An analysis of entry-then-privatization model: welfare and policy implications

Abstract

This study formulates a new model of mixed oligopolies in free entry markets. A state-owned public enterprise is established before the game, private enterprises enter the market, and then the government chooses the degree of privatization of the public enterprise (termed the entry-then-privatization model herein). We find that under general demand and cost functions, the timing of privatization does not affect consumer surplus or the output of each private firm, while it does affect the equilibrium degree of privatization, number of entering firms, and output of the public firm. The equilibrium degree of privatization is too high (low) for both domestic and world welfare if private firms are domestic (foreign).

Appendix

In the following proofs, we suppress the arguments of functions.

Proof of Result 1

First, note that $\sum _{i \ne 0}q_i = n q = Q-q_0$. By differentiating (2), (3), and (7), we obtain

$$\begin{aligned} H \begin{pmatrix} dq_0 \\ dq_1 \\ dQ \end{pmatrix} = - \begin{pmatrix} (1-\theta )p^{\prime }q_0+\theta p^{\prime }Q \\ 0 \\ 0 \end{pmatrix} d \alpha , \end{aligned}$$

(18)

where

$$\begin{aligned} H:= \begin{pmatrix} (1-(1-\alpha )(1-\theta ))p^{\prime } - c_0^{{\prime }{\prime }} &{} 0 &{} (1-(1-\alpha )\theta )p^{\prime } + (1-(1-\alpha )(1-\theta ))p^{{\prime }{\prime }}q_0 - (1-\alpha )\theta p^{{\prime }{\prime }}Q \\ 0 &{} p^{\prime }-c^{{\prime }{\prime }} &{} p^{\prime }+p^{{\prime }{\prime }} q \\ -1 &{}-n &{} 1 \end{pmatrix}. \end{aligned}$$

From (6) and the second-order condition for $q_0$, we obtain

$$\begin{aligned} |H|= & {} (1-(1-\alpha )(1-\theta ))p^{\prime } - c_0^{{\prime }{\prime }})(p^{\prime }-c^{{\prime }{\prime }}) \\&+ \biggr ((1-(1-\alpha )\theta )p^{\prime } + (1-(1-\alpha )(1-\theta ))p^{{\prime }{\prime }}q_0 -(1-\alpha )\theta p^{{\prime }{\prime }}Q\biggr ) (p^{\prime }-c^{{\prime }{\prime }}) \\&+\,n(p^{\prime }+p^{{\prime }{\prime }} q)((1-(1-\alpha )(1-\theta ))p^{\prime } - c_0^{{\prime }{\prime }}) \\= & {} \biggr ((1+\alpha )p^{\prime } - c_0^{{\prime }{\prime }} + \alpha p^{{\prime }{\prime }}q_0 - (1-\alpha ) \theta p^{{\prime }{\prime }}(Q-q_0)\biggr )(p^{\prime } - c^{{\prime }{\prime }}) \\&+\,n(p^{\prime } + p^{{\prime }{\prime }}q)((1-(1-\alpha )(1-\theta ))p^{\prime } - c_0^{{\prime }{\prime }})>0. \end{aligned}$$

By applying Cramer’s rule to (18), we obtain

$$\begin{aligned} \frac{dq_0}{d \alpha }= & {} - \frac{((1-\theta )p^{\prime }q_0+\theta p^{\prime }Q)(p^{\prime }-c^{{\prime }{\prime }}) +n(p^{\prime } +p^{{\prime }{\prime }}q)((1- \theta )p^{\prime }q_0 +\theta p^{\prime }Q)}{|H|}\\< & {} 0, \\ \frac{dq}{d \alpha }= & {} \frac{((1-\theta )p^{\prime }q_0+\theta p^{\prime }Q)(p^{\prime } +p^{{\prime }{\prime }}q)}{|H|} >0, \\ \frac{dQ}{d \alpha }= & {} - \frac{((1-\theta )p^{\prime }q_0+\theta p^{\prime }Q)(p^{\prime } -c^{{\prime }{\prime }})}{|H|}<0. \end{aligned}$$

$\square$

Proof of Result 2

By using (8), (2), and Result 1(ii), we obtain

$$\begin{aligned} \frac{d W}{d \alpha }\Bigl |_{\alpha =0} = n \Bigl (\frac{d q}{d \alpha }\Bigr )p^{\prime }(-\theta (Q-q_0) - (1-\theta )q)>0. \end{aligned}$$

This implies Result 2(i).

By substituting $\theta =1$ into (8) and using (2), we obtain

$$\begin{aligned} \frac{d W}{d \alpha }\Bigl |_{\alpha =1} = \Bigl (\frac{d q_0}{d \alpha }\Bigr )(-p^{\prime }Q) + n \Bigl (\frac{d q}{d \alpha }\Bigr )p^{\prime }(-Q+q_0). \end{aligned}$$

(19)

From Result 1(iii), we obtain $dq_0/d\alpha + n (dq/d\alpha ) <0.$ Because $p^{\prime }<0$ and $q_0 >0$, we find that (19) is negative and thus $\alpha =1$ is not an equilibrium outcome. $\square$

Proof of Result 3

Substituting $\theta =0$ into (8) yields

$$\begin{aligned} \frac{d q_0}{d \alpha } (p -c_0^{\prime }) -n\frac{d q}{d \alpha } p^{\prime }q=0. \end{aligned}$$

Because $d q_0/d \alpha <0$, $d q/d \alpha > 0$ (Result 1(ii)) and $p^{\prime }<0$, we find that $p-c_0^{\prime }$ is positive.

Substituting $\theta =1$ into (8) yields

$$\begin{aligned}&\frac{dq_0}{d \alpha } (-p^{\prime }Q+p-p^{\prime }q_0-c_0^{\prime }) - n \frac{d q}{d \alpha } (p^{\prime }(Q-q_0))\\&\quad = -\Bigl (\frac{dq_0}{d \alpha }+n \frac{d q}{d \alpha }\Bigr ) (p^{\prime }(Q-q_0)) +\frac{dq_0}{d \alpha } (p-c_0^{\prime })=0. \end{aligned}$$

From Result 1(iii), we obtain $dq_0/d\alpha + n (dq/d\alpha ) <0.$ From Result 1(i), we obtain $dq_0/d \alpha <0.$ Under these conditions, $p-c_0^{\prime }$ must be negative. $\square$

Proof of Result 4

In Eqs. (15) and (16), there are only two unknown variables, $Q^{**}$ and $q^{**}$. Thus, these two equations determine $Q^{**}$ and $q^{**}$. Because neither $\alpha ^{**}$ nor $\theta$ appears in these equations, $Q^{**}$ and $q^{**}$ must not depend on these two. This implies (i).

Because these two equations are common with equation system (9)–(13), $Q^{**}=Q^{*}$ and $q^{**}=q^{*}.$ This implies (ii).

By differentiating (14), we obtain

$$\begin{aligned} \frac{d q^{**}_0}{d \alpha ^{**}} = - \frac{p^{\prime }(q^{**}_0 + \theta n^{**}q^{**})}{(1-(1-\alpha ^{**})(1-\theta ))p^{\prime } - c_0^{{\prime }{\prime }}}<0. \end{aligned}$$

This implies (iii). Note that $Q^{**}$ is independent of $\alpha ^{**}$ and that $d q_0^{**}/d\alpha ^{**} = - q^{**}(d n^{**}/d\alpha ^{**})$ (because $Q^{**}=q_0^{**}+ n^{**}q^{**}$).

Because neither $Q^{**}$ nor $q^{**}$ depends on $\alpha ^{**}$ and $q_0^{**}$ is decreasing in $\alpha ^{**}$, (17) implies that $n^{**}$ is increasing in $\alpha ^{**}$. This implies (iv).

Obviously, if $\alpha ^{*}=\alpha ^{**}$, then $n^{*}=n^{**}$. Thus, Result 4(iv) implies Result 4(v). $\square$

Proof of Result 5

(i) is proven in the proof of Result 4(i).

Because (15) and (16) determine $q^{**}$ and $Q^{**}$, the remaining unknown variables $q_0^{**}$ and $n^{**}$ are determined by (14) and (16). By differentiating (14) and (17), we obtain

$$\begin{aligned} \begin{pmatrix} \alpha p^{\prime } - c_0^{{\prime }{\prime }} &{} -(1-\alpha )\theta p^{\prime }q^{**} \\ -1 &{} q^{**} \end{pmatrix} \begin{pmatrix} dq^{**}_0 \\ dn^{**} \end{pmatrix} = - \begin{pmatrix} (1-\alpha ^{**})n^{**}q^{**} \\ 0 \end{pmatrix} d \theta . \end{aligned}$$

(20)

By applying Cramer’s rule to (20), we obtain

$$\begin{aligned} \frac{dq^{**}_0}{d \theta }= & {} \frac{(1-\alpha ^{**})n^{**}q^{**}}{(1-(1-\alpha ^{**})(1-\theta ))p^{\prime } - c_0^{{\prime }{\prime }}} \ge 0, \end{aligned}$$

(21)

$$\begin{aligned} \frac{dn^{**}}{d \theta }= & {} \frac{(1-\alpha ^{**})n^{**}}{((1-(1-\alpha ^{**})(1-\theta ))p^{\prime } - c_0^{{\prime }{\prime }})} \le 0, \end{aligned}$$

(22)

and the equalities hold if and only if $\alpha ^{**}=1$. This implies (ii) and (iii). $\square$

Proof of Proposition 2

$$\begin{aligned} \frac{dW^{**}}{d \alpha ^{**}}\Bigl |_{\alpha ^{**} =\alpha ^{*}}= & {} \frac{d Q^{**}}{d \alpha ^{**}} \frac{\partial W^{**}}{\partial Q} + \frac{d q_0^{**}}{d \alpha ^{**}} \frac{\partial W^{**}}{\partial q_0}+ \frac{d q^{**}}{d \alpha ^{**}} \frac{\partial W^{**}}{\partial q} + \frac{d n^{**}}{d \alpha ^{**}} \frac{\partial W^{**}}{\partial n} \nonumber \\= & {} \frac{d q_0^{**}}{d \alpha ^{**}} (p-c_0^{\prime }) + \frac{d n^{**}}{d \alpha ^{**}} n^{**}(1-\theta ) (pq^{**} -c -K) \nonumber \\= & {} \frac{d q_0^{**}}{d \alpha ^{**}} (p-c_0^{\prime }), \end{aligned}$$

(23)

where we use Result 4 and (16). Thus, (23) is positive if and only if $p-c_0^{\prime } <0$. Result 3 implies Proposition 2. $\square$

Proof of Proposition 3

We show that $q^*_0$ is increasing in $\theta$. In the proof of Proposition 2, we showed that a marginal increase in $\alpha ^{**}$ from $\alpha ^{*}$ improves welfare if and only if $p-c_0^{\prime }<0$ when $\alpha ^{**}=\alpha ^{*}.$ Because p is independent of $\alpha$ and $c_0^{\prime }$ is increasing in $q_0$, the result that $q^*_0$ is increasing in $\theta$ implies Proposition 3.

By substituting $p(Q) = a-bQ$, $c_0(q_0) = (k_0/2)q^2_0$, and $c(q_i) = (k/2)q^2_i$ into (3) and using (7), we obtain

$$\begin{aligned} q = \frac{a- bq_0}{(n+1)b + k}, \end{aligned}$$

and thus

$$\begin{aligned} \frac{dq}{d\alpha } = -\frac{b}{(n+1)b+k}\frac{dq_0}{d\alpha }. \end{aligned}$$

(24)

We now consider the first-order condition for $\alpha$ in the privatization stage. By using (24), the first-order condition (8) for $\alpha$ is rewritten as

$$\begin{aligned}&\frac{dq_0}{d\alpha }\left( a - bq_0 - k_0q_0 - b(1-\theta )nq - \frac{nb}{(n+1)b+k}b\left( \theta (Q-q_0) \right. \right. \nonumber \\&\quad \left. \left. + (1-\theta )q\right) \right) =0. \end{aligned}$$

Since $dq_0/d\alpha <0$, and $q^*$ and $Q^*$ are independently determined by (11) and (12), the following two equations determine $q^*_0$ and $n^*$:

$$\begin{aligned}&a - bq^*_0 - k_0q^*_0 - b(1-\theta )n^*q^* - \frac{n^*b}{(n^*+1)b+k}b\left( \theta \left( Q^*-q^*_0\right) \right. \nonumber \\&\quad \left. +(1-\theta )q^*\right) =0 \end{aligned}$$

(25)

$$\begin{aligned}&Q^{*} = n^*q^* + q^*_0. \end{aligned}$$

(26)

By substituting $n^* = (Q^*-q^*_0)/q^*$ into (25), we obtain

$$\begin{aligned}&\left( a-bq^*_0 - k_0q^*_0 - b(1-\theta )(Q^* -q^*_0)\right) \left( \left( \frac{Q^*-q^*_0}{q^*}\right) b+k\right) \\&\quad -\frac{Q^*-q^*_0}{q^*}b^2\left( \theta \left( Q^*-q^*_0\right) + (1-\theta )q^*\right) = 0. \end{aligned}$$

By rearranging it, we obtain the quadratic equation

$$\begin{aligned} \begin{aligned}&bk_0 (q^*_0)^2 + \biggr [(b\theta + k_0)((Q^* + q^*)b + kq^*) \\&\quad +b(a-(1-\theta )bQ^*) b^2(2\theta Q^* * (1-\theta )q^*)\biggr ] q^*_0\\&\quad - (a-b(1-\theta )Q^*)((Q^*+q^*)b + kq^*) - Q^*b^2(\theta Q^* + (1-\theta )q^*) = 0. \end{aligned} \end{aligned}$$

We obtain the equilibrium output of the public firm $q^*_0$ from

$$\begin{aligned} q^*_0 = \frac{-E + \sqrt{E^2 - 4bk_0F}}{2bk_0}, \end{aligned}$$

(27)

where

$$\begin{aligned} E&:= (b\theta + k_0)((Q^* + q^*)b + kq^*) \\&+ b(a-(1-\theta )bQ^*) b^2(2\theta Q^* * (1-\theta )q^*) > 0 \end{aligned}$$

and

$$\begin{aligned} F :=- (a-b(1-\theta )Q^*)((Q^*+q^*)b + kq^*) - Q^*b^2(\theta Q^* + (1-\theta )q^*) <0. \end{aligned}$$

We obtain

$$\begin{aligned} \frac{\partial E}{\partial \theta } = b(2q^*b + kq^*) >0 \end{aligned}$$

and

$$\begin{aligned} \frac{\partial F}{\partial \theta } = -bQ^*(2q^*b + kq^*) = -Q^* \frac{\partial E}{\partial \theta }. \end{aligned}$$

Finally, we obtain

$$\begin{aligned} \frac{\partial q^*_0}{\partial \theta }= & {} \frac{-\frac{\partial E}{\partial \theta } + \left( \frac{\partial E}{\partial \theta }2E - 4bk_0 \frac{\partial F}{\partial \theta }\right) \frac{1}{2}\left( E^2 - 4bk_0F\right) ^{-1/2} }{2bk_0} \\= & {} \frac{\left( E^2 - 4bk_0F\right) ^{-1/2}\frac{\partial E}{\partial \theta }}{2bk_0}\left( 2bk_0Q^* + E - \sqrt{E^2 - 4bk_0}\right) \\= & {} \left( E^2 - 4bk_0F\right) ^{-1/2}\frac{\partial E}{\partial \theta }\left( Q^* - q^*_0\right) > 0. \end{aligned}$$

Thus, $q^*_0$ is increasing in $\theta$. $\square$