {5 - 3\alpha + b\left( {3 - 2\alpha } \right)} \right)\left( {1 - \gamma } \right)}}. \\ \end{aligned} $$

It can be shown that $\partial \beta^{k} /\partial \gamma > 0$, and that $\partial \beta^{k} /\partial \alpha < 0$, k = U, M. By substituting γ = α in $\beta^{U}$ and $\beta^{M}$, we find that $\beta^{U}$ = $\beta^{M}$ for $\alpha = \alpha_{{}}^{*}$. It can be verified that the same result is obtained for values of α and γ, such that γ = Kα, Κ > 0.