Corporate social responsibility and privatization policy in a mixed oligopoly

Abstract

This article formulates a mixed oligopoly in which a public firm competes with two private firms that may adopt corporate social responsibility (CSR). We then determine the optimal privatization policy and find that, depending on CSR level and cost differences among firms, either nationalization or full privatization can be optimal. For identical cost functions, we also show the optimal degree of privatization is decreasing with the level of CSR if private firms are homogeneous, while it is non-monotone with the CSR level in a significant heterogeneity of objectives among the firms. Our analysis suggests that CSR activities that affect the magnitude of heterogeneity among firms are crucial in choosing an optimal privatization policy in a mixed oligopoly.

Appendices

Appendix A: Proofs

1.1 A1. Proof of Proposition 1

We consider the corner solutions because of the constraint $0 \le \theta \le 1$. The analysis of maximizing welfare provides three following cases.

• (1-1) For the optimality condition of nationalization, we have $\left. {\frac{{\partial W^{*} (\theta )}}{\partial \theta }} \right|_{\theta = 0} = \frac{{A^{2} (2 - \alpha_{1} - \alpha_{2} ) \cdot f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )}}{{2\left[ {2 + (4 - \alpha_{1} - \alpha_{2} )k_{0} } \right]^{3} }}.\;{\text{Thus}},\;{\text{if}}\;f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) < 0,\;\theta^{*} = 0.$

• (1-2) For the optimality condition of full privatization, we have $\left. {\frac{{\partial W^{*} (\theta )}}{\partial \theta }} \right|_{\theta = 1} = \frac{{A^{2} (2 - \alpha_{1} - \alpha_{2} ) \cdot g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )}}{{2\left[ {6 - \alpha_{1} - \alpha_{2} + (4 - \alpha_{1} - \alpha_{2} )k_{0} } \right]^{3} }}.\;{\text{Thus}},\;{\text{if}}\;g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) > 0,\;\theta^{*} = 1.$

• (1-3) Otherwise, we have an interior solution from the first-order condition under certain parameter ranges. Then, the optimal degree of partial privatization in (7) is derived from $\frac{{\partial W^{*} (\theta )}}{\partial \theta } = 0$. Thus, we have $\theta^{*} \in (0,\;1)$ if $k_{0}^{*} \le k_{0} \le k_{0}^{**}$ where $k_{0}^{*} = \frac{{2\left[ {\alpha_{1} (1 - \alpha_{1} ) + \alpha_{2} (1 - \alpha_{2} )} \right]}}{{3(\alpha_{1}^{2} + \alpha_{2}^{2} ) + 4 + 2\alpha_{1} \alpha_{2} - 6(\alpha_{1} + \alpha_{2} )}} < k_{0}^{**} = \frac{{12 + 2\alpha_{1} \alpha_{2} - 3(\alpha_{1}^{2} + \alpha_{2}^{2} ) - 4(\alpha_{1} + \alpha_{2} )}}{{3(\alpha_{1}^{2} + \alpha_{2}^{2} ) + 4 + 2\alpha_{1} \alpha_{2} - 6(\alpha_{1} + \alpha_{2} )}}.$ In this case, we will check the second-order condition with the optimal degree of partial privatization in (7). Then, we can show that $\frac{{\partial^{2} W(\theta^{*} )}}{{\partial \theta^{2} }} = - \frac{{A^{2} \left[ {(\alpha_{1} - \alpha_{2} )^{2} + 6(\alpha_{1} + \alpha_{2} - 2)} \right]^{4} }}{{16(2 - \alpha_{1} - \alpha_{2} )^{2} \cdot z(k_{0} ,\alpha_{1} ,\alpha_{2} )}} \le 0\;{\text{where}}$

  $$\begin{aligned} z(k_{0} ,\alpha_{1} ,\alpha_{2} ) &= \left\{ {k_{0} \left[ {(\alpha_{1} + \alpha_{2} )^{2} - 8(\alpha_{1} + \alpha_{2} ) + 16} \right] + \alpha_{1}^{2} + \alpha_{2}^{2} - 2(\alpha_{1} + \alpha_{2} ) + 6} \right\}^{3} > 0 \\ &\quad {\text{for}}\;k_{0} \ge 0.\end{aligned} $$

  □

1.2 A2. Proof of Proposition 2

From our definitions, necessary calculations provide that $k_{0}^{*} = \frac{{2\left[ {\alpha_{1} (1 - \alpha_{1} ) + \alpha_{2} (1 - \alpha_{2} )} \right]}}{{D(\alpha_{1} ,\alpha_{2} )}}$ and $k_{0}^{**} = \frac{{12 + 2\alpha_{1} \alpha_{2} - 3(\alpha_{1}^{2} + \alpha_{2}^{2} ) - 4(\alpha_{1} + \alpha_{2} )}}{{D(\alpha_{1} ,\alpha_{2} )}}$ where numerators in both equations are positive, i.e., $2\left[ {\alpha_{1} (1 - \alpha_{1} ) + \alpha_{2} (1 - \alpha_{2} )} \right] > 0$ and $12 + 2\alpha_{1} \alpha_{2} - 3(\alpha_{1}^{2} + \alpha_{2}^{2} ) - 4(\alpha_{1} + \alpha_{2} ) > 0$ for $0 \le \alpha_{i} < 1$. Then, we can show that $k_{0}^{**} < k_{0}^{*} < 0$ if $D \le 0$ while $0 < k_{0}^{*} < k_{0}^{**}$ if $D > 0$. First, if $D \le 0$, we have $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) < 0$ for any $k_{0}$. Proposition 1 implies that nationalization is always optimal. However, if $D > 0$, we have the following relations: If $k_{0} \frac{ > }{ < }k_{0}^{*}$, $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )\frac{ > }{ < }0$ and if $k_{0} \frac{ > }{ < }k_{0}^{**}$, $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )\frac{ > }{ < }0$. Proposition 1 implies the following optimal policies:

• (2-1) If $k_{0} < k_{0}^{*}$, $\theta^{*} = 0$ since $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) < 0$,

• (2-2) If $k_{0} > k_{0}^{**}$, $\theta^{*} = 1$ since $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) > 0$,

• (2-3) If $k_{0}^{*} \le k_{0} \le k_{0}^{**}$, $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) > 0$ and $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) < 0$. Then, the optimal degree of partial privatization is $\theta^{*} \in (0,\;1)$. □

1.3 A3. Proof of Corollary 1

From Proposition 2, if $\alpha_{i} \ge 0$, we have $k_{0}^{**} \ge 3$.

1.4 A3. Proof of Proposition 3

We provide the general results on the optimal policies in Appendix B. Then, using the Appendix B, we can have the special results in Proposition 3 by assuming $k_{0} = 1$. In particular, we can define $\;\alpha_{2}^{*} (1,\alpha_{1} )$ that satisfies $f(1,\;\alpha_{1} ,\;\alpha_{2} ) = 0$, which is either $\alpha_{2}^{ + } = \frac{{4 - \alpha_{1} + 2\sqrt {8\alpha_{1} - 6\alpha_{1}^{2} - 1} }}{5}$ or $\alpha_{2}^{ - } = \frac{{4 - \alpha_{1} - 2\sqrt {8\alpha_{1} - 6\alpha_{1}^{2} - 1} }}{5}$ for $\alpha_{1} \ge \frac{{4 - \sqrt {10} }}{6}$. Thus, we have two following relations:

• (3-1) If $\alpha_{1} > \frac{1}{{2(1 + 2k_{0} )}}$, then $0 < \alpha_{2}^{ - } < 1 < \alpha_{2}^{ + }$. Thus, if $\alpha_{2} \frac{ < }{ > }\alpha_{2}^{ - }$, $f(1,\;\alpha_{1} ,\;\alpha_{2} )\frac{ > }{ < }0$ (see Proposition B1(2) in Appendix B).

• (3-2) If $\frac{{4 - \sqrt {10} }}{6} \le \alpha_{1} \le \frac{1}{{2(1 + 2k_{0} )}}$, then $0 < \alpha_{2}^{ - } < 1 < \alpha_{2}^{ + }$. Thus, $f(1,\;\alpha_{1} ,\;\alpha_{2} ) < 0$ if $\alpha_{2} \in (\alpha_{2}^{ - } ,\alpha_{2}^{ + } )$ (see Proposition B1(3) in Appendix B).

We can also define $\;\alpha_{2}^{**} (1,\alpha_{1} )$ that satisfies $g(1,\;\alpha_{1} ,\;\alpha_{2} ) = 0$, which is either $\tilde{\alpha }_{2}^{ + } (k_{0} ,\alpha_{1} ) = \frac{{1 + \sqrt {49 + 12\alpha_{1} - 36\alpha_{1}^{2} } }}{6}$ or $\tilde{\alpha }_{2}^{ - } (k_{0} ,\alpha_{1} ) = \frac{{1 - \sqrt {49 + 12\alpha_{1} - 36\alpha_{1}^{2} } }}{6}$ where $C(1,\;\alpha_{1} ) = 49 + 12\alpha_{1} - 36\alpha_{1}^{2} > 0$. Then, $\tilde{\alpha }_{2}^{ + } > 1$ since $C(1,\;\alpha_{1} ) > 25$ and $\tilde{\alpha }_{2}^{ - } < 0$ since $C(1,\;\alpha_{1} ) > 1$ (see Proposition B2 in Appendix B). Thus, $g(1,\;\alpha_{1} ,\;\alpha_{2} ) < 0$. As a result, we have $\;\alpha_{2}^{**} < 0 < \alpha_{2}^{*}$. □

1.5 A4. Proof of Proposition 4

From Proposition 3, if $k_{0} = 1$ and $\alpha_{2} < \alpha_{2}^{*}$, we have $0 < \theta^{*} < 1$. Then, we can derive $\frac{{\partial \theta^{*} }}{{\partial \alpha_{1} }} = \frac{{h(\alpha_{1} ,\alpha_{2} )}}{{(12 - \alpha_{1}^{2} - (6 - 2\alpha_{2} )\alpha_{1} - \alpha_{2}^{2} - 6\alpha_{2} )^{2} }}\;\,\frac{ > }{ < }\;\,0$ where $h(\alpha_{1} ,\alpha_{2} ) = (12\alpha_{2} - 38)\alpha_{1}^{2} + (128 - 76\alpha_{2} )\alpha_{1} - 12\alpha_{2}^{3} + 42\alpha_{2}^{2} + 16\alpha_{2} - 72$, which is a convex function of $\alpha_{1}$. Define $\hat{\alpha }_{1} (\alpha_{2} ) = \frac{{32 - 19\alpha_{2} \pm 2\sqrt {(1 - \alpha_{2} )^{2} (9\alpha_{2}^{2} - 42\alpha_{2} + 85)} }}{{19 - 6\alpha_{2} }}$ which satisfies $h(\alpha_{1} ,\alpha_{2} ) = 0$. Then, $0 < \hat{\alpha }_{1}^{ - } (\alpha_{2} ) = \frac{{32 - 19\alpha_{2} - 2\sqrt {(1 - \alpha_{2} )^{2} (9\alpha_{2}^{2} - 42\alpha_{2} + 85)} }}{{19 - 6\alpha_{2} }} < 1$ and $\hat{\alpha }_{1}^{ + } (\alpha_{2} ) = \frac{{32 - 19\alpha_{2} + 2\sqrt {(1 - \alpha_{2} )^{2} (9\alpha_{2}^{2} - 42\alpha_{2} + 85)} }}{{19 - 6\alpha_{2} }}$ > 1 for $\alpha_{2} \in (0,1)$. Therefore, $\alpha_{1} \frac{ < }{ > }\hat{\alpha }_{1}$ if $\frac{{\partial \theta^{*} }}{{\partial \alpha_{1} }}\frac{ < }{ > }0$ with $0 < \hat{\alpha }_{1} (\alpha_{2} ) < 1$ for all $\alpha_{2}$. □

1.6 A5. Proof of Proposition 5

Using the same procedures in Proposition 1, we show that condition $\left. {\frac{\partial W(\theta )}{\partial \theta }} \right|_{\theta = 1} \ge 0$ never holds. Also, condition $\left. {\frac{\partial W(\theta )}{\partial \theta }} \right|_{\theta = 0} \le 0$ shows that $\alpha \ge \frac{1}{3}\;$. Otherwise, the first-order condition for an interior solution should satisfy the following condition: $\frac{\partial W}{\partial \theta } = \frac{{A^{2} (1 - \alpha )(1 - 3\alpha - 3\theta )}}{{2(3 + 2\theta - \alpha - \alpha \theta )^{3} }} = 0$. Then, we have $\theta^{*} = \frac{1}{3} - \alpha$. The second-order condition, $\frac{{\partial^{2} W(\theta^{*} )}}{{\partial \theta^{2} }} \le 0$, is satisfied. □

1.7 A6. Proof of Proposition 6

The first-order condition for an interior solution should satisfy the following condition: $\frac{\partial W}{\partial \theta } = \frac{{A^{2} (2 - \alpha_{1} )[(\alpha_{1}^{2} + 6\alpha_{1} - 12)\theta + 5\alpha_{1}^{2} - 8\alpha_{1} + 4]}}{{2(6 + 4\theta - \alpha_{1} - \alpha_{1} \theta )^{3} }} = 0$. Note that the second-order condition is satisfied. From (13), we obtain $\theta^{*} = \frac{{5\alpha_{1}^{2} - 8\alpha_{1} + 4}}{{12 - (6 + \alpha_{1} )\alpha_{1} }}$ where $0 < \theta^{*} < 1\;$ for $\alpha_{1} \in [0,1]$. We can also derive $0 < \hat{\alpha } = \frac{{32 - 2\sqrt {85} }}{19} = 0.713$, which satisfies $\frac{{\partial \theta^{*} }}{{\partial \alpha_{1} }} = 0$. Then, we have $\frac{{\partial \theta^{*} }}{{\partial \alpha_{1} }} = - \frac{{2(19\alpha_{1}^{2} - 64\alpha_{1} + 36)}}{{(\alpha_{1}^{2} + 6\alpha_{1} - 12)^{2} }}\tfrac{ < }{ > }0$ when $\alpha_{1} \frac{ < }{ > }\hat{\alpha }$. □

Appendix B: Alternative Approach on Optimal Privatization Policies

We will provide the comparable analysis with Proposition 2 in which the optimal policies are determined by $\alpha_{i}$, given $k_{0}$:

Proposition B1

Define$B(k_{0} ,\;\alpha_{1} ) = 4(3k_{0} + 1)\alpha_{1} - 4(2k_{0} + 1)\alpha_{1}^{2} - 3k_{0} + 1$. Then, if$B(k_{0} ,\;\alpha_{1} ) \le 0$, $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) \ge 0$. However, if$B(k_{0} ,\;\alpha_{1} ) > 0$, we have$f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) < 0$if one of the following conditions holds:

• (B1-1) For any$k_{0}$, $\alpha_{1}^{**} \le \alpha_{1} \le \alpha_{1}^{*}$and$\alpha_{2}^{ - } < \alpha_{2} < \alpha_{2}^{ + }$;

• (B1-2) For$k_{0} > 1/3$, $\alpha_{1} > \alpha_{1}^{*}$and$\alpha_{2} > \alpha_{2}^{ - }$;

• (B1-3) For$k_{0} \le 1/3$, (i)$\alpha_{1}^{ - } \le \alpha_{1} \le \alpha_{1}^{ + }$and$0 < \alpha_{2} < 1$, (ii)$\alpha_{1}^{*} < \alpha_{1} < \alpha_{1}^{ - }$or$\alpha_{1} > \alpha_{1}^{ + }$and$\alpha_{2} > \alpha_{2}^{ - }$.

Proof of Proposition B1

Since $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )$ is a convex function of $\alpha_{2}$, we can easily show that $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) \ge 0$ if $B(k_{0} ,\;\alpha_{1} ) \le 0$. If $B(k_{0} ,\;\alpha_{1} ) > 0$, we can define $\;\alpha_{2}^{*} (k_{0} ,\alpha_{1} )$ that satisfies $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) = 0$, which is either $\alpha_{2}^{ + } = \frac{{1 + (3 - \alpha_{1} )k_{0} + \sqrt {(k_{0} + 1) \cdot B(k_{0} ,\;\alpha_{1} )} }}{{2 + 3k_{0} }}$ or $\alpha_{2}^{ - } = \frac{{1 + (3 - \alpha_{1} )k_{0} - \sqrt {(k_{0} + 1) \cdot B(k_{0} ,\;\alpha_{1} )} }}{{2 + 3k_{0} }}$. Then, $\alpha_{2}^{ + } \,\frac{ > }{ < }\;\,1$ if $B(k_{0} ,\;\alpha_{1} )\;\,\frac{ > }{ < }\;\,\frac{{(1 + \alpha_{1} k_{0} )^{2} }}{{1 + k_{0} }}$ and $\alpha_{2}^{ - } \;\,\frac{ < }{ > }\;\;\,0$ if $B(k_{0} ,\;\alpha_{1} )\;\,\frac{ > }{ < }\;\,\frac{{[1 + (3 - \alpha_{1} )k_{0} ]^{2} }}{{1 + k_{0} }}$ where $\frac{{(1 + \alpha_{1} k_{0} )^{2} }}{{1 + k_{0} }} < \,\frac{{[1 + (3 - \alpha_{1} )k_{0} ]^{2} }}{{1 + k_{0} }}$ for $\alpha_{1} \in (0,1)$. Thus, we have the following relations:

• (1) If $B(k_{0} ,\;\alpha_{1} ) \ge \frac{{[1 + (3 - \alpha_{1} )k_{0} ]^{2} }}{{1 + k_{0} }}$, then $\alpha_{2}^{ - } < 0$ and $\alpha_{2}^{ + } > 1$. Thus, $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) < 0$. This case happens when $\alpha_{1}^{ - } = \frac{{1 + 3k_{0} - \sqrt {(k_{0} + 1)(1 - 3k_{0} )} }}{{2 + 3k_{0} }} \le \alpha_{1} \le \alpha_{1}^{ + } = \frac{{1 + 3k_{0} + \sqrt {(k_{0} + 1)(1 - 3k_{0} )} }}{{2 + 3k_{0} }}$ for $k_{0} \le 1/3$.

• (2) If $\frac{{(1 + \alpha_{1} k_{0} )^{2} }}{{1 + k_{0} }} < B(k_{0} ,\;\alpha_{1} ) < \,\frac{{[1 + (3 - \alpha_{1} )k_{0} ]^{2} }}{{1 + k_{0} }}$, then $0 < \alpha_{2}^{ - } < 1 < \alpha_{2}^{ + }$. Thus, $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )\frac{ > }{ < }0$.if $\alpha_{2} \frac{ < }{ > }\alpha_{2}^{ - }$. This case happens when (i) $\alpha_{1} > \alpha_{1}^{*} = \frac{{k_{0} }}{{2 + 3k_{0} }}$ for $k_{0} > 1/3$ and (ii) either $\alpha_{1}^{*} < \alpha_{1} < \alpha_{1}^{ - }$ or $\alpha_{1} > \alpha_{1}^{ + }$ for $k_{0} \le 1/3$.

• (3) If $B(k_{0} ,\;\alpha_{1} ) \le \frac{{(1 + \alpha_{1} k_{0} )^{2} }}{{1 + k_{0} }}$, then $0 < \alpha_{2}^{ - } < \alpha_{2}^{ + } < 1$. Thus, $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) < 0$ if $\;\alpha_{2} \in (\alpha_{2}^{ - } ,\alpha_{2}^{ + } )$. This case happens when $\alpha_{1}^{**} = \frac{{1 + 3k_{0} - \sqrt {(k_{0} + 1)(3k_{0} + 2)} }}{{2(1 + 2k_{0} )}} \le \alpha_{1} \le \alpha_{1}^{*}$ for any $k_{0}$.

Therefore, summarizing the cases for $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )$ < 0 yields the Proposition B1. □

Proposition B1 states (i) nationalization is never optimal if $k_{0}$ is sufficiently large and $\alpha_{i}$ is small, that is, $B(k_{0} ,\;\alpha_{1} ) \le 0$. However, (ii) nationalization is always optimal if $k_{0}$ is sufficiently small and $\alpha_{i}$ respects certain conditions (see Fig. 4). Note that we define: $\alpha_{1}^{*} = \frac{{k_{0} }}{{2 + 3k_{0} }},\;\alpha_{1}^{**} = \frac{{1 + 3k_{0} - \sqrt {(k_{0} + 1)(3k_{0} + 2)} }}{{2(1 + 2k_{0} )}},\;\alpha_{1}^{ + } = \frac{{1 + 3k_{0} + \sqrt {(k_{0} + 1)(1 - 3k_{0} )} }}{{2 + 3k_{0} }},\;\alpha_{1}^{ - } = \frac{{1 + 3k_{0} - \sqrt {(k_{0} + 1)(1 - 3k_{0} )} }}{{2 + 3k_{0} }},\;\alpha_{2}^{ + } = \frac{{1 + (3 - \alpha_{1} )k_{0} + \sqrt {(k_{0} + 1) \cdot B(k_{0} ,\;\alpha_{1} )} }}{{2 + 3k_{0} }},\;{\text{and}}\;\alpha_{2}^{ - } = \frac{{1 + (3 - \alpha_{1} )k_{0} - \sqrt {(k_{0} + 1) \cdot B(k_{0} ,\;\alpha_{1} )} }}{{2 + 3k_{0} }}.$

Fig. 4

The value of $\alpha_{2}$ under $f(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )$

Proposition B2

Define$C(k_{0} ,\alpha_{1} ) = 40 - 4(2k_{0} + 1)(k_{0} + 2)\alpha_{1}^{2} + 4(3k_{0} - 2)(k_{0} + 2)\alpha_{1} - 3k_{0} (k_{0} - 4)$. Then, if$C(k_{0} ,\;\alpha_{1} ) > 0$,$g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) > 0$. However, if$C(k_{0} ,\;\alpha_{1} ) \le 0$, we have$g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) \le 0$if$0 < k_{0} \le \frac{{7 - 2\alpha_{1} }}{{3 - 2\alpha_{1} }}$while$g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) > 0$if one of the following conditions holds:

• (B2-1) For$\frac{{7 - 2\alpha_{1} }}{{3 - 2\alpha_{1} }} < k_{0} < \frac{{6 + 2\sqrt {39} }}{3}$, $\hbox{max} [0,\;\,\alpha_{1}^{*} ] < \alpha_{1} \le \tilde{\alpha }_{1}^{ - }$or$\alpha_{1} > \tilde{\alpha }_{1}^{ + }$and$\alpha_{2} < \tilde{\alpha }_{2}^{ - }$;

• (B2-2) For$k_{0} \ge \frac{{6 + 2\sqrt {39} }}{3}$, (i)$\alpha_{1} > \tilde{\alpha }_{1}^{*}$and$\alpha_{2} < \tilde{\alpha }_{2}^{ - }$, (ii)$\tilde{\alpha }_{1}^{**} \le \alpha_{1} \le \tilde{\alpha }_{1}^{*}$and$\alpha_{2} < \tilde{\alpha }_{2}^{ - }$or$\alpha_{2} > \tilde{\alpha }_{2}^{ + }$.

Proof of Proposition B2

Since $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )$ is a convex function of $\alpha_{2}$, we can easily show that $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) > 0$ if $C(k_{0} ,\;\alpha_{1} ) < 0$. If $C(k_{0} ,\;\alpha_{1} ) \ge 0$, we can define $\;\alpha_{2}^{**} (k_{0} ,\alpha_{1} )$ that satisfies $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) = 0$, which is either $\tilde{\alpha }_{2}^{ + } (k_{0} ,\alpha_{1} ) = \frac{{(3 - \alpha_{1} )k_{0} + \alpha_{1} - 2 + \sqrt {C(k_{0} ,\alpha_{1} )} }}{{3(1 + k_{0} )}}$ or $\tilde{\alpha }_{2}^{ - } (k_{0} ,\alpha_{1} ) = \frac{{(3 - \alpha_{1} )k_{0} + \alpha_{1} - 2 - \sqrt {C(k_{0} ,\alpha_{1} )} }}{{3(1 + k_{0} )}}$. Then, $\tilde{\alpha }_{2}^{ + } \;\,\frac{ > }{ < }\;\,1$ if $C(k_{0} ,\alpha_{1} )\;\,\frac{ > }{ < }\;\,\left[ {\alpha_{1} (k_{0} - 1) + 5} \right]^{2}$ and $\tilde{\alpha }_{2}^{ - } \;\,\frac{ > }{ < }\;\,0$ if $C(k_{0} ,\alpha_{1} )\;\,\;\frac{ < }{ > }\;\;\,\left[ {(3 - \alpha_{1} )k_{0} + \alpha_{1} - 2} \right]^{2}$ where $\left[ {\alpha_{1} (k_{0} - 1) + 5} \right]^{2} \;\,\frac{ > }{ < }\,\;\,\left[ {(3 - \alpha_{1} )k_{0} + \alpha_{1} - 2} \right]^{2}$ if $\alpha_{1} \frac{ > }{ < }\,\frac{{3k_{0} - 7}}{{2(k_{0} - 1)}}$. Thus, we have the following relations:

• (1) If $\alpha_{1} > \,\frac{{3k_{0} - 7}}{{2(k_{0} - 1)}}$, equivalently, if $1 < k_{0} < \frac{{7 - 2\alpha_{1} }}{{3 - 2\alpha_{1} }}$, we have $C(k_{0} ,\alpha_{1} )\; \ge \left[ {\alpha_{1} (k_{0} - 1) + 5} \right]^{2}$ and $C(k_{0} ,\alpha_{1} )\;\,\; > \;\;\,\left[ {(3 - \alpha_{1} )k_{0} + \alpha_{1} - 2} \right]^{2}$. Then, $\tilde{\alpha }_{1}^{ - } = \frac{{3k_{0} - 2 - \sqrt {40 - 3k_{0}^{2} + 12k_{0} } }}{{3(1 + k_{0} )}} < 0$ and $\tilde{\alpha }_{1}^{ + } = \frac{{3k_{0} - 2 + \sqrt {40 - 3k_{0}^{2} + 12k_{0} } }}{{3(1 + k_{0} )}} > 1$. Thus, $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) < 0$ for $0 < \alpha_{1} < 1$.

• (2) If $\alpha_{1} < \,\frac{{3k_{0} - 7}}{{2(k_{0} - 1)}}$, equivalently, if $0 < k_{0} < 1$ or $k_{0} > \frac{{7 - 2\alpha_{1} }}{{3 - 2\alpha_{1} }}$, we have $C(k_{0} ,\alpha_{1} )\; \ge \left[ {\alpha_{1} (k_{0} - 1) + 5} \right]^{2}$. We have three cases.

• (2-1) If $C(k_{0} ,\alpha_{1} )\; \ge \left[ {(3 - \alpha_{1} )k_{0} + \alpha_{1} - 2} \right]^{2}$, then $\tilde{\alpha }_{2}^{ - } < 0$ and $\tilde{\alpha }_{2}^{ + } > 1$. Thus, $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) < 0$. This happens when $\tilde{\alpha }_{1}^{ - } \le \alpha_{1} \le \tilde{\alpha }_{1}^{ + }$.

• (2-2) If $\,\left[ {\alpha_{1} (k_{0} - 1) + 5} \right]^{2} < C(k_{0} ,\alpha_{1} ) < \,\left[ {(3 - \alpha_{1} )k_{0} + \alpha_{1} - 2} \right]^{2}$, then $0 < \tilde{\alpha }_{2}^{ - } < 1 < \tilde{\alpha }_{2}^{ + }$. Thus, $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )$$\frac{ > }{ < }0$ if $\alpha_{2} \frac{ < }{ > }\tilde{\alpha }_{2}^{ - }$. This case happens when (i) either $\hbox{max} [0,\,\;\alpha_{1}^{*} ] < \alpha_{1} < \tilde{\alpha }_{1}^{ - }$ or $\alpha_{1} > \tilde{\alpha }_{1}^{ + }$ for $\frac{{7 - 2\alpha_{1} }}{{3 - 2\alpha_{1} }} < k_{0} < \frac{{6 + 2\sqrt {39} }}{3}$ and (ii) $\alpha_{1} > \tilde{\alpha }_{1}^{*}$ for $k_{0} > \frac{{6 + 2\sqrt {39} }}{3}$.

• (2-3) If $C(k_{0} ,\alpha_{1} )\; \le \;\,\left[ {\alpha_{1} (k_{0} - 1) + 5} \right]^{2}$, then $0 < \tilde{\alpha }_{2}^{ - } < \tilde{\alpha }_{2}^{ + } < 1$. Thus, $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) > 0$ either if $\alpha_{2} < \tilde{\alpha }_{2}^{ - }$ or $\alpha_{2} > \tilde{\alpha }_{2}^{ + }$. This case happens when $\,\tilde{\alpha }_{1}^{**} = \frac{{3k_{0}^{2} + 4k_{0} - 4 - (4 + k_{0} )\sqrt {3(k_{0} + 2)(k_{0} + 1)} }}{{2(k_{0} + 2)(2k_{0} + 1)}} \le \alpha_{1} \le \tilde{\alpha }_{1}^{*} .$

Therefore, summarizing the cases for $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} ) > 0$ yields the Proposition B2. □

Proposition B2 states (i) full privatization is never optimal if $k_{0}$ is sufficiently small, that is, $0 < k_{0} \le \frac{{7 - 2\alpha_{1} }}{{3 - 2\alpha_{1} }}$. However, (ii) full privatization is always optimal if $k_{0}$ is sufficiently large and $\alpha_{i}$ follows certain conditions (see Fig. 5). Note that we define: $\tilde{\alpha }_{1}^{*} = \frac{{k_{0} - 5}}{{3k_{0} + 1}}$, $\tilde{\alpha }_{1}^{**} \, = \frac{{3k_{0}^{2} + 4k_{0} - 4 - (4 + k_{0} )\sqrt {3(k_{0} + 2)(k_{0} + 1)} }}{{2(k_{0} + 2)(2k_{0} + 1)}},\;\tilde{\alpha }_{1}^{ + } = \frac{{3k_{0} - 2 + \sqrt {40 - 3k_{0}^{2} + 12k_{0} } }}{{3(1 + k_{0} )}},\;\tilde{\alpha }_{1}^{ - } = \frac{{3k_{0} - 2 - \sqrt {40 - 3k_{0}^{2} + 12k_{0} } }}{{3(1 + k_{0} )}},\;\tilde{\alpha }_{2}^{ + } = \frac{{(3 - \alpha_{1} )k_{0} + \alpha_{1} - 2 + \sqrt {C(k_{0} ,\alpha_{1} )} }}{{3(1 + k_{0} )}},\;{\text{and}}\;\tilde{\alpha }_{2}^{ - } = \frac{{(3 - \alpha_{1} )k_{0} + \alpha_{1} - 2 - \sqrt {C(k_{0} ,\alpha_{1} )} }}{{3(1 + k_{0} )}}.$

Fig. 5

The value of $\tilde{\alpha }_{2}$ under $g(k_{0} ,\;\alpha_{1} ,\;\alpha_{2} )$

2.1 Appendix C: Rigorous explanations on Proposition 6

We present a more detailed explanation. We first examine the sign of $\frac{{\partial \theta^{*} }}{{\partial \alpha_{1} }}$. Let $W_{\theta } (\alpha_{1} )$ satisfy the optimal condition in $\frac{\partial W}{\partial \theta } = 0$. Then, from the implicit function theorem and the second-order condition of optimality, the sign of $\frac{{\partial \theta^{*} }}{{\partial \alpha_{1} }}$ is the same as that of $\frac{{\partial W_{\theta } }}{{\partial \alpha_{1} }}$. Then, we obtain the following relations:

$$ \frac{{\partial W_{\theta } }}{{\partial \alpha_{1} }}\frac{ > }{ < }0\;\;{\text{if}}\;\alpha_{1} \frac{ > }{ < }\theta_{W} = \frac{{2\theta^{2} + 13\theta + 22 - \sqrt {28\theta^{4} + 164\theta^{3} + 333\theta^{2} + 260\theta + 52} }}{{2(2\theta^{2} + 8\theta + 9)}} $$

The result confirms Proposition 4 in that the relation between $\alpha_{1}$ and $\theta^{*}$ is U-shaped, depending on the degree of CSR, $\alpha_{1}$. Using this fact, we can provide the economic explanation with regard to Proposition 6.

When $\alpha_{1}$ increases, $q_{1}$ increases but both $q_{2}$ and $q_{0}$ decrease. Then, we have $\left| {\frac{{\partial q_{1} }}{{\partial \alpha_{1} }}} \right| > \left| {\frac{{\partial q_{ - 1} }}{{\partial \alpha_{1} }}} \right|$. Again, a higher degree of CSR activities by firm 1 leads to an increase in the total industry output, and thus, the production substitution effect is beneficial to consumer surplus. However, when $\alpha_{1}$ increases, firm 1 has to pay a higher cost while its rival firms can save their costs. This represents the existence of the cost reallocation effect. Therefore, the welfare effect also depends on the trade-off between the production substitution effect and the cost reallocation effect. In particular, if $\alpha_{1} < \theta_{W}$, the relation between $\alpha_{1}$ and $\theta^{*}$ is negative, and thus the production substitution effect is dominated by the cost reallocation effect. However, the production substitution effect dominates the cost reallocation effect if $\alpha_{1} < \tilde{\theta }_{1}$, while the production substitution effect is outweighed by the cost reallocation effect if $\alpha_{1} > \tilde{\theta }_{1}$ in Fig. 3.

Therefore, when deciding the optimal degree of privatization, the welfare effect depends on $\alpha_{1}$. On the one hand, when $\alpha_{1} < \tilde{\theta }_{1}$, and thus when the public firm produces more output than the other private firms, welfare increases as $\alpha_{1}$ increases because the production substitution effect dominates cost reallocation effect. Thus, the government decreases the degree of privatization in order to increase the outputs of firm 0 and the industry, that is, $\;\left| {\frac{{\partial q_{0} }}{\partial \theta }} \right| > \left| {\frac{{\partial q_{ - 0} }}{\partial \theta }} \right|$. This will decrease the marginal output substitution effect but increase the marginal cost reallocation effect, so that the two marginal benefits will be the same at a lower level of privatization.

On the other hand, when $\alpha_{1} > \tilde{\theta }_{1}$ and thus the CSR-firm 1 produces more output than the other private and public firms, welfare decreases as $\alpha_{1}$ increases because the production substitution effect is outweighed by cost reallocation effect. In that case, we have two cases.

First, when $\tilde{\theta }_{1} < \alpha < \hat{\alpha }$, at which $\hat{\alpha }$ satisfies $\theta^{*} = \theta_{W}$, welfare decreases as $\alpha_{1}$ increases. In that case, the output difference between CSR-firm 1 and the public firm is relatively small. Thus, the government still decreases the degree of privatization in order to increase the outputs of firm 0 and the industry, that is, $\;\left| {\frac{{\partial q_{0} }}{\partial \theta }} \right| > \left| {\frac{{\partial q_{ - 0} }}{\partial \theta }} \right|$. This will increase the marginal production substitution effect, which is larger than the increment of the marginal cost reallocation effect. Thus, the two marginal benefits will be the same at a lower level of privatization.

Second, however, when $\hat{\alpha }$ < $\alpha_{1}$, welfare increases as $\alpha_{1}$ increases. In that case, the output difference between CSR-firm 1 and the public firm is relatively large. Then, the government increases the degree of privatization in order to decrease the output of firm 0, which induces an increase in other firms’ outputs but a decrease in industry outputs, that is, $\left| {\frac{{\partial q_{ - 0} }}{\partial \theta }} \right| < \left| {\frac{{\partial q_{0} }}{\partial \theta }} \right|$. This will decrease marginal benefit of the production substitution effect. In that case, however, the CSR-firm is less sensitive to the change in the public firm, and thus, it will decrease the marginal cost reallocation effect even further. Thus, the two marginal benefits will be the same at a higher degree of privatization. Thus, the positive relations between $\alpha_{1}$ and $\theta^{*}$ becomes steeper, as $\alpha_{1}$ increases. Therefore, $\theta^{*}$ takes a U-shape in $\alpha_{1}$.