Liberalisation, competition and ownership in the presence of vertical relations

Abstract

This contribution analyses a market with an upstream bottleneck monopoly and a downstream activity that may either be vertically integrated or separated. Separation always reduces the consumer surplus, and the total surplus unless there are large cost reductions. Downstream competition from a public or private network monopoly would crowd out other firms, also when public ownership is associated with more modest objectives than welfare-maximisation. A market is therefore less likely to remain a mixed oligopoly than without vertical relations. However, private firms would survive in a moderately welfare-improving mixed oligopoly with cross-subsidisation and access charges equal to marginal costs.

Appendix

Proof of Proposition 1 (a)

The prices in a vertically integrated monopoly and an n-firm Cournot-oligopoly are r ^M = (a + zc)/2, and r ^C = [(n + 2)a + nzc]/2(n + 1) respectively (see Sect. 2); it is obvious that r ^C > r ^M. Separation and competition then unambiguously reduce the total surplus:

$$ TS^{C}-TS^{M}=-\frac{(2n+3)(a-cz)^{2}}{8(n+1)^{2}}-(n-1)F. $$

(A.1)

As for free entry, it is obvious that the sign of (A.1) does not change by substituting $[(a-cz)/2\sqrt{F}]-1$ for n.

(b) The result follows from straightforward manipulation of (5) and (11) and substituting n* for $[(a-cz)/2\sqrt{F}].$ □

Proof of Proposition 2 (a)

Suppose that sunk costs in the public firm are denoted F ^G and F ^G₀ respectively, so that (6) becomes

$$ y^{PM}=\frac{a-cz}{2}+\sqrt{\left( {\frac{a-cz}{2}} \right)^{2}-F_0^G -F^{G}}. $$

(A.2)

As the consumer surplus is equal to y ²/2, it is sufficient to compare (A.2) to the output levels y ^M, y ^C, and y ^D respectively. It is obvious that a sufficient condition for part a) to hold true is that the square root in (A.2) greater than zero.

(b) Suppose that n firms would be able to break even after privatisation, vertical separation and free entry, so that the downstream sunk costs are $F\approx(a-cz)^{2}/4(n+1)^{2}$ and that but n(a − cz)²/4(n + 1)² in the integrated public monopoly. Use this to eliminate F from (6) and (10) and rearrange so that we get the following expressions for the total surplus:

$$ TS^{PM}=\frac{1}{2}\left[{\frac{a-cz}{2}+\sqrt{\frac{(n^{2}+n+1) (a-cz)^{2}}{4(n+1)^{2}}-F_0 }}\right]^{2}. $$

(A.3)

$$ TS^{D}=\frac{(3n^{2}+2n)(a-cz)^{2}}{8(n+1)^{2}}-F_0. $$

(A.4)

Note that the public monopoly would still be able to break even, because the expression in the square root in (A.3) is positive if (A.4) is positive (or if a private monopolist can break even if n = 1). Suppose now that Proposition 2a is false and rearrange the antithesis as

$$ \frac{(n^{2}-n-2)(a-cz)^{2}}{4(n+1)^{2}}-F_0 -(a-cz)\sqrt{\frac{(n^{2}+n+1)(a-cz)^{2}}{4(n+1)^{2}}-F_0} > 0, $$

(A.5)

which is equivalent to:

$$ \left[{\sqrt{\frac{(n^{2}+n+1)(a-cz)^{2}}{4(n+1)^{2}}-F_0 }} \right]\left[ {-(a-cz)+\sqrt{\frac{(n^{2}+n+1)(a-cz)^{2}}{4(n+1)^{2}}-F_0 }} \right]-\frac{(4n-1)(a-cz)^{2}}{4(n+1)^{2}} > 0. $$

(A.6)

However, this cannot be true, for it can easily be verified that the second bracket from the left must be negative, so the antithesis is false. □

To see that the total surplus is higher in the vertically integrated public monopoly than in a profit-maximising monopoly or after separation and free entry, suppose that TS ^PM < TS ^M, as expressed by (5) and (9). This would mean

$$ \left[{\left({\sqrt{\left({\frac{a-cz}{2}}\right)^{2}-F-F_0}} \right)-(a-cz)}\right]\sqrt{\left({\frac{a-cz}{2}} \right)^{2}-F-F_0} > 0, $$

(A.7)

which cannot hold true. The fact that TS ^M > TS ^C and TS ^M > TS ^C (Proposition 1) mean that TS ^PM > TS ^C and TS ^PM > TS ^C. Part (a) is thereby proved.

Proof of Proposition 3

Inserting the solutions for y ₀ and ŷ given p into (18) and differentiating with respect to p yields p = [(1 − ρ)a + cz]/(2 − ρ)z, for the Cournot-case, so the public firm would produce y ₀ = (a − cz)/(2 − ρ), whereas the private firms would produce a zero output. Routine calculations yield the same solution for the Stackelberg case. □

Proof of Proposition 4 (a)

Note that the total surplus in a conventional oligopoly with n = m can be written

$$ TS^{C}\approx \frac{3m^{2}+2m}{2}F-F_0 $$

(A.8)

because of (25). As for the mixed oligopoly, rearrange the consumer surplus $(\hat{{y}}^{MO}+y_0^{MO})^{2}/2$ using (25):

$$ CS^{MO}\approx \frac{(2m+1)^{2}}{2}F. $$

(A.9)

Note that the profits are $(y_0\hat{{y}}/m)-F-F_0,$ because r − cz = y _i . Using (25) yields:

$$ \pi_0^{MO}\approx mF-F_0. $$

(A.10)

Add (A.9) and (A.10) and compare to (A.8). It follows that the mixed oligopoly increases the total surplus as compared to a conventional downstream oligopoly if

$$ m^{2}+4m+1 > 0, $$

(A.11)

which is true for all positive m. Part (a) is thereby proved. Part (b) is obvious because of (26) and the fact that profits are nonnegative; part (c) follows directly from (A.10). □