Post-merger internal organization in multitier decentralized supply chains

Abstract

This paper analyses the effects upon firms and consumers of horizontal mergers in a multitier decentralized supply chain with a finite number of players in each tier, when firms may opt for two different post-merger internal organization forms: multidivisional, in which separate divisions are kept, or traditional, with cost synergies. To this effect, we develop and solve a formal game theory-based Cournot model. The main results are: independently of the tier in which the merger takes place, higher synergies do not always lead to higher consumer welfare; despite the fact that the proposal of a traditional merger reveals significant cost savings consumer welfare may still decrease with the merger; traditional downstream mergers tend to be more profitable than traditional upstream ones; multidivisional mergers are always profitable.

Appendices

Appendix 1

Proof of Lemma 1

Making use of the results in Brito and Catalão-Lopes (2019) and observing that unitary costs in the downstream tier are equal to $v_{1}+P_{2}$, we know that after $t_{1}$ traditional mergers and $m_{1}$ multidivisional mergers the equilibrium quantity in tier 1 for the outsiders, $q_{1i}(P_{2})$, insider to a traditional merger, $q_{1t}(P_{2})$, and leader and follower in a multidivisional merger ($q_{1L}(P_{2})$ and $q_{1F}(P_{2})$, respectively) are:

$$\begin{aligned} q_{1i}(P_{2})&=q_{1F}(P_{2})=\frac{A_{1}-v_{1}-P_{2}-t_{1}s_{1}}{b_{1}\left( n_{1}+m_{1}-t_{1}+1\right) }\\ q_{1L}(P_{2})&=\frac{2\left( A_{1}-v_{1}-P_{2}-t_{1}s_{1}\right) }{b_{1}\left( n_{1}+m_{1}-t_{1}+1\right) }\\ q_{1t}(P_{2})&=\frac{A_{1}-v_{1}-P_{2}+s_{1}\left( n_{1}+m_{1} -2t_{1}+1\right) }{b_{1}\left( n_{1}+m_{1}-t_{1}+1\right) } \end{aligned}$$

from where total output is obtained:

$$\begin{aligned} Q_{1}(P_{2})&=(n_{1}-2t_{1}-2m_{1})q_{1i}(P_{2})+t_{1}q_{1t}(P_{2} )+m_{1}(q_{1F}(P_{2})+q_{1L}(P_{2}))\\&=\frac{\left( n_{1}+m_{1}-t_{1}\right) \left( A_{1}-v_{1}-P_{2}\right) +t_{1}s_{1}}{b_{1}\left( n_{1}+m_{1}-t_{1}+1\right) } \end{aligned}$$

Knowing that $Q_{1}=Q_{2}$ we can write $Q_{2}$ as

$$\begin{aligned} Q_{2}(P_{2})=\frac{\left( A_{1}-v_{1}\right) \left( m_{1}+n_{1} -t_{1}\right) +s_{1}t_{1}}{b_{1}\left( m_{1}+n_{1}-t_{1}+1\right) } -P_{2}\frac{m_{1}+n_{1}-t_{1}}{b_{1}\left( m_{1}+n_{1}-t_{1}+1\right) } \end{aligned}$$

or, inverting,

$$\begin{aligned} P_{2}=\frac{\left( m_{1}+n_{1}-t_{1}\right) \left( A_{1}-v_{1}\right) +s_{1}t_{1}}{m_{1}+n_{1}-t_{1}}-\frac{m_{1}+n_{1}-t_{1}+1}{m_{1}+n_{1}-t_{1} }b_{1}Q_{2}. \end{aligned}$$

Thus, the demand in tier 2 is given by $P_{2}=A_{2}-b_{2}Q_{2}$, with $A_{2}=\frac{\left( m_{1}+n_{1}-t_{1}\right) \left( A_{1}-v_{1}\right) +s_{1}t_{1}}{m_{1}+n_{1}-t_{1}}$ and $b_{2}=\frac{m_{1}+n_{1}-t_{1}+1}{m_{1}+n_{1}-t_{1}}b_{1}$.

Replacing into the equilibrium expressions for the equilibrium outputs of the different types of firms in the second tier, we obtain

$$\begin{aligned} q_{2i}&=q_{2F}=\frac{A_{2}-v_{2}-t_{2}s_{2}}{b_{2}\left( n_{2} +m_{2}-t_{2}+1\right) }\\ q_{2L}&=\frac{2\left( A_{2}-v_{2}-t_{2}s_{2}\right) }{b_{2}\left( n_{2}+m_{2}-t_{2}+1\right) }\\ q_{2t}&=\frac{A_{2}-v_{2}+s_{2}\left( n_{2}+m_{2}-2t_{2}+1\right) }{b_{2}\left( n_{2}+m_{2}-t_{2}+1\right) } \end{aligned}$$

which simplifies to:

$$\begin{aligned} q_{2i}&=q_{2F}=\frac{\left( m_{1}+n_{1}-t_{1}\right) \left( A_{1} -v_{1}-v_{2}-s_{2}t_{2}\right) +s_{1}t_{1}}{b_{1}\left( m_{1}+n_{1} -t_{1}+1\right) \left( m_{2}+n_{2}-t_{2}+1\right) }\\ q_{2L}&=2\frac{\left( m_{1}+n_{1}-t_{1}\right) \left( A_{1}-v_{1} -v_{2}-s_{2}t_{2}\right) +s_{1}t_{1}}{b_{1}\left( m_{1}+n_{1}-t_{1} +1\right) \left( m_{2}+n_{2}-t_{2}+1\right) }\\ q_{2t}&=\frac{\left( m_{1}+n_{1}-t_{1}\right) \left( A_{1}-v_{1} -v_{2}+s_{2}\left( m_{2}+n_{2}-2t_{2}+1\right) \right) +s_{1}t_{1}}{b_{1}\left( m_{1}+n_{1}-t_{1}+1\right) \left( m_{2}+n_{2}-t_{2}+1\right) } \end{aligned}$$

As for aggregate output,

$$\begin{aligned} Q_{2}=\frac{\left( n_{2}+m_{2}-t_{2}\right) \left( A_{2}-v_{2}\right) +t_{2}s_{2}}{b_{2}\left( n_{2}+m_{2}-t_{2}+1\right) }, \end{aligned}$$

we obtain after inserting the expressions for $A_{2}$ and $b_{2},$

$$\begin{aligned} Q_{2}= & {} \frac{\left( n_{2}+m_{2}-t_{2}\right) \left( m_{1}+n_{1} -t_{1}\right) }{b_{1}\left( n_{2}+m_{2}-t_{2}+1\right) \left( m_{1} +n_{1}-t_{1}+1\right) }\\&\left( \left( A_{1}-v_{1}-v_{2}\right) +\frac{t_{1}s_{1}}{\left( m_{1}+n_{1}-t_{1}\right) }+\frac{t_{2}s_{2}}{\left( n_{2}+m_{2}-t_{2}\right) }\right) \end{aligned}$$

Given that $P_{2}=A_{2}-b_{2}Q_{2}$ and replacing for $A_{2}$, $b_{2}$ and $Q_{2}$ we get

$$\begin{aligned} P_{2}= & {} \left( \frac{(A_{1}-v_{1})}{\left( n_{2}+m_{2}-t_{2}+1\right) }+\frac{\left( n_{2}+m_{2}-t_{2}\right) v_{2}}{\left( n_{2}+m_{2} -t_{2}+1\right) }\right. \\&\quad \left. +\frac{t_{1}s_{1}}{\left( n_{2}+m_{2}-t_{2}+1\right) \left( n_{1}+m_{1}-t_{1}\right) }-\frac{t_{2}s_{2}}{\left( n_{2} +m_{2}-t_{2}+1\right) }\right) \end{aligned}$$

Replacing $P_{2}$ into the outputs of the firms in the first tier we obtain

$$\begin{aligned} q_{1i}&=q_{1F}=\frac{q_{1L}}{2}\\&=\frac{\left( m_{2}+n_{2}-t_{2}\right) \left( m_{1}+n_{1}-t_{1}\right) \left( A_{1}-v_{1}-v_{2}\right) }{b_{1}\left( m_{1}+n_{1}-t_{1}+1\right) \left( m_{2}+n_{2}-t_{2}+1\right) \left( m_{1}+n_{1}-t_{1}\right) }\\&\quad + \frac{s_{2}t_{2}\left( m_{1}+n_{1}-t_{1}\right) -s_{1}t_{1}\left( \left( m_{2}+n_{2}-t_{2}+1\right) \left( m_{1}+n_{1}-t_{1}\right) +1\right) }{b_{1}\left( m_{1}+n_{1}-t_{1}+1\right) \left( m_{2} +n_{2}-t_{2}+1\right) \left( m_{1}+n_{1}-t_{1}\right) }\\ q_{1t}&=\frac{\left( A_{1}-v_{1}-v_{2}\right) \left( m_{2}+n_{2} -t_{2}\right) -\frac{t_{1}s_{1}}{\left( n_{1}+m_{1}-t_{1}\right) } +t_{2}s_{2}+s_{1}\left( n_{2}+m_{2}-t_{2}+1\right) \left( n_{1} +m_{1}-2t_{1}+1\right) }{b_{1}\left( n_{1}+m_{1}-t_{1}+1\right) \left( n_{2}+m_{2}-t_{2}+1\right) } \end{aligned}$$

For total output, $Q_{1}=\frac{\left( n_{1}+m_{1}-t_{1}\right) \left( A_{1}-v_{1}-P_{2}\right) +t_{1}s_{1}}{b_{1}\left( n_{1}+m_{1}-t_{1} +1\right) }$, we obtain the same expression as for $Q_{2}$:

$$\begin{aligned} Q_{1}=\frac{(A_{1}-v_{1}-v_{2})\left( n_{1}+m_{1}-t_{1}\right) (n_{2} +m_{2}-t_{2})+t_{1}s_{1}(n_{2}+m_{2}-t_{2})+t_{2}s_{2}\left( n_{1} +m_{1}-t_{1}\right) }{b_{1}\left( n_{1}+m_{1}-t_{1}+1\right) \left( n_{2}+m_{2}-t_{2}+1\right) } \end{aligned}$$

so $P_{1}=A_{1}-b_{1}Q_{1}$ yields

$$\begin{aligned} P_{1}=\frac{A_{1}\left( n_{1}+m_{1}-t_{1}+n_{2}+m_{2}-t_{2}+1\right) +(v_{1}+v_{2})\left( n_{1}+m_{1}-t_{1}\right) (n_{2}+m_{2}-t_{2})-t_{1} s_{1}(n_{2}+m_{2}-t_{2})-t_{2}s_{2}\left( n_{1}+m_{1}-t_{1}\right) }{\left( n_{1}+m_{1}-t_{1}+1\right) \left( n_{2}+m_{2}-t_{2}+1\right) }. \end{aligned}$$

Let us denote $n_{i}+m_{i}-t_{i}$ by $r_{i}$ ($i=1,2$). Then we can simply write:

$$\begin{aligned} Q_{1}&=Q_{2}=\frac{(A_{1}-v_{1}-v_{2})r_{1}r_{2}+t_{1}s_{1}r_{2} +t_{2}s_{2}r_{1}}{b_{1}\left( r_{1}+1\right) \left( r_{2}+1\right) }\\ P_{1}&=\frac{A_{1}\left( r_{1}+r_{2}+1\right) +(v_{1}+v_{2})r_{1} r_{2}-t_{1}s_{1}r_{2}-t_{2}s_{2}r_{1}}{\left( r_{1}+1\right) \left( r_{2}+1\right) }\\ P_{2}&=\frac{r_{1}\left( A_{1}-v_{1}+r_{2}v_{2}-t_{2}s_{2}\right) +t_{1}s_{1}}{r_{1}\left( r_{2}+1\right) } \end{aligned}$$

We now address the conditions needed for positive quantities. As we only deal with a single two-firm merger, we will only consider this case. When there is a multidivisional merger, all outputs are symmetrically divided across firms except for the leader division of the merged firm, that produces twice as much as any other. Hence, if aggregate output is positive (which is always the case), so will be the output of each individual firm. When there is a traditional merger in tier i, then the same happens in tier j: firms are symmetric in the tier in which the merger did not occur and if total output is positive, so will be the output of each individual firm in tier j. In tier i, however, the outsiders will have a lower output because they do not benefit from synergies. Thus we only need to ensure that the outsiders to a traditional merger have a positive equilibrium output. If the merger occurs in the first tier, we need that $q_{1i}(n_{1},0,1)>0$ and if the merger occurs in the second tier, we need that $q_{2i}(n_{2},0,1)>0$. Using the expressions above:

$$\begin{aligned}&\frac{n_{2}\left( n_{1}-1\right) \left( A_{1}-v_{1}-v_{2}\right) -s_{1}\left( n_{1}-n_{2}+n_{1}n_{2}\right) }{b_{1}n_{1}\left( n_{1}-1\right) \left( n_{2}+1\right) }>0\Leftrightarrow z_{1}<\frac{n_{2}\left( n_{1}-1\right) }{\left( n_{1}-n_{2}+n_{1}n_{2}\right) }\\&n_{1}\frac{A_{1}-v_{1}-v_{2}-s_{2}}{b_{1}n_{2}\left( n_{1}+1\right) } >0\Leftrightarrow z_{2}<1 \end{aligned}$$

$\square$

Proof of Lemma 2

If the multidivisional merger occurs in tier 1, the variation in the downstream price is

$$\begin{aligned} \Delta P_{1}&=\frac{A_{1}\left( n_{1}+n_{2}+2\right) +n_{2}\left( v_{1}+v_{2}\right) \left( n_{1}+1\right) }{\left( n_{1}+2\right) \left( n_{2}+1\right) }-\frac{(n_{1}+n_{2}+1)A_{1}+n_{1}n_{2}(v_{1}+v_{2})}{(n_{1}+1)(n_{2}+1)}\\&=\frac{-n_{2}\left( A_{1}-v_{1}-v_{2}\right) }{\left( n_{1}+2\right) \left( n_{1}+1\right) \left( n_{2}+1\right) } \end{aligned}$$

If the multidivisional merger occurs in tier 2, the variation in the downstream price is

$$\begin{aligned} \Delta P_{1}&=\frac{A_{1}\left( n_{1}+n_{2}+2\right) +n_{1}\left( v_{1}+v_{2}\right) \left( n_{2}+1\right) }{\left( n_{1}+1\right) \left( n_{2}+2\right) }-\frac{(n_{1}+n_{2}+1)A_{1}+n_{1}n_{2}(v_{1}+v_{2})}{(n_{1}+1)(n_{2}+1)}\\&=\frac{-n_{1}\left( A_{1}-v_{1}-v_{2}\right) }{\left( n_{2}+2\right) \left( n_{2}+1\right) \left( n_{1}+1\right) } \end{aligned}$$

Price drops more when the merger occurs in tier 1 if and only if$\ n_{2} >n_{1}{:}$

$$\begin{aligned}&\frac{n_{2}\left( A_{1}-v_{1}-v_{2}\right) }{\left( n_{1}+2\right) \left( n_{1}+1\right) \left( n_{2}+1\right) }-\frac{n_{1}\left( A_{1}-v_{1} -v_{2}\right) }{\left( n_{2}+2\right) \left( n_{2}+1\right) \left( n_{1}+1\right) }\\&\quad =-\left( n_{1}-n_{2}\right) \frac{\left( n_{1} +n_{2}+2\right) \left( A_{1}-v_{1}-v_{2}\right) }{\left( n_{2}+2\right) \left( n_{2}+1\right) \left( n_{1}+2\right) \left( n_{1}+1\right) } \end{aligned}$$

$\square$

Proof of Lemma 3

• (a) If the traditional merger occurs in tier 2, the final price change is

  $$\begin{aligned} \Delta P_{1}&=\frac{A_{1}\left( n_{1}+n_{2}\right) +n_{1}\left( v_{1}+v_{2}\right) \left( n_{2}-1\right) -s_{2}n_{1}}{\left( n_{1}+1\right) n_{2}}\\&\quad -\frac{(n_{1}+n_{2}+1)A_{1}+n_{1}n_{2}(v_{1}+v_{2} )}{(n_{1}+1)(n_{2}+1)}\\&=n_{1}\frac{A_{1}-v_{1}-v_{2}-s_{2}\left( n_{2}+1\right) }{n_{2}\left( n_{2}+1\right) \left( n_{1}+1\right) } \end{aligned}$$

• (b) If the traditional merger occurs in tier 1, the final price change is

  $$\begin{aligned} \Delta P_{1}&=\frac{A_{1}\left( n_{1}+n_{2}\right) +n_{2}\left( v_{1}+v_{2}\right) \left( n_{1}-1\right) -s_{1}n_{2}}{n_{1}\left( n_{2}+1\right) }\\&\quad -\frac{(n_{1}+n_{2}+1)A_{1}+n_{1}n_{2}(v_{1}+v_{2})}{(n_{1}+1)(n_{2}+1)}\\&=n_{2}\frac{A_{1}-v_{1}-v_{2}-s_{1}\left( n_{1}+1\right) }{n_{1}\left( n_{2}+1\right) \left( n_{1}+1\right) } \end{aligned}$$

• (c) Price falls more when the merger occurs in tier 1 if and only if

  $$\begin{aligned}&\frac{A_{1}\left( n_{1}+n_{2}\right) +n_{2}\left( v_{1}+v_{2}\right) \left( n_{1}-1\right) -s_{1}n_{2}}{n_{1}\left( n_{2}+1\right) }\\&\quad <\frac{A_{1}\left( n_{1}+n_{2}\right) +n_{1}\left( v_{1}+v_{2}\right) \left( n_{2}-1\right) -s_{2}n_{1}}{\left( n_{1}+1\right) n_{2}}\\&\qquad \Leftrightarrow \frac{\left( n_{1}-n_{2}\right) \left( n_{1}+n_{2}\right) \left( A_{1}-v_{1}-v_{2}\right) -n_{1}^{2}s_{2}\left( n_{2}+1\right) +n_{2} ^{2}s_{1}\left( n_{1}+1\right) }{n_{1}n_{2}\left( n_{2}+1\right) \left( n_{1}+1\right) }>0\\&\qquad \Leftrightarrow z_{1}>\frac{n_{1}^{2}}{n_{2}^{2}}\frac{n_{2}+1}{n_{1}+1}z_{2}-\left( n_{1}-n_{2}\right) \frac{\left( n_{1}+n_{2}\right) }{n_{2}^{2}\left( n_{1}+1\right) } \end{aligned}$$

$\square$

Proof of Corollary 1

• (a) For an equal number of firms in both tiers, $n_{1}=n_{2}$, the downstream price is lower with the traditional merger that involves the largest cost reductions. This follows from Lemma 3, making $n_{1}=n_{2}:$ $z_{1}>z_{2}$.

• (b) For an equal degree of cost savings, the price downstream is lower when the merger takes place at the downstream tier if and only if $n_{1}>n_{2}$ and $z<\frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}}$, or $n_{1}<n_{2}$ and $z>\frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}}$. This follows from Lemma 3, making $z_{1}=z_{2}=z$:

  $$\begin{aligned} z\left( n_{1}-n_{2}\right) <\left( n_{1}-n_{2}\right) \frac{n_{1}+n_{2} }{n_{1}+n_{2}+n_{1}n_{2}} \end{aligned}$$

• (c) In the absence of synergies, $z_{1}=z_{2}=0$ and the expression in Lemma 3 becomes

  $$\begin{aligned} 0>0-\left( n_{1}-n_{2}\right) \frac{\left( n_{1}+n_{2}\right) }{n_{2} ^{2}\left( n_{1}+1\right) }\Leftrightarrow n_{1}>n_{2}. \end{aligned}$$

$\square$

Proof of Lemma 4

Consumers prefer a traditional to a multidivisional downstream merger if and only if

$$\begin{aligned} n_{2}\frac{\left( n_{1}-1\right) +z_{1}}{n_{1}\left( n_{2}+1\right) }>\frac{n_{2}\left( n_{1}+1\right) }{\left( n_{1}+2\right) \left( n_{2}+1\right) }\Leftrightarrow z_{1}>z_{1tm}:=\frac{2}{n_{1}+2} \end{aligned}$$

Consumers prefer a traditional to a multidivisional upstream merger if and only if

$$\begin{aligned} n_{1}\frac{\left( n_{2}-1\right) +z_{2}}{n_{2}\left( n_{1}+1\right) }>\frac{n_{1}\left( n_{2}+1\right) }{\left( n_{2}+2\right) \left( n_{1}+1\right) }\Leftrightarrow z_{2}>z_{2tm}:=\frac{2}{n_{2}+2} \end{aligned}$$

$\square$

Proof of Proposition 1

The merger that leads to higher consumer surplus is the one that leads to higher output. Recall that the equilibrium outputs in the four different mergers are:

Equilibrium Output
Merger	Upstream	Downstream
Multidivisional	$Q_{M2}=n_{1}\frac{\left( n_{2}+1\right) }{\left( n_{2}+2\right) \left( n_{1}+1\right) }$	$Q_{M1}=n_{2}\frac{\left( n_{1}+1\right) }{\left( n_{1}+2\right) \left( n_{2}+1\right) }$
Traditional	$Q_{T2}=n_{1}\frac{\left( n_{2}-1\right) +z}{n_{2}\left( n_{1}+1\right) }$	$Q_{T1}=n_{2}\frac{\left( n_{1}-1\right) +z}{n_{1}\left( n_{2}+1\right) }$

(a) Consider initially the case of $n_{1}-n_{2}>0$. Then:

$$\begin{aligned}&Q_{M2}>Q_{M1}\Leftrightarrow \frac{n_{1}\left( n_{2}+1\right) }{\left( n_{2}+2\right) \left( n_{1}+1\right) }>\frac{n_{2}\left( n_{1}+1\right) }{\left( n_{1}+2\right) \left( n_{2}+1\right) }\\&\quad \Leftrightarrow \frac{\left( n_{1}-n_{2}\right) \left( n_{1}+n_{2}+2\right) }{\left( n_{2}+2\right) \left( n_{2}+1\right) \left( n_{1}+2\right) \left( n_{1}+1\right) }>0\\&Q_{M2}>Q_{T2}\Leftrightarrow \frac{n_{1}\left( n_{2}+1\right) }{\left( n_{2}+2\right) \left( n_{1}+1\right) }>n_{1}\frac{\left( n_{2}-1\right) +z}{n_{2}\left( n_{1}+1\right) }\\&\quad \Leftrightarrow n_{1}\frac{2-z\left( n_{2}+2\right) }{n_{2}\left( n_{2}+2\right) \left( n_{1}+1\right) }>0\Leftrightarrow z<\frac{2}{n_{2}+2}\\&Q_{M2}>Q_{T1}\Leftrightarrow \frac{n_{1}\left( n_{2}+1\right) }{\left( n_{2}+2\right) \left( n_{1}+1\right) }>n_{2}\frac{\left( n_{1}-1\right) +z}{n_{1}\left( n_{2}+1\right) }\\&\quad \Leftrightarrow \frac{2n_{2}+n_{1}^{2} +n_{2}^{2}-zn_{2}\left( n_{2}+2\right) \left( n_{1}+1\right) }{n_{1}\left( n_{2}+2\right) \left( n_{2}+1\right) \left( n_{1}+1\right) }>0\Leftrightarrow z<\frac{2n_{2}+n_{1}^{2}+n_{2}^{2}}{n_{2}\left( n_{2}+2\right) \left( n_{1}+1\right) }\\&Q_{T2}>Q_{T1}\Leftrightarrow n_{1}\frac{\left( n_{2}-1\right) +z}{n_{2}\left( n_{1}+1\right) }>n_{2}\frac{\left( n_{1}-1\right) +z}{n_{1}\left( n_{2}+1\right) }\\&\quad \Leftrightarrow \left( n_{1}-n_{2}\right) \frac{-\left( n_{1}+n_{2}\right) +z\left( n_{1}+n_{2}+n_{1}n_{2}\right) }{n_{1}n_{2}\left( n_{2}+1\right) \left( n_{1}+1\right) }>0\Leftrightarrow z>\frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}} \end{aligned}$$

It is easy to show that

$$\begin{aligned} \frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}}<\frac{2}{n_{2}+2}<\frac{2n_{2}+n_{1}^{2}+n_{2}^{2}}{n_{2}\left( n_{2}+2\right) \left( n_{1}+1\right) } \end{aligned}$$

This results from the following inequalities:

$$\begin{aligned}&\frac{2n_{2}+n_{1}^{2}+n_{2}^{2}}{n_{2}\left( n_{2}+2\right) \left( n_{1}+1\right) }-\frac{2}{\left( n_{2}+2\right) } =\frac{\left( n_{1}-n_{2}\right) ^{2}}{n_{2}\left( n_{2}+2\right) \left( n_{1}+1\right) }>0\\&\frac{2}{n_{2}+2}-\frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}} =\frac{n_{2}\left( n_{1}-n_{2}\right) }{\left( n_{2}+2\right) \left( n_{1}+n_{2}+n_{1}n_{2}\right) }>0 \end{aligned}$$

So, for different values of z the output ranking is as follows:

$z<\frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}}$	$\frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}}<z<\frac{2}{n_{2}+2}$	$\frac{2}{n_{2}+2}<z<\frac{2n_{2}+n_{1}^{2}+n_{2}^{2}}{n_{2}\left( n_{2}+2\right) \left( n_{1}+1\right) }$	$z>\frac{2n_{2}+n_{1}^{2}+n_{2}^{2}}{n_{2}\left( n_{2}+2\right) \left( n_{1}+1\right) }$
$Q_{M2}>Q_{T1}>Q_{T2}$	$Q_{M2}>Q_{T2}>Q_{T1}$	$Q_{T2}>Q_{M2}>Q_{T1}$	$Q_{T2}>Q_{T1}>Q_{M2}$

(b) Consider now the case of $n_{1}-n_{2}<0$. Then:

$$\begin{aligned}&Q_{M1}>Q_{M2}\Leftrightarrow \frac{n_{2}\left( n_{1}+1\right) }{\left( n_{1}+2\right) \left( n_{2}+1\right) }>\frac{n_{1}\left( n_{2}+1\right) }{\left( n_{2}+2\right) \left( n_{1}+1\right) }\\&\quad \Leftrightarrow \frac{-\left( n_{1}-n_{2}\right) \left( n_{1}+n_{2}+2\right) }{\left( n_{2}+2\right) \left( n_{2}+1\right) \left( n_{1}+2\right) \left( n_{1}+1\right) }>0\\&Q_{M1}>Q_{T2}\Leftrightarrow \frac{n_{2}\left( n_{1}+1\right) }{\left( n_{1}+2\right) \left( n_{2}+1\right) }>n_{1}\frac{\left( n_{2}-1\right) +z}{n_{2}\left( n_{1}+1\right) }\\&\quad \Leftrightarrow \frac{2n_{1}+n_{1}^{2} +n_{2}^{2}-zn_{1}\left( n_{2}+1\right) \left( n_{1}+2\right) }{n_{2}\left( n_{1}+2\right) \left( n_{1}+1\right) \left( n_{2}+1\right) }>0\Leftrightarrow z<\frac{2n_{1}+n_{1}^{2}+n_{2}^{2}}{n_{1}\left( n_{2}+1\right) \left( n_{1}+2\right) }\\&Q_{M1}>Q_{T1}\Leftrightarrow \frac{n_{2}\left( n_{1}+1\right) }{\left( n_{1}+2\right) \left( n_{2}+1\right) }>n_{2}\frac{\left( n_{1}-1\right) +z}{n_{1}\left( n_{2}+1\right) }\\&\quad \Leftrightarrow n_{2}\frac{2-z\left( n_{1}+2\right) }{n_{1}\left( n_{2}+1\right) \left( n_{1}+2\right) }>0\Leftrightarrow z<\frac{2}{n_{1}+2}\\&Q_{T2}>Q_{T1}\Leftrightarrow n_{1}\frac{\left( n_{2}-1\right) +z}{n_{2}\left( n_{1}+1\right) }>n_{2}\frac{\left( n_{1}-1\right) +z}{n_{1}\left( n_{2}+1\right) }\\&\quad \Leftrightarrow \left( n_{1}-n_{2}\right) \frac{-\left( n_{1}+n_{2}\right) +z\left( n_{1}+n_{2}+n_{1}n_{2}\right) }{n_{1}n_{2}\left( n_{2}+1\right) \left( n_{1}+1\right) }>0\Leftrightarrow z<\frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}} \end{aligned}$$

As before,

$$\begin{aligned} \frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}}<\frac{2}{n_{1}+2}<\frac{2n_{1}+n_{1}^{2}+n_{2}^{2}}{n_{1}\left( n_{2}+1\right) \left( n_{1}+2\right) } \end{aligned}$$

So, for different values of z, the output ranking is as follows:

$z<\frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}}$	$\frac{n_{1}+n_{2}}{n_{1}+n_{2}+n_{1}n_{2}}<z<\frac{2}{n_{1}+2}$	$\frac{2}{n_{1}+2}<z<\frac{2n_{1}+n_{1}^{2}+n_{2}^{2}}{n_{1}\left( n_{2}+1\right) \left( n_{1}+2\right) }$	$z>\frac{2n_{1}+n_{1}^{2}+n_{2}^{2}}{n_{1}\left( n_{2}+1\right) \left( n_{1}+2\right) }$
$Q_{M1}>Q_{T2}>Q_{T1}$	$Q_{M1}>Q_{T1}>Q_{T2}$	$Q_{T1}>Q_{M1}>Q_{T2}$	$Q_{T1}>Q_{T2}>Q_{M1}$

$\square$

Proof of Lemma 5

• (a) A multidivisional downstream merger is profitable if and only if

  $$\begin{aligned}&\frac{2n_{2}^{2}}{\left( n_{1}+2\right) ^{2}\left( n_{2}+1\right) ^{2} }+\frac{n_{2}^{2}}{\left( n_{1}+2\right) ^{2}\left( n_{2}+1\right) ^{2}}>\frac{2n_{2}^{2}}{(n_{2}+1)^{2}(n_{1}+1)^{2}}\\&\quad \Leftrightarrow \frac{n_{2}^{2}\left( -2n_{1}+n_{1}^{2}-5\right) }{\left( n_{1}+2\right) ^{2}\left( n_{1}+1\right) ^{2}\left( n_{2}+1\right) ^{2}} >0 \end{aligned}$$

  A multidivisional upstream merger is profitable if and only if

  $$\begin{aligned}&\frac{n_{1}}{\left( n_{1}+1\right) \left( n_{2}+2\right) ^{2}} +\frac{2n_{1}}{\left( n_{1}+1\right) \left( n_{2}+2\right) ^{2}}>\frac{2n_{1}}{(n_{1}+1)(n_{2}+1)^{2}}\\&\quad \Leftrightarrow \frac{n_{1}\left( -2n_{2}+n_{2}^{2}-5\right) }{\left( n_{2}+2\right) ^{2}\left( n_{2}+1\right) ^{2}\left( n_{1}+1\right) } >0 \end{aligned}$$

  We have that $\left( -2n_{i}+n_{i}^{2}-5\right) >0$ for any $n_{i}\ge 4$.

• (b) A traditional upstream merger is profitable if and only if

  $$\begin{aligned} n_{1}\frac{\left( 1+z_{2}\left( n_{2}-1\right) \right) ^{2}}{n_{2} ^{2}\left( n_{1}+1\right) }>\frac{2n_{1}}{(n_{1}+1)(n_{2}+1)^{2} }\Leftrightarrow z_{2}>z_{2TN}:=\frac{\left( \sqrt{2}-1\right) n_{2} -1}{n_{2}^{2}-1} \end{aligned}$$

• (c) A traditional downstream merger is profitable if and only if

  $$\begin{aligned}&\frac{\left( n_{2}\left( n_{1}-1\right) +z_{1}\left( n_{2}+n_{1}\left( n_{2}+1\right) \left( n_{1}-2\right) \right) \right) ^{2}}{n_{1} ^{2}\left( n_{2}+1\right) ^{2}\left( n_{1}-1\right) ^{2}}>\frac{2n_{2} ^{2}}{(n_{2}+1)^{2}(n_{1}+1)^{2}}\\&\quad \Leftrightarrow z_{1}>z_{1TN}:=n_{2}\frac{n_{1}-1}{n_{1}+1}\frac{\left( \sqrt{2}-1\right) n_{1}-1}{n_{1}\left( n_{2}+1\right) \left( n_{1}-2\right) +n_{2}} \end{aligned}$$

$\square$

Proof of Proposition 2

A pair of firms in any tier will propose the merger that is more profitable.

• (a) A multidivisional downstream merger is more profitable than a traditional downstream merger if and only if

  $$\begin{aligned}&\frac{2n_{2}^{2}}{\left( n_{1}+2\right) ^{2}\left( n_{2}+1\right) ^{2} }+\frac{n_{2}^{2}}{\left( n_{1}+2\right) ^{2}\left( n_{2}+1\right) ^{2}} \\&\quad >\frac{\left( n_{2}\left( n_{1}-1\right) +z_{1}\left( n_{2} +n_{1}\left( n_{2}+1\right) \left( n_{1}-2\right) \right) \right) ^{2} }{n_{1}^{2}\left( n_{2}+1\right) ^{2}\left( n_{1}-1\right) ^{2}}\\&\qquad \Leftrightarrow z_{1} <z_{1TM}:=\frac{n_{2}(n_{1}-1)\left( \left( \sqrt{3}-1\right) n_{1}-2\right) }{n_{1}(n_{1}^{2}(n_{2}+1)-3n_{2}-4)+2n_{2}} \end{aligned}$$

• (b) A multidivisional upstream merger is more profitable than a traditional upstream merger if and only if

  $$\begin{aligned}&\frac{n_{1}}{\left( n_{1}+1\right) \left( n_{2}+2\right) ^{2}} +\frac{2n_{1}}{\left( n_{1}+1\right) \left( n_{2}+2\right) ^{2}} >n_{1}\frac{\left( 1+z_{2}\left( n_{2}-1\right) \right) ^{2}}{n_{2} ^{2}\left( n_{1}+1\right) }\\&\quad \Leftrightarrow z_{2} <z_{2TM}:=\frac{\left( \sqrt{3}-1\right) n_{2}-2}{\left( n_{2}+2\right) \left( n_{2}-1\right) } \end{aligned}$$

$\square$

Proof of Proposition 3

The result follows directly from Lemma 4 and Proposition 2. The intervals always exist:

$$\begin{aligned} z_{1TM}-z_{1tm}=n_{1}\frac{2\left( n_{1}-2\right) +\left( 3-\sqrt{3}\right) n_{2}\left( n_{1}-1\right) }{\left( n_{1}+2\right) \left( -n_{1}^{2}\left( n_{2}+1\right) +n_{1}\left( 2n_{2}+2\right) -n_{2}\right) }<0 \end{aligned}$$

and

$$\begin{aligned} z_{2TM}-z_{2tm}=\frac{\left( \sqrt{3}-3\right) n_{2}}{\left( n_{2} -1\right) \left( n_{2}+2\right) }<0. \end{aligned}$$

Note that $\left( -n_{1}^{2}\left( n_{2}+1\right) +2n_{1}\left( n_{2}+1\right) -n_{2}\right)$ is maximized at $n_{1}=1$ and at $n_{1}=2$ takes value $-n_{2}<0$. Thus, it is always negative for $n_{1}\ge 4$. $\square$

Proof of Proposition 4

The proof follows from combining the results in Lemma 3 and Proposition 2. The intervals for $z_{i}$ are non-empty for all parameter values because

$$\begin{aligned} z_{1tn}-z_{1TM}&=\frac{1}{n_{1}+1}-\frac{n_{2}(n_{1}-1)\left( \left( \sqrt{3}-1\right) n_{1}-2\right) }{n_{1}(n_{1}^{2}(n_{2}+1)-3n_{2}-4)+2n_{2}}\\&=n_{1}\frac{\left( 2-\sqrt{3}\right) n_{2}\left( n_{1}-1\right) \left( n_{1}+2\sqrt{3}+5\right) +\left( n_{1}-2\right) \left( n_{1}+2\right) }{\left( n_{1}+2\right) \left( n_{1}+1\right) \left( n_{2}+n_{1}\left( n_{2}+1\right) \left( n_{1}-2\right) \right) }>0\\ \frac{1}{n_{2}+1}-z_{2TM}&=\frac{1}{n_{2}+1}-\frac{\left( \sqrt{3}-1\right) n_{2}-2}{n_{2}^{2}+n_{2}-2}\\&=\frac{\left( 2-\sqrt{3}\right) n_{2}\left( n_{2}+2\sqrt{3}+5\right) }{\left( n_{2}-1\right) \left( n_{2}+1\right) \left( n_{2}+2\right) }>0 \end{aligned}$$

$\square$

Appendix 2

In this appendix we present most of our results for the case in which a given number of traditional mergers or multidivisional mergers may have occurred in any tier before the additional merger under analysis. We start with four useful remarks that include expressions that will be relevant in the proofs, as well as parameter constraints.

Remark 1

For any $n_{i}$ we have $r_{i}=n_{i}+m_{i}-t_{i}$. Given that we are assuming that at the current industry structure it is possible to have an additional two-firm merger and that there is at least one surviving outsider after the merger, the maximum number of previous multidivisional or traditional two-firm mergers is $(n_{i}-3)/2$. So $n_{i}-(n_{i} -3)/2<r_{i}<n_{i}+(n_{i}-3)/2,$ where the lower bound corresponds to all previous mergers being traditional and the upper bound corresponds to all previous mergers being multidivisional. As we assume $n_{i}\ge 4$, we have that $r_{i}>4-(4-3)/2=3.5$. As the number of firms is an integer, we assume $r_{i}\ge 4$. After any number of traditional mergers in tier i, it is possible that there is at least an additional merger between previously non-merged firms and at least a single outsider. So, $r_{i}>t_{i}$.

Remark 2

From Lemma 1, the equilibrium output and prices are given by:

$$\begin{aligned} Q_{1}(r_{1},r_{2},t_{1},t_{2})&=Q_{2}(r_{1},r_{2},t_{1},t_{2})=\frac{r_{1}r_{2}+r_{2}t_{1}z_{1}+r_{1}t_{2}z_{2}}{\left( r_{1}+1\right) \left( r_{2}+1\right) }\\ P_{1}(r_{1},r_{2},t_{1},t_{2})&=\frac{A_{1}\left( r_{1}+r_{2}+1\right) +(v_{1}+v_{2})r_{1}r_{2}-t_{1}s_{1}r_{2}-t_{2}s_{2}r_{1}}{\left( r_{1}+1\right) \left( r_{2}+1\right) } \\ P_{2}(r_{1},r_{2},t_{1},t_{2})&=\frac{r_{1}\left( A_{1}-v_{1}+r_{2} v_{2}-t_{2}s_{2}\right) +t_{1}s_{1}}{r_{1}\left( r_{2}+1\right) } \end{aligned}$$

(2)

where $r_{i}=n_{i}+m_{i}-t_{i}$ and $s_{i}=z_{i}(A_{1}-v_{1}-v_{2})$ ($i=1,2$)

Individual outputs, divided by $\frac{A_{1}-v_{1}-v_{2}}{b_{1}}$ are:

$$\begin{aligned} q_{1i}&=q_{1F}=\frac{q_{1L}}{2}=q_{1t}-z_{1}=\frac{r_{1}\left( r_{2}\left( 1-t_{1}z_{1}\right) +t_{2}z_{2}\right) -r_{1}t_{1}z_{1} -t_{1}z_{1}}{r_{1}\left( r_{2}+1\right) \left( r_{1}+1\right) }\\ q_{2i}&=q_{2F}=\frac{r_{1}\left( 1-t_{2}z_{2}\right) +t_{1}z_{1} }{\left( r_{1}+1\right) \left( r_{2}+1\right) }\\ q_{2L}&=2\frac{r_{1}\left( 1-t_{2}z_{2}\right) +t_{1}z_{1}}{\left( r_{1}+1\right) \left( r_{2}+1\right) }\\ q_{2t}&=\frac{r_{1}\left( 1-t_{2}z_{2}\right) +t_{1}z_{1}+r_{1} z_{2}\left( r_{2}+1\right) }{\left( r_{1}+1\right) \left( r_{2}+1\right) } \end{aligned}$$

We need $t_{2}z_{2}<1$ to ensure a positive $q_{2i}=q_{2F}$ in the absence of downstream traditional mergers $(t_{1}=0)$. This implies that all outputs in tier 2 are positive, regardless of the numbers of firms or previous mergers.

So, we only need to ensure that $q_{1i}>0,$ that is:

$$\begin{aligned} r_{1}\left( r_{2}\left( 1-t_{1}z_{1}\right) +t_{2}z_{2}\right) -r_{1} t_{1}z_{1}-t_{1}z_{1}>0\Leftrightarrow z_{1}<\frac{r_{1}\left( r_{2} +t_{2}z_{2}\right) }{t_{1}\left( r_{1}+r_{1}r_{2}+1\right) } \end{aligned}$$

It is easy to check that this threshold increases with $r_{1}$ and $r_{2}$, meaning that additional multidivisional mergers make this condition easier to verify.

With an additional traditional downstream merger the condition becomes

$$\begin{aligned} z_{1}<\frac{(r_{1}-1)\left( r_{2}+t_{2}z_{2}\right) }{\left( (r_{1} -1)+(r_{1}-1)r_{2}+1\right) (t_{1}+1)} \end{aligned}$$

which is a stronger condition. With an additional traditional upstream merger the condition becomes

$$\begin{aligned} z_{1}<\frac{r_{1}\left( (r_{2}-1)+(t_{2}+1)z_{2}\right) }{\left( r_{1}+r_{1}(r_{2}-1)+1\right) t_{1}} \end{aligned}$$

So, the following constraint on cost savings ensures that all outputs are positive:

$$\begin{aligned} z_{1}&<\overline{z_{1}}:=\min \left\{ \frac{(r_{1}-1)\left( r_{2} +t_{2}z_{2}\right) }{\left( (r_{1}-1)+(r_{1}-1)r_{2}+1\right) (t_{1} +1)},\frac{r_{1}\left( (r_{2}-1)+(t_{2}+1)z_{2}\right) }{\left( r_{1} +r_{1}(r_{2}-1)+1\right) t_{1}}\right\} \\ z_{2}&<\overline{z_{2}}:=1/t_{2} \end{aligned}$$

Note that these imply

$$\begin{aligned} t_{1}z_{1}&<\frac{r_{1}\left( r_{2}+t_{2}z_{2}\right) }{\left( r_{1}+r_{1}r_{2}+1\right) }<1+\frac{t_{2}z_{2}}{r_{2}}\\ t_{2}z_{2}&<1+\frac{t_{1}z_{1}}{r_{1}} \end{aligned}$$

Remark 3

Individual profits, divided by $\frac{\left( A_{1} -v_{1}-v_{2}\right) ^{2}}{b_{1}}$, are:

$$\begin{aligned} \pi _{1i}(r_{1},r_{2},t_{1},t_{2})&=\frac{\left( -r_{1}r_{2}+t_{1} z_{1}\left( r_{1}+r_{1}r_{2}+1\right) -r_{1}t_{2}z_{2}\right) ^{2}}{r_{1}^{2}\left( r_{1}+1\right) ^{2}\left( r_{2}+1\right) ^{2}}\\ \pi _{1m}(r_{1},r_{2},t_{1},t_{2})&=3\left( r_{1}r_{2}\left( t_{1} z_{1}-1\right) +t_{1}z_{1}+r_{1}t_{1}z_{1}-r_{1}t_{2}z_{2}\right) \\&\quad *\frac{-r_{1}r_{2}+\left( r_{1}+r_{1}r_{2}+1\right) t_{1}z_{1}-r_{1} t_{2}z_{2}}{r_{1}^{2}\left( r_{1}+1\right) ^{2}\left( r_{2}+1\right) ^{2}}\\ \pi _{1t}(r_{1},r_{2},t_{1},t_{2})&=\frac{\left( z_{1}\left( r_{1} ^{2}\left( r_{2}+1\right) -r_{1}\left( t_{1}-1\right) \left( r_{2}+1\right) -t_{1}\right) +r_{1}\left( r_{2}+t_{2}z_{2}\right) \right) ^{2}}{r_{1}^{2}\left( r_{1}+1\right) ^{2}\left( r_{2}+1\right) ^{2}}\\ \pi _{2i}(r_{1},r_{2},t_{1},t_{2})&=\frac{\left( -r_{1}-t_{1}z_{1} +r_{1}t_{2}z_{2}\right) ^{2}}{r_{1}\left( r_{1}+1\right) \left( r_{2}+1\right) ^{2}}\\ \pi _{2m}(r_{1},r_{2},t_{1},t_{2})&=\frac{3\left( -r_{1}-t_{1}z_{1} +r_{1}t_{2}z_{2}\right) ^{2}}{r_{1}\left( r_{2}+1\right) ^{2}\left( r_{1}+1\right) }\\ \pi _{2t}(r_{1},r_{2},t_{1},t_{2})&=\frac{\left( r_{1}+r_{1}z_{2} +t_{1}z_{1}+r_{1}r_{2}z_{2}-r_{1}t_{2}z_{2}\right) ^{2}}{\left( r_{1}+1\right) \left( r_{2}+1\right) ^{2}r_{1}} \end{aligned}$$

Remark 4

Throughout the proofs it will be necessary to solve several inequations of the type $f(x)=ax^{2}+bx+c>0$ with $x\in \left[ 0,\overline{x}\right] ,$ where the sign of a is uncertain, $c<0$ and $b>0$. Note that under these circumstances $f(0)=c<0$ and $\frac{\partial f}{\partial x}(x=0)=b>0$. If $f({\overline{x}})>0$ the solution is: if $a>0$: $x>\max \left\{ \frac{-b+\sqrt{b^{2}-4ac}}{2a},\frac{-b-\sqrt{b^{2}-4ac}}{2a}\right\} =\frac{-b+\sqrt{b^{2}-4ac}}{2a}$. If $a<0,$ $x>\min \left\{ \frac{-b+\sqrt{b^{2}-4ac}}{2a},\frac{-b-\sqrt{b^{2}-4ac}}{2a}\right\} =\frac{-b+\sqrt{b^{2}-4ac}}{2a}$. So, whenever we have this type of inequality, the solution is $x>\frac{-b+\sqrt{b^{2}-4ac}}{2a}$.

Lemma 2B

A multidivisional merger decreases final consumer price, independently of the tier where the merger takes place. $\square$

Proof of Lemma 2B

Downstream price is given by

$$\begin{aligned} P_{1}(r_{1},r_{2},t_{1},t_{2})=\frac{A_{1}\left( r_{1}+r_{2}+1\right) +(v_{1}+v_{2})r_{1}r_{2}-t_{1}z_{1}(A_{1}-v_{1}-v_{2})r_{2}-t_{2}z_{2} (A_{1}-v_{1}-v_{2})r_{1}}{\left( r_{1}+1\right) \left( r_{2}+1\right) } \end{aligned}$$

Let $\Delta P_{1}^{mi}$ denote the change in downstream price after an additional multidivisional merger at tier i.

An additional multidivisional two-firm merger at tier i increases $r_{i}$ by 1. Thus, the change in downstream price is given by $\Delta P_{1}^{mi} =P_{1}(r_{i}+1,r_{j},t_{i},t_{j})-P_{1}(r_{i},r_{j},t_{i},t_{j}).$

After a merger in tier 1, the change in price is then

$$\begin{aligned} \Delta P_{1}^{m1}=-\left( A_{1}-v_{1}-v_{2}\right) \frac{r_{2}\left( 1-t_{1}z_{1}\right) +t_{2}z_{2}}{\left( r_{1}+2\right) \left( r_{1}+1\right) \left( r_{2}+1\right) }<0 \end{aligned}$$

as $r_{2}\left( 1-t_{1}z_{1}\right) +t_{2}z_{2}>0.$

After a merger in tier 2, the change in price is then

$$\begin{aligned} \Delta P_{1}^{m2}=-\left( A_{1}-v_{1}-v_{2}\right) \frac{r_{1}\left( 1-t_{2}z_{2}\right) +t_{1}z_{1}}{\left( r_{2}+2\right) \left( r_{2}+1\right) \left( r_{1}+1\right) }<0 \end{aligned}$$

as $r_{1}\left( 1-t_{2}z_{2}\right) +t_{1}z_{1}>0$. $\square$

Lemma 3B

(a) In the presence of a traditional upstream merger, the downstream price falls if and only if $z_{2}>z_{2tn}:=\frac{r_{1}+t_{1}z_{1} }{r_{1}\left( r_{2}+t_{2}+1\right) }$. (b) In the presence of a traditional downstream merger, the downstream price falls if and only if $z_{1} >z_{1tn}:=\frac{r_{2}+t_{2}z_{2}}{r_{2}\left( r_{1}+t_{1}+1\right) }$. (c) The downstream price is lower after the traditional downstream merger than after the traditional upstream merger if and only if $z_{1}>z_{2} \frac{r_{1}^{2}\left( r_{2}+1\right) +t_{2}\left( r_{2}+r_{1}^{2}\right) }{r_{2}^{2}\left( r_{1}+1\right) +t_{1}\left( r_{1}+r_{2}^{2}\right) }-\left( r_{1}-r_{2}\right) \frac{r_{1}+r_{2}}{r_{2}^{2}\left( r_{1}+1\right) +t_{1}\left( r_{1}+r_{2}^{2}\right) }$. $\square$

Proof of Lemma 3B

Likewise, let $\Delta P_{1}^{ti}$ denote the change in downstream price after an additional traditional merger at tier i. A traditional merger in tier i means increasing $t_{i}$ by 1 which decreases $r_{i}$ by 1. Thus,

$$\begin{aligned} \Delta P_{1}^{ti}=P_{1}(r_{i}-1,r_{j},t_{i}+1,t_{j})-P_{1}(r_{i},r_{j} ,t_{i},t_{j}). \end{aligned}$$

(a) If $i=2$ (upstream merger), the change in price is given by

$$\begin{aligned} \Delta P_{1}^{t2}=\left( A_{1}-v_{1}-v_{2}\right) \frac{r_{1}+t_{1} z_{1}-r_{1}z_{2}\left( r_{2}+t_{2}+1\right) }{r_{2}\left( r_{2}+1\right) \left( r_{1}+1\right) } \end{aligned}$$

which is negative if and only if

$$\begin{aligned} z_{2}>\frac{r_{1}+t_{1}z_{1}}{r_{1}\left( r_{2}+t_{2}+1\right) } \end{aligned}$$

(b) If $i=1$ (downstream merger), the change in price is given by

$$\begin{aligned} \Delta P_{1}^{t1}=\left( A_{1}-v_{1}-v_{2}\right) \frac{r_{2}+t_{2} z_{2}-r_{2}z_{1}\left( r_{1}+t_{1}+1\right) }{r_{1}\left( r_{2}+1\right) \left( r_{1}+1\right) } \end{aligned}$$

which is negative if and only if

$$\begin{aligned} z_{1}>\frac{r_{2}+t_{2}z_{2}}{r_{2}\left( r_{1}+t_{1}+1\right) } \end{aligned}$$

(c) This follows from

$$\begin{aligned}&P_{1}(r_{1},r_{2}-1,t_{1},t_{2}+1)>P_{1}(r_{1}-1,r_{2},t_{1}+1,t_{2} )\\&\quad \Leftrightarrow z_{1}>z_{2}\frac{r_{1}^{2}\left( r_{2}+1\right) +t_{2}\left( r_{2} +r_{1}^{2}\right) }{r_{2}^{2}\left( r_{1}+1\right) +t_{1}\left( r_{1}+r_{2}^{2}\right) }-\left( r_{1}-r_{2}\right) \frac{r_{1}+r_{2}}{r_{2}^{2}\left( r_{1}+1\right) +t_{1}\left( r_{1}+r_{2}^{2}\right) } \end{aligned}$$

$\square$

Corollary 1B

• (a) For an equal industry structure in both tiers, $r_{1}=r_{2}$ and $t_{1}=t_{2}$, the downstream price is lower after the traditional merger (upstream or downstream) that involves the largest cost reductions.

• (b) For an equal degree of cost savings, $z_{1}=z_{2}=z,$ the downstream price is lower after the traditional downstream merger than after the traditional upstream merger if and only if $\left( r_{1}-r_{2}\right) \left( r_{1}+r_{2}+r_{1}r_{2}\right) +t_{2}\left( r_{2}+r_{1}^{2}\right) -t_{1}\left( r_{1}+r_{2}^{2}\right) >0$ and $z<\frac{\left( r_{1} -r_{2}\right) \left( r_{1}+r_{2}\right) }{\left( r_{1}-r_{2}\right) \left( r_{1}+r_{2}+r_{1}r_{2}\right) +t_{2}\left( r_{2}+r_{1}^{2}\right) -t_{1}\left( r_{1}+r_{2}^{2}\right) }$ , or

  $\left( r_{1}-r_{2}\right) \left( r_{1}+r_{2}+r_{1}r_{2}\right) +t_{2}\left( r_{2}+r_{1}^{2}\right) -t_{1}\left( r_{1}+r_{2}^{2}\right) <0$ and $z>\frac{\left( r_{1}-r_{2}\right) \left( r_{1}+r_{2}\right) }{\left( r_{1}-r_{2}\right) \left( r_{1}+r_{2}+r_{1}r_{2}\right) +t_{2}\left( r_{2}+r_{1}^{2}\right) -t_{1}\left( r_{1}+r_{2}^{2}\right) }$.

• (c) In the absence of synergies, the downstream price is lower after the traditional downstream merger than after the traditional upstream merger if and only if there are more “equivalent firms” downstream $(r_{1}>r_{2})$.

$\square$

Proof of Corollary 1B

• (a) Making $r_{1}=r_{2}$ and $t_{1}=t_{2}$ in the expression of Lemma 3(c) it results immediately that the downstream price is lower after the traditional downstream merger than after the traditional upstream merger if and only if $z_{1}>z_{2}$.

• (b) This follows from Lemma 3(c), making $z_{1}=z_{2}=z$.

• (c) This results immediately from Lemma 3(c), making $z_{1}=z_{2} =0$.

$\square$

Lemma 4B

Consumers prefer a traditional to a multidivisional merger in tier i if and only if $z_{i}>z_{itm}:=\frac{r_{j}+t_{j}z_{j}}{r_{j}} \frac{2}{r_{i}+2t_{i}+2}$. $\square$

Proof of Lemma 4B

This happens when

$$\begin{aligned} P_{1}(r_{i}-1,r_{j},t_{i}+1,t_{j})-P_{1}(r_{i}+1,r_{j},t_{i},t_{j})<0 \end{aligned}$$

If $i=1$ (downstream merger), this is equivalent to:

$$\begin{aligned}&\left( A_{1}-v_{1}-v_{2}\right) \frac{2\left( r_{2}+t_{2}z_{2}\right) -r_{2}z_{1}\left( r_{1}+2t_{1}+2\right) }{r_{1}\left( r_{2}+1\right) \left( r_{1}+2\right) }<0\\&\quad \Leftrightarrow z_{1}>\frac{r_{2}+t_{2}z_{2}}{r_{2}}\frac{2}{r_{1}+2t_{1}+2} \end{aligned}$$

If $i=2$ (upstream merger), this is equivalent to:

$$\begin{aligned}&\left( A_{1}-v_{1}-v_{2}\right) \frac{2\left( r_{1}+t_{1}z_{1}\right) -r_{1}z_{2}\left( r_{2}+2t_{2}+2\right) }{r_{2}\left( r_{2}+2\right) \left( r_{1}+1\right) }<0\\&\quad \Leftrightarrow z_{2}>\frac{r_{1}+t_{1}z_{1}}{r_{1}}\frac{2}{r_{2}+2t_{2}+2} \end{aligned}$$

$\square$

Lemma 5B

(a) A multidivisional merger in any tier is profitable. (b) A traditional upstream merger is profitable if and only if $z_{2}>z_{2TN}$. (c) A traditional downstream merger is profitable if and only if $z_{1}>z_{1TN}$. $\square$

Proof of Lemma 5B

(a) If the multidivisional merger takes place at tier 1, it is profitable if

$$\begin{aligned} \pi _{1m}(r_{1}+1,r_{2},t_{1},t_{2})-2\pi _{1i}(r_{1},r_{2},t_{1},t_{2})>0 \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{\left( \begin{array}{c} \left( r_{2}^{2}r_{1}^{2}\left( r_{1}^{2}-2r_{1}-5\right) -r_{2}\left( -2r_{1}^{4}+2r_{1}^{3}+20r_{1}^{2}+16r_{1}\right) +\left( -4r_{1}+r_{1} ^{2}-2\right) \left( r_{1}+2\right) ^{2}\right) z_{1}^{2}t_{1} ^{2}\\ -2r_{1}\left( r_{1}^{3}\left( r_{2}+1\right) -r_{1}^{2}\left( 2r_{2}+1\right) -r_{1}\left( 5r_{2}+10\right) -8\right) \left( r_{2}+t_{2}z_{2}\right) z_{1}t_{1}\\ +r_{1}^{2}\left( r_{2}+t_{2}z_{2}\right) ^{2}\left( r_{1}^{2} -2r_{1}-5\right) \end{array} \right) }{r_{1}^{2}\left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{2}+1\right) ^{2}}>0 \end{aligned}$$

(3)

The coefficient of $z_{1}^{2}t_{1}^{2}$ is always positive: It is a U-shaped parabola in $r_{2}$ (because $r_{1}^{2}\left( r_{1}^{2}-2r_{1}-5\right) >0,$ $\forall r_{1}\ge 4)$ which, evaluated at $r_{2}=4,$ takes value $-88r_{1}-174r_{1}^{2}-40r_{1}^{3}+25r_{1}^{4}-8>0$. It is also increasing at that point because

$$\begin{aligned}&2*4*r_{1}^{2}\left( r_{1}^{2}-2r_{1}-5\right) -\left( -2r_{1} ^{4}+2r_{1}^{3}+20r_{1}^{2}+16r_{1}\right) \\&\quad =2r_{1}\left( -30r_{1}-9r_{1} ^{2}+5r_{1}^{3}-8\right) >0 \end{aligned}$$

which proves that

$$\begin{aligned}&\left( r_{2}^{2}r_{1}^{2}\left( r_{1}^{2}-2r_{1}-5\right) -r_{2}\left( -2r_{1}^{4}+2r_{1}^{3}+20r_{1}^{2}+16r_{1}\right) \right. \\&\quad \left. +\left( -4r_{1}+r_{1} ^{2}-2\right) \left( r_{1}+2\right) ^{2}\right) >0 \end{aligned}$$

and hence the numerator in (3) is a U-shaped parabola in $t_{1}z_{1}$.

Evaluating the numerator of (3) at $t_{1}z_{1}=\frac{r_{1}\left( r_{2}+t_{2}z_{2}\right) }{\left( r_{1}+r_{1}r_{2}+1\right) }$ we obtain $\frac{3r_{1}^{2}\left( r_{2}+t_{2}z_{2}\right) ^{2}}{\left( r_{1} +r_{1}r_{2}+1\right) ^{2}}>0$. Moreover, (3) is decreasing at that point: the derivative is $-6r_{1}^{2}\left( r_{2}+t_{2}z_{2}\right) \frac{r_{1}+r_{2}+r_{1}r_{2}+2}{r_{1}+r_{1}r_{2}+1}<0.\$Thus, (3) is always positive for $z_{1}t_{1}\in \left[ 0,\frac{r_{1}\left( r_{2} +t_{2}z_{2}\right) }{\left( r_{1}+r_{1}r_{2}+1\right) }\right]$.

If the multidivisional merger takes place at tier 2, it is profitable if

$$\begin{aligned} \pi _{2m}(r_{1},r_{2}+1,t_{1},t_{2})-2\pi _{2i}(r_{1},r_{2},t_{1},t_{2})>0. \end{aligned}$$

This is equivalent to

$$\begin{aligned} \frac{\left( r_{2}^{2}-2r_{2}-5\right) \left( -r_{1}-t_{1}z_{1}+r_{1} t_{2}z_{2}\right) ^{2}}{r_{1}\left( r_{2}+2\right) ^{2}\left( r_{2}+1\right) ^{2}\left( r_{1}+1\right) }>0 \end{aligned}$$

which is always true.

(b) If the traditional merger takes place at tier 2, it is profitable if

$$\begin{aligned} \pi _{2t}(r_{1},r_{2}-1,t_{1},t_{2}+1)-2\pi _{2i}(r_{1},r_{2},t_{1},t_{2})>0 \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{\left( \begin{array}{c} r_{1}^{2}\left( 2t_{2}+2r_{2}t_{2}^{2}-2r_{2}^{2}t_{2}-2r_{2}^{3}t_{2} -2r_{2}^{2}+r_{2}^{4}+t_{2}^{2}-r_{2}^{2}t_{2}^{2}+2r_{2}t_{2}+1\right) z_{2}^{2}\\ +2r_{1}\left( t_{2}\left( -2r_{2}+r_{2}^{2}-1\right) +r_{2}^{2}-r_{2} +r_{2}^{3}-1\right) \left( r_{1}+t_{1}z_{1}\right) z_{2}\\ +\left( 2r_{2}-r_{2}^{2}+1\right) \left( r_{1}+t_{1}z_{1}\right) ^{2} \end{array} \right) }{r_{1}r_{2}^{2}\left( r_{2}+1\right) ^{2}\left( r_{1}+1\right) } \end{aligned}$$

At $z_{2}=0$ this is negative and increasing, respectively because $\left( 2r_{2}-r_{2}^{2}+1\right) <0$ and

$$\begin{aligned} t_{2}\left( -2r_{2}+r_{2}^{2}-1\right) +r_{2}^{2}-r_{2}+r_{2}^{3}-1>0. \end{aligned}$$

At $z_{2}=\frac{1+\frac{t_{1}z_{1}}{r_{1}}}{t_{2}}$ (the maximum admissible value), it is positive ($\left( r_{2}-1\right) ^{2}\frac{\left( r_{1} +t_{1}z_{1}\right) ^{2}}{r_{1}r_{2}^{2}t_{2}^{2}\left( r_{1}+1\right) }>0$).

Then, from Remark 4,

$$\begin{aligned} z_{2}>z_{2TN}:=\frac{r_{1}+t_{1}z_{1}}{r_{1}}\frac{\left( r_{2}-1\right) \left( r_{2}+1\right) \left( r_{2}(\sqrt{2}-1)-1\right) -t_{2}\left( r_{2}^{2}-2r_{2}-1\right) }{t_{2}^{2}\left( 2r_{2}-r_{2}^{2}+1\right) -2t_{2}\left( r_{2}-1\right) \left( r_{2}+1\right) ^{2}+\left( r_{2}-1\right) ^{2}\left( r_{2}+1\right) ^{2}} \end{aligned}$$

(c) If the traditional merger takes place at tier 1, it is profitable if

$$\begin{aligned} \pi _{1t}(r_{1}-1,r_{2},t_{1}+1,t_{2})-2\pi _{1i}(r_{1},r_{2},t_{1},t_{2})>0 \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{\left( \begin{array}{c} \left( \begin{array}{c} t_{1}^{2}\left( 2r_{1}^{2}r_{2}+6r_{1}^{3}r_{2}-2r_{1}^{4}r_{2}+5r_{1} ^{2}+2r_{1}^{3}+r_{2}^{2}-r_{1}^{4}-4r_{1}^{2}r_{2}^{2}+4r_{1}^{3}r_{2} ^{2}-r_{1}^{4}r_{2}^{2}-6r_{1}r_{2}-2\right) +\\ -2t_{1}\left( r_{1}+1\right) ^{2}\left( r_{2}\left( r_{1}-1\right) ^{2}+r_{1}\left( r_{1}-2\right) \right) \left( r_{1}-r_{2}+r_{1} r_{2}\right) +\\ \left( r_{1}+1\right) ^{2}\left( -2r_{1}+r_{2}+r_{1}^{2}r_{2}+r_{1} ^{2}-2r_{1}r_{2}\right) ^{2} \end{array} \right) z_{1}^{2}\\ +2\left( r_{2}+t_{2}z_{2}\right) \left( r_{1}-1\right) \left( \begin{array}{c} t_{1}\left( r_{2}+r_{1}^{2}r_{2}\left( r_{1}-3\right) +r_{1}^{2}\left( r_{1}-2\right) +r_{1}\left( r_{2}-3\right) \right) \\ +\left( r_{1}+1\right) ^{2}\left( r_{2}+r_{1}\left( r_{1}r_{2}-2\right) +r_{1}\left( r_{1}-2r_{2}\right) \right) \end{array} \right) z_{1}\\ +\left( 2r_{1}-r_{1}^{2}+1\right) \left( r_{1}-1\right) ^{2}\left( r_{2}+t_{2}z_{2}\right) ^{2} \end{array} \right) }{r_{1}^{2}\left( r_{1}-1\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{2}+1\right) ^{2}} \end{aligned}$$

At $z_{1}=0$ this is negative and increasing, respectively because $\left( 2r_{1}-r_{1}^{2}+1\right) <0$ and

$$\begin{aligned} \left( \begin{array}{c} t_{1}\left( r_{2}+r_{1}^{2}r_{2}\left( r_{1}-3\right) +r_{1}^{2}\left( r_{1}-2\right) +r_{1}\left( r_{2}-3\right) \right) \\ +\left( r_{1}+1\right) ^{2}\left( r_{2}+r_{1}\left( r_{1}r_{2}-2\right) +r_{1}\left( r_{1}-2r_{2}\right) \right) \end{array} \right) >0. \end{aligned}$$

At the maximum $z_{1}$ ($z_{1}=\frac{r_{1}\left( r_{2}+t_{2}z_{2}\right) }{\left( r_{1}+r_{1}r_{2}+1\right) t_{1}}$) this is positive: $\left( r_{2}+t_{2}z_{2}\right) ^{2}\frac{\left( -t_{1}+r_{1}^{3}r_{2}-2r_{1} ^{2}r_{2}-2r_{1}^{2}+r_{1}^{3}+r_{1}r_{2}\right) ^{2}}{r_{1}^{2}t_{1} ^{2}\left( r_{1}-1\right) ^{2}\left( r_{2}+1\right) ^{2}\left( r_{1}+r_{1}r_{2}+1\right) ^{2}}>0.$

Then, from remark 4,

$$\begin{aligned} z_{1}&>z_{1TN}\\&:=\left( r_{2}+t_{2}z_{2}\right) \left( r_{1}-1\right) \\&\quad \times \frac{ \begin{array}{c} -\left( \begin{array}{c} t_{1}\left( r_{2}+r_{1}^{2}r_{2}\left( r_{1}-3\right) +r_{1}^{2}\left( r_{1}-2\right) +r_{1}\left( r_{2}-3\right) \right) +\\ \left( r_{1}+1\right) ^{2}\left( r_{2}+r_{1}\left( r_{1}r_{2}-2\right) +r_{1}\left( r_{1}-2r_{2}\right) \right) \end{array} \right) \\ +\left( r_{1}+1\right) \left( -t_{1}-2r_{1}^{2}r_{2}+r_{1}^{3}r_{2} -2r_{1}^{2}+r_{1}^{3}+r_{1}r_{2}\right) \sqrt{2} \end{array} }{\left( \begin{array}{c} t_{1}^{2}\left( 2r_{1}^{2}r_{2}+6r_{1}^{3}r_{2}-2r_{1}^{4}r_{2}+5r_{1} ^{2}+2r_{1}^{3}+r_{2}^{2}-r_{1}^{4}-4r_{1}^{2}r_{2}^{2}+4r_{1}^{3}r_{2} ^{2}-r_{1}^{4}r_{2}^{2}-6r_{1}r_{2}-2\right) \\ -2t_{1}\left( r_{1}+1\right) ^{2}\left( r_{2}\left( r_{1}-1\right) ^{2}+r_{1}\left( r_{1}-2\right) \right) \left( r_{1}-r_{2}+r_{1} r_{2}\right) \\ +\left( r_{1}+1\right) ^{2}\left( -2r_{1}+r_{2}+r_{1}^{2}r_{2}+r_{1} ^{2}-2r_{1}r_{2}\right) ^{2} \end{array} \right) } \end{aligned}$$

In order to discuss in which tier a traditional merger is more likely to be profitable, we now compare the conditions for merger profitability when tiers are symmetric, that is, when $z_{i}=z$, $t_{i}=t$ and $r_{i}=r$.

In this case, the condition that for a traditional upstream merger to be profitable, $z_{2}>z_{2TN},$ is equivalent to

$$\begin{aligned} z&>\frac{r+tz}{r}\frac{\left( r-1\right) \left( r+1\right) \left( r(\sqrt{2}-1)-1\right) -t\left( r^{2}-2r-1\right) }{t^{2}\left( 2r-r^{2}+1\right) -2t\left( r-1\right) \left( r+1\right) ^{2}+\left( r-1\right) ^{2}\left( r+1\right) ^{2}}\Leftrightarrow \\ z&>\frac{-r+r^{2}\left( \sqrt{2}-1\right) }{\left( r-1\right) \left( r+r^{2}+\left( \sqrt{2}-1\right) t\left( r-\sqrt{2}-1\right) \right) } \end{aligned}$$

(4)

and the condition that for a traditional downstream merger to be profitable, $z_{1}>z_{1TN}$, is equivalent to

$$\begin{aligned} z&>\left( r+tz\right) \left( r-1\right) \\&\quad \times \frac{-\left( t\left( r\left( -r-2r^{2}+r^{3}-2\right) \right) +\left( r+1\right) ^{2}r\left( -r+r^{2}-1\right) \right) +\left( r+1\right) \left( -t+r^{2}\left( -r+r^{2}-1\right) \right) \sqrt{2} }{\left( t^{2}\left( 4r^{3}+r^{4}+2r^{5}-r^{6}-2\right) -2t\left( r+1\right) ^{2}\left( r\left( -r+r^{2}-1\right) \right) r^{2}+\left( r+1\right) ^{2}\left( r\left( -r+r^{2}-1\right) \right) ^{2}\right) }\Leftrightarrow \\ z&>\frac{\left( \sqrt{2}-1\right) r\left( r-1\right) \left( r-\sqrt{2}-1\right) }{\left( -r-2r^{2}+r^{4}+\left( \sqrt{2}-1\right) t\left( -r^{2}\left( \sqrt{2}+2\right) +r\left( \sqrt{2}+2\right) +r^{3}-2\sqrt{2}-3\right) \right) } \end{aligned}$$

As

$$\begin{aligned}&\frac{\left( \sqrt{2}-1\right) r\left( r-1\right) \left( r-\sqrt{2}-1\right) }{\left( -r-2r^{2}+r^{4}+\left( \sqrt{2}-1\right) t\left( -r^{2}\left( \sqrt{2}+2\right) +r\left( \sqrt{2}+2\right) +r^{3}-2\sqrt{2}-3\right) \right) }\\&\quad -\frac{-r+r^{2}\left( \sqrt{2}-1\right) }{\left( r-1\right) \left( r+r^{2}+\left( \sqrt{2}-1\right) t\left( r-\sqrt{2}-1\right) \right) }\\&\quad =-\frac{r\left( r-\sqrt{2}-1\right) \left( r+1\right) }{\left( r+r^{2}+\left( \sqrt{2}-1\right) \left( r-\sqrt{2}-1\right) t\right) \left( r-1\right) }\\&\qquad \times \frac{\left( \sqrt{2}-1\right) r\left( r-2\right) +\left( 3-2\sqrt{2}\right) \left( r-\sqrt{2}-2\right) t}{-r-2r^{2}+r^{4}+\left( \sqrt{2}-1\right) \left( r^{3}-r^{2}\left( \sqrt{2}+2\right) +r\left( \sqrt{2}+2\right) -2\sqrt{2}-3\right) t}<0 \end{aligned}$$

because all terms in brackets are positive, we conclude that there are values of z for which only the downstream merger is profitable and not the opposite. $\square$

Proposition 2B

(a) Downstream firms propose a traditional merger for $z_{1}>z_{1TM}$ and a multidivisional one otherwise. (b) Upstream firms propose a traditional merger for $z_{2}>z_{2TM}$ and a multidivisional one otherwise. $\square$

Proof of Proposition 2B

(a) Downstream firms will propose the most profitable merger. It is the traditional merger if and only if

$$\begin{aligned} \pi _{1t}(r_{1}-1,r_{2},t_{1}+1,t_{2})>\pi _{1m}(r_{1}+1,r_{2},t_{1},t_{2}) \end{aligned}$$

which is equivalent to:

$$\begin{aligned} \frac{\left( \begin{array}{c} z_{1}^{2}\left( \begin{array}{c} r_{1}^{8}\left( r_{2}+1\right) ^{2}-2r_{1}^{7}\left( r_{2}+1\right) ^{2}\left( t_{1}-1\right) -r_{1}^{6}\left( r_{2}+1\right) \left( 5r_{2}+8t_{1}+2r_{2}t_{1}^{2}+2t_{1}^{2}+6r_{2}t_{1}+7\right) \\ +2r_{1}^{5}\left( r_{2}+1\right) \left( -4r_{2}-t_{1}+2r_{2}t_{1} ^{2}+2r_{2}t_{1}-8\right) +\\ r_{1}^{4}\left( 18r_{2}+28t_{1}+32r_{2}t_{1}^{2}+20r_{2}^{2}t_{1}+11r_{2} ^{2}+22t_{1}^{2}+8r_{2}^{2}t_{1}^{2}+46r_{2}t_{1}+8\right) \\ -2r_{1}^{3}\left( -18r_{2}-20t_{1}-2r_{2}t_{1}^{2}+r_{2}^{2}t_{1}-5r_{2} ^{2}-12t_{1}^{2}+4r_{2}^{2}t_{1}^{2}-13r_{2}t_{1}-16\right) \\ -r_{1}^{2}\left( 8r_{2}-16t_{1}+28r_{2}t_{1}^{2}+22r_{2}^{2}t_{1}+11r_{2} ^{2}+8t_{1}^{2}+10r_{2}^{2}t_{1}^{2}+32r_{2}t_{1}-16\right) \\ +4r_{1}r_{2}\left( t_{1}+1\right) \left( -r_{2}-2t_{1}+r_{2}t_{1}-4\right) +4r_{2}^{2}\left( t_{1}+1\right) ^{2} \end{array} \right) \\ +2z_{1}\left( r_{1}-1\right) \left( r_{1}+1\right) \left( r_{2} +t_{2}z_{2}\right) \left( \begin{array}{c} r_{1}^{5}\left( r_{2}+1\right) +r_{1}^{4}\left( 2t_{1}+3\right) \left( r_{2}+1\right) \\ -r_{1}^{3}\left( r_{2}+2t_{1}+4r_{2}t_{1}+2\right) -r_{1}^{2}\left( 7r_{2}+14t_{1}+6r_{2}t_{1}+12\right) \\ +4r_{1}\left( -t_{1}+r_{2}t_{1}-2\right) +4r_{2}\left( t_{1}+1\right) \end{array} \right) \\ +2\left( 2r_{1}-r_{1}^{2}+2\right) \left( r_{1}-1\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{2}+t_{2}z_{2}\right) ^{2} \end{array} \right) }{r_{1}^{2}\left( r_{1}-1\right) ^{2}\left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{2}+1\right) ^{2}} \end{aligned}$$

At $z_{1}=0$ this is negative and increasing, respectively because $\left( 2r_{1}-r_{1}^{2}+2\right) <0$ and

$$\begin{aligned}&r_{2}\left( r_{1}-1\right) \left( r_{1}+1\right) \left( 2t_{1}\left( -2r_{1}+r_{1}^{2}-2\right) +3r_{1}^{2}+r_{1}^{3}-4\right) \\&\quad +r_{1}\left( r_{1}+2\right) \left( 2t_{1}\left( r_{1}^{2}-3r_{1}-1\right) +r_{1}\left( r_{1}-4\right) +r_{1}^{3}-4\right) >0. \end{aligned}$$

At $z_{1}=\frac{r_{1}\left( r_{2}+t_{2}z_{2}\right) }{\left( r_{1} +r_{1}r_{2}+1\right) t_{1}}$ this is equal to

$$\begin{aligned} \frac{\left( r_{2}+t_{2}z_{2}\right) ^{2}\times \left( \begin{array}{c} r_{2}^{2}r_{1}^{2}\left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{1}-1\right) ^{4}+\\ 2r_{2}r_{1}\left( r_{1}-1\right) ^{2}\left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{1}^{3}-2r_{1}^{2}-t_{1}\right) +\\ 2t_{1}^{2}\left( -r_{1}^{4}+6r_{1}^{3}+5r_{1}^{2}+6r_{1}+2\right) +\\ r_{1}^{3}\left( r_{1}-2\right) \left( r_{1}^{2}-2r_{1}-2\right) \left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}+\\ 2r_{1}^{2}\left( r_{1}-2\right) \left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{1}-t_{1}\right) \end{array} \right) }{r_{1}^{2}\left( r_{1}-1\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{1}+2\right) ^{2}\left( r_{2}+1\right) ^{2}t_{1}^{2}\left( r_{1}+r_{1}r_{2}+1\right) ^{2}} \end{aligned}$$

(5)

If $-r_{1}^{4}+6r_{1}^{3}+5r_{1}^{2}+6r_{1}+2>0$ then this is clearly positive. If $-r_{1}^{4}+6r_{1}^{3}+5r_{1}^{2}+6r_{1}+2<0$ then the second term in the numerator of (5) is larger than (replacing $r_{2}$ by 4 in the first line and $t_{1}^{2}$ by $r_{1}^{2}$ in the third line)

$$\begin{aligned} \left( \begin{array}{c} 4^{2}r_{1}^{2}\left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{1}-1\right) ^{4}+\\ 2r_{2}r_{1}\left( r_{1}-1\right) ^{2}\left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{1}^{3}-2r_{1}^{2}-t_{1}\right) +\\ 2r_{1}^{2}\left( -r_{1}^{4}+6r_{1}^{3}+5r_{1}^{2}+6r_{1}+2\right) +\\ r_{1}^{3}\left( r_{1}-2\right) \left( r_{1}^{2}-2r_{1}-2\right) \left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}+\\ 2r_{1}^{2}\left( r_{1}-2\right) \left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{1}-t_{1}\right) \end{array} \right) \end{aligned}$$

which can be written as

$$\begin{aligned}&r_{1}^{2}\left( -36r_{1}-110r_{1}^{2}+232r_{1}^{3}+180r_{1}^{4}-152r_{1} ^{5}-89r_{1}^{6}+34r_{1}^{7}+17r_{1}^{8}+68\right) \\&\quad +2r_{1}^{2}\left( r_{1}-2\right) \left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{1}-t_{1}\right) \\&\quad +2r_{2}r_{1}\left( r_{1}-1\right) ^{2}\left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{1}^{3}-2r_{1}^{2}-t_{1}\right) >0. \end{aligned}$$

Then, from Remark 4,

$$\begin{aligned} z_{1}&>z_{1TM}\\&:=\left( r_{1}-1\right) \left( r_{1}+1\right) \left( r_{2}+t_{2}z_{2}\right) \\&\quad \times \frac{\left( \begin{array}{c} \left( \begin{array}{c} -\left( r_{1}+1\right) \left( r_{1}+2\right) ^{2}\left( r_{1}\left( r_{2}+1\right) \left( r_{1}-2\right) +r_{2}\right) \\ -2t_{1}\left( -2r_{1}+2r_{2}-3r_{1}^{2}r_{2}-2r_{1}^{3}r_{2}+r_{1}^{4} r_{2}-7r_{1}^{2}-r_{1}^{3}+r_{1}^{4}+2r_{1}r_{2}\right) \end{array} \right) \\ +\left( r_{1}+2\right) r_{1}\left( \left( r_{1}+1\right) \left( -2r_{1}+r_{2}+r_{1}^{2}r_{2}+r_{1}^{2}-2r_{1}r_{2}\right) -2t_{1}\right) \sqrt{3} \end{array} \right) }{\left( \begin{array}{c} -2t_{1}^{2}\left( \begin{array}{c} r_{1}^{6}\left( r_{2}+1\right) ^{2}-2r_{1}^{5}r_{2}\left( r_{2}+1\right) -r_{1}^{4}\left( 16r_{2}+4r_{2}^{2}+11\right) \\ +2r_{1}^{3}\left( 2r_{2}+3\right) \left( r_{2}-2\right) +r_{1}^{2}\left( 14r_{2}+5r_{2}^{2}+4\right) \\ -2r_{1}r_{2}\left( r_{2}-2\right) -2r_{2}^{2} \end{array} \right) \\ -2t_{1}\left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( r_{1}\left( r_{2}+1\right) \left( r_{1}-2\right) +r_{2}\right) \left( r_{1}-r_{2}+r_{1}r_{2}\right) \\ +\left( r_{1}+2\right) ^{2}\left( r_{1}+1\right) ^{2}\left( -2r_{1} +r_{2}+r_{1}^{2}r_{2}+r_{1}^{2}-2r_{1}r_{2}\right) ^{2} \end{array} \right) } \end{aligned}$$

(b) Upstream firms will propose the most profitable merger. It is the traditional merger if and only if

$$\begin{aligned} \pi _{2t}(r_{1},r_{2}-1,t_{1},t_{2}+1)>\pi _{2m}(r_{1},r_{2}+1,t_{1},t_{2}) \end{aligned}$$

which is equivalent to:

$$\begin{aligned} \frac{\left( \begin{array}{c} r_{1}^{2}\left( -2t_{2}^{2}\left( r_{2}^{2}-2r_{2}-2\right) -2t_{2}\left( r_{2}-1\right) \left( r_{2}+2\right) ^{2}+\left( r_{2}+2\right) ^{2}\left( r_{2}-1\right) ^{2}\right) z_{2}^{2}+\\ 2r_{1}\left( -4t_{2}+2r_{2}^{2}t_{2}+3r_{2}^{2}+r_{2}^{3}-4r_{2} t_{2}-4\right) \left( r_{1}+t_{1}z_{1}\right) z_{2}+\\ 2\left( 2r_{2}-r_{2}^{2}+2\right) \left( r_{1}+t_{1}z_{1}\right) ^{2} \end{array} \right) }{r_{1}r_{2}^{2}\left( r_{2}+2\right) ^{2}\left( r_{1}+1\right) } \end{aligned}$$

At $z_{2}=0$ this is negative and increasing, respectively because $-2\left( -2r_{2}+r_{2}^{2}-2\right) <0$ and

$$\begin{aligned} 2r_{1}\left( 2t_{2}\left( r_{2}^{2}-2r_{2}-2\right) +3r_{2}^{2}+r_{2} ^{3}-4\right) >0. \end{aligned}$$

At $z_{2}=\frac{1+\frac{t_{1}z_{1}}{r_{1}}}{t_{2}}$ it is positive: $\left( r_{2}-1\right) ^{2}\frac{\left( r_{1}+t_{1}z_{1}\right) ^{2}}{r_{1} r_{2}^{2}\left( r_{1}+1\right) t_{2}^{2}}>0.$

Then, from remark 4,

$$\begin{aligned} z_{2}>z_{2TM}:=\frac{r_{1}+t_{1}z_{1}}{r_{1}}\frac{\left( \begin{array}{c} -2t_{2}\left( -2r_{2}+r_{2}^{2}-2\right) -\left( r_{2}-1\right) \left( r_{2}+2\right) ^{2}\\ +r_{2}\left( r_{2}+2\right) \left( r_{2}-1\right) \sqrt{3} \end{array} \right) }{\left( \begin{array}{c} -2t_{2}^{2}\left( r_{2}^{2}-2r_{2}-2\right) -2t_{2}\left( r_{2}-1\right) \left( r_{2}+2\right) ^{2}\\ +\left( r_{2}+2\right) ^{2}\left( r_{2}-1\right) ^{2} \end{array} \right) } \end{aligned}$$

$\square$

Proposition 3B

In tier i, final consumers’ and merging firms’ interests diverge for $z_{iTM}<z_{i}<z_{itm}$. $\square$

Proposition 4B

In tier i, the most profitable merger lowers consumer welfare for $z_{iTM}<z_{i}<z_{itn}$. $\square$